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Students get through AP Inter 2nd Year Chemistry Important Questions 9th Lesson Biomolecules which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 9th Lesson Biomolecules

Very Short Answer Questions

Question 1.
Define Carbohydrates.
Answer:
The compounds which are primarily produced by plants and form a very large group of naturally occurring organic compounds are called Carbohydrates.
Eg: Glucose, Fructose, Starch. Carbohydrates are the polyhydroxy aldehydes (or) ketones.

Question 2.
Name the different types of carbohydrates on the basis of their hydrolysis. Give one example for each.
Answer:
On the basis of hydrolysis, carbohydrates are classified as

1. Monosaccharides, Eg : Glucose, fructose No saccharides
2. Oligosaccharides, Eg : Sucrose, maltose two monosaccharides
3. Polysaccharides, Eg : Starch, cellulose large number of monosaccharides.

Question 3.
How is Glucose prepared ? [IPE – 2014]
Answer:
Glucose is prepared by the hydrolysis of starch in presence of a little acid.

Question 4.
Why are sugars classified as reducing and non-reducing sugars ?
Answer:

• Carbohydrates that reduce Fehling’s reagent, Tollen’s reagent are called reducing sugars.
  Eg: Glucose.
• Carbohydrates that doesnot reduce Fehling’s reagent, Tollen’s reagent are called non reducing sugars.
  Eg : Sucrose.

Question 5.
What do you understand by invert sugars ?
Answer:
During the hydrolysis of sucrose there is a change in the sign of rotation, from dextro (+) to laevo (-) and the product is named as invert sugar.

Question 6.
What do you mean by essential amino acids ? Give two examples for non essential amino acids ? [IPE 2014] [T.S. Mar. 16]
Answer:
Essential amino acids : The amino acids which cannot be synthesized in the body and must be obtained through diet are known as essential aminoacids.
Eg : valine, leucine etc.
Examples of non essential amino acids are Alanine, Glycine, Aspartic acid.

Question 7.
What is zwitter ion ? Give an example. [IPE 2015 (AP)]
Answer:
Zwitter ion: In aqueous solution of amino acids, the carboxyl group can lose a proton and amino acid can accept that proton to form a dipolar ion. This ion is called as zwitter ion.

Question 8.
What are reducing sugars ?
Answer:
Carbohydrates that reduce Fehling’s reagent, Tollen’s reagent are called reducing sugars.
Eg: glucose.

Question 9.
What are proteins ? Give an example.
Answer:
Proteins : A poly peptide with more than hundred amino acid residues, having molecular mass higher than 10,000 units is called a protein.
Eg : keratin, myosin, insulin.

Question 10.
What are the components of a nucleic acid ?
Answer:

• Nucleic acids are long chain polymers of nucleotides i.e., poly nucleotides.
• Nucleic acids are constituted by pentose sugar, phosphoric acid, and nitrogenous hetero cyclic base (purine (or) pyrimidine).

Question 11.
What are essential and non – essential amino acids ? Give one example for each. [IPE – 2016 (TS)]
Answer:

• Essential amino acids : The amino acids that cannot be synthesised by the body and must be supplied in the diet are called essential amino acids.
  Eg : valine, leucine, phenyl alanine etc.
• Non-essential amino acids : Other amino acids synthesised by the tissues of the body are called non-essential amino acids..
  Eg: Glycine, alanine etc.

Question 12.
Differentiate between globular and fibrous proteins.
Answer:
Globular proteins

1. In this proteins the chains of poly peptides coil around to give a spherical shape.
2. hese are soluble in water.
3. Eg : Insulin

Fibrous proteins

1. In this proteins the poly peptides run parallel and are held together by hydrogens and disulphite – bonding.
2. These are insoluble in water.
3. Eg: keratin

Question 13.
Why are vitamin A and vitamin C essential to us ? Give their important sources.
Answer:
Vitamin A and Vitamin C are essential to us.
Explanation :

• Deficiency of vitamin A causes night blindness, redness in eyes, xerophthalnia.
• Deficiency of vitamin C causes pernicious anaemia (RBC deficient in haemoglobin).

Sources:

• Vitamin – A : Fish liver oil, carrots, butter and milk.
• Vitamin – C : Citrous fruits, amla, green leafy vegetables

Question 14.
What do you understand from the names (a) aldo pentose and (b) ketoheptose ?
Answer:
a) Aldo pentose : If a monosaccharide contains 5 carbon atoms with aldehyde group then it is known as aldo pentose.
b) Ketoheptose : If a monosaccharide contain seven carbons with a ketone group then it is called ketoheptose.

Question 15.
What are anomers ?
Answer:
Anomers : The two isomeric structures of a compound which differ in configuratiofi at C-l only are called Anomers.

Question 16.
What are amino acids ? Give two examples.
Answer:
The organic compounds which contain amino (-NH₂) functional group and carboxyl (-COOH) functional group are called amino acids.
Eg : Glycine, Alanine etc.

Question 17.
What are fibrous proteins ? Give examples.
Answer:
When the poly peptide chains run parallel and are held together by hydrogen and disulphide bonds then fibre – like structure is formed. These are called fibrous proteins. These are insoluble in water.
Eg : keratin, myosin.

Question 18.
What are globular proteins ? Give examples.
Answer:
When the chains of polypeptides coil around to give a spherical shape then globular proteins are formed. These are usually soluble in water.
Eg : insulin and albumins.

Short Answer Questions

Question 1.
Explain the denaturation of proteins.
Answer:
The phenomenon of disorganization of native protein structure is known as denaturation. Denaturation results in the loss of secondary, tertiary and quaternary structure of proteins. This involves a change in physical, chemical and biological properties of protein molecules.
Agents of denaturation:
Physical agents – Heat, violent shaking, X – rays, UV – radiation.
Chemical gents – Acids, alkalies, organic solvents, urea, salts of heavy metals.

Question 2.
What are enzymes ? Give examples ?
Answer:
The group of complex proteinoid organic compounds, elaborated by living organism which catalyse specific organic reactions are called enzymes.
Eg.: Lipases, Rennin, Maltase, Invertase etc. Practically all biological processes such as digestion, respiration etc., are carried on through the agency of enzymes.
Enzymes may be defined as biocatalysts synthesised by living cells.
The functional unit of enzyme is known as holo enzyme made up at apo enzyme (protein part) and co enzyme (non-protein part).

Question 3.
Write notes on vitamins.
Answer:
Vitamin is defined as an “accessory food factor which is essential for growth and healthy maintenance of the body.”
Classification : Vitamins are broadly classified into two major groups.
a) the fat soluble Eg : vitamin A, D, E and K.
b) water soluble Eg : vitamin B – complex and C.

Question 4.
What are harmones ? Give one example for each. [IPE – 2016 (TS)] [Mar. 14]

1. steroid hormones
2. Poly peptide hormones and
3. amino acid derivatives.

Answer:
Hormones: Hormone is defined as an “organic compound synthesised by the ductless glands of the body and carried by the blood stream to another part of the body for its function”.
Eg : testosterone, oestrogen.

1. Example for steroid hormones : Testosterone, oestrogen
2. Example for poly peptide hormones: Insulin
3. Example for Amino acid derivative : Thyroidal hormones thyroxine.

Question 5.
Write the importance of carbohydrates.
Answer:
Importance of carbohydrates:

• Carbohydrates are essential for life of plants.
• Carbohydrates are used as storage molecules as starch in plants.
• Cell wall of plants is made up of cellulose.
• Carbohydrates are also essential for life of animals. Carbohydrates are used as storage molecules as glycogen in animals.
• Carbohydrate source honey is used for a long time as an instant source of energy.

Question 6.
Give the sources of the following vitamins and name the diseases caused by their- deficiency [T.S. Mar. 17] [IPE AP & TS (Mar. 15) BMP, 2016 (AP]
(a) A
(b) D
(c) E and
(d) K
Answer:
Vitamin : A
Sources : Fish oils, liver, kidney
Deficiency diseases : Night blindness, Redness in eyes.

Vitamin : D
Sources : Fish oils, butter, milk, egg
Deficiency diseases : Rickets in children, osteomalacia in adults

Vitamin : E
Sources : Wheat germ oil, egg yolk, green vegetables
Deficiency diseases : Sterility

Vitamin : K
Sources : Green vegetables, Intestinal flora.
Deficiency diseases : Blood coagulation is prevented

Question 7.
Write notes on proteins. [IPE – 2016 (TS)]
Answer:
Proteins are polypeptide chains of amino acids.
Classification of Proteins : Proteins can be classified into two types on the basis of their molecular shape.
a) Fibrous Proteins : These are fibre like proteins, the polypeptide chains are parallel which are held together by hydrogen and disulphide bonds. These are insoluble in water.
Ex : Keratin present in hair, wool, silk etc., and myosin present in muscles.

b) Globular proteins: In these proteins, the polypeptide chains coil around to give a spherical shape. These are soluble in water.
Ex : insulin and albumin.

The structure of proteins is explained in four different levels
a) Primary structure
b) Secondary structure
c) Tertiary structure
d) Quaternary structure

Denaturation of proteins : A protein in a biological system with a specific structure and biological activity is called a native protein. The process in which a protein loses its activity when subjected to heating, change in pH, addition of reagents is called denaturation of protein. Denaturation may be reversible or irreversible.
Ex : Coagulation of egg white on boiling is an irresersible denaturation.
Renaturation is the reverse process of denaturation.

Question 8.
Write about polysaccharides with starch and cellulose as examples.
Answer:
Polysaccharides : The saccharides which on hydrolysis to form large number of monosaccharides are called polysaccharides.
Eg : Starch and cellulose

Starch :

• Starch is the most important dietary source for human beings.
• Vegetables, roots, cereals are important sources of starch.
• It is a polymer ofα – glucose.
• It is constituted by two components Amylose and amylopectin.

Amylose :

• It constitutes 15 – 20% of starch.
• Amylose is water soluble component.
• Amylose is a branched chain with 200 – 1000 α – D – glucose units held by C – 1 to C – 4 glycosidic linkage.

Amylopectin :

• Amylopectin constitutes 80 – 85% of starch.
• It is a branched chain polymer of a – glucose units in which chain is formed by C. – 1 to C – 4 glycosidic linkage whereas branching occurs by C – 1 to C – 6 glycosidic linkage.

Cellulose :

• Cellulose occurs in plants and it is the most abundant organic substance.
• It is a major constituent of cell wall of plant cells.
• Cellulose is a straight chain polysaccharide composed only of β – D – glucose units . which are joined by glycosidic linkage between C – 1 of one glucose and C – 4 of the next glucose.

Question 9.
Write notes on the functions of different hormones in the body. [IPE 2014]
Answer:
Functions of Hormones:

• Hormones help to maintain the balance of biological activities in the body.
• Insulin maintains the blood glucose level within the limit.
• Growth hormones and sex hormones play role in growth and development.
• Low level of thyroxine (produced from thyroid gland) causes hypothyroidism. High level of thyroxine causes hyper thyroidism.
• Gluco corticoids control the carbohydrate metabolism, modulates the inflammatory reactions.
• The mineralo corticoids control the level of excretion of water and salt by the kidney.
• Adrenal cortex does not function properly then results in Addison’s disease.
• Hormones released by gonads are responsible for development of secondary sex characters.
• Testosterone is responsible for development of secondary sex hormone produced in male.
• Estradiol is the main female sex hormone responsible for development of secondary female characterstics like control of menstrual cycle.
• Progesterone is responsible for preparing the uterus for implantation of fertiised egg.

Question 10.
Write the sources of vitamin and diseases due to vitamin deficIency. [AP Mar. 20]
Answer:

Students get through AP Inter 2nd Year Chemistry Important Questions 8th Lesson Polymers which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 8th Lesson Polymers

Very Short Answer Questions

Question 1.
What are polymers ? Give example.
Answer:
Polymer : A large molecular weight complex compound which is formed by the repeated combination of simple units (monomers) is called polymer. E.g.: Nylon 6, 6, Buna-S,. rubber etc.

Question 2.
What is polymerization ? Give an example of polymerization reaction. [IPE 2014 Mar. 14]
Answer:
Polymerization: The process of formation of polymers from respective monomers is called polymerization.
(or)
A large molecular weight complex compound which is formed by the repeated combination of simple units is called polymer. This process is called polymerisation.
E.g. : Formation of polyethene from ethene and reaction of hexamethylene diamine and adipic acid leading to the formation of Nylon 6, 6 are examples of two different types of polymerisation reactions.

Question 3.
What is addition polymer ? Give example. [IPE Mar. 2015 (TS)]
Answer:
Addition Polymer: The polymer which is formed by the addition of molecules of monomers of same type (or) different type containing double bonds is called addition polymer.
E.g. : Polyethene, poly acrylonitrile.

Question 4.
What is condensation polymer ? Give example.
Answer:
Condensation polymer : The polymer which is formed by the condensation reaction between molecules having more than one functional group is called condensation polymer.
E.g. : Nylon 6, 6, Poly ethylene terephthalate.

Question 5.
What are homopolymers ? Give example.
Answer:
Homopolymers : The polymers which are formed by the polymerisation of a single monomeric species are known as homopolymers.
E.g.: Polyethene, Poly styrene.

Question 6.
What are copolymers ? Give example. [IPE 2014]
Answer:
Copolymers: A polymer which is formed by the polymerisation of two (or) more chemically different types of monomer units is called copolymer.
E.g.: Butadiene – Styrene polymer (Buna-S)

Question 7.
What are elastomers. Give example.
Answer:
Elastomers : There are rubber like solids with elastic properties. In elastomers the polymer chains are held together by the weak enter molecular forces.
E.g. : Buna – S, Buna – N etc.

Question 8.
What are fibres ? Give example.
Answer:
Fibres : Fibres are the thread forming solids which possess high tensile strength.
E.g. : Nylon 6, 6, polyesters.

Question 9.
What are thermoplastic polymers ? Give example.
Answer:
Thermoplastic Polymers: These are the linear (or) slightly branched long chain molecules capable of softening on heating and hardening on cooling.
E.g. : Polystyrene, polythene.

Question 10.
What are thermosetting polymers ? Give example.
Answer:
Thermo setting polymers: These polymers are cross linked (or) heavily branched molecules which on heating undergo extensive cross linking in moulds and again become infusible.
E.g. : Bakelite, urea-formaldehyde resin etc.

Question 11.
What is Ziegler – Natta catalyst ?
Answer:
A mixture of Tri alkyl aluminium and titanium chloride is called Ziegler – Natta catalyst
E.g. : (C₂H₅)₃ Al + TiCl₄

Question 12.
What are the repeating monomeric units of Nylon 6 and Nylon 6, 6 ?
Answer:
The repeating monomeric units of Nylon – 6 is Capro lactum.

The repeating monomeric units of Nylon 6, 6 are hexamethylene diamine and Adipic acid.
H₂N-(CH₂)₆ – NH₂
Hexamethylene

HOOC – (CH₂)₄ – COOH
Adipic acid

Question 13.
What is the difference between Buna – N and Buna – S ?
Answer:
Buna – N : Buna – N is the copolymer which is formed by the polymerisation of 1, 3 – Butadiene and acrylonitrile.

Buna – S : Buna – S is the copolymer which is formed by the polymerisation of 1, 3 – Butadiene and Styrene.

Question 14.
What is PDI (Poly Dispersity Index) ?
Answer:
Poly Dispersity Index (PDI) : The ratio between weight average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{w}}\)) and the number average molecular mass (\(\overline{\mathrm{M}}_{\mathrm{n}}\)) of a polymer is called Poly Dispersity Index (PDI).

Question 15.
What is vulcanization of rubber ? [A.P. Mar. 17] [IPE – 2016 (TS)]
Answer:
Vulcanization of rubber : The process of heating the raw rubber with sulphur (or) with sulphur compounds to improve it’s physical properties is called vulcanization of rubber.

Question 16.
What is biodegradable polymer ? Give one example of a biodegradable polyester ?
Answer:
Biodegradable polymers : The polymers degradable by enzymatic hydrolysis and to some extent by oxidation are called biodegradable polymers.
Eg.: Nylon – 2, Nylon – 6, PHBV, Polyglycolic acid, Polylactic acid etc.

Question 17.
What is PHBV ? How is it useful to man ? [IPE Mar – 2015 (TS), B.I.E, 2016 (TS)]
Answer:
Poly β – hydroxy butyrate – CO – β – hydroxy Valerate (PHBV) : It is a Copolymer of 3 -hydroxy butanoic acid and 3 – hydroxy pentanoic acid.

Properties & Uses : The properties of PHBV vary according to the ratio of both the acids, 3-hydroxy butanoic acid provides stiffness and 3-hydroxy pentanoic acid imparts flexibility to copolymer.
It is used in medicine for making capsules.
PHBV also undergoes degradation by bacteria.

Short Answer Questions

Question 1.
Write the names and structures of the monomers of the following polymers. [A.P. Mar. 17]
i) Buna – S
ii) Buna – N
iii) Dacron
iv) Neoprene
Ans:
i) Buna – S
Monomers : 1, 3 – Butadiene, Styrene
Structure :

ii) Buna-N
Monomers : 1, 3 – Butadiene, Acrylonitrile
Structure : CH₂ = CH – CH = CH₂, CH₂ = CH – CN

iii) Dacron
Monomers : Ethylene glycol, Terephthalic acid
Structure : HOCH₂ – CH₂OH, HOOC COOH

iv) Neoprene .
Monomers : 2 – chloro – 1, 3 – Butadiene
Structure :

Question 2.
Define thermoplastics and thermosetting polymers with two examples of each.
Answer:
Thermoplastic Polymers: These are the linear (or) slightly branched long chain molecules capable of softening on heating and hardening on cooling. E.g.: Polystyrene, polythene.
Thermo setting polymers: These polymers are cross linked (or) heavily branched molecules which on heating undergo extensive cross linking in moulds and again become infusible.
Eg.: Bakelite, urea-formaldehyde, resin etc.

Question 3.
Distinguish between the terms homo polymer and co polymer. Give one example of each.
Answer:
Homopolymers : The polymers which are formed by the polymerisation of a single monomeric species are known as homopolymers.
E.g.: Polyethene, Poly styrene.
Copolymers: A polymer which is formed by the polymerisation of two (or) more chemically different types of monomer units is called copolymer.
E.g.: Butadiene – Styrene polymer (Buna-S)

Question 4.
Explain the purpose of vulcanization of rubber. [T.S. Mar. 17]
Answer:

1. Natural rubber becomes soft at high temperatures and brittle at low temperatures. It shows high water absorption capacity.
2. Natural rubber is soluble in non-polar solvents and is non-resistant to oxidising agents.
3. To improve these physical properties rubber can be vulcanised.
4. The process of heating the raw rubber with sulphur (or) with sulphur compounds to improve it’s physical properties is called vulcanisation of rubber.
5. The vulcanized rubber has improved properties like elasticity, minimum water absorbing tendency, high resistance to chemical oxidation as well as organic solvents.

Question 5.
Write the names and structures of the monomers used for getting the following polymers [A.P. Mar. 17; Mar. 14]
i) Polyvinyl chloride
ii) Teflon
iii) Bakelite
iv) Polystyrene.
Answer:
i) Polyvinyl chloride .
Monomer: Vinyl chloride
Structure : CH₂ = CH – Cl

ii) Teflon
Monomer: Tetrafluoro ethylene
Structure : CF₂ = CF₂

iii) Bakelite [IPE 2015 (AP), 2014, B.M.P, 2016 (AP)
Monomers : Phenol, Formaldehyde
Structure :

iv) Polystyrene
Monomer : Styrene
Structure :
q

Question 6.
Name the different types of molecular masses of polymers.
Answer:
The different types of important molecular masses of polymers are

1. Number average molecular mass (\(\overline{\mathbf{M}}_{\mathrm{n}}\))
2. Weight average molecular mass (\(\overline{\mathbf{M}}_{\mathrm{w}}\))

Question 7.
What is natural rubber ? How does it exhibit elastic properties ?
Answer:

1. Natural rubber is a polymer and possesses elastic properties.
2. It is an elastomer and it is manufactured from rubber latex. Latex is a colloidal dispersion of rubber in water.
3. Natural rubber may be considered as a linear polymer of isoprene. It is also called as cis-1, 4-Poly isoprene.
4. The cis poly isoprene molecule consists of various chains held together by weak vander waals interactions and has a colloid structure. Thus it stretches like a spring and exhibits elastic properties.

Question 8.
Give the structure of nylon 2 – nylon 6 ?
Answer:
Nylon 2 – Nylon 6 :
It is an alternating polyamide copolymer of glycine and amino caproic acid (H₂N – (CH₂)₅ – COOH). It is a biodegradable polymer.
Structure of Nylon 2 – Nylon – 6

Question 9.
Explain copolymerization with an example. [IPE Mar – 2015 (AP), B.I.E.]
Answer:
Copolymers: A polymer which is formed by the polymerisation of two (or) more chemically different types of monomer units is called copolymer. E.g.: Butadiene-Styrene polymer(Buna-S) The process of formation of copolymer is called copolymerisation.
E.g. : Buna – S : Buna – S is the copolymer which is formed by the polymerisation of 1, 3 – Butadiene and Styrene.

Question 10.
What are LDP and HDP ? How are they formed ?
Answer:
Polythenes are two types :

1. LDP (Low Density Polythene),
2. HDP (High Density Polythene)

1) Low Density Polythene : LDP is formed by the polymerisation of ethene under high pressure of 1000 to 2000 atm. at a pressure of 350 to 570K in the presence of traces of dioxygen (or) a peroxide initiator.
Properties :
a) This is obtained through the free radical additon.
b) LDP is chemically inert and tough.
c) LDP is flexible and a poor conductor of electricity.
Uses:
a) It is used in the insulation of electric cables.
b) It is used in the manufacture of pipes in agriculture irrigation.

2) High Density Polythene : HDP is formed by the polymerisation of ethene in a hydro carbon solvent in presence of Ziegler Natta catalyst at a temperature of 333K to 343 K and under a pressure of 6 – 7 atm.
Properties :
a) HDP consists of linear molecules and has high density due to close packing.
b) It is chemically inert and more tough, hard.
Uses:
a) It is used in manufacture of house hold articles like buckets, dustbins etc.
b) It is used in manufacture of pipes.

Question 11.
What are natural and synthetic polymers ? Give two examples of each type.
Answer:
Natural polymers : The polymers which are obtained from natural sources such as plants and animals are called natural polymers.
E.g. : Starch, cellulose, rubber etc.

Synthetic polymers : The polymers which are artificially prepared i.e., man-made are called synthetic polymers.
These have wide applications in daily life as well as in industry.
E.g. : Plastics, Nylon 6, 6, synthetic rubbers.

Students get through AP Inter 2nd Year Chemistry Important Questions 7th Lesson d and f Block Elements & Coordination Compounds which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 7th Lesson d and f Block Elements & Coordination Compounds

Very Short Answer Questions

Question 1.
What are transition elements? Give examples.
Answer:
Transition elements are the elements that contain partially filled d-subshells in their ionic state (or) in their elementary state. Eg : Mn, Co, Ag etc.

Question 2.
Write the general electronic configuration of transition elements.
Answer:
General electronic configuration of transition elements is (n – 1)d^1-10 ns^1-2.

Question 3.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their +3 state ?
Answer:
Mn⁺² has electronic configuration [Ar] 4s⁰ 3d⁵.
Fe⁺² has electronic configuration [Ar] 4s⁰ 3d⁶.

• Mn⁺² has half filled d-subshell which is more stable.
  Hence Mn⁺² compounds are more stable than Fe⁺² toward oxidation to their +3 state.

Question 4.
Why Zn²⁺ is diamagnetic whereas Mn²⁺ is paramagnetic ?
Answer:

• Zn⁺² electronic configuration is [Ar] 4s⁰3d¹⁰. It has no unpaired electrons. So it is dia magnetic.
• Mn⁺² electronic configuration is [Ar] 4s⁰3d⁵. It has five unpaired electrons so it is paramagnetic.

Question 5.
Write ‘spin only’ formula to calculate the magnetic moment of transition metal ions.
Answer:
Spin only formula to calculate the magnetic moment of transition metal ions is
μ = \(\sqrt{n(n+2)}\) BM BM = Bohr Magneton.

Question 6.
Claculate the ‘spin only’ magnetic moment of \(\mathrm{Fe}^{2+}(\mathrm{aq})\) ion. (Board Model Paper) (AP Mar. ’17)
Answer:
Fe⁺² ion has electronic configuration [Ar] 4s⁰3d⁶.
It has four unpaired electrons n = 4.
Spin only magnetic moment μ = \(\sqrt{n(n+2)}\) BM = \(\sqrt{4(4+2)}\) = \(\sqrt{24}\) BM = 4.89 BM

Question 7.
Aqueous Cu²⁺ ions are blue in colour, where as aqueous Zn²⁺ ions are colourless. Why ?
Answer:

• Electronic configuration of Cu⁺² ion is [Ar] 4s⁰3d⁹. It contains one unpaired electron due to presence of this unpaired electron aq. Cu⁺² ions are blue in colour.
• Electronic configuration of Zn⁺² ion is [Ar] 4s⁰3d¹⁰. It contains no unpaired electrons, due to absence of unpaired electrons aq. Zn⁺² ions are colourless.

Question 8.
What are complex compounds ? Give examples.
Answer:
Complex compounds : Transition metal atoms or ions form a large number of compounds in which anions or neutral groups are bound to metal atom or ion through co-ordinate covalent bonds. Such, compounds are called co-ordination compounds (or) complex compounds.
Eg. : [Fe(CN)₆]^4-, [Co(NH₃)₆]³⁺

Question 9.
What is an alloy? Give example.
Answer:
Alloy : An intimate mixture having physical properties similar to that of the metal formed by a metal with other metals or metalloids or sometimes a non metal is called as an alloy.
Eg.: Invar — 64% Fe, 35% Ni, Mn 8cc in traces
Nichrome — 60% Ni,, 25% Fe, 15% Cr.

Question 10.
What is lanthanoid contraction ? (IPE 2016(TS))
Answer:
The slow decrease of atomic and ionic radii in lanthanides with increase in atomic number is called lanthanide contraction.

Question 11.
What is mischmetall ? Give its composition and uses. (AP Mar. ’16)
Answer:
Mischmetall is an alloy which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces of S, C, Ca and Al.

• It is used in Mg- based alloy to produce bullets, shell and lighter flint.

Question 12.
What is a coordination polyhedron ?
Answer:
The spatial arrangement of the ligands which are directly bonded to the central atom or ions defines the geometry about the central atom is called co-ordination polyhedron.
Eg.: Octahedral, tetrahedral etc.

Question 13.
What is a ligand ? (TS Mar. ’18) (IPE 2014)
Answer:
Ligand : A co-ordinating entity which is bound to the central atom by donating electron pairs is called a ligand. Eg.: Cl^–, NH₃, CN^– etc.

Question 14.
What is a chelate ligand ? Give example.
Answer:
The ligands which can form two co-ordinate covalent bonds through two donor atoms are called bidentate ligands. These bidentate ligands are also called chelate ligands.
Eg.: C₂\(\mathrm{O}_4^{-2}\), \(\mathrm{CO}_3^{-2}\) etc.

Question 15.
What is an ambidentate ligand ? Give example. (TS Mar. ’18) (IPE 2016(AP))
Answer:
A unidentate ligand containing two possible donor atoms can co-ordinate through either of donor atoms. Such ligands are called ambidentate ligands. Eg : \(\mathrm{NO}_2^{-}\)

Question 16.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Why ?
Answer:

• [Cr(NH₃)₆]³⁺ is paramagnetic due to the presence of three unpaired electrons.
• [Ni(CN)₄]^2- is diamagnetic due to the absence of unpaired electrons.

Question 17.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why ?
Answer:
Copper exhibits +1 oxidation state most frequently because Cu⁺ has electronic configuration [Ar] 4s⁰3d¹⁰ which has a fulfilled d-subshell which is more stable.

Question 18.
Why do transition elements exhibit more than one Oxidation state (variable oxidation states) ?
Answer:
Transition elements exhibits more than one oxidation state.
Reason :

• The energy difference between (n – 1) d subshell and ns subshell is very low. So both of these subshells complete to lose the electrons.

Question 19.
CuSO₄.5H₂O is blue in colour where as anhydrous CuSO₄ is colourless. Why ? (IPE 2014)
Answer:
CuSO₄.5H₂O is blue in colour whereas anhydrous CuSO₄ is colourless because in the absence of ligand, crystal field splitting doesnot occurs.

Question 20.
Give the formula the following complexes.
a) Potassium hexacyano ferrate (III)
b) Tetra carbonyl nickel (O) (IPE 2016 (TS))
Answer:
a) K₃ [Fe (CN)₆]
b) [Ni(CO)₄]

Question 21.
Scandium is a transition element. But Zinc is not. Why ? (IPE 2014)
Answer:
Scandium has electronic configuration [Ar] 4s²3d¹.
Zinc has electronic configuration [Ar] 4s²3d¹⁰.
Scandium has one unpaired d-electron where as Zinc has zero unpaired d-electrons so Scandium is transition element but Zinc is not.

Question 22.
What are interstitial compounds ? (IPE 2015)
Answer:
Interstitial compounds are non-stichiometric compounds which are formed when non-metals like H, C, N etc., are heated with transition elements. In these compounds the non-metal atoms occupy the interstitial species of metal atoms. Ex : TiH_0.54

Question 23.
In what way is the electronic configuration of transition elements different from non transition elements ?
Answer:

• The general electronic configuration of transition elements is (n – 1) d^1-10 ns^1-2.
• The general electronic configuration of non-transition elements is (n – 1)d¹⁰ ns².

Question 24.
Even though silver has d¹⁰ configuration, it is regarded as transition element. Why ?
Answer:
The outer electronic configuration of silver is – 4d¹⁰5s¹. It is having general electronic configuration of a transition element (n – 1) d^1-10 ns^1-2.
So silver is a transition element.

Question 25.
Though Sc is a transition element, it does not exhibit variable oxidation state. Why ?
Answer:
Scandium has electronic configuration [Ar] 4s²3d¹. It has only one unpaired d-electron. So it does not exhibits variable oxidation state. It exhibits +3 stable oxidation state.

Question 26.
Why is it difficult to obtain M³⁺ oxidation state in Ni, Cu and Zn ?
Answer:

• Ni has electronic configuration [Ar] 4s²3d⁸.
  Ni⁺² has electronic configuration [Ar] 4s⁰3d⁸.
  It is difficult to remove an electron from 3d⁸. So, Ni⁺³ is difficult to obtain. (Ni has high negative enthalpy of hydration).
• Cu has electronic configuration [Ar] 4s¹3d¹⁰.
  Cu⁺ has electronic configuration [Ar] 3d¹⁰.
  It is difficult to remove the electrons from 3d¹⁰ (stable). So, Cu⁺³ is difficult to obtain.
• Zn has electronic configuration [Ar] 4s²3d¹⁰.
  Zn⁺² has electronic configuration [Ar] 4s⁰3d¹⁰.
  It is difficult to remove the electron from 3d¹⁰ (stable). So Zn⁺³ is difficult to obtain.

Question 27.
Cu⁺² forms halides like CuF₂, CuCl₂ and CuBr₂ but not CuI₂. Why ?
Answer:
Cu⁺² forms halides like CuF₂, CuCl₂ and CuBr₂ but not CuI₂ because Cu⁺² oxidises I^– to I₂.
2Cu⁺² + 4I^– → Cu₂I₂ + I₂.

Question 28.
Why is Cr²⁺ reducing and Mn³⁺ oxidizing even though both have the same d⁴ electronic configuration ?
Answer:
Cr⁺² is reducing as its configuration changes from d⁴ to d³, the latter having a half filled t_2g level. On the other hand the change from Mn⁺² to Mn⁺³ results in the half filled (d⁵) configuration which has extra stability.

Question 29.
What is meant by ‘disproportionation’ ? Give an example of disproportionation reaction in aqueous solution.
Answer:
The reactions in which only one element undergo both oxidation as well as reduction are called disproportionation reactions.

• When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation.
  Eg: Cu⁺ ion is not stable in aqueous solution because it undergo disproportionation in aqueous solution.

Question 30.
Why do the transition metals readily form alloys?
Answer:
Because of similar atomic or ionic radii and similar characterstic properties of transition elements alloys are readily formed by these elements.

Question 31.
How do the Ionic character and acidic nature vary among the oxides of first transition series?
Answer:

• As the oxidation number of a metal increases ionic character decreases in case of transition elements. Eg: Mn₂O₇ is a covalent green oil.
• In CrO₃ and V₂O₅ the acidic character is predominant.
• V₂O₅ is however amphoteric though mainlý acidic V₂O₅ reacts with alkali as well as acids to give \(\mathrm{VO}_4^{-3}\) and \(\mathrm{VO}_4^{-3}\).
• CrO is basic.
• Mn₂O₇ gives HMnO₄ and CrO₃ gives H₂CrO₄ and H₂Cr₂O₇.

Question 32.
What are the different oxidation s.tates exhibited by the lanthanoids?
Answer:

• Lanthanoids exhibits +2, +3 states majorly. Sometimes +2 and +4 states exhibited in solid compounds.
• The common oxidation state of lanthanoids is +3.

Question 33.
What is difference between a double salt and a complex compound?
Answer:
Double salts dissociate into simple ions completely when dissolved in water while complex compounds dissociate to give complex ion and the counter ions.

Question 34.
What is meant by coordination number.
Answer:
Co-ordination Number is the number of ligands present around a central metal atom or ion in a complex Ex : The co-ordination number of Co in [Co(NH₃)₃ Cl₃] is six.

Question 35.
Using IUPAC norms write the systematic names of the following.

1. [CO(NH₃)₆]Cl₃
2. [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
3. [Ti(H₂O)₆]³⁺ and
4. [NiCl₄]^2-

Answer:

1. Hexa amine cobalt (III) chloride
2. Diamine chloride (methyl amine) platinum (II) chloride
3. Hexa aquo titanium (III) ion
4. Tetra chloro nickelate (III) ion

Question 36.
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25. (IPE 2016 (AP)) (AP Mar. ’16)
Solution:
With atomic number 25, the divalent ion in aqueous solution will have d⁵ configuration (five unpaired electrons). The magnetic moment, μ is
μ = \(\sqrt{5(5+2)}\) = 5.92 BM

Short Answer Questions

Question 1.
Write the characteristic properties of transition elements.
Answer:
Transition elements exhibits typical characteristic properties.

• Electronic configurations.
• Para and ferro magnetic properties.
• Alloy forming ability.
• Complex forming ability.
• Interstitial compounds.
• Variable oxidation states.
• Formation of coloured hydrated ions.
• Catalytic property.
• Metallic character.

Question 2.
What is lanthanoid contraction ? What are the consequences of lanthanoid contraction ? (AP Mar. ’17) (IPE Mar. 15 (TS), 14 B.I.E)
Answer:
Lanthanoid contraction : The overall decrease in atomic and ionic radii from lanthanum to leutetium is observed in the lanthanoids. This phenomenon is called lanthanoid contraction. It is due to the fact that with every additional proton in the nucleus, the corresponding electron goes into a 4f-subshell. This is too diffused to screen the nucleus as effectively as the more localised inner shell. Hence, the attraction of the nucleus for the outermost electrons increases steadily with the atomic number.

Consequences of Lanthanoid Contraction : The important consequences of lanthanoid contraction are as follows :

i) Basic character of oxides and hydroxides : Due to the lanthanoid contraction, the covalent nature of La-OH bond increases and thus, the basic character of oxides and hydroxides decreases from La(OH)₃ to Lu(OH)₃.

ii) Similarity in the size of elements of second and third transition series : Because of lanthanoid contraction, elements which follow the third transition series are considerably smaller than would otherwise be expected. The normal size increases from Sc → Y → La and disappears after lanthanides. Thus, pairs of elements such as Zr/Hf, Nb/Ta and Mo/w are almost identical in size.
Due to almost similar size, such pairs have very similar properties which makes their separation difficult.

iii) Separation of lanthanoids : Due to lanthanoid contraction, there is a difference in some properties of lanthanoid like solubility, degree of hydration and complex formation. These difference enable the separation of lanthanoids by ion exchange method.

Question 3.
Explain Werner’s theory of coordination compounds with suitable examples. (Mar. ’14) (TS Mar. ’15) (IPE – May. 2015 (TS), ’14 B.I.E)
Answer:
Werner’s theory :
Postulates :

1) Every complex compound has a central metal atom (or) ion.

2) The central metal shows two types of valencies namely primary valency and secondary valency.

A) Primary valency : The primary valency is numerically equal to the oxidation state of the metal. Species or groups bound by primary valencies undergo complete ionization. These valencies are identical with ionic bonds and are non-directional. These valencies are represented by discontinuous lines (….)
Eg. : CoCl₃ contains Co³⁺ and 3Cl^– ions. There are three Primary Valencies or three ionic bonds.

B) Secondary Valency : Each metal has a characteristic number of Secondary Valencies. They are directed in space around the central metal.
The number of Secondary Valencies is called Coordination number (C.N.) of the metal. These valencies are directional in nature.
For example in CoCl₃. 6NH₃
Three Cl^– ions are held by Primary Valencies and 6NH₃ molecules are held by Secondary Valencies. In CuSO₄.4NH₃ complex \(\mathrm{SO}_4^{2-}\) rion is held by two Primary Valencies and 4NH₃ molecules are held by Secondary Valencies.

3) Some negative ligands, depending upon the complex, may satisfy both primary and secondary valencies. Such ligands, in a complex, which satisfy both primary as well as secondary valencies do not ionize.

4) The primary valency of a metal is known as its outer sphere of attraction or ionizable valency while the Secondary valencies are known as the inner sphere of attraction or coordination sphere. Groups bound by secondary valencies do not undergo ionization in the complex.
Example to Clarify Werner’s Theory

Question 4.
Using IUPAC norms write the formulas for the following.

1. Tetrahydroxozincate (II)
2. Hexaaminecobalt (III) sulphate
3. Potassium tetrachloropalladate (II) and
4. Potassium tri(oxalato) chromate (III) (IPE – May. 2015 (AP, TS), 14)

Answer:

1. Tetrahydroxozincate (II) ion – [Zn(OH)₄]^-2
2. Hexa amine cobalt (III) sulphate – [Co(NH₃)₆]₂ (SO₄)₃
3. Potassium tetrachioro palladate (II) – K₂[PdCl₄]
4. Potassium tri(oxalato) chromate (III) – K₃[Cr(C₂O₄)₃]

Question 5.
What are homoleptic and heteroleptic complexes ? Give one example for each.
Answer:
Homoleptic complexes : These are the complexes in which a metal is bound by only ore kind of ligands, eg. : [Co(NH₃)₆]³⁺
Heteroleptic complexes: These are the complexes in which a metal is bound by more than one kind of ligands, eg. : [Co(NH₃)₄Cl₂]⁺

Question 6.
Give the geometrical shapes of the following complex entities

1. [CO(NH₃)₆]³⁺
2. [Ni(CO)₄]
3. [Pt Cl₄]^2- and
4. [Fe(CN)₆]^4-.

Answer:

1. Geometrical shape of [CO(NH₃)₆]³⁺ is octahedral.
2. Geometrical shape of [Ni(CO)₄] is tetrahedral.
3. Geometrical shape of [PtCl₄]^2- is square planar.
4. Geometrical shape of [Fe(CN)₆]^-4 is octahedral.

Question 7.
Using IUPAC norms write the systematic names of the following.

1. K₄[Fe(CN)₆]
2. [Cu(NH₃)₄]SO₄
3. [Ti (H₂O)₆]³⁺ and
4. [NiCl₄]^2-.

Answer:

1. K₄[Fe(CN)₆] – Potassium hexa cyanao ferrate (II)
2. [Cu(NH₃)₄]SO₄ – Tetra amine copper (II) sulphate
3. [Ti (H₂O)₆]³⁺ and – Hexa aquo titanium (III) ion
4. [NiCl₄]^2- – Tetra chloro nickelate (II) ion

Question 8.
Explain the terms
(i) Ligand
(ii) Coordination number
(iii) Coordination entity
(iv) Central metaf atom/ion.
Answer:
i) Ligand : The ions, or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be

a) simple ions such as Cl^–
b) small molecules such as H₂O or NH₃
c) large molecules such as H₂N.CH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃ or even
d) macro-molecules, such as proteins.

On the basis of the numberof donor atoms available for coordination, the ligands can be classified as :

a) Unidentate : One donor atom, Eg. etc.
b) Bidentate : Two donor atoms, Eg.: H₂NCH₂CH₂NH₂
(ethane-1, 2-diamine or en), C₂\(\mathrm{O}_4^{2-}\) (oxalate), etc.
c) Polydentate : More than two donor atoms, Eg. : N(CH₂CH₂NH₂)₃ EDTA, etc.

ii) Coordination number : The coordination number (CN) of metal ion in a complex can be defined as the number of ligands or donor atoms to which the metal is directly bonded.
Eg : In [Ptcl₆]^2-, CN of Pt = 6, In [Ni(NH₃)₄]²⁺, CN of Ni = 4.

iii) Coordination entity: A central metal atoms or ions bonded to a fixed number of molecules or ions (ligands) is known as coordination entity. For example [CoCl₃(NH₃)₃]. Ni(CO)₄], etc.

iv) Central metal atom/ion : In a coordination entity, the atom/ion to which a fixed no. of ions/groups are bound in a definite geometrical arrangement around it is called central metal atom or ion. Eg : K₄[Fe(CN)₆] ‘Fe’ is central metal.

Question 9.
Explain the terms

1. Unidentate ligand
2. Bidentate ligand
3. Polydentate ligand and
4. Ambidentate ligand giving one example for each.

Answer:

1. Unidentate : The negative ion or neutral molecule having only one donor atom is called unidentate ligand
   Eg : etc.
2. Bidentate (or didentate) : The ions or molecules having two donor atoms are called
   
3. Polydentate ligands : Ligands having more than one donor atom in the coordinating group and are capable of forming two or more coordinate bonds with same central atom simultaneously are called poly dentate ligands. Eg : C₂\(\mathrm{O}_4^{-2}\).
4. Ambidentate: Ligand which can ligate through two different atoms is called ambidentate ligand. Eg : \(\mathrm{NO}_2^{-}\), \(\mathrm{SCN}^{-}\) ions. \(\mathrm{NO}_2^{-}\) ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion. Similarly, \(\mathrm{SCN}^{-}\) ion can coordinate through the sulphur or nitrogen atom.
   

Question 10.
Explain the colour and para magnetic property of transition elements.
Answer:
Transition elements having unpaired electrons in d-orbitals are coloured. The colour is due to d-d transitions of electrons. Electrons are excited from lower set of d-orditals to higher set of d- orbitals by absorbing energy from visible light and emits complementary colour which is the colour of the ion. Transition elements are ions having atleast one unpaired electron in d-orbitals are para magnetic. The extent of para magnetic property is expressed in terms of magnetic moment (μ). μ = \(\sqrt{n(n+2)}\) where n is number of unpaired electrons present in d- orbitals.

Question 11.
Explain different types of isomerism exhibited by Co-ordination compounds, giving suitable examples.
Answer:
Isomerism in Co-ordination compounds : Isomers are compounds that have the same chemical formula but different arrangement of atoms. Two principal types of isomerism are known among co-ordination compounds namely stereo isomerism and structural isomerism.

a) Stereoisomerim : Stereoisomerism is a form of isomerism in which two substances have the same composition and structure but differ in the relative spatial positions of the ligands. This can be sub divided into two classes namely.
i) Geometrical isomerism and
ii) optical isomerism

b) Structural isomerism :
i) Linkage isomerism
ii) Co-ordination isomerism
iii) Ionisation isomerism
iv) Hydrate isomerism

a) (i) Geometrical isomerism :

• Geometrical isomerism arises in Co-ordination complexes due to different possible geometric arrangements of the ligands.
• This isomerism found in complexes with Co-ordination numbers 4 and 6.
• In a square planar complex of formula [Mx₂L₂] the two ligands may be arranged in adjacent to each other in a cis isomer (or) opposite to each other in a trans isomer.
  
• Square planar complex of type [MAB XL] shows three isomers two-cis and one trans. (A, B, X, L are unidentate ligands is square planar complex).
• Geometrical isomerism is not possible in tetrahedral geometry.
• In octahedral complex of formula [MX₂L₄] in which two ligands X may be oriented cis or trans to each other.
  
• Another type of geometrical isomerism occurs in octahedral Co-ordination compounds of type [Ma₃b₃] if three donor atoms of the same ligands occupy adjacent positions at the comers of an octahedral face then it is facial (fac) isomer. When the positions occupied are around the meridian of the octahedran then it is meridonial (mer) isomer.
  

a(ii) Optical isomerism : Optical isomerism arises when two isomers of a compound exist such that one isomer is a mirror image of the other isomer. Such isomers are called optical isomers or enantiomers. The molecules or ions that cannot be superimposed are called chiral.

The two forms are called dextro (d) and laevo (l) depending upon the direction they rotate the plane of polarised light in a polarimeter (d rotates to the right, l to the left). Optical isomerism is common in octahedral complexes involving bidentate ligands.

b(i) Linkage isomerism: Linkage isomerism arises in Co-ordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand-NCS^–, which may bind through the nitrogen to give M-NCS or through sulphur to give M-SCN.
eg. : [Mn(CO)₅SCN] and [Mn(CO)₅NCS]

(ii) Co-ordinate isomerism : This type isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
eg. : [CO(NH₃)₆] [Cr(CN)₆] and [Co(CN)₆] [Cr(NH₃)₆]

(iii) Ionisation isomerism: This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion.
eg. : [Co(NH₃)₅SO₄] Br and [Co(NH₃)₅Br]SO₄

(iv) Hydrate isomerism : This form of isomerism is known as ‘hydrate isomerism since water is involved as a solvent. Hydrate isomers differ by whether or not a hydrate molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice.

Question 12.
What is meant by chelate effect ? Give example.
Answer:
When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metql ion, a 5-or 6-membered ring is formed, the effect is known as chelate effect. Example

Question 13.
Discuss the nature of bonding and magnetic behaviour in the following co-ordination entities on the basis of valence bond theory.
(i) [Fe(CN)₆]^4-
(ii) [FeF₆]^3-
(iii) [Co(C₂O₄)₃]^3- and
(iv) [CoF₆]^3-
Answer:
i) [Fe(CN)₆]^4- : In this complex Fe is present as Fe²⁺.
Fe = [Ar] 3d⁶4s²
Outer configuration of Fe²⁺ = 3d⁶4s⁰

CN^– being strong field ligand, pair up the unpaired d electrons Thus, two 3d-orbital are now available for CN^– ions.

Since, all the electrons are paired, the complex is diamagnetic. Moreover (n – 1) d-orbitals are involved in bonding, so, it is an inner orbital or low spin complex.

ii) [FeF₆]^3- : In this complex, the oxidation state of Fe is + 3.

F- is not a strong field ligand. It is a weak field ligand, so no pairing occurs. Thus, 3d- orbitals are not available to take part in bonding.

Because of the presence of five unpaired electrons, the complex is paramagnetic. Moreover, nd-orbitals are involved in bonding, so it is an outer orbital or high spin complex.

iii) [Co(C₂O₄)₃]^3- : In this complex, the oxidation state of Co is + 3.
Outer configuration of Co = 3d⁷ 4s²
Co³⁺ = 3d⁶4s⁰

Oxalate ion being a strong field ligand pair up the 3d electrons, thus two out of the five 3d-orbitals are available for oxalate ions.

Since, all the electrons are paired, this complex is diamagnetic. It is an inner orbital complex because of the involvement of (n – 1) d-orbital for bonding,

iv) [CoF₆]^3- : In this complex, Co is present as Co³⁺
Outer configuration of Co³⁺ = 3d⁶ 4s⁰

Because of the presence of four unpaired electrons, the complex is paramagnetic. Since, nd orbitals take part in bonding, it is an outer orbital complex or high spin complex.

Question 14.
What is spectrochemical series ? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strengths, i.e., increasing crystal field splitting energy (CFSE) values is called spectrochemical series.

The ligands with a small value of CFSE (Δ₀) are called weak field ligands. For such ligands Δ₀ < P where P is the pairing energy. Whereas the ligands with a large value of CFSE are called strong field ligands. In case of such ligands Δ₀ > R

When ligands approach a transition metal ion, the d-orbitals split into two sets (t_2g and e_g), one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy, Δ₀ for octahedral field and Δ_t for tetrahedral field.

If Δ₀ < P (pairing energy), the 4^th electron enters one of the e_g orbitals giving the configuration \(t_{2 g}^3 e_g^1\) thereby forming high spin complex. Such ligands for which Δ₀ < P are known as weak field ligands.
If Δ₀ > P, the 4^th electron pairs up in one of the t_2g orbitals giving the configuration \(t_{2 g}^3 e_g^1\), thus forming low spin complexes. Such ligands for which Δ₀ > P are called strong field ligands.

Question 15.
Explain the terms
(i) Ligand
(ii) Coordination number
(iii) Coordination entity
(iv) Central metal atom/ion.
Answer:
i) Ligand : The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be
a) simple ions such as Cl^–
b) small molecules such as H₂O or NH₃
c) large molecules such as H₂NCH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃ or even
d) macro-molecules, speh as proteins.
On the basis of the number of donor atoms available for coordination, the ligands can be classified as :

a) Unidentate : One donor atom, Eg.: etc.
b) Bidentate : Two donor atoms, Eg.: H₂NCH₂CH₂NH₂
(ethane-1, 2-diamine or en), C₂\(\mathrm{C}_2 \mathrm{O}_4^{2-}\) (oxalate), etc.
c) Polydentate: More than, two donor atoms, Eg.: N(CH₂CH₂NH₂)₃ EDTA etc.

ii) Coordination number: The coordination number (CN) of metal ion in a complex can be defined as thê number of ligands or donor atoms to which the metal is directly bonded.
Eg: In [PtCl₆]^2-, CN of Pt = 6, In [Ni(NH₃)₄]²⁺, CN of Ni = 4.

iii) Coordination entity : A central metal atoms or ion bonded to a fixed number of molecules or ions (ligands) is known as coordination entity. For example [COCl₃(NH₃)₃]. Ni(CO)₄], etc.

iv) Central metal atom/ion: In a coordination entity, the atom/ion to which a fixed no. of ions/groups are bound in a definite geometrical arrangement around it is called central metal atom or ion. Eg: K₄[Fe(CN)₆] ‘Fe’ is central metal.

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(c) Group-17 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(c) Group-17 Elements

Very Short Answer Questions

Question 1.
Interhalogen compounds are more reactive than the constituent halogens except for fluorine – Explain.
Answer:
Inter halogen compounds are more reactive than halogens. This is because X – X’ bond in interhalogens is weaker than X — X bond in halogens except F – F bond.

Question 2.
What is the use of ClF₃?
Answer:
ClF₃ is very useful fluorinating agent and it is used for the production of VF₆ in the enrichment of U²³⁵.
U_(s) + 3ClF_3(l) → UF_6(g) + 3ClF_(g)

Question 3.
Why are halogens coloured?
Anšwer:
Halogens are coloured due to the absorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. Halogens absorb different quanta of radiation and display different colours.

Question 4.
Write the reactions of F₂ and Cl₂ with water. (TS Mar. 17) (IPE Mar. ‘14)
Answer:
Fluorine produces O₂ and O₃ on passing through water.
3F₂ + 3H₂O → 6HF + O₃
2F₂ + 2H₂O → 4HF + O₂

Question 5.
Electron gain enthalpy of fluorine is less than that of chlorine – explain.
Answer:
The negative electron gain enthalpy of the fluorine is less than that of chlorine. This is due to small size of fluorine atom which results in strong inter electronic repulsions in the relatively small 2p orbitals of fluorine and thus the incoming electron does not experience much attraction.

Question 6.
Bond dissociation enthalpy of F₂ is less than that of Cl₂ – Explain.
Answer:
Bond dissociation enthalpy of F₂ is less than that of Cl₂
Explanation:
In F₂ molecule electron repulsions are greater among lone pairs because these lone pairs are much closer to each other than in case of Cl₂.

Question 7.
Write the formulae of the compounds, in which oxygen has positive oxidation states and mention the oxidation states of oxygen in them.
Answer:
In OF₂ and O₂F₂ oxygen has positive oxidation states.

• In OF₂, oxygen oxidation state is + 2.
• In O₂F₂ oxygen oxidation state is + 1.

Question 8.
What is the use of O₂F₂ and I₂O₅?
Answer:
Uses of O₂F₂

• O₂F₂ is a fluorinating agent. O₂F₂ oxidises plutonium to PUF₆ and the reaction is used in removing plutonium as PUF₆ from spent nuclear fuel.

Use of I₂O₅:

• I₂O₅ is a good oxidising agent and is used in the estimation of carbon monoxide (CO).

Question 9.
Explain the reactions of Cl₂ with NaOH. (IPE – 2015 (AP), 2016 (TS), (AP))
Answer:
i) Reaction with cold dilute NaOH: Chlorine reacts with cold dilute NaOH to give sodium hypochlorite and sodium chloride.

ii) Reaction with hot concentrated NaOH : Chlorine reacts with hot concentrated NaOH to give sodium chlorate and sodium chloride.

Question 10.
What happens when Cl₂ reacts with dry slaked lime ? (AP Mar. 17: IPE 16, 15 (AP))
Answer:
Chlorine reacts with dry slaked lime and forms bleaching powder.
Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O.

Question 11.
What is aqua regia ? Write its reaction with gold and platinum.
Answer:
A mixture of 3 parts of conc. HCl and one part of Conc. HNO₃ constitutes aqùa regia. It is used for dissolving noble metals.
It’s reaction with gold:
Au + 4H⁺ + \(\mathrm{NO}_3^{-}\) + 4Cl^– → \(\mathrm{AuCl}_4^{-}\) + NO + 2 H₂O
It’s reaction with Platinum:
3Pt + 16H⁺ + \(4 \mathrm{NO}_3^{-}\) + 18Cl^– → 3PtC\(l_6^{-2}\) + 4NO + 8 H₂O

Question 12.
How is chlorine manufactured by Deacon’s method? (AP Mar.’ 17: IPE ’16 (TS))
Answer:
Deacon’s process: [n Deacon’s process. Chlorine is obtained by the oxidation of hydrogen chloride gas by atmospheric oxygen in the presence of CuCl₂ (catalyst) at 723 K.

Question 13.
Why is dry chlorine cannot act as a bleaching agent.
Answer:
Dry chlorine cannot produce nascent oxygen. Hence it cannot act as a bleaching agent.

Question 14.
HF is a liquid while HCl is a gas — explain.
Answer:
HF is a liquid due to the presence of inter molecular hydrogen bonding where as HCl is a gas as there is no such type of bonding in it.

Question 15.
Write two uses of hydrogen chloride.
Answer:
Uses of hydrogen chloride:

• It is used in medicines and as a laboratory reagent.
• It is used in the manufacturing of Cl₂, NH₄Cl and glucose (from corn starch).
• It is used in extracting glue from bones and purifying bone black.

Question 16.
Chlorine acts as an oxidizing agent – explain with two examples.
Answer:
Chlorine acts as oxidising agent.
Example – 1: Cl₂ oxidises Iodine to Iodate.
I₂ + 6 H₂O + 5 Cl₂ → 2HIO₃ + 10 HCl

Example – 2: Cl₂ oxidises Sodium Sulphite to Sodium Sulphate.
Cl₂ + Na₂SO₃ + H₂O → Na₂SO₄ + 2 HCl

Question 17.
Write the reaction of chlorine with hypo (Na₂S₂O₃). (IPE Mar.2015 (AP, TS), 2016 (TS), (AP))
Answer:
Reaction of chlorine with hypo (Na₂S₂O₃). Na₂S₂O₃ + Cl₂ + H₂O → Na₂SO₄ + 2HCl + S. In this reaction hypo acts an “antichlor”.

Question 18.
Give the bond dissociation order of halogens.
Answer:
Bond dissociation order of halogens is, Cl₂ > Br₂ > F₂ > I₂.

Question 19.
I₂ is more soluble in KI give reason.
Answer:
I₂ forms soluble KI₃ with KI solution. I₂ + KI → KI₃.

Question 20.
Chlorine acts as a bleaching agent only in the presence of moisture — explain.
Answer:
Moist chlorine is a powerful bleaching agent. This bleaching property is due to oxidation.
Cl₂ + H₂O → 2HCl + (O)
Ex: Coloured substance + (0) → colourless substance.

Question 21.
The decreasing order of acidic character among hypohalogen acids is HClO > HBrO > HIO. Give reason.
Answer:
Given the decreasing order of acidic character among hypohalogen acids is
HClO > HBrO > HIO

From the above mentioned Ka values the given order of hypohalogen acids – acid strength order is
HClO > HBrO > HIO

Question 22.
fluorine exhibits only-1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Explain.
Solution:
Fluorine is the most electronegative element and cannot exhibit any positive oxidation state. Other halogens have d- orbitais and therefore, can expand their octets and show +1, +3, +5 and +7 oxidation states also.

Question 23.
What are the inter halogen compounds ? Give two examples.
Answer:
The compounds formed between different halogens are called inter halogen compounds.
Ex: ClF₃, BrF₃, IF₇, ICl₃.

Short Answer Questions

Question 1.
Explain the structures of
a) BrF₅ and
b) IF₇.
Answer:
a) Structure of BrF₅.

• Central atom in BrF₅ is ‘Br’
• ‘Br’ undergoes sp³d² hybridisation in 2^nd excited state.
  
• Shape of the molecule is octahedral with one position occupied by a lone pair (or) square pyramidal.
  

b) Structure of IF₇:

• Central atom in IF₇ is 1’.
• ‘I’ undergoes sp³d³ hybridisation iñ 3^rd excited state.
  
• Shape of the molecule is Pentagonal bipyramid structure.
  

Question 2.
How is chlorine obtained in the laboratory? How does it react with the following?
a) cold dil. NaOH
b) excess NH₃
c) KI (IPE Mar & May – 2015 AP TS), BMP)
Answer:
In the laboratory chlorine is prepared by the oxidation of HCl with MnO₂.
4 HCl + MnO₂ → MnCl + 2 H₂O + Cl₂↑
a) Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O
b) Cl₂ reacts with excess of NH₃, Nitrogen and ammonium chloride are formed.
8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂↑
c) Cl₂ reacts with KI to liberate iodine.
Cl₂ + 2KI → 2KCl + I₂↑

Question 3.
How is ClF₃ prepared? How does it react with water? Explain its structure.
Answer:
Preparation of ClF₃: Chlorine reacts with excess of fluorine to form ClF₃.

Reaction with H₂O : ClF₃ reacts with water explosively and oxidises water to give oxygen or in controlled quantitieš oxygen diflouride (OF₂) as well as HF and HCl.
ClF₃ + 2 H₂O → 3 HF + HCl + O₂
ClF₃ + H₂O → HF + HCl + OF₂

Structure of ClF₃:

• Central atom in ClF₃ is ‘Cl’
• Excited state electronic configuration of ‘Cl’ is
  
• Cl’ atom undergoes sp³d hybridisation.
• It is a bent T-shaped molecule (or) trigonal bipyramidal with 2 – positions occupied by lone pairs.
  

Question 4.
How can you prepare Cl₂ from HCl and HCl from Cl₂ ? Write the reactions.
Answer:
Preparation of Cl₂ from HCl:

• On heating MnO₂ with conc. HCl, Cl₂ gas is liberated.
  MnO₂ + 4 HCl → MnCl₂ + Cl₂ + 2 H₂O
• By the oxidation of HCl gas by atmospheric oxygen in presence of CuCl₂ catalyst at 723K.
  
  Preparation of HCl from Cl₂:
• Cl₂ when reacted with H₂ to form HCl,
  H_2(g) + Cl_2(g) → 2HCl_(g)

Question 5.
How is chlorine prepared by electrolytic method? Explain Its reaction with
a) NaOH and
b) NH₃ under different conditions. (IPE Mar & May – 2015 (AP, TS))
Answer:
Preparation of chlorine by electrolytic method: Chlorine is obtained by the electrolysis of brine solutions (Cone. NaCl). Cl₂ gas is liberated at anode.
2 NaCl → 2 Na⁺ + 2Cl^–
2 H₂O + 2e^– → 2 OH^– + H₂ (cathode)
2 Cl^– → Cl₂ + 2e^– (anode) .

a)

i)NaOH:
Chlorine reacts with cold dil. NaOH to form sodium hypochlorite.
Cl₂ + 2 NaOH → NaCl + NaOCl + H₂O
ii) Cl₂ reacts with hot conc. NaOH to form sodium chloride and sodium chlorate.
3 Cl₂ + 6 NaOH → 5 NaCl + NaClO₃ + 3 H₂O

b)

i) Cl₂ reacts with excess of NH₃, Nitrogen and ammonium chlorate are formed.
8 NH₃ + 3 Cl₂ → 6 NH₄Cl + N₂↑
ii) NH₃ reacts with excess of Cl₂ to form NCl₃ and HCl.

Question 6.
What are interhalogen compounds? Give some examples to illustrate the definition. How are they classified?
Answer:
The binary diamagnetic compounds of halogens which are formed by the reaction of halogens among themselves are called interhalogen compounds.
E.g.: IF₇, ClF₃, BrF₃, ClF, IF₃ etc.
The above examples are binary diamagnetic compounds and formed by combination of halogens only.
Inter halogen compounds are classified into four types.

1. AX – Týpe :Eg: ClF, BrF
2. AX₃ – Tÿpe : Eg : ClF₃, IF₃
3. AX₅ – Type : Eg: ClF₅, BrF₅
4. AX₇ – Type : Eg : IF₇
   • ‘A’ is less electronegative halogen.
   • X is more electronegative halogens.

Question 7.
Write the names and formulae of the oxoacids of chlorine. Explain their structures and relative acidic nature.
Answer:
Four oxyacids of chlorine are known. They are
Hypochlorous acid – HOCl
Chlorous acid — HClO₂
Chloric acid – HClO₃
Perchloric acid – HClO₄
Structure of HClO: In this chlorine atom is sp³ hybridised. Outer electronic configuration of Cl in ClO^– after sp³ hybridisation.

Shape is tetrahedral with 3 lone pairs (or) linear
No π — bonds.
Chlorous acid: (HClO₂) : Chlorine is in sp³ hybrid state, in first excited state. Shape is tetrahedral with 2 lone pairs (or) angular one π_d-p bond is present.
First excited state

Chloric acid (HClO₃) : The central chlorine atom undergoes sp³ hybridisation in second excited state.
Second excited state

Shape is tetrahedral with one lone pair (or) pyramidal.
Two π_d-p bonds present.

Perchloric acid (HClO₄) : The central chlorine atom undergoes sp³ hybridisation in third excited state.

Shape is perfect tetrahedral. No lone pairs.
Three π_d-p bonds present.

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(b) Group-16 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(b) Group-16 Elements

Very Short Answer Questions

question 1.
Write any two compounds, in which oxygen shows an oxidation state different from -2. Give the oxidation states of oxygen in them.
Answer:
OF₂ and O₂ F₂ are two compounds in which oxygen shows an oxidation state different from -2.

• In OF₂ the oxidation state of oxygen is +2.
• In O₂F₂ the oxidation state of oxygen is +1.

Question 2.
Why H₂O a liquid while H₂S Is a gas? ( IPE May – 2014)
Answer:
H₂O is liquid due to the presence of intermolecular hydrogen bonding. While H₂S is gas because it is not having such type of bonding.

Question 3.
H₂O is neutral while H₂S is acidic — explain.
Answer:
H₂O is neutral while H₂S is acidic.
Reason: The O-H bond dissociation enthalpy is greater than the S — H bond dissociation enthalpy.

Question 4.
Explain the structures of SF₄ and SF₆.
Answer:
Structure of SF₄:

• In SF₄ ‘S’ undergoes sp³d hybridisation.
• It has trigonal bipyramidal structure in which one of the equitorial positions is occupied by a lone pair of electrons. This geometry is also known as see – saw geometry.
  Structure of SF₆:
• In SF₆, ‘S’ undergoes sp³d² hybridisation,
• It has octahedral structure.

Question 5.
Give one example each for
a) a neutral oxide
b) a peroxide
c) a super oxide
Answer:
a) CO, N₂O are neutral oxides.
b) Na₂O₂, BaO₂ are peroxides.
c) KO₂, RhO₂ are super oxides.

Question 6.
What is tailing of mercury? How is it removed? (IPE Mar – 2015 (TS))
Answer:
Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury
2Hg + O₃ → Hg₂O + O₂
It is removed by shaking it with water which dissolves Hg₂O.

Question 7.
How does ozone react with Ethylene?
Answer:
Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to form formaldehyde.

Question 8.
Which form of sulphur shows paramagnetism?
Answer:
In vapour state sulphur partly exists as S₂ molecule which has two unpaired electrons in the antibonding π (π*)orbitals like O₂. Hence it exhibits paramagnetism.

Question 9.
Why are group — 16 elements called chalçogens ?
Answer:
Chalcogens means mineral forming (or) ore forming elements. Most of elements exist in earth crust as oxides, sulphides, selinides, telurids etc. So Group – 16 elements are called as chalcogens.

Question 10.
Write any two uses each for O₃ and H₂SO₄.
Answer:
Uses of O₃:

• Ozone is used in sterilisation of water.
• Ozone is used in manufacture of artificial silk and camphor etc.
• Ozone is used to identify unsaturation in carbon compounds.

Uses of H₂SO₄:

• H₂SO₄ is used in the manufacture of fertilisers.
• H₂SO₄ is used in petrol refining.
• H₂SO₄ is used in detergent industry.

Short Answer Questions

Question 1.
Write a. short note on the allotropy of sulphur.
Answer:
The important allotropes of sulphur are
a) yellow rhombic (α. sulphur).
b) Monoclinic (β – sulphur).

The stable from is α-sulphur (at room temperature).
Rhombic sulphur (α – Sulphur):

• Colour : Yellow.
• Melting point : 3858K.
• Specific gravity: 2.06.
• It is insoluble in water and partially soluble in alcohol, benzene etc., and readily soluble in CS₂.

Monoclinic sulphur (β – Sulphur):

• Melting point: 392K
• Specific gravity: 1.98.
• It is soluble in Cs₂.
• Rhombic sulphur transforms to monoclinic sulphur by heating above 369K. This temperature is called transition temperature.

Question 2.
Which is used for drying ammonia?
Answer:
For drying ammonia quick lime (CaO) is used.

• For drying ammonia conc. H₂SO₄ , P₄O₁₀ and anhydrous CaCl₂ cannot be used because they react with ammonia and forms (NH₄)₂SO₄, (NH₄)₃PO₄ and CaCl₂. 8NH₃

Question 3.
Explain the conditions favourable for the formation of SO₃ from SO₂ in the contact process of H₂SO₄.
Answer:
Le Chatlier’s principle — Application to produce SO₃:
The oxidation of SO₂ to SO₃ in the presence of a catalyst is a reversible reaction. The thermochemical equation for the conversion is written as

The equation reveals the following points:

1. 3 volumes of the reactants convert into 2 volumes of SO₃. i.e., a decrease of volume accompanies the reaction.
2. the reaction is an exothermic change.
3. the catalyst may be present to increase the SO₃ yields.

According to Le Chattier’s principle,

i. a decrêase in volume of the system is favoured at high pressures. But in practice only about 2 bar pressure is used. The reason for not using high pressures is acid resisting towers which can withstand high pressures cannot be built.

ii. exothermic changes are favoured at low temperatures. It is not always convenient in the industry to work at low temperatures. In such situations an optimum temperature is maintained. At this temperature considerable amounts of the product are obtained, in the manufacture of H₂SO₄, the optimum temperature suitable for the conversion of SO₂ into SO₃ is experimentally found to be 720K.

iii. The rate of formation of SO₃ is enhanced by the use of a catalyst. (V₂O₅ (or) Pt – asbestos).
Favourable Conditions:
Temperature: 720K
Pressure :2 bar
Catalyst : V₂O₅ (or) platinized asbestos.

Question 4.
Which oxide of sulphur can act as both oxidizing and reducing agent? Give one example each.
Answer:
Sulphur dioxide (SO₂) acts as both oxidising as well as reducing agent.
SO₂ as Oxidising agent:
Sodium sulphide oxidises to hypo with SO₂.
2Na₂S + 3SO₂ → 2Na₂S₂O₃ + S
SO₂ as Reducing agent:
SO₂ reduces Fe⁺³ ions to Fe⁺² ions.
2Fe⁺³ + SO₂ + 2H₂O → 2Fe²⁺ + \(\mathrm{SO}_4^{-2}\) + 4H⁺

Long Answer Questions

Question 1.
Explain in detail the manufacture of sulphuric acid by contact process. ( IPE 2016 (TS))
Answer:
Manufacture of H₂SO₄ by contact process:
Manufacturing of H₂SO₄ involves three main steps.

Step-I:
SO₂ production : The required SO₂ for this process is obtained by burning S(or) Iron
pyrites in oxygen.
S + O₂ → SO₂
4FeS₂ + 15O₂ → 2Fe₂O₃ + 8SO₃

Step—2
SO₃ formation: SO₂ is oxidised in presence of catalyst with atmospheric air to form SO₃.

Le Chatliers principle — Application to produce SO₃ ;
The oxidation of SO₂ to SO₃ in the presence of a catalyst is a reversible reaction. The
thermochemical equation for the conversion is written as

The equation reveals the following points:

1. 3 volumes of the reactants convert into 2 volumes of SO₃. i.e., a decrease of volume accompanies the reaction.
2. The reaction is an exothermic change.
3. The catalyst may be present to increase the SO₃ yields.

According to Le Chatlier’s principle,

1. a decrease in volume of the system is favoured at high pressures. But in practice only about 2 bar pressure is used. The reason for not using high pressures is acid resisting towers which can withstand high pressures cannot be built.
2. exothermic changes are favoured at low temperatures. It is not always convenient in the industry to work at low temperatures. in such situations an optimum temperature is maintained. At this temperature considerable amounts of the product are obtained. In the manufacture of H₂SO₄, the optimum temperature suitable for the conversion of SO₂ into SO₃ is experimentally found to be 720K.
3. The rate of formation of SO₃ is enhanced by the use of a catalyst. (V₂ O₅ (or) Pt – asbestos).
   Favourable Conditions:
   Temperature: 720K :
   Pressure : 2 bar
   Catalyst : V₂O₅ (or) platinized asbestos.
   

Step-3

• Formation of H₂SO₄: SO₃ formed in the above step absorbed in 98% H₂SO₄ to get oleum (H₂S₂O₇). This oleum is diluted to get desired concentration of H₂SO₄.
  SO₃ + H₂SO₄ → H₂S₂O₇
  H₂S₂O₇ + H₂O → H₂SO₄

Question 2.
How is ozone prepared from oxygen? Explain its reaction with
a) C₂H₄
b) KI
c) Hg
d) PbS. (T.S. & A.P. Mar. ’17) (A.P.Mar. ’16) (Mar. ’14)
Answer:
Preparation of Ozone:
A slow dry stream of oxygen under silent electric discharge to form ozone (about 10%). The product obtained is known as ozonised oxygen.
3O₂ → 2O₃ ΔH° = 142kJ/mole .

• The formation of ozone is an endothermic reaction.
• It is necessary to use silent electric discharge in the preparation of O₃ to prevent its decomposition

a) Reaction with C₂H₄ : Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to for formaldehyde.

b) Reaction with KI : Moist KI is oxidised to Iodine in presence of ozone.
2KI + H₂O + O₃ → 2KOH + I₂ + O₂
c) Reaction with Hg: Mercury loses it’s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.
2Hg + O₃ → Hg₂O + O₂
It is removed by shaking it with water which dissolves Hg₂O.
d) Reaction with PbS : Black lead sulphide oxidised to white lead sulphate in presence of ozone.
PbS + 4O₃ → PbSO₄ + 4O₂.

Question 3.
Write the structures of oxoacids of sulphur. (IPE 2016 (TS))
Answer:
Oxoacids of sulphur: Sulphur forms a number of oxoacids such as H₂SO₃. H₂S₂O₃, H₂S₂O₄, H₂S₂O₅, H₂S₂O₆ (x = 2 to 5), H₂SO₄, H₂S₂O₇, H₂SO₅, H₂S₂O₈.

Question 4.
Write any two oxidation and any two reduction properties of ozone with equations.
Answer:
Oxidation properties:

1. Ozone oxidises moist potassium idodide and liberates I₂.
   2KI + H₂O + O₃ → 2KOH + I₂ + O₂
2. Ozone oxidises black lead sulphide to white lead sulphate. PbS + 4O₃ → PbSO₄ + 4O₂

Reduction properties:

1. Ozone reduces H₂O₂ to H₂O.        H₂O₂ + O₃ → H₂O + 2O₂
2. Ozone reduces Ag₂O to Ag.         Ag₂O + O₃ → 2Ag + 2O₂

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 6(a) Group-15 Elements which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 6(a) Group-15 Elements

Very Short Answer Questions

Question 1.
Nitrogen molecule is highly stable – Why? (IPE 2014)
Answer:
The nitrogen Molecule is more stable because in between two nitrogen atoms of N₂, a triple bond is present. To break this triple bond high energy is required (941.4KJ/mole).

Question 2.
Why are the compounds of bismuth more stable in +3 oxidation state?
Answer:
Bismuth compounds are more stable in +3 oxidation state because ‘Bi’ exhibits +3 stable oxidation state instead of +5 due to inert pair effect.

Question 3.
What is allotropy ? Explain the different allotropic forms of phosphorus. (IPE 2016(TS))
Answer:
Allotropy: The existence of an element in different physical forms having similar chemical properties is called allotropy.
→ Allotropes of ‘P’: → White ‘P’ (or) Yellow P.

• Red P
• Scarlet ‘P’
• Violet P
• α – black P’
• β – black ‘P’.

White phosphorus :

• It is poisonous and insoluble in water and soluble in carbon disulphide and glows in dark.
• It is a translucent white waxy solid.
• It dissolves in boiling NaOH solution and gives PH₃.
  
• It is more reactive than other solid phases.
• Bond angle is 60° and it readily catches fire,

Red phosphorus :

• Red P possesses iron grey lustre.
• In it odour less, non poisonous and insoluble in water as well as CS₂.
• Red P is much less reactive than white P.

Black P’:

• α – Black P : It is formed when red P is heated in a sealed tube 803K.
• β – Black P : It is prepared by heating white P at 473 K under high pressure.

Question 4.
Explain the difference in the structures of white and red phosphorus.
Answer:
White ‘P’ molecule has tetrahedral structure (discrete molecule). Discrete ‘P’ molecules are held by vander waal’s forces.

Red ‘P’ is polymeric consisting of chains of P₄ tetrahedron linked together through covalent bonds.

Question 5.
What is inert pair effect ?
Answer:
Inert pair effect: The reluctance of ns pair of electrons to take part in bond formation is called inert pair effect.

• Bi exhibits +3 oxidation state instead of +5 due to inert pair effect.

Question 6.
How do calcium phosphide and heavy water react ?
Answer:
Calcium phosphide reacts with heavy water to form Deutero phosphine.
Ca₃P₂ + 6D₂O → 3 Ca (OD)₂ + 2PD₃

Question 7.
Ammonia is a good complexing agent – Explain with an example.
Answer:
NH₃ is a lewis base and it donates electron pair to form dative bond with metal ions. This results in the formation of complex compound.
Eg :

Question 8.
NO is paramagnetic in gaseous state but diamagnetic in liquid and solid states – Why ?
Answer:
In gaseous state NO₂ exists as a Monomer and contains one unpaired electron but in solid state it dimerises to N₂O₄ so it doesnot contain unpaired electron.
Hence NO₂ is para magnetic in geseous state but diamagnetic in solid state.

Question 9.
Iron becomes passive in cone. HNO₃ – Why ?
Answer:
Iron becomes passive in cone. HNO₃ due to formation of a passive film of oxide on the surface of iron.

Question 10.
Give the neutral oxides of nitrogen.
Answer:
Nitrous oxide (N₂O) and Nitric oxide (NO) are neutral oxides of nitrogen.

Question 11.
Give the paramagnetic oxides of nitrogen.
Answer:
Nitric oxide (NO) and nitrogen dioxide (NO₂) are the paramagnetic oxides of nitrogen due to the presence of odd number of electrons.

Question 12.
Why is white phosphorus is more reactive than red phosphorus ?
Answer:
In white phosphorus the P – P – P bond angle is 60° hence the bonds are in strain. As a result the bonds are broken easily. In red phosphorus the bond angle is 120°, hence it is stable and less reactive.

Question 13.
What happens when white phosphorus is heated with conc. NaOH solution in an inert atmosphere of CO₂ ?
Answer:
When white phosphorus heated with con. NaOH solution in an inert atmosphere of CO₂ forms PH₃.
P₄ + 3NaoH + 3H₂O → PH₃ + 3NaH₂PO₂.

Question 14.
Give the uses of
a) nitric acid and
b) ammonia.
Answer:
Nitric acid : It is used in the manufacture of fertilizers, explosives, nitro glycerine, nitro toluence etc.,
Ammonia : It is used to produce fertilizers like urea liquid NH₃ is used as a refrigerant.

Question 15.
Give the disproportionation reaction of H₃PO₃.
Answer:
Orthophosphoric acid (H₃PO₃) on heating disproportionates to give orthophosphoric acid and phosphine.
4H₃PO₃ → 3H₃PO₄ + PH₃.

Question 16.
Nitrogen exists as diatomic molecule and phosphorus as P₄ – Why ?
Answer:
Nitrogen exists as diatomic molecule :

• Nitrogen has small size and high electronegativity and nitrogen atom forms Pπ – Pπ multiple bonds with itself (triplebond). So it exists as a discrete diatomic molecule in elementary state.

Phosphorus exists as tetra atomic molecule :

• Phosphorus has large size and less electronegative and it forms P-P single bonds. So it exists as tetra atomic i.e., P₄.

Question 17.
Arrange the hydrides of group – 15 elements in the increasing order of basic strength and decreasing order of reducing character.
Answer:

• Increasing order of basic strength of Group – 15 elements hydrides is
  BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃.
• Decreasing order of reducing character of Group – 15 elements hydrides is
  BiH₃ > SbH₃ > AsH₃ > PH₃ > NH₃.

Question 18.
PH₃ is a weaker base than NH₃ – Explain.
Answer:
PH₃ is a weaker base than NH₃.

• In NH₃ nitrogen atom undergoes sp³ hybridisation and due to small size it has high electron density than in’P of PH₃.
• Due to large size of ‘P’ atom and availability of large surface area, lone pair of electron spread in PH₃. Hence PH₃ is weaker base than NH₃.

Question 19.
A mixture of Ca₃P₂ and CaC₂ is used in making Holme’s signal – Explain.
Answer:
A mixture of Ca₃P₂ and CaC₂ is used in Holme’s signal. This Mixture containing containers are pierced and thrown in the sea, when the gas is evolved bum and serve as a signal.
The spontaneous combustion of PH₃ is the technical use of Holme’s signal.

Question 20.
Which chemical compound is formed in the brown ring test of nitrate ions ?
Answer:
In the brown ring test of nitrate salts a brown ring is formed. It’s formula is [Fe(H₂O)₅NO]⁺².

Question 21.
Why does NH₃ act as a Lewis base ?
Answer:
Nitrogen atom in NH₃ has one lone pair of electrons with is available for donation. Therefore, it acts as a Lewis base.

Question 22.
Why does NO₂ dimerise ?
Answer:
NO₂ contains of odd number of electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N₂O₄ molecule with even number of electrons.

Question 23.
Why does PCl₃ fume in moisture ?
Answer:
PCl₃ hydrolyses in the presence of moisture giving fumes of HCl.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 24.
Are all the five bonds in PCl₅ molecule equivalent ? Justify your answer.
Answer:
PCl₅ has a trigonal bipyramidal structure and the three equatorial P-Cl bonds are equivalent. While the two axial bonds are different and longer than equatorial bonds.

Question 25.
How is nitric oxide (NO) prepared ?
Answer:
Nitric oxide (NO) is prepared by the action of dilute nitric acid on copper.

Question 26.
Explain the following .
a) reaction of alkali with red phosphorus.
b) reaction between PCl₃ and H₂O.
Answer:
a) Red phosphorus reacts with alkalies slowly and forms phosphine and hypophosphite.
P₄ + 3NaOH + 3 H₂O → PH₃ + 3 NaH₂ PO₂
b) PCl₃ is hydrolysed by water and gives H₃PO₃

Question 27.
Write the oxidation states of phosphorus in solid PCl₅.
Answer:
In solid state PCl₅ exists as an ionic solid [PCl₄)⁺ [PCl₆]^–

• ‘P’ exhibits +5 oxidation state in [PCl₄)⁺ and [PCl₆]

Question 28.
Illustrate how copper metal can give different products on reaction with HNO₃.
Answer:

Question 29.
H₃PO₂ is a good reducing agent – Explain with an example.
Answer:
In H₃PO₂, two H-atoms are bonded directly to P-atom which imparts reducing character to the acid.

Question 30.
NH₃ forms hydrogen bonds but PH₃ does not – Why ?
Answer:
NH₃ forms hydrogen bonds but PH₃ does not.
Reason : Ammonia forms hydrogen bonds because it is a polar molecule and N-H bond is highly polar. Nitrogen has more electronegativity than phosphorus. In case of PH₃, P-H bond polarity decreases.

Question 31.
PH₃ has lower boiling point than NH₃. Why ?
Answer:
Unlike NH₃, PH₃ molecules are not associated through inter molecular hydrogen bonding. That is why the boiling point of PH₃ is lower than NH₃.

Long Answer Questions

Question 1.
How is ammonia manufactured by Haber’s process ? Explain the reactions of ammonia with
a) ZnSO_4(aq)
b) CuSO_4(aq)
c) AgCl_(s) (TS Mar. ’17 IPE 2015(AP))
Answer:
In Haber process ammonia is directly synthesised from elements (nitrogen and hydrogen). The principle involved in this is

This is a reversible exothermic reaction.
According to Le Chatelier’s principle favourable conditions for the better yield of ammonia are low temperature and high pressure. But the optimum conditions are :
Temperature : 720k
Pressure : 200 atmospheres
Catalyst: Finely divided iron in the presence of molybdenum (Promoter).
Procedure : A mixture of nitrogen and hydrogen in the volume ratio 1 : 3 is heated to 725 – 775K at a pressure of 200 atmospheres is passed over hot finely divided iron mixed with small amount of molybdenum as promotor. The gases coming out of the catalyst chamber consists of 10 – 20% ammonia. Gases are cooled and compressed, so that ammonia gas is liquified and the uncondensed gases are sent for recirculation.

a) Aq. ZnSO₄ reacts with ammonia aqueous solution to form white ppt of Zinc hydroxide.

Question 2.
How is nitric acid manufactured by Ostwald’s process ? How does it react with the following ? (A.P. Mar. ’17)
a) Copper
b) Zn
c) S₈
d) P₄
Answer:
Ostwald’s process : Ammonia, mixed with air in 1 : 7 or 1 : 8, when passed over a hot platinum gauze catalyst, is oxidised to NO mostly (about 95%).
The reaction is

The liberated heat keeps the catalyst hot. The ’NO’ is cooled and is mixed with oxygen to give the dioxide, in large empty towers (oxidation chamber). The product is then passed into warm water, under pressure in the presence of excess of air, to give HNO₃.
4NO₂ + O₂ + 2H₂O → 4HNO₃.
The acid formed is about 61% concentrated.

a) Copper reacts with dil. HNO₃ and cone. HNO₃ and liberates Nitric Oxide and Nitrogen dioxide respectively
3 Cu + 5 HNO₃ (dil) → 3 Cu (NO₃)₂ + 2 NO + 4 H₂O
Cu + 4 HNO₃ (conc.) → Cu (NO₃)₂ + 2 NO₂ + 2 H₂O

b) Zn reacts with dil. HNO₃ and Cone. HNO₃ and liberates Nitrous oxide and Nitrogen
dioxide respectively.
4 Zn + 10 HNO₃ (dil) → 4 Zn (NO₃)₂ + 5 H₂O + N₂O.
Zn + 4 HNO₃ (conc.) → Zn (NO₃)₂ + 2 H₂O + 2 NO₂

c) S₈ reacts with cone, nitric acid to form Sulphuric acid, NO₂ gas.
S₈ + 48 HNO₃ → 8 H₂SO₄ + 48 NO₂ + 16 H₂O
d) P₄ reacts with cone, nitric acid to form phosphoric acid and NO₂ gas.
P₄ + 20 HNO₃ → 4 H₃PO₄ + 20 NO₂ + 4 H₂O

Students get through AP Inter 2nd Year Chemistry Important Questions 5th Lesson General Principles of Metallurgy which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 5th Lesson General Principles of Metallurgy

Very Short Answer Questions

Question 1.
What is the role of depressant in froth floatation?
Answer:
By using depressants in the froth floatation process, it is possible to separate a mixture of two sulphide ores.
Eg: In the ore containing ZnS and PbS, the depressant used is NaCN. It prevents ZnS from coming to the froth but allows PbS to come with the froth.

Question 2.
Explain “poling”. (AP Mar. ’16, ’15; IPE ’16, 15 (AP))
Answer:
When the metals are having metal oxides as impurities this method is employed. The impure metal is. melted and is then covered by carbon powder. Then it is stirred with green wood poles. The reducing gases formed from the green wood and the carbon, reduce the oxides to the metal.
Eg : Cu & Sn metals are refined by this method.

Question 3.
Decribe a method for the refining of nickel.
Answer:
Mond’s process:

• In Mond’s process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra carbonyl.
  
• Nickel tetra carbonyl is strongly heated to decompose and gives the pure Nickel.
  

Question 4.
What is the role of cryolite in the metallurgy of aluminium ? (IPE 2015 (TS), BMP, 2016 (TS))
Answer:
By adding the cryolite to the pure Alumina, the melting point of pure Alumina is lowered (which is very high 2324K) and electrical conductivity of pure alumina is increased.

Question 5.
Give the composition of the following alloys (IPE ’16, ’14 (TS)) (AP & TS Mar. ’17)
a) Brass
b) Bronze
c) German Silver
Answer:
a) Composition of Brass : 60 – 80% Cu, 20 – 40% Zn
b) Composition of Bronze : 75 – 90% Cu, 10 – 25% Sn
c) Composition of German silver : 50 – 60% Cu, 10 – 30% Ni, 20 – 30% Zn.

Question 6.
What is matte ? Give its composition.
Answer:
During the extraction of ’Cu’ from copper pyrites the product of the blast furnace consists mostly of Cu₂S and a little of FeS. This product is known as “Matte”. It is collected from the outlet at the bottom of the furnace.

Question 7.
What is flux ? Give an example.
Answer:
Flux : An outside substance added to. ore to lower its melting point is known as flux.

• Flux combines with gangue and forms easily fusible slag.
  gangue + flux → slag
  

Question 8.
How is aluminium useful in the extraction of chromium and manganese from their oxides ?
Answer:

• ‘Al’ is used as reducing agent.
• By Alumino thermite process Cr, Mn are extracted from their oxides.
• The reactions are highly exothermic.

Cr₂O₃ + 2Al → 2Cr + Al₂O₃
3Mn₃O₄ + 8Al → 4Al₂O₃ + 9Mn

Question 9.
What is a mineral ?
Answer:
The chemical compound from which a metal can be extracted is called a mineral. Bauxite, crayolite are the minerals of aluminium.

Question 10.
What is an ore ?
Answer:
An ore is a mineral from which the metal can be extracted easily and economically. Bauxite is the ore of aluminium.

Question 11.
What is a ore. Give the ores of Al, Zn, Fe, Cu.
Answer:
The mineral from which metal can be extracted economically is called ore.
Aluminium ores : Bauxite = Al₂ O₃. 2H₂O; Cryolite = Na₃Al F₆
Zinc ores : Zinc blende = Zns; Calamine = ZnCO₃
Copper ores : Cuprite: CuO; Copper pyrites = Cu₂S
Iron ores : Haematite = Fe₂O₃ Magnetite = Fe₃O₄

Question 12.
What is the role of SiO₂ in the extraction of copper.
Answer:
In the extraction of copper SiO₂ acts as flux. It combines with FeO and removes as slag.
FeO + SiO₂ (flux) → FeSiO₃ (Slag)

Short Answer Questions

Question 1.
Outline the principles of refining of metals by the following methods.
(a) Zone refining
(b) Electrolytic refining
(c) Poling
(d) Vapour phase refining.
Answer:
a) Zone refining :

• Zone refining is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal.
• A circular mobile heater is fixed at one end of a rod of impure metal.
• The molten zone moves along with the heater moves forward the pure metal crystallises out of the melt and the impurities pass into the adjacent molten zone.

• The above process is repeated several times and the heater is moved in the same direction form one end to the other end. At one end impurities get concentrated. This end is cut off.
• This method is very useful for producing semiconductor grade metals of very high purity.
  Eg : Ge, Si, B, Ga etc…

b) Electrolytic refining: This process is used for less reactive metals like Cu, Ag, AZ, Au etc.

• In this process anode is made by impure metal and a thin strip of pure metal acts as cathode.
• On electrolysis metal dissolves from anode and pure metal gets deposited at cathode.

Impurities settle down below anode in the form of anode mud.

c) Poling : When the metals are having the metal oxides as impurities this method is employed. The impure metal is melted and is then covered by carbon powder. Then it is stirred with green wood poles. The reducing gases formed from the green wood and the carbon, reduce the oxides to the metal. Eg : Cu & Sn metals are refined by this method.

d) Vapour phase refining: In this method the metal is converted into its volatile compound and coll Ted. It is then decomposed to give pure metal.

1. The metal should form a volatile compound with an available reagent.
2. The volatile compound should be easily decomposable. So the recovery is easy.
   E.g : Mond’s process :
   • In Mond’s process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetra carbonyl,
     
   • Nickel tetra carbonyl is strongly heated to decompose and gives the pure Nickel.
     

Question 2.
Explain the purification of sulphide ore by froth floatation method. (Mar ’15) (A.P. Mar. ’17)
Answer:
Froth floatation method :

• This method is used to concentrate sulphide ores.
• In this process a suspension of the powdered ore is made with water.
• A rotating paddle is used to agitate the suspension and air is blown into the suspension in presence of an oil.
• Froth is formed as a result of blown of air, which carries the mineral particles.
  
• To the above slurry froth collectors and stabilizers are added.
• Collectors like pine oil enhance non-wettability of the mineral particles.
• Froth stabilizers like cresol stabilize the froth.
• The mineral particles wet by oil and gangue particles wet by water.
• The broth is light and is skimmed off. The ore particles are then obtained from the froth. By using depressants in froth floatation process, it is possible to separate a mixture of two suplhide ores. Eg: In the ore containing ZnS and PbS, the depressant used is NaCN. It prevents ZnS from coming to the froth but allows PbS to come with the froth.

Question 3.
What is Ellingham diagram ? What information can be known from this in the reduction of oxides ?
Answer:
The graphical representation of Gibbs energy which provides a sound basis for considering the choice of reducing agent in the reduction of oxides. This graphical representation is known as Ellingham diagram.
This diagram helps us in predicting the feasibility of thermal reduction of an ore.

• The Criterion of feasibility of a reaction is that at a given temperature, Gibbs energy of the reaction must be negative.
• Ellingham diagram normally consists plots of ΔG° vs T for formation of oxides of elements.
• The graph indicates whether a reaction is possible or not, i.e., the tendency of reduction with a reducing agent is indicated.
• The reducing agent forms its oxide when the metal oxide is reduced. The role of reducing agent is to make the sum of ΔG° values of the two reactions negative.
• Out of C and CO, Carbon monoxide (CO) is a better reducing agent at 673K.
• At 983K (or) above coke(C) is better reducing agent.
• The above observations are from Ellingham diagram.
  Zinc is not extracted from zinc oxide through reduction by using CO.

Explanation :
2Zn + O₂ → 2ZnO, ΔG° = -650 kJ
2CO + O₂ → 2CO₂, ΔG° = -450 kJ
2ZnO + 2CO → 2Zn, 2CO₂, ΔG° = 200 kJ
ΔG° = Positive indicates that the reaction is not feasible.
The above fact is explained on the basis of Ellingham diagram.

Question 4.
Give examples to differentiate roasting and calcination. (IPE ’14, B.M.P. 2016) (A.P. & T.S. Mar. ’16, ’15)
Answer:
Roasting: Removal of the volatile components of a mineral by heating mineral either alone (or) mixed with some other substances to a high temperature in the presence of air is called Roasting.

• It is applied to the sulphide ores.
• SO₂ gas is producted along with metal oxide.
  

Calcination: Removed of the volatile components of a mineral by heating in the absence of air is called calcination.

• It is applied to carbonates and bicarbonates.
• CO₂ gas is produced along with metal oxide.

Question 5.
How is copper extracted from copper pyrites ?
Answer:
Extraction of copper from copper pyrites :
Copper pyrite is the main source of copper metal. Various steps involved in the extraction of copper are discussed below.

Step -I:

Concentration of ore by froth floatation process :
The ore is first crushed in ball mills. The finely divided ore is suspended in water. A little pine oil is added and the mixture is vigorously’ agitated by a current of air. The froth formed carries the ore particles almost completely. The gangue sinks to the bottom of the tank. The froth is separated and about 95% concentrated ore is obtained.

Step -II:

Roasting : To remove the volatile impurities like As (or) Sb, the ore is roasted in a free supply of air. A mixture of sulphides of copper and iron are obtained and these are partially oxidised to respective oxides.
Cu₂S. Fe₂S₃ + O₂ → Cu₂S + 2FeS + SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2FeS + 3O₂ → 2FeO + 2SO₂

Step -III: .

The roasted ore is mixed with a little coke and sand (Silica) and smelted in a blast furnace and fused. A blast of air, necessary for the combustion of coke, is blown through the tuyeres present at the base of the furnace.’ The oxidation of the sulphides of copper and iron will be completed further. A slag of iron silicate is formed according to the reactions given below :

Step -IV :

After smelting the copper ore in blast furnace, the product of the blast furnace consists mostly of Cu₂S and a little of ferrous sulphide. This product is known as “Matte.” It is collected from the outlet at the bottom of the furnace. After then the following processes are carried out for getting the pure copper.

Bessemerization : The matte is charged into a Bessemer converter. A bessemer converter is a pear-shaped furnace. It is made of steel plates. The furnace is given a basic lining with lime or magnesium oxide (obtained from dolomite or magnesite). The converter is held in position by trunnions and can be tilted in any position. A hot blast of air and sand is blown through the tuyeres present near the bottom. Molten metal, the product in the furnace, collects at the bottom of the converter.

Reactions that took place in blast furnace go to completion. Almost all of iron is eliminated slag. Cuprous oxide combines with cuprous sulphide and forms Cu metal.
2Cu₂O + Cu₂S → 6Cu + SO₂
The molten metal is cooled in sand moulds. SO₂ escapes. The impure copper metal is known as ‘Blister copper” and is about 98% pure.

Step -V:

Refining: The Blister copper is purified by electrolysis. The impure copper metal is made into plates. They are
suspended into lead — lined tanks containing Copper (II) Sulphate solution. Thin plates of pure copper serve as
cathode. The cathode plates are coated with graphite. On electrolysis, pure copper is deposited at the cathode. The
copper obtained is almost 100% pure Cu.

Question 6.
Explain the process of leacing of aluminium from bauxite.
Answer:
Extraction of Aluminium from Bauxite:
For the purpose of extraction of Al, Bauxite is the best source.
Purification of Bauxite: Bauxite containing Fe₂O₃ as impurity is known as Red Bauxite.
Bauxite containing SiO₂ as impurity is known as White Bauxite and can be purified by “Set-peck’s Process. Red Bauxite is purified by Bayer’s process and Hall’s process.

Bayer’s Process: Red bauxite is roasted and digested in concentrated NaOH at 423 K. Bauxite dissolves in NaOH to form sodium meta aluminate while impurity Fe₂O₃ does not dissolve which can be removed by filtration.
Al₂O₃.2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O
The solution which contains sodium meta aluminate is diluted and crystals of Al(OH)₃, are added which serves as seeding orgent. Sodium meta aluminate undergoes hydrolysis to precipitate Al(OH)₃.
2NaAlO₂ + 4H₂O → 2NaOH + 2Al(OH)₃↓
Al(OH)₃ is filtered and ignited at 1200°C to get anhydrous alumina.

Halls’ Process: Red Bauxite is fused with sodium carbonate to form sodium meta aluminate
which is extracted with water. The impurity Fe₂O₃ is filtered out.
Al₂O₃ + Na₂CO₃ → 2NaAlO₂ + CO₂
Into the solution of sodium meta aluminate, CO₂ gas is passed to precipitate Al(OH)₃.

The precipitated Al(OH)₃ is ignited at 1200°C to get anhydrous alumina.
2Al(OH)₃ → Al₂O₃ + 3H₂O

Serpeck’s Process: Powdered bauxite is mixed with coke and heated to 2075 K in a current of nitrogen gas. Aluminium Nitride is formed while SiO₂ is reduced to Si which escapes out.

Aluminium nitride is hydrolysed to get aluminium hydroxide which on ignition gives anhydrous alumina.

Electrolytic Reduction of Alumina: Pure Alumina (Al₂O₃) is a bad conductor of electricity and it has high melting point (2050°C). So it cannot be electrolysed. Alumina is electrolysed by dissolving in fused cryolite to increase the conductivity and small amount of Fluorspar is added to reduce its melting point. Thus the electrolyte is a fused mixture of Alumina, Cryolite and Fluorspar.

Electrolysis is carried out in an iron tank lined inside with graphite (carbon) which functions as cathode. A number of carbon rods (or) copper rods suspended in the electrolyte functions as anode.

An electric current of 100 amperes at 6 to 7 volts is passed through the electrolyte. Heat produced by the current keeps the mass in fused state at 1175 to 1225K. The following reactions take place in the electrolytic cell under these conditions.
Na₃AlF₆ → 3NaF + AlF₃
Cryolite
4AlF₃ → 4Al³⁺ + 12F^–
At cathode : 4Al³⁺ + 12e^– → 4Al
At anode : 12F^– → 6F₂ + 12e^–
F₂ formed at the anode reacts with alumina and forms Aluminium fluoride.
2Al₂O₃ + 6F₂ → 4AlF₃ + 3O₂
Aluminium, produced at the cathode, sinks to the bottom of the cell. It is removed from time to time through topping hole.

Purification of Aluminium: (Hoope’s Process)

The impurities present are Si, Cu, Mn etc.,
The electrolytic cell used for refining of aluminium consists of iron tank lined inside with
carbon. This acts as anode. The tank contains three layers of fused masses. The bottom layer
contains impure aluminium. Middle layer contains mixture of AlF₃, NaF and BaF₂ saturated with Al₂O₃. Top layer contains pure aluminium and graphite rods kept in it act as cathode.

On passing current aluminium ions from the middle layer are discharged at the cathode as pure aluminium. Equivalent amount of aluminium from the bottom layer passes into middle layer.

Question 7.
Explain the reactions occuring in the blast furnace in the extraction of iron.
Answer:
In the Blast furnace, reduction of iron oxides takes place in different temperature ranges. Hot air is blown from the bottom of the furnace and coke is burnt to give temperature upto about 2200K in the lower portion itself, The burning of coke therfore supplies most of the heat required in the process. The CO and heat moves to upper part of the furnace. In upper part, the temperature is lower and the iron oxides (Fe₂O₃ and Fe₃O₄) coming from the top are reduced in steps to FeO. Thus, the reduction reactions taking place in the lower temperature range and in the higher temperature range, depend on the points of corresponding intersections in the Δ_rG^θ vs T plots. These reactions can be summarised as follows:

At 500 — 800 K (lower temperature range in the blast furnace)
3 Fe₂O₃ + CO → 2 Fe₃O₄ + CO₂
Fe₃O₄ + 4CO → 3 Fe + 4 CO₂.
Fe₂O₃ + CO → 2 FeO + CO₂
At 900- 1500 K (higher temperature range in the blast furnace)
C + CO₂ → 2 CO
FeO + CO → Fe + CO₂
Lime stone is also decomposed to CaO which removes silicate impurity of the ore as CaSiO₃ slag. The slag is in molten state and separåtes out from iron.

The iron obtained from blast furnace contains about 4% carbon and many impurities in smaller amount (e.g., S, P, Si, Mn). This is known as pig iron. Cast iron is different from and is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%) and is extremely hard and brittle.
Fe₂O₃ + 3 C → 2 Fe + 3 CO

Students get through AP Inter 2nd Year Chemistry Important Questions 4th Lesson Surface Chemistry which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 4th Lesson Surface Chemistry

Very Short Answer Questions

Question 1.
What is adsorption? Give one example.
Answer:
Adsorption: The accumulation (or) concentration of a substance on the surface rather than in the bulk of solid (or) liquid is known as adsorption.
Eg : Adsorption of gases like O₂, H₂, Cl₂ etc., on charcoal.

Question 2.
What is absorption ? Give one example.
Answer:
Absorption : The uniform distribution of a substance through out the bulk of the solid substance is known as absorption.
Eg : Chalk stick dipped in ink.

Question 3.
What is desorption ?
Answer:
Desorption : The process of removing an adsorbed substance from a surface on which it is adsorbed is called desorption.

Question 4.
What is sorption ? (or) What is the name given to the phenomenon when both absorption and adsorption take place together ?
Answer:
The process in which both adsorption and absorption takes place simultaneously is called sorption.

Question 5.
Give any two applications of adsorption.
Answer:
Applications of adsorption :
a) Separation of inert gases : Different noble gases adsorb at different temperatures on coconut charcoal. By this principle (adsorptipn) mixture of noble gas is separated by adsorption on coconut charcoal.

b) Gas masks : Gas mask is a device which consists of activated charcoal (or) mixture of adsorbents is used by coal miners to adsorb poisonous gases during breathing.

Question 6.
What is an adsorption isotherm ? Write the equation of Freundlich adsorption isotherm.
Answer:
Adsorption Isotherm : The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature can be expressed by means of a curve known as adsorption isotherm.

• Freundlich adsorption isotherm equation is \(\frac{\mathrm{x}}{\mathrm{m}}\) = k. P^1/n
  x = mass of the gas adsorbed
  m = mass of the adsorbent
  P, k and n are constants.

Question 7.
Define “promoters” and “poisons” in the phenomenon of catalysis. ‘
Answer:
Promoters: The substances which enhance the activity of catalyst are known as promoters.
Poisons : The substances which decrease the activity of a catalyst are known as poisons.

Question 8.
What is homogeneous catalysis ? How is it different from heterogeneous catalysis ?
Answer:
Homogeneous Catalysis : The catalysis in which reactants and catalyst are in same phase is called Homogeneous catalysis.

• In case of heterogeneous catalysis, catalyst and reactants are present in different phases where as in case of homogeneous catalysis catalyst and reactants are present in same phase.

Question 9.
What are enzymes ? What is their role in human body ?
Answer:
Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals.

• These act as specific catalysts in biological reactions.
• These catalyse the numerous reactions that occur in the bodies of animals and plants to maintain the life process.

Question 10.
Name any two enzyme catalyzed reactions. Give the reactions.
Answer:
1) Inversion of Cane Sugar :

2) Decomposition of urea into ammonia and CO₂ :

Question 11.
What is critical micelle concentration (CMC) and kraft temperature (T_k) ?
Answer:
The formation of micelles takes place only above a particular temperature called Kraft temperature (T_k) and above a particular concentration called Critical micelle concentration (CMC).

Question 12.
What is Peptization ?
Answer:
Peptization : The process of converting a precipitate into colloidal sol by shaking it with the dispersion medium in the presence of a small amount of electrolyte is called Peptization.

Question 13.
What is Brownian movement.
Answer:
Brownian movement: This is a kinetic property of colloidal solution.

When a colloidal solution is examined by ultramicroscope, the colloidal particles are seemed to be moving in a rapid zig-zag motion.
This rapid motion of colloidal particles is called Brownian movement.
This motion is due to unequal bombardment of colloidal particles by molecules of dispersion medium. Smaller the colloidal particles the more rapid is the Brownian movement.

Question 14.
What is Tyndall effect ?
Answer:
Tyndall effect : When light enters a colloidal solution, it is scattered by the large sized colloidal dispersed phase particles. Therefore when light passes through a solution we will be able to see the path of the light as a luminous beam. This is called Tyndall effect.

This is an optical property exhibited by colloidal solution. This phenomenon is clearly seen with a microscope placed at right angles to the path of light. Colloidal particles become self luminous due to absorption of light. A true solution does not show Tyndall effect.

Question 15.
What is electrokinetic potential or zeta potential ?
Answer:
In a colloidal sol the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers. The potential difference between the fixed layer and the diffused layer of opposite charge is called electro kinetic potential (or) zeta potential.

Question 16.
What is electrophoresis ?
Answer:
When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied emf is called “electrophoresis”.

Question 17.
What is coagulation ?
Answer:
The stability of the lyophilic sols is due to the presence of charge on the colloidal particles. If this charge is neutralised the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. This process of settling downward colloidal particles is called coagulation (or) flocculation (or) precipitation.

Question 18.
State Hardy – Schulze rule.
Answer:
Greater the valence of the coagulating ion added, the greater is its power to cause coagulation. This is known as Hardy-Schulze rule.

Question 19.
What is protective colloid ?
Answer:
Lyophilic colloids used for the prevention of coagulation of lyophobic colloids are called protective colloids.

• Lyophilic colloids protect the lyophobic colloids. .

Question 20.
What is an emulsion ? Give two examples. (AP Mar. 17)
Answer:
Emulsion : The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called emulsion. Eg : Milk, Vanishing cream, Cold cream.

Question 21.
What is an emulsifying agent ?
Answer:
Emulsifying agent: The third substance which is added in small amounts to an emulsion to stabilize the emulsion is called emulsifying agent.

Question 22.
Name any two applications of colloidal solutions.
Answer:
Applications of colloidal solutions :
Rubber: Plant latex is a colloidal solution of rubber particles which are negatively charged. Rubber is obtained from latex by coagulation.
Industrial Products: Paints, inks, synthetic plastics, rubber, graphite, lubricants, cement, etc., are all colloidal in nature.

Question 23.
Define the terms adsorbate and adsorbent.
Answer:
The substance which is adsorbed is called adsorbate. The substance on whose surface the adsorption takes place is called adsorbent.

Question 24.
What is Gold number ?
Answer:
The number of milligrams of a protective colloid required to prevent the coagulation of 10ml Gold sol when 1 ml of 10% NaCl solution is added is called Gold number.

Short Answer Questions

Question 1.
What are different types of adsorption ? Give any four differences between characteristics of these different types. (IPE Mar & May 2015 (AP), (TS), 2016 (TS))
Answer:
Adsorption process is divided into two types.

1. Physisorption
2. Chemisorption.

Distinguishing characteristics of Physisorption and Chemisorption are given in the following table:

Physisorption

1. This process is weak, due to Vander Waals forces.
2. The process is reversible.
3. This is a quick process i.e., takes place quickly.
4. The process decreases with increase of temperature.
5. This is a multilayered process.
6. The process depends mainly on the nature of the adsorbent.

Chemisorption

1. This process is strong, due to chemical forces.
2. The process is irreversible.
3. This is a slow process.
4. The process increases with increase of temperature.
5. This is a unilayered process.
6. The process depends both on the nature of adsorbent and adsorbate.

Question 2.
What is catalysis ? How is catalysis classified? Give two examples for each type of catalysis. (IPE 2015 (AP), 14, BMP. 2016 (AP))
Answer:
Catalysis : A substance which alters the rate of a chemical reaction without itšelf being consumed in the process, is called a catalyst.
The action of catalyst in altering the rate of a chemical reaction is called catalysis
Types of catalysis : Catalysis is classified into two týpes as
a) Homogeneous catalysis and
b) Heterogeneous catalysis.

Homogeneous catalysis: The catalytic process in which the catalyst is present in the same phase as that of reactants, is known as homogeneous catalysis.

Heterogeneous catalysis : The catalytic process in which the catalyst is present in a phase different from that of the reàctants is known as heterogeneous catalysis.

Question 3.
How can the constants k and n of the Freundlich adsorption equation be calculatéd?
Answer:
Freundlich adsorption isotherm equation is
\(\frac{x}{m}\) = k. P^1/n ⇒ x/m = Extent of adsorption ⇒ P = Pressure
k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.
Applying logarithm to the above equation
log \(\frac{x}{m}\) = log k + \(\frac{1}{n}\) log P.

• A graph is plotted taking log \(\frac{x}{m}\) on y – axis and log P on x – axis. If the graph is a straight line then Freundlich isotherm is valid.
• The slope of the straight line gives \(\frac{1}{n}\) value.
• The intercept on the y-axis gives value of log k.
• \(\frac{1}{n}\) has values between 0 and 1.
  When \(\frac{1}{n}\) = 0, \(\frac{x}{m}\) = constant, the adsorption is independent of pressure.
  \(\frac{1}{n}\) = 1, \(\frac{x}{m}\) = kP i.e., \(\frac{x}{m}\) ∝ p.

Question 4.
How are colloids classified on thé basis of interaction between dispersed phase and dispersion medium?
Answer:
Lyophilic colloid : The colloidal solution in which the dispersed phase has great affinity to the dispersion medium is called a Lyophilic colloid or Lyophilic solution.
Ex: Starch solution.

The starch paste when dissolved in hot water, with stirring, the starch solution is formed. The starch particles (dispersed phase) has great affinity to water molecules (dispersion medium). So starch solution is a lyophilic solution or lyophilic colloid.

Lyophobic colloid : The colloidal solution in which there exists not much affinity between the dispersed phase and dispersion medium, it is called a Lyophobic colloid or Lyophobic solution.
Ex: Gold solution.

Gold rods are placed in water containing alkali. Electric arc is applied between gold rods. The gold particles dissolves in water, to give gold solution.
Gold particles (dispersed phase) have not much affinity towards water (dispersion medium). So this is a Lyophobic solution or Lyophobic colloid.

Question 5.
Explain any 2 methods for the preparation of colloids.
Answer:
Method – 1: Bredig’s arc Method : This process consists of dispersion and condensation colloids of Gold, Platinum, Silver arc prepared by this method. In this method an electric arc is struck between the electrodes of the metal immersed in the dispersion medium. The heat produced vapourises the metal which then condensed to colloidal size.

Method – 2 : Colloids can be prepared by chemical reactions like double decomposition, oxidation, reduction and hydrolysis.
Double decomposition : As₂O₃ + 3H₂S → AS₂S₃ (sol) + 3H₂O
Oxidation : SO₂ + 2H₂S → 3S(sol) + 2H₂O
Reduction : 2 AuCl₃ + 3 HCHO + 3 H₂O → 2 Au (sol) + 3HCOOH + 6HCl
Hydrolysis : FeCl₃ + 3H₂O → Fe (OH)₃ (sol) + 3HCl

Question 6.
Discuss the use of colloids in
i) Purification of drinking water
ii) Tanning
iii) Medicines.
Answer:
i) Purification of drinking water: The water obtained from natural sources often contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.

ii) Tanning : Animal skins are colloidal in nature. When a skin, which has positively charged particles, is soaked in tannin, which contains negatively charged colloidal particles, mutual coagulation takes place. This results in the hardening of skin (leather). This process is termed as tanning. Chromium salts are also used in place of tannin.

iii) Medicines : Most of the medicines are colloidal in nature. For example argyrol is a silver sol used as an eye lotion. Colloidal antimony is used in curing kalaazar. Colloidal gold is used as intramuscular injection. Milk of magnesia an emulsion, is used for stomach disorders. Colloidal medicines are more effective because they have large surface area and are therefore easily assimilated.

Question 7.
What do you mean by activity and selectivity of catalysts?
Answer:
Activity:
The ability of a catalyst in increasing the rate of reaction is defined as activity of catalyst.

• The activity of a catalyst depends upon the strength of chemisorption to a large extent.
• The reactants must get adsorbed reasonably strongly onto the catalyst to become reactive.
  Eg: The catalystic activity increases from Group – 5 to Group – 11 for hydrogenation reactions.
  The maximum activity being shown by 7 – 9 group metals.
  

Selectivity:
The selectivity of a catalyst is its ability to direct a reacti6n to form specific products. The following reactions indicate the selectivity of heterogeneous catalysis.

• Starting with H₂ and CO, and using different catalysts, we get different products,
  

The action of a catalyst is highly sélective in nature. A substance which acts as a catalyst in one reaction may fail to catalyse another reaction.

Question 8.
What are emulsions? How are they classified? Describe the applications of emulsion. ( AP Mar. 2017; IPE 2016 (TS))
Answer:
Emulsion : The colloidal system in which a dispersion of finely divided droplets of a liquid in another liquid medium is called emulsion.
Ex: Milk.
In Milk, the droplets of liquid fat are dispersed in water. This is an example for oil in water type emulsion.
Classification of emulsions: Emulsions are classified into two classes. These are
a) Oil in Water (O/W) and
b) Water in Oil (W/O), (O = Oil; W = Water).

These emulsions are classified as such depending on which is dispersed phase and which is dispersion medium.

a) Oil in Water (O/W) type emulsions : In this type of emulsions, the dispersed phase is oil and the dispersion medium is water.
Ex: Milk, liquid, fat (oil) in water.
Vanishing cream; fat in water.

b) Water In Oil (W/O) type emulsions : In this type of emulsions, the dispersed phase is water and the dispersion medium is oil.
Ex : Stiff greases : water in lubrication oils
Cod liver oil : water in cod liver oil

Applications of Emulsions : Emulsions are useful

• In the digestion of fats in intestines.
• In washing processes of clothes and crockery.
• In the preparation of lotions, creams, ointments in pharmaceutical and cosmetics.
• In the extraction of metals (froth floatation).
• In the conversion of cream into butter by churning.
• To break oil and water emulsions in oil wells.
• In the preparation of oily type of drugs for easy adsorption to the body.

Question 9.
What is adsorption? Explain different types of adsorptions with suitable examples.
Answer:
The process of concentration of molecules of a gas (or) liquid on the surface of another substance is called adsorption. Adsorption is of two types,
(a) Physical adsorption
(b) Chemical adsorption.

a) Physical adsorption: It is also called Vander waals adsorption. A very weak Vander forces of attraction exists between adsorbate and adsorbent. It is multi layered and non-selective.
Ex: Adsorption of inert gases on activated coconut charcoal.

b) Chemical adsorption: It is also called chemisorption. A very strong chemical forces of attraction exists between adsorbate and adsorbent. It is mono-layered and highly selective.
Ex: Adsorption of H₂ gas on nickel surface.

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 3(b) Chemical Kinetics which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 3(b) Chemical Kinetics

Very Short Answer Questions

Question 1.
What is the Rate of a reaction?
Solution:
Rate of reaction: The rate [speed or velocity] of a reaction is the change in concentration of reactants or products per unit of time.
Rate = \(\frac{\Delta x}{\Delta t}\) Where, Δx is the change in concentration, At is the time interval.
The rate of reaction is always positive and it decreases as the reaction proceeds.
For the reaction, R → P, rate may be expressed as

Question 2.
What is Rate equation (or) Rate expression (or) Rate Law ?
Answer:
Rate equation or Rate expression or Rate Law is the mathematical expression in which aA + bB → cC + dD reaction rate is given in terms of molar concentration of reactants. Where a, b, e and d are the stoichiometric coefficients of reactants and products.
The rate expression for this reaction is Rate ∝ [A]^x [B]^y
Exponents x and y may or may not equal to stiochiometric coefficients a and b.
Rate = K [A]^x [B]^y Where K is proportionality constant called rate constant.

Question 3.
Write the differences between Order and Molecularity of a reaction. (IPE 2014, BIE) (A.P. Mar. ’18)
Solution:
Differences between Order and Molecularity of a reaction :

Order of reaction

1. Order is the sum of powers of concentration terms of reactants in the rate law.
   aA + bB → cC + dD Rate = K [A]^x [B]^y
   x + y is the total order of the reaction. x and y may or may not be equal to a and b respectively.
   Ex : 2H₂O₂ → 2H₂O + O₂ is an example for first order reaction.
2. It is determined by experiment.
3. It may be zero, positive, fractional 0, 1, 2, 1.5 etc.,

Molecularity of reaction

i) Molecularity is the no. of reacting species taking part in an elementary reaction.
Ex:
i) NH₄NO₂ → N₂ + 2H₂O
(Unimolecular)
Ex : ii) 2HI → H₂ + I₂ (Bimolecular)
Ex: iii) 2NO + O₂ → 2NO₂ (Trimolecular)

ii) It is determined by reaction mechanism.

iii) It cant be zero, fractional or negative. It is always a whole number.

Question 4.
The rate constant of a first order reaction (K) is 5.5 × 10^-14 sec^-1. Find the half life. (IPE 2015(AP))
Solution:
Half life for a first order reaction is, t_1/2 = \(\frac{0.693}{\mathrm{~K}}\)
t_1/2 = \(\frac{0.693}{5.5 \times 10^{-14} \mathrm{~S}^{-1}}\) = 1.26 × 10¹³S

Question 5.
What is Zero Order reaction ?
Solution:
Zero Order reaction: The rate of reaction is independent of the concentration of reactants.
For the reaction R → P, Rate = –\(\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = k[R]
[R] – Concentration of reactant
K – First order rate constant
Ex:
1) Decomposition of NH₃ on hot Pt surface ;
2) Decomposition of HI on Gold surface is a zero order reaction

Question 6.
What is First Order reaction ? Give example. (IPE 2014)
Solution:
First Order reaction: The rate of reaction which depends only on one concentration term.
For the reaction R → P, Rate = \(-\frac{d[R]}{d t}\) = k[R]
[R] – Concentration of reactant
K – First order rate constant
Ex:

1. Decomposition of N₂O₅ N₂O₅ (g) → 2NO₂(g) + 1/2 O₂ (g)
2. Decomposition of N₂O N₂O_(g) → N_2(g) + 1/2 O_2(g).

Question 7.
What are pseudo first order reactions ? Give one example.
Answer:
First order reactions whose molecularity is more than one are called pseudo first order reactions.

Order = 1
molecularity = 2

Question 8.
What is Half life of a reaction ?
Solution:
Half life of a reaction: The time in which the concentration of a reactant is reduced to one half of its initial concentration is called half life period. It is represented by t_1/2.
t_1/2 of n^th order reaction is given by t_1/2 ∝ [R₀]^1-n
Where R₀ is initial concentration
t_1/2 = \(\frac{\left[R_{0}\right]}{2 k}\) for a zero order reaction; t_1/2 = \(\frac{0.693}{k}\) for first order reaction

Question 9.
Define the speed or rate of a reaction.
Answer:
The change in the concentration of a réactant (or) product in unit time is called speed or rate of a reaction.
(or)
The decrease in the concentration of a reactant (or) increase in the concentration of product per unit time.

Question 10.
Define Order of a reaction. Illustrate your answer with an example. (IPE 2015; T.S. Mar. ’15)
Answer:
Order of a reaction : The sum of the powers of the concentration terms of the reactants present in the rate equation is called order of a reaction.

• Order of a reaction can be 0, 1, 2, 3 and even a fraction.

Eg:

1. N₂O₅ → N₂O₄ + \(\frac{1}{2}\)O₂
   rate ∝ [N₂O₅]
   ∴ It is a first order reaction.
2. 2N₂ → 2N₂ + O₂
   rate ∝ [N₂O]²
   ∴ It is 2^nd first order reaction.

Question 11.
What are elementary reactions?
Answer:
The reactions taking place in one step are called elementary reactions.

Question 12.
Give two examples for zero Order reactions.
Answer:
Examples for zero order reactions

Question 13.
Write the integrated equation for a first order reaction in terms of [R], [R₀] and ‘t’.
Answer:
[R] = Concentration of reaction after time T
[R₀] = Initial concentrations of reactant
∴ k = \(\frac{2.303}{t}\)log\(\frac{\left[R_0\right]}{[R]}\)
This is the integrated equation for a first order reaction

Question 14.
Give two examples for gaseous first order reactions.
Answer:
The following are the examples for gaseous first order reactions

Question 15.
What is a second order reaction? Give one example.
Solution:
Second order reaction : The reactions in which the rate of reaction depends on changing concentration of two substances is called second order reaction.
Ex : Alkaline hydrolysis of ester.

Question 16.
A reaction has a half – life of 10 minutes. Calculate the rate constant for the first order reaction. (IPE 2016 (TS)
Solution:
In case of first order reaction t_1/2 = \(\frac{0.693}{k}\)
∴ k = \(\frac{0.693}{\mathrm{t}_{1 / 2}}\) = \(\frac{0.693}{10}\) = 0.0693 min^-1

Question 17.
In a first order reaction, the concentration of the reactant is reduced from 0.6 mol/L to 0.2 mol/L in 5 min. Calculate the rate constant (k).
Solution:
a = 0.6mol L^-1; a – x = 0.2 mol L^-1; t = 5 min.
Since it is a first order reaction.
k = \(\frac{2.303}{\mathrm{t}}\) log₁₀ \(\frac{a}{(a-x)}\)
k = \(\frac{2.303}{t}\) log \(\frac{0.6}{0.2}\) = 0.2197 min^-1

Short Answer Questions

Question 1.
Define and explain the order of a reaction. How is it obtained experimentally ?
Answer:
Order of a reaction : The sum of the powers of the concentration terms of the reactants present in the rate equation is called order of a reaction.

• Order of a reaction can be 0, 1, 2, 3, and even a fraction.

Eg:

1) N₂O₅ → N₂O₄ + \(\frac{1}{2}\)O₂
rate ∝ [N₂O₅]
∴ It is a first Order reaction.

2) 2N₂ → 2N₂ + O₂
rate ∝ [N₂O]²
∴ It is 2^nd order reaction

• Order of a reaction can be determined experimentally.

Half-Time (t_1/2) method: The time required for the initial concentration (a) of the reactant to become half its value (a/2) during the progress of the reaction is called half-time (t_1/2) of the reaction.
A general expression for the half life, (t_1/2) is given by
t_1/2 ∝ \(\frac{1}{a^{n-1}}\)

Therefore, for a given reaction two halftime values (t’_1/2 and t”_1/2) with initial concentrations a’ and a” respectively are determined experimentally and the order is established from the equation.
\(\left(\frac{\mathrm{t}_{1 / 2}^{\prime}}{\mathrm{t}_{1 / 2}^{\prime}}\right)\) = \(\left(\frac{a^n}{a^{\prime}}\right)^{n-1}\)
Where ‘n’ is the order of the reaction.

Question 2.
Describe the salient features of the collision theory of reaction rates of bimolecular reactions.
Answer:
Collision theory of reaction rate bimolecular reactions salient features.

• The reaction molecules are assumed to be hard spheres.
• The reaction is postulated to occur when molecules collide with each other.
• The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z).
• For a bimolecular elementary reaction.

A + B → products
Rate = Z_AB. e^-E_a/RT ; Z_AB = collision frequency.

• All collisions do not lead to product formation.
• The collisions with sufficient kinetic energy (Threshold energy) are responsible for product formation. These are called as effective collisions.
• To account for effective collisions a factor p called to probability factor or steric factor is introduced.
  Rate = P Z_AB. e^-E_a/RT

Question 3.
What is “moleculartiy” of a reaction ? How is it different from the ‘order’ of a reaction ? Name one bimolecular and one trimolecular gaseous reactions.
Answer:
The number of reacting species (atoms, ions or molecules) taking parts in an elementary reaction, which must colloid simultaneously to bring about a chemical reaction is called molecularity of a reaction.
NH₄NO₂ → N₂ + 2H₂O (Unimolecular)
2HI → H₂ + I₂ (Bimolecular)
2NO + O₂ → 2NO₂ (Trimolecular)

• Molecularity has only integer values (1, 2, 3, ….)
• It has non zero, non fraction values while order has zero, 1, 2, 3, ….. and fractional values.
• It is determined by reaction mechanism, order is determined experimentally.

Question 4.
What is half-life (t_1/2) of a reaction ? Derive the equations for the ‘half-life’ value of zero and first order reactions.
Answer:
The time required for the initial concentration of the reactants to become half of it’s value during the progress of the reaction is called half life (t_1/2) of reaction.
Eg : The radio active of C-14 is exponential with a half life of 5730 years.
Half life of zero order reaction :

Question 5.
Explain the terms
a) Activation energy (E_a)
b) Collision frequency (Z)
c) Probability factor (P) with respect to Arrhenius equation.
Answer:
a) Activation Energy: The energy required to form an intermediate called activated complex (C) during a chemical reaction is called activation energy.

b) Collision frequency: The number of collisions per second per unit volume of the reaction mixture is called collision frequency (Z). For a bimolecular elementary reactions
A + B → products

c) Probability factor (P) with respect to Arrhenius equation : To account for effective collisions a factor p called to probability factor or steric factor is introduced.

Question 6.
Explain the factors influencing rate of reaction. (IPE 2015 (TS))
Solution:
Factors affecting rate of a reaction: The rate of a chemical reaction depends on the following.

i) Concentration of reactants : The rate of a chemical reaction is directly proportional to the concentration of the reactants. As the concentration of reactants increases the number of molecules increase, collisions between molecules increases as a result number of fruitful collisions increases, hence rate of reaction increases.

ii) Nature of reactants : Reactions between ionic compounds are fast when compared to reactions between the covalent compounds. In case of ionic reactions only exchange of ions takes place, hence the reactions between ionic compounds in solutions are very fast. In case of covalent molecules breaking and formation of bonds involved hence the reactions between covalent molecules are very slow.

iii) Temperature : As temperature increases, the rate of reaction increases. For every 10°C rise of temperature the rate of reaction is almost doubled. At high temperatures maximum number of molecules are activated. The collisions between activated molecules are fruitful, hence the rate of reaction increases.

iv) Catalyst: In the presence of catalyst also the rate of reaction increases. Catalyst takes the reaction in a new path having low activation energy.

Question 7.
Show that in the case of first order reaction, the time required for 99.9% completion of the reaction is 10 times that required for 50% completion. (log 2 = 0.3010)
Solution:
Initial concentration, a = 100; Half life period, t_1/2 = 0.693/K
Concentration after time ‘t’ = a – x = 100 – 99.9 = 0.1
For first order reaction, K = \(\frac{2.303}{t}\)log\(\frac{a}{(a-x)}\)

Question 8.
Derive an integrated rate equation for a first order reaction.
Answer:
In first order reactions rate depends on only one concentration term.
R → P
Rate = k[R]; \(\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}\) = -k. dt
Integration on both sides
ln[R] = -kt + I ——- (1)
I = Integration constant
At t = 0, [R] = [R₀] ⇒ l_n[R₀] = I
Substituting I = ln [R] in the above equation (1)
lñ [R] = -kt + ln[R₀]
ln \(\frac{[\mathrm{R}]}{\left[\mathrm{R}_0\right]}\) =-kt . ——— (2)
k = \(\frac{1}{\mathrm{t}} \ln \frac{\left[\mathrm{R}_0\right]}{[\mathrm{R}]}\)
Taking antilog on both sides of eq. (2)
R = [R_o]. e^-kt
This is first order rate equation.

Students get through AP Inter 2nd Year Chemistry Important Questions Lesson 3(a) Electro Chemistry which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions Lesson 3(a) Electro Chemistry

Very Short Answer Questions

Question 1.
What is a galvanic cell or a voltaic cell? Give one example.
Answer:
Galvanic cell: A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (or) voltaic cell.
Eg.: Daniell cell.

Question 2.
What is standard hydrogen electrode ?
Answer:
The electrode whose potential is known as standard electrode (or) standard hydrogen electrode.
To determine the potential of a single electrode experimentally it combine with standard hydrogen electrode and the EMF of cell so constructed is measured with potentiometer.

Question 3.
What is Nernst equation ? Write the equation for an electrode with electrode reaction
Mⁿ⁺ (aq) + ne^– \(\rightleftharpoons\) M_(s).
Answer:
The electrode potential at any concentration measured with respect to standard hydrogen electrode is represented by Nernst equation.
Nernst equation is

Here, E_(Mⁿ⁺/M) = Electrode potential
E⁰_(Mⁿ⁺/M) = Standard electrode potential
R = gas constant = 8.3 14 J/k.mole
F = Faraday = 96487 c/mole
T = temperature .
[Mⁿ⁺] = concentration of species Mⁿ⁺

Question 4.
How is E⁰ cell related mathematically to the equilibrium constant K_c of the cell reaction?
Answer:
Relation between E₀ cell and equilibrium constant K_c of the cell reaction
\(E_{\text {cell }}^0\) = \(\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{c}}\)
n = number of electrons involved
F = Faraday = 96500 C mol^-1
T = Temperature
R = gas constant

Question 5.
How is Gibbs energy (G) related to the cell emf (E) mathematically ?
Answer:
Relation between Gibb’s energy (G) and emf (E) mathematically
ΔG° = – nFE_(cell)
ΔG = change in Gibb’s energy
n = number of electrons involved
F = Faraday = 96500 C mol^-1

Question 6.
Define conductivity of a material. Give its SI units.
Answer:
The reciprocal of specific resistance (or) resistivity is called conductivity. It is represented by (K)
(Or)
The conductance of one unit cube of a conductor is also called conductivity.
SI units : ohm^-1 m^-1 (or) Sm^-1 S = Siemen

Question 7.
What is cell constant of a conductivity cell ?
Answer:
Cell constant:

The cell constant of a conductivity cell is the product of resistance and specific conductance.

Question 8.
State Faraday’s second law of electrolysis.
Answer:
The amounts of different substances liberated when the same quantity of electricity is passed through the electrolytic solution are proportional to their chemical equivalent weights.
m ∝ E

Question 9.
State Kohlrausch’s law of independent migration of ions.
Answer:
Kohlrausch’s law of independent migration of ions : The limiting molar conductivity of the electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.
\(\lambda_{m(A B)}^0\) = \(\lambda_{\mathrm{A}^{+}}^0\) + \(\lambda_{\mathrm{B}^{-}}^0\)
\(\lambda_m^0\) = Limiting molar conductivity
\(\lambda_{\mathrm{A}^{+}}^0\) = Limiting molar conductivity of cation ‘
\(\lambda_{\mathrm{B}^{-}}^0\) = Limiting molar conductivity of anion

Question 10.
State Faraday’s first law of electrolysis.
Answer:
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte.
(Or)
The mass of the substance deposited at an electrode during the electrolysis of electrolyte is directly proportional to quantity of electricity passed through it.
m ∝ Q; m ∝ c × t
m = ect; m = \(\frac{\text { Ect }}{96,500}\)
e = electrochemical equivalent
t = time in seconds
c = Current in amperes
E = Chemical equivalent

Question 11.
What is a primary battery? Give one example. (AP Mar. ’17)
Answer:
The batteries which after their use over a period of time, becomes dead and the cell reaction is completed and this cannot be reused again are called primary batteries.
Eg: Leclanche cell, dry cell.

Question 12.
Give one example for a secondary battery. Give the cell reaction.
Answer:
Lead storage battery is an example of secondary battery.
The cell reactions when the battery is in use are

Question 13.
What is metallic corrosion? Give one example. (TS Mar. 2017, IPE ‘15 (AP))
Answer:
Metallic corrosion : The natural tendency of conversion of a metal into its mineral compound form on interaction with the environment is known as metallic corrosion.
Eg:

1. Iron converts itself into its oxide.
   [Rusting] (Fe₂O₃)
2. Silver converts itself into it’s sulphate
   [tarnishing] [Ag₂S]

Question 14.
Define electrochemical equivalènt (e.c.e).
Answer:
The weight of the substance deposited or liberated when one ampere of current is passed for 1 second (1 coloumb) is called electrochemical equivalent (e.c.e).

Question 15.
A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? (IPE 2015 (AP), 14)
Solution:
t =600 s charge = current × time = 1.5A × 600 s = 900 C
According to the reaction:

We require 2F or 2 × 96487 C to deposit 1 mol or 63 g of Cu.

Question 16.
Can you store copper sulphate solutions in a zinc pot ?
Solution:
No, zinc pot cannot store copper sulphate solutions because the standard electrode potential (E⁰) value of zinc is less than that of copper. So, zinc is stronger reducing agent than copper.

So, zinc will loss electrons to Cu²⁺ ions and redox reaction will occur as follows.

Question 17.
Write the cell reaction taking place in the cell, Cu(s) / Cu²⁺ (aq) //Ag⁺ (aq) / Ag (s).
Answer:
Cu_(s)/Cu⁺²_(aq)//Ag⁺_(aq)/Ag
In the above notation copper electrode acts as anode and silver electrode acts as cathode.

Question 18.
Write the Nernst equation for the EMF of the cell

Answer:

Question 19.
Write the cell reaction for which

Answer:

Short Answer Questions

Question 1.
What are galvanic cells ? Explain the working of a galvanic cell with a neat sketch taking Daniell cell as example.
Answer:
Galvanic cell: A device which converts chemical energy into electrical energy by the use of spontaneous redox reaction is called Galvanic cell (or) voltaic cell. Eg : Daniell cell.

Daniell cell: It is a special type of galvanic cell. It contains two half cells in the same vessel. The vessel is divided into two chambers. Left chamber is filled with ZnSO₄ (aq) solution and Zn – rod is dipped into it. Right chamber is filled with aq. CuSO₄ solution and a copper rod is dipped into it. Process diaphragm acts as Salt bridge. The two half cell’s are connected to external battery.

Cell reactions :
Ion Zn/ZnSO₄ half cell, oxidation reaction occurs.
Zn → Zn²⁺ + 2e^–
Ion Cu/CuSO₄ half cell, reduction reaction occurs.
Cu⁺² + 2e^– → Cu
The net cell reaction is
Zn + Cu⁺² \(\rightleftharpoons\) Zn⁺² + Cu
Cell is represented as Zn / Zn⁺² || Cu²⁺ / Cu

Question 2.
Give the construction and working of a standard hydrogen electrode with a neat diagram.
Answer:
To determine the potential of a single electrode experimentally, it is combined with a standard hydrogen electrode (electrode whose potential is known) and the EMF of the cell so constructed is measured with a potentiometer. Standard Hydrogen Electrode is constructed and is used as standard electrode or reference electrode. Standard Hydrogen Electrode (SHE) or Normal Hydrogen Electrode (NHE).

Pure hydrogen gas is bubbled into a solution of 1M HCl along a platinum electrode coated with platinum block. A platinum block electrode placed in the solution at atmospheric pressure.

Generally the electrode is fitted into the tube. The tube will have two circular small holes. This tube is immersed in the acid solution such that one half of the circular hole is exposed to air and another half in the solution.

The following equilibrium exists at the electrode.
\(\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}\left(1_{\mathrm{atm}}\right) \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}(\mathrm{M})+\mathrm{e}^{-}\)

Question 3.
State and explain Kohlrausch’s law of independent migration of ions. (IPF. 2015, BMP, 2016 (TS)
Answer:
Kohlrausch’s law of independent migration of ions : The limiting molar conductivity of the electrolytes can be represented as the sum of the individual contributions of the anion and the cation of the electrolytes.
\(\lambda_{\mathrm{m}(\mathrm{AB})}^0=\lambda_{\mathrm{A}^{+}}^0+\lambda_{\mathrm{B}^{-}}^0\)
\(\lambda_{\mathrm{m}}^0\) = Limiting molar conductivity
\(\lambda_{\mathrm{A}^{+}}^0\) = Limiting molar conductivity of cation
\(\lambda_{\mathrm{B}^{-}}^0\) = Limiting molar conductivity of anion,

Applications :

1. Kohlrausch’s law is used in the calculation of the limiting molar conductivity of weak electrolytes.
   
2. This law is used in the calculation of degree of dissociation of a weak electrolyté.
3. This law is used in the calculation of solubility of sparingly soluble salts like AgCl, BaSO₄ etc.

Question 4.
What is electrolysis? Give Faraday’s first law of electrolysis. (IPE 2014)
Answer:
Electrolysis : The decomposition of a chemical compound in the molten state or in the solution state into its constituent elements under the influence of an applied EMF is called electrolysis.
The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of current passing through the electrolyte.
(Or)
The mass of the substance deposited at an electrode during the electrolysis of electrolyte is directly proportional to quantity of electricity passed through it.

e = electrochemical equivalent
t = time in seconds
c = Current in amperes
E = Chemical equivalent

Question 5.
Calculate the emf of the cell at 25°C Cr | Cr³⁺ (0.1 M) || Fe²⁺ (0.01M)| Fe, given that

Answer:
Given cell is

Question 6.
Determine the values of K_c for the following reaction.

Solution:

Question 7.
If a current of 0.5 ampere flows through a metallic wire for 2h, then how many electrons would flow through the wire ?
Solution:
Quantity of charge (Q) passed = Current (C) × Time (t)
= (0.5 A) × (2 × 60 × 60 s)
= (3600) Ampere sec = 3600 C
Number of electrons flowing through the wire on passing charge of one Faraday (96500 C) = 6.022 × 10²³
Number of electrons flowing through the wire on passing a charge of 3600 C
= \(\frac{6.022 \times 10^{23} \times(3600 \mathrm{C})}{(96500 \mathrm{C})}\)
= 2.246 × 10²²
Number of electrons = 2.246 × 10²²

Question 8.
Define emf. Calculate the emf of the following galvanic cell:

Solution:
EMF : The difference in electrode potentials between reduction potential of cathode and reduction potential of anode is called emf.
emf = reduction potential of cathode – reduction potential of anode
emf = + 0.34 – (-0.76) ⇒ + 0.34 + 0.76 = 1.1V

Question 9.
Write Nernst equation for a metal and non metal eletrode.
Solution:
For a metal electrode M : M_+n (aq) + ne^– → M(s)

Students get through AP Inter 2nd Year Chemistry Important Questions 2nd Lesson Solutions which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 2nd Lesson Solutions

Very Short Answer Questions

Question 1.
Define molarity. [AP IPE ’15] [TS Mar. ’17 ’11]
Answer:
Molarity : The number of moles of solute dissolved in one litre of solution is called molarity.

Units : moles / kg.

Question 2.
Define molality. [AP IPE 2015, May ’11]
Answer:
Molality : The number of moles of solute present in one kilogram of solvent is called molality.

Units : moles / kg.

Question 3.
Define mole fraction. (AP IPE 2015, Mar. ’14)
Answer:
Mole fraction : The ratio of number of moles of the one component of the solution to the total number of moles of all the components of the solution is called mole fraction.
Mole fraction of solute X_s = \(\frac{n_{\mathrm{s}}}{\mathbf{n}_0+\mathrm{n}_{\mathrm{s}}}\)
Mole fraction of solvent X₀ = \(\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_0+\mathrm{n}_{\mathrm{s}}}\)
[n_s = number of moles of solute
n₀ = number of moles of solvent]
→ It has no units.

Question 4.
State Raoult’s law. (AP Mar. ’17, ’14 IPE ‘16, 14 (AP, TS)
Answer:
Raoult’s law for volatile solute: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
Raoults law for non-válatlle solute : The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal to the mole fraction of solute.

Question 5.
State Henrýs law. (TS IPE ’16)
Answer:
At constant temperature, the solubility of a gas in a liquid -is directly proportional to the partial pressure of the gas present above the surface of liquid.
The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution.
P = K_H × X
K_H = Henry’s constant
P = Partial pressure
X = Mole fraction of gas

Question 6.
What is Ebullioscopic constant ?
Answer:
Ebullioscopic constant : The elevation of boiling point observed in one molal solution containing non-volatile solute is called Ebullioscopic constant (or) molal elevation constant.

Question 7.
What are isotonic solutions ?
Ans. Isotonic solutions : Solutions having same osmotic pressure at a given temperature are called isotonic solutions.
e.g.: Blood is isotonic with 0.9% (\(\frac{\mathrm{W}}{\mathrm{V}}\)) NaCl [Saline]

Question 8.
What is Cryoscopic constant?
Answer:
Cryoscopic constant : The depression in freezing point observed in one molal solution containing non-volatile solute is called cryoscopic constant (or) molal depression constant.

Question 9.
Define osmosis and osmotic pressure. (IPE 15, (AP) ‘16 (TS), (AP))
Answer:
Osmosis : The spontaneous flow of solvent particles from a solution of lower concentration in to higher concentration through semi permeable membrane is called osmosis.

Osmotic pressure : The pressure required to prevent the flow of solvent particles from pure
solvent into solution through semipermeable membrane is called osmotic pressure.
π = (nRT) / V Here π = Osmotic pressure; V = Volume of solution;
n = No. of moles of solute; R = Universal gas constant: T = Absolute temperature

Question 10.
What is Van’t Hoff’s factor ‘i’ and how is it related to ‘α’ in the case of a binary electrolyte (1:1)?
Answer:
Van’t Hoff’s factor (i): ‘It is defined as the ratio of the observed value of colligative property to the theóretical value of colligative property”.

Solute dissociation or ionization process: If a solute on ionization gives ‘n ions and ‘α’ is degree of ionization at the given concentration, we will have [1 + (n – 1) α] particles.
(1 – α) \(\rightleftharpoons\) nα
Total 1 – α + nα = [1 + (n – 1)α]
∴ i = \(\frac{[1+(n-1) \alpha]}{1}\)
α = \(\frac{\mathrm{i}-1}{\mathrm{n}-1}\)
α_ionization = \(\frac{\mathrm{i}-1}{\mathrm{n}-1}\)
Solute association process:
If ‘n’A molecules combine to give A, we have
nA \(\rightleftharpoons\) A_n
If ‘α’ is degree of association at the given concentration.

Question 11.
What is relative lowering of vapour pressure? [IPE ’16, (TS)]
Answer:
Relating lowering of vapour pressure : The ratio of lowering of vapour pressure of a solution containing non-volatile solute to the vapour pressure of pure solvent is called relative lowering of vapour pressure.
R.L.V.P. = \(\frac{\mathrm{P}_0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_0}\)
P₀ – P_S = lowering of vapour pressure
P_o = Vapour pressure of pure solvent

Question 12.
What is vapour pressure of a liquid?
Answer:
The pressure exerted by vapour over liquid when it is in equilibrium with the liquid is called vapour pressure.

Question 13.
What is elevation of boiling point? (IPE 16, (TS) )
Answer:
The boiling point of a solution containing non-volatile solute is higher than the boiling point of pure solvent. The difference in boiling points between solution and pure solvent is called elevation of boiling point.

Question 14.
What is depression of freezing point?
Answer:
The freezing point of a solution containing non-volatile solute is always lower than the freëzing point of puré solvent. The difference in freezing points between solution and pure solvent is called depression of freezing point.

Question 15.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 ml of
0.15 M solution in methanol.
Answer:
Given
Molarity = 0.15 M
Volume V = 250 ml
Molecular weight of benzoic acid (C₆H₅COOH) = 122
Molarity (M) = \(\frac{\text { Weight }}{\mathrm{GMw}} \times \frac{1000}{\mathrm{~V}(\mathrm{~m} l)}\)
0.15 = \(\frac{\mathrm{W}}{122} \times \frac{1000}{250}\)
W = \(\frac{122 \times 0.15}{4}\) = 4.575 gms.

Question 16.
Calculate the mole fraction of H₂SO₄ in a solution containing 98% H₂SO₄ by mass. (T.S. Mar. ’18; A.P Mar. ‘17, IPE ‘15, (AP))
Answer:
Given a solution containing – 98% H₂SO₄ by mass.
It means 98 gms of H₂SO₄ and 2 gms of H₂O mixed to form a solution.

Question 17.
Calculate the molarity of a solution containing 5g of NaOH in 450 ml solution. (T.S. Mar. ’19, ’17 )
Solution:
Moles of NaOH = \(\frac{5 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.125 mol
Volume of the solution in litres = 450 mL / 1000 mL L^-1
Using equation

Question 18.
200 cm³ of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10^-3 bar. Calculate the molar masš of the protein.
Sólution:
The various quantities known to us are as follows : π = 2.57 × 10^-3 bar.
V = 200 cm³ = 0.200 litre
T = 300 K
R = 0.083 L bar moL^-1 K^-1
Substituting these values in equation,
we get M₂ = \(\frac{\mathrm{w}_2 \mathrm{RT}}{\pi \mathrm{V}}\)
M₂ = \(\frac{1.26 \mathrm{~g} \times 0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{2.57 \times 10^{-3} \mathrm{bar} \times 0.200 \mathrm{~L}}\)
= 61.022 g mol^-1

Question 19.
Calculate the molality of 10g of glucose in 90g of water.
Answer:

Question 20.
Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (Ccl₄) if 22g of benzene is dissolved in 122g of carbon tetrachloride.
• Then, calculate the mass percentage from the formula

Solution:
Mass of benzene = 22g; Mass of CCl₄ = 122g
Mass of solution = 22 + 122 = 144g
Mass % of benzene = \(\frac{22}{144}\) × 100 = 15.28%
Mass of CCl₄ = 100 – 15.28 = 84.72%
Note: Mass percent of CCl₄ can also be calculated by using the formula as:
Mass % of CCl₄ = \(\frac{122}{144}\) × 100 = 84.72%

Question 21.
What are colligative properties? Give their names.
Answer:
The properties of a solution which depend on number of solute particles but not on the nature are called colligative properties.
Names:

1. Lowering of vapour pressure
2. Elevation of boiling point.
3. Depression of freezing point.
4. Osmotic pressure.

Question 22.
Calculate the weight of Glucose required to prepare 500 ml of 0.1 M solution. (IPE ‘16, (TS))
Answer:
Molarity = 0.1; Molecular weight of glucose = 180; Volume of solution = 500 ml

Short Answer Questions

Question 1.
What is an ideal solution?
Answer:
A solution of two or more components which obeys Raoult’s law at all concentrations and at all temperatures is called ideal solution. In ideal solution there should not be any association between solute and solvent, (Le.) no chemical interaction between solute and solvent of solution.

Ex : The following mixtures form ideal solutions.

• Benzene + Toluéne ‘
• n – hexane + n – heptane
• ethyl bromide + ethyl iodide

Question 2.
What is relative lowering of vapour pressure? How is it useful to determine the molar mass of a solute?
Answer:
Raoult’s law for volatile solute: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
Raoult’s law for non-volatile solute : The relative lowering of vapour pressure of dilute solution containing non-volatile solute is equal to the mole fraction of solute.

Relative lowering of vapour pressure \(\frac{\mathrm{P}_0-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_0}\) = X_s (mole fraction of solute)
\(\frac{\mathrm{P}_0-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_0}\) = \(\frac{\mathbf{n}_{\mathrm{s}}}{\mathrm{n}_0+\mathrm{n}_{\mathrm{s}}}\)
For very much dilute solutions n_s < < < ….. n₀
∴ \(\frac{\mathrm{P}_0-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_0}\) = \(\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_0}\) = \(\frac{\mathrm{w}}{\mathrm{m}}\) × \(\frac{\mathrm{M}}{\mathrm{W}}\)
W = Weight of solute
m = Molar mass of solute
w = Weight of solvent
M = Molar mass of solvent
Molar mass of solute m = \(\frac{w \times M}{W} \times \frac{P_0}{P_0-P_5}\)

Question 3.
The vapour pressure of a solution containing non volatile solute Is less than the vapour pressure of pure of solvent. Give reason.
Answer:
In pure solvent the surface is occupied by solvent molecules only. Number of molecules evapoarating will be more, hence vapour pressure is more. In case of solution the surface is occupied by both solute and solvent molecules. The number of molecules evaporating will be less in case of solution containing non volatile solute. Hence the vapour pressure of solution containing non volatile solute is less than the vapour pressure of pure solvent.

Question 4.
An antifreeze solution is prepared from 222.6g of ethylene glycol [(C₂H₆O₂)] and 200g of water (solvent). Calculate the molality of the solution.
Solution:
Weight of Ethylene glycol = 222.6 gms
G.mol wt = 62

Question 5.
Vapour pressure of water at 293K is 17.535 mm Hg. Calculate the vapour pressure of the solution at 293K when 25g of glucose is dissolved in 450g of water? (BMP)
Answer:
Raoult’s law formula

Question 6.
The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5g when added to 39.0 g of benzene (molar mass 78 g mol^-1). Vapour pressure-of the solution, then, is 0.845 bar. What is the molar mass of the solid substance? (IPE 2016 (AP))
Solution:
The various quantities known to us are as follows.
\(\mathrm{p}_1^0\) = 0.850 bar
P = 0.845bar
M₁ = 78 g mol^-1
w₂ = 0.5 g
w₁ = 39 g
Substituting these values in equation \(\frac{\mathrm{P}^0-\mathrm{P}}{\mathrm{P}_1^0}\) = \(\frac{\mathrm{w}_2 \times \mathrm{M}_1}{\mathrm{M}_2 \times \mathrm{w}_1}\) we get

Therefore, M₂ = 170 g mol^-1

Question 7.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50g urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
• Consider Raoults law and formula for relative lowering in vapour pressure,
\(\frac{\mathbf{p}_{\mathbf{A}}^0-\mathbf{p}_{\mathrm{s}}}{\mathbf{p}_{\mathbf{A}}^0}\) = \(\frac{\mathbf{n}_{\mathbf{B}}}{\mathbf{n}_{\mathbf{A}}}\) = \(\frac{\mathbf{W}_{\mathrm{B}}}{\mathbf{M}_{\mathrm{B}}}\) × \frac{\mathbf{M}_{\mathbf{A}}}{\mathbf{W}_{\mathbf{A}}}\(\)
Where, \(\frac{\mathbf{p}_{\mathrm{A}}^0-\mathbf{p}_{\mathrm{s}}}{\mathbf{p}_{\mathrm{A}}^0}\) is called relative lowering in vapour pressure.
Solution:

Step 1: Calculation of vapour pressure of water for this solution.
According to Raoult’s law,

(Pure water) \(\mathrm{p}_{\mathrm{A}}^0\) = 23.8 mm; .
W_B (urea) = 50 g; W_A (water) = 850 g
M_B (urea) = 60 g mol^-1; M_A (water) = 180 g mol^-1
Placing the values in eq. (i)

Step II: Calculation of relative lowering of vapour pressure
Relative lowering in vapour pressure = \(\frac{\mathrm{p}_{\mathrm{A}}^0-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{A}}^0}\) = \(\frac{(23.8-23.38) \mathrm{mm}}{(23.8 \mathrm{~mm})}\) = 0.0176

Question 8.
A solution of sucrose in water is labelled as 20% w/W. What would be the mole fraction of each component in the solution? (IPE ‘2014)
Answer:
20% \(\frac{w}{W}\) Sucrose in water solution means
20 gms of Sucrose and 80 gms of water.
Number of moles of sucrose (n_s) = \(\frac{20}{342}\) = 0.0584
Number of moles of water (n₀) = \(\frac{80}{18}\) = 4.444
Mole fraction of sucrose X_s = \(\frac{n_s}{n_0+n_s}\)
= \(\frac{0.0584}{4.444+0.0584}\) = \(\frac{0.0584}{4.50284}\) = 0.01296
Mole fraction of water X_o = \(\frac{\mathbf{n}_0}{\mathrm{n}_0+\mathrm{n}_{\mathrm{s}}}\) = \(\frac{4.444}{4.50284}\) = 0.9869
X_s + X₀ = 1
X₀ = 1 – X_s = 1 – 0.01296 = 0.987

Question 9.
Calculate the vapour pressure of a solution containing 9g of glucose in 162g of water at 293K. The vapour pressure of water of 293K is 17.535mm Hg. (IPE ‘15, ’14, BOARD MODEL PAPER)
Solution:
Weight of solute (w) = 9g; Weight of solvent (W) = 162g
Molecular weight of solute (mw) = 180; molecular weight of solvent Mw = 18
Vapour pressure of pure solvent = 17.535 vapour pressure of solution P_s = ?
\(\frac{P_0-P_3}{P_0}\) = \(\frac{W}{m w} \times \frac{M W}{W}\)
⇒ \(\frac{17.535-P_s}{17.535}\) = \(\frac{9}{180} \times \frac{18}{162}\)
17.535 – P_s = 17.535 × \(\frac{9}{180}\) × \(\frac{18}{162}\)
⇒ 17.535 – P_s = 0.0972
∴ P_s = 17.535 – 0.0972 = 17.4378 mm

Students get through AP Inter 2nd Year Chemistry Important Questions 1st Lesson Solid State which are most likely to be asked in the exam.

AP Inter 2nd Year Chemistry Important Questions 1st Lesson Solid State

Very Short Answer Questions

Question 1.
What is meant by the term coordination number?
Answer:
The number of nearest neighbouring particles of a particle is defined as the coordination number.
(Or)
The number of nearest oppositely charged ions surrounding a particular ion is also called a coordination number.
E.g.: Co-ordination no. of Na⁺ in NaCl lattice is ‘6’.

Question 2.
What is the coordination number of atoms in a cubic close-pack structure?
Answer:
The coordination number of atoms in a cubic close pack structure is ’12’.

Question 3.
What is the coordination number of atoms in a body-centered cubic structure?
Answer:
The coordination number of atoms in a body-centered cubic structure is ‘8’.

Question 4.
How do you distinguish between crystal lattice and unit cells? [Board Model Paper]
Answer:
Crystal lattice: A regular arrangement of the constituent particles of a crystal in the three-dimensional space is called a crystal lattice.
Unit cell: The simple unit of a crystal lattice that when repeated again and again gives the entire crystal of a given substance is called a unit cell.

Question 5.
What is Schottky defect ? [A.P. IPE 2015]
Answer:
Schottky defect:

1. “It is a point defect in which an atom or ion is missing from its normal site in the lattice”.
2. In order to maintain electrical neutrality, the number of missing cations and anions are equal.
3. This sort of defect occurs mainly in highly ionic compounds, where cationic and anionic sizes are similar.
   In such compounds the co-ordination number is high.
   Ex.: NaCl, CsCl etc.
4. Illustration :
   
5. This defect decreases the density of the substance.

Question 6.
What is Frenkel defect ? [A.P. IPE 2015]
Answer:
Frenkel defect:

1. “It is a point defect in which an atom or ion is shifted from its normal lattice position”. The ion or the atom now occupies an interstitial position in the lattice.
2. This type of a defect is favoured by a large difference in sizes between the cation and anion. In these compounds co-ordination number is low.
   E.g.: Ag – halides, ZnS etc.
3. Illustration :
   
4. Frenkel defect do not change the density of the solids significantly.

Question 7.
What are f – centers ?
Answer:

• f – centers are the anionic sites occupied by unpaired electrons.
• These import colour to crystals. This colour is due to the excitation of electrons when they absorb energy from the visible light.
• f – centres are formed by heating alkyl halide with excess of alkali metal.
  E.g. : NaCl crystals heated in presence of Na – vapour, yellow colour is produced due to f – centres.

Question 8.
Why X – rays are needed to probe the crystal structure ?
Answer:
According to the principles of optics, the wavelength of light used to observe an object must be no greater than the twice the length of the object itself. It is impossible to see atom s using even the finest optical microscope. To see the atoms we must use light with a wavelength of approximately 10^-10 m. X – rays are present with in this region of electromagnetic spectrum. So X – rays are used to probe crystal structure.

Question 9.
Explain Ferromagnetism with suitable example.
Answer:
Ferromagnetic Substances : Some substances containing more number of unpaired electrons are very strongly attracted by the external magnetic field. In Ferromagnetic substances the magnetic moments in individual atoms are all alligned in the same direction. Such substances are called Ferromagnetic Substances. In ferromagnetic substances the field strength B > > > H.
E.g.: Fe, Co and Ni.

Question 10.
Explain Ferrimagnetisms with suitable example.
Answer:
Ferrimagnetism is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti parallel directions in unequal numbers.

• These are weakly attracted by magnetic field as compared to ferromagnetic substances.
• These lose ferrimagnetism on heating and becomes paramagnetic.

Question 11.
Explain Antiferromagnetism with suitable example.
Answer:
Substances like Mno showing anti-ferromagnetism having domain structure similar to ferromagnetic substance, but their domains are oppositely oriented and cancel out each others magnetic moment.

Question 12.
What are Tetrahedral voids ?
Answer:
The second layer spheres over the first layer arrangement, the spheres of second layer are placed in the depressions of the first layer. All the triangle voids of the first layer are covered by the spheres of second layer. These are called ‘Tetrahedral voids’.

Question 13.
What are Octahedral voids ?
Answer:
The triangular voids in the second layer are above the triangular voids in the first layer. Such voids are surrounded by six spheres and are called ‘Octahedral voids’.

Question 14.
What are n-type semiconductors ?
Answer:
Silicon and Germanium belong to IVA group and have four valence electrons. When these elements are doped with VA group like P or As which have 5 valence electrons some of lattice sites of Si are replaced by VA group element. Each VA group element forms four bonds with four Si atoms and fifth electron is extra and becomes delocalised. These delocalised electrons increase the conductivity. The increase in conductivity is due to negatively charged electrons. Hence it is called n – type semi conductor.

Question 15.
What are p-type semi conductors ?
Answer:
Silicon and Germaium when doped with IIIA group elements like B or Al which have only 3 valence electrons. These electrons are bonded to three silicon atoms and fourth valence electron place is vacant. It is called hole. Under the influence of electric field, electrons move towards positive electrode through holes. Hence this type of semi-conductors are called p – type semi conductors.

Question 16.
How many lattice points are there in one unit cell of face – centered tetragonal lattice ?
Answer:
In face centered’tetragonal unit cell
Number of face centered atoms per unit cell
= 6 face centered atoms × \(\frac{1}{2}\) atom per unit cell
6 × \(\frac{1}{2}\) = 3 atoms
∴ Total no. of lattice points = 1 + 3 = 4.

Question 17.
How many lattice points are there in one unit cell of body centered cubic lattice ?
Answer:
In body – centered cubic unit cell
The number of comer atoms per unit cell
= 8 comers × \(\frac{1}{8}\) per corner atom
= 8 × \(\frac{1}{8}\) = 1 atom
Number of atoms at body center = 1 × 1 = 1 atom
∴ Total no. of lattice points = 1 + 1 = 2.

Short Answer Questions

Question 1.
Calculate the efficiency of packing in case of a metal of simple cubic crystal.
Answer:
Packing efficiency in case of metal of simple cubic crystal:

The edge length of the cube
a = 2r. (r = radius of particle)
Volume of the cubic unit cell = a³ = (2r)³
= 8r³
∵ A simple cubic unit cell contains only one atom
The volume of space occupied = \(\frac{4}{3}\) πr³
∴ Packing efficiency
= \(\frac{\text { Volume of one atom }}{\text { Volume of cubic unit cell }}\) × 100
= \(\frac{4 / 3 \pi r^{3}}{8 r^{3}}\) × 100 = \(\frac{\pi}{6}\) × 100 = 52.36%.

Question 2.
Calculate the efficiency of packing in case of a metal of body centered cubic crystal.
Answer:
Packing efficiency in case of a metal of body centred cubic crystal:

in B.C.C. Crystal
\(\sqrt{3}\)a = 4r
a = \(\frac{4 \mathrm{r}}{\sqrt{3}}\)
In this structure total no. of atoms is ‘2’ and their volume = 2 × (\(\frac{4}{3}\)) πr³
Volume of the cube = a³ = (\(\frac{4}{\sqrt{3}}\)r)³

Question 3.
Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer:
The solids which are having moderate conductivity between insulators and conductors are called semi conductors.

• These have the conductivity range from 10^-6 to 10⁴ Ohm^-1m^-1.
• By doping process the conductivity of semi conductors increases.
  E.g.: Si, Ge, crystal.

Semi conductors are of two types. They are :
1. Intrinsic semi-conductors : In case of semi-conductors, the gap between the valence band and conduction band is small. Therefore, some electrons may jump to conduction band and show some conductivity. Electrical conductivity of semi-conductors increases with rise in “temperature”, since more electrons can jump to the conduction band. Substances like silicon and germanium show this type of behaviour and are called intrinsic semi-conductors.

2. Extrinsic semi – conductors : Their conductivity is due to the presence of impurities. They are formed by “doping”.
Doping: Conductivity of semi-conductors is too low to be of pratical use. Their conductivity is increased by adding an appropriate amount of suitable impurity. This process is called “doping”.
Doping can be done with an impurity which is electron rich or electron deficient.

Extrinsic semi-conductors are of two types.
a) n-type semi-conductors : It is obtained by adding trace amount of V group element (P, As, Sb) to pure Si or Ge by doping.
When P, As, Sb (or) Bi is added to Si or Ge, some of the Si or Ge in the crystal are replaced by P or As atoms and four out of five electrons of P or As atom will be used for bonding with Si or Ge atoms while the fifth electron serve to conduct electricity.

b) p-type semi-conductors : It is obtained by doping with impurity atoms containing less electrons i.e., Ill group elements (B, Ai, Ga or In).
When B or AZ is added to pure Si or Ge, some of the Si or Ge in the crystal are replaced by B or AZ atoms and four out of three electrons of. B or AZ atom will be used for bonding with “Si” or Ge atoms while the fourth valence electron is missing is called electron hole (or) electron vacancy. This vacancy on an atom in the structure migrates from one atom to another. Hence it facilitates the electrical conductivity.

Question 4.
Derive Bragg’s equation. [A.P. & T.S. Mar. 17, 16; IPE Mar & May 15]
Answer:
Derivation of Bragg’s equation: When X-rays are incident on the crystal or plane, they are diffracted from the lattice points (lattice points may be atoms or ions or molecules). In the crystal the lattice points are arranged in regular pattern. When the waves are diffracted from these points, the waves may be constructive or destructive interference.

The 1^st and 2^nd waves reach the crystal surface. They undergo constructive interference. Then from the figure 1^st and 2^nd rays are parallel waves. So, they travel the same distance till the wave form AD. The second ray travels more than the first by an extra distance (DB + BC) after crossing the grating for it to interfere with the first ray in a constructive manner. Then only they can be in the same phase with one another. If the two waves are to be in phase, the path difference between the two ways must be equal to the wavelength (X) or integral multiple of it (nλ, where n = 1, 2, 3, ………..)
(i.e.,) nλ = (DB + BC) [where n = order of diffraction]
DB = BC = d sin θ [θ = angle of incident beam,]
(DB + BC) = 2d sin θ [d = distance between the planes]
nλ = 2d sin θ
This relation is known as Bragg’s equation.