AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 14 Carbon and its Compounds.

AP State Syllabus SSC 10th Class Chemistry Important Questions 14th Carbon and its Compounds

10th Class Chemistry 14th Lesson Carbon and its Compounds 1 Mark Important Questions and Answers

Question 1.
Define Isomerism. (AP March 2016)
Answer:
The phenomenon of possessing same molecular formula but different properties by the compounds is known as “Isomerism”.

Question 2.
Give the names of the functional groups. (AP March 2018)
a) – COOR
b) – OH
Answer:
a) Ester
b) Alcohol

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 3.
How do you explain the role of Oxygen in combustion process? (TS March 2015)
Answer:
Oxygen helps the combustion (or) No combustion will take place without oxygen.
Ex : C + O2 → CO2

Question 4.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 1 AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 2
Predict and write the products. (TS March 2016)
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 2

Question 5.
Write two uses of nano tubes. (TS June 2017)
Answer:

  1. Nano tubes are used as molecule wires.
  2. In intigrated circuits nano tubes are used to connect the components together.
  3. Nano tubes are used to incert Bio-molecules into the single cell.

Question 6.
Write two uses of Ethanol in day to day life. (TS March 2018)
Answer:
Ethanol is used in
i) Preparation of Alchoholic drinks
ii) Preparing tincture iodine
iii) Preparing cough syrup and tonics

Question 7.
Write the atomic structure of the following carbon compound. 3, 7-dibromo-4, -6 dichloro – oct-5-ene-l, 2-diol. (TS March 2019)
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 3

Question 8.
Thanish added acetic acid along with concentrated sulphuric acid to ethanol what would be his observation during the experiment? (AP SCERT: 2019-20)
Answer:

  1. He may observe that the resulting mixture is a sweet odoured substance.
  2. The substance is ethyl acetate, an ester.

Question 9.
Why do the various micelles present in water do not come together to form a precipitate? Guess the reason. (TS June 2019)
Answer:
The various micelles present in water do not come together to form a precipitate as each micelle repels the other because of the ion-ion repulsion.

Question 10.
Mention any two uses of graphite in day to day life. (TS June 2019)
Answer:
Uses of graphite in day to day life :

  1. Pencil lead.
  2. Lubricant.

Question 11.
What is “Allotropy”?
Answer:
The property of an element to exist in two or more different forms due to the difference in their atomic arrangement is called “Allotropy” and the different forms are called allotropes.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 12.
‘Diamond is a bad conductor of heat.’ Why?
Answer:
Diamond is a bad conductor of heat due to lack of free electrons.

Question 13.
What is ‘cleavage’?
Answer:
Cleavage is a property of splitting of crystals of some minerals in certain directions to produce a flat, even surface.

Question 14.
“Diamond is the hardest natural substance but is brittle.” Why?
Answer:
Diamond is the hardest natural substance but is brittle and can be broken due to the property of cleavage.

Question 15.
Explain about high refractive index of diamond.
Answer:
Diamond has a high refractive index, due to which most of the light that enters the diamond gets reflected back internally. This internally reflected light is responsible for the brilliance of a diamond.

Question 16.
What is catenation?
Answer:
Catenation is the phenomenon in which atoms of same element join together to form long chains.

Question 17.
What is an alkyl group?
Answer:
If one hydrogen is removed from an alkane, it is called alkyl group.
Ex : CH4 → methane
CH3 → methyl group

Question 18.
What is polymerization?
Answer:
The reaction in which a large number of identical and simple molecules join together to form a large molecule is called ‘polymerization’.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 4

Question 19.
What do you understand by a ‘Functional group’?
Answer:
A group of atoms in carbon compounds showing characteristic properties is called a functional group.

Question 20.
Name some functional groups.
ANswer:
Alcohol – OH, Aldehyde – CHO, Ketone – > C = O, Carboxylic acid (- GOOH), ester (-COOR), and amine – NH2 are some important functional groups.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 21.
What is pyrolysis?
Answer:
Decomposition of a compound on heating in the absence of air is called pyrolysis.

Question 22.
What is hydrocarbon?
Answer:
Compounds containing only carbon and hydrogen are called ‘hydrocarbons’.
Ex : Alkanes (Saturated hydrocarbons),
Alkenes and Alkynes (Unsaturated hydrocarbons).

Question 23.
What is ‘Saturated hydrocarbon’? (Or) What is an alkane?
Answer:
The valency of carbon is 4, of all the valencies of carbon, are satisfied, the resultant hydrocarbons are referred to as ‘saturated hydrocarbons’ or alkanes. Their general formula is CnH2n+2.

Question 24.
What are ‘Unsaturated hydrocarbons’?
Answer:
The hydrocarbons containing one or more double bonds or triple bonds between two carbon atoms are called ‘unsaturated hydrocarbons’.
Ex : C2H6 and C3H6, etc.

Question 25.
What are alkenes?
Answer:
Alkenes are unsaturated hydrocarbons having at least one (C = C) double bond in their structures, Alkenes are also called olefins. Their general formula is CnH2n.
Ex : Ethylene (C2H4) and propene (C3H6), etc.

Question 26.
What are alkynes?
Answer:
Alkynes are unsaturated hydrocarbons having at least one (\(C \equiv C\)) triple bond in their structures. Their general formula is CnH2n-2.
Ex: Acetylene (\(\mathrm{HC} \equiv \mathrm{HC}\))

Question 27.
Mention the natural sources of carbon compounds.
Answer:
Plants, wood, natural gas, coal, petroleum, etc. are the natural sources of carbon compounds.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 28.
Explain about methanol (or) methyl alcohol.
Answer:
Methanol is the simplest alcohol, It is the first member of the homologous series of alcohol. It is also known as wood alcohol, as it was initially obtained by the destructive distillation of wood.

Question 29.
What is organic chemistry?
Answer:
The chemistry of carbon compounds (excluding the carbonates, bicarbonates, carbides, cyanides, carbon dioxide, and carbon monoxide) is called organic chemistry. The large number of organic compounds necessitated their study in separate branch of chemistry, known as organic chemistry,

Question 30.
What is halogenation?
Answer:
Alkanes react with halogens in the presence of sunlight. For example, when a mixture of methane and chlorine is exposed to sunlight, a hydrogen atom of methane is replaced by a chlorine atom,
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 5

Question 31.
How many rings are there in buckminsterfullerene?
Answer:
In buckminsterfullerene, there are 32 rings, of them 12 are pentagonal rings and 20 are hexagonal rings.

Question 32.
Give example for homologous series.
Answer:
CH4 and C2H6 → These differ by a – CH2 unit.
and C2H6 and C3H8 → These differ by a – CH2 unit.

Question 33.
What is hybridisation?
Answer:
The intermixing of orbitals to form equivalent new orbitals is called hybridisation.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 34.
What are nanotubes?
Answer:
Nanotubes are allotropic form of carbon.

Question 35.
What are homologous series?
Answer:
The ierles of carbon compound in which successive compounds differ by -CH2 unit is called homologous series.

Question 36.
Write the molecular formula of the fourth member of the homologous series of alcohols.
Answer:
CH3 – CH2 – CH2 – CH2 – OH

Question 37.
What is a catalyst?
Answer:
The substance which does not take part in chemical reaction but changes the rate of reaction.

Question 38.
Why are oils liquids at room temperature?
Answer:
Oils are unsaturated compounds so they are in liquid state.

Question 39.
Why are fats solids at room temperature?
Answer:
They are saturated compounds so they are in solid state.

Question 40.
Do you know the police detect whether suspected drivers have consumed alcohol or not? Explain.
Answer:
Orange Cr2O72- changes bluish green Cr3+ during the process of the oxidation of alcohol. The length of die tube that turned into green is the measure of die quantity of alcohol that had been drunk.

Question 41.
What is pka?
Answer:
The negative value of logarithm of dissociation constant of an acid.

Question 42.
What is Saponification?
Answer:
Alkaline hydrolysis of triesters of higher fatty acids producing soaps is called saponification.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 43.
What is a soap?
Answer:
Sodium or potassium salt of fatty acid.

Question 44.
What is micelle?
Answer:
A spherical aggregate of soap molecules in water is called micelle.

Question 45.
What change will you observe if you test soap with litmus papers?
Answer:
Red litmus turns into blue.

Question 46.
Write the valency of carbon in CH3 – CH3, CH2 = CH2 and \(\mathrm{HC} \equiv \mathrm{CH}\)?
Answer:
The valency of carbon in CH3 – CH3 is 4.
The valency of carbon in CH2 = CH2 is 3.
The valency of carbon in \(\mathrm{HC} \equiv \mathrm{C}\) – H is 2.

Question 47.
Out of butter and groundnut oil which is unsaturated in nature?
Answer:
Groundnut oil is unsaturated in nature.

Question 48.
What are hydrophobic and hydrophilic parts in soap?
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 6

Question 49.
Name the carboxylic acid used as preservative.
Answer:
Acetic acid is used as preservative.

Question 50.
Why does graphite act as a good conductor of electricity?
Answer:
Graphite is a good conductor of electricity because of delocalized x electron system.

Question 51.
Among objects made of glass and diamond, which one shines more? Why?
Answer:
Diamond shines more because of low conical angle of 24,4° and also high refractive

Question 52.
Write IUPAC names of the following compounds.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 7
Answer:
a) 2, 2, 3, 3 – tetra methyl butane
b) 3-chloro butan-l-oic acid.

Question 53.
What is the difference between combustion and oxidation reaction?
Answer:
Combustion is an oxidation reaction where a compound is burnt in the presence of oxygen, whereas oxidation is addition of oxygen which does not require any burning.

Question 54.
Write the order of priority of functional groups for naming carbon compounds.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 8

Question 55.
What is glycerol?
Answer:
The trihydroxy alcohol is called glycerol.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 9

Question 56.
What do you mean by CMC?
Answer:
CMC means Critical Micelle Concentration.

Question 57.
Name the simplest chloride of saturated hydrocarbon.
Answer:
Chloro methane or methyl chloride.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 10

Question 58.
Write the IUPAC name of next homolog of CH3CH2CHO.
Answer:
The next homolog of CH3CH2CHO is CH3CH2CH2CHO (its IUPAC name is butanol). Since homologs differ by – CH2.

Question 59.
How do physical properties like boiling point and melting point vary as the number of carbon atoms increases in a homologous series?
Answer:
There is regular gradation in physical properties of homologous series. So the physical properties like boiling point and melting point vary.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 60.
What is meant by Hybridisation?
Answer:
Mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals.

Question 61.
Write any two uses of Graphite.
Answer:
i) Conductor
ii) Lubricant

Question 62.
Write any two examples to Amorphous form of carbon.
Answer:
i) Coke
ii) coal
iii) charcoal.

Question 63.
Write any two examples to crystalline forms of carbon.
Answer:
i) Diamond
ii) graphite

Question 64.
What are the applications of Buckminster fullerene?
Answer:
i) Antioxidants
ii) Anti aging and damage agent in cosmetic sector.

Question 65.
What is meant by catenation?
Answer:
Binding of an element to itself through covalent bonds to form chain or ring molecules.

Question 66.
Write any one use of nanotubes.
Answer:
i) Used as molecular wires.
ii) Used in integrated circuits.

Question 67.
On which reason, graphite is used as lubricant and as the lead in pencils?
Answer:
Graphite has free electrons.

Question 68.
How many isotopes are there for C4H10, what are they?
Answer:
i) n – Butane
ii) Iso – Butane

Question 69.
CH3 – CH = CH – CH3, how many sigma bonds are present in the above compound?
Answer:
11

Question 70.
Write the IUPAC name of Ethyle alcohol.
Answer:
Ethanol.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 71.
Classify the following into alkanes, alkenes and alkynes.
C12 H22, C10 H22, C11 H22
Answer:
i) C10 H22 – Alkanes
ii) C11 H22 -Alkenes
iii) C12 H22-Alkynes

Question 72.
Hi ……… I am carboxylic acid. I am used in the making vinegar, who am I?
Answer:
Acetic acid.

Question 73.
What does IUPAC represent?
Answer:
International Union of Pure and Applied Chemistry.

Question 74.
Write any one example for esterification reaction.
Answer:
CH3COOH + C2H5OH → CH3COOC2H5 + H2O

Question 75.
A compound with molecules formula C2H6O is used in cough syrup. Identify the compound.
Answer:
Ethyl Alcohol.

Question 76.
Which substance is added for the denaturation of ethyl alcohol?
Answer:
Pyridine.

Question 77.
What is the abbreviation of CMC?
Answer:
Critical Micelle Concentration.

Question 78.
Write the names of polar end and non-polar end in a soap.
Answer:
Polar end – COO Na+, Non-polar end – R.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 79.
Write the IUPAC name of the alcohol which one carbon atom.
Answer:
Methanol.

Question 80.
Write the chemical equation which indicates the preparation of ethanol industrially?
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 11

Question 81.
What is the formula of chloroform? Write its one use.
Answer:
CHCl3, Anesthetic.

Question 82.
Which type of hydrocarbons are participate in addition reaction?
Answer:
Unsaturate Hydrocarbons.

Question 83.
What are the oxidising agents used in oxidisation of C2H5?
Answer:
K2Cr2O7, KMn04.

Question 84.
What is meant by catalyst?
Answer:
To change die rate of reaction without itself undergoing any permanent chemical change.

Question 85.
What are the main constituents of LPG?
Answer:
Butane, Methane.

Question 86.
What is the difference between saturated and unsaturated hydrocarbons?
Answer:
Saturated house single bonds, unsaturated have multiple bonds.

Question 87.
Describe a test for carboxylic acid.
Answer:
React with metals liberate hydrogen gas.

Question 88.
What is meant by denatured alcohol?
Answer:
Unfit for human consumption by adding one or more chemicals.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 89.
Complete the following equation.
CH4 + 2O2
Answer:
CH4 + 2O2 → CO2 + 1H2O

Question 90.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 12
i) In the above substance, what is the hybridisation of 3rd carbon?
Answer:
sp²

ii) What is the hybridisation of 4th carbon?
Answer:
sp³

Question 91.
What is the main misuse of Ethanol?
Answer:
Drinking.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 92.
What is gasohol?
Answer:
10% Ethyl alcohol with gasoline.

Question 93.
Write any two uses of Ethyle alcohol.
Answer:
i) Good solvent
ii) Additive to automotive gasoline.

Question 94.
Write two IUPAC name
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 13
Answer:
3 – Chloro 1 – Butane

Question 95.
Name the following functional groups.
i) – COOR
ii) R – COOH
Answer:
i) – COOR (Ester)
ii) R – COOH (Carboxylic acid)

Question 96.
Name the crystalline allotrope of carbon which conducts electricity.
Answer:
Graphite.

Question 97.
Ravi gets confused while understanding the between R – COOH and R – OH functional groups, ask him one question to classify it.
Answer:
i) What is carboxylic acid?
ii) What is Alcohol?

Question 98.
Formic acid (HCOOH)
Farmaldehyde (HCHO)
Methanol (CH3OH), then answer the following questions.
i) Which is present in ants?
Answer:
HCOOH (Formic acid).

ii) Which is used to preservation of dead bodies?
Answer:
HCHO (Farmaldehyde).

Question 99.
Write the symbolic representation showing the functional groups.
i) amine
ii) amide
Answer:
i) R – NH2
ii) R – CONH2

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 100.
How many sigma and pi-bonds present in Acetylene?
Answer:
\(\mathrm{HC} \equiv \mathrm{CH}\) ; σ bonds – 3 ; π bonds – 2

Question 101.
Which of the following will give substitution reactions?
CH4, C3H6, C3H4, C5H12, C4H8
Answer:
CH4, C5 H12

Question 102.
Which of the following will give addition reactions?
CH4, C3H6, C3H4, C5H12, C4H10
Answer:
C3H6, C3H4

Question 103.
What is a homologous series?
Answer:
Same functional group, difference between successive members is a simple structural unit – CH2.

Question 104.
Name the hydrocarbon which is used in the artificial ripening of fruits?
Answer:
C2H4

Question 105.
Define fermentation process.
Answer:
Chemical break down of a substance by bacteria, yeast or other microorganisms.

Question 106.
Define functional group.
Answer:
They are specific substituents within molecules that are responsible for die characteristic chemical reactions.

Question 107.
Which hydrocarbons participate in sp² hybridisation?
Answer:
C2H4

Question 108.
Name the following compounds,
i) CH3 – CH2 – Br
Answer:
1 – Bromo Ethane

ii)
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 14
Answer:
Ethanol

iii)
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 15
Answer:
2 – Butanone

iv)
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 16
Answer:
2, 3 – dichloro Butane

Question 109.
Which constituents are present in tincture Iodine?
Answer:
i) Iodine
ii) Alcohol.

Question 110.
Write the uses of esters in daily life.
Answer:
i) Solvents
ii) Plasticizers

Question 111.
Name the gas evolved when acetic acid reacts with sodium hydrogen carbonate.
Answer:
The gas liberated is carbon dioxide.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 112.
Name the organic acid present in vinegar. Write its chemical formula.
Answer:
The acid present in vinegar is acetic acid. Its formula is CH3COOH.

Question 113.
Why is graphite a good conductors’of electricity?
Answer:
Graphite has free electrons.

Question 114.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Tbtravalency.

Question 115.
Why are alkanes called as paraffins?
Answer:
Low reactivity.

Question 116.
Draw two possible structures with formula C3HgO and what they are called?
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 17

Question 117.
Draw structure of 3 – methyl pentan-3-ol.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 18

Question 118.
Draw the shape of soap molecule.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 119.
Draw the shape of Micelle.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 19

Question 120.
Draw the shape of methane.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 20

Question 121.
Draw the structure of pentanoic acid.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 21

Question 122.
How do you appreciate the role of diamond in space probes?
Answer:
Since it has the ability to filter out harmful radiations, it is used in making protective windows for space probes.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 123.
How do you appreciate the role of acetic acid as a preservative?
Answer:

  • Dilute acetic acid is used as a food preservative in the preparation of pickles and sauces,
  • As vinegar, it is also used as an appetiser for dressing food dishes.

Question 124.
How do you appreciate the role of diamond in surgery?
Answer:
A sharp edged diamond is used as a tool to remove cataract in eye surgery.

10th Class Chemistry 14th Lesson Carbon and its Compounds 2 Marks Important Questions and Answers

Question 1.
Draw the simple figure of a soap molecule. (AP March 2016)
Answer:
Structure of soap molecule :
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 2.
Draw the structure of the methane molecule. Write its bond angle. (TS June 2015)
Answer:
The bend angle in methane is 109°2 8′.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 22

Question 3.
a) Why are vegetable oils healthy as compared to vegetable ghee? (TS March 2015)
b) Write the IUPAC name of
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 25
Answer:
a) Because vegetable oils contain unsaturated fatty acids or vegetable oils are easily digestible.
b) 3 – Mono chloro butene (or) 3 Chloro butene

Question 4.
What are alkenes? Write the general formula of alkenes. Give an example for alkenes. (TS June 2017)
Answer:

  • Unsaturated hydrocarbons those are having carbon * carbon double bond are known as alkenes.
  • The general formula of Alkenes is CHH2h.
  • Example : Ethelene (C2H4).

Question 5.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 23
Based on the diagram, answer the following.
1) Write the name of the compound.
2) Write the name of functional group in the structure. (AP March 2019)
Answer:

  1. The compound is 2, 3-di ethyl-cycle hexan-1-ol.
  2. Alcohol (OH) is the functional group in the structure.

Question 6.
Identify the functional groups in the following compounds and write IUPAC names.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 24
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 26
The IUPAC name of the compound Is 2 – Chloro-Butan 1-ol.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 27
The IUPAC name of the compound is 3 – Methyl-2-Butan-one.

Question 7.
Draw the structure of butanoic acid C3H7COOH.
Answer:
Formula of butanoic acid is C4H5O2.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 28

Question 8.
What is ‘Isomerism’?
Answer:
Compounds having same molecular formula but different structures are called isomers, and the phenomenon is called isomerism.
Ex: C4H10 exists an n-hutane and iso-butane.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 29

Question 9.
How do you detect leakage in the cylinder?
Answer:

  • To detect any leakage of gas from die cylinder, a strong-smelling substance like ethyl mercaptan (C2H5 SH) is added to die gas.
  • Then the leakage can be easily detected by the foul smell of die ethyl mercaptan.

Question 10.
How is LPG gas useful for environment?
Answer:

  • Because of its heat producing capacity (calorific value), it is considered to be a good fuel.
  • It bums without producing smoke. Hence, it does not cause any pollution.
  • It is a dean fuel and can be conveniently handled.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 11.
How is ethanol useful in pharmaceutical industry?
Answer:

  • Solutions in ethanol are often prepared in pharmaceutical industry, these solutions are known as tinctures.
  • For example, a solution of Iodine and potassium iodide in ethanol is called tincture of iodine.
  • It is also used as an important raw material for the synthesis of many organic compounds, for example, ethanol, ethanoic acid, ethanoie anhydride, esters, chloroform, etc.

Question 12.
How are synthetic detergents harmful for environment?
Answer:

  • Some synthetic detergents resist biodegradation, i.e. they are not decomposed by micro-organisms such as bacteria.
  • Hence, they cause water pollution in lakes and rivers.
  • They tend to persist for a long time, making the water unfit for aquatic life.

Question 13.
Explain about allotropic forms of carbon.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 30

Question 14.
Diamond is considered to be the purest form of carbon. How can we prove it?
Answer:
When diamond is heated in oxygen alone, it bums at about 800° C and forms carbon dioxide leaving no residue. This proves that diamond to be the purest form of carbon.

Question 15.
Why does carbon not form C4+? Why?
Answer:

  • Electronic configuration of carbon is 1s²2s²2p².
  • If carbon loses four electrons from the outer shell, it will form C4+ ions.
  • This requires huge amount of energy which is not available normally.
  • Therefore C4+ formation is not possible.

Question 16.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Carbon is unable to form C4+ ion as well as C4- ion. So carbon has to satisfy its tetra- valency by sharing electrons with other atoms. So it mainly forms covalent bonding.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 17.
Define Allotropy. What are the allotropic forms of carbon?
Answer:
The property of an element to exist in two or more physical forms having more or less similar chemical properties but different physical properties is called allotropy. The allotropic forms of carbon are graphite, diamond, etc.

Question 18.
Identify the unsaturated compounds of the following.
a) CH3 – CH2 – CH2
b) CH3 – CH = CH3
c)
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 31
Answer:
a) CH3 – CH2 – CH2 saturated compound.
b) CH3 – CH = CH3 unsaturated compound.
c)
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 32

Question 19.
Define Isomers. Write structural formula of isomers of butane.
Answer:
Compounds having same molecular formula but different properties are called isomers.
Isomers of butane :
1) CH3 – CH2 – CH2 – CH3
Butane
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 33

Question 20.
What happens when a small piece of sodium is dropped into ethanol?
Answer:
When a small piece of sodium is dropped into ethanol it releases hydrogen gas and forms sodium ethoxide.
2C2H5OH + 2 Na → 2C2H5ONa + H2

Question 21.
What type of reaction takes place between ethane and chlorine?
Answer:
Substitution reaction takes place between ethane and chlorine in die presence erf sunlight
CH4 + Cl2 → CH3Cl + HCl
CH3Cl +Cl2 → CH2Cl2 + HCl
CH2Cl2 + Cl2 → CHCl3 + HCl
CHCl3 + Cl2 → CCl4 + HCl

Question 22.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:

  1. Catenation
  2. Isomerism.

Question 23.
How could you name the following compounds?
a) CH3 – CH2 – CH2 – Br
b) CH3 – CH2 – CH2 – CH2
Answer:
a) Bromo propane
b) Hexyne

Question 24.
Give examples for primary, secondary and tertiary amines.
Answer:
Primary amine – CH3NH2
Secondary amine – CH3 – NH – CH3
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 34

Question 25.
Write the following conversions.
1) Ethanol to Ethene
2) Ethene to Ethanol
3) Methane to carbon tetra chloride.
Answer:
1) Ethanol reacts with cone. H2SO4 at about 170°C to give ethene.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 35

2) Ethanol is prepared from ethene by the addition of water vapour in the presence of catalyst P2O5.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 36

3) Methane reacts with chlorine in the presence of sunlight. Hydrogen atoms of CH4 are replaced by chlorine atAP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 37

Question 26.
Name the following compounds and which one is saturated among them.
a) CH3 – \(\mathrm{C} \equiv \mathrm{H}\) – CH3
b) CH3 – CH = CH – CH3
c) CH3 – CH2 – CH2 – CH3
Answer:
a) 2-Butyne
b) 2 – Butene
c) Butane

Butane does not show any double or triple bonds. Its valency is completely satisfied with formation of single bond. So it is a saturated compound.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 27.
How do you identify the given organic compound contains carboxylic acid functional group?
Answer:

  • On adding carbonates and bicarbonates the compound containing carboxylic acid group evolves carbon dioxide gas.
  • When warmed with alcohol and cone. H2SO4 a pleasant fruity smell is produced due to formation of ester.

Question 28.
Explain briefly about the structure of “Diamond”.
Answer:

  • In a diamond, each carbon atom is surrounded by four other carbon atoms.
  • In these carbon atoms, each carbon atom undergoes in its excited state sp3 hybridisation.
  • These are placed at the four corners of a regular tetrahedron.
  • This results in a 3-dimensional network of carbon atoms.
  • So diamond is in three dimensional structure.

Question 29.
Explain briefly about the structure of “Graphite”.
Answer:

  • In graphite, each ‘C’ is surrounded by three other ‘C’ atoms.
  • The ‘C’ atoms are arranged in layers.
  • In the layer structure, the carbon atoms are in trigonal planar environment.
  • Each layer consists of a 2-dimensional hexagonal network.

Question 30.
Diamond is an extremely bad conductor of electricity.” Why?
Answer:
1) In diamond, each carbon atom is covalently bonded with four other carbon atoms.
2) So, the four outermost electrons of a carbon atom are engaged or trapped in the covalent bonds, having no free electrons making it a bad conductor of electricity.

Question 31.
Why is diamond hard but graphite is smooth and slippery?
Answer:
Diamond has sp³ hybridisation with tetrahedral environment. As C – C bonds are very strong any attempt to distort the diamond structure requires large amount of energy. Hence diamond is one of the hardest material.

Whereas graphite has sp² hybridisation with layer structure with trigonal planar environment. The layers tend to slide on one another. So graphite is smooth and slippery.

Question 32.
An organic compound X with a molecular formula C2H6O undergoes oxidation within presence of alkaline KMnO4 to form a compound Y. X on heating in presence of con. H2SO4 at 443 K gives Z. Which on reaction with Br2 and decolorizes it? Identify X, Y, and Z and write the reactions involved.
Answer:
X is ethanol.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 38

Question 33.
Complete the following reactions.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 39
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 40

Question 34.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 41
What are A and B?
Answer:
1) Alkynes undergo addition reaction in the presence of nickel catalyst and hydrogen to form Alkene.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 42

Question 35.
Draw the structure for the following compounds.
a) Propanoic acid
b) Chlorobutane
c) Hexanone
d) Pentanal
Answer:
a) CH3CH2COOH
b) CH3CH2CH2CH2Cl
c) CH3CH2CH2CH2COCH3
d) CH3CH2CH2CH2CHO

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 36.
Give IUPAC names of the following compounds. If more than one compound is possible, name all of them.
i) A chloride derived from butane.
ii) A ketone derived from pentane.
Answer:
i) The following chlorides are possible for butane.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 43
ii) The following ketones are possible for pentane.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 44

Question 37.
a) What are the various possible structural formulae of a compound having molecular formula C3H6?
b) Give IUPAC names of the above possible compounds and represent them in structure.
c) What is the difference between those
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 45

b) The IUPAC names of compounds are propene and cycle propane.
c) The main difference Is that the first compound Is alkene-an unsaturated compound and second is cyclo alkane-a saturated compound.

Question 38.
Draw isomeric forms of C6H14.
Answer:
Isomers of hexane :
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 46
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 47

Question 39.
How do you appreciate the role of carbon in everyday life?
Answer:

  • Major components of our daily food have carbohydrates, proteins, fats, etc. which are all made up of carbon compounds.
  • The fibres of cloth are made up of cellulose and other types of materials, which are all carbon compounds.
  • Cement and steel form the core of any of the modern buildings. Carbon bestows steel with hardness, while limestone (CaCO3) a major constituent of cement also contains carbon.

Question 40.
How do you appreciate the role of oxygen in combustion process?
Answer:

  • When the oxygen supply is insufficient, the fuels burn incompletely producing mainly a yellow flame.
  • When the oxygen supply is sufficient, the fuels burn completely producing a blue flame.

Question 41.
How do you appreciate the role of Ethanol as a fuel?
Answer:

  • A material which is burnt to obtain heat is called a fuel. Since ethanol burns with a clear flame giving a lot of heat, it is used as a fuel.
  • Some countries add ethanol to petrol to be used as a fuel in cars. Thus ethanol is used as an additive in petrol.
  • Ethanol alone can also be used as a fuel for cars.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 42.
What are the uses of fullerenes?
Answer:
Fullerenes are under study for potential medical use such as specific antibiotics to target resistant bacteria and even target cancer cells such as melanoma.

Question 43.
Write the HJPAC names of the following compounds.
i) CH3 – CH0 – CH2 – CH2 – CH2 – CH2 – CH2 – CH2 – CH2OH
ii) CH3 – CH2 – CH = CH- CH2 – \(\mathrm{C} \equiv \mathrm{CH}\)
iii) CH3 – CH2 – CH2 – CH2 – CHO
iv) CH3 – CH2 – CH2 – CH2 – COOH
Answer:

  1. nananol
  2. 4- ene – 1 heptyne
  3. pentanal
  4. pentanoic acid

Question 44.
What are the uses of alcohol?
Answer:

  • Alcohols are goods solvent for resin and gums.
  • Ethanol is used in the thermometers because of its low freezing point.
  • One of the products of ethyl alcohol is chloroform, which is used as an aesthetic.
  • 10% ethanol in gasoline is a good motor fuel.
  • It is used in medicines such as tincture iodine, cough syrups and many tonics.

Question 45.
What are the uses of acetic acid?
Answer:

  • 5 to 8% solution of acetic acid in water is called vinegar and is used widely as a preservative in pickles.
  • Used as a laboratory reagent.
  • Used in the production of perfumes, dyes, esters, etc.
  • Used in medicine.

10th Class Chemistry 14th Lesson Carbon and its Compounds 4 Marks Important Questions and Answers

Question 1.
Write IUPAC names for the following carbon compounds.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 48
Answer:
A) 2 – methyle pentane – 3 – ol
B) 3 – chloro, 4 – Methyle hexanoic acid
C) 2 Bromo – Bute – 2 – ene
D) 2, 5 Dimethyle hexane

Question 2.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 49 (AP June 2017)
Observe the given carbon compound and answer the following questions.
a) Give numbering to the carbons in the given compound according to IUPAC rules.
b) Name the functional group present in the given compound.
c) Name the word root for the given carbon compound.
d) Write the IUPAC name of the given compound.
Answer:
a)
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 50
b) The given compound contains functional group – OH. It is an alcohol.
c) Word root: The number of carbon atoms present in the molecules is called word root. Here the word root is (C5) – pent.
d) IUPAC name of the given compound is pent 4 – ene 2 – ol.

Question 3.
Alkanes are considered as Paraffins. So, they undergo substitution reactions but not addition reactions. Explain with suitable example. (AP March 2017)
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 51

Question 4.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 52
Observe the structure and answer the following.
a) Write the name of principal functional group present in the compound.
b) Identify the parental chain in the compound.
c) What are the substituents in the above compound?
d) Name the above compound as per IUPAC nomenclature. (AP June 2018)
Answer:
a) Ketone
b)
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 53
c) Methyl group ; Hydroxy group
d) 7 – hydroxy – S – methyl heptan – 2 – one

Question 5.
In the table given below, fill the information in the empty boxes and give answers to the following questions. (TS June 2015)
a Write the general formula of alkanes from the table.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 54
b) How many a bonds are there in C3H6?
c) What sequential order did you notice in the molecular formulae?
d) There exist single bonds between carbon atoms of alkanes. Do you agree with this statement? Give reasons.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 55
a) The general formula of Alkanes is CnH2n+2
b) The number of o bonds in C2H6 are 7.
c) Two successive alkanes are differed by – CH2 group.
d) Except Methane all other alkanes have single bonds between carbon atmos because it is a saturated hydro carbon.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 6.
Why do we call alkanes as paraffins? Explain the substitution reactions of alkanes. (TS June 2016)
Answer:
a) 1. Alkanes are saturated hydrocarbons with least reactivity.
2. Therefore they are called paraffins.
3. Parum = little and affins = affinity.

b) 1. A reaction in which one atom or a group of atoms in a given compound is replaced by other atom or group of atoms is called a substitution reaction.
2. Alkanes have single bonds and undergo substitution reactions.

3. For example :
Methane (CH4) reacts with chlorine in the presence of sunlight.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 56

Question 7.
Write the types of Allotropes of Carbon. Give any three examples of each. (TS March 2016)
Answer:
The allotropes of carbon are classified into two types. They are
i) Amorphous forms,
ii) Crystalline forms.
Examples :
Amorphous forms :
Coal, coke, wood charcoal, animal charcoal, lamp black, gas carbon, petroleum coke, sugar charcoal, etc.

Crystalline forms :
Diamond, graphite, and buckminsterfullerene.

Question 8.
Write any 4 characteristic features of homologous series of Organic compounds. (TS March 2016)
Answer:
Homologous series :
The series of carbon compounds in which two successive compounds differ by – CH2 unit is called homologous series.

Characteristic features of homologous series :

  1. They have one general formula.
    Ex : Alkane (C4H2n + 2), Alkene (C4H2n), Alkyne (C4H2n-2)
  2. Successive compounds in their series possess a difference of (- CH2) unit.
  3. They possess similar chemical properties due to the same functional group.
  4. They show a regular gradation in their physical properties.

Question 9.
List out the materials required to conduct the experiment to understand the esterification reaction. Explain the procedure of the experiment. How can you identify that an ester is formed in this reaction?(TS March 2017)
Answer:
Required Material :
Test tube, beaker, tripod* burner, water, wire guage, ethanol (absolute alcohol), glacial acetic acid, concentrated sulphuric acid.

Procedure :

  1. Take 1 ml of ethanol and 1 ml of glacial acetic acid along with a few drops of concentrated sulphuric acid in the test tube.
  2. Warm it in a water bath or in a beaker containing water for atleast five (5) minutes.
  3. Pour the warm contents into a beaker containing 20-50 ml of water and observe the odour of the resulting mixture.
    If we smell sweet odour from the beaker, we can confirm that ester is formed.

Question 10.
Explain the Isomerism and Catenation properties of carbon. (TS March 2018)
Answer:
Catenation properties of carbon :
i) Carbon has ability to form longest chains with its own atoms. This special property of carbon is called catenation.
ii) Due to catenation property of carbon it can form largest chain containing millions of carbon atoms, branches and cyclic compounds.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 57

Isomerism of carbon :
The phenomenon of possessing some molecular formula but different properties by the compounds is known as isomerism.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 58

Molecular formula of above two molecules is C4 H10 but they have different structure. These two are isomers.

By there two special properties of carbon it can make number of compounds.

Question 11.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 59
Observe the above table and answer the following questions. (TS March 2019)
1) Write the general formula of Alkanes.
2) Mention the names of unsaturated hydrocarbons.
3) Write the homologous series of Alkynes.
4) Write the formula of Hexyne.
Answer:
1) General formula for Alkanes : CnH2n+2.
2) Unsaturated Hydrocarbons in the list are :
Propene C3H6, Butene C4H6, Pentyne C5H8, Hexyne C6H10.

3) Homologous series of Alkynes is C2H2 (Ethyne), C3H4 (Propyne), C4H6 (Butyne), C5H8 (Pentyne), C6H10 (Hexyne).

4) Formula of Hexyne is C6H10.

Question 12.
Complete the following table based on functional groups of organic compounds, their structural formulas and respective suffixes. (AP SCERT: 2019-20)
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 60
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 61

Question 13.
Explain the occurrence of carbon.
Answer:
Carbon occurs in nature in free state as well as in combined state.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 62

Question 14.
What is sp hybridisation? Explain.
Answer:

  • Each carbon is only joining to two other atoms rather than four or three.
  • Here the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridise two of the orbitals.
  • They use the ‘s’ orbital (2s) and one of the 2p orbitals, but leave the other 2p orbitals unchanged.
  • The new hybrid orbitals formed are called sp-hybrid orbitals, because they are made by an s-orbital and a p-orbital reorganizing themselves.

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 15.
Write the characteristics of homologous series of organic compounds.
Answer:
Characteristics of homologous series :

  1. They have one general formula.
    e.g.: Alkanes (CnH2n+2)
  2. Successive compounds in the series possess a difference of – CH2 unit.
  3. They possess similar chemical properties due to same functional group.
    e.g.: C – OH
  4. They show a regular gradation in their physical properties.

Question 16.
What is sp³ hybridisation with diagram? Explain.
Answer:
The excited carbon atom allows its one s-orbital (2s) and three p-orbitals (2px, 2py, 2pz) to intermix and reshuffle into four identical orbitals known as sp³ orbitals. Thus, carbon atom undergoes sp³ hybridization. The four electrons enter the new four identical hybrid orbitals known as sp³ hybrid orbitals, one each as per Hu nd’s rule.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 63

1) Since carbon has four unpaired electrons, it is capable of forming bonds with four other atoms.

2) When carbon reacts with hydrogen, four hydrogen atoms allow their ‘s’ orbitals containing one electron each to overlap with four sp³ orbitals of carbon atom which are oriented at an angle of 109°. 28’.

3) Four orbitals of an atom in the outer shell orient along the four corners of a tetrahedron to have minimum repulsion between their electrons. ‘The nucleus of the atom is at the centre of the tetrahedron.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 64

4) This leads to form four sp³ – s sigma bonds between carbon atom and four hydrogen atoms, All these bonds are of equal energy,

Question 17.
What is sp² hybridisation? Explain.
Answer:
Consider ethene molecule
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 65

  • In the formation of CH2 – CH2 each carbon atom in its excited state undergoes sp² hybridisation by intermixing one s-orbital (2s) and two p-orbitals (say 2px and 2py) and reshuffling to form three sp² orbitals.
  • Mow each carbon atom is left with one ‘p’ orbital (say 2pz) unhybridised,
  • The three sp² orbitals having one electron each get separated around the nucleus of carbon atoms at an angle of 120°.
  • When carbon is ready to form bonds one sp² orbital of one carbon atom overlaps the sp² orbital of the other carbon atom to form sp² – sp² sigma (σ) bond,
  • The remaining two sp² orbitals of each carbon atom get overlapped by ‘s’ orbitals of two hydrogen atoms containing unpaired electrons.
  • The unhybridised pz orbitals on the two carbon atoms overlap laterally as shown in figure to form a π (pi) bond.
  • Hence, there exist a sigma (σ) bond and a pi π (pi) bond between two carbon atoms in ethene molecule. Hence, the molecule ethene (C2H4) is

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 66

Question 18.
The list of some organic compounds is given below.
Ethanol, ethane, methanol, methane, ethyne and ethene.
From the above list name the compound …………..
a) formed by the dehydration of ethanol by cone. H2SO4.
b) which forms methanoic acid on oxidation?
c) which forms chloroform on halogination in the presence of light?
d) which are unsaturated compounds?
e) which have compounds containing alcohol group?
Answer:
a) Dehydration ethanol in the presence of Cone. H2SO4 forms ethene,
b) Methanol on oxidation turns to methanoic acid,
c) Methane in the presence of light forms chloroform,
d) Unsaturated compounds are ethene and ethyne.
e) The compounds containing alcohol group are methanol, ethanol,

Question 19.
Give the IUPAC names of the following compounds.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 67
Answer:

  1. 1 – propyne
  2. 3 – pentanel (or) pentamSml
  3. 2 – methyl propane
  4. 1, 2 dichloro ethane

Question 20.
Give one example of each of the following.
i) Saturated hydrocarbon
ii) Cyclic compounds
iii) Unsaturated hydrocarbon
iv) Functional group
v) Homologous series
Answer:
i) Saturated hydrocarbons are Alkanes, So the examples are methane (CH4), Ethane, (C2H6).
ii) Cyclic compounds are cycle alkanes, eg : Cyclo propane (C3H6), Cycle butane (C4H6).
iii) Unsaturated hydrocarbons are Aikynes, eg : Ethene (C2H4), Propene
iv) The examples for functional groups are ‘ 1. Aldehyde – CHO, 2. Alcohol = OH
v) A series of carbon compounds that differ by – CH2 with similar chemical properties is called homologous series.
eg: 1, Alkane, 2, Alkene

AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 21.
Write the differences between saturated and unsaturated hydrocarbons.
Answer:

Saturated hydrocarbons Unsaturated hydrocarbons
1) All the four valencies of each carbon atom are satisfied by forming single covalent bonds with carbon and hydrogen atoms, 1) The valencies of at least two carbon atoms are not fully satisfied by the hydrogen atoms.
2) Carbon atoms are joined only by single bonds. 2) Carbon atoms are joined by at least one double bond or by a triple bond.
3) They are less reactive due to non­availability of electrons in the single covalent bond therefore they undergo substitution reactions, 3) They are more reactive because of the presence of electrons in the double or triple bond and therefore undergo addition reactions.

Question 22.
Answer the following.
a) What are the first three members of carboxylic acid series?
b) Name the compounds which can be oxidised directly or in stages to produce ethanoic acid.
c) Write one equation each when acetic acid reacts with a metal, a base, and a carbonate.
d) Name the organic compound formed when acetic acid and ethanol react together.
Answer:
a) The first three members of carboxylic acids are :
i) Methanoic acid – HCOOH
ii) Ethanoic acid – CH3COOH
iii) Propanoic acid – CH3CH2COOH

b) Ethanol in stages oxidises to acetic acid whereas ethanol directly oxidises to ethanoic acid.

c) i) 2 CH3COOH + 2 Na → 2CH3COONa + H2
ii) CH3COOH + NaOH → CH3COONa + H2O
iii) CH3COOH + Na2CO3 → CH3COONa + H2O + CO2

d) When ethanol reacts with ethanoic acid it forms an ester namely ethyl acetate.

Question 23.
What are the rules to be followed to name a carbon compound?
Answer:
Rules to be followed
i) Longest carbon chain is selected,
ii) Chain is numbered in such a way that the branched chain or substituent gets the smallest number,
iii) If the functional group is present, it is given the. lowest number,
iv) Substituents are named in the alphabetical order,
v) The position of substituents are prefixed with hyphen,
vi) Multiple substituents are written with numerical prefixes such as di or tri,

Question 24.
Write suffixes and prefixes for some important characteristic functional group in a tabular form.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 68

Question 25.
Correct the following statements.
1) Alkenes undergo substitution reactions.
2) Alkanes are polar in nature.
3) When sodium piece is added to ethanol oxygen gas liberates.
4) On complete combustion of carbon compound it gives carbon monoxide and water.
Answer:

  1. Alkenes are unsaturated hydrocarbons. So they undergo addition reactions.
  2. Alkanes are covalent compounds. So they are non-polar in nature.
  3. When sodium piece is added to ethanol it releases hydrogen gas.
  4. On complete combustion of carbon compound it forms carbon dioxide and water.

Question 26.
Copy and complete the following table which relates to three homologous series of hydrocarbons.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 69
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 70

Question 27.
Draw the structures of isomers of butane.
Answer:
Isomers of butane are n-butane, iso butane and cyclo butane :
Structures :
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 71

Question 28.
Draw the structures of the following.
a) Ethanoic acid
b) Propanal
c) Propene
d) Chloro propene
Answer:
Structures:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 72

Question 29.
Draw the structures of the following compounds
a) 2 – bromo pentane
b) 2 – methyl propane
c) butanal
d) 1 – hexyne
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 73

Question 30.
Write the molecular formula of the first four compounds of the homologous series of aldehydes.
Answer:
Homologous series of aldehydes ate Formaldehyde, Acetaldehyde, Propionaldehyde and Butanaldehyde.
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 74
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 75

Question 31.
How many isomers can be drawn for pentane with molecular formula C-H(2? What are they? Draw their structures and mention theii common names.
Answer:
Isomers of pentane are three. These are
1) Pentane
2) Iso pentane
3) Neo pentane.
Structures :
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 76

Question 32.
Draw the Allotropes of Carbon. Diamond
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 77

Question 33.
Draw the Graphite.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds

Question 34.
Draw the Buckminsterfullerene (60C).
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 78

Question 35.
Draw the Nanotubes. A.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 79

Question 36.
Draw the structures of Methane :
Answer:
Methane :
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 64

Question 37.
Draw the structures of Ethyne :
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 80

Question 38.
Draw structures of the Ethane and electron dot structure of Chlorine.
Answer:
Ethane:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 66
Chlorine:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 81

Question 39.
Draw the electron dot structures of Ethanoic acid arid Ethyne (Acetylene).
Answer:
Ethanoic acid (Acetic acid) :
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 82
Structure of Ethyne (Acetylene) :
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 80

Question 40.
Draw the electronic dot structure of ethane molecule.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 83

Question 41.
Write the structures of the following compounds.
a) prop-l-ene
b) 2, 3-dimethyl butane
c) 3-hexene
d) 2-methyl prop-l-ene
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 14 Carbon and its Compounds 84

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 5 Refraction of Light at Plane Surfaces.

AP State Syllabus SSC 10th Class Physics Important Questions 5th Lesson Refraction of Light at Plane Surfaces

10th Class Physics 5th Lesson Refraction of Light at Plane Surfaces 1 Mark Important Questions and Answers

Question 1.
Take a bright metal ball and make it black with soot on a candle flame. Immerse it in the water. Mention one observation. (AP June 2015)
Answer:
1) The ball shines.
2) The ball appears to raise up in water.

Question 2.
What is critical angle? (AP March 2015)
Answer:
The angle of incidence at which the light ray propagates from denser to rarer graze along interface is called critical angle of denser medium.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 3.
If refractive index of glass is \(\frac{3}{2}\), then what is speed of light in glass? (AP June 2016)
(OR)
Find the speed of light in a transparent medium, whose refractive index is 3/2.
Answer:
The refractive index of glass or transparent medium = \(\frac{3}{2}\)
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 1

Question 4.
Write any two questions about the ‘formation of mirages’. (AP June 2017)
Answer:

  1. When does a mirage form?
  2. How does a mirage form?

Question 5.
Optical Fibre Cable (OFC) are oftenly used in tele-communications. What is the working principle behind the OFC? (AP March 2017)
Answer:
Total Internal Reflection.

Question 6.
Among objects made of glass and diamond, which one shines more? Why? (AP June 2015)
Answer:
Diamond shines more because of low conical angle of 24.4°.

Question 7.
Suggest reasons for the phenomenon associated with the following : Twinkling of stars. (TS March 2015)
Answer:
Refraction of light is the reason for the twinkling of stars.

Question 8.
Draw the diagram showing the path of the ray when it travels from denser medium to rarer medium when the incident angle is more than the critical angle. (TS June 2016)
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 2

Question 9.
Why does the light ray deviate in refraction? (AP SA-1:2019-20)
Answer:
Light ray always chooses the path of least time to travel. Hence speed of light changes at interface of two media. So, the light ray deviate in refraction.

Question 10.
Name the phenomenon involved in the function of optical fibre. (AP SCERT : 2019-20)
Answer:
Total Internal Reflection.

Question 11.
What is Fermat’s principle?
Answer:
The light ray always travels in a path which needs shortest possible time to cover the distance between the two given points.

Question 12.
What happens when light travels from one medium to another medium?
Answer:
It bends towards or away from normal.

Question 13.
When does speed of light decrease?
Answer:
When it travels from rarer to denser medium.

Question 14.
What do you mean by denser medium?
Answer:
The medium which has more optical density.

Question 15.
What is refraction?
Answer:
The process of changing speed when light travels from one medium to another is called refraction of light.

Question 16.
Which quantity will compare the refractive indices of two media?
Answer:
Relative refractive index.

Question 17.
What is relative refractive index?
Answer:
It is the ratio of refractive index of second medium to refractive index of first medium.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 3

Question 18.
When does a light ray bend away from normal?
Answer:
When a light ray moves from denser to rarer medium it bends away from normal.

Question 19.
When will angle of refraction be equal to 90°?
Answer:
When angle of incidence is equal to critical angle then angle of refraction will be equal to 90°.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 20.
When angle of incidence of light ray is greater than critical angle, what happens?
Answer:
Light ray undergoes total internal reflection.

Question 21.
What happens to refractive index of air with height?
Answer:
Refractive index of air increases with height.

Question 22.
Which has greater refractive index between these?
1) cool air at the top
2) hotter air just above the road
Answer:
Cooler air has greater refractive index due to more density.

Question 23.
What is mirage?
Answer:
The virtual images of distant high objects cause the optical illusion called mirage.

Question 24.
What happens to a light ray when it falls perpendicular to one side of the slab surface?
Answer:
It comes out without any duration.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 25.
What are the conditions for total internal reflection?
Answer:

  1. The rays of light must travel from denser to rarer medium.
  2. The angle of incidence of denser medium must be greater than critical angle.

Question 26.
What is meant by a vertical shift?
Answer:
When a ray emerges out of a glass slab, it is parallel to the incident ray but is displaced laterally relative to incident ray. This shift of emergent ray is called vertical shift.

Question 27.
A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer:
It bends towards the normal. This is because it travels from an optically rarer to , optically denser medium.

Question 28.
When does Snell’s law fail?
Answer:
Snell’s law fails when light is incident normally on the surface of a refracting medium.

Question 29.
Why does a ray of light bend when it travels into another medium?
Answer:
It bends because its velocity changes when it moves from one medium to the other.

Question 30.
A pencil when dipped in a glass tumbler containing water appears to be bent at the interface of air and water. Explain why.
Answer:

  • When light travels obliquely from one transparent medium to another, the direction of propagation of light changes due to refraction of light.
  • In this case, light travels from denser medium to rarer medium, hence it bends away from the normal and the pencil appears to be bent.

Question 31.
Why does light travel in vacuum?
Answer:
Light travels in vacuum because it does not require medium for its propagation.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 32.
What are the factors on which refractive index depends?
Answer:

  1. Nature of material
  2. Wavelength of light used.

Question 33.
What does the ratio of sine of angle of incidence and sine of angle of refraction give?
Answer:
The ratio of sine of angle of incidence and sine of angle of refraction gives refractive index.

Question 34.
What is the relationship between critical angle and refractive index?
Answer:
The relationship between critical angle and refractive index
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 4

Question 35.
What is the relationship between angle of incidence and shift?
Answer:
As the angle of incidence increases, the shift also increases.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 36.
Why does a coin placed in a water appear to be raised?
Answer:
It is due to refraction of light.

Question 37.
Can you guess what happens when light travels from denser medium to rarer medium?
Answer:
The light ray bends away from normal.

Question 38.
A ray of light falls normally on a face of a glass slab. What are the values of angle of incidence and angle of refraction of this ray?
Answer:
Both angles are zero.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 39.
When does light ray from slab not undergo any deviation?
Answer:
The light ray that incidents perpendicular to one side of the slab surface comes out without any deviation.

Question 40.
What is the factor on which refraction depends?
Answer:
Refraction depends on optical density.

Question 41.
What is absolute refractive index?
Answer:
It is the ratio of speed of light in vacuum to speed of light in medium.
\(\mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}}\)

Question 42.
In water filled vessel, the coin of the bottom can be seen at a height. Give reasons.
Answer:
Rising of coin when water is poured in a cylindrical transparent vessel.

Question 43.
Write one activity in showing the process of ‘total internal refraction’.
Answer:
Due to refraction of light speed of light changes when it travels from one medium to another medium.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 44.
Define “Glass Slab”.
Answer:
A thin glass slab is formed when a medium is isolated from its surroundings by two plane surfaces parallel to each other.

Question 45.
A ray of light is incident normally on a plane glass slab. What will be the angle of refraction and angle of deviation for the ray?
Answer:
The ray is incident normally on a plane glass slab. So there is no deviation of light ray. Therefore the angle of refraction and angle of deviation both have 0° values.

Question 46.
A light ray in passing from water to a medium (a) speeds up, (b) slows down. In each case get one example of the medium.
Answer:
a) Air, because its optical density is less than water,
b) Glass, because its optical density is more than water.

Question 47.
If an angle of refraction is 90°, what is the corresponding angle of incidence called?
Answer:
The angle of incidence is called critical angle.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 48.
If the angle of incidence is more than critical angle, what happens to light ray if the light ray travels from denser to rarer medium?
Answer:
The light ray undergoes total internal reflection.

Question 49.
The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?
Answer:
It means that light travels 2.42 times faster in vacuum than in diamond.

Question 50.
For the same angle of incidence 45°, the angle of refraction in two transparent media, I and II is 20° and 30° respectively. Out of I and II, which medium is optically denser and why?
Answer:
Medium I is optically dense as angle of refraction is lesser in it, hence light bends towards normal.

Question 51.
For which colour of white light is the refractive index maximum and for which colour of white light is the refractive index minimum?
Answer:
The refractive index is maximum for violet because its wavelength is least.
The refractive index is minimum for red because its wavelength is maximum.

Question 52.
Correct the statement. “If the angle of incidence is greater than the critical angle the light is refracted when it falls on the surface from a denser medium to rarer medium”.
Answer:
If the angle df incidence is greater than the critical angle the light undergoes total internal reflection when it falls on the surface from a denser medium to rarer medium.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 53.
A light ray passes from medium 1 to medium 2. Which of the following quantities of refracted ray will differ from that of the incident ray?
Speed, intensity, frequency, wavelength.
Answer:
Speed, intensity and wavelength will differ from that of incident ray.

Question 54.
The refractive indices of alcohol and turpentine oil with respect to air are 1.36 and 1.47 respectively. Find the refractive index of turpentine oil with respect to alcohol. Which one of these permits the light to travel faster?
Answer:
The refractive index of turpentine oil with respect to alcohol = \(\frac{1.47}{1.36}\) = 1.08.

The refractive index increases when the speed of light decreases. So light travels faster in alcohol as its refractive index is less.

Question 55.
Light enters from air to diamond which has refractive index of 2.42. Calculate the speed of light in diamond, if speed of light in air 3 × 108 ms-1.
Answer:
Absolute refractive index = \(\frac{c}{v}\)
2.42 = \(\frac{3 \times 10^{8}}{\mathrm{v}}\) ⇒ v = 1.24 × 108 ms-1.

Question 56.
A glass block 3.0 cm thick is placed over a stamp. Calculate the height through which image of stamp is raised. Refractive index of glass is 1.54.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 5

Question 57.
The refractive index of water is 4/3. Calculate the critical angle for water – air interface (sin 49 = 3/4).
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 6

10th Class Physics 5th Lesson Reflection of Light by Different Surfaces 2 Marks Important Questions and Answers

Question 1.
A ray of light enters from air to a medium X. The speed of light in the medium is 1.5 × 108 m/s and the speed of light in air is 3 × 108 m/s.
Find the refractive index of the medium X. (TS March 2015)
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 7

Question 2.
What are the applications of optical fibres?
(OR)
Write two uses of fibre optics in daily life. (TS June 2016)
Answer:
Applications oruses of optical fibres :

  1. Light pipes using optical fibres may be used to see places which are difficult to reach things such as inside of a human body.
  2. The other important application of fibre optics is to transmit communication signals through light pipes.

Question 3.
Focal length of the lens depends on its surrounding medium. What happens, if we use a liquid as surrounding media of refractive index, equal to the refractive index of lens? (TS June 2018)
Answer:

  • When the refractive index of surrounding media is equal to the refractive index of lens, the lens looses its characteristics.
  • Lens do not diverge or converge the light.
  • Light do not get refracted when it passes through that lens.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 4.
Why does the light ray travel slowly in diamond when compared to vacuum? (AP SA-I : 2019-20)
Answer:

  • Refractive index of diamond (2.42) is greater than that of vacuum (1).
  • Speed of light is inversely proportional to refractive index of substances.
  • Hence, light ray travel slowly in diamond when compared to vacuum.

Question 5.
Write about laws of refraction.
(OR)
Write the laws of refraction.
Answer:
Laws of refraction :

  1. The incident ray, the refractive ray and the normal to interface of two transparent media at a point of incidence lie in the same plane.
  2. During refraction light follows Snell’s law, i.e., the ratio of sine of angled of incidence to sine of angle of refraction is constant.
    n1 sin i = n2 sin r (OR) \(\frac{\sin \mathrm{i}}{\sin \mathrm{r}}\) = constant.

Question 6.
What is total internal reflection ? What are the applications of total internal reflection?
Answer:
When the angle of incidence is greater than critical angle, the light ray is reflected into denser medium at interface i.e., light never enters rarer medium. This phenomenon is called total internal reflection.
1) Brilliance of diamonds :
Total internal reflection is the main cause for brilliance of diamonds. The critical angle of diamonds is very low (24.4°). So if a light ray enters a diamond it is very likely to get total internal reflection which makes the diamond shine brilliant.

2) Optical fibres:
Total internal reflection is the basic principle for working of optical fibre.

Question 7.
What are optical fibres? How do they work?
Answer:

  • An optical fibre is very thin fibre made of glass or plastic having radius about a micrometer.
  • A bunch of such thin fibres forms a light pipe.

Working :

  1. Because of the small radius of the fibre, light going into it makes a nearly glancing incidence on the wall.
  2. The angle of incidence is greater than the critical angle and hence total internal reflection takes place.
  3. The light is thus transmitted along the fibre.

Question 8.
How can a patient’s stomach be viewed by using optical fibres?
(OR)
How do you observe patient’s stomach by using a light pipe?
Answer:

  • The patient’s stomach can be viewed by inserting one end of a light pipe into the stomach through the mouth.
  • Light is sent down through one set of fibres in the pipe.
  • This illuminates the inside of the stomach.
  • The light from the inside travels back through another set of fibres in the pipe and the viewer gets the image at the outer end.

Question 9.
State four differences between reflection and total internal reflection.
Answer:

Reflection Total internal reflection
1) Smooth polished surface is required for reflection. 1) No smooth polished surface is required for total internal reflection.
2) It takes place for all angles of incidence. 2) It takes place only, when angle of incidence is greater than critical angle.
3) It takes place when the rays of light travel from rarer to denser medium to an opaque medium. 3) It takes place when rays of light travel from denser to rarer medium.
4) Some amount of light is absorbed by reflecting surface. 4) No light is absorbed by reflecting surface.

Question 10.
The figure shows refraction and emergence of a ray of light incident on a rectangular glass slab. Copy the diagram and mark the lateral displacement of the incident ray. Name the two factors on which the lateral displacement depends.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 8
Answer:
The lateral displacement depends on

  1. The angle of incidence of the incident ray PQ, on the slab and
  2. The thickness of the glass slab.

The perpendicular distance between the emergent forward.

Question 11.
1) What happens to a ray of light when it travels from one equal refractive indices?
2) State the cause of refraction of light.
Answer:
1) No refraction or bending would take place. The light will travel in a straight line.

2) The refraction occurs due to change in speed of light as it enters from one medium to another.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 12.
A coin placed at the bottom of a tank appears to be raised when water is poured into it. Explain.
Answer:

  • It happens due to the phenomenon of refraction of light.
  • When the rays of light from the coin, in the denser medium fall on the interface separating the two media, the rays of light move away from the normal after refraction.
  • The point from which the refracted rays appear to come gives the apparent position of the coin.
  • As the rays appear to come from a point above the coin, therefore, the coin seems to be raised.

Question 13.
Define refractive index. Explain the relationship between the refractive index of the medium and to the speed of light in the medium.
Answer:
The ratio of speed of light in vacuum to the speed of light in that medium is defined as refractive index ‘n’ with respect to the vacuum. It is also called absolute refractive index.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 9
When refractive index of a medium is high, then the speed of light is low and vice-versa.

Question 14.
Explain lateral shift and vertical shift.
Answer:
Lateral shift:
The distance between incident and emergent ray is called lateral shift.

Vertical shift :
The perpendicular distance between object and its image is called vertical shift.

Question 15.
During refraction of light, which of the following quantities does not change.
(1) velocity,
(2) wavelength,
(3) frequency,
(4) amplitude.
Answer:
During refraction of light velocity of light changes and also wavelength and amplitude. Frequency does not change during refraction.

Question 16.
The upper surface of water contained in a beaker and held above the eye level appears silvery. Why?
Answer:
Critical angle for water is 48°. The rays of light entering in water from below, suffer refraction. If these rays strike the water-air surface at an angle which is greater than 48°, they get totally internally reflected. These rays on emerging out of water, appear to come from the upper surface of water, which in turn appear silvery.

Question 17.
Why don’t the planets twinkle?
Answer:

  • The planets are much closer to the earth, and are thus seen as extended sources.
  • We can consider a planet as a collection of a large number of point-sized sources of light.
  • The total variation in the amount of light entering our eye from all the individual point-sized sources will average out to zero, thereby nullifying twinkling effect

Question 18.
Why did an empty test tube placed obliquely in water, appears filled with mercury, when seen from above?
Answer:
When the rays of light travelling through water they strike the water glass interface of test tube at an angle, which is more than critical angle for water, they suffer total internal reflection. When these totally reflected rays reach eye, then to the eye they appear as they come from surface of test tube, which in turn appears filled with mercury.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 19.
Why are the bubbles rising up the fish tank appear silvery?
Answer:
When the rays of light travelling through water they strike the water air interface of the bubble at an angle, which is greater than critical angle for water, they get totally internally reflected. These reflected rays on reaching the eye appear to come from air bubble, which in turn appears silvery.

Question 20.
Why does a crack in a window pane appear silvery?
Answer:
There is always some amount of air present in the crack. When the rays of light travelling through glass, strike the glass, the glass air interface at an angle, greater than critical angle of glass, they are totally internally reflected. When these reflected rays reach eye, then to the eye they appear to come from the crack, which in turn appears silvery.

Question 21.
Explain why a straight stick appears to be bent when dipped in water.
Answer:

  • Suppose two rays originate from the end of the stick in water.
  • As these rays get refracted into the air, they bend away from the normal.
  • When these two refracted rays are produced backwards they seem to meet at a point higher than the end of stick.
  • This point gives the apparent position of the end of the stick. Thus, the stick appears to be bent.

Question 22.
A pond appears to be shallower than it really is when viewed obliquely. Why?
Answer:

  • Suppose two rays originate from the bottom of the pond. As these rays get refracted into the air, they bend away from the normal.
  • When these two refracted rays are produced backwards they seem to meet at a point higher than the bottom of the pond.
  • This point gives the apparent position of the bottom of the pond.
  • Thus, the pond appears to be shallower.
  • This effect is absent if the pond is viewed normally.

Question 23.
Frame some questions to know about the formation of mirages.
Answer:

  1. What are mirages?
  2. What is the principle involved in mirages?
  3. Can mirages be photographed?
  4. Where does the water on the road go?

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 24.
A glass slab is placed over a piece of paper on which VIBGYOR is printed with each letter into corresponding colours.
1) Will the image of all the letters be in the same place?
2) The letter of which colour appears to be raised maximum and which colour minimum? Explain your answer.
Answer:

  1. The image of all letters will not be in the same place.
  2. The letter of violet colour appears to be raised maximum, while the letter of red colour appears to be raised minimum because refractive index of glass is most for the violet light while least for the red light, therefore the apparent depth is least for violet and most for red.

Question 25.
Why does sun appear bigger during the sunset or the sunrise?
Answer:

  1. We already know that the apparent position of sun is higher than actual position in the horizon.
  2. Moreover, due to refraction, the apparent image of sun is closer to eye than the actual position. Since during sunset or sunrise, the rays of light travel through maximum length of atmosphere therefore the refraction is also maximum.
  3. Hence apparent image of sun is very much closer to eye. Thus it appears bigger.

Question 26.
Write the material required in finding out the relation between angle of incidence and angle of refraction.
Answer:
Material required :
A plank, white chart, protractor, scale, small black painted plank, a semi-circular glass disc of thickness nearly 2cm, pencil and laser light.

Question 27.
Write the aim and apparatus experiment in finding the refractive index of the glass slab.
Answer:
Aim :
Finding the refractive index of the glass slab.

Apparatus :
Glass slab, white chart, pin.

Question 28.
Observe the following table and answer the following question.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 10
Questions :
1) Find out from the table the medium having highest optical density and the medium with lowest optical density.
2) You are given kerosene, turpentine oil and water. In which of these does the light travel fast? Use the information given in the table.
3) The refractive index of diamond is 2.42. What is the meaning of this statement?
4) When light travels from water to crown glass, what happens?
5) When light travels from diamond to air, what happens?
Answer:
1) The medium with highest optical density is diamond as its refractive index is maximum, i.e. 2.42.
The medium with lowest optical density is air, as its refractive index is minimum, i.e. 1.0003.

2) The refractive index of medium is given by the expression, n = \(\frac{c}{v}\) or v = \(\frac{c}{n}\)
This expression shows that light travels faster in the medium whose refractive index is minimum. From the table, we can find that water has the minimum value of refractive index. Therefore light travels faster in water.

3) This statement means that light travels 2.42 times faster in vacuum than in diamond.

4) The light bends towards normal.

5) The light bends away from the normal.

Question 29.
A ray of light enters from a medium A into a slab made up of a transparent substance B. Refractive indices of medium A and B are 2.42 and 1.65 respectively. Complete the path of ray of light till it emerges out of slab.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 11
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 12

Question 30.
A glass slab made of material of refractive index n1 is kept in medium of refractive index n2. A light ray is incident on the slab. Complete the path of rays of light emerging from glass slab, if a) n1 > n2 b) n1 = n2 c) n1 < n2.
Answer:
a) n1 > n2
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 13

b) n1 = n2
There is no deviation of light ray
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 14
c) n1 < n2
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 15

Question 31.
How do you appreciate the process of total internal reflection in nature?
Answer:

  1. Total internal reflection is responsible for brilliance of diamond.
  2. Total internal reflection is basic pruxiplo behind working of optical fibres which are used in getting the images ol internal ouaiis and also used in telecommunications. So the role of total internal reflection is thoroughly appreciated.

Question 32.
Write the application of optical fibres in communication.
Answer:

  • Optical fibres are used to transmit communication signals through light pipes.
  • For example, about 2000 telephone signals, approximately mixed with lightwaves, may be simultaneously transmitted through a typical optical fibre.
  • The clarity of the signals transmitted in this way is much better than other conventional methods.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 33.
Write the applications of total internal reflection.
Answer:
Application of total internal reflection :

  1. Brilliance of diamonds,
  2. Optical fibres.

Question 34.
A monochromatic ray of light strikes the surface of transparent medium at an angle of incidence 60° and gets refracted into the medium at an angle of refraction 45°. What is the refractive index of the medium?
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 16

Question 35.
A light ray enters a liquid at an angle of incidence 45° and it gets refracted on liquid at angle of refraction 30°. Calculate the refractive index of the liquid.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 17

Question 36.
Refractive index of water is 4/3. Calculate the speed of light in water.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 18

Question 37.
A postage stamp placed under glass appears raised by 8 mm. If refractive index of glass is 1.5, calculate the actual thickness of glass slab.
Answer:
Let real thickness of glass = x.
Vertical shift = 8 mm.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 19

Question 38.
Refractive index of glass is 1.5. Find its critical angle.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 20

Question 39.
What is the advantage of using prism in place of plane mirror in periscope or binocular?
Answer:

  • When total internal reflection occurs from a prism, the entire incident light is reflected back into the denser medium.
  • Whereas in ordinary reflection from a plane mirror, some light is refracted and absorbed. So the reflection is partial.
  • This is the reasons why total reflecting prism is used in place of a plane mirror to deviate the light ray by 90° in a periscope and 180° in a binocular.

10th Class Physics 5th Lesson Reflection of Light by Different Surfaces 4 Marks Important Questions and Answers

Question 1.
What is the angle of deviation produced by the glass slab? Explain with ray diagram. (AP June 2015)
(OR)
Which angle of deviation is produced by glass slab? Write your explanation with a ray diagram.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

  1. Angle of deviation is the angle between incident ray and emergent ray.
  2. The angle of deviation produced by a glass slab is ‘O’, because the incident ray and emergent ray are parallel to each other that can be seen in the figure.

Question 2.
Explain the phenomenon of total internal reflection with two examples. (AP June 2018)
(OR)
What is total internal reflection? Explain with examples. (AP SA-I:2019-20)
Answer:

  • When the angle of incidence is greater than the critical angle, the light ray is reflected into denser medium at interface. This phenomenon is called total internal reflection.
  • Total internal reflection is the main reason for brilliance of diamonds. The critical angle of a diamond is very low. So if a light ray enters a diamond it is very likely to undergo total internal reflection which makes the diamond shine.
  • Total internal reflection is the basic principle behind working of optical fibre. Because of the small radius of the fibre light going into it makes a nearly glancing incidence is greater than the critical angle and hence total internal reflection takes place. The light is thus transmitted along the fibre.

Question 3.
Explain the relation between angle of incidence and angle of refraction with an experiment. (AP March 2018)
Answer:
Aim :
To verify the relation between angle of incidence and angle of refraction.

Material required :
A plank, white chart, protractor, semicircular glass disc, pencil and leser light.

Procedure :

  1. Take a drawing sheet on a cardboard and mark different angles (on both side of MM line)
  2. Place a semi circular glass disc, so that its diameter coincides with the line “MM”.
  3. Send a laser light along a line with makes 15° with NN.
  4. Let it is incident angle.
  5. Measure its corresponding angle of refraction by observing light coming from outside of the glass slab.
  6. Repeat this experiment with various values of angle of incidence, refraction and not in the table.

7.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 21
8. From the above table we observe that \(\frac{\sin \mathrm{i}}{\sin r}\) = constant.
AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 5AP SSC 10th Class Physics Solutions Chapter 5 Refraction of Light at Plane Surfaces 6

Question 4.
Give some daily life consequences of refraction of light.
Answer:

  • A star appears twinkling in the sky.
  • The sun is seen a few minutes before it rises above the horizon in the morning and in the evening few minutes longer after it sets.
  • A coin kept in a vessel not visible when seen from just below the edge of the vessel, can be viewed from the same position when water is poured into the vessel.
  • A print appears to be raised when a glass block placed over it.
  • A piece of paper stuck at the bottom of a glass block appears to be raised when seen from above.
  • A tank appears shallow than its actual depth.
  • A person’s legs appear to be short when standing in a tank.
  • An object placed in a denser medium when viewed from a rarer medium appears to be at a lesser depth.
  • An object in a rarer medium, when viewed from a denser medium, appears to be at a greater distance than its real distance.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 5.
What are the factors which influence refractive index of material?
Answer:

  • Nature of medium, i.e. its optical density. Smaller the speed of light in a medium relative to air, higher is the refractive index of the medium.
  • Physical condition such as temperature. With rise in temperature the speed of light in medium increases, so the refractive index of medium decreases.
  • The colour or wavelength of light (refractive index increases with decrease in wavelength, eg : µv > µR).

Question 6.
What is the advantage of total internal reflection over reflection?
Answer:

  • In the process of total internal reflection, 100% energy is reflected back.
  • No other device such as plane mirror, etc. produces 100% reflection due to absorption and refraction of some part of light.
  • Due to this property the phenomenon total internal reflection is of great practical application in the construction of periscope, binocular and certain type of camera.

Question 7.
The diagram below shows a glass block suspended in a liquid. A beam of light of single colour is incident from liquid on one side of block.
1) Draw diagrams to show how light bends when it travels from liquid to glass and then to liquid if (i) the light slows down in glass (ii) the light speeds up in glass.
2) State two conditions under which the light ray moving from liquid to glass passes straight without bending. Will the glass be visible them?
Answer:
1) If light slows down in going from liquid to glass (i.e., µglass > µliquid), it will bend towards the normal at the point of incidence in passing from liquid to glass at the first surface, while it is bent away from normal at the second surface in passing from glass to liquid. In the ray diagram, the light beam suffers lateral shift.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 22

2) If light speeds up in going from liquid to glass (i.e., µglass < µliquid). It will bend away from the normal at the point of incidence on the first surface in passing from liquid to glass, while it bends towards the normal at the second surface in passing from glass to liquid. The light beam suffers lateral shift in direction opposite to that
Note that in both cases, the emergent ray is parallel to the incident ray.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 23

Question 8.
A ray of light is incident on a rectangular glass block PQRS, which is silvered at the surface RS. The ray is partly reflected and partly refracted.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 24
1) Trace the path of reflected and refracted rays.
2) Show at least two rays emerging from the surface PQ after reflection from the surface RS.
3) How many images are formed in the above case? Which image is the brightest?
Answer:
1) In the figure OB is reflected ray and OC is the refracted ray for the incident ray AO.
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 25
2) Two rays emerging from surface PQ after reflections for the surface RS are labelled as 1 and 2.

3) Multiple (or infinite) images are formed. The second image formed due to first reflection at C at the silvered surface RS is the brightest. It is seen in the direction of ray 1.

Question 9.
What are the factors which affect critical angle? The critical angle for a given pair of media depends on their refractive index which is affected by the following factors.
Answer:
1. Effect of colour of light:
The refractive index of transparent medium is more for violet light and less for red light, therefore the critical angle for pair of media is less for the violet light and more for the red light. Thus critical angle increases with increase in wavelength of light.

2. Effect of temperature :
On increasing the temperature of medium, its refractive index decreases, so the critical angle for that pair of media increases. Thus critical angle increases with increase in temperature.

Question 10.
The table shows the refractive index of some material media.

Material Medium Refractive Index
Air 1.0003
Ice 1.31
Water 1.33
Kerosene 1.44
Fused quartz 1.46
Turpentine oil 1.47
Crown glass 1.52
Benzene 1.50

Answer the following questions with the help of the above table.
1) Find the speed of light in Benzene.
2) Write the relationship between mass density and optical density of kerosene and water.
3) What are the factors that refractive index depends on?
4) Write the relative refractive index of kerosene with water.
Answer:
1)
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 26
Speed of light in benzene = 2 × 108 m/s

2) Optical denser medium may not possess greater mass density. Kerosene with high refractive index is optically denser than water although its mass density is lesser than water.

3) Refractive index depends on
1) nature of material,
2) wavelength of light used.

4) Relative refractive index of kerosene with water = \(\frac{1.44}{1.33}\) = 1.08

Question 11.
Red light of wavelength 6600A travelling in air gets refracted in water. If the speed of light in air is 3 × 108 ms-1 and refractive index of water is 4/3, find the
(i) frequency of light in air,
(ii) the speed of light in water,
(iii) the wavelength of light in water.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 27

Question 12.
Draw the ray diagram which shows the ray takes curved path because of total internal reflection.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 28

Question 13.
Give some daily life consequences of total internal reflection?
Answer:

  • On a hot sunny day, a driver may see a pool of water on the road before him. It is the phenomenon of mirage which is often observed in desert.
  • An empty test tube placed in a beaker with mouth outside the water surface shines like a mirror.
  • A crack in a glass vessel often shines like a mirror.
  • A piece of diamond sparkles when viewed from certain directions.
  • An optical fibre is used to transmit a light signal over a long distance with negligible loss of energy.

AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces

Question 14.
Light travels from air to water, then the refraction index of water is 1.33. Hence find the refractive index when light travels from water to air.
Answer:
Refractive index of water (n21) = 1.33
Refractive index of air (n12) = \(\frac{1}{1.33}\) = 0.75

Question 15.
The refractive index of diamond is 2.42 and the refractive index of glass is 1.5; compare the critical angle between them. (Diamond 24°, glass 42°)
Answer:
Refractive index of diamond (µ1) = 2.42
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 29

Question 16.
A ray of light travels from an optically denser to rarer medium. The critical angle of the two media is ‘C’. What is the maximum possible deviation of the ray?
Answer:
The relation between angle of deviation and angle of incidence, angle of emergence and angle of prism is given by
Angle of deviation = i1 + i2 – A
For maximum deviation, Angle of incidence (i1) = 90°
Angle of emergence (i2) = 90°
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 30
∴ Maximum deviation = i1 + i2 – A = 90 + 90 – 2C = 180 – 2C = n – 2C.

Question 17.
A ray of light strikes a glass slab 5 cm thick making an angle of incidence equal to 30°.
a) Construct the ray diagram showing emergent ray and refracted ray through the glass block. The refractive index of glass is 1.5.
b) Measure the lateral shift of the ray.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 5 Refraction of Light at Plane Surfaces 31

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 13 Principles of Metallurgy.

AP State Syllabus SSC 10th Class Chemistry Important Questions 13th Principles of Metallurgy

10th Class Chemistry 13th Lesson Principles of Metallurgy 1 Mark Important Questions and Answers

Question 1.
Which method is suitable to enrich sulphide ores? (AP June 2016)
Answer:
Froth flotation method is suitable to enrich sulphide ores.

Question 2.
We use P.V.C. pipes for water supply instead of metal pipes. Why? (AP March 2017)
Answer:
PVC pipes do not rust. So they are used as water pipes instead of metal.

Question 3.
Arrange the metals Fe, Na, Ag and Zn in increasing order of their chemical reactivity. (TS March 2017)
Answer:
Ag < Fe < Zn < Na
(OR)
Ag, Fe, Zn, Na

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 4.
Write the deferences between Roasting and Calcination. (TS June 2018)
Answer:
1) Burning of ore in the presence of air or oxygen is called “Roasting”.
So in the roasting air is present.

2) Burning of ore in the absence of air or oxygen is called “Calcination.”
So in the calcination air is absent.

Question 5.
What are the preventive methods do you take for rusting iron materials? (TS March 2018)
Answer:

  1. Covering the surface of iron materials with paint or by some chemicals.
  2. Electroplating.

Question 6.
Mention the application of thermite process in daily life. (AP SCERT: 2019-20)
Answer:
1) Joining railing of railway tracks,
2) Joining cracked machine parts.

Question 7.
What are the essential condition that iron articles get rust? (TS June 2019)
Answer:
The essential condition that iron articles get rust is presence of water and air both.

Question 8.
What is metallurgy?
Answer:
The process of extraction of metals from their ores is called metallurgy.

Question 9.
What is the Bronze an alloy of?
Answer:
Bronze is an alloy of copper and tin.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 10.
What are ores?
Answer:
The minerals from which the metals are extracted without economical loss are called ores.

Question 11.
What is the percentage of Aluminium oxide in Bauxite?
Answer:
50-70%.

Question 12.
Why is 16th group called chalcogen family?
Answer:
Chaleo means ore and genus means produce. We notice that the ores of many metals are oxides and sulphides. This is why oxygen-sulphur (16th group) group as chalcogen family.

Question 13.
Which metals form oxides, sulphides and carbonates?
Answer:
Moderate reactive metals.

Question 14.
Based on the reactivity arrange the metals.
Answer:
Based on reactivity we can arrange metals in descending order of their reactivity as shown below:
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 1

Question 15.
What is gangue?
Answer:
The unwanted material in the ore is called gangue.

Question 16.
What is activity series?
Answer:
Arrangement of the metals in decreasing order of their reactivity is known as activity series.

Question 17.
Why do we add some impurities to ore?
Answer:
We add some suitable impurities to ore in order to decrease its melting point.

Question 18.
What is roasting?
Answer:
Roasting is a pyrochemical process in which ore is heated in the presence of oxygen or air below its melting point.

Question 19.
What is thermite process?
Answer:
The reaction of metal oxides with aluminium is called thermite process.

Question 20.
Write the chemical equations involving thermite reaction.
Answer:
2 Al + Fe2O3 → Al2O3 + 2 Fe + heat
2 Al + Cr2O3 → Al2O3 + 2 Cr + heat

Question 21.
How do you convert cinnabar into mercury?
Answer:
When cinnabar (HgS) is heated in air, it is first converted into (HgO), then reduced to mercury on further heating.

Question 22.
What is distillation of metals?
Answer:
The extracted metal in the molten state is distilled to obtain the pure metal as distillate by distillation of metals. Here the impurities are high boiling point metals.

Question 23.
What is poling?
Answer:
The molten metal is stirred with logs (poles) of greenwood and impurities are removed either as gases or they get oxidized and form slag over surface of the molten metal is called poling.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 24.
What is liquation?
Answer:
A low melting metal can be made to flow on a slope surface to separate it from high melting impurities is called liquation.

Question 25.
What is calcination?
Answer:
Calcination is a pyrochemical process in which the ore is heated in the absence of air.

Question 26.
What is flux?
Answer:
Flux is a substance added to the ore to remove the gangue from it by reacting with ore.

Question 27.
What are the ores of iron?
Answer:
Haematite (Fe2O3), Magnetite (Fe3O4).

Question 28.
What are the ores of zinc?
Answer:
Zinc blende (ZnS), Zincite (ZnO).

Question 29.
What is the formula of gypsum and metal present in gypsum?
Answer:
The formula of gypsum is CaSO4.2H2O. The metal present in gypsum is calcium.

Question 30.
What is a furnace?
Answer:
The furnace is one which is used to carry out pyrochemical process in metallurgy.

Question 31.
Arrange the following chlorides in ascending order of reactivity of respective metals. MgCl2, NaCl, PbCl2, HgCl2.
Answer:
The ascending order is HgCl2, PbCl2, MgCl2, NaCl.

Question 32.
What are the fuel and flux for haematite ore?
Answer:
The coke is used as fuel and limestone (CaCO3) is used as flux for haematite ore.

Question 33.
Why can copper not displace zinc from its compound?
Answer:
Copper is less reactive than zinc. So copper cannot displace zinc from its compound or salt.

Question 34.
How do various metals in activity series react with chlorine on heating?
Answer:

  1. All the metals react with chlorine on heating to form their respective chlorides.
  2. But the reactivity decreases from top to bottom.

Question 35.
How do you know the reactivity of metals with chlorine decreases from top to bottom?
Answer:
We know that the reactivity of metals with chlorine decreases from top to bottom the heat evolved when the metal reacts with one mole of chlorine gas to form chloride.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 36.
Give some examples for corrosion.
Answer:
Examples for corrosion :

  1. The rusting of iron (Iron oxide)
  2. Tarnishing of silver (Silver sulphide)
  3. Development of green coating on copper (Copper carbonate) and bronze.

Question 37.
Why are potassium, sodium, calcium never found in free state?
Answer:
The metals potassium, sodium and calcium are so reactive that is why they never exist in free state.

Question 38.
How do you extract metals at the top of activity series?
Answer:
The metals at the top of activity series are extracted by electrolysis of their fused compounds.

Question 39.
What is meant by enrichment of ore?
Answer:
Some physical methods are useful in removing unwanted rocky material from ore. It is called enrichment of ore.

Question 40.
How do you extract metal from the crude metal?
Answer:
To extract metal from enriched ore it is converted into metallic oxide by reduction reaction. Then this metallic oxide is further reduced to get metal with certain impurities.

Question 41.
What are the impurities you get in the refining of copper?
Answer:
Antimony, selenium, tellurium, silver, gold and platinum.

Question 42.
What is slag?
Answer:
The substance formed due to reaction of gangue and flux,
Eg : CaSiO3, FeSiO3

Question 43.
What is meant by pyrochemical reactions?
Answer:
Pyre means heat. So the chemical reactions involving heat are called pyrochemical reactions.

Question 44.
Can you mention some articles that are made up of metals?
Answer:
Jewellery, conducting wires and utensils.

Question 45.
Why are we mixing small amount of carbon to iron?
Answer:
To make iron hard and strong.

Question 46.
What is the main difference between steel and stainless steel?
Answer:
Steel will rust whereas stainless steel will not rust.

Question 47.
Give some examples for corrosion.
Answer:
The rusting of iron, tarnishing of silver, development of green coating on copper.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 48.
Do metals exist in the same form as that we use in our daily life?
Answer:
No, they exist as ores and minerals and some may exist in the form of metals.

Question 49.
Do you know how metals are obtained?
Answer:
The metals are extracted from their ores.

Question 50.
Have you ever heard the words like ore, mineral and metallurgy?
Answer:
Yes, these words are related to extraction of metal.

Question 51.
Where do we carry out pyrochemical processes in metallurgy?
Answer:
Pyrochemical processes can be carried out inside the furnace.

Question 52.
Which process converts sulphide ore into oxide ore?
Answer:
Roasting is the process which converts sulphide ore into oxide ore.

Question 53.
Is silver mineral or ore? Justify your answer.
Answer:
Silver is neither mineral nor ore. It is a metal.

Question 54.
Give two examples for corrosion.
Answer:
1) Rusting of iron
2) Green coating on copper.

Question 55.
Name the form of carbon used in the blast furnace for the extraction of iron.
Answer:
Carbon is used in the form of coke to reduce iron in blast furnace.

Question 56.
Give name and formulae of sulphide ore of lead and mercury.
Answer:
a) Sulphide ore of lead is Galena. Its formula is PbS.
b) Sulphide ore of mercury is Cinnabar. Its formula is HgS.

Question 57.
What are the various pyrochemical processes used in metallurgy?
Answer:
The various pyrochemical processes used are
a) Smelting,
b) Roasting,
c) Calcination.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 58.
What is the gas released at anode when fused sodium chloride is electrolysed?
Answer:
When sodium chloride is electrolysed, sodium metal is formed at cathode and chlorine gas is formed at anode.
NaCl → Na+ + Cl
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 2

Question 59.
Which pyrochemical process is useful to convert zinc blende into oxide ore?
Answer:
Zinc blende is sulphide ore of zinc. Its formula is ZnS. So it can be converted into oxide ore by heating strongly in excess of air known as roasting.
2 ZnS + 3 O2 → 2 ZnO + 2 SO2

Question 60.
Which pyrochemical process is useful to convert Magnesite into oxide ore?
Answer:
Magnesite is carbonate ore of magnesium. Its formula is MgCO3. So it can be converted into oxide ore by heating in the absence of air.
MgCO3 → MgO + CO2

Question 61.
What is the main impurity present in iron when it is removed from the blast furnace?
Answer:
The main impurity that can be removed is slag because it is formed when gangue in the ore reacts with flux.
CaO + SiO2 → CaSiO3

Question 62.
Name two metals normally manufactured by the electrolysis of fused compounds.
Answer:
Metals with high reactivity can be extracted by electrolysis of their fused compounds.
The examples are potassium, sodium and calcium.

Question 63.
What are the examples of corrosion?
Answer:

  1. Rusting of iron.
  2. Tarnishing of silver.
  3. Development of green coating on copper (CuCO3).
  4. Green coating on Bronze.

Question 64.
What is importance of prevention of corrosion?
Answer:

  1. Save the money.
  2. Preventing accidents such as a bridge collapse.
  3. Failure of a key component.

Question 65.
What is meant by galvanisation?
Answer:
Preventing the rust on metals by using layer of zinc. This phenomena is called galvanisation.

Question 66.
Write the example of electroplating in daily life.
Answer:

  1. Rold gold.
  2. Copper coating on cookware.

Question 67.
What is formula and name of iron rust?
Answer:
Iron rust is equal to hydrated ferric oxide.
Formula : Fe2O3 × H2O

Question 68.
What is importance of alloying?
Answer:

  1. Improving the properties of metal.
  2. To avoid the rust.
  3. To increase the hardness.

Question 69.
Which one used as flux extracting of iron from heamatite?
Answer:
Limestone or calcium carbonate (CaCO3).

Question 70.
Marne the two metals which corrode easily?
Answer:
Iron and copper.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 71.
Atmospheric air always contains moisture. Then, how can you protect iron articles from the affect of atmosphere?
Answer:
By painting, oiling and greasing, etc.

Question 72.
Explain the terms gangue and flux.
Answer:
The impurity present in the ore is called gangue. The substance added to the ore to remove gangue from it is called flux.

Question 73.
What are the metals are present in carnallite?
Answer:
Potassium (K) and Magnesium (Mg).

Question 74.
Write the elements are present in high reactivity series.
Answer:
Na, Mg, Al, K, Ca
(11 12 13 19 20)

Question 75.
Write the elements that are in moderate reactivity series.
Answer:
Fe, Cu, Zn, Pb.

Question 76.
Name the two metals which do not corrode easily?
Answer:
Gold and platinum.

Question 77.
Mention some important methods of refining.
a) Distillation
b) Poling
c) Liquation
d) Electrolysis
Answer:
d) Electrolysis

Question 78.
What is the role of furnace in metallurgy?
Answer:
Furnace is the one which is used to carry out pyrochemical process in metallurgy.

Question 79.
What is meant by calcination?
Answer:
It is the process of heating the concentrated ore in the absence of air.

Question 80.
Write the equation of heating of one sulphide ore in the process of roasting.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 3

Question 81.
Mention two methods which produce very pure metals?
Answer:
a) Electrolytic reduction.
b) Smelting.

Question 82.
What are the applications of thermite reaction in daily life?
Answer:
a) To join railings of railway tracks.
b) To join cracked machine parts.

Question 83.
Arrange the metals Ag, Mg, K in reactivity series.
Answer:
K > Mg > Ag.

Question 84.
How do you extract highly reactive metals?
Answer:
Highlyreactive metals can be extracted by electrolysis of their fused compounds.

Question 85.
What is dressing of an ore?
Answer:
The process of removal of impurities from an ore is called dressing of the ore or concentration of the ore.

Question 86.
Write the equation of example of calcination.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 4

Question 87.
Write the some properties of metals.
Answer:
Malleability, Ductility, Sonarity and Electrical conductivity.

Question 88.
Mention the stages involved in extraction of a metal from its ore.
Answer:

  1. Dressing or concentration.
  2. Extraction of crude metal.
  3. Refining or purification of the metal.

Question 89.
How do you extract moderately reactive metals?
Answer:
These metals are generally sulphides and carbonates. They are converted into oxides before reducing them to metals.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 90.
Give an example for reduction of metal oxide with carbon.
Answer:
The oxides are reduced by coke in a furnace which gives the metal and carbon monoxide.
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 5

Question 91.
Give an example for reduction of oxide ore with CO.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 6

Question 92.
What is flux?
Answer:
Flux is a substance added to the ore to remove the gangue from it by reacting with ore. If the impurity is acidic substance, basic substance is used as flux and vice – versa.

Question 93.
How do various metals in activity series react with chlorine on heating?
Answer:

  1. All the metals react with chlorine on heating to form their respective chlorides.
  2. But the reactivity decreases from top to bottom.

Question 94.
What are the substances to be added if the gangue is acidic or basic?
Answer:
If the gangue (impurity) is acidic substance like SiO2, basic substance like CaO is used as flux and if the impurity is of basic nature like FeO acidic flux like SiO2 is added to the gangue.
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 7

Question 95.
Why will stainless steel not rust?
Answer:
Stainless steel is prepared by mixing iron with nickel and chromium. Nickel and chromium are less reactive with oxygen. So stainless steel will not rust.

Question 96.
Why is sodium metal stored in kerosene?
Answer:
Sodium is highly reactive with both air (oxygen) and water. So it should be stored in kerosene.

Question 97.
Which metal gets covered with protective film of oxide when exposed to air?
Answer:
The metal is aluminium. When aluminium is exposed to air it forms a protective layer of aluminium oxide (Al2O3).

Question 98.
All ores are minerals, but all minerals need not be ores. Why?
Answer:
A mineral from which a metal can be extracted and economical loss is called ore.

Question 99.
Why is carbon not used for reducing aluminium from aluminium oxide?
Answer:
The oxide of Aluminium is very stable and can be reduced by electrolytic process.

Question 100.
Name few metals which occur in native state in nature. Why?
Answer:
Gold, Platinum, Silver are the metals which occur in native state, because of their low chemical reactivity.

Question 101.
Why do we call oxygen – sulphur group is chalcogen family?
Answer:
Chaleo means ore. We know that most of ores of many metals are oxides and sulphides. That’s why oxygen – sulphur group is called chalcogen family.

Question 102.
Aluminium occurs in combined state in nature whereas gold is in free state. Why?
Answer:
Gold has low reactivity and so occurs in free state. Aluminium is electropositive metal and high reactivity. So it is oxide or chloride.

Question 103.
What are the uses of thermite reaction?
Answer:
Thermite reaction is used to join railings of railway tracks or cracked machine parts.

10th Class Chemistry 13th Lesson Principles of Metallurgy 2 Marks Important Questions and Answers

Question 1.
Define mineral. Mention any two ores of ‘magnesium’. (AP June 2017)
Answer:
1) Minerals :
The elements or compounds of the metals which occur in nature in the earth crust are called ‘minerals’.

2) Two ores of magnesium :
Magnesite – MgCO3
Carnalite – KCl MgCl2 6H2O

Question 2.
Potassium, Sodium, Magnesium are high reactive metals and occur as chlorides in nature. Suggest and explain the suitable method for the extraction of the above metals from their ores. (AP March 2017)
Answer:

  • The suitable method to extract these metals from their chlorides is electrolysis of their fused compounds.
  • It is not feasible for method of reduction, electrolysis of their aqueous solutions.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 3.
Predict, what happens in the field of domestic use of metals if alloys were not discovered. (TS June 2016)
Answer:
If alloys were not discovered,

  1. All the vessels/utensils made of single metal like iron, copper, aluminium, etc. may be used for cooking purpose.
  2. We may face problems like rusting of iron, tarnishing of silver and copper, etc.
  3. We may face the problems of corrosion of home appliances.
  4. We may face difficulties in cleaning of the vessels due to rusting and tarnishing.
  5. Cost of the utensils may be risen, because of less availability of the metals like copper.
  6. Using of the plastic ware may be risen for storage due to lack of steel containers.
  7. Brass, steel, bronze, etc. utensils are not available to use.
  8. Making of jewellary is also difficult.

Question 4.
Give an example with the chemical equation for the reduction of ores using more reactive metals. (TS March 2017)
Answer:
The reaction of Iron oxide with aluminium.
Fe2O3 + 2Al → 2Fe + Al2O3 + Heat
(Or)
Reaction of Titanium Chloride with Magnesium.
TiCl4 + 2Mg → Ti + 2MgCl2
(Or)
Reaction of Titanium Chloride with Sodium.
HCl4 + 4Na → Ti + 4NaCl
(Or)
Reaction of Cromium oxide with aluminium.
Cr2O3 + 2Al → 2Cr + Al2O3 + Heat

Question 5.
Write two precautions to prevent corrosion of metals in your daily life. (TS June 2018)
Answer:
Precautions to be taken to prevent corrosion of metals.
i) Painting the metals.
ii) By keeping the metals in the dry places.
iii) Cover the surface by other metals that are inert or non reactive to the atmosphere.
iv) Applying oil/grease to the metals.
v) Making of alloys.

Question 6.

High reactivity Moderate reactivity Low reactivity
K, Na, Ca, Mg, Al Zn, Fe, Pb, Cu Ag, Au

Observe the table and answer the following questions. 4jt*y (June 2019
i) Which of the above metals found even in free state in nature ?
ii) Which of the above metal’s ore are concentrated by using magnetic separation?
Answer:
i) Ag, Au.
ii) Fe.

Question 7.
Silicon is a metalloid. How do you support this?
Answer:
Silicon exhibits following properties, so I conclude that it is a metalloid.

  1. It is metallic lustre in nature.
  2. It exists in several metallic and non-metallic compounds.
  3. It has brittle nature.
  4. All metalloids usually occur in combined states both metals and non-metals.

Question 8.
Explain the reaction of various metals in activity with cold water.
Answer:
1) From potassium to magnesium displace hydrogen from cold water with decreasing reactivity. Potassium reacts with cold water violently but reaction of Magnesium is very slow. The reactivity order is given below.
Mg < Ca < Na < K

2) From aluminium to gold do not displace hydrogen from cold water.

Question 9.
How do various metals in activity series react with steam?
Answer:

  • The metals from potassium to iron displace H2 (Hydrogen gas) from steam with decreasing reactivity. That means the reaction of potassium with steam is voilent but the reaction of iron is very slow.
  • The metals from lead to gold do not displace hydrogen from steam.

Question 10.
How do various metals in activity series react with dilute strong acids?
Answer:
1) The metals from potassium to lead displace hydrogen from dilute strong acids with decreasing reactivity.
a) The reaction of potassium is explosive.
b) The reaction of magnesium is vigorous.
c) The reaction of iron is steady.
d) The reaction of lead is slow.

2) The metals from copper to gold do not displace H2 from strong dilute acids.

Question 11.
What are the preventive techniques used in corrosion of metals?
Answer:
Prevention of corrosion of metals :

  • Covering the surface of metal with paint or by some chemicals like bisphenol which prevent the surface of metallic object to come in contact with atmosphere.
  • Covering the surface of metal by other metals like tin or zinc that are inert or react themselves with atmosphere to save the metal.
  • An electrochemical method in which a sacrificial electrode of another metal like magnesium and zinc, etc. corrodes itself to save the metal.

Question 12.
What are the chemical reactions that take place inside blast furnace?
Answer:
The chemical reactions that take place inside the blast furnace.
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 8

Question 13.
What are the various types of furnaces? Explain.
Various types of furnaces :
1) Blast furnace:
Blast furnace has both fire box and hearth combined in big chamber which accommodates both ore and fuel.

2) Reverberatory furnace :
It has both fire box and hearth separated, but the vapours (flame) obtained due to burning of the fuel touch the ore in the hearth and heat it.

3) Retort furnace :
In this furnace there is no direct contact between the hearth or fire box and even the flames do not touch the ore.

Question 14.
Why is alloying preferred for metals? Explain with examples.
Answer:

  • Alloying is a method of improving properties of a metal. We can get desired properties by this method.
  • For example, iron is the most widely used metal. But it is never used in its pure state.
  • This is because pure iron is very soft and stretches easily when hot.
  • But, if it is mixed with a small amount of carbon, it becomes hard and strong.
  • When iron is mixed with nickel and chromium we get stainless steel which will not rust.

Question 15.
What is 22 carat gold? Why is it preferred for making jewellery?
Answer:

  • Pure gold, known as 24 carat gold is very soft.
  • So it is not suitable for making jewellery.
  • It is alloyed with either silver or copper to make it hard.
  • So they use 22 carat gold in which pure gold is alloyed with 2 parts of either silver or copper for making gold jewellery.

Question 16.
Write about electrolysis of NaCl.
Answer:
1) Fused NaCl is electrolysed with steel cathode and graphite anode.

2) The metal sodium (Na) will be deposited at cathode and chlorine gas liberates at the anode.
At Cathode : 2 Na+ + 2e → 2 Na
At Anode : 2 Cl → Cl2 + 2e

Question 17.
Identify the metal present in the following ores.
i) Epsom Salt
ii) Horn Silver
iii) Cinnabar
iv) Galena
Answer:
i) Magnesium
ii) Silver
iii) Mercury
iv) Lead

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 18.
What is meant by extraction of metals? Write the main stages of extraction of metals from its Ore.
Answer:
Separation of metals from ores is called extraction of metals. Extraction of metals involves mainly three stages.

  1. Concentration or dressing
  2. Extraction of crude metal
  3. Refining or purification of the metal.

Question 19.
Write differences between roasting and calcination.
Answer:

Roasting Calcination
1. Ore is heated in the presence of oxygen or air. 1 Ore is heated in the absence of air.
2. Sulphide ore is converted into oxide ore. 2. Carbonate ore is converted into oxide ore.

Question 20.
What are the differences between minerals and ores?
Answer:

Minerals Ores
1) Minerals contain a low percentage of metal. 1) Ores contain a large percent of metal.
2) Metals cannot be extracted from minerals. 2) Ores can be used for the extraction of metals.
3) All minerals cannot be called ores. 3) All ores are minerals.

Question 21.
What are the different types of reduction?
Answer:
The different types of reduction are
a) Chemical reduction,
b)Auto reduction,
c) Electrolytic reduction.

Question 22.
How do potassium and sodium react with oxygen?
Answer:
a) Potassium and sodium form oxides of type M2O in limited supply of oxygen.
4 K + O2 → 2 K2O
4 Na + O2 → 2 Na2O

b) In excess of oxygen they form peroxides of type M2O2.
2 Na + O2 → Na2O2
2 K + O2 → K2O2

Question 23.
How does reactivity of chlorine vary in the reactivity series?
Answer:

  1. All metals react with chlorine on heating to form their respective chlorides but with decreasing reactivity in the reactivity series.
  2. This is understood from the heat evolved when metal reacts with one mole of chlorine gas to form chloride.

Question 24.
Name two ores of calcium and give their formulae.
Answer:
The ores of calcium are

  1. Gypsum (CaSO4 • 2H2O)
  2. Limestone (CaCO3)

Question 25.
Which method is useful to separate sand from iron? Explain.
Answer:

  • Sand can be separated from iron by using magnetic separation method.
  • This can be done by using electromagnet. Iron being a magnetic material is attracted by electromagnet whereas sand is not attracted by electromagnet.
  • So these materials are separated.

Question 26.
Which metals do not displace hydrogen from dilute strong acids?
Answer:

  • Copper, mercury, silver, platinum, gold do not displace hydrogen from dilute strong acids like HCl, H2SO4, etc.
  • The reactivity of these metals are less than hydrogen. So, they are unable to displace hydrogen from dilute acids.

Question 27.
Which metals are not found in free state? Why?
Answer:

  • The metals like potassium, sodium, calcium, magnesium and aluminium are never found in free state in nature.
  • The reason is that these metals have high reactivity. So, they exist as compounds.

Question 28.
Why do silver and gold exist even in free state?
Answer:

  • Silver and gold are least reactivity metals. So, they are also called noble metals.
  • Due to least reactivity they are unable to react with other elements.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 29.
How do moderately reactive metals occur in nature?
Answer:

  • The metals like zinc, iron, lead are moderately reactive.
  • They are found in the earth’s crust mainly as oxides, sulphides and carbonates.

Question 30.
Mention the most important metals and non-metals from the following products.
a) Annapurna salt
b) Liquid used in thermometer
c) Lead of the pencil
d) Chlorophyll
e) Filament in electric bulb
f) Enamel layer on teeth
Answer:
a) Annapurna salt : Iodine, chlorine – Non-metals
b) Liquid used in thermometer : Mercury – Metal
c) Lead of the pencil : Graphite – Non-metal
d) Chlorophyll : Magnesium – Metal
e) Filament in electric bulb : Tungsten – Metal
f) Enamel layer on teeth : Calcium phosphate – Non-metal

Question 31.
What is a furnace? Explain various parts of furnace.
Answer:
Furnace :
Furnace is the one which is used to cany out pyrochemical processes in metallurgy.

Furnace has mainly three parts :
1) Hearth :
Hearth is the place inside the furnace where the ore is kept for heating.

2) Chimney:
Chimney is the outlet through which flue (waste) gases go out of the furnace.

3) Fire box :
Fire box is the part of the furnace where the fuel is kept for burning.

10th Class Chemistry 13th Lesson Principles of Metallurgy 4 Marks Important Questions and Answers

Question 1.
What is a furnace? Draw Reverberatory furnace and label it parts. (AP March 2018)
Answer:
1) Furnace :
Furnace is the one which is used to carry out pyrochemical processes in metallurgy.
2) Diagram of Reverberatory furnace.
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 9

Question 2.
What are the various techniques used in purification of the crude metals? Explain.
(OR)
State the methods used for the purification of crude metals. Explain in which context these methods are used. (TS June 2015)
Answer:
1) The process of obtaining the pure metal from the impure metal is called refining of the metal.
2) Some of the processes of refining are
i) Distillation
ii) Poling
iii) Liquation
iv) Electrolytic refining.

3) The process that has to be adopted for purification of a given metal depends on the nature of the metal and its impurities.

Various methods adopted in purification of metals :
1) Distillation :
This method is very useful for purification of low boiling metals like zinc and mercury containing high boiling metals as impurities. The extracted metal in the molten state is distilled to obtain the pure metal as distillate.

2) Poling :
The molten metal is stirred with logs (poles) of greenwood. The impurities are removed either as gases or they get oxidized and form slag (Scum) over the surface of molten metal.

3) Liquation :
Low melting metal like tin can be made to flow on a slopey surface to separate it from high melting impurities.

4) Electrolytic refining :

  1. In this method, the impure metal is made to act as anode.
  2. A strip of the same metal in pure form is used as cathode.
  3. They are put in a suitable electrolytic bath containing soluble salt of the same metal.
  4. The required metal gets deposited on the cathode in the pure form.
  5. The metal constituting impurity, goes as the anode mud.

The reactions are :
At anode : M → Mn+ + ne
At cathode : Mn+ + ne → M. ; (M = pure metal, n = 1, 2, 3, …….)

Question 3.
Four metals A, B, C and D are in turn added to the following solutions one by one. The observations made are tabulated below. (TS March 2015)
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 10
Answer the following based on the given information.
i) Which is the most reactive metal? Why?
ii) What would be observed, if ‘B’ is added to a solution of Copper (II) sulphate and why?
iii) Arrange the metals A, B, C and D in order of increasing reactivity.
iv) Which one among A, B, C and D metals can be used to make containers that can be used to store any of the above solutions safely?
Answer:
i) Metal ‘B’ is more reactive.
– Metal ‘B’ is replacing iron from iron sulphate.
– Metal ‘A’ is replacing copper from copper sulphate.
– Metal ‘C’ is replacing silver from silver nitrate.

ii) Metal B displaces Cu from CuS04 solution. Because metal B is more reactive than Cu.

iii) D < C < A < B.

iv) The container made up of metal D can be used to store any solution mentioned above.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 4.
Write the physical methods used for the concentration of the ore. Explain the method used for concentration of the sulphide ore. (TS June 2017)
Answer:
Physical methods used for the concentration of the ore is,
i) Hand Picking
ii) Washing
iii) Froth floatation
iv) Magnetic Separation.

Concentration of sulphide ore :

  • Sulphide ore is concentrated by using froth floatation Method.
  • The ore with impurities is tinely powdered and kept in water taken in a flotation cell.
  • Air under pressure is blown to produce froth in water.
  • Froth so produced, taken the ore particles to the surface whereas impurities settle at the buttom.
  • Froth is separated and washed to get ore particles.

Question 5.
Draw a neat diagram of froth floatation process for the concentration of sulphide ore why we add pine oil to the mixture in this process? (AP SCERT 7201 9-20)
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy
Froth floatation process for the concentration of sulphide ores

  1.  The mineral particles in the ore are preferentially wetted by the oil and float on the top of the froath.
  2. The gangue particles are wetted by water and settle down.
  3. Thus, the minerals can be separated from the gangue by adding pine oil.

Question 6.
Describe the reaction of various metals in activity series with oxygen.
Answer:

  • The metals which are at the bottom of activity series have very low reactivity and do not burn or oxidase even on surface.
    Eg : Ag, Pt, Au.
  • The metals like Pb, Cu and Hg do not burn but only form a surface layer of oxide, i.e., PbO, CuO, HgO.
  • The metals like Al, Zn, Fe react with oxygen to form respective oxides.
  • The metals like Ca and Mg burn with decreasing vigorousity to form oxides.
  • The metals like K, Na burn vigorously to form Na2O, K2O in limited supply of oxygen but form peroxides in excess of oxygen.

Question 7.
How do you reduce purified ore to the metal of the top of activity series? Explain.
Answer:
The reduction of ore to particular metal mainly depends on the position of metal in the activity series.

Extraction of metals at the top of activity series :

  1. Simple chemical reduction methods like heating C, CO, etc. to reduce the ores of the metals are not possible with metals like K, Na, Ca, Mg and Al.
  2. The temperature required for the reduction is too high and more expensive.
  3. The only method available is to extract these metals by electrolysis of their fused compounds.

Question 8.
How do you extract metals in the middle of activity series?
Answer:
Extraction of metals in the middle of the activity series :
The ores of these metals are generally present as sulphides or carbonates. Therefore prior to reduction of ores of these metals, they must be converted into metal oxides.

The metal oxides are then reduced to the corresponding metals by using the following methods :
1) Reduction of metal oxides with carbon :
The oxides are reduced by coke in closed furnace which gives the metal and carbon monoxide (CO).
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 11

2) Reduction of oxide ores with CO :
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 12

3) Auto (Self) reduction of sulphide ores:
In the extraction of copper from its sulphide ore, the ore is subjected to partial roasting in air to give its oxide.
2 Cu2S + 3O2 → 2 Cu2O + 2SO2

When the supply of air is stopped and temperature is raised, it results in the reaction of rest of the sulphide ore with oxide to form metal and S02. ‘
2 Cu2O + Cu2S → 6 Cu + SO2

4) Reduction of ores (compounds) by more reactive metals :
When highly reactive metals such as sodium, calcium, aluminium, etc. are used as reducing agents, they displace metals of low reactivity from the compound.
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 13

Question 9.
How do you extract metals at the bottom of the activity series?
Answer:
1) Metals at the bottom of the activity series are often found in free state.

2) The oxides of these metals can be reduced to metals by heat alone and sometimes by displacement from their aqueous solutions.

3) When cinnabar (HgS) is heated in air, it is converted into HgO, then reduced to mercury on further heating.
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 14

4) Displacement from aqueous solution :
When Ag2S is dissolved in KCN solution, it forms dicyanoargentate ions. When these ions are treated with Zn dust powder then Ag is precipitated.
Eg : Ag2S + 4 CN → 2 [Ag(CN)2] + S2-
2 [Ag(CN)2](aq) + Zn(s) → [Zn(CN)4]2-(aq) + 2 Ag(s)

Question 10.
Explain the process involved in corrosion.
Answer:
1) Corrosion is an electrochemical phenomenon.
2) In corrosion, a metal is oxidised by loss of electrons generally to oxygen and results in the formation of oxides.
3) During corrosion at a particular spot on the surface of an object made of iron, oxidation takes place and that spot behaves as anode.
Anode : 2 Fe(s) → 2 Fe2+ + 4e
4) Electrons released at this anodic spot move through the metal and go to another spot and reduce oxygen at that spot in the presence of H+.

5) This spot behaves as cathode.
Cathode : O2(g) + 4 H+(aq) + 4e → 2H2O(l)
Net reaction : 2 Fe(s) + O2(g) + 4 H+(aq) → 2 Fe2+(aq) + 2 H2O(l)

6) The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe2O3 . XH2O) and with further production of hydrogen ions.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 11.
Explain electrolytic refining with an example.
Answer:

  • The impure metal is taken as anode and pure metal is taken as cathode.
  • They are put in a suitable electrolytic bath containing soluble salts of same metal.
  • The required metal gets deposited on the cathode in the pure form.
  • The metal constituting the impurity goes as the anode mud.

Electrode reactions :
At Anode : M → Mn+ + ne
At Cathode : Mn+ + ne → M (M pure metal); where n = 1, 2, 3,………….

Examples :

  1. In order to refine copper, impure copper is taken as anode and pure copper strips are taken as cathode.
  2. The electrolyte is an acidified solution of copper sulphate.
  3. As a result of electrolysis copper in pure form is transferred from anode to cathode.
    At Anode : Cu → Cu2+ + 2e
    At Cathode : Cu2+ + 2e → Cu
  4. The soluble impurities go into the solution, whereas insoluble impurities from the blister copper deposited at the bottom of anode as anode mud.

Question 12.
What is activity series? Give two examples each of them.
i) Low reactivity metals
ii) Moderate reactivity metals
iii) High reactivity metals
Answer:
Arranging metals in descending order of their reactivity is called activity series,
e.g.:
i) Low reactivity : Ag (Silver), Au (Gold)
ii) Moderate reactivity : Zn (Zinc), Fe (Iron)
iii) High reactivity : K (Potassium), Na (sodium)

Question 13.
Write the balanced chemical equations, extraction of iron from haematite in the Blast furnace.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 15

Question 14.
How do you prevent corrosion of various metals?
Answer:
Prevention of corrosion :The corrosion can be prev ented by the following methods.
1) Barrier protection :
In this method the metal surface is not allowed to come in contact with moisture, oxygen and carbon dioxide. This can be achieved by the following.
a) By coating iron with oils, paints, coal tar, grease, pitch, etc.
b) By blowing steam over red hot iron to form protective coating of Fe3O4.
c) By alloying iron with Ni, Cr, Si, etc.

2) Sacrificial protection:
Sacrificial protection means covering the iron surface with a layer of metal which is more electropositive than iron thus prevents iron from losing electrons. It is done by following methods.
a) By galvanisation (by dipping iron in a bath of molten zinc).
b) By tinning (by dipping iron in molten tin).
c) By the coating of copper.
d) Decorative coating : By using Zn, Mg and A/ powders mixed with paints.

3) Electrical protection:
In this method, the iron object to be protected from corrosion is connected to more active metal eg. magnesium, zinc or aluminium directly or through a wire. The iron object acts as cathode and the protective metal acts as anode. The anode is gradually used up to the oxidation of metal to its ions due to loss of electrons. Hydrogen ions collect at cathode and prevent rust formation.

4) Using anti-rust solution:
On applying alkyl phosphates and alkyl chromates to the iron objects corrosion can be prevented.

Question 15.
What are the salient features of the activity series?
Answer:
Salient features:

  1. Any metal which is placed higher up in the series can displace any metal below it in order to from the salt solution of the later metal.
  2. The larger the difference in the position of metals in the series, the more rapidly does the displacement take place.
  3. Metals which are placed above hydrogen in the series have the ability to reduce ions from dilute sulphuric acid to liberate hydrogen gas.
  4. Oxides of metals K, Na, Ca and Mg cannot be reduced by H2, CO or C.
  5. Oxides and nitrates of less reactive metals Hg, Ag and Au decompose to give metals on being strongly heated.
  6. Metals below copper such as mercury, silver, platinum and gold do not rust easily.
  7. Hydrogen though a non-metal, has been included in the series.
    It occupies the position based on its formation of positive ions.

Question 16.
How do you extract metals based on activity series?
Answer:

  • Highly active metals like potassium, sodium, calcium, magnesium and aluminium are obtained by the electrolysis of their fused halides or oxides, that is, by electrolytic reduction because their oxides cannot be reduced by common reducing agents like carbon, carbon monoxide and hydrogen.
  • Zinc is obtained only by heating its oxide with carbon.
  • Iron, lead and copper are obtained by reduction of their oxides with carbon, carbon monoxide and hydrogen.
  • Copper is obtained by reducing black copper oxide with carbon or by air reduction.
  • Mercury and silver are obtained by heating their respective oxides to temperature above 300°C when they lose oxygen and are reduced to free metals.
  • However, less active mercury can also be obtained by merely heating its sulphide in air.
  • Silver and Gold are obtained by displacement from solutions containing their ions by more electropositive metal zinc.

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 17.
X is an element in the form of a powder. X burns in oxygen and the product is soluble in water. The solution is tested with litmus.
Write down the answers for the following questions from the above information and give reasons.
i) If X is a metal, then which colour will litmus turn ?
ii) If X is a non-metal, then which colour will litmus turn ?
iii) If X is a reactive metal, what gas will be released with dilute sulphuric acid ?
Answer:
i) If X is a metal, then the litmus turns into blue because metal reacts with oxygen and forms metallic oxide and aqueous solution of metallic oxide ore basic in nature.

ii) Mf X is a non-metal, then the litmus turns into red because non-metal reacts with oxygen and forms non-metallic oxide and aqueous solution of non-metallic oxide ore acidic in nature.

iii) If X is a reactive metal, then it releases hydrogen gas from sulphuric acid because more reactive metal displaces hydrogen from acid.

Question 18.
Complete the missing statements and give reasons.
i) Metals are ……………………….., while non-metals are poor conductors of heat.
ii) Metals are malleable, while non-metals are ……………………….. .
iii) Metals form positive ions, while non-metals form ……………………….. .
iv) Non-metals form acidic oxides, while metals form ……………………….. .
Answer:
i) Good conductors.
Reason :
Metals containing free electrons are very good conductors of electricity whereas non-metals are bad conductors of electricity because they do not have free electrons.

ii) Non-malleable.
Reason :
Metals are hard. So, they can be beaten into sheets whereas non-metals are soft, so they are non-malleable.

iii) Negative ions.
Reason :
Metals are electropositive in nature. They easily lose electrons to form positive ions, whereas non-metals are electronegative in nature. So, they gain electrons to form negative ions.

iv) Basic oxides.
Reason :
Non-metallic oxide solutions turn blue litmus into red. They are acidic in nature. So, they are called acidic oxides whereas metallic oxide solutions turn red litmus into blue. They are basic in nature. So, they are called basic oxides.

Question 19.
Answer the following questions.
a) i) Name two naturally occurring compounds of zinc other than carbonate and give their formulae.
ii) Give equations for the extraction of zinc from zinc carbonate.
b) Write equations for the action of zinc on the following.
i) dil. H2SO4
ii) Copper (II) sulphate solution.
Answer:
a) i) The ores of zinc other than carbonate ore are zinc blende (ZnS) and Zincite (ZnO).
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 16

b) i) Zn + dil. H2SO4 → ZnSO4 + H2
Zr(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)

Question 20.
i) The ore zinc blende is an important source of the metal zinc. What is the name of zinc compound in zinc blende?
ii) What is the compound obtained by roasting zinc blende?
iii) What is the type of chemical reaction carried out after roasting in order to obtain zinc?
iv) What is the name of the alloy formed between zinc and copper?
Answer:
i) The zinc compound in zinc blende is ZnS (zinc sulphide).
ii) By roasting zinc blende it converts into zinc oxide.
2 ZnS + 3O2 → 2 ZnO + 2 SO2
iii) The chemical reaction carried out to convert zinc oxide to zinc metal is reduction in the presence of coke.
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 17

iv) The alloy of copper and zinc is bronze.

Question 21.
The basic material used for the production of iron in the blast furnace are limestone, coke and air in addition to iron ore.
a) Name one iron ore and write its formula.
b) Hot air is blown at the base of furnace where it reacts with coke. Give the chemical equations for the reactions that take place.
c) Higher up in the furnace the iron ore is reduced to iron by one of the gases produced in the furnace. Give the chemical equation for the reaction by which the gas is produced and give a balanced equation to show how the ore is reduced to iron.
d) Which compound produced from limestone takes part in forming the slag?
Answer:
a) The iron ore is Haematite (Fe2O3).
b) Coke bums partially to produce carbon monoxide gas.
2 C(s) +O2(g) → 2 C0(g)

c) Iron oxide reacts with carbon monoxide gas and reduces to iron.
Fe2O3 + 3 CO → 2 Fe + 3 CO2

d) Calcium carbonate (limestone) undergoes calcination to produce calcium oxide which takes part in the reaction to form slag.
CaCO3(s) → CaO(s) + CO2
CaC(s) + SiO2(s) → CaSiO3(l)

AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy

Question 22.
What information do you get from metal activity series given below.
K > Na > Ca > Mg > Al > Zn > Fe > Pb > [H] > Cu > Ag > Pt > Au
Answer:

  • Metals below hydrogen [H] cannot displace hydrogen from acids and above hydrogen can displace hydrogen from acids.
  • Metals which are higher in the series, can displace metals below it from the salt solution.
  • The higher the position, the more active is the metal.
  • Hydrogen has electropositive character, so it is placed among the metals.

Question 23.
The results of reactions of metals A, B, C, D, E with different solutions are given in the table below. Observe the table and write answers.
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 18
1) Which is the highly reactive metal? Why?
2) Which is the least reactive metal? Why?
3) Which metals form brown layer?
4) Arrange the metals A, B, C, D, E in the order of their reactivity?
5) Among these identify the silver, copper, iron, zinc and aluminium.
Answer:

  1. Metal ‘E’ is more reactive among all the metals given because it displaces all the elements from the compounds given in the table,
  2. Metal ‘C’ is the least reactive metal because it does not displace any other metal from the compounds given in the table.
  3. Metals B and E will form brown layer.
  4. The ascending order is as follows C < A < D < B < E.
  5. C is silver, A is copper, D is iron, B is zinc and E is aluminium.

Question 24.
Draw the diagram of blast furnace and label its parts.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 13 Principles of Metallurgy 19

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 1 Heat.

AP State Syllabus SSC 10th Class Physics Important Questions 1st Lesson Heat

10th Class Physics 1st Lesson Heat 1 Mark Important Questions and Answers

Question 1.
What is humidity? (AP March 2015)
Answer:
Humidity :
The amount of water vapour present in air is called humidity.

Question 2.
Define latent heat of Fusion. (AP Morch 2016)
Answer:
Latent heat of Fusion :
At constant temperature, the heat energy required to convert one gram of solid completely into liquid is called latent heat of Fusion.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 3.
While drinking water, Rama spilled some water on the floor. After some time, the water disappeared from the floor. What happened to the water? (TS June 2015)
Answer:

  • The water disappeared due to evaporation because we know that as the surface area increases rate of evaporation also increases.
  • So water molecules escape from the floor to air.

Question 4.
Give an example to explain that evaporation is a cooling process. (TS March 2016)
Answer:
The examples to explain that evaporation is a cooling process are

  1. Drying of wet clothes
  2. When the floor is washed with water, the water on the floor disappears.
  3. Sweating, etc.

Question 5.
Let heat is not lost by any other process between two objects in thermal contact, “Net heat lost (by hot body) = Net heat gain (by cold body).” above statement indicates a principle. Write the name of that principle. (AP March 2019)
Answer:
Principle of method of mixtures.

Question 6.
Convert 25°C into Kelvin scale. (AP SCERT: 2019-20)
Answer:
25°C = (273 + 25) K = 298 K

Question 7.
Given a beaker with water, a thermometer and a stand, draw the arrangement of an experiment to measure boiling point of water. (AP SA-1:2019-20)
Answer:
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 1

Question 8.
Define heat.
Answer:
Heat: Heat is a form of energy which is transferred from one body to the other body due to the difference in their temperature.

Question 9.
What is meant by thermal equilibrium?
Answer:
Thermal Equilibrium :
Two bodies are at the same temperature then they are said to be in thermal equilibrium.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 10.
Define dew.
Answer:
Dew :
The water droplets condensed on surface are known as dew.

Question 11.
What is boiling?
Answer:
Boiling is a process in which the substance changes from liquid to gas.

Question 12.
What is melting?
Answer:
Melting :
The process in which the substance changes from solid to liquid state is called melting.

Question 13.
What are the different energies possessed by system (body or material)?
Answer:

  1. Linear kinetic energy
  2. Rotational kinetic energy
  3. Vibrational energy
  4. Potential energy and Internal energy (I.E).

Question 14.
Why does samosa seem to be cool but hot when we eat?
Answer:
Because the curry inside samosa contains ingredients with higher specific heats.

Question 15.
On which factors does rate of evaporation depend?
Answer:

  1. Surface area
  2. temperature
  3. the amount of vapour already present in the surrounding air.

Question 16.
What is the value of latent heat of vapourization of water?
Answer:
Latent heat of vapourization of water = 540 cal/gm. (or) 2.26 × 106 J/kg

Question 17.
What is the value of latent heat of fusion of ice?
Answer:
Latent heat of fusion of ice is 80 cal/gm. (or) 3.26 × 105 J/kg

Question 18.
Give some liquids which solidify (convert into solid) in winter season.
Answer:
Coconut oil, ghee are some liquids which solidify in winter season.

Question 19.
What is freezing?
Answer:
Freezing :
The process in which the substance changes from liquid to solid state by losing some energy from it is called freezing.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 20.
Which will have lower temperature when we take out a wooden piece and a metal piece from a fridge?
Answer:
The metal piece will have lower temperature as compared to the wooden piece when they are taken out of the fridge.

Question 21.
When do you say there is thermal equilibrium between two bodies?
Answer:
It is said that there is thermal equilibrium between two bodies when there is no transfer of heat energy between them.
(OR)
When temperature between two bodies is same it is said that there is thermal. equilibrium between them.

Question 22.
What is absolute temperature?
Answer:
Temperature measured in Kelvin scale is called absolute temperature.

Question 23.
What is latent heat of vapourisation?
Answer:
At constant temperature the heat energy required to change one gram of liquid into gaseous state.

Question 24.
What is boiling point?
Answer:
The temperature at which the substance changes from liquid to gaseous state at the fixed temperature is called boiling point.

Question 25.
What is melting point?
Answer:
Melting point :
The temperature at which the substance changes from solid to liquid state at constant temperature is called melting point.

Question 26.
How is aquatic animal able to live at poles?
Answer:
The ice has less density compared to water. So it forms a layer on the top of water which prevents the solidification of water.

Question 27.
What are the phases of water present at 0° C?
Answer:
Two phases namely, ice and water.

Question 28.
What happens if external pressure of liquid increases?
Answer:
The boiling point of the liquid will increase.

Question 29.
Does ice melt below 0° C?
Answer:
Yes, if the external pressure increases it melts at low temperature.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 30.
What happens when two objects of same temperature are in contact with each other?
Answer:
Heat does not flow between two objectives.

Question 31.
What is the principle involved in pressure cooker?
Answer:
Boiling point of liquid increases with external pressure.

Question 32.
What happens to kinetic energy of particles if we increase the temperature?
Answer:
Kinetic energy of particles increases with increase of temperature.

Question 33.
Why does transfer of heat energy take place between systems?
Answer:
When heat energy gives to the system, internal energy increases. Similarly, internal energy decreases when heat energy flows out of the system.

Question 34.
What is internal energy?
Answer:
Internal energy :
It is the energy possessed by the system by virtue of its molecular motion and molecular configuration. It is a stored energy. It depends on the temperature of the system.

Question 35.
What is transit energy?
Answer:
Transit energy:
Energy possessed by a system which can cross its boundary is called transit energy. Heat and work are transit energies.

Question 36.
Where does air get? vapour from?
Answer:
The vapour may come from evaporation of water from the surfaces of rivers, lakes, ponds, and from the drying of wet clothes, sweat, and so on.

Question 37.
Why do pigs toil in the mud during hot summer days?
Answer:
They do not have sweat glands for evaporation process. So pigs toil in the mud.

Question 38.
Why is it easy to cook food in a pressure cooker?
Answer:
We know as the atmospheric pressure increases the boiling point of water increases. So we can increase the boiling point of water to 120°C in a pressure cooker. So it is \ easy to cook in a pressure cooker.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 39.
Why is water used as coolant?
Answer:
Water has the highest specific heat. So it takes lot of time to become hot. So it is used as coolant.

Question 40.
How is fog formed?
Answer:
The water molecules present in vapour condense on the dust particles in air and form small droplets of water which form a thick mist called fog.

Question 41.
Equal amounts of water is kept in a cup and in a disc. Which will evaporate faster? Why?
Answer:
The water present disc evaporates faster because of greater surface area.

Question 42.
Explain why dogs pant during hot summer days using the concept of evaporation.
Answer:
Dog does not have pores on its body. The only place where a dog can sweat is on its foot pads and the rest of the body is covered in a fur coat. So it cannot sweat; that’s why dogs pant to keep cool themselves.

Question 43.
Same amount of heat is supplied to two liquids A and B. The liquid A shows a greater rise in temperature. What can you say about the specific heat of A?
Answer:
The specific heat of A is less than that of B because rise in temperature is inversely proportional to temperature.

Question 44.
What is the specific heat of-water at boiling point?
Answer:

  1. Generally, the specific heat of water is 1. (or) 4.187 KJ / Kg K
  2. Specific heat of water at 100° C = 4.219 KJ / KgK

Question 45.
What is the equation of heat energy when change the state?
Answer:
Q = mL
Where m = mass of body, L = latent heat.

Question 46.
Convert 212°F into Kelvin scale.
Answer:
212°F= 100°C. So 100 + 273 = 373 K.

Question 47.
Convert 310 K into centigrade system.
Answer:
310-273 = 37°C.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 48.
Are the processes of evaporation and boiling the same?
Answer:
No. Evaporation takes place at any temperature, while boiling occurs at a definite temperature called the boiling point.

Question 49.
Define latent heat of vaporization?
Answer:
The heat energy required to change one unit mass of liquid to gas at constant temperature is called latent heat of vaporization.
\(L=\frac{Q}{m}\)
The value of latent heat of vapourization of water is 540 cal/gm.

Question 50.
What is meant by internal energy?
Answer:
Combination of linear kinetic energy, rotational kinetic energy, vibrational energy, and potential energy of molecules is known as internal energy of the system.

Question 51.
Write the formula for resultant temperatures of a mixture, when V1 ml of water at T1°C is mixed with V2 ml of water at T2° C.
Answer:
Resultant temperature \(T=\frac{V_{1} T_{1}+V_{2} T_{2}}{V_{1}+V_{2}}\)

Question 52.
Write the equation of heat energy when change the temperature.
Q = mS∆T
m == mass, S = specific heat, AT = change in temperature

Question 53.
The figure shows change in state of ice from – 5°C to 110°C with temperature. What are the melting and vaporization curves?
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 2
Answer:
BC = melting
DE = vaporization curve.

Question 54.
Write principle of method of mixtures.
Answer:
When two or more bodies are brought into thermal contact, then heat lost by hot body is equal to heat gain by cold body. Until they attain thermal equilibrium.

Question 55.
Evaporation is a cooling process. Why?
Answer:
During evaporation process, the energy of the molecules inside the liquid decreases and they slow down.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 56.
Which factors are influence the rate of evaporation of a liquid?
Answer:
Rate of evaporation of a liquid depends on surface area, temperature, pressure, and amount of vapour present in surrounding air.

Question 57.
What is meant by fog?
Answer:
The droplets keep floating in the air and form a thick mist which restricts visibility. This thick mist is called fog.

Question 58.
Why do we sweat while doing a work?
Answer:
When we work our body produces heat. As a result, the temperature of the skin becomes higher and the water in the sweat glands starts evaporating. This evaporation cools the body.

Question 59.
A samosa appears to be cool when touched outside but it is hot when we eat it. Why?
Answer:
A samosa appears to be cool outside but it is hot when we eat it because the curry inside the samosa contain ingredients with higher specific heats. Hence they remain hot for a long time.

Question 60.
Equal amounts of water are kept in a cup and in a dish. Which will evaporates faster? Why?
Answer:
Dish evaporates faster, because dish has large surface area. Evaporation of liquids depends on surface area.

Question 61.
Why water is used as coolant in the cooling system of automobile engines?
Answer:
Due to high specific heat, water absorbs large amount of heat and temperature does not rise quickly. So water used as coolant in cooling system of automobile engines.

Question 62.
Why do pigs toil around in the mud?
Answer:
Pigs do not have sweat glands. Water in the mud evaporates and helps the pig to be cool from heat. So pigs toil in the mud during summer.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 63.
Take small glass bottle with a tight lid. Fill it with water completely without any gaps and fix the lid tightly in such a way that water should not come out of it. Put the bottle into the deep freezer for a few hours. Take it out from the fridge. You observe the glass bottle is broken. Why?
Answer:
We know, the volume of the water poured into the glass bottle is equal to the volume of the bottle. When the water freezes to ice, the bottle is broken. Because the volume of the ice is greater than the volume of the water filled in bottle.

Question 64.
From the given figure, in which the thermometer mercury level is increases and decreases?
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 3
Answer:
Thermometer A(in oil) is increases.
Thermometer B(water) is decreases.

Question 65.
What are the materials are used in to find the specific heat of solid?
Answer:
Calorimeter, thermometer, stirrer, water steam heater, wooden box, and lead shots.

Question 66.
What is the value of following temperatures in Kelvin scale?
(a) 30° C b) 70° C
Answer:
a) 30° C = 30 + 273 = 303 K
b) 70° C = 70 + 273 = 343 K

Question 67.
How much heat energy is required to raise the temperature of unit mass of material by 1° C?
Answer:
1 cal/g – °C = lk cal / kg – K = 4.2 x J/kg – K = 4.2 kJ/kg – K.

Question 68.
How much energy is required to turn 1 g of ice of 0°C into 1 gm of water at 0°C?
Answer:
The energy required to convert 1 g of ice at 0°C into lg of water at 0°C is latent heat of fusion that is 80 cal/g.

Question 69.
What is the temperature of mixture if 10 g of steam at 100°C is mixed with 50 g of ice at 0°C?
Answer:
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 4

Question 70.
Boiling water at 100°C and cold water at t°C are mixed in the ratio of 3 : 5 and the resultant temperature is 40°C. Find the value of t.
Answer:
Suppose the quantities of water is 3x and 5x.
Given that
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 5

Question 71.
What amount of ice can be melted by 4000 cal of heat?
Answer:
Latent heat of fusion of ice Lf = 80 cal/g
Given that Q = 4000 cal
Q = mLf ⇒ 4000 = m × 80
∴ m = \(\frac{4000}{80}\) = 50 g

Question 72.
5 gm of ice is at (J°C. It is converted into water at same temperature. How much heat energy is required?
Answer:
In change the state Q = mL
m = 5 gm, L = Latent heat of fusion = 80 Cal/gm
Q = mL = 5 × 80 = 400 cal.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 73.
What would be the final temperature of mixture 50 g. of water at 20°C and 50 gm of water at 40°C?
Answer:
If masses are equal, then resultant temperature of mixture = \(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}\)
∴ T = \(\frac{20+40}{2}\) ⇒ T =30

10th Class Physics 1st Lesson Heat 2 Marks Important Questions and Answers

Question 1.
Observe the following table regarding the values of specific heat of substances and answer the following questions : (AP SA-I: 2018-19)
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 6
i) Which material is suitable as the base of the cooking vessel?
Answer:
Copper. Because, it has low specific heat value.

ii) Why do we prefer water as a coolant?
Answer:
Due to high specific heat value of water, it can take more time to heat itself and acts as a coolent.

Question 2.
Why do water drops (dew) form on flowers and grass during morning hours of winter season? (AP March 2016)
Answer:
During winter nights, the atmospheric temperature goes down. The flowers, grass, etc. become still colder. The air near them becomes saturated with vapour and condensation begins. The water droplets condensed on such surfaces are known as dew.

Question 3.
Temperatures of two cities at different times are given as follows : (AP March 2019)
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 7
On the basis of above table, answer the following questions.
1) In which city, the morning temperature at 6 O’clock is relatively high?
Answer:
In ‘B’ city, the morning temperature at 6 O’clock is relatively high.

2) At what time, both cities are having the equal temperature?
Answer:
At 11 : 30 AM, both cities are having the equal temperature.

Question 4.
A student took the same quantity of water and petrol in two different tumblers. He kept them on a table. When observed after a day there was water in a glass but petrol was completely evaporated. Give reasons why water was not evaporated completely but petrol was completely evaporated.
Answer:
Petrol has a lower vapour point than water. And evaporation depends on the nature of the liquid so petrol evaporates quickly than water at room temperature.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 5.
What are the differences between dew and fog?
Answer:

Dew Fog
1) Dew is the droplets that appear on the exposed objects in the morning or evening. 1) Fog is nothing but cloud on ground.
2) Dew does not effect visibility. 2) Due to fog visibility is greatly effected.
3) Dew is formed when relative humidity higher than temperature. 3) Fog is formed when island area is warmer than the ocean or large body of water.

Question 6.
Why is spirit evaporated in petri dish quickly under a fan when compared to that kept in closed room?
Answer:

  • The blowing air increases the rate of evaporation.
  • This is because any molecule escaping from the surface is blown away from the vicinity of the liquid.
  • This increases the rate of evaporation.
  • This is the reason why the spirit in petri dish evaporates quickly when compared to that kept in closed room.

Question 7.
Does the temperature of water rise continuously if heat is supplied continuously?
Answer:
Yes. If heat is supplied to water its temperature rises continuously till it reaches 100°C. At 100°C there would be no further rise of temperature, because the heat is sterilized to convert water to water vapour. So if heat is supplied beyond 100°C, all the water is converted into vapour.

Question 8.
Why does the mercury level of thermometer rise up when it is placed in hot water and fall down when it is placed in cold water?
Answer:

  • We know that bodies which are in contact achieve thermal equilibrium due to transfer of heat energy.
  • If we keep thermometer in hot water, its mercury level rises because heat is transformed from hot body to cold body.
  • Similarly, we observe that mercury level comes down when it is placed in cold water.

Question 9.
When we place thermometer in hot water, there is a rise in mercury level, thereafter it stops. What is the reason for steadiness of mercury level? What does reading of thermometer give at that time?
Answer:

  • The steadiness of the mercury column of the thermometer indicates that, flow of heat between the thermometer liquid (mercury) and water has stopped and thermal equilibrium has been attained between the water and thermometer liquid.
  • The reading of thermometer gives thermal equilibrium state that is temperature.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 10.
What is the relationship between temperature and kinetic energy?
Answer:

  • The average kinetic energy of molecules/particles of the hotter body is more than the colder body.
  • So we can say that the temperature of a body is an indicator of the average kinetic energy of molecule of that body.
  • So the average kinetic energy of molecules is directly proportional to the absolute temperature. [KE<sub>avg</sub> ∝ T]

Question 11.
What is the relationship between rise in temperature and specific heat of material?
Answer:

  • Temperature depends on nature of the material, hence the specific heat depends on its nature.
  • If the specific heat, is high, the rise in temperature is low. [Q = mSΔt]
  • It gives us an idea of degree of reluctance of a material to rise in temperature.

Question 12.
What is the principle of method of mixtures?
Answer:
When two or more bodies at different temperatures are mixed with each other, then net heat lost by the hot bodies, is equal to net heat gained by the cold bodies until they attain thermal equilibrium or equal temperature.

Net heat lost = Net heat gained

This is known as principle of method of mixtures.

Question 13.
Why is evaporation of a liquid faster under a fan?
Answer:

  • If air is blown over the liquid surface in an open petri dish, a number of molecules evaporate from the surface of liquid.
  • Because any molecule escaping from the surface is blown away from the vicinity of liquid.
  • This increases the rate of evaporation.
  • So evaporation of a liquid is faster under a fan.

Question 14.
Why do we get dew on the surface of a cold soft drink bottle kept in open air?
Answer:

  • The temperature of surrounding air is higher than the temperature of cold soft drink.
  • Air contains water molecules in the form of water vapour which tend to condense by losing their kinetic energy to form water droplets.

Question 15.
Why does water take more time to become hot and take more time to become cool?
Answer:
Water has the highest specific heat among all liquids. So it takes more time to become hot and takes more time to become cool.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 16.
What is calorimeter? Name the material of which it is made of. Give two reasons for using the material stated by you.
Answer:

  • The vessel used for measurement of heat is calorimeter.
  • It is made of thin sheet of copper.
  • The reason is that the specific heat of copper is low and by making the vessel thin, its thermal capacity becomes low so that it takes a negligible amount of heat from its contents to attain the temperature of contents.

Question 17.
Why are burns caused by steam at 100°C more painful than that of water at 100°C?
Answer:

  • Water at 100°C takes additional heat energy to convert from liquid state to vapour (steam) state. This energy is called latent heat of vapourisation.
  • Hence, steam at 100°C contains more heat energy than that of water at 100°C.
  • So, burns caused by steam at 100°C are more painful than that of water at 100°C.

Question 18.
Why is cooking fast in a pressure cooker compared to open vessel?
Answer:

  • Boiling point of liquid increases with external pressure.
  • Pressure cooker gives external pressure to the water in it.
  • So, while cooking in pressure cooker it increases boiling point of water more than 100°C.
  • So that, cooking is fast in pressure cooker compared to open vessel.

Question 19.
What happens to the water when wet clothes dry?
Answer:
When wet clothes dry, the water molecules from wet clothes, after evaporation, change into water vapour and mix with water molecules present in surrounding air, in the form of water vapour.

Question 20.
Why do we use hot water bottles for fomentation?
Answer:
The reason is that water does not cool quickly due to its large specific heat, so hot water bottle provides heat energy for fomentation for a long time.

Question 21.
Why do all plants and animals have a high content of water in their bodies?
Answer:
All plants and animals have nearly 80% to 90% of water in their bodies. So it helps in maintaining the body temperature in all seasons due to high specific heat.

Question 22.
Water is used as an effective coolant. Give reason.
Answer:
1) Water is used as an effective coolant because by allowing water to flow in pipes around the heated parts of machine, heat energy from such parts is removed (e.g. radiators in car and generator are filled with water).
2) Water in pipes extracts more heat from surroundings without much rise in its temperature because of its large specific heat.

Question 23.
Why is the base of cooking pan made thick?
Answer:

  • By making the base of cooking pan thick, its thermal capacity becomes large and it imparts sufficient heat energy at a lower temperature to the food for its proper cooking.
  • Further, it keeps the food warm for a long time, after cooking.

Question 24.
Water in lakes and ponds in cold countries does not freeze all at once. Give reason.
Answer:

  • The latent heat of fusion of ice is sufficiently high.
  • So to freeze water, a large quantity of heat has to be withdrawn, hence it freezes slowly and thus keeps the surroundings moderate.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 25.
Why do drinks get more quickly cooled by adding pieces of ice at 0°C than ice-cold water at 0°C?
Answer:

  • This is because 1 g of ice at 0°C takes 336 J of heat energy from the drink to melt into water at 0°C.
  • Thus drink loses an additional 336 J of heat energy for 1 g of ice at 0°C than for 1 g ice cold water at 0°C. Therefore cooling produced by 1 g of ice at 0°C is more than that by 1 g of water at 0°C.

Question 26.
When ice in a frozen lake starts melting, its surroundings become very cold? Why?
Answer:
The reason is that the heat energy required for melting the frozen lake is absorbed from the surrounding atmosphere. As a result, the temperature of surroundings falls and it becomes very cold.

Question 27.
Why is it more cold after the hail-storm than during or before the hail-storm?
Answer:
The reason is that after the hail-storm, the ice absorbs the heat energy required for melting from the surroundings, so the temperature of the surroundings falls further down and we feel more cold.

Question 28.
Which of the substances A, B, and C has the least specific heat? The temperature versus time graph as shown below.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 8
The substance ‘A’ has least specific heat because the rise in temperature is more for substance. We know that specific heat and rise in temperature are inversely proportional. So A has the least specific heat.

Question 29.
Why do we need pressure cooker to cook food at higher altitudes?
Answer:

  • At higher altitudes, such as hills and mountains, atmospheric pressure is low, therefore water boils at a temperature lower than 100°C and so it does not provide the required heat energy for cooking.
  • Thus cooking there becomes very difficult and it takes a much longer time.
  • So we require a pressure cooker to cook the food at a faster rate.

Question 30.
Lalitha wants to determine the specific heat of Aluminium shots. What apparatus of material is required to do this experiment?
Answer:
The apparatus required is calorie meter, thermometer, stirrer, water, steam heater, wooden box, and aluminium shots.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 31.
What are the material required in order to find specific heat of soild?
Answer:
Calorimeter, thermometer, stirrer, water, steam, heater, wooden box, and lead shots.

Question 32.
Your teacher made an experiment to show the formation of dew and frost. Explain how you show the formation of dew and frost.
Answer:
Place a water bottle in a deep fridge of refrigerator. After some time remove bottle from the refrigerator. We can observe ice in the bottle and water droplets are formed outside the bottle. This experiment is useful in formation of dew and frost.

Question 33.
Why a bottle completely filled with water and closed with a tight cap break after freezing?
Answer:

  • Density of ice is less than that of water.
  • This means water expands on freezing and converts into ice.
  • So, excess space is required to expand the water.
  • The bottle completely filled with water and closed with a tight cap has no excess space to expand water.
  • So, bottle breaks while freezing water in it.

Question 34.
Which of the following substances take more time to raise its temperature for a certain degree Celsius? Give reason.
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 9
Answer:
Water takes more time to raise temperature because it has greater specific heat is, i. e., 1 cal/g-°C. As the specific heat of substance increases, it takes more time to raise its temperature.

Question 35.
The graph shows variations of temperature (T) of one kilogram of material with the heat (H) supplied to it.
At ‘O’ the substance is in the solid state. From the graph can conclude that
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 10
i) The melting point of the solid is …………….
Answer:
The melting point of the solid is T1

ii) The latent heat of fusion is …………….
Answer:
The latent heat of fusion is (H2 – H1)

iii) The latent heat of vaporisation is …………….
Answer:
The latent heat of vaporisation is (H4 – H3)

iv) The boiling point of the substance is ……………
Answer:
The boiling point of the substance is T3

Question 36.
Which of the substances A, B, and C has the highest specific heat? The temperature versus time graph as shown given below.
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 11
Answer:

  • The substance ‘C’ has the highest specific heat.
  • Because ‘C’ does not rise its temperature quickly. In other words ‘C’ takes more time to rise its temperature.

Question 37.
Can the average kinetic energy of a body be even zero?
Answer:
The average kinetic energy of a body can be zero.
Reason :
K.E = \(\frac{1}{2}\) mv2
m is never zero
if v = 0; then body is not in the motion.

Question 38.
A slab of ice at -50°C is constantly heated till the steam attains a temperature of 150°C. Draw a graph showing the change in temperature with time. Label the various parts of the graph properly.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 12

Question 39.
Iron of weight 2 kg was supplied with 12000 calories of heat. Initial temperature of iron was 20°C. Its specific heat is 0.1 cal/g-°C. What is the final temperature of iron?
Answer:
Mass of iron (m) = 2 kg = 2 × 1000 g. = 2000 g.
Quantity of heat supplied (Q) = 12,000 cal.
Initial temperature = θ(i) = 20°C
Final temperature = θf = ?
Specific heat of iron (s) = 0.1 cal / g / °C.
Heat = Q = ms∆θ
Q = ms (θf – θi)
θf – θi = Q/ms
θf – 20 = \(\frac{12000}{2000 \times 0.1}=\frac{12}{2 \times 0.1}\) = 60
θf = 60 + 20 = 80°C
∴ The final temperature of iron = θf = 80°C

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 40.
What is the heat energy required to rise 20 kg of water from 25° C to 75° C?
Answer:
Given m = 20 kg = 20,000 gm
t1 = 25° C ; t2 = 75° C ;
S = 1 cal/gm °C.
Q = mS∆T = 20000 × 1 × (75 – 25) = 20000 × 50
Q = 1000000 calories

Question 41.
If you drink 200 ml of water at 20° C, what is the heat gained by water from your body? (Body temperature is 37° C)
Answer:
m = 200 g (1 ml of water = 1 gm of water)
S = 1 cal/gm °C ;
t1 = 20° C; t2 = 37° C
Q = mS∆T (∆T = t2 – t1)
= 200 × 1 × (37 – 20)
= 200 × 17
Q = 3400 calories

Question 42.
What would be the final temperature of a mixture of 60 gm of water at 30°C temperature and 60 gm of water at 60°C temperature?
Answer:
m1 = 60 g ;
T1 = 30°C ;
m2 = 60 g ;
T2 = 60°C
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 13

Question 43.
The quantity of heat which can rise the temperature of ‘x’ gram of a substance through t1°C and the quantity of heat which can rise the temperature of ‘y’ grams of water through t2°C is same. What is ratio of specific heats? What is ratio of specific heats if rise in temperatures are same and if amount of substances are same?
Answer:
Suppose specific heats of substance and water are s1 and s2 respectively.
Heat absorbed by x gram of substance to rise its temperature to t1°C.
Q1 = ms∆T = x × s1 × t1
Heat absorbed by y gram of water to rise its temperature to t2°C.
Q2 = ms∆T2 = y × s2 × t2
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 14

10th Class Physics 1st Lesson Heat 4 Marks Important Questions and Answers

Question 1.
Write the factors that effect the process of evaporation. Explain with suitable examples. (AP March 2017)
Answer:
Process of evaporation is effected by surface area, wind speed, humidity, and temperature.
Ex:

  • The water kept in a china dish evaporates faster than in a cup because of more surface area.
  • Water in wet the clothes are kept under fan evaporates faster than in normal conditions.
  • Water in wet clothes evaporates faster on a less humid day than on a more humid day.

Question 2.
A) Write the principle of method of Mixtures.
B) What would be the final temperature of a mixture of 60 gms of water at 50°C and 50 gms of water at 70°C? (AP March 2018)
Answer:
A) Principle of method of mixtures :
Net heat lost by the hot body = Net heat gain by the cold body.

B) m1 = 60 gms.,
T1 = 50°C ;
m2 = 50 gms.,
T2 = 70°C
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 15

Question 3.
Answer the following questions by using the data given in the table. (AP March 2018)

Substance Specific heat (cal / g°C)
Lead 0.031
Aluminium 0.21
Copper 0.095
Water 1.00
Iron 0.115

a) Write SI units for specific heat.
Answer:
Joule / Kilogram-Kelvin

b) Based on specific heat values, arrange the substances given in the table in ascending order.
Answer:
Lead, Copper, Iron, Aluminium, Water

c) If we supply same quantity of heat, which substance will heat up faster?
Answer:
Lead

d) Calculate the amount of heat required to raise the temperature of 1 kg of Iron through 10°C.
Answer:
Q = mS∆T = 1000 × 0.115 × 10 = 1150 cal.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 4.
Suggest an experiment to show that when ice is converted into water, its temperature does not change. How much heat is required to convert 5 grams of ice at 0°C to water, at the same temperature? (Latent heat of fusion of ice is 80 cal/gram). (TS June 2015)
Answer:
Procedure :

  1. Take small ice cube in a beaker. Insert the thermometer in the beaker.
  2. Now start heating the beaker and note down readings of thermometer every one minute till the ice completely melts and gets converted into water.
  3. Before heating the temperature of ice is 0° C or less than 0° C.

Observation :

  1. We will observe that the temperature of ice at the beginning is equal to or below 0°C.
  2. If the temperature of ice is below 0°C, it goes on changing till it reaches 0° C.
  3. When ice starts melting, we will observe no change in temperature though you are supplying heat continuously.

Explanation :

  1. Given heat energy is used to break the bonds (H2O) in ice and melts.
  2. So, temperature is constant while melting.

Conclusion :

  1. This process is called melting. In this process heat converts solid phase to liquid phase.
  2. The temperature of the substance does not change until all the ice melts and converts into water.
  3. The heat given to melting is called latent heat of fusion.
  4. The heat required to convert 1 gm of solid completely to liquid at constant temperature is called “latent heat of fusion”.
    m = 5 gm; Lf = 80 cal/g

The amount of heat absorbed Q = MLf = 5 × 80 = 400 cal /g.

Question 5.
The graph shows the values of temperature, when ice is heated till it becomes water vapour. Observe the graph and answer the following questions. (TS March 2016)
(Note that the figure is not completely quantitative and also not to the scale. It is purely qualitative)
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 16
a) At what temperature, ice converts into water?
Answer:
Ice converts into water at 0°C and above.

b) What does \(\overline{\mathrm{DE}}\) represent?
Answer:
\(\overline{\mathrm{DE}}\) represents the latent heat of vapourisation.

c) What is the range of temperature of liquid water?
Answer:
The range of temperature of liquid water is 0°C to 100°C

d) Which part of the graph represents change of state from ice to water?
Answer:
\(\overline{\mathrm{BC}}\) represents the change of state of ice to water.

Question 6.
Write the differences between heat and temperature.
Answer:

Heat Temperature
1) It is a thermal energy. 1) It is the measurement of hotness or coldness.
2) Heat is an extensive property, means it does not depend on amount of the substance that is present. 2) Temperature is an intensive property means that the substance present will not change the specific characteristic.
3) Heat is the amount of energy of the system. 3) Temperature is the measure of the average molecular motions in a system.
4) Its S.I unit is Joules. 4) Its S.I. unit is degrees C or K.
5) Heat is energy itself that flows. 5) It decides the direction of heat flow.

Question 7.
Why is the specific heat different for different substances?
Answer:

  • We know that the temperature of a body is directly proportional to the average kinetic energy of particles of the body.
  • The molecules of the system have different forms of energies. The total energy of the system is called internal energy of the system.
  • When we supply heat energy to the system, the heat energy given to it will be shared by the molecules among the various forms of energy.
  • This sharing will vary from substance to substance.
  • The rise in temperature is high for a substance, if the maximum share of heat energy is utilised for its linear K.E.
  • This sharing of heat energy also varies with temperature. That is why the specific heat is different for different substances.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 8.
Explain the process of evaporation.
Answer:

  • The molecules of a liquid that kept in a dish, continuously move with random speeds in various directions. As a result, these molecules collide with other molecules.
  • During this collision they transfer energy to other molecules. Hence, the molecules at the surface acquire energy and may fly off from the surface.
  • Some of these escaping molecules may be directed back into liquid when they collide with the particles of air.
  • If the number of escaping molecules is greater than the number returned, then the number of molecules in the liquid decreases.
  • Thus when a liquid is exposed to air, the molecules at the surface keep on escaping from the surface till the entire liquid disappears into air. This process is called evaporation.

Question 9.
Define evaporation. Explain what are the affecting factors of evaporation and how they effect the rate of evaporation.
Answer:
Evaporation :
The process of escaping of molecules from the surface of liquid at any temperature is called evaporation.

The affecting factors of evaporation :

  1. Temperature,
  2. Surface area,
  3. Wind speed,
  4. Humidity.

The affection on the rate of evaporation :
1) Temperature :
As the temperature increases evaporation increases.

2) Surface area :
As the surface area of liquid increases, more molecules tend to leave the surface. So rate of evaporation increases.

3) Wind speed :
As the wind speed increases rate of evaporation increases.

4) Humidity :
As the humidity increases rate of evaporation decreases.

Question 10.
Why is climate near the seashore moderate?
Answer:

  1. The specific heat of water is very high. It is about five times as that of sand.
  2. Hence the heat energy required for the same rise in temperature by certain mass pf water will be nearly five times than that required by same mass of sand.
  3. Similarly, a certain mass of water will give out nearly five times more heat energy than that given by sand of the same mass for the same fall in temperature.
  4. As such, sand (or earth) gets heated or cooled more rapidly as compared to water under similar conditions.
  5. Thus, a large difference in temperature is developed between the land and sea due to which land and sea breezes are formed.
  6. These breezes make the climate near seashore moderate.

Question 10.
Why do farmers fill their fields with water on a cold winter night?
Answer:

  • In the absence of water, if on a cold winter night, the atmospheric temperature falls below 0°C, the water in the fine capillaries of plants will freeze, so the veins will burst due to the increase in volume of water on freezing.
  • As a result, plants will die and the crop will be destroyed.
  • In order to save crop on such cold nights, farmers fill their field with water because water has high specific heat, so it does not allow the temperature in the plants of surrounding area to fall up to 0°C.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 11.
Explain the factors effecting boiling.
Answer:
The factors effecting boiling are
I) Pressure :

  1. The boiling point of pure water at one atmospheric pressure is 100°C.
  2. Water boils at a temperature higher than 100°C, if the atmospheric pressure is higher than one atmosphere pressure, and boils at a temperature lower than 100°C, if the atmospheric pressure is less than 1 atmosphere.

II) Impurities :
The boiling point of liquid increases by the addition of impurities to it. If a little common salt is added to water, the water boils at a temperature higher than 100°C.

Question 12.
A, B and C are the three liquids at 20°C, 30°C and 40°C respectively. If equal masses of A and B are mixed, the resultant temperature is 26°C. If equal masses of A and C are mixed, the resultant temperature is 33°C. Find the ratio of specific heats of A, B and C.
Answer:
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 17
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 18
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 19

Question 13.
A refrigerator converts 5kg of water at 40°C into ice at 0°C in 20 minutes. Find the power of refrigerator.
Answer:
m = 5kg = 5000gr
∆t = 40°C – 0°C = 40°C
S = 1 (water) .
Q = m.s.∆t = 5000.1.40 = 200000 cal.
We know the relation between heat and work done as
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 20

Question 14.
Snow on mountains does not melt all at once. Why?
Answer:

  • Snow on mountains does not melt all at once because the ice has a high specific latent heat of fusion.
  • It is due to this fact that it changes into water slowly as it gets heat energy from the sun.
  • If latent heat would not have been so high, the snow would have melted quickly even with a small amount of heat energy and there would have been floods in rivers.

Question 15.
Collect specific heats of various substances.
Answer:

Substance Specific heat
In cal/g – °C In J/kg – K
Lead 0.031 130
Mercury 0.033 139
Brass 0.092 380
Zinc 0.093 391
Copper 0.095 399
Iron 0.115 483
Glass(flint) 0.12 504
Aluminium 0.21 882
Kerosene oil 0.50 2100
Ice 0.50 2100
Water 1 4180
Sea water 0.95 3900

Question 16.
The graph given below represents a cooling curve for a substance being cooled from a higher temperature to a lower temperature.
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 21
a) What is the boiling point of the substance?
Answer:
The boiling point of the substance is 150°C (because the part BC represents condensation where the vapour changes into the liquid without the change in temperature).

b) What happens in the region DE?
Answer:
The region DE represents freezing of the substance where the liquid changes into solid at a constant temperature that is 100°C.

c) What is the melting point of the substance?
Answer:
The melting point of substance is 100°C.

Question 17.
You’ve taken water in vessel at 0°C and closed it with a glass vessel as shown in the figure. You used and created a vacuum inside.
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 22
a) Explain what happens.
b) A part of water condenses; what is the amount of water that gest condensed?
Answer:
a) At 0°C also water is available in liquid state (generally at 0°C ice is also available) because the air in vacuum rise the temperature. Here evacuation is possible so it allows evaporation.

b) Let y ml of water is taken at 0°C.
’x’ ml of water is evaporated
Latent heat of vapourisation = Lsteam = 540 Cal/g.
Latent heat of ice = Lice = 80 Cal/g.
After sometime conversion process stops. So equilibrium is possible.
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 23

Question 18.
What are the applications of specific heat capacity?
Answer:
1. The oceans behave like heat store houses for the earth. They absorb large amounts of heat at the equator without rise in temperature due to high specific heat capacity of water. So, oceans moderate the surrounding temperature near the equator. Ocean water transports the heat away from the equator to areas closer to the north and south poles. This transported heat helps moderate the climates in parts of the Earth that are far from the equator.

2. Watermelon brought out from the refrigerator retains its coolness for a long time than any other fruit because it contains a large amount of water (water has greater specific heat).

3. The samosa seems to be cool outside but it is hot when we eat it because the curry inside the samosa contains ingredients with higher specific heats.

AP SSC 10th Class Physics Important Questions Chapter 1 Heat

Question 19.
Some hot water Is added to three times its mass of cold water at 10°C. The resulting temperature is found to be 20°C. Find the initial temperature of hot water.
Answer:
Let the initial temperature of hot water be t°C.
Mass of hot water = mg
Mass of cold water = 3 mg
Initial temperature of cold water = 10°C
And resultant temperature = 20°C
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 24
∴ Initial temperature of hot water = 50°C.

Question 20.
40 g of water at 60°C is poured into vessel of 200 g mass containing 50 g of water at 20°C. The final temperature of mixture is 30°C. Calculate the specific heat of vessel.
Answer:
Mass of hot water at 60°C = 40 g
Mass of cold water at 20°C = 50 g
Mass of vessel = 200 g
Suppose the specific heat of vessel is Sv
Heat energy given by hot water = mSw∆T
= 40 × 1 × (60 – 30) [∵ T1 = 60°C, T2 = 30°C]
= 40 × 30 = 1200 cal
Heat energy taken by cold water = mSw∆T
= 50 × 1 × (30 – 20) = 50 × 10 = 500 cal
Heat energy taken by vessel = mSv∆T = 200 × Sv × (30 – 20) = 2000 Sv
According to the principle of method of mixtures,
Heat lost by hot water = heat gained by cold water + heat gained by vessel 1200 = 500 + 2000 Sv
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 25

Question 21.
A, B and C are three liquids at 20°C, 30°C and 40°C respectively. If equal masses of A and B are mixed, the resultant temperature is 29°C. The equal masses of A and C are mixed, the resultant temperature is 33°C. Find the specific heats of A, B and C.
Answer:
Suppose specific heats of liquids A, B and C are s1, s2 and s3 respectively.
Given that the temperatures of liquids are 20°C, 30°C and 40°C.
Given that equal mass of A and B are mixed, the resultant temperature is 29°C.
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 26

Further given that equal masses of A and C are mixed, the resultant temperature is 33°C.
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 27

Question 22.
A refrigerator converts 5 kg of water at 40°C into ice at 0°C. Find the total energy released in Joules.
Answer:
m = 5 kg = 5000 g
The heat energy released to convert 5 kg of water at 40°C to 5 kg of water at 0°C.
Q1 = ms∆T
= 5000 × 1 × (40 – 0) = 5000 × 40 = 200000 cal [s = 1 cal/g-°C for water]
The heat energy released to convert 5 kg of water at 0°C to 5 kg of ice at 0°C.
Q2 = mLf
= 5000 × 80 = 400000 cal [∵ Lf = 80 cal/g]
Total energy released = 200000 + 400000
= 600000 cal
= 600 kcal
= 142.86 kJ.

Question 23.
The quantity of heat which can rise the temperature ‘x’ grams of a substance through t1°C can rise the temperature of ‘y’ grams of water through t2°C is same.
What is the ratio of specific heats?
Answer:
Given,
AP SSC 10th Class Physics Important Questions Chapter 1 Heat 28

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 9 Classification of Elements- The Periodic Table.

AP State Syllabus SSC 10th Class Chemistry Important Questions 9th Classification of Elements- The Periodic Table

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table 1 Mark Important Questions and Answers

Question 1.
What is modern periodic law? (AP June 2015)
Answer:
Modern periodic law :
The physical and chemical properties of the elements are periodic functions of their electronic configurations.

Question 2.
Define Moseley’s periodic law. (AP June 2015)
Answer:
Moseley’s periodic law: The physical and chemical properties of elements are periodic functions of their atomic numbers.

Question 3.
Which group elements are called Carbon family? (AP Mareh 2016)
Answer:
14 (or) IVA Group of elements are called Carbon family.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 4.
Which atom is bigger in size, Ne or Ar? Why? (AP June 2018)
Answer:
Ar. In groups as we go down number of shells increases due to the formation of new shell.

“O Group”
He
Ne
Ar
Kr
Xe
Rn

Question 5.
A and B are two elements. The compound formed with A and B is A2 B. What are the valencies of A and B. (TS March 2018)
Answer:
The valency of A is 1 and B is 2.

Question 6.
A teacher asked to give an example for Dobereiner’s triad. Ramu wrote them as “Li, Na, Mg”. In these three, identify which element does not belongs to this triad? (AP March 2019)
Answer:
Mg or Magnesium do not belongs to this triad.

Question 7.
Write the difference between Mendeleeff’s periodic law and modern periodic law. (AP SCERT: 2019-20)
Answer:
Mendeleeff’s periodic table is prepared based on atomic mass whereas modem periodic table is prepared based on atomic number (electronic configuration).

Question 8.
What is Dobereiner Triad? Give two examples to it.
Answer:
A group of three elements in which atomic weight of middle element is average of first and third element is called Dobereiner triad with similar propertion.
Eg: 1) U, Na, K
2) Cl, Br, I

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 9.
What is Newlands’ law of octaves?
Answer:
When elements are arranged in the ascending order of their atomic weights, every eighth element starting from a given element resembles in its properties to that of starting element. This is called Newlands’ law of octaves.

Question 10.
What is MendeleefFs periodic law?
Answer:
MendeleefFs periodic law:
The physical and chemical properties of the elements are the periodic functions of their atomic weight.

Question 11.
What is the name given to horizontal rows and vertical columns in MendeleefFs periodic table?
Answer:
Horizontal rows are periods and vertical columns are groups.

Question 12.
What is the property on which MendeleefFs periodic table depends upon?
Answer:
Mendeleeff’s periodic table depends upon atomic weight.

Question 13.
What is the name given to I(A) group elements?
Answer:
Alkali metal family, because aliquili = plant ashes. Na, K, etc. were obtained from plant ash.

Question 14.
Why are VI A group elements called chalcogens?
Answer:
Chalcogeneous = Ore product. As the elements in group 16 (VI A) form ores with metals, they are called chalcogeneous family.

Question 15.
Why are VII A group elements called halogens?
Answer:
Halos – sea salt, genus – produced. So VII A (17) are obtained from nature as sea salt. So they are called halogen family.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 16.
What are halogens?
Answer:
Fluorine, Chlorine, Bromine, Iodine, and Astatine of VIIA group elements are called halogens, which are obtained from sea salt.

Question 17.
What are noble gases? What is the general electronic configuration of noble gases?
Answer:
The elements of group VIII A (18) are chemically least reactive so they are called noble gases. Their group electronic configuration is ns²np6 (except) for helium it is 1s².

Question 18.
What are Lanthanides?
Answer:
Elements acquiring same properties are called lanthanides, i.e. 4f elements. They are from 58Ce (Cerium) to 71Lu (Lutetium).

Question 19.
What are Actinides?
Elements acquiring different properties are called actinides, i.e. 5f elements. They are from 90Th (Thorium) to 103Lr (Lawrensium).

Question 20.
What are metals and non-metals?
Answer:
The elements with three or less electrons in the outer shell are considered to be metals and the ore with five or more electrons in the outer shell are considered to be non-metals.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 21.
What are metalloids?
Answer:
The properties of elements which are intermediate between the properties of metals and non-metals are called metalloids.

Question 22.
Which will behave like semi-conductors?
Answer:
Metalloids or semi-metals behave like semi-conductors.

Question 23.
What is valency?
Answer:
The combining power of element with respect to hydrogen, oxygen or indirectly any other element through hydrogen and oxygen is called valency.

Question 24.
What is the latest definition of valency?
Answer:
The number of electrons lost or gained or shared during a chemical reaction.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 25.
How do we measure atomic radius of solids?
Answer:
It is half of the distance of radius of each atom.

Question 26.
What is covalent radius?
Answer:
Half of the distance between length of covalent bond is called covalent radius.

Question 27.
In which units is atomic radius measured?
Answer:
Atomic radius is measured in pico meter (pm) units.
1 pm = 10-12 m.

Question 28.
What is the method given by Milliken to calculate electronegativity of an element?
Answer:
According to Milliken, electronegativity of element is average value of its ionization energy and electron affinity.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 1

Question 29.
What is electropositive character?
Answer:
The tendency of metals to remain positive ions in compounds is called electropositive character. (OR) The tendency of an atom to lose electrons to form positive ions.

Question 30.
What is screening effect or shielding effect?
Answer:
More the shells with electrons between the nucleus and the valence shell, they act as screens to decrease nuclear attraction over valence electron. This is called screening effect or shielding effect.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 31.
What do you mean by negative or positive electron gain enthalpy?
Answer:
The negative sign indicates that energy is liberated or lost, and the positive sign indicates that the energy is gained or absorbed.

Question 32.
What is a triad?
Answer:
A group of three elements with similar properties in which atomic weight of middle element is average of other two elements.

Question 33.
Chlorine, bromine, iodine are Dobereiner’s triads. How do you justify?
Answer:
Chlorine, bromine and iodine have similar properties and atomic weight of bromine is average of chlorine and iodine.

Question 34.
Why are lanthanides and actinides placed separately at the bottom of the periodic table?
Answer:
Lanthanides and actinides belong to f – block elements with different properties so they are placed at the bottom of periodic table.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 35.
Lithium, Sodium, and Potassium were put in one group on the basis of their similar properties.
1) What are those similar properties?
2) What is the common name of this group of family?
Answer:

  1. They have same number of valence electrons that is 1 and valency 1. So they have similar chemical properties.
  2. They are called alkali metals.

Question 36.
What are the following groups known as?
1) group VIIA elements
2) Zero group elements.
Answer:

  1. Group VII A elements are called Halogens.
  2. Zero group elements are called Noble gases.

Question 37.
An element Barium lies in 2nd group; then answer the following.
1) What is its valency?
2) What will be the formula of its Phosphate?
Answer:

  1. The element lies in second group. So its valency is 2.
  2. The formula of Phosphate is Ba3(PO4)2 [since the valency of Phosphate is 3],

Question 38.
A, B, C are three elements having their atomic numbers equal to 2,10 and 5 respectively.
a) Which of these elements belong to same period?
b) Which of these elements belong to same group?
Answer:
The electronic configurations of A, B, C are as follows
A – 2, B – 2, 8, C – 2, 3.
a) So, B and C belong to same period because valence electron enters same orbit.
b) A and C belong to same group because both are noble gases.

Question 39.
Which element of 3rd period will form a chloride of Cl4?
Answer:
It would be Silicon because its electronic configuration is 2, 8, 4. So, it lies in third period and its valency is 4.

Question 40.
Which two elements of 3rd period will form a covalent compound?
Answer:
The two elements are phosporous and chlorine.

Question 41.
An element has atomic number 12. State whether it is metal or non-metal. Why?
Answer:
Its electronic configuration is 2, 8, 2. It lies in 2nd group. The elements towards left of periodic table are generally metals. So the element is metal.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 42.
Elements X, Y, and Z belong to IA group of the periodic table. Their atomic radii are as follows.
X → 1.33 Å,
Y → 0.95 Å,
Z → 0.60 Å.
Arrange the elements in the increasing order of atomic number by giving reason.
Answer:
As we move from top to bottom in a group, atomic size increases and atomic number also increases.
So the correct increasing order is Z, Y, X.

Question 43.
An element has an atomic number 16. State
i) period to which it belongs
ii) the number of valence electrons.
Answer:
Its electronic configuration is 2, 8, 6.
i) So it belongs to 3rd period (orbit number).
ii) The number of valence electrons is 6.

Question 44.
Why is energy absorbed when electron is added to uni-negative ion?
Answer:
It is difficult to add an electron to uni-negative ion. In order to overcome the repulsion between the electrons, actually energy should be supplied to add another electron to uni-negative ion.

Question 45.
When do you observe liberation of energy?
Answer:
Atoms of some elements gain electrons while forming ionic compounds. An atom is able to gain electron when the electron is attracted by the nucleus. Attraction involves the liberation of energy.

Question 46.
Why does nitrogen have less electron affinity value compared to oxygen?
Answer:
The electron affinity of nitrogen is less than oxygen because of stable configuration of nitrogen (i.e., 2p³ configuration).

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 47.
Which one between Na and Na+ would have more size? Why?
Answer:
Na has more size because when one electron is removed from Sodium atom the nucleus attraction over outermost electron increases so atomic size decreases.

Question 48.
Second ionization energy of an element is higher than its first ionization energy. Why?
Answer:
It is difficult to remove an electron from unipositive ion when compared with neutral atom. So second ionization energy is always greater than first ionization energy.

Question 49.
Hydrogen can be placed in group’1 and group 7 periodic table. Why?
Answer:
Hydrogen has both +1 as well as -1 oxidation states. So still there is some ambiguity in position of hydrogen.

Question 50.
Why do inert gases have zero valency value?
Answer:
Inert gases show zero valency because they do not take part in chemical reactions due to stable configuration.

Question 51.
Element ’Z’ belongs to (second) 2nd group in the periodical table. Write the formula of oxide.
Answer:
The formula of oxide of the element is ZO.

Question 52.
Do the atom of an element and its ion have same atomic size?
Answer:
No, generally cation has smaller size and anion has greater size.

Question 53.
The electronegativities of the elements in period 3 of the periodic table are as. follows.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 2
Arrange the elements in which they occur in the periodic table from left to right.
Answer:
Na Mg Al Si P S Cl

10th Class Chemistry 9th Lesson Classification of Elements – The Periodic Table 2 Marks Important Questions and Answers

Question 1.
An element has atomic number 17. Where would you expect this element in the Periodic Table? Why? (AP June 2018)
Answer:

  • Electronic configuration of the given element is 1s² 2s² 2p6 3s² 3p5.
  • So, it is in 3rd period and 17th group of periodic table.
  • Due to the valency electronic configuration of 3s² 3p5 it belongs to 3rd period and 17th group.

Question 2.
How do you appreciate the special nature of inert gases?
Answer:
I appreciate the special nature of inert gases because it helps us in explaining the formation of chemical bonds among the atoms of elements and their stability.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 3.
The atomic number of an element is 35. Where would you expect the position of this element in the periodic table? Why? (TS June 2015)
Answer:

  • The Electronic configuration of element with atomic number 35 is 2, 8, 18, 7.
  • So it has seven valence electrons.
  • That’s why it is present in 17th group or VII A group and 4th period.
  • The element is Bromine.

Question 4.
Why were Dobereiner, Newlands and Mendeleeff not 100% successful in their classification of elements? Why is the modern table relatively a better classification? (TS March 2016)
Predict the reason.
Answer:

  • All the known elements at the time of Dobereiner could not be arranged in the form of triads.
  • Newlands’ periodic table was restricted only for 56 elements.
  • As Mendeleeffs classification is based on atomic weight, his classification led to two defects like anomalous pair of elements and dissimilar elements placed together.
  • Modern periodic table was prepared on the basis of atomic number. So the periods and groups are clearly defined.

Hence Dobereiner, Newlands, and Mendeleeffs classifications were not 100% successful, but modern classification is successful.

Question 5.
Observe the electronic configurations given below and write the group and period numbers of those elements. (TS March 2016)
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 3
Answer:
a) The period number is 3 and group number is 1.
b) The period number is 3 and group number is 15.

Question 6.
Observe the information provided in the table and answer the questions given below it. (TS June 2017)
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 4
i) What are the s-block elements in the table?
ii) What are the ‘p’ block and ‘d’ block elements in the table?
Answer:
i) s-block elements : Na, Ca

ii) p-block elements : C, P
d-block elements : Ti, Ni.

Question 7.
Imagine, which one in each of the following pairs is large in size relatively with other? Explain. (AP March 2019)
(X) Na, Al (Y) Na, Mg+2
Answer:
(X) 1. Na is large in size than Al.
2. Atomic size gradually decreases from left to right in a period.

(Y) 1. Na is large in size than Mg2+.
2. Na is larger than Mg and Mg is larger than Mg2+.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 8.
What are the limitations of Dobereiner triad?
Answer:

  • All the known elements could not be arranged in the form of triads.
  • The law failed for very low mass or for very high mass elements.
    Eg : In case of F, Cl, Br the atomic mass of Cl is not an arithmetic mean of atomic masses of F and Br.
  • As the techniques improved for measuring atomic masses accurately, the law was unable to remain strictly valid.

Question 9.
Distinguish between electron affinity and electronegativity.
Answer:

Electron affinity Electronegativity
1. It is the property of an isolated gaseous atom. 1. It is the property of a bonded atom.
2. It is the energy released and is measured in ev/atom or kJ/mole. 2. It is relative quantity and has no units.
3. It is the attraction of an atom for a single electron. 3. It is the attraction of an atom for a pair of electrons.

Question 10.
What is electronegativity? What are the various methods used to determine electronegativity? Explain.
Answer:
Electronegativity :
The electronegativity of an element is defined as the relative tendency of its atom to attract electrons towards itself when it is bounded to the atom of another elements.

Various methods to calculate Electronegativity :
1) Milliken Scale :
According to Milliken, the electronegativity of an element is the average value of its ionization energy and electron affinity.

2) Pauling Scale :
Pauling scale is based on bond energies. The electronegativity of hydrogen is assumed as 2.20. Electronegativity of other elements is calculated with respect to hydrogen.

Question 11.
Give the electronic configurations of following elements. What do say about these elements by writing their electronic configurations?
a) Na
b) Al
c) Sc
d) Ce
Answer:
a) Na : 1s² 2s² 2p6 3s¹
b) M : 1s² 2s²2 2p6 3s² 3p¹
c) Sc : 1s² 2s² 2p6 3s² 3p6 4s² 3d¹
d) Ce : 1s² 2s² 2p6 3s² 3p6 3d10 4s² 4p6 4d10 5s² 5p6 6s² 4f²

Inference :
The valence electron enters different orbitals. So these elements belong to different blocks in modern periodic table, i.e. s, p, d, and f respectively.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 12.
Do you think that Newlands’ law of octaves is correct? Justify.
Answer:
No, Newlands’ law of octaves was restricted to only 56 elements and did not leave any room for new elements. Elements that were discovered later could not be filled into Newlands’ table in accordance with properties.

Question 13.
Why did Mendeleeff have to leave certine blank spaces in his periodic table?
Answer:
Mendeleeff predicted that some elements were missing in the table so he left blank spaces at the appropriate places in the table.

Question 14.
Give reason for the need of classification of elements.
Answer:
Classification is necessary because it is difficult to remember the properties of all the elements separately. It is easy to identify the properties of elements by making them groups with similar properties.

Question 15.
x, y, and z are the elements of a Dobereiner’s triad. If the atomic mass of ‘x’ is 7 and that of ‘z’ is 39, what should be the atomic mass of ‘y’?
Answer:
The atomic mass of x = 7 ;
The atomic mass of z = 39
x, y, z form Dobereiner triad
∴ Atomic mass oi y = average of x and z = \(\frac{7+39}{2}\) = \(\frac{46}{2}\) = 23

Question 16.
Name the two elements that would expect to have chemical properties similar to element with atomic number 11. What is the base for your choice?
Answer:
The element with atomic number 11 is sodium and its electronic configuration is 1s² 2s² 2p6 3s¹ or 2, 8, 1.

So it has one valence electron, i.e. present in I group. We know that the elements present in same group have same valence electrons. So they show similar properties.

Therefore the other two elements are Lithium and Potassium.

Question 17.
An element X belongs to 3rd period and group 14 of the periodic table. State
a) the number of valence electrons
b) the valency
c) the name of the element.
Answer:
a) The number of valence electrons are 4.
b) Its valency = 8 – 4 = 4.
c) The electronic configuration of element is 2, 8, 4.
(Because 3rd period means third orbit, group 14 has 4 valence electrons). So, the element with atomic number 14 is Silicon.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 18.
Why is it easier to remove 4f electron than 4s?
Answer:
Orbitals belonging to the same main shell have different penetration power towards the nucleus. In fourth main shell the order of penetration is like this 4s > 4p > 4d > 4f. So, it is easier to remove 4f electron than 4s.

Question 19.
How is screening effect responsible for low ionization of cesium?
Answer:

  • More the shells with electrons between the nuclear and the valence shell, they act as screens and decrease nuclear attraction over valence electron. This is called the screening effect.
  • More the screening effect, less is the ionization energy.
  • Cesium with more inner shells has less ionization energy.

Question 20.
Why does Boron have less ionization energy when compared with Beryllium?
Answer:

  • The electronic configuration of Be and B are 1s² 2s² and 1s² 2p² 2p¹.
  • The element Boron has less ionization energy due to less penetration power of 2p compared to 2s.

Question 21.
We know that as we move from left to right ionization energy increases. But ionization energy Nitrogen is more than Oxygen. Why?
Answer:

  • It is easier to remove an electron from Oxygen when compared to Nitrogen.
  • This is because Nitrogen has stable 1s² 2s² 2p³ electronic configuration which contains half filled 2p orbitals whereas Oxygen has 1s² 2s² 2p4 configuration.

Question 22.
Why is it difficult to remove an electron from Mg+ when compared with Mg?
Answer:

  • The energy required to remove the first electron outermost orbit of a neutral gaseous atom of the element is called first ionization energy.
  • The energy required to remove from unipositive ion of the element is called second ionization energy.
  • Second ionization energy is always more than first ionization energy because it is difficult to move electron from unipositive ion due to greater nuclear attraction.
  • So it is difficult to remove an electron from Mg+ when compared with Mg.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 23.
Using the periodic table predict formula of compound formed between an element ‘X’ of group 2 and another element of group 17.
Answer:

  • The element X belongs to group 2. So, the number of valence electrons are 2 and its valency is 2.
  • The element Y belongs to group 17 or VII. So, the number of valence electrons are 7 and its valency = 8 – 7= 1.

During formation compound elements exchange their valencies.
∴ The formula of compound is XY2.

Question 24.
How do electronegativity values vary in period and group?
Answer:
Period :
When we move from left to right in period, the electronegativity increases due to decrease in atomic size.

Group:
When we move from top to bottom in a group, the electronegativity decreases due to increase in atomic size.

Question 25.
How does metallic and non-metallic characters vary in a period and group?
Answer:
Period:
When we move from left to right in a period, the metallic character decreases and non-metallic character increases.

Group :
When we move form top to bottom in a group, non-metallic character decreases and metallic character increases.

Question 26.
How do valency vary in period and group?
Answer:
Period :
When we move from left to right in a period, the valency does not follow a regular trend. For example, in second period the valency starts from 1 and increases to 4, then thereafter decreases to ‘O’.

Group :
When we move top to bottom in a group, the valency remains the same because in a group the valence electrons are same.

Question 27.
How does electron affinity vary in a period and group?
Answer:
Period :
When we move from left to right in period, electron affinity increases due to greater nuclear attraction over electron.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Group :
When we move from top to bottom, the electron affinity decreases in atomic size. As the size of the atom increases, the nuclear attraction over outermost electron decreases. So electron affinity decreases.

Question 28.
An element has atomic number 19. Where would you expect this element in the periodic table anti why?
Answer:
The electronic configuration of element is 1s²2s²2p63s²3p64s¹. So the element is in 4th period and I group.

Question 29.
The electronic configuration of the element X, Y, and Z are given below,
a) X = 2, 5
b) Y = 2, 8, 1
c) Z = 2, 8
i) Which element belongs to 18th group?
ii) Which element belongs to 15th or V group?
iii) Which element belongs to third period?
Answer:
i) Z belongs to 18th group because it is a noble gase (i.e. Ne).
ii) X belongs to 15th or V group because it has 5 valence electrons.
iii) Y belongs to 3rd period because the valence electron is present in 3rd orbit.

Question 30.
Referring the part of periodic table given below answer the questions that follow.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 5
1) What happens to the atomic size if moved from left to right? Support your answer.
2) What changes do you observe in the metallic properties of the elements when moved from left to right?
Answer:
1) When we move from left to right in a periodic table atomic radii of elements decrease, as a result the size of the,atom decreases,

2) When we move from left to right in a periodic table electronegativity values of elements increase, as a result the metallic properties of the elements decrease.

Question 31.
State the name of element, number of valence electrons, valency, the group number and the period number of each element given in the following table.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 6
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 7

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table 4 Marks Important Questions and Answers

Question 1.
What is Ionization Energy? Explain the factors that effect Ionization Energy. (AP June 2017)
(OR)
What is ionization energy? What are the factors which influence ionization energy? Explain.
(OR)
Write the factors that influence ionization energy and explain any three of them. (TS March 2019)
Answer:
Ionization energy :
The energy required to remove an electron from the outermost orbit or shell of a neutral gaseous atom is called ionization energy.
Factors influencing ionization energy :
1) Nuclear charge :
As nuclear charge increases, ionization energy increases.

2) Screening effect or shielding effect:
More the screening effect, less is the ionization energy.

3) Penetrating power of the orbitals :
If the orbitals have less penetrating power, then the ionisation energy is less. Generally, the penetrating power of orbits are like this : s > p > d > f.

4) Stable configuration :
The elements having half-filled or completely filled orbitals have more stability. So the ionization energy is more when the element has stable configuration.

5) Atomic size :
As the atomic size increases, the nucleus attraction over outermost electron decreases. So ionization energy decreases.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 2.
Elements of one short period of the Periodic Table are given below in the order from left to right. (AP March 2017)
Li, Be, B, C, N, F, Ne
Answer the following:
(i) To which period, do these elements belong?
(ii) One element of this period is missing. Which is the missing element and where it should be placed?
(iii) Which of the above elements belong to the family of halogens? What is its electronegativity value?
(iv) How does the metallic character varies in the Period?
Answer:
(i) 2nd period.
(ii) Oxygen.
It should be placed between Nitrogen and Flourine.

(iii) Flourine
Electro negativity 4.0

(iv) Decreases from left to right.

Question 3.
In the table given below, names of some elements of families are given. Based on this, fill the information in the empty boxes. (TS June 2015)
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 8
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 9

Question 4.
Two elements X and Y belong to Groups 1 and 2 respectively in the same period of the Periodic Table. Compare these elements with respect to : (TS March 2015)
i) number of electrons in their outermost orbit.
Answer:
The number of electrons in the outermost orbit of element X = 1
The number of electrons in the outermost orbit of element Y = 2

ii) their atomic size and their valencies.
Answer:
The atomic size of the Y is lesser than X
Valence of X = 1 ; Valence of Y = 2

iii) their ionisation energy and metallic character.
Answer:
The ionization energy of Y is greater than X, X has higher metallic character than Y.

iv) formulae of their chlorides and sulphates.
Answer:
Chloride of X …. XCl
Chloride of Y …. YCl2
Sulphate of X …. X2SO4
Sulphate of Y …. YSO4

Question 5.
How are the elements arranged into groups and periods in the Modern Periodic Table? Elements in a group possess similar properties, but elements in a period do not show similarities in their properties. Why? (TS June 2017)
Answer:

  • The Modern periodic table is arranged in groups and periods based on the electronic configuration of the atoms of elements.
  • Physical and Chemical Properties of elements are related to their electronic con-figurations particularly the outershell configurations.
  • The atoms of the elements in a group posses similar electronic configurations.
  • The elements in a group should have similar chemical properties and there should be regular gradation in their physical properties from top to bottom.
  • Across the table from left to right in any period, elements gets an increase in the atomic number by 1 unit between any two successive elements.
  • Therefore the electronic configuration of valence shell of any two elements in a given period is not same.
  • Due to this reason elements along a period posses different chemical properties with regular gradation in their physical properties from left to right.

Question 6.
Explain any four factors which influence the electron affinity (Electron Gain Enthalpy). (TS March 2017)
Answer:
Factors effecting of electron affinity
1. Nuclear Charge :
If nuclear charge increases electron affinity increases, similarly it decreases if nuclear charge decreases.

2. Screening effect:
If screening effect value increases electron affinity increases, if it decreases electron affinity decreases.

3. Penetration power of the orbitals :
If penetration power of the orbitals increases electron affinity increases. If it decreases electron affinity decreases.

4. Stable configuration :
If an atom has stable electron configuration electron affinity will decreases.

5. Atomic radius :
If atomic radius increases electronic affinity will be increases. If atomic radius decreases electron affinity decreases.

6. Metallic property :
If metallic property increases electron affinity decreases.

7. Non-Metallic property :
Non-Metallic property increases electron affinity value increases.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 7.
Observe the information and answer the following questions. (TS June 2018)

Name of the Element Atomic Number Electronic Configuration
Sodium 11 [Ne] 3s1
Magnesium 12 [Ne] 3s2
Potassium 19 [Ar] 4s1
Calcium 20 [Ar] 4s2

1) What is valency of Magnesium?
Answer:
Valency of magnesium is two.

2) Which element has more electro-positivity?
Answer:
Potassium (K) has more electro-positivity.

3) Write the elements which belongs to (third) 3rd Period.
Answer:
The elements which belongs to 3rd period are Sodium (Na), Magnesium (Mg).

4) Write the elements which belongs to 1st Group.
Answer:
Sodium (Na), Potassium (K) belong to 1st Group.

Question 8.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 10
Answer the following from the above in brmation. (TS March 2018)
i) Which element posses the higher atomic radius in the above table?
Answer:
The element having higher atomic radius is ‘K’ (Potassium)

ii) Mention two plair of element which forms ionic bond.
Answer:
Na, Cl Mg, CL

iii) Name the two elements having valency 2.
Answer:
Elements having valency 2 are Be, Mg, Ca, 0, S, Se.

iv) Which element has electronic configuration of 1s² 2s² 2p4.
Answer:
Oxygen.

Question 9.
Explain the significance of three quantum numbers in predicting the position of an electron in an atom. (AP SCERT: 2019-20)
Answer:
Each electron in an atom is described by a set of three quantum numbers n, 1 and ml.
1. Principal quantum number (n):
The principal quantum number is used to describe the size and energy of the main shell. It is denoted by ‘n’. ‘n’ has positive integer values of 1, 2, 3, It is used to know the number of orbitals (n²) and electrons in an orbit. (2n²).
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 11

As ‘n’ increases the shells becomes larger and the electrons in those shells are farther from the nucleus and their energies increases.

2. The angular – momentum quantum number (l) :
‘l’ has integer values from O’ to n – 1, for each value of ‘n’. Each ‘l’ value represents one sub-shell. It is used to describe the shape of an orbit.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 12

3. The magnetic quantum number (ml) :
The magnetic quantum number (ml) has integer values between -l and +l including zero.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 13
If l = 0, the possible ml value is 1.
l = 1, the possible ml value is -1, 0 and 1.
Thus for a certain value of 1, there are (2l + 1) integer values of ml.

These values describe the orientation of the orbital in space relative to the other orbitals in the atom.

Ex: When l = 1, (2l + 1) = 3, that means ml has 3 values namely -1, 0, 1 or three p orbitals, with different orientations along x, y, z axes, labelled as px, py and pz orbitals.

Predicting the position of an electron in an atom :
If the values of n, l, and ml are 2, 1,-1 respectively the electron is present in 2px orbital in L – shell.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 10.
Answer the following question based on the values of the atomic radii of the elements of one of the periods in modern periodic table (AP SCERT: 2019-20)
Li (152), Be (111), B (88), C (77), N (74), O (66) and F (64)
a) What is the trend of atomic radii of given elements?
b) In the numerical listing of periods in the modern periodic table, what number was given to above elements?
c) Mention the unit of atomic radius.
d) Why the values of atomic radius varied along the period?
Answer:
a) Atomic radii of elements decrease while going left to right in the periodic table.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 14
b) Period – 2
c) Unit of atomic radius is ‘pm’ (picometer).
d) There should be no change in distance between nucleus and outer most shell for the elements in one period.

But, nuclear charge increases because of the increase in the atomic number of elements in a period.

Hence, the nuclear attraction on the outer shell electrons increases.

As a result the size of the atoms decreases while going left to right in a period.

Question 10.
Mendeleeff classified the then known 63 elements in the form of a periodic table. Mention any two things that benefitted study of chemistry, to support the above statement.
Answer:

  • Mendeleeff accepted minor inversions in the order of increasing atomic weights as these inversions resulted in elements being placed in the correct group.
  • It was the extraordinary thinking of Mendeleeff that made the chemists to accept the periodic table and recognise Mendeleeff more than anyone else as the originator of the periodic law.
  • At the time when Mendeleeff introduced his periodic table even electrons were not discovered.
  • Even then the periodic table was prepared to provide a scientific base for the study of chemistry of elements.
  • In his honour the 101 element was named “Mendelevium”.

Question 11.
How do these properties vary in period and group?
1) Valency
2) Atomic radius
3) Ionisation energy
4) Electron affinity
5) Electronegativity
6) Electropositivity
7) Metallic nature
8) Non-Metallic nature.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 15

Question 12.
Explain the salient features and achievements of the Mendeleeffs periodic table.
Answer:
Mendeleeffs periodic table is based on atomic weight.
1) Periodic law:
The physical and chemical properties of the elements are the periodic functions of their atomic weights.

2) Groups and sub-groups :
The vertical columns in Mendeleeffs periodic table are called groups. There are eight groups and elements in each group have similar properties. Each group is divided into sub-groups A and B.

3) Periods :
The horizontal rows are called periods. There are ‘seven’ periods in Mendeleeffs periodic table.

4) Predicting the properties of missing elements :
Based on the arrangement of elements in table, Mendeleeff predicted that some elements were missing and left blank spaces at appropriate places in the table. Later they were discovered.

5) Correction of atomic weight :
It is useful in correcting atomic weights of elements.

6) Anomalous series :
More atomic weight element like Tellurium (Ti) is placed before the less atomic weight element like Iodine in order to place these elements in the correct group.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 13.
How does atomic radius vary la period and group? Explain.
Answer:
Period :

  1. As we move from left to right the atomic radius decreases because the electrons enter the same main shell.
  2. The nuclear charge increases because of increase in atomic number of elements in period.
  3. Hence, the nuclear attraction on the outer shell electron increases. As a result, the size of atom decreases.

Group:

  1. Atomic radius increases from top to bottom in a group of the periodic table.
  2. As we go down in a group, the atomic number of element increases. In order to accommodate more number of electrons, there are more additional shells.
  3. As a result, the distance between the nucleus and the outer shell of atom increases.
  4. So atomic size increases.

Question 14.
What is electron affinity? What are the factors which influence electron affinity?
Answer:
Electron affinity :

  1. The electron affinity of an element is defined as the energy liberated when an electron is added to its neutral gaseous atom.
  2. Electron affinity of an element is also called electron gain enthalpy of that element.
  3. M(g) + e → M(g) + EA1 (M = Atom of element, EA1 = First Electron affinity)
    M(g) + e → M-2(g) + EA2 (EA2 = Second Electron affinity)

Factors influencing Electron affinity :
1) Nuclear charge :
Greater the nuclear charge, greater the electron affinity value because of greater attraction for incoming electron.

2) Atomic size :
As the atomic size increases, the attractive force of the nucleus on the electron decreases. So electron affinity decreases.

3) Electronic configuration :
The elements having stable electronic configurations of half filled or completely filled valence sub-shells show very small tendency to accept additional electron. So the electron affinity is low or almost zero for these elements.

4) Penetrating power of the orbitals :
As the penetrating power of the orbitals increases, the electron affinity increases.

5) Screening effect or shielding effect:
More the screening effect of orbitals, less is the electron affinity value.

Question 15.
How did Mendeleeff correct atomic weights of various elements?
Answer:

  • Atomic weight = Equivalent weight x Valency
  • By using the formula, the atomic weight of Beryllium was calculated as 13.5 (Equivalent weight of Be = 4.5, valency = 3)
  • With this atomic weight the element should be placed in wrong group.
  • So Mendeleeff predicted its valency is only 2. From that he calculated the atomic weight of Beryllium as 9.
  • Now it fitted into correct group.
  • Similarly, Mendeleeff corrected atomic weights of Indium and Gold.

Question 16.
Answer the following questions if atomic number of element is 15.
1) What is the name of the element?
2) What is the electronic configuration of element?
3) Which period and group does it belong to?
4) How many valence electrons are there in the element?
5) What is the valency of the element?
Answer:

  1. The element is phosporous,
  2. The electronic configuration of element is 1s² 2s² 2p6 3s² 3p³ or 2, 8, 5.
  3. It belongs to 3rd period (orbit number is 3) and V or 15 group (Number of electrons in valence orbit is 5.)
  4. Number of valence electrons are 5.
  5. Its valency is 8 – 5 = 3.

Question 17.
If an element belongs to 3rd period and 17th group, then answer the following questions.
1) What is its electronic configuration?
2) How many valence electrons are there in the element?
3) What is the valency of element?
4) What is atomic number of element?
5) What is the name of the element?
6) Give two more elements which have similar properties as this element?
Answer:

  • The element belongs to 3rd period and 17th group. So the valence orbit is 3rd and number of valence electrons in that orbit is 7. So its electron configuration is 2, 8, 7.
  • The number of valence electrons are 7.
  • The valency of element = 8 – 7 = 1.
  • The atomic number of element is 17.
  • Name of the element is chlorine.
  • Chlorine belongs to Halogen family. So Fluorine, Bromine, Iodine, and Astatine have similar properties as chlorine.

Question 18.
The elements of a periodic table are given below in the order from left to right.
Li Be B C O F Ne
1) To which period do these elements belong?
2) One element of this period is missing. Which is the missing element and where should it be placed?
3) Which one of the elements in this period shows the property of catenation?
4) Place the three elements fluorine, beryllium, and oxygen in the order of increasing electronegativity.
5) Which one of the above elements belongs to halogen series?
Answer:

  1. The elements belong to 2nd period.
  2. The element which is missing is Nitrogen which is placed in between carbon and oxygen.
  3. Carbon shows the property of catenation.
  4. The ascending order of electronegativity for these element is Beryllium < Oxygen < Fluorine.
  5. Fluorine belongs to halogen family.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 19.
A group of elements in periodic table is given below.
Boron, Aluminium, Gallium, Indium, and Thallium.
(Boron is the first element and Thallium is the last element)
Answer the following questions in relation to the above group of elements.
1) Which element has the most metallic character?
2) Which element would be expected to have the highest electronegativity?
3) If the electronic configuration of Aluminium is 2, 8, 3, how many electrons are there in outer shell of thallium?
4) The atomic number of Boron is 5. Write the chemical formula of the compound formed when Boron reacts with Chlorine.
5) Do the elements in the group to the right of this Boron group have more metallic or less metallic character? Justify your answer.
Answer:
1) Thallium has the most metallic character because as we move from top to bottom in a group the metallic character increases.

2) Boron has the highest electronegativity because as we move from top to bottom in a group electronegativity decreases.

3) Thallium is in the same group as Boron. So, the number of electrons in outermost shell of Thallium is 3.

4) The atomic number of Boron is 5. So, its electronic configuration is 2, 3. Therefore its valency is 3.

Whereas the atomic number of Chlorine is 17. So, its electronic configuration is 2, 8, 7. Therefore its valency is 1.

The formula of compound formed between Boron and Chlorine is BCl3.

5) The elements in the group right to Boron group have lesser metallic character because as we move from left to right in a period metallic character decreases.

Question 20.
The following questions refer to the periodic table.
1) Name the first and the last element in period 2.
2) What happens to the atomic size of elements moving from top to bottom of a group?
3) Which of the elements has the highest electron affinity among the halogens?
4) What is common feature of the electronic configurations of the elements in group 16?
Answer:
1) The first and the last elements of 2nd period or Lithium and Neon.

2) The atomic size decreases as we move from top to bottom in a group because there is an addition of shell each time as we move down the group.

3) Chlorine has the highest electron affinity. We know as move from top to bottom the electron affinity values decrease. But due to small size of fluorine there would be more electron-electron repulsions, if we add electron. So Chlorine has more electron affinity.

4) All these have same general outermost electronic configuration that is ns² np4.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 21.
Answer the following.
1) Elements of which groups have low ionization energy?
2) What is your guess about atomic size of an element with seven electrons among all the elements in the same period?
3) Which element has the highest electronegativity? Why?
4) Which element has the highest electropositivity? Why?
Answer:
1) Group IA, IIA elements have lower ionization energy values because they have metallic character.

2) As we move from left to right in a period atomic size decreases. So element with seven outermost electrons has least size among all the elements in the same period.

3) Fluorine has the highest electronegativity because when we move from left to right in a period atomic size decreases and electronegativity values increase. So Fluorine has the highest electronegativity.

4) Cesium has the highest electropositivity or positive character because when we move from top to bottom in a group atomic size decreases. So electropositive character increases. Therefore Cesium has the highest electropositive character.

Question 22.
Given below is the electronic configuration of A, B, C, D.

A) 1s2 2s2 2p¹ a) Which are the elements coming within the same period?
B) 1s2 2s2 2p6 b) Which are the elements coming within the same group?
C) 1s2 2s2 2p6 3s2 3p6 c) Which are the noble gas elements?
D) 1s2 d) Which group and period does the element C belong to?

Answer:
a) A and B belong to same period because the valence electrons of both the elements lie in the same orbit.
b) Elements A and C and elements B and D.
c) B and D are noble gas elements.
d) C belongs to 3rd period (orbit number) and III group (Number of valence electrons).

Question 23.
Write down the characteristics of the element having atomic number 16.
i) Electronic configuration
ii) Period number
iii) Group number
iv) Element family
v) Number of valence electrons
vi) Valency
vii) Metal or non-metal
viii) Name of the element
Answer:
i) Electronic configuration of element is 1s² 2s² 2p6 3s² 3p4 or 2, 8, 6.
ii) Period number is 3 because valence electron lies in 3rd orbit.
iii) Group number is 6 because the number of valence electrons are 6.
iv) Element belongs to chalcogen family.
v) Number of valence electrons are 6.
vi) Valency = 8 – 6 = 2.
vii) It is a non-metal because in a period when we move from left to right non-metallic character increases.
viii) Name of the element is sulphur.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 24.
The second period element ‘F has electron gain enthalpy than the third period elements of same group ‘Cl’. Why?
Answer:

  • In a group of elements, the electron gain enthalpy decreases from top to bottom.
  • But in general the second element in a group, i.e. 3rd period element has greater electron gain enthalpy than the first element, i.e. 2nd period element.
    Ex : E.A of F < E.A of Cl.
  • This is because Fluorine atom is smaller in size than Chlorine atom.
  • F2 also has strong inter electronic repulsions.
  • In the addition of an electron to fluorine atom, the electronic repulsions overcome at the expense of a part of the energy liberated.

Hence the overall energy liberated is less than that of Chlorine atom.

Question 25.
Differentiate the metals and non-metals.
Answer:

Metals Non-Metals
1. Metals have lustrous surface. 1. Non-metals do not have lustrous surface.
2. They show malleability. 2. They do not show malleability.
3. They show ductility. 3. They do not show ductility.
4. They produce sonorous sound. 4. They do not produce sonorous sound.
5. Generally they are hard. 5. Generally they are soft.
6. They are good conductors of electricity. 6. They are bad conductors of electricity.
7. Generally they liberate hydrogen gas when they are treated with acids. 7. They do not liberate hydrogen gas.

Question 26.
The electronic configuration of atom A is 2, 8, 6.
a) What is the atomic number of element A?
b) State whether the atomic size of element A is bigger or smaller than the atom having atomic number 14. Why?
c) Which of the elements exhibits similarity in chemical properties as element A 0(8), C(6), N(7), AV(18). Why?
d) How does the element form inert gas configuration?
Answer:
The electronic configuration of atom – A is 2, 8, 6.
a) Atomic number of element ‘A’ is 16, i.e. Sulphur.

b) The atom which has atomic number – 14 is Silicon (Si).

Atomic size of element decreases across period from left to right. So the atomic size of element ‘A’ is smaller than the atom having atomic number 14.

c) Element oxygen O8 – exhibits similarity in chemical properties as element A, because they belong to the same group.

d) Given element – A becomes inert gas, i.e. Argon configuration by gaining ‘2’ electrons.

AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table

Question 27.
Select the correct answers from the choices A, B, C, D which are given with reference to the variation of properties in the periodic table. Which of the following is generally true?
A : Atomic size increases from left to right across a period.
B : Ionisation energy increases from left to right across a period.
C : Electropositive character increases going down a group.
D : Electronegativity increases going down a group.
Answer:
1) A is wrong because when we move from left to right the atomic number increases. So, the nuclear attraction over outermost orbital increases. Therefore the atomic size decreases.

2) B is correct but it does not follow a regular trend in a period.

3) C is correct. As move from top to bottom in a group atomic size increases. Therefore it is easy to lose electrons. So electropositive character increases.

4) D is wrong because as we move from top to bottom in a group atomic size increases. So electronegativity decreases.

Question 28.
Some elements belonging to second period of periodic table, and their atomic radii are given below. Observe them and write answers.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 16
1) Write the elements in the ascending order of their atomic radii.
2) Which of the 2nd period elements closer to the configuration of inert gas?
3) Which is the outermost orbit of all these elements?
4) Which element’s atomic size is bigger, Beryllium or Carbon? Why?
Answer:

  1. The ascending order of atomic sizes is O, N, C, B, Be and Li.
  2. Lithium has closest inert gas configuration, i.e. 1s² 2s¹. Its nearest inert gas is Helium.
  3. The outermost orbit for all these elements is second orbit.
  4. Beryllium has more atomic size than Carbon. Because when we move across a period the atomic number increases. So nuclear attraction over outermost orbit increases and atomic size decreases. So carbon has lesser size than Beryllium.

Question 29.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 17
Refer the above part of periodic table and answer the following questions.
a) Element with the least atomic size.
b) Write the electronic configuration of the elements B and E.
c) Identify the elements that have similar physical and chemical properties as the element Y.
d) Arranged elements increasing order of their electronegativity values.
Answer:
a) The element with least atomic size is E. Because when we move from left to right in a period the atomic size decreases,

b) Electronic configuration of B is 1s² 2s² 2p6 3s² 3p¹ Because the element belongs to 13th group its general configuration is ns² np¹ and the element belongs to third period and its atomic number is 13. Similarly electronic configuration of E is 1s² 2s² 2p6 3s² 3p¹. Because the element belongs to 16th group. Its general configuration is ns2np4 and it is in third period. So its atomic number is 16.

c) The elements which have similar physical and chemical properties with Y are X and Z. Because they lie in a single group, i.e. 1st group. In a group, elements are having similar physical and chemical properties.
d) Z < Y < X < B < C < D < E.

Question 30.
Consider the section of the periodic table given below.
AP SSC 10th Class Chemistry Important Questions Chapter 9 Classification of Elements- The Periodic Table 18
1) Which is the most electronegative?
2) How many valence electrons are present in G?
3) Write the formula of the compound between B and H.
4) Which element has similar properties as J?
5) Which element has greater size-either D or E?
Answer:

  1. J is the most electronegative. In a period electronegative values increase.
  2. G is present in V group. So the number of valence electrons is 5.
  3. B is present in first group. So its valency is 1 and hydrogen also has valency 1. Therefore the compound is BH.
  4. K lies in same group as J. Elements belonging to same group have similar properties. So, K has similar properties as J.
  5. E has greater size because as we move from top to bottom the atomic size increases.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 8 Structure of Atom.

AP State Syllabus SSC 10th Class Chemistry Important Questions 8th Structure of Atom

10th Class Chemistry 8th Lesson Structure of Atom 1 Mark Important Questions and Answers

Question 1.
Write the electronic configuration of chromium. (AP June 2016)
Answrr:
The electronic configuration of chromium is
1s²2s²2p63s²3p64s¹3d5 or [Ar] 4s¹3d5

Question 2.
Out of 3d and 4s, which has more (n + l) value? Explain. (AP June 2017)
Answer:
1) 3d ⇒ n + l ⇒ 3 + 2 ⇒ 5 (energy)
4s ⇒ n + l ⇒ 4 + 0 ⇒ 4 (energy)

2) Hence, ‘3d’ has more (n + l) value than ‘4s’.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 3.
Prepare a question on nlx method. (AP SA-I:2018-19)
Answer:

  • How is nlx method useful.
  • Explain the nlx method with an example.

Question 4.
Which colours do you observe when an iron rod is gradually heated to higher tem-peratures? (TS June 2015)
Answer:
First iron turns into red (lower energy corresponding to higher wavelength) and as the temperature rises it glows and turns into orange, yellow, blue or even white respectively (higher energy and lower wavelength).

Question 5.
Which principle is not followed in writing the electronic configuration of 1s² 2s¹ 2p4? Give reasons. (TS June 2015)
Answer:
1) Principle :
Aufbau principle is not followed in writing the electronic configuration of 1s² 2s¹ 2p4.

2) Reasons :
i) According to Aufbau principle electron enters into orbital of lowest energy.
ii) Between 2s and 2p, 2s has least energy. So 2s must be filled before the electron has to enter 2p.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 6.
Write the symbol of the outermost shell of magnesium (Z = 12) atom. How many electrons are present in the outermost shell of magnesium? (TS June 2017)
Answer:
Symbol of the outermost shell of magnesium (3rd shell) = M
No. of electrons in outermost shell of Magnesium = 2.

Question 7.
The four quantum number values of the 21st electrons of scandium (Sc) are given in the following table. (TS March 2017)
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 1
Write the values of the four quantum numbers for the 20th electron of scandium (Sc) in the form of the table.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 2

Question 8.
If n = 3, mention the orbitals present in the shell and write maximum number of electrons in the shell. (TS March 2018)
(OR)
Write the maximum number of electrons and number of orbitals in the shell, when n = 3.
Answer:

  1. When n = 3, number of subshells = 3 (3s1 3p1 3d)
  2. Number of orbitals = 9 (3s(1) 3p(3) 3d(5))
  3. Maximum number of electrons (3s² 3p6 3d10) = 18

Question 9.
What is dispersion?
Answer:
The splitting of light into different colours is called dispersion.

Question 10.
What is an electromagnetic wave?
Answer:
When electric field and magnetic fields are perpendicular to each other and at right angles to the direction of propagation of wave is formed. Such a wave is called electromagnetic wave.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 11.
What is a Zeeman effect?
Answer:
The splitting of spectral lines in the presence of magnetic field is called Zeeman effect.

Question 12.
What is a spectrum?
Answer:
Group of wavelengths is called spectrum (or) A collection of dispersed light giving its wavelength composition is called a spectrum.

Question 13.
What is speed of electromagnetic wave?
Answer:
It is equal to speed of light, i.e. 3 x 108 ms-1

Question 14.
Which colour has highest wavelength and which colour has least wavelength on visible spectrum?
Answer:
The colour that has highest wavelength in visible spectrum is red and least wavelength is violet.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 15.
If n = 5, then what is the maximum value for l?
Answer:
The maximum value for l is 4.

Question 16.
If l = 4, what is the number erf values for ml?
Answer:
ml = 2l + 1 = 2(4) +1=9.

Question 17.
What are the values of ms?
Answer:
½ or – ½

Question 18.
What is electronic configuration?
Answer:
Distribution of electrons in shells, sub-shells and orbital in an atom is known as electronic configuration.

Question 19.
What is Heisenberg’s principle of uncertainty?
Answer:
It is not possible to find the exact position and velocity of electron simultaneously.

Question 20.
Give ascending order of various atomic orbitals according to Moeller diagram.
Answer:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4d < 5d < 6p < 7s < 5f < 6d < 7p < 8s.

Question 21.
What is Hund’s Rule?
Answer:
Electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 22.
What is Planck’s equation?
Answer:
Planck’s equation is E = hv.
E = Energy of the radiation
h = Planck’s constant = 6.625 × 10-34 J
v = Frequency of radiation.

Question 23.
What is electromagnetic spectrum?
Answer:
Electromagnetic waves can have a wide variety of wavelengths. The entire range of wavelengths is known as the electromagnetic spectrum.

Question 24.
What is wavelength?
Answer:
The distance from one wave peak to the next is called wavelength (λ).

Question 25.
What is frequency?
Answer:
The number of wave peaks that pass by a given point per unit time is called frequency.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 26.
When are electromagnetic waves produced?
AnElectromagnetic waves are produced when an electric charge vibrates.

Question 27.
Which is the example for line spectrum?
Answer:
The atomic spectrum of hydrogen atom.

Question 28.
Which model explains fine spectrum of atom?
Answer:
Bohr – Sommerfeld model.

Question 29.
How are wavelength and velocity of light related?
Answer:
c = vλ
where
c = velocity of light,
v = frequency of light,
λ = wavelength of light.

Question 30.
Give the equation which gives electromagnetic energy (light) that can have only Certain discrete energy values.
Answer:
E = hv
E = Energy of light
h = Planck’s constant = 6.62 5 × 10-27 erg sec or 6.625 × 10-34 Joule-sec
v = Frequency of radiation
This equation is called Planck’s equation.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 31.
Which group elements are called Noble gases?
Answer:
VIII A group or 18th group elements are called inert gases (or) Noble gases.

Question 32.
Which elements are highly stable?
Answer:
Noble gases are highly stable.

Question 33.
Write the set of quantum numbers for the electrons in a 3pz orbital.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 3

Question 34.
What is the difference between an orbit and orbital?
Answer:
An orbit is a well defined path of electron that revolves around the nucleus.

An orbital is the space around the nucleus, where the probability of finding electrons is maximum.

Question 35.
What are the factors which influence electromagnetic energy?
Answer:
Electromagnetic energy depends on two factors

  1. wavelength
  2. frequency.

Question 36.
What is a wave?
Answer:
The disturbance occurred in a medium is called wave.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 37.
When cupric chloride is kept in non-luminous flame then what is the colour of flame?
Answer:
Green colour.

Question 38.
If the colours gradually changes there are no sharp boundaries in between them, then what is the name given to that type of spectrum?
Answer:
Continuous spectrum of emission.

Question 39.
What is the information given by magnetic orbital quantum numbers?
Answer:
Orientation of orbitals in space.

Question 40.
How many orbitals are present in a sub-shell?
Answer:
The number of orbitals are present in a sub-shell is n² (where n is principal quantum number).

Question 41.
What happens when an object is suitably excited by heating?
Answer:
Light is emitted by the object.

Question 42.
What is meant by Aufbau?
Answer:
The German word Aufbau means building up.

Question 43.
Which elements are examples for Noble gases?
Answer:
Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) and Radon (Rn) are examples for Noble gases.

Question 44.
Which element has duplet configuration?
Answer:
Helium (1s²)

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 45.
Write the set of quantum number for the added electron of oxygen atom.
Answer:
Configuration of oxygen is 1s² 2s² 2p4.
The added electron is 4th in the 2p.
The set of quantum numbers (2, 1, -1, -½)

Question 46.
Can we apply c = vλ, to sound waves?
Answer:
Yes. It is a universal relationship and applies to all waves.

Question 47.
What is the value of Planck’s constant?
Answer:
The value of Planck’s constant is 6.626 × 10-34 Js.

Question 48.
Why do different elements emit different flame colours when heated by the same non-luminous flame?
Answer:’
The light emitted by different kind of atoms is different because the excited states electrons will go are different.

Question 49.
How many values can ‘l’ have for n = 4?
Answer:
If n = 4, l can take values 0, 1, 2, 3. So there are four values.

Question 50.
Write the four quantum numbers for the differentiating electrons of lithium (Li) atom.
Answer:
The electronic configuration of lithium is 1s² 2s¹. So differentiating electron enters into 2s. The values of four quantum numbers are as given below.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 4

Question 51.
Write four quantum numbers for 2p¹ electrons.
Answer:
The four quantum numbers for 2p are
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 5

Question 52.
Which rule is violated in the following electronic configuration?
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 6
Answer:
The rule violated is Hund’s rule.

Question 53.
How many maximum number of electrons that can be accommodated in N principle energy shell?
Answer:
For N shell n = 4.
The maximum number of electrons accommodated in a shell is 2n².
∴ A maximum of 32 electrons can be filled in N shell.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 54.
How many maximum number of electrons that can be accommodated in a ‘l’ sub-shell?
Answer:
l sub-shell has 3 orbitals. Each orbital accommodates 2 electrons. So 6 electrons can be filled in l sub-shell.

Question 55.
How many maximum number of electrons can be accommodated in ‘d’ sub-shell?
Answer:
d sub-shell has 5 orbitals. So 10 electrons can be filled in d sub-shell.

Question 56.
How many sub-shells present in a ‘M’ principal energy shell?
Answer:
For M shell n = 3
The number of sub-shells in M shell is 3.

Question 57.
How many spin orientations are possible for an electron in s-orbital?
Answer:
The spin quantum number values for electrons are ½ or -½. So 2 spin orientations are possible.

Question 58.
Write valence electronic configuration of element which has the following set of quantum numbers.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 7
n = 3 indicates 3rd orbit and l = 1 indicates p orbital and there is one electron in p orbital. So the valence electron configuration is 3p¹.

Question 59.
How many unpaired electrons are present in chromium?
Answer:
The electronic configuration of chromium 1s² 2s² 2p6 3s² 3p6 4s¹ 3d5.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 8
The number of unpaired electrons = 6.

Question 60.
Find the four quantum number values of 3rd and 4th electrons of Beryllium.
The electronic configuration of Beryllium is 1s² 2s².
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 9

Question 61.
What is the n + l value of 4f orbital?
Answer:
For 4f orbital n = 4 and f orbital l = 3.
∴ n + l = 4 + 3 = 7

Question 62.
When you heat iron rod first it turns red. Why?
Answer:
Iron turns into red because red has higher wavelength. So it has lower energy which is emitted by iron.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 63.
What is the significance of Planck’s proposal?
Answer:
Electromagnetic energy can be gained or lost in discrete values and not in a continuous manner.

Question 64.
When do you see an emission line?
The energy emitted by an electron is seen in the form of an electromagnetic energy and when the wavelength is. in the visible region it is visible as an emission line.

Question 65.
How many elliptical orbits are there in 4th orbit of Sommerfeld?
Answer:
The number of elliptical orbits in 4th orbit of Sommerfeld is 3.

Question 66.
Why is spin quantum number introduced?
Answer:
When we observe spectrum of yellow light by using high resolution spectroscope it has very closely spaced doublet. Similar patterns are shown by Alkali and Alkaline earth metals. In order to account this spin quantum number is introduced.

Question 67.
Which of the following magnetic quantum number values is not possible for 3d orbital?
a) – 2
b) – 1
c) 0
d) 4
Answer:
For d orbital the possible m; values – 2, – 1, 0, 1, 2. So the value 4 is not possible.

Question 68.
If an element has 11 electrons in its M shell, then what is the name of element and its atomic number?
Answer:
The electronic configuration of element is 1s² 2s² 2p6 3s² 3p6 4s² 3d3.
[M shell electrons = 2 + 6 + 3=11]

So the element is Titanium.

Question 69.
The wave length of a wave is 100 nm. Find its frequency.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 10

10th Class Chemistry 8th Lesson Structure of Atom 2 Marks Important Questions and Answers

Question 1.
Explain Hund’s Rule with an example. (TS March 2016) (AP SA-1:2018-19)
Answer:
Hund’s Rule :
According to this rule electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

(1) Ex :

  1. The configuration of carbon atom (Z = 6) is 1s² 2s² 2p².
  2. The first four electrons go into the 1s and 2s orbitals.
  3. The next two electrons go into 2px and 2py orbitals.
    AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 11
  4. But, they do not pair in 2px orbital.

(2) Ex :

  1. In oxygen atom (8O), distribution of electrons is given below.
  2. AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 12
  3. Here, pairing of electrons in 2px orbital takes place after, 2px, 2py and 2pz orbitals are filled with a single electron.

Question 2.
The electronic configuration of Sodium is 1s² 2s² 2p6 3s¹. (AP March 2017)
What information that it gives?
Answer:

  1. Its atomic number is 11
  2. It is s-block element
  3. It is in 3rd period ’
  4. It is in 1st group
  5. Its valency is 1
  6. Number of valency electrons are 1
  7. It can form uni positive ion
  8. It can form ionic bond, 9. It is metal.

Question 3.
Explain the principle which describes the arrangement of electrons in degenerate orbitals.
Answer:
According to Hund’s rule the degenerate orbitals are occupied with one electron each before pairing of electron starts.
Ex : Electronic configuration of carbon is 1s² 2s² 2p²
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 13
the last two electrons will enter into separate 2p orbitals.

Question 4.
Name the principle, which says an orbital can hold only 2 electrons and explain. (AP March 2018)
Answer:

  • Name of the principle : Pauli’s exclusion principle.
  • No, two electrons in an orbital can have all four quantum numbers same.
  • It says there is a chance to hold only 2 electrons, one rotates in clockwise direction (+½)other rotates in anticlock wise direction (-½).

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 5.
For a better understanding about the electronic configuration in an atom, the teacher wrote shorthand notation nlx on the blackboard.
Looking at this notation, what could be the probable questions that generate in the student’s mind? Write any two of them. (TS March 2015)
Answer:

  1. What n, l, x indicates related to atoms?
  2. How nlx indicates the position of the electrons in the atom?

Question 6.
Write the ‘Octet Rule’. How does Mg (12) get stability while reacting with chlorine as per this rule? (TS June 2017)
Answer:
Octet Rule :
The atoms of elements tend to undergo chemical changes that help to leave their atoms with eight outer – shell electrons.

Magnesium atom looses 2 electrons and get 8 electrons in its outermost shell as Neon. So that it gets stability.

Question 7.
Write the electronic configuration of the atom of an element having atomic number 11. Write the names of the rules and the laws followed by you in writing this electronic configuration. (TS March 2017)
Answer:
1s²2s²2p63s¹.
(OR)
Principles followed :

  1. Aufbau principle.
  2. Hund’s Rule.
  3. Pauli Exclusion Principle

Question 8.
The electron enters into 4s orbital after filling 3p orbital but not into 3d. Explain the reason (TS March 2018)
Answer:
Based on (n+l) values energy value of 3d orbitals is 3 + 2 = 5, energy value of 4s orbitals is 4 + 0 = 4
The energy level of 4s orbital is less than the 3d orbital according to the Aufbau principle electron enters into lower energy orbital first.
Thats why electrons enters into 4s orbital after filling 3p, but not into 3d.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 9.
Write the electronic configuration of Na+ and Cl.
Answer:
Electronic Configuration of Na+ is 1s² 2s² 2p6 and Cl is 1s² 2s² 2p6 3s² 3p6.

Question 10.
Observe the given table and answer the following questions. (TS March 2019)

Sl.No. Electron Configuration
1. 1s2 2s2 2p6 3s2 3p3
2. 1s2 2s2 2p6 3s2 3p6 4s2
3. 1s2 2s2 2p6 3s2 3p6

1) Mention the divalent element name.
2) Name the element belongs to 3rd period and VA Group.
Answer:

  1. Name of the divalent element in the table is Calcium.
  2. Name of the element which belongs to 3rd period and VA Group is Phosphorous.

Question 11.
Your friend is unable to understand nlx. What questions will you ask him to understand nlx method? (AP SCERT: 2019-20)
Answer:

  1. What is nlx method?
  2. Where does it use ?
  3. What is meant by ‘n’, 7′ and ‘x’?
  4. How can we use nlx method in the writing of electronic configuration?

Question 12.
Why do valency electrons involve in bond formation, than electrons of inner shells? (AP SCERT: 2019-20)
Answer:

  1. When two atoms come sufficiently close together the valence electrons of each atom experience the attractive force of the nucleus in the other atom.
  2. The nucleus and the electrons in the inner shell remain unaffected when atoms come close together.
  3. The electrons in outer most shell of an atom get affected.
  4. Thus electrons in valence shell are responsible for the formation of bond between atoms.

Question 13.
Explain Pauli’s exclusion principle with an example. (AP SA-I:2019-20)
Answer:
Pauli’s exclusion principle :
No two electrons of the same atom can have all four quantum numbers the same. If n, l and ml are same for two electrons, then ms must be different.
Suppose take the example of Helium atom.
The four quantum numbers for two electrons in the Helium atom given below.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 14

We can observe from the table that three of the quantum numbers are same but fourth quantum number is different. The electronic configuration of Helium atom is² ↑↓. So the maximum number of electrons filled in an orbital is 2.

Question 14.
Explain Aufbau principle. (AP SA-I:2019-20)
Answer:
Aufbau principle :

  1. In the ground state the electronic configuration can be built up by placing electrons in the lowest available orbitals until the total number of electrons added is equal to atomic number.
  2. Thus orbitals are filled in the order of increasing energy.
  3. Electrons are assigned to orbitals in order of increasing value of n + l.
  4. For sub-shells with the same value of n + l, electrons are assigned first to the subshell with lower n.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 15.
The electronic configuration of an atom is as follows 1s² 2s² 2p².
a) Which element’s atom is it?
b) Which orbital is the last electron in?
c) When excited what could be the number of lone / single electrons in this atom?
d) What is the value of principal quantum numbers of two electrons in the first box?
Answer:
Given electronic configuration of atom is 1s² 2s² 2p².
a) The element is carbon.
b) The valence electron enters 2p orbital.
c) In excited state the electron in 2s orbital enters 2p orbital. So it has 4 unpaired electrons.
d) The value of principal quantum number is 1.

Question 16.
Draw the table which gives the information about the quantum numbers and the number of the quantum states.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 15

Question 17.
Explain briefly about spin quantum number.
Answer:

  • This gives spin of the electrons about their own axes. It is denoted by ms.
  • This quantum number refers to the two possible orientations of the spin of an electron, one clockwise and the other anti-clockwise spin.
  • These are represented by +½ -½ and .

Question 18.
Write electronic configurations of following elements,
a) Hydrogen
b) Helium
c) Lithium
d) Beryllium
e) Boron
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 16

Question 19.
What does a line spectrum tell us about the structure of an atom?
Answer:
The electrons in ground state i.e. lowest, energy state absorb energy and move into excited state where they are unable to stay for long periods so lose the energy and come back to the ground state. The emitted radiation appears as line in line spectrum.

Question 20.
What are the spins of electrons in Helium atom?
Answer:
The quantum numbers for two electrons of Helium are given below as per Pauli’s exclusive principle.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 17
Three of quantum numbers are same. So fourth must be different so the two electrons have anti-parallel spins.

Question 21.
1s² 2s² 2p6 3s² 3p6 3d10 4s¹ is the electronic configuration of Cu (Z = 29). Which rule is violated while writing this configuration? What might be the reason for writing this configuration?
Answer:
The rule violated is Aufbau principle. The elements which have half filled or completely filled orbitals have greater stability. So copper can get stability by transferring one electron from 4s to 3d (their energies are close to each other).

So the electronic configuration of copper is 1s² 2s² 2p6 3s² 3p6 3d10 4s¹, not 1s² 2s² 2p 3s² 3p6 3d9 4s².

Question 22.
Why are chromium and copper exceptions to electronic configuration?
Answer:
Elements which have half-filled or completely filled orbitals have greater stability. So in chromium and copper the electrons in 4s and 3d redistribute their energies to attain stability by acquiring half-filled and completely filled d orbitals.

Hence the actual electronic configurations of chromium and copper are as follows.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 18

Question 23.
Distinguish between emission and absorption spectrum.
Answer:

Emission spectrum Absorption spectrum
1) The spectrum produced by emitted radiation is called emission spectrum. 1) The spectrum produced by absorption of radiation is called absorption spectrum.
2) The emission spectrum contains bright lines on the dark back-ground. 2) The absorption spectrum contains dark lines on the bright back-ground.
3) The emission spectrum corresponds the radiation emitted when an excited electron comes back to the ground state. 3) The absorption spectrum corresponds the radiation absorbed in exciting an electron from lower to higher energy levels.

Question 24.
Draw a table which will give the relationship between l values and number of orbitals and name of sub-shell and maximum number of electrons.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 19

Question 25.
Distinguish between line and band spectrum.
Answer:

Line spectrum Band spectrum
1) The spectrum has sharp distinct lines. 1) The spectrum has many closed, spaced lines.
2) The spectrum is characteristic of atoms and is also called atomic spectrum. 2) The spectrum is characteristic of molecules is also called molecular spectrum.
3) The spectrum is given by inert gases, metal vapours and atomised non-metals. 3) The spectrum is given by hot metals and molecular non-metals.

Question 26.
What is Dispersion of light? Explain a natural example of dispersion of light.
Answer:
Dispersion:
Splitting of white light into colours (VIBGYOR) is called Dispersion of light.

The natural example for dispersion of light is formation of Rainbow. It is caused by dispersion of sunlight by tiny water droplets present in atmosphere which act as small prisms.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 27.
Wien is an electromagnetic wave produced? Write about characteristics of electromagnetic wave.
Answer:
Electromagnetic wave is produced when an electric charge vibrates (moves back and fortn).

Characteristics of electromagnetic waves :

  1. Efectric field and magnetic fields are perpendicular to each other and at right angles to direction of propagation of wave.
  2. It travels with speed of light i.e., 3 × 108 ms-1.
  3. Electromagnetic energy is characterized by wavelength (λ) and frequency (v). The relation is given by c = vλ.

Question 28.
The valence electron configuration of element is given as 4s¹. Then give the following information.
1) What is the name of that element?
2) What is the outermost orbit of element?
3) What is ‘l’ value of outermost sub-shell?
4) What is the atomic number of element?
Answer:

  1. Potassium.
  2. N
  3. The outermost sub-shell is 4s its l value is ‘O’.
  4. Its electronic configuration is 1s² 2s² 2p6 3s² 3p6 4s¹. So its atomic number is 19.

Question 29.
Write all the quantum numbers for valance electron of sodium.
Answer:

  1. The electronic configuration of sodium is 1s² 2s² 2p6 3s¹.
  2. The valance orbital is 3s.
  3. The quantum numbers for this orbital is
    AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 20

Question 30.
Give electronic configurations of following elements.
a) Sodium
b) Phosphorous
Answer:
Sodium – 1s² 2s² 2p6 3s¹
Phosphorous – 1s² 2s² 2p6 3s² 3p³

Question 31.
Why does nitrogen has more chemical stability when compared with oxygen?
Answer:
The electronic configurations of Nitrogen and Oxygen are as follows.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 21
Nitrogen has half filled 2p³ configuration. So it has greater chemical stability when compared with Oxygen.

Question 32.
Ramu gave electronic configuration of potassium as 1s² 2s² 2p6 3s² 3p6 3d¹ whereas Ravi expressed the configuration as 1s² 2s² 2p6 3s² 3p6 4s¹. Who gave the correct configuration? Why?
Answer:
Ravi gave the correct configuration because according to Aufbau principle after completion of 3p orbital electron may enter either 4s or 3d.
Their n + l values are given below.

Orbital n + l Values
4s 4 + 0 = 4
3d 3 + 2 = 5

So 4s orbital has lower n + l value when compared with 3d orbital. So electron enters into 4s.

Question 33.
Given the valence electron configuration of an element is 4s¹. Then what are its quantum number values. Which element does it represent?
Answer:
Given valence electronic configuration is 4s¹. So n = 4 for ‘s’ sub shell l = 0 and if l is ‘O’ then ml is also zero. ms takes only two values that is +½ or -½ for convenience we can take ms as +½.
∴ The quantum number values are like this
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 22
The element is potassium.

Question 34.
The atomic number of an element is 17, then calculate the total number of electrons present in its s and p orbitals.
Answer:
The element with atomic number 17 is chlorine.
Its electronic configuration is 1s² 2s² 2p6 3s² 3p5.
So, the total number of electrons present in s orbitals = 2 + 2 + 2 = 6.
The total number of electrons present in p orbitals = 6 + 5 = 11.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 35.
Based on Aufbau’s principle, in which of the three 4d, 5p and 5s orbitals the electrons will be filled first? Why?
Answer:
According to Aufbau’s principle, the electron enters the orbital having lower n + l value. If both orbitals have same n + l values, electron enters the orbital with lower ‘n’ value.

Oprbital n + l value
4d 4 + 3 = 7
5p 5 + 1 = 6
5s 5 + 0 = 5

So, 3s has least n + l value. Therefore the electron enters 5s orbital first.

Question 36.
Find the following.
1) Number of orbitals present in M orbit.
Answer:
Number of orbitals present in an orbit = n² For M orbit n = 3.
∴ So the number of orbitals = 3² = 9.

2) The maximum and minimum possible ml values for 4f orbital.
Answer:
For f orbital ‘l’ value is 3. If l is 3, then ml takes values from – 3 to + 3.
So the minimum value for ml is – 3 and maximum value is + 3.

3) The possible values of l if n = 4.
Answer:
If n = 4, then l take values from 0 to 3 i.e., 0, 1, 2, 3.

4) The maximum number of electrons that can be filled in ‘N’ energy level.
Answer:
For N orbit n = 4. The maximum number of electrons present in a shell = n².
∴ The maximum number of electrons filled in N shell = 4² = 16.

Question 37.
How do the vibrating electric and magnetic fields around the charge become a wave that travel through space?
Answer:

  • A vibrating electric charge creates a change in the electric field.
  • The changing electric field creates a changing magnetic field.
  • This process continues, with both the created fields being perpendicular to each other and at right angles to the direction of propagation of the wave.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 38.
What information do the quantum numbers provide?
Answer:
The quantum numbers describe the space around the nuclear where the electrons are found and also their energies.

Question 39.
An electron in an atom has the following set of four quantum numbers. Which orbital does it belong to?

n l ml ms
3 0 0

Answer:
The quantum numbers of an atom is given below.

n l ml ms
3 0 0

By using nlx method n = 3, if l = 0 then the sub-shell is s.
So the electron belongs to 3s.

Question 40.
Look at the following table and answer the following questions.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 23
a) What is the law that is voilated in the above table? State that law.
b) Write the correct table using that law.
Answer:
a) The law violated is Pauli’s Exclusive Principle. Pauli’s law states that no two electrons of the same atom can have all the four quantum numbers same.
b)
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 24

10th Class Chemistry 8th Lesson Structure of Atom 4 Marks Important Questions and Answers

Question 1.
Draw Moeller chart of filling order of atomic orbitals.
(OR)
Draw a diagram showing the increasing value of (n + l) of orbitals. (AP June 2017)
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 25
The filling order of atomic orbitals (Moeller Chart)

Question 2.
Complete the following table based on quantum numbers related to atomic orbitals and electron of an atom. (AP SA-I-2018-19)
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 26
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 27

Question 3.
Based on the information given in the table, answer the questions given below.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 28
i) For the 4th main shell, how many values are there for ml? What are they? (TS June 2016)
Answer:
1) There are 16 m( values in the 4th main shell.
2) They are

Orbital ml values Total
4s 0 1
4p -1, 0, + 1 3
4d -2, -1, 0, + 1, +2 5
4f -3, -2, -1, 0, +1, +2, +3 7
Total 16

ii) For sub-shell with n = 3, l = 1, write the m, values.
Answer:
For sub – shell with n = 3, l = 1 the m, values are -1, 0, 1.

iii) Write the principal quantum number value for ‘N’ shell. How many sub-shells are there in the main shell?
Answer:

  1. The principal quantum number value for ‘N’ shell is 4.
  2. The number of sub-shells is 4. They are 4s, 4p, 4d, 4f.

iv) In the above table ml and l values are given. Write a formula that gives the relationship between ml and l based on those values.
Answer:
ml (No. of values) = 2l + 1.

Question 4.
Observe the information provided in the table about quantum numbers. Then answer the questions given below it. (TS June 2017)

n l ml
1 0 0
2 0 0
1 -1, 0, +1
3 0 0
1 -1, 0, +1
2 -2, -1, 0, +1, +2

i) Write the ‘l’ value and symbol of the spherical shaped sub-shell.
ii) How many values that ‘ml‘ takes for 1 = 2? What are they?
iii) Write the symbols of the orbitals for l = 1 sub-shell.
iv) What is the shape of the sub-shell for l = 2? What is the maximum number of electrons that can occupy this sub-shell?
Answer:
i) Spherical shaped sub-shell “l” value is zero and symbol is ‘s’.
ii) Number of m; values for 1 = 2 is 5, those are -2, -1, 0, 1, 2.
iii) Symbols of the orbitals for l = 1 sub-shell are px, py, pz.
iv) Shape of the subshell l = 2 is double dumbel.
The maximum number of electrons that can occupy in this sub-shell is 10.

Question 5.
Write postulates and limitations of Bohr Hydrogen atomic model. (TS June 2018)
Answer:
Postulates :

  1. Niels Bohr proposed that electrons in an atom occupy ‘stationary’ orbits of fixed energy at different distances from the nucleus.
  2. When an electron jumps from a lower energy state (ground state) to higher energy state it absorbs energy or emits energy when such a jump occurs from a higher energy state.
  3. The energies of an electron in an atom can have only certain values E1, E2, E5…. that is, the energy is quantized.

Limitations :

  1. Bohr’s model failed to account for splitting of line spectra of hydrogen atom into liner lines.
  2. Bohr theory could not explain the Zeeman and Stark effect.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 6.
Explain Bohr’s model of hydrogen atom and its limitations. (March 2019)
Answer:
Niels Bohr proposed that,
a) electrons in an atom occupy stationary orbits of fixed energy (K, L, M, N,…) at different distances from the nucleus.

b) when an electron jumps from a lower energy state to higher energy state, it ab¬sorbs energy or emits energy when such a jump occurs from a higher energy state to lower energy state.

c) the energies of an electron in an atom can have only certain values E1, E2, E3, ……… i.e. the energy is quantized. The states corresponding to these energies are called stationary states and the possible values of the energy are called energy levels.

d) the angular momentum of electron is multiple integral of \(\frac{L}{2 \pi}\).
∴ L = mvr = \(\frac{\mathrm{nh}}{2 \pi}\)
m = mass of electron ;
v = velocity of electron ;
r = radius of circular path;
h = plank constant

Limitations :

  1. Bohr’s model failed to account for splitting of line spectra (Zeeman effect).
  2. This model failed to account for the atomic spectra of atoms of more than one electron.
  3. Bohr theory was not explained the quantisation of angular momentum of an electron.
  4. It was not explained the formation of chemical bonds.

Question 7.
Explain four quantum numbers with an example. (AP SA-I; 2019-20)
Answer:
Each electron iii an atom is described by a set of three numbers called Quantum numbers.

1) Principal quantum number (n) :
It is used to know the size and energy of the main shell The values of ‘n’ are 1, 2, 3 …..
energy of the shell, n = 1 < energy of the Shell n = 2.

2) Angular momentum of quantum number (l) :
It is used to know the shape of a particular sub shell.
The values of ‘l’ are 0, 1, 2, 3
l = 0 = s orbital = spherical in shape
l = 1 = p orbital = dumbel in shape
l = 2 = d orbital = double dumbel Shape

3) i) Magnetic quantum number (ml) :
It is used to describe the orientation of the orbital in space relative to the other orbitals in the atom.
The values of mt are – ‘l’ to ’+l’ including zero,

ii) Spin quantum number (ms) :
It is used to know the orientation of the spins of electrons.
The values of ms are +½ and – ½.

Question 8.
How does Hund’s rule helps in writing electronic configuration of an atom? Explain with a suitable example. (TS June 2019)
Answer:
Hund’s rule :
According to this rule electron pairing in orbitals starts only when all available empty orbitals of the same energy are singly occupied.

This Hund’s rule helps in writing of electronic configuration of an element.

Example :

  1. The electronic configuration of carbon (C) atom (Z = 6) is 1s² 2s² 2p6.
  2. The first four electrons go into the Is and 2s orbitals.
  3. The next two electrons go into separate 2p orbitals, with both electrons having the same spin.
    AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 29

Another example :

  1. The electronic configuration of oxygen (8O) is 1s² 2s² 2p4.
  2. The first four electrons go into the 1s, 2s orbitals.
  3. The next four electrons go into 2p orbits as 2 in 2px, 1 in 2py and 1 in 2pz orbital.
    AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 30
  4. Here pairing of electron in 2px starts after filling of electron in each 2px, 2py, 2pz orbitals.
  5. But electrons do not occupy like this
    AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 31

Question 9.
Your father asked you to go to the market and purchase an electric lamp. The shop-keeper displayed two lamps – one is violet and another is red. Which coloured lamp do you purchase to put in your bedroom? Support your choice of solution.
Answer:
Red has highest wavelength and violet has lowest wavelength. We know that relationship between energy as follows.
E = hv = h\(\frac{c}{\lambda}\)

h and c are remains constant.
∴ E ∝ \(\frac{1}{\lambda}\)

Energy is inversely proportional to wavelength. Red coloured light has lower energy and least intensity. So red coloured lamp is preferable as bed light.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 10.
Heisenberg contradicts Neils Bohr. Explain in what way he contradicts.
Answer:

  • According to Bohr, electrons revolve around nucleus in definite paths or orbits. So the exact position of the electron at various times will be known to us.
  • In order to explain Bohr’s postulate we have to know the velocity and exact position of electron.
  • In order to find the position of electron we have to take the help of suitable light to find the position. As the electrons are very small, light of very short wavelength is required for this task.
  • This short wavelength light interacts with the electron and disturbs the motion of the electron.
  • Hence, it is not possible to find the exact position and velocity of electron simultaneously. This was stated by Heisenberg which is called Heisenberg’s principle of uncertainty.
  • In this way Heisenberg contradicts Neils Bohr.

Question 11.
Explain Bohr-Sommerfeld model of an atom. What is the merit of this model? What are its limitations?
Answer:

  • In an attempt to account for the structure of line spectra, Sommerfeld modified Bohr’s atomic model by adding elliptical orbits.
  • While retaining the first of Bohr’s circular orbit as such, he added one elliptical orbit to Bohr’s second orbit, two elliptical orbits to Bohr’s third orbit, etc.
  • Nucleus of the atom is one of the principal foci of these elliptical orbits because periodic motion under the influence of a central force will lead to elliptical orbits with the force situated at one of the foci.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 32

Merit:
Bohr-Sommerfeld model is successful in accounting for the fine line structure of hydrogen atomic spectra.

Limitations :

  1. This model failed to account for the atomic spectra of atoms of more than one electron.
  2. It did not explain Zeeman and Stark effects.

Question 12.
In an atom the number of electrons in N-shell is equal to the number of electrons in K, L and M shells. Answer the following questions.
i) Which is the outermost shell?
Answer:
The outermost shell is p(n = 6).

ii) How many electrons are there in its outermost shell?
Answer:
Two electrons are there in outermost shell.

iii) What is the atomic number?
Answer:
Its atomic number is 56.

iv) Write the electronic configuration of the elements.
Answer:
The electronic configuration of element is
1s² 2s² 2p6 3s² 3p6 4s² 3d10 4p6 5s² 4d10 5p6 6s².

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 13.
Explain the following electron configurations by using nlx method.
a) 2p¹
b) 3d5
c) 4f9
d) 6s²
Answer:
In nlx method n is the principle quantum number and l is the angular momentum quantum number and x is number of electrons. Now let us explain the following configurations
a) 2p¹ – It indicates that there is one electron in ‘p’ sub-shell of second orbit or shell.
b) 3d5 – It indicates that there are five electrons in ‘d’ sub-shell of third orbit or shell.
c) 4f9 – It indicates that there are nine electrons in ‘f’ sub-shell of fourth orbit or shell.
d) 6s² – It indicates that there are two electrons in ‘s’ sub-shell of sixth orbit or shell.

Question 14.
In an atom the number of electrons in L shell is equal to three times of K shell. Then answer the following.
1) Which is the outermost shell?
2) How many electrons are there in outermost shell?
3) What is the atomic number of element?
4) Write electronic configuration of element.
5) Write name of element.
Given that the number of electrons in L shell is three times of K shell.
Answer:
We know that number of electrons in K shell is 2.
Therefore number of electrons in L shell = 3 × 2 = 6

  1. So the outermost shell is L.
  2. The number of electrons in outermost shell is 6.
  3. The atomic number of element is 8.
  4. The electronic configuration of element is 1s² 2s² 2p4.
  5. The element is oxygen.

Question 15.
We know that the electron configuration of copper is [Ar] 4s¹ 3d10. Is it against to Aufbau principle or not. If so, why is the configuration violated?
Answer:

  • The atomic number of copper is 29. So its electron configuration should be [Ar] 4s² 3d4.
  • But if one electron from 4s orbital jumps into 3d orbital, then copper gets half filled ‘d’ orbitals which gives stability to the atom.
  • The energy difference between 4s and 3d is very less. So one electron can easily jump from 4s to 3d which gives half-filled 3d5 configuration.
  • So, in order to get additional stability, Aufbau principle is violated i.e., electron enters the orbital of higher energy before the completion ortrital of lower energy.

Question 16.
Electronic configurations of following elements are written wrongly. Correct those configurations with proper explanation.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 33
Answer:
1) The electron configuration of oxygen is 1s² 2s² 2p4 because the maximum number of electrons that can be filled in s orbital is 2 and so the extra electron should be entered in 2p.

2) The correct electron configuration of nitrogen is
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 34
The reason is that the pairing of electrons does not take place until each degenerate orbital is filled with one electron each (Hund’s principle).

3) The correct electronic configuration of scandium is 1s² 2s² 2p6 3s² 3p6 4s¹ because after completion of 3p orbital electron enters in 4s because the energy of 4s orbital is less than 3d (Aufbau principle).

4) The correct electronic configuration of chromium is 1s² 2s² 2p6 3s² 3p6 4s¹ 3d5. Because atoms having half filled or completely filled orbitals are more stable. So by transferring one electron from 4s to 3d the atom gets extra stability.

Question 17.
Here is set of quantum numbers. Do they form correct values of quantum numbers or not. If not, give reason.
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 35
The values l, ml, ms are not correct values for given ‘n’.
Answer:

  • The maximum value for l is n – 1. If n = 3, then l takes values from 0 to 2. That is 0, 1, 2.
  • ml values depend on ‘l’. ml take values from – l to + l including zero. So, the possible values for m7 may be from – 2 to + 2.
  • Spin quantum number has only two values, i.e. and –\(\frac{1}{2}\) and \(\frac{1}{2}\). So \(\frac{1}{4}\) value is not possible.

AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom

Question 18.
Answer the following questions.
a) If n = 4, then what energy level does it represent?
Answer:
If n = 4, then it represents N energy level.

b) If n = 5, then what is the maximum value for l and why?
Answer:
If n = 5, then the maximum value of l for 4 because the maximum value for l is n- 1.

c) If l = 3, then what are the maximum possible values for ml?
Answer:
Given l = 3.
Then possible values for ml is 2l + 1.
∴ Maximum possible values = 2(3) +1 = 7

d) What is the number of electrons present in M energy level?
Answer:
For M energy level n = 3.
The maximum number of electrons in an orbit = 2n² = 2 × 3² = 2 × 9 = 18

Question 19.
Draw the shapes of s, p and d orbitals.
Answer:
s – orbital (Spherical) :
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 36
p – orbital (Dumbell) :
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 37
All P:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 38
d – orbital (Double Dumbell):
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 39

Question 20.
Draw electromagnetic wave and label its parts.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 40

Question 21.
Draw the diagram of electromagnetic spectrum.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 8 Structure of Atom 41

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 10 Chemical Bonding

AP State Syllabus SSC 10th Class Chemistry Important Questions 10th Chemical Bonding

10th Class Chemistry 10th Lesson Chemical Bonding 1 Mark Important Questions and Answers

Question 1.
What is a chemical bond? (AP March 2015)
Answer:
An attractive force between two atoms in a molecule is called a chemical bond.

Question 2.
Write the names of any two compounds which have ionic bond. (TS June 2016)
Answer:
1) NaCl
2) MgCl2

Question 3.
Draw the structural diagram of Ammonia molecule as per the valence-shell electron pair repulsion theory. (TS March 2016)
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 1

Question 4.
Show the formation of HC/ molecule with Lewis dot structures using the information given below. (TS March 2017)
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 2
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 3

Question 5.
Explain, why bonding angle (HOH) in water is 104° 31′ instead of 109° 28′? (TS June 2018)
Answer:
Due to the repulsion between the lone pair of electrons and bond pair of electrons in water molecule the bond angle will be 104° 31′ rather than 109° 28′.

Question 6.
Imagine and write what type of ion can be formed generally by an atom of element with low ionisation energy, low electron affinity with high atomic size? (AP March 2019)
Answer:
Cation or positivity charged ion can be formed generally by an atom of element with low ionisation energy, low electron affinity with high atomic size.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 7.
Draw the structure of Methane molecule and mention bond angle. (TS March 2019)
Answer:
Structure of methane CH4. Tetrahedral structure with bond angle 109°28′.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 4
Bond Angle is 109°28′

Question 8.
Why do elements form chemical bonds? (TS SCERT: 2019-20)
Answer:

  • Elements are unstable if they contain less than eight electrons (octet) in their valency shell.
  • Hence, they form chemical bond with other elements to get octet in their valency shell.

Question 9.
How is covalent bond formed?
Answer:
A covalent bond is formed by the sharing of electrons between two atoms.
Ex : Bonding in H2 molecule, O2 molecule, N2 molecule, etc.

Question 10.
What is ‘Octet rule’?
Answer:
Octet rule :
Presence of 8 electrons in the outermost shell of an atom or a molecule is called ‘octet rule’.

Question 11.
What is ‘Bond length’?
Answer:
Bond length :
It is the inner-nucleus distance between the two atoms in a molecule. It is measured in Angstrom, 1 Å = 10-8 cm.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 12.
How is a cation formed?
Answer:
A cation is formed when an atom loses electrons.

Question 13.
How is an anion formed?
Answer:
An anion is formed when an atom gains electrons.

Question 14.
Which type of compounds are more soluble in polar solvents?
Answer:
Ionic compounds are more soluble in polar solvents.

Question 15.
Which compounds exhibit high melting and boiling points?
Answer:
Ionic compounds exhibit high melting and boiling points.

Question 16.
What is electronic configuration?
Answer:
A systematic arrangement of electrons in the atomic orbits is called electronic configuration.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 17.
Why are molecules more stable than atoms?
Answer:
Molecules have lower energy than that of the combined atoms. Molecules are more stable than atoms since chemical species with lower energy are more stable.

Question 18.
What is ionic linkage?
Answer:
When two oppositely charged ions are engaged in a bond, it is known as ionic linkage.

Question 19.
On which factors do anions depend?
Answer:

  1. Atomic size
  2. Ionisation potential
  3. Electron affinity
  4. Electronegativity

Question 20.
How do you know the valence of a metal?
Answer:
The number of electrons lost from a metal atom is the valence of its element which is equal to its group number.
Ex : Na and Mg have valence 1 and 2 respectively.

Question 21.
How do you know the valence of a non-metal?
Answer:
The number of electrons gained by a non-metal element for its atom is its valency, which is equal to 8 – its group number. Ex : The valency of chlorine is (8 – 7) = 1.

Question 22.
What is Tonic bond’?
Answer:
The electrostatic attractive force that keeps cation and anion together to form a new electrically neutral compounds is called ionic bond.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 23.
Why do atoms combine and form molecules?
Answer:
The energy of molecule is less than the total energy of constituent atoms. Therefore atoms combine and go to a stable state of lower energy.

Question 24.
What is ‘orbital concept of bond formation’?
Answer:
Atoms with half-filled or vacant orbitals try to get paired electrons in those orbitals by bond formation, i.e. by losing, gaining or sharing of electrons.

Question 25.
Name the bonds present in the molecules
i) BaCl2
ii) C2H4.
Answer:
i) In BaCl2 – Ionic.
ii) In C2H4 – Covalent (double bond H2C = CH2).

Question 26.
Why are ionic compounds good electrolytes?
Answer:

  • Electrolytes produce ions in solution, which carry current.
  • Ionic compounds in the fused state and aqueous solutions contain ions moving freely. Hence they conduct electricity.

Question 27.
When is ionic bond formed between atoms?
Answer:
Ionic bond is readily formed between atoms of elements with a low ionisation energy and atoms of elements with high electronegativity.

Question 28.
What is crystal lattice’?
Answer:
In a crystal of an ionic compound, each ion is surrounded by oppositely charged ions. The ions arrange themselves at an optimum distance with regular periodicity in a well-defined three-dimensional network called crystal lattice.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 29.
What is Lattice energy’?
Answer:
The energy released when gaseous positive and negative ions are brought together from infinity to form one mole ionic crystals is called lattice energy.

Question 30.
How is a σ (sigma) bond formed?
Answer:
By the axial or head on overlap of pure orbitals or hybrid orbitals of two atoms.

Question 31.
How is a π (pi) bond formed?
Answer:
By the lateral or parallel overlap of only pure atomic orbitals after the a bond formation.

Question 32.
Which type of atoms easily enter ionic bonding?
Answer:
An atom with low ionization potential and another with high electron affinity.

Question 33.
What is a polar bond?
Answer:
A covalent bond in a heteroatomic molecule.
Eg : HCl.

Question 34.
What is meant by inter-nuclear axis?
Answer:
The hypothetical line joining the centre of nuclei of two atoms in a molecule.

Question 35.
What are multiple bonds?
Answer:
Double and triple bonds are multiple bonds.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 36.
How many a and π bonds are in O2 molecule?
Answer:
One σ and one π bond.

Question 37.
What are Lewis structures?
Answer:
The symbol of the atom gives the core (or kernel) of the atom, in which valence electrons are shown as dots (•), circles (O) or crosses (x). Thus in Na, Na represents the core of sodium atom and the cross (x) represents the valence electron of sodium.

Question 38.
What are the structures of sodium chloride and calcium chloride crystals?
Answer:
Sodium chloride has face centered cubic structure. Calcium chloride has body centered cubic structure.

Question 39.
NaCl dissolves in water but not in benzene. Explain.
Answer:
NaCl dissolves in water because of hydration. Water being a polar molecule has positive and negative ends which hydrate Na+ and Cl ions. Benzene being non-polar cannot solvate the ions of NaCl.

Question 40.
What are the bond angles in H2O and NH3 molecule?
Answer:
Bond angle in H2O molecule is 104° 30′.
Bond angle in NH3 molecule is 107°.

Question 41.
What are the forces present in an ionic bond?
Answer:
Electrostatic forces of attraction are present in ionic bonds.

Question 42.
Which forces are weaker forces and where are they operative?
Answer:

  1. Van der Waal’s forces are very weak forces.
  2. They are operative between non-polar molecules.

Question 43.
Which compounds exhibit low melting and boiling points?
Answer:
Covalent compounds exhibit low melting and boiling points when compared to ionic compounds.

Question 44.
How are cations formed?
Answer:
The metal atoms lose electrons to form positively charged ions or cations.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 45.
Why is a molecule of hydrogen more stable than the uncombined atoms?
Answer:
When a molecule of hydrogen is formed from the atoms, energy is released (104 Kcal/mol). Thus the molecules possessing lower energy are more stable than the atoms.

Question 46.
How many sigma and pi bonds are present in acetylene molecule between carbon atoms?
Answer:
\(\mathrm{HC} \equiv \mathrm{CH}\), one sigma and two pi bonds are present.

Question 49.
In case of ionic substances, a more appropriate term is formula weight, rather than ‘molecular weight’. Why?
Answer:
Molecules are not present in ionic substances. Only ions are present. The formula of an ionic substance represents the simpler ratio of ions in one mole of crystal. Hence a more appropriate term is ‘formula weight’.

Question 50.
Write a short note on bond angles.
Answer:
It is the average angle between two adjacent atoms bonded to the central atom in a molecule. Molecules with larger bond angles are more stable than those with smaller angles.

Question 51.
Sulphur dioxide is a diamagnetic molecule. Explain.
Answer:
Sulphur dioxide (SO2) is a diamagnetic molecule because it has all electrons paired (no free electrons).

Question 52.
What is an ion?
Answer:
An ion is an electrically charged atom (or group of atoms).

Question 53.
What is a Coordination number?
Answer:
The number of ions of opposite charge that surround a given ion of given charge is known as coordination number of that given ion. –

Question 54.
What factors affect or which factors influence the formation cation or anion?
Answer:
Factors affecting formation of anion or cation are :

  1. Atomic size,
  2. Ionization potential,
  3. Electron affinity,
  4. Electronegativity.

Question 55.
Two elements X and Y have the following configurations.
X = 1s² 2s² 2p6 3s² 3p6 4s²
Y = 1s² 2s² 2p6 3s² 3p5
What is the formula of the compound?
Answer:
The electronic configuration of X is 1s² 2s² 2p6 3s² 3p6 4s².
So its valency is 2.
The electronic configuration of Y is 1s² 2s² 2p6 3s² 3p5.
So its valency is 1.
∴ The formula of the compound is XY2.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 56.
The electronegativities of two elements are 1 and 3. What type of bond is formed between the two elements? Why?
Answer:
The electronegativity difference of elements = 3 – 1 = 2.
So the electronegativity difference is more than 1.9. Therefore the bond formed between the elements is ionic in nature.

Question 57.
What do you mean by Doublet configuration?
Answer:
If two electrons are present in valence shell, then it is called doublet configuration.
Eg.: Helium exhibits doublet configuration.

Question 58.
Write Lewis symbol of potassium and calcium.
Answer:
Potassium – \(\dot{\mathrm{K}}\) and Calcium – •Ca•

Question 59.
Which of the following is / are true?
1) Ionic compounds exist as solids.
2) Ionic compounds have high melting and boiling points.
3) Ionic compounds conduct electricity in solid state.
4) Ionic compounds have low melting and boiling points.
Answer:
Ionic compounds generally exists as solids and they have very high melting and boiling points and they conduct electricity in aqueous or molten state.
So first and second statements are true.

Question 60.
How many a and n bonds are there in acetylene molecule?
Answer:
The structure of acetylene is given by
\(\mathrm{H}-\mathrm{C} \equiv-\mathrm{H}\)
So they are 3a and 2n bonds.

Question 61.
What is the valency of carbon in CO2?
Answer:
The valency of oxygen is 2. So the valency of carbon = 2 × 2 = 4.

Question 62.
Which of the following does not get Helium configuration during formation of chemical bonding?
1) Hydrogen,
2) Lithium,
3) Beryllium,
4) Oxygen.
Answer:
Oxygen attains Neon configuration when it undergoes chemical bondings and the rest will attain Helium configuration.

Question 63.
Write the valencies of following elements.
a) Be
b) N and what is the compound formed when these two reacted?
Answer:
a) The electronic configuration of Be is 1s² 2s². So its valency is 2.
b) The electronic configuration of Nitrogen is 1s² 2s² 2p³. So its valency is 3.
∴ Formula of the compound is Be3N2.

Question 64.
Which of the following solutions are conductors of electricity?
a) Sugar solution,
b) Alcohol solution,
c) Glucose solution,
d) Salt solution.
Answer:
Salt solution is good conductor of electricity because common salt (NaCl) is an ionic compound and the rest are covalent compounds.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 65.
Why are noble gases (inert gases) stable?
Answer:

  1. Noble gases are stable because their outermost orbit contains 8 or 2 electrons.
  2. Noble gases exist an individual atoms.

10th Class Chemistry 10th Lesson Chemical Bonding 2 Marks Important Questions and Answers

Question 1.
Represent each of the following molecules using Lewis notation. (TS March 2015)
(i) Calcium and Chlorine to form Calcium chloride.
(ii) Formation of Oxygen molecule from Oxygen atoms.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 5

Question 2.
Between a neutral atom and its cations which has bigger size? Why? (TS June 2016)
Answer:

  • A neutral atom has bigger in size than its cation.
  • A cation has more protons and nucleus of the cation attracts the electrons in the outermost orbital more.
  • Hence, the radius of the anion decreases. It means size of the anion decreases.
  • So, a neutral atom is bigger than its cation.

Question 3.
Draw the diagram to show the formation of Oxygen molecule by Valence bond theory. (TS March 2017)
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 6

Question 4.
Explain Ionic bond with suitable example. (TS June 2018)
Answer:
1) Sodium (Na) looses one electron and forms Sodium ion (Na+).
Na → Na+ + 1e

2) Chlorine (Cl) gains one electron and forms chloride ion (Cl).
Cl + 1e → Cl

3) Positive sodium ion (Na+) and negative chloride (Ct) ion come together due to electrostatic forces, participates in ionic bond and form NaCl.
Na+ + Cl → NaCl

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 5.
Give two examples of each to ionic and covalent compounds. (AP SCERT: 2019-20)
Answer:
1) Ionic compounds eg : NaCl, MgCl2
2) Covalent compounds eg : Cl2, HCl

Question 6.
Distinguish between a sigma and a pi bond.
Answer:

Sigma bond Pi bond
1. It is formed by the end-end on overlap of orbitals. 1. It is formed by the lateral overlap of orbitals.
2. It has independent existence. 2. It has no independent existence.
3. It is a strong bond. Because axial overlap is more. 3. It is a weak bond. Because lateral overlap is less.
4. There can be only one a bond. 4. There can be one or two 7t bonds between two atoms.
5. All orbitals form ‘o’ bond. 5. Only p, d orbitals form a bond.

Question 7.
Write about ‘Hydrogen bond’
Answer:

  • Hydrogen bond is formed between molecules in which hydrogen atom is attached to an atom of an element with large electronegativity and very small size (F, O, N). Because in hydrogen bond the molecules associate themselves and hence possess higher B.P’s and M.P’s.
  • The hydrogen bond formed between two molecules is called inter-molecular hydrogen bond.
  • The hydrogen bond formed between different groups of the same molecule is called intra-molecular hydrogen bond.

Question 8.
Bring out the difference between ionic and covalent bonds.
Answer:

Ionic bond Covalent bond
1. It is formed by transference of electrons from one atom to the other. 1. It is formed by the sharing of electron pairs by two atoms.
2. Electrostatic. 2. Not electrostatic, but rigid.
3. Ionic substances are formed by ionic bonds. 3. Molecules are formed by covalent bonds.
4. Non-directional. 4. Directional.

Question 9.
What is hybridisation?
Answer:
In the formation of molecules, the atomic orbitals of the atoms may hybridise.

  1. It is the process of mixing up of atomic orbitals of an atom to form identical hybrid orbitals.
  2. This takes place only during the formation of bond.
  3. There should not be much difference in the energies of the orbitals that hybridise.
  4. The number of hybrid orbitals formed is equal to the number of hybridising atomic orbitals.
  5. Hybrid orbitals form a bonds only not n bonds.

Question 10.
What is ionisation? Give one example.
Answer:
1) The process of removal of electron (s) from an atom or molecule is termed as ionisation.
Eg : Na → Na+ + e

2) Dissociation of an ionic solid into constituent ions upon its dissolution in a suitable solvent, is also called ionisation.
Eg : NaCl(aq) → Na+(aq) + Cl(aq)

Question 11.
Write a short note on octet rule with example.
Answer:
Atoms try to get 8 electrons in their outermost shell (inert gas structure) by combining with other atoms. In order to get 8 outer electrons, atoms may lose to, gain from or share electrons with other atoms.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 7

Question 12.
What are the important characteristic features of hybridisation?
Answer:

  • Orbitals on a single atom only would undergo hybridization.
  • The orbitals involved in hybridisation should not differ largely in their energies.
  • The number of hybrid orbitals formed is equal to the number of hybridising orbitals.
  • The hybrid orbitals form stronger directional bonds than the pure s, p, d atomic orbitals.
  • It is the orbitals that undergo hybridisation and not the electrons.
  • Concept of hybridisation is useful in explaining the shape of molecules.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 13.
Why is a large amount of energy needed to remove an electron from a neutral gaseous neon atom than the energy needed to remove an electron from gaseous sodium atom?
Answer:
Na (g) +118.4 K.cal → Na+ + e
Ne (g) + 497.0 K.cal → Ne+ + e

Orbitals which are fully-filled are very stable, so need large amount of energy to remove an electron from them.

Question 14.
What are the drawbacks of electronic theory of valence?
Answer:
1) When covalent bond formed between any two atoms, irrespective of the nature of the atoms, the bond lengths and bond energies’ are expected to be same. But practically it was observed that bond lengths and bond energies are not same when the atoms that form the bond are different.

2) This theory fails to explain the shapes of molecules.

Question 15.
What is the structure of NaCl and write its coordination number of its constituent ions?
Answer:
The structure of NaCl is face centred cubic lattice.
Its coordination number is 6 for Na+ and Cl.

Question 16.
Why is there absorption of energy in certain chemical reactions and release of energy in other reactions?
Answer:
If bond dissociation energy of reactants is more than bond energy of products, then energy is absorbed in the chemical reaction.

If bond dissociation energy of reactants is less than bond energy of products, then energy is released in the chemical reaction.

Question 17.
Why do ionic compounds dissolve in polar solution and covalent compounds dissolve in non-polar solution?
Answer:
Ionic compounds are polar in nature so they are soluble in polar solvents whereas covalent compounds are non-polar in nature so they are soluble in non-polar solvents.

Question 18.
Identify the following whether they are either oxidation or reduction reactions.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 8
Answer:
i) Oxygen atom is gaining electrons. So it is reduction reaction.
ii) Potassium atom is losing electron. So it is oxidation reaction.
iii) Ferric ion is gaining electron. So it is reduction reaction.

Question 19.
Elements A, B, and C have atomic numbers 9, 20, and 10 respectively.
a) State which one is (i) a metal, (ii) a non-metal, (iii) chemically inert.
b) Write down the formula of the compound formed by two of the above elements.
Answer:
a) i) Element with atomic number 20 is a metal because its electronic configuration is 2, 8, 8, 2. It belongs to 2nd group and the element is calcium.
ii) Element with atomic number 9 is a non-metal because its electronic configuration is 2, 7. So it belongs to 17th group and it is Fluorine.
iii) Element with atomic number 10 is inert gas because its electronic configuration is 2, 8 and the element is Neon.

b) The formula of compound formed between elements Calcium and Fluorine is CaF2.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 20.
In the formation of the compound XY2, an atom X gives one electron to each Y atom. What is the nature of bond in XY2? Give two properties of XY2.
Answer:
Two electrons are transferred from X to Y. X forms positive ion and Y forms a negative ion. So the bond formed is ionic in nature.
Properties of ionic compounds :
a) They are hard crystals.
b) They have high boiling point and melting point.
c) They are soluble in water.

Question 21.
Element X is a metal with valency 2. Element Y is a non-metal with valency 3.
i) Write equations to show X and Y form ions.
Answer:
X is a metal with valency 2.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 9

ii) If Y is a di-atomic gas, write the equation for the direct combination of X and Y to form a compound.
Answer:
4X + 3Y2 → 2X2Y3

Question 22.
What are the differences between sodium atom and sodium ion?
Answer:

  • Sodium atom is neutral whereas sodium ion has unipositive charge.
  • The size of sodium ion is smaller than sodium atom.
  • The properties of sodium ion is different when compared with sodium.

Question 23.
What is the percent of p-character in sp, sp² and sp³ hybrid orbitals?
Answer:
In sp the p-character is 50%.
In sp² the p-character is 66.66%.
In sp³ the p-character is 75%.

Question 24.
What is the percent of s-character in sp, sp² and sp³ hybrid orbitals?
Answer:
In sp hybrid orbitals s-character is 50%.
In sp² hybrid orbitals s-character is 33.33%.
In sp³ hybrid orbitals s-character is 25%.

Question 25.
A chemical compound has following Lewis notation.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 10
a) How many valence electrons does element Y have?
b) What is the valency of element Y?
c) What is the valency of element X?
d) How many covalent bonds are there in the molecule?
e) Suggest a name for the element X and Y.
Answer:
a) The valence electrons in Y are 5.
b) The valency of Y is 3.
c) The valency of X is 1.
d) There are three covalent bonds.
e) The element Y is Nitrogen and X is Hydrogen.

Question 26.
The electronic configurations of three elements.
X is 2, 6; Y is 2, 8, 7; Z is 2, 8,1.
In each ease given below state whether the bonding is ionic or covalent and give the formula of molecules of the compound formed,
a) Between two atoms of X.
Answer:
As we move from left to right in a period non-metallic character increases. So Z is a metal and remaining two are non-metals.
Therefore between two atoms of X, there would be covalent bond. The formula of compound is X2.

b) Between the atom X and atom Z.
Answer:
X is a non-metal and Z is a metal. So between these two atoms, there would be an ionic bond.
The valency of X is 2 and the valency of Z is 1.
So the formula of compound is Z2X.

c) Between the atom Y and atom Z.
Answer:
Y is also non-metal and we know Z is metal.
The valency of Y is 1.
So the compound formed is ZY.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 27.
Electronic configuration of X is 2, 8, 1 and electronic configuration of Y is 2, 8, 7. Explain what type of bond is formed between them.
Answer:
The element with electronic configuration 2, 8,1 is sodium which is a metal. Similarly the element with electronic configuration 2, 8, 7 is chlorine which is a non-metal. The electronegativity difference between these two elements is more than 1.9 so they form ionic bond.

Question 28.
Complete the following statements.
i) The bond which is formed by transfer of an electron from one atom A to another atom B is called ……………
Answer:
Ionic bond

ii) The atoms A and B become ions. Thus A – 2e = …………….. and B + e = ………….
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 11

iii) The formula of the compound is
Answer:
The valency of A is 2 and valency of B is 1. So the compound formed is AB2.

Question 29.
Answer the following.
i) What is the valency of Nitrogen in Ammonia?
Answer:
The formula of Ammonia is NH3. We know valency of hydrogen is 1. So valency of Nitrogen is 3.

ii) What is the valency of Oxygen in Water?
Answer:
The formula of water is H2O. The valency of hydrogen is 1. So the valency of oxygen is 2.

iii) What is the valency of Carbon in Carbon tetra chloride?
Answer:
The formula of carbon tetra chloride is CCl4. The valency of chlorine is 1. So the valency of carbon is 4.

10th Class Chemistry 10th Lesson Chemical Bonding 4 Marks Important Questions and Answers

Question 1.
What is Hybridisation? Explain the formation of BeF, molecule using hybridisation. (AP June 2017)
Answer:
Hybridisation :
Hybridisation is a phenomenon of intermixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

Formation of BeF2 molecule :

  1. 4Be has electronic configuration 1s² 2s².
  2. In excited state electronic configuration of 4Be is 1s² 2s¹ 2px¹
  3. 4Be allows its 2s orbitals and 2px orbitals which contain unpaired electrons to intermix and redistribute to two identical orbitals.
  4. The hybridisation is called ‘sp’ hybridisation and form two ‘sp’ hybrid orbitals. Each orbital contains one electron.
  5. 9F has electronic configuration 1s² 2s² 2p5 (or) 1s² 2s² 2px² 2py² 2pz¹
  6. F contains one unpaired electron.
  7. Two Fluorine atoms come closure to Be and form two covalent bonds.
    σsp – p and σsp – p
  8. Now the BeF2 has (F \(\widehat{\mathrm{Be}}\) F) angle of 180°.
    AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 12

Question 2.
The arrangement of electrons in different shells of atoms of 18th Group elements is given in the table.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 13
Answer the following : (AP March 2017)
(i) What is the general electronic configuration of the above elements except He?
(ii) What is the valency of Argon?
(iii) Write Lewis dot structure of Neon.
(iv) Why the above elements do not take part in bond formation?
Answer:
(i) ns² np6
(ii) Zero
(iii)
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 14

(iv) They are stable as they have 8 electrons (except Helium) in their outer most orbit.

Question 3.
“Nitrogen and Hydrogen react to form a molecule of Ammonia (NH3). Carbon and Hydrogen react to form a molecule of Methane (CH4).”
For each reaction :
a) What is the valency of each of the atom involved in the reaction?
b) Draw the dot structure of the products that are formed. (AP June 2018)
Answer:
a) In ammonia : Nitrogen valency – 3, Hydrogen valency – 1
In methane : Carbon valency – 4, Hydrogen valency – 1

b)
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 15

Question 4.
Explain the formation of Boron tri-fluoride molecule by Hybridization. (TS June 2018)
Answer:
1) Electron configuration of Boran at ground state.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 16
2) Electron configuration of Boran at excited state is
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 17
3) As it forms three identical B-F bonds in BF3.
4) In excited state of Boran, one ‘s’ orbital and two ‘p’ orbitals undergoes hybridization and forms three sp² orbitals.
1s + 2p → 3sp²
5) Three sp² orbitals make end – on – end overlap with three ‘p’ orbitals of fluorine atoms.
6) They form three sp² – p sigma bonds.
7) In this way Boran tri-fluoride molecule is formed.
8)

AP SSC 10th Class Chemistry Solutions Chapter 10 Chemical Bonding 22
BF3

Question 5.
Explain the formation of BF3 molecule with the help of Valence Bond theory? (TS March 2018)
Answer:
Formation of BF3 molecule :

  1. 5B has electronic configuration 1s² 2s² 2px.
  2. The excited electronic configuration of 5B is 1s² 2s¹ 2px¹ 2py¹.
  3. As it forms three identical B-F bonds in BF3
  4. It is suggested that excited ‘B’ atom undergoes hybridisation.
  5. There is an intermixing of 2s, 2px, 2py orbitals and their redistribution into three identical orbitals called sp² hybrid orbitals.
  6. For three sp² orbitals to get separated to have minimum repulsion the angle between any two orbitals is 120° at the central atom and each sp² orbital gets one electron.
  7. Now three fluorine atoms overlap their 2pz orbitals containing unpaired electrons (F9 1s² 2s² 2px² 2py² 2pz²) the three sp² orbitals of ‘B’ that contain unpaired electrons to form three ssp²-p bonds.
AP SSC 10th Class Chemistry Solutions Chapter 10 Chemical Bonding 22
BF3

Question 6.
Explain the formation of N2 molecule using Valence Bond theory. (TS June 2019)
Answer:
Formation of N, molecule (Valence Bond theory) :

  1. 7N has electronic configuration 1s² 2s² 2px¹ 2py¹ 2pz.
  2. Suppose that px orbital of one ‘N’ atom overlaps the ‘px‘ orbital of the other ‘N’ atom giving σ px – px bond along the inter – nuclear axis.
  3. The py and pz orbitals of one ‘N’ atom overlap the py and pz orbital of other ‘N’ atom laterally, respectively perpendicular to inter – nuclear axis giving π py – py and π pz-pz bonds.
  4. Therefore, N2 molecule has a triple bond between two nitrogen atoms.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 18

Question 7.
If the electronic configurations of atoms A and B are 1s², 2s², 2p6, 3s², 3p¹ and 1s², 2s², 2p4 respectively, then
a) which atom forms negative ion?
Answer:
Given electronic configuration of atom A is 1s² 2s² 2p6 3s² 3p¹, i.e. Aluminium and B is 1s² 2s² 2p4, i.e. Oxygen.

The atom ‘B’ tends to form negative ion by gaining two electrons in order to get nearest inert gas. Neon’s configuration is 1s² 2s² 2p6. So its valency is 2.

b) which atom forms positive ion?
Answer:
The atom ‘A’ tends to form positive ion by losing three electrons in order to get nearest inert gas. Neon’s configuration is 1s² 2s² 2p6.

c) what is the valency of atom A?
Answer:
Valency of atom ‘A’ is ‘3’.

d) what is the molecular formula of the compound formed by atoms A and B ?
Answer:
According to Criss-Cross method, the molecular formula of the compound formed by atoms both A and B is A2B3, i.e. Al2O3.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 8.
Writs the salient features of VSEPRT.
(OR)
Explain VSEPR theory.
Answer:
The full form of VSEPRT is Valence Shell Electron Pair Repulsion Theory.

  1. VSEPRT considers electrons in the valence shells which are in covalent bonds and in lone pairs as charge clouds that repel one another and stay as far apart as possible which will give specific shapes to molecules.
  2. The presence of lone pair in the central atom causes slight distortion of bond angles from expected regular shape.
  3. If the angle between lone pair and bond pair increases at the central atom due to more repulsion, the actual bond angles between atoms must be decreased.
  4. If two bond pairs are present without any lone pair, the shape of the molecule is linear with bond angle 180°.
  5. If three bond pairs are present without any lone pair, the shape of the molecule is trigonal planar with bond angle 120°.
  6. If there are four bond pairs in the valency shell of central atom without lone pair, the shape is tetrahedron and the bond angle is 109°28′.
  7. If there are three bond pairs and one lone pair, the shape of the molecule is pyramidal.
  8. If there are two bond pairs and two lone pairs, the shape is V.

Question 9.
Do you think that a pair of Na+Cl as units would be present in the solid crystal? Explain.
Answer:

  • No, the electrostatic forces are non-directional.
  • Therefore, it is not possible for one Na+ to be attracted by one Cl and vice-versa.
  • Depending upon the size and charge of particular ion, number of oppositely charged ions get attracted by it, but, in a definite number.
  • In sodium chloride crystal each Na+ is surrounded by 6 Cl and each Cl by six Na+ ions.
  • Ionic compounds in the crystalline state consists of orderly arranged cations and anions held together by electrostatic forces of attractions in three dimensions.

Question 10.
Explain the valence bond theory.
Answer:
Valence bond theory :

  1. A covalent bond between two atoms is formed when the two atoms approach each other closely and one atom overlaps its valence orbital containing unpaired electron with the valence orbital of other atom that contains the unpaired electron of opposite spin.
  2. The greater the overlapping of the orbitals, the stronger will be bond.
  3. Valence bond theory gives a directional character to the bond when other than ‘s’ orbitals are involved.
  4. Electron pair in the overlapping orbitals is shared bv both the atoms involved in the overlapping.
  5. If orbitals overlap along inter-nucleus axis, they form a strong bond called sigma (σ) bond.
  6. If the orbitals overlap laterally, they form a weak bond called pi (π) bond.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 11.
Answer the following.
a) Name the charged particles which attract one another to form ionic or electro-valent compounds.
Answer:
The charged particles are ions which are formed due to transfer of electrons from atom of one element to atom of other element.

b) In the formation of ionic compounds, electrons are transferred from one element to another. How are electrons involved in the formation of covalent compounds?
Answer:
In the formation of covalent bond electrons are mutually shared between atoms of elements.

c) The electronic configuration of Nitrogen is 2, 5. How many electrons in the outer- shell of a nitrogen atom are not involved in the formation of Nitrogen molecule?
Answer:
In the formation of Nitrogen molecules three pairs of electrons are mutually shared by the two nitrogen atoms. So each Nitrogen has two electrons which do not take part in bonding. Those two electrons are called lone pair.

d) In the formation of magnesium chloride, name the substance that is oxidized and the substance that is reduced.
Answer:
Two electrons are transferred from Magnesium atom to Chlorine atom in the formation of Magnesium chloride.

So Magnesium is oxidised (loss of electrons is oxidation) and Chlorine is reduced (gain of electrons is reduction).

Question 12.
The symbols of two elements with their atomic numbers are given below. 16A and 20B
i) Which element will form a cation?
Answer:
The atomic number of B is 20. So its electronic configuration is 1s² 2s² 2p6 3s² 3p6 4s² or 2, 8, 8, 2. So it easily loses two electrons to form cation thereby attain inert gas stability.

ii) Which element will form an anion?
Answer:
The atomic number of A is 16. So its electronic configuration is 1s² 2s² 2p6 3s² 3p4 or 2, 8, 6. So it easily gains two electrons to form anion. Therefore it attains inert gas stability.

iii) Show how each element forms an ion.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 19

iv) Which element will be oxidised in forming the ion?
Answer:
B is losing electrons. So it is oxidised. .

v) When the elements react, what will be the nature of the bond formed?
Answer:
When the elements react they form ionic bond because B is metal and A is non-metal.

Question 13.
Give one example for each of the following.
i) A polar covalent bond formed from two dissimilar atoms one of which is oxygen.
Answer:
In the formation water molecule, polar covalent bond is formed between two dissimilar atoms hydrogen and oxygen.

ii) A non-polar covalent compound formed from two similar atoms.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 20
Answer:
The examples for a non-polar covalent compound formed from two similar atoms are Fluorine (F2), Chlorine (Cl2).

iii) A solid non-polar covalent compound.
Answer:
The example for solid non-polar covalent compound is Iodine (I2).

iv) A polar covalent compound containing three shared pairs of electrons.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 21
Answer:
The polar covalent compound containing three shared pairs of electrons is Ammonia.

Question 14.
The list of some substances is given below.
HCl, H2O, Cl2, NaBr, CH4, NH3, N2, O2, CaO, HF, I2 and Br2.
From the list above choose the substance / substances.

i) Which have only ionic bond in the molecules?
Answer:
NaBr, CaO have only ionic bond in the molecules,

ii) Which has a triple covalent bond in its molecule?
Answer:
N2 has a triple covalent bond in its molecule (\(\mathbf{N} \equiv \mathbf{N}\)).

iii) Which are solids?
Answer:
Generally ionic compounds are solids. So NaBr, CaO are solids. I2 being a covalent compound still it is solid.

iv) Which has a double covalent bond in its molecule?
Answer:
O2 has a double covalent bond in its molecule (O = O).

v) Which is non-polar covalent liquid?
Answer:
The non-polar covalent liquid is Br2.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 15.
There are elements with atomic numbers 4, 14, 8, 15 and 19. From this information answer the following questions.
a) A solid non-metal of valency 3.
Answer:
The element with atomic number 15 is phosphorus. Its electronic configuration is 2, 8, 5. So its valency is 3 and it is a solid non-metal.

b) A gas of valency 2.
Answer:
The element with atomic number 8 is oxygen. Its electronic configuration is 2, 6. So its valency is 2 and it is a gas.

c) A metal of valency 1.
Answer:
The element with atomic number 19 is potassium. Its electronic configuration is 2, 8, 8, 1. So its valency is 1 and it is a metal.

d) A non metal of valency 4.
Answer:
The element with atomic number 14 is silicon. Its electronic configuration is 2, 8, 4. Its valency is 4 and it is a non-metal.

Question 16.
Explain the following.
a) Ionic compounds conduct electricity.
Answer:
Ionic compounds can be electrolysed to give their constituent ions. The ions obtained by dissociation move freely in solution and hence conduct electricity. When ionic compound is dissolved in water or melted, it becomes a good conductor of electricity.

b) Ionic compounds have high melting and boiling points while covalent compounds have low melting and boiling points.
Answer:
The forces existing between ionic compounds are electrostatic in nature. They are strong forces. So in order to break these forces lot of energy is required. Therefore ionic compounds have higher boiling and melting points.

c) Ionic compounds dissolve in water whereas covalent compounds do not.
Answer:
Ionic compounds are polar in nature. So they are soluble in polar solvent like water whereas covalent compounds are non-polar in nature. So they are insoluble in water.

d) Ionic compounds are usually hard crystals.
Answer:
Due to strong attractive forces they are usually hard crystals.

Question 17.
If A, B, C and D are elements given that B is an inert gas (not Helium).

Element Atomic Number
A Z- 1
B Z
C Z + 1
D Z + 4

What type of bonding would take place between (i) A and C and (ii) D and A and write their formulae?
Answer:
1) Given that B is inert gas.
So the valency of A is 1 and it is a non-metal and valency of C is 1 and it is metal and valency of D is 4 and it is a non-metal.

2) The bond formed between C and A is ionic nature and its formula is CA.
The bond formed between D and A is covalent nature and its formula is DA4.

Question 18.
State whether the following statements are true or false. If statement is false, correct it.
i) In polar compounds the electropositive atom attracts the electron pair towards it.
Answer:
This statement is wrong. In polar compounds the more electronegative atom attracts the electron pair towards it which is known as electronegativity.

ii) Hydrogen chloride gas is a di-polar molecule.
Answer:
Chlorine is more electronegative than Hydrogen. So the electron pair shifts more towards chlorine atom. Thus partial positive charge is formed on Hydrogen and partial negative charge is formed on Chlorine. So it is dipolar molecule. So the statement is true.

iii) Covalent compounds are generally gases due to presence of weak electrostatic forces of attraction.
Answer:
This statement is false. The forces are not electrostatic and also they are weak.

iv) Atoms achieve stable electronic configuration only by transfer of electrons from one atom to another.
Answer:
This statement is false because they can acquire by not only electron transfer but also mutually sharing of electrons.

v) Ionic compounds are soft, solids or liquids at ordinary temperature.
Answer:
This statement is false because ionic compounds are generally hard solids.

Question 19.
Five atoms are labelled from A to E.

Atoms Atomic Number Mass Number
A 20 40
B 9 19
C 3 7
D 8 16
E 7 14

a) Which one of these atoms (i) contain 7 protons, (ii) has an electronic configuration 2, 7?
Answer:’
i) The atom E contains 7 protons because its atomic number is 7.
ii) The atom with electronic configuration 2, 7 is B because its atomic number is 9.

b) Write down the formula of the compound formed between C and D.
Answer:
The electronic configuration of C is 2, 1 since its atomic number is 3. The electronic configuration of D is 2, 6 since its atomic number is 8.
So the valency of C is 1. –
The valency of D is 2 (8 – 6 = 2).
∴ The compound formed between C and D is C2 D.

c) Predict which are (i) metals, (ii) non-metals?
Answer:
Metals are A, C.
Non-metals are B, D, E.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding

Question 20.
Can you suggest an experiment to prove that ionic compounds possess strong bonds when compared to that of covalent bonds? Explain the procedure.
Answer:

  • Take a small amount of sodium chloride (NaCl) on a metal spatula (having an insulated handle).
  • Heat it directly over the flame of a burner.
  • We will see that sodium chloride (NaCl) does not melt easily.
  • Sodium chloride melts (and becomes a liquid) only on strong heating.
  • This shows that sodium chloride which is an ionic compound possesses strong bonds.
  • So it has a high melting point.
  • Whereas covalent compound like naphthalene and carbon tetra chloride has low boiling points such as 80° C and 77° C respectively.
  • The force of attraction between the molecules of a covalent compound is very weak.
  • Only a small amount of heat energy is required to break these weak molecular forces, due to which covalent compound has low melting points and low boiling points.

Question 21.
The electronic configurations of following elements are given below.
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 22
From these values complete the table.

Compound Type of bonding
Lithium chloride
Lithium hydride
Hydrogen chloride
Carbon tetrachloride

Answer:
a) Lithium is metal and chlorine is non-metal. So the bond is ionic in nature.
b) Lithium is metal and hydrogen is non-metal. So the bond is ionic in nature.
c) Hydrogen and chlorine are two dissimilar non-metals. So the bond formed is polar covalent bond.
d) Carbon and chlorine are dissimilar non-metals but electronegativity difference is less. So they form non-polar covalent bond.

Compound Type of bonding
Lithium chloride Ionic bond
Lithium hydride Ionic bond
Hydrogen chloride Polar covalent bond
Carbon tetrachloride Non-polar covalent bond

Question 22.
Draw the structure of O2 by using valence bond theory.
Answer:
O2 Formation :
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 23

Question 23.
Draw the structure of N2.
Answer:
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 24

Question 24.
Draw the structure of molecules by Lewis method.
Answer:
Formation of F2:
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 25
Formation of O2 :
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 26

Formation of N2 :

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 27
Formation of CH4 :
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 28

Formation of NH3 :
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 29
Formation of H2O :
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 30
Formation of BeCl2 :
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 31
Formation of BF3 :
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 32

Question 24.
Explain the formation of the following molecules using Lewis theory.
a) N2
b) O2
(OR)
Write the formation of double bond and triple bond according to Lewis theory.
Answer:
a) Formation of N2 molecule by Lewis theory :

  1. The electronic configuration of ’N’ atom is 2, 5 and to have octet in the valence shell it requires three more electrons.
  2. When two nitrogen atoms approach each other, each atom contributes 3 electrons for bonding.
  3. There are six electrons shared between two nitrogen atoms in the form of three pairs.
  4. Therefore, there is a triple bond between two nitrogen atoms in N2 molecule.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 27
AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 33

b) Formation of O2 molecule by Lewis theory :

  1. The electronic configuration of 8O is 2, 6.
  2. Oxygen atom has six electrons in its valence shell.
  3. It requires two more electrons to get octet in its valence shell.
  4. Therefore oxygen atoms come close and each oxygen atom contributes two electrons for bonding.
  5. Thus, there exist two covalent bonds between two oxygen atoms in O2 molecule as there are two pairs of electrons shared between them.
  6. Two pairs of electrons are distributed between two oxygen atoms.
  7. So, we can say that a double bond is formed between two oxygen atoms in O2 molecule.
  8. By viewing the following diagram, both the oxygen atoms have octet in the valence shell.

AP SSC 10th Class Chemistry Important Questions Chapter 10 Chemical Bonding 34

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AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna

SCERT AP Board 6th Class English Solutions 1st Lesson Clever Tenali Ramakrishna Questions and Answers.

AP State Syllabus 6th Class English Solutions 1st Lesson Clever Tenali Ramakrishna

6th Class English 1st Lesson Clever Tenali Ramakrishna Textbook Questions and Answers

Look at the picture and answer the questions that follow.
AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 1

Question 1.
What do you see in the picture?
Answer:
I see a king, two poets / two courtiers, a man covering his head with a pot and a maid-servant.

Question 2.
Who is sitting on the throne?
Answer:
A king (Srikrishna Devaraya)

Question 3.
Can you guess the person who covered his head with a pot?
Answer:
Yes, he might be Tenali Ramakrishna.

Question 4.
Do you know any funny incidents related to Tenali Ramakrishna?
Answer:
Yes, I know one funny incident related to Tenali Ramakrishna. The following is a funny incident related to Tenali Ramakrishna.

Reading Comprehension

A. Answer the following questions

Question 1.
Who was Chitra Varma?
Answer:
Chitra Varma was a great painter in Srikrishna DeVaraya’s court. The king made him a chief advisor.

Question 2.
How did Chitra Varma feel when the emperor decided to reward him?
Answer:

The king’s reward was to make Chitra Varma his chief advisor. But Chitra Varma was afraid that he was not suitable for that post.

Question 3.
Why did Chitra Varma fail as a chief advisor?
Answer:
Chitra Varma was good at painting. He was not good at giving good advice. So he failed as a chief advisor.

Question 4.
Why was the emperor angry with Tenali Ramakrishna?
Answer:
Tenali Ramakrishna served the emperor horrible dishes. He made a carpenter cook food. So the emperor was angry with him.

Question 5.
Discuss in pairs what will happen when people are offered to do work that is not meant for them.
Answer:
Pah 1 : It is not a good idea to offer to do work that is not meant for them because they spoil the work.
Pah 2 : If we offer people to do work that is not meant for them, people will suffer.
Pah 3 : They may take improper decisions and they may lead to some problems.
Pair 4 : It may lead to the loss of time and money.

B. Put the following sentences in the order of events

1. People went to Tenali Ramakrishna and told him their problem.
2. Chitra Varma was appointed a chief advisor to the emperor.
3. Tenali Ramakrishna implemented his plan by engaging a carpenter as a cook.
4. Emperor Srikrishna Devaraya realized his mistake at the end.
5. Chitra Varma was a famous painter in the emperor’s court.
Answer:
1. Chitra Varma was a famous painter in the emperor’s court. (5)
2. Chitra Varma was appointed a .chief advisor to the emperor. (2)
3. People went to Tenali Ramakrishna and told him their problem. (1)
4. Tenali Ramakrishna implemented his plan by engaging a carpenter as a cook. (3)
5. Emperor Srikrishna Devaraya realized his mistake at the end, (4)

C. Say whether the following statements are True / False:

1. Chitra Varma was a famous carpenter.
Answer:
False

2. Emperor Srikrishna Devaraya awarded a lot of gold coins to Chitra Varma.
Answer:
False

3. Tenali Ramakrishna was a poet and jester in emperor Srikrishna Devaraya’s court.
Answer:
True

4. Tenali Ramakrishna prepared a delicious dinner for the emperor.
Answer:
False

5. Emperor Srikrishna Devaraya realized his mistake at the end.
Answer:
True

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 2
Correction of False Statements :
1. Chitra Varma was a famous painter.
2. Emperor Srikrishna Devaraya appointed Chitra Varma as his chief advisor.
4. Tenali Ramakrishna prepared horrible, spicy, salty, and undercooked food for the emperor.

Vocabulary

Look at these sentences. Observe the underlined words.
a. He took some irrelevant decisions.
b. With this the people grew impatient.
‘Here the prefixes such as “ir”, “il”, and “im” are fixed before the root/stem;
“relevant – irrelevant”
“logical – illogical”
“patient – impatient”

These prefixes change the meanings of words. By adding a prefix to a word we get another word. See more examples below.
AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 3
happy – unhappy
sufficient – insufficient
mature – immature
agree – disagree
responsible – irresponsible

The prefixes ir, il, im, un, in, and dis are used to form the opposites in meaning. Now, write the opposite words using the prefixes mentioned above.

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 4
Word – Opposite Word
balance – imbalance
legal – illegal
regular – irregular
acceptable – unacceptable
comfort – discomfort
valid – invalid

In the story we have come across words such as encouragement, announcement, arrangement, confusion, etc. You might have observed that the part ‘-ment’ or ‘-sion ’ is added to the words encourage, announce, arrange and confuse in the end. The suffixes, ‘-ment’ and ‘-sion’ are used to form nouns out of verbs. Now, write the noun forms of the following using ‘-ment’, ‘-sion’.

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 5
Word – Noun Form
improve – improvement
discuss – discussion
depress – depression
confuse – confusion
enrich – enrichment
treat – treatment

You have come across certain words in the lesson such as “painter, grateful” etc; You can extract some other words from the parent words like

painter – pain
grateful – rate

Now make meaningful words from the following words.
1. reward – read, red, raw, war, ward, draw, drew, rear, rare, dew, wed, drawer, dear, ear, rear, rare
2. blessings – less, sing, sin, big, bin, leg, gel, bliss, sign
3. snapped – nap, pan, pen, pad, sad, sap, pane, sand, span
4. personal – son, reason, pearl, nose, slope, pole, sole, soap, lose, also, rope, rose
5. matter – mat, rat, art, arm, ram, mar, meat, team, tame, rate, tear, mate, mart, tram
6. continue – tin, cone, toe, tie, cut, cute, tune, cot, coin, none, note, nut, tone, not, ton, on, one, once
7. assure – sure, use, urea, ass
8. gold – old, go, god
9. properly – rope, role, pore, pole
10. slowly – low, sly, owl, sow

Grammer

A. i) Read the following sentences.

1. The emperor was pleased with Chitra Varma’s paintings.
2. Oh! What a beautiful picture!
3. Make an announcement.
4. Are you crazy, Ramakrishna?

The first one is a “Declarative sentence” as it makes a statement.
The second one is an “Exclamatory sentence” as it expresses a strong feeling.
The third one is an “Imperative sentence” as it gives an order.
The fourth one is ari “Interrogative sentence” as it asks a question.

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna

ii) Read these sentences and write “D” for declarative, “E” for exclamatory, “In” for interrogative and “Im” for imperative in the brackets given.
1. Please show me the way to the library. (Im)
2. What a great painting this is! (E)
3. Where can I find a sweet shop? (In)
4. Mother looks after her children carefully. (D)
5. What happened? (In)

iii) Read the following sentences and put full stops, question marks and exclamation marks wherever necessary.
AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 6
1. He got a good job
2. Did he get good job
3. What a great job
4. Get a great job first
5. What is your job
6. Have you finished your job
7. I am ready to finish my job
8. Is he ready to finish his job
Answers:
1. He got a good job. (full stop)
2. Did he get good job? (question mark)
3. What a great job! (exclamation mark)
4. Get a great job first. (full stop)
5. What is your job? (question mark)
6. Have you finished your job? (question mark)
7. I am ready to finish my job. (full stop)
8. Is he ready to finish his job? (question mark)

B. Conjunctions:

Read the following sentences and observe how the words in bold are used.
1. It is all your affection and encouragement.
2. People grew impatient but the emperor continued to support him.
3. Bring him at once or you will be punished.

The word “and” joins two words or phrases.
The word “but” is used to suggest a contrast.
The word “or” presents an alternative or a choice. These are called conjunctions. Conjunctions “and, but, or, yet, so, ” etc are joining words. They help to join two or more words, phrases or clauses.

Ex : He rushed into the hall and sat in the sofa.
It is raining heavily, so we won’t go out to play.

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna

C. Read the passage carefully and underline the conjunctions.

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 7
Meena wants to go on holidays. Everyone is happy but Meena is not. She is unhappy because there are some baby sparrows in the nest on a corner of a room. If all the members of the family went out of the house, it would-be locked up, then how would the mother sparrow or the father sparrow feed their babies?

Meena’s mother said that they would remove all the things from the room, open the windows and lock the door from outside. Thus the house would be safe and sparrows could come anifl go freely. So Meena is happy again.
Answer
Meena wants to go on holidays. Everyone is happy but Meena is not. She is unhappy because there are some baby sparrows in the nest on a corner of a room. If all the members of the family went out of the house, it would be locked up, then how would the mother sparrow or the father sparrow feed their babies?

Meena’s mother said that they would remove all the things from the room, open the windows and lock the door from outside. Thus the house would be safe and sparrows could come and go freely. So Meena is happy again.

Writing

I. Conversation : Read the following conversation and write in the form of a narrative/story.

Ramakrishna : Good morning gentlemen.
People : Good morning Raman.
Ramakrishna : Happy to see you all. But.. .what made you come here?
People : You know everything. The emperor appointed Chitra Varma as his chief advisor.
Ramakrishna : Yes, I know.
People : The decisions.. ..the decisions taken by him confused us.
Ramakrishna : Yes, I’m also thinking about it.
People : Please do something and save us.
Ramakrishna : Don’t worry. I have an idea. Your problem will be solved soon.
People : Thank you Ramakrishna, we know you can do this.
Answer:
One day some people of Vijayanagara kingdom came to meet Tenali Ramakrishna. They both greeted each other. Ramakrishna told them that he was happy to see them all and asked them what was the reason for their arrival. The people told him that the emperor had appointed Chitra Varma as his chief advisor. They also informed him that the decisions taken by Chitra Varma were confusing them.

Then Ramakrishna told them that he was also thinking about it. They requested Ramakrishna to do something and save them. Ramakrishna told them not to worry. He also said that he had an idea. He assured them that their problem would be solved soon. The people felt happy and thanked Ramakrishna.

II. Read the following part from your lesson and write it in the form of a story.

Chitra Varma : Welcome, Your Highness.

Srikrishna Devaraya : Oh! What a beautiful picture! It looks so natural.

The courtiers : Yes, Your Majesty, really it is wonderful.

Srikrishna Devaraya : No painter on earth can compete with you, Chitra Varma. You are a unique painter. You shall be rewarded. What do you say Ramakrishna?

Ramakrishna : Yes, My Lord! Such an artist he is! He deserves your encouragement.

Srikrishna Devaraya : That’s great! Let him ask for anything. He shall get whatever he wants.

Chitra Varma : Thank you very much Your Majesty. It’s all your affection and encouragement.

Srikrishna Devaraya : Your presence on the council of advisors will be a boon to the Vijayanagar empire.

Chitra Varma : I am afraid I am not, Your Majesty! Your blessings are worth the strength of thousand elephants.

Srikrishna Devaraya : No, a great painter like you should be rewarded in a fitting manner.

Chitra Varma : Your highness is merciful.

Srikrishna Devaraya : (Thinking for a while, he snapped his fingers) You will be my chief advisor from today onwards.

Chitra Varma : What! I can’t understand, how I am blessed, Your Majesty.

Ramakrishna : Excuse me, Your Majesty! He is talented in painting but…would it be a right decision to make him…?

Srikrishna Devaraya : The emperor of Vijayanagar knows what is right. Minister, please make an announcement that Chitra Varma will be the chief advisor hereafter in our court.

Minister : Yes, Your Majesty. 1 will make an announcement in our empire.
Answer:
One day the king Srikrishna Devafaya visited the house of Chitra Varma, a great painter in his empire. Chitra Varma welcomed the king in great respect. The king saw a beautiful picture painted by Chitra Varma and praised that the painting was so natural. The courtiers who came along with the king appreciated that the painting , was wonderful. The king praised that Chitra Varma was a unique painter and no painter ‘ on earth could compete with him. He should be rewarded. The king asked for the opinion of Tenali Ramakrishna. Ramakrishna said that Chitra Varma deserved the king’s encouragement.

The king said that he would get whatever he wanted. Chitra Varma thanked the king for his affection and encouragement. The king felt that Chitra Varma’s presence on the council of advisors would be a boon to the Vijayanagar empire. But Chitra Varma was not willing and said that the king’s blessings were worth the strength of a thousand elephants. But the king wanted to reward him in a fitting manner as he was a great painter. The king appointed him his chief advisor. Chitra Varma felt that he was greatly blessed.

Ramakrishna opposed it by saying that Chitra Varma was talented in painting but he could nqt be a good advisor. It might not be a good decision to make him the chief advisor. But the king said that he knew what was right and asked the Minister to make an announcement that Chitra Varma would be the chief advisor thereafter in his court. The Minister said that he would make an announcement in the empire.

Talking Time

A. Announcement

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 8
Where do we listen to announcements?
At the bus station, at the railway station, at some public places, etc.
Here are some model announcements usually made at the Railway Station.

1. Your attention please… Train Number one, seven, four, eight, seven… the Tirumala Express… from Visakhapatnam to Kadapa is arriving on platform Number 7.
2. A Key Chain of a car has been found on the platform No.l. Whoever has lost it can collect it from Mrs Bharathi at the counter No 2.
3. Departmental announcement… Mr. Rajasekhar, the Electrical Engineer, is requested to attend to A-4 compartment of the Konark Express immediately.

B. Role play

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 9
Help me please…

Policeman : Who are you, little girl?

Girl : My name is Rishika.

Policeman : Why are you standing alone?

Rishika : I am lost.

Policeman : Oh, is it? Why did you come here?

Rishika : I came here with my parents to see the exhibition.

Policeman : Come, I will help you find your parents.

Rishika : Sorry, my parents and teachers advised me not to go with strangers Policeman : I will not harm you. You know, I am a policeman.

Rishika : OK. Then, I will give you my father’s phone number. Please call him.

Policeman : That’s good. 1 will call your father.

Rishika : Thank you sir.

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna

C. Pronunciation (‘P”, “Ph” (/p/, /f/))

Read the following words aloud.

/p/ sound /f /sound
parrot photo
parcel phone
pencil phrase
happy alphabet
paper elephant
apply dolphin
cap graph
trip epitaph
tap triumph

Read the following sentences aloud.
Put the photo on the pad.
Pay up the penalty.
Post the parcel to the photographer.
Paste the graph and paint it.
Pay the price for a dolphin show.

Listeing

Listen to the following story carefully.

An Unexpected Reward

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 10
A boy and his grandfather were walking along a road. They passed a farmer working hard on his land. The farmer had taken his ragged shirt and worn-out shoes and placed them at the road’s edge. The boy felt he could play a little mischief. He said to his granddad, “Grandpa, why don’t we take the man’s shoes and shirt and hide behind the tree? When he finds them missing, he will be so angry. And we’ll have the fun of watching his expressions.”

But the grandfather replied, “Grandson, never make fun of a farmer or hurt his heart. If you just want to feel happy, then let’s do this instead. Take some coins and place them in each of the farmer’s shoes. Then we’ll hide behind the tree and see what happens.”

The boy did as his grandfather told him. After a little while, the farmer came for his clothes. He put on his ragged old shirt and his feet into his shoes. He felt something hard inside. Annoyed, he took the shoes off to see what it was. There were coins. He was astonished.

He looked up at the sky and fell to his knees.

“Oh! Great God! You are so kind. When I needed money the most, you sent me this gift. Thank you, God! And thank you to whoever you made your messenger.”

Then he walked home, overjoyed. The boy felt proud and happy. His grandfather said, “Grandson, the path to happiness is to make others happy.”

Answer the following questions:
1. Identify the characters in the story.
Answer:
A boy, his grandfather and a farmer are the characters in the story.

2. What did the boy and his grandfather observe while walking along the road?
Answer:
The boy and his grandfather observed a farmer working hard on his land. The farmer had taken his ragged shirt and worn-out shoes and placed them at the road’s edge.

3. Why did the boy want to hide the farmer’s clothes?
Answer:
The boy wanted to hide the farmer’s clothes to have fun of watching his expressions.

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna

4. What advice did the grandfather give to the boy?
Answer:
The grandfather advised the boy not to make fun of a farmer or hurt his heart. He asked the boy to keep some coins in each of the farmer’s shoes to make him happy.

5. What was the unexpected reward to the farmer?
Answer:
The unexpected reward to the farmer was some coins kept by the boy on his grandfather’s advice.

The Thousand Gold Coins

Comprehension

Question 1.
Why did the neighbour throw a bag of coins to Nasruddin?
Answer:
The neighbour wanted to know whether Nasruddin would accept the coins or not if the coins were lesser than one thousand.

Question 2.
Who was the clever man in the story?
Answer:
The clever man was Nasruddin.

AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna

Question 3.
The neighbour said sorry. Why?
Answer:
The neighbour lost the case. Nasruddin gave him his money, coat and horse. The neighbour said that he was sorry for interfering in his business.

Fun Time

Laugh and make others laugh.
AP Board 6th Class English Solutions Chapter 1 Clever Tenali Ramakrishna 11
Friend : You are so lucky, no mosquito can bite you now.

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Joke:
Teacher: If I give an apple to you and your sister, and ask you to share it equally, which part of the apple will you get?

I : The whole apple.

Teacher : How?

I : My sister doesn’t like apples.

You can use any two-letter word in the first step, three- letter word next step, four letter word last.
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Study Skills

Study the bill and answer the questions.
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1. What is the name of the shop?
Answer:
Mahalakshmi Stores

2. What is the customer’s name?
Answer:
Srinivas

3. When did he buy the above things?
Answer:
On 21st of July, 2020

4. How much money did he pay to the shopkeeper?
Answer:
Rs. 275

5. How many pens did the customer buy?
Answer:
10

6. W hat is the price of one notebook?
Answer:
Rs. 40

Clever Tenali Ramakrishna Summary

Chitra Varma was a great painter. On seeing his great painting, one day Srikrishna Devaraya appointed him the chief advisor. Chitra Varma accepted it unwillingly. Tenali Ramakrishna also opposed it. Thus the announcement was made in the kingdom.
As Chitra Varma had no experience in handling the court matters, the people of the kingdom became impatient and they met and requested Tenali Ramakrishna for solving the problem.

According to his plan, Ramakrishna invited the emperor to bless his son on his naming ceremony. He made all the arrangements. The emperor attended the ceremony and blessed the baby boy. Later, lunch was arranged and dishes were served. The dishes were horrible. They were not good to eat. They were not cooked properly. Then Ramakrishna brought the cook before the emperor and said that he appointed the best carpenter in the kingdom to cook dishes.

The emperor asked Ramakrishna how an excellent carpenter could become a cook. Ramakrishna replied that if a painter could be the emperor’s chief advisor, why a carpenter could not be a cook. The emperor thought and realized his mistake in appointing Chitra Varma his chief advisor. He sent Chitra Varma back to his previous job and thanked Ramakrishna. All praised Tenali Ramakrishna’s presence of mind.

The Thousand Gold Coins Summary

Nasruddin was clever. He had a habit of praying aloud. Every day he used to pray for one thousand gold coins. In his prayer he said that if God gave him one coin lesser, he would not accept the money.

A neighbour who heard the prayer of Nasruddin wanted to test Nasruddin and threw a bag of 999 coins into Nasruddin’s house. Nasruddin counted the coins. Though there were no 1000 coins, he accepted the money. The neighbour asked him how he could accept the money when they were 999 coins only. He said that he was the person who had thrown the money into his house.

Nasruddin said that he would accept the money as it was given to him by God. But the neighbour wanted his money back. Nasruddin refused to give it back. The neighbour asked Nasruddin to come to the court for justice. He offered his coat and his horse to come to the court. He told everything to the judge. Nasruddin played a trick and made the judge believe that the money, the coat and the horse belonged to him.

The neighbour felt sorry and said that he would never interfere in Nasruddin’s matters. Then Nasruddin gave him his money, horse and coat back.