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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 = [latex]\frac{\Delta x}{\Delta t}[/latex] 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 = [latex]\frac{0.693}{\mathrm{~K}}[/latex]
t_1/2 = [latex]\frac{0.693}{5.5 \times 10^{-14} \mathrm{~S}^{-1}}[/latex] = 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 = -[latex]\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}[/latex] = 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 = [latex]-\frac{d[R]}{d t}[/latex] = 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 = [latex]\frac{\left[R_{0}\right]}{2 k}[/latex] for a zero order reaction; t_1/2 = [latex]\frac{0.693}{k}[/latex] 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 + [latex]\frac{1}{2}[/latex]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 = [latex]\frac{2.303}{t}[/latex]log[latex]\frac{\left[R_0\right]}{[R]}[/latex]
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 = [latex]\frac{0.693}{k}[/latex]
∴ k = [latex]\frac{0.693}{\mathrm{t}_{1 / 2}}[/latex] = [latex]\frac{0.693}{10}[/latex] = 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 = [latex]\frac{2.303}{\mathrm{t}}[/latex] log₁₀ [latex]\frac{a}{(a-x)}[/latex]
k = [latex]\frac{2.303}{t}[/latex] log [latex]\frac{0.6}{0.2}[/latex] = 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 + [latex]\frac{1}{2}[/latex]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 ∝ [latex]\frac{1}{a^{n-1}}[/latex]

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.
[latex]\left(\frac{\mathrm{t}_{1 / 2}^{\prime}}{\mathrm{t}_{1 / 2}^{\prime}}\right)[/latex] = [latex]\left(\frac{a^n}{a^{\prime}}\right)^{n-1}[/latex]
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 = [latex]\frac{2.303}{t}[/latex]log[latex]\frac{a}{(a-x)}[/latex]

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]; [latex]\frac{\mathrm{d}[\mathrm{R}]}{\mathrm{dt}}[/latex] = -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 [latex]\frac{[\mathrm{R}]}{\left[\mathrm{R}_0\right]}[/latex] =-kt . ——— (2)
k = [latex]\frac{1}{\mathrm{t}} \ln \frac{\left[\mathrm{R}_0\right]}{[\mathrm{R}]}[/latex]
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^– [latex]\rightleftharpoons[/latex] 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
[latex]E_{\text {cell }}^0[/latex] = [latex]\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{c}}[/latex]
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.
[latex]\lambda_{m(A B)}^0[/latex] = [latex]\lambda_{\mathrm{A}^{+}}^0[/latex] + [latex]\lambda_{\mathrm{B}^{-}}^0[/latex]
[latex]\lambda_m^0[/latex] = Limiting molar conductivity
[latex]\lambda_{\mathrm{A}^{+}}^0[/latex] = Limiting molar conductivity of cation ‘
[latex]\lambda_{\mathrm{B}^{-}}^0[/latex] = 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 = [latex]\frac{\text { Ect }}{96,500}[/latex]
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 [latex]\rightleftharpoons[/latex] 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.
[latex]\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}\left(1_{\mathrm{atm}}\right) \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}(\mathrm{M})+\mathrm{e}^{-}[/latex]

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.
[latex]\lambda_{\mathrm{m}(\mathrm{AB})}^0=\lambda_{\mathrm{A}^{+}}^0+\lambda_{\mathrm{B}^{-}}^0[/latex]
[latex]\lambda_{\mathrm{m}}^0[/latex] = Limiting molar conductivity
[latex]\lambda_{\mathrm{A}^{+}}^0[/latex] = Limiting molar conductivity of cation
[latex]\lambda_{\mathrm{B}^{-}}^0[/latex] = 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
= [latex]\frac{6.022 \times 10^{23} \times(3600 \mathrm{C})}{(96500 \mathrm{C})}[/latex]
= 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 = [latex]\frac{n_{\mathrm{s}}}{\mathbf{n}_0+\mathrm{n}_{\mathrm{s}}}[/latex]
Mole fraction of solvent X₀ = [latex]\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_0+\mathrm{n}_{\mathrm{s}}}[/latex]
[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% ([latex]\frac{\mathrm{W}}{\mathrm{V}}[/latex]) 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) [latex]\rightleftharpoons[/latex] nα
Total 1 – α + nα = [1 + (n – 1)α]
∴ i = [latex]\frac{[1+(n-1) \alpha]}{1}[/latex]
α = [latex]\frac{\mathrm{i}-1}{\mathrm{n}-1}[/latex]
α_ionization = [latex]\frac{\mathrm{i}-1}{\mathrm{n}-1}[/latex]
Solute association process:
If ‘n’A molecules combine to give A, we have
nA [latex]\rightleftharpoons[/latex] 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. = [latex]\frac{\mathrm{P}_0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_0}[/latex]
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) = [latex]\frac{\text { Weight }}{\mathrm{GMw}} \times \frac{1000}{\mathrm{~V}(\mathrm{~m} l)}[/latex]
0.15 = [latex]\frac{\mathrm{W}}{122} \times \frac{1000}{250}[/latex]
W = [latex]\frac{122 \times 0.15}{4}[/latex] = 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 = [latex]\frac{5 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}[/latex] = 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 = [latex]\frac{\mathrm{w}_2 \mathrm{RT}}{\pi \mathrm{V}}[/latex]
M₂ = [latex]\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}}[/latex]
= 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 = [latex]\frac{22}{144}[/latex] × 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 = [latex]\frac{122}{144}[/latex] × 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 [latex]\frac{\mathrm{P}_0-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_0}[/latex] = X_s (mole fraction of solute)
[latex]\frac{\mathrm{P}_0-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_0}[/latex] = [latex]\frac{\mathbf{n}_{\mathrm{s}}}{\mathrm{n}_0+\mathrm{n}_{\mathrm{s}}}[/latex]
For very much dilute solutions n_s < < < ….. n₀
∴ [latex]\frac{\mathrm{P}_0-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}_0}[/latex] = [latex]\frac{\mathrm{n}_{\mathrm{s}}}{\mathrm{n}_0}[/latex] = [latex]\frac{\mathrm{w}}{\mathrm{m}}[/latex] × [latex]\frac{\mathrm{M}}{\mathrm{W}}[/latex]
W = Weight of solute
m = Molar mass of solute
w = Weight of solvent
M = Molar mass of solvent
Molar mass of solute m = [latex]\frac{w \times M}{W} \times \frac{P_0}{P_0-P_5}[/latex]

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.
[latex]\mathrm{p}_1^0[/latex] = 0.850 bar
P = 0.845bar
M₁ = 78 g mol^-1
w₂ = 0.5 g
w₁ = 39 g
Substituting these values in equation [latex]\frac{\mathrm{P}^0-\mathrm{P}}{\mathrm{P}_1^0}[/latex] = [latex]\frac{\mathrm{w}_2 \times \mathrm{M}_1}{\mathrm{M}_2 \times \mathrm{w}_1}[/latex] 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,
[latex]\frac{\mathbf{p}_{\mathbf{A}}^0-\mathbf{p}_{\mathrm{s}}}{\mathbf{p}_{\mathbf{A}}^0}[/latex] = [latex]\frac{\mathbf{n}_{\mathbf{B}}}{\mathbf{n}_{\mathbf{A}}}[/latex] = [latex]\frac{\mathbf{W}_{\mathrm{B}}}{\mathbf{M}_{\mathrm{B}}}[/latex] × \frac{\mathbf{M}_{\mathbf{A}}}{\mathbf{W}_{\mathbf{A}}}[latex][/latex]
Where, [latex]\frac{\mathbf{p}_{\mathrm{A}}^0-\mathbf{p}_{\mathrm{s}}}{\mathbf{p}_{\mathrm{A}}^0}[/latex] 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) [latex]\mathrm{p}_{\mathrm{A}}^0[/latex] = 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 = [latex]\frac{\mathrm{p}_{\mathrm{A}}^0-\mathrm{p}_{\mathrm{s}}}{\mathrm{p}_{\mathrm{A}}^0}[/latex] = [latex]\frac{(23.8-23.38) \mathrm{mm}}{(23.8 \mathrm{~mm})}[/latex] = 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% [latex]\frac{w}{W}[/latex] Sucrose in water solution means
20 gms of Sucrose and 80 gms of water.
Number of moles of sucrose (n_s) = [latex]\frac{20}{342}[/latex] = 0.0584
Number of moles of water (n₀) = [latex]\frac{80}{18}[/latex] = 4.444
Mole fraction of sucrose X_s = [latex]\frac{n_s}{n_0+n_s}[/latex]
= [latex]\frac{0.0584}{4.444+0.0584}[/latex] = [latex]\frac{0.0584}{4.50284}[/latex] = 0.01296
Mole fraction of water X_o = [latex]\frac{\mathbf{n}_0}{\mathrm{n}_0+\mathrm{n}_{\mathrm{s}}}[/latex] = [latex]\frac{4.444}{4.50284}[/latex] = 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 = ?
[latex]\frac{P_0-P_3}{P_0}[/latex] = [latex]\frac{W}{m w} \times \frac{M W}{W}[/latex]
⇒ [latex]\frac{17.535-P_s}{17.535}[/latex] = [latex]\frac{9}{180} \times \frac{18}{162}[/latex]
17.535 – P_s = 17.535 × [latex]\frac{9}{180}[/latex] × [latex]\frac{18}{162}[/latex]
⇒ 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 × [latex]\frac{1}{2}[/latex] atom per unit cell
6 × [latex]\frac{1}{2}[/latex] = 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 × [latex]\frac{1}{8}[/latex] per corner atom
= 8 × [latex]\frac{1}{8}[/latex] = 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 = [latex]\frac{4}{3}[/latex] πr³
∴ Packing efficiency
= [latex]\frac{\text { Volume of one atom }}{\text { Volume of cubic unit cell }}[/latex] × 100
= [latex]\frac{4 / 3 \pi r^{3}}{8 r^{3}}[/latex] × 100 = [latex]\frac{\pi}{6}[/latex] × 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
[latex]\sqrt{3}[/latex]a = 4r
a = [latex]\frac{4 \mathrm{r}}{\sqrt{3}}[/latex]
In this structure total no. of atoms is ‘2’ and their volume = 2 × ([latex]\frac{4}{3}[/latex]) πr³
Volume of the cube = a³ = ([latex]\frac{4}{\sqrt{3}}[/latex]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.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 13th Lesson Organic Compounds Containing Nitrogen Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 13th Lesson Organic Compounds Containing Nitrogen

Very Short Answer Questions

Question 1.
Write the IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
i) (CH₃)₃CHNH₂
ii) CH₃ (CH₂)₂ NH₂
iii) (CH₃CH₂)₂ NCH₃
Answer:
i) (CH₃)₃CHNH₂ :

IUPAC name : 2 – methyl 2 – propananine ^ – NH
It is a 1° – amine

ii) CH₃ (CH₂)₂ NH₂:

IUPAC name : 1 – Propananine
It is a 1° – amine

iii) (CH₃CH₂)₂ NCH₃

IUPAC name : N – Elthyl N – Methyl Ethanamine
It is a 3° – amine

Question 2.
Explain why ethylamine is more soluble in water, whereas aniline is not soluble.
Answer:
Elthyl amine is more soluble in water due to the presence of hydrogen bonding. Where are in case of aniline due to presence of bulky hydrocarbon part, the extent of hydrogen bonding is less and it is not soluble in water.

Question 3.
Why aniline does not undergo Friedel – Crafts reaction ?
Answer:
Aniline is a lewis base and AlCl₃ is a Lewis acid. In Friedal Craft’s reaction both of these combined to form a complex

Lewis box Lewisacid Due to formation of complex the electrophilic substitution tendency decreases in aniline and it does not undergo these reaction.

Question 4.
Gabriel Phthalimide synthesis exclusively forms primary amines only. Explain.
Answer:
Gabriel Phthalinide synthesis exclusively forms primary amines only.
Reason : In this reaction primary amines are formed without the traces of 2° (or) 3° amines.

Question 5.
Arrange the following bases in decreasing order of pK_b values. C₂H₅NH₂, C₆H₅NHCH₃, (C₂H₅)₂ NH and C₆H₅NH₂.
Answer:
The decreasing order of pK_b values.of gives amines is
C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH

Question 6.
Arrange the following bases in increasing order their basic strength. Aniline. P – nitroaniline and P – toluidine.
Answer:
The increasing order of basic strength of given compounds is

Question 7.
Write equations for carbylamine reaction of any one aliphatic amine. [A.P.& T.S. Mar. 16]
Answer:
When Ethyl amine (1° – amine) reacts with chloro form in presence of alkali to form ethyl isocyanide.
CH₃ – CH₂ – NH₂ + CHCl₃ + 3KOH CH₃ – CH₂ – NC + 3KCl + 3H₂O

Question 8.
Give structures of A, B and C in the following reactions.

Answer:

A – Phenyl Cyanide B – Benzoic acid C – Benzamide

Question 9.
Accomplish the following conversions. [Mar. 14]
i) Benzoic acid to benzamide
ii) Aniline to P – bromoaniline.
Answer:
Conversion of benzoic acid to benzamide

ii) Conversion of Aniline to P – bromo aniline

Question 10.
Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Answer:
Aromatic 1° – amines cannot be prepared by Gabriel phthalinide synthesis beczause aryl halides do not undergo nucleophilic substitution with the an ion formed by phthalinide.

Short Answer Questions

Question 1.
Write the IUPAC names of the following compounds.
i) CH₃CH₂NHCH₂CH₂CH₃
ii) PhCH₂CN

Answer:
i) CH₃CH₂NHCH₂CH₂CH₃
N – Ethyl 1 – Propanamine

ii) PhCH₂CN
Phenyl Ethane nitrile (Benzyl cyanide)

Question 2.
Give one chemical test to distinguish between the following pairs of compounds.
i) Methylamine and dimethylamine
ii) Aniline and N.Methylaniline .
iii) Ethylamine and aniline
Answer:
i) Methyl amine (1° – amine) and dimethyl amine (2° – amine) are distinguished by iso cyanide test (or) Carbylamine (est. Methyl amine responds to carbylamine reaction to produce methyl isocyanide where as dimethyl amine does not respond to the iso cyanide test

ii) Aniline (1° – amine) and N.methyl (2° – amine) aniline are distinguished, by carbylamine test (or) isocyanide test. Aniline responds to carbyl amine test to give foul smelling phenyl iso cyanide where as N – methyl aniline does not responds to carbyl amine Test.

iii) Ethyl amine (1° – aliphafic amine) and aniline (1° – aromatic amine) are distinguished by Diazotisation reaction. Aniline under go diazotisation reaction to form benzene diazonium salt where as ethyl amine form highly unstable alkyl diazonium salt

Question 3.
Account for the following :

1. pK_b of aniline is more then of methylamine.
2. Reduction of alkylcyanide forms primary amine whereas alkyl isocyanide forms secondary amine.

Answer:

1. In aniline there exist conjugation between the electron pair on nitrogen and benzene ring and is less available for protonation than in nethyl amine.
   ∴ pK_b value of aniline is more than that of methylamine and aniline is less basic.
2. The reduction of alkyl cyanides forms 1°- amines. In alkyl cyanides alkyl group is attached to carbon atom of cyanide, Where as in alkyl iso cyanides alkyl group is attached to nitrogen atom of isocyanide. So reduction of alkyl isocyanide forms 2°- amines.
   

Question 4.
How do you prepare the following ?
i) N, N-Dimethyl proponamine from ammonia
ii) Propanamine from chloroethane
Answer:
i) Preparation of N, N – Di nethyl propanamine from ammonia : Chloro propane reacts with ammonia followed by the reaction of methylchloride to form N, N – Di methyl propananine.

ii) Preparation of propanamine from chloroethane : Chloro ethane reacts with KCN followed by the reduction forms propanamine

Question 5.
Compare the basicity of the following in gaseous and in a queous state and arrange them in increasing order of basicity.
Answer:
Givein compounds CH₃NH₂, (CH₃)₂NH, (CH₃)₃N and NH₃.
In the above compounds methly substituted ammonium ion gets stabilised due to dispersal of the positive charge by the +1 effect of the methyl group. Hence methyl amines are stronger bases than ammonia. Basic nature of amines increase with increase of methyl groups. This trend is followed in gaseous phase.
(CH₃)₃ N > (CH₃)₂ NH > CH₃ NH₂ > NH₃
In the aqueous phase the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group but also by solvation with water molecules. Due to steric hinderance the basic strength of methyl amines in the aqueous state changed as follows
(CH₃)₂ NH > CH₃ NH₂ > (CH₃)₃ N > NH₃

Question 6.
How do you carryout the following conversions?
i) N – Ethylamine to N, N – Diethyl propanamine
ii) Aniline to Benzene suiphonamide
Answer:
Conversion of
i) N – Ethyl amine to N, N – Diethyl propanamine Ethyl amine reacts with ethyl chloride and propyl chloride to from N N – Di ethyl propanamine.

ii) Conversion of Aniline to Benzene sulphonamide : Aniline reacts with benzene sulphonyl chloride to form N – Phenyl benzene suiphonamide.
C₆H₅NH₂ + C₆H₅SO₂Cl → C₆H₅NHSO₂ C₆H₅ + HCl Phenyl benzene sulphonamide

Question 7.
Explain with a suitable example how benzene sulphonylchioride can distinguish primary secondary and tertiary amines.
Answer:
Benzene suiphonyl chloride is called Hinsbergs reagent. This is used to distinguish the 1°, 2°,. 3° – amines.
— with le – amine: Benzene suiphonyl chloride reacts with 1° amine and produce N – Alkyl benzene sulphonamide which is soluble in alkali.

—, with 2° – amine: Benzene suiphonyl chloride reacts with 2° – amine and produce N, N – DiaLkyl benzene sulphonamide which is insoluble in alkali

—, with 3° – amine : Benzene sulphonyl chloride does not react with benzene sulphonyl chloride.

Question 8.
Write the reactions of
i) aromatic and
ii) aliphatic primaiy amines with nitrous acid.
Answer:
i) Reaction of aromatic 1° – amine with nitrous acid : Aromatic 1° – amine react with , nitrous acid at low temperature (0 – 5°C) to form diazonium salts.

ii) Reaction of Aliphatic 1° – amine with nitrous acid : Aliphatic 1° – amine react with nitrous acid to form highly unstable diazonium salts which gives nitrogen gas and alcohols after decomposition.

Question 9.
Explain why amines are less acidic than alcohols of comaparable molecular masses.
Answer:
Amines are less acidic than alcohols of comparable molecular massey
Explaination : In alcohols the O – H bond is more polar than N – H bond of amines. Hence Amines releates H⁺ ion with more difficulty as compared to alcohol.

Question 10.
How do you prepare Ethyl cyanide and Ethyl isocyanide from a common alkylhalide ?
Answer:
Preparation of ethyl cyanide : Ethyl chloride reacts with aq. Ethanolic KCN to form Ethyl cyanide as a major product

Preparation of Ethyl isocyanide : Ethyl chloride reacts with aq. Ethanolic AgCN to form Ethyl iso cyanide as a major product

Long Answer Questions

Question 1.
An aromatic compound A’ on treatment with aqueous ammonia and heating forms compound B which on heating with Br₂ and KQH forms compound ‘C’ of molecular formula C₆H₇N. Write the structures and IUPAC names of compounds A, B and C.
Answer:
Given that an aromatic compound A on treatment with aq. NH₃ and heating forms compound ‘B’. Which on heating with Br₂ and KOH forms compound ‘C’ of molecular formula C₆H₇N.
1. From the above information ‘B is amide and ‘C’ is amine
2. Molecular formula of ‘C’ is C₆H₇N, So ‘C’ is Amiine (C₆H₅NH₂)
3. Compound A on treatment with aq. NH₃ forms B.
So ‘A’ is Benzoic acid (C₆H₅ – COOH)
and B’ is Benzamide (C₆H₅ – CONH₂)

Question 2.
Complete the following conversions.
i) CH₃NC + HgO → ?
ii) ? 2H₂O → CH₃NH₂ + HCOOH

Answer:

Question 3.
i) Write the structures of different isomeric amines corresponding to the molecular formula C₉H₁₃N.
ii) What reducing agents can bring out reduction of nitrobenzene?
iii) Write the product formed when benzyl chloride is reacted with ammonia followed by treatment with methyl and ethyl chlorides. Write the product
Answer:
i) Given compound molecular formula C₉H₁₃N
The structures of different isomeric amines corresponding to the above formula C₉H₁₃N

ii) The reducing agents that bring out the reduction of nitro benzene are :

1. H₂/Pd (or) Pt (or) Ni [Ethanol]
2. Sn + HCl (or) Fe + HCl
3. Li AlH₄
4. Zn + alc.KOH
5. Zn + NH₄Cl

iii) a) Benzyl chloride reacts with ammonia to form benzyl amine followed by the reaction with methyl chloride forms N, N – Dimethyl phenyl methanamine

b) Benzyl chloride reacts with ammonia to form benzyl amine followed by the reaction with Ethyl chloride form

Question 4.
i) Identify the amide and cyanide which on reduction with appropriate reducing agent gives n – butylamlne.
ii) Write the mechanism of Hoffmann bromamide reaction.
Answer:
i) a) Propyl cyanide on reduction gives n – Butylamine

ii) Hoffmann bromamide reaction : It is a simple way of converting an amide to an amine having one carbon atom less than the starting amide. The reaction is a rearrangement which is brought about by bromide in presence of alkali. It is believed to proceed through the steps shown below.

Question 5.
How do you make the following convertions ?
i) Chlorophenylmethane to phenylacetic acid
ii) Chlorophenylmethane to 2 – phenylethanamine.
Answer:

Question 6.
Identify the starting amide which gives p – methyl aniline on reaction with bromine and sodium hydroxide and write all the steps involved in the reaction. .
Answer:
P – methyl Acetanilide reacts with bromine and sodium hydroxide to form P – methyl aniline.

Question 7.
Explain why the order of basicity for methyl amine, N, N – dimethyl amine and N, N, N – tnmethyl amine changes in gasous and a aqueous medium.
Answer:
Given compounds
CH₃NH₂, (CH₃)₂NH, (CH₃)₃N and NH₃
In the above compounds methyl substituted, ammonium ion gets stabilised due to dispersal of the positive charge by the + 1 effect of the methyl group. Hence methyl amines are stronger bases than ammonia. Basic nature of amines increase with in increase of methyl groups. This trend is followed in gaseous phase.
(CH₃)₃ N> (CH₃)₂ NH > CH₃NH₂ > NH₃.
In the aqueous phase the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group but; also by solvation with water molecules. Due to sterichindrance the basic strength of methyl amines in the aqueous state changed as follows
(CH₃)₂ NH > CH₃ NH₂> (CH₃)₃ N > NH₃

Question 8.
Write the equations involved in the reaction of Nitrous acid with Ethylamine and aniline.
Answer:
Reaction of Ethylamine with nitrous acid : Ethyl amine reacts with nitrous acid to form highly unstable diazonium salt which gives nitrogeñ gas and Ethyl alcohol after decomposition.

Reaction of Aniline with nitrous acid : Aniline reacts with nitrous acid at low temperatures (0- 5°C) to form diazomum salts. (Benzene diazonium salt)

Question 9.
Explain with equations how methylamine, N, N – dimethylamine and N, N, N – trimethylamine react with benzenesulphonyl chloride and how this reaction Is useful to separate these amines.
Answer:
Benzene suiphonyl chloride is called Hinsbergs reagent. This is used to distinguish the 1°, 2°, 3° – amines.
-. with 1°- amine : Benzene sulphonyl chloride reacts with 1°-. amine and produce N – Alkyl benzene sulphonamide which is soluble in alkali.

.- with 2° – amine: Benzene sulphonyl chloride reacts with 2° – amine and produce N, N – Dialkyl benzene sulphonamide which is insoluble in alkali

—, with 3° – amine: Benzene sulphonyl chloride does not react with benzene sulphonyl chloride.

Question 10.
Explain why aniline In strong acidic medium gives a mixture of Nitro anilines and what steps need to be taken to prepare selectively p – nitro aniline.
Answer:
In strong acidic medium anime under go nitration to form mixture of nitro animes. In strongly acidic medium aniline is protonated to form the anilinium ion which is metadirecting. So besides the ortho and para derivatives meta derivative also formed.

By protecting – NH₂ group by acetylation reaction with acetic an hydride the nitration reaction can be controlled and the – P – nitro derivative can be formed as major product.

Question 11.
i) Account for the stability of aromatic diazoniun ions when compared to aliphatic diazonium ions.
ii) Write the equations showing the conversion of aniline diazoniumchloride to
a) chlorobenzene, b) Iodobenzene and c) Bromobenzene
Answer:
i) —> Aliphatic diazonium salts which are formed from 1° – aliphatic amines are highly unstable and liberate nitrogen gas and alcohols.
—> Aromatic diazonium salts formed from 1° – aromatic amines are stable for a short time in solution at low temperatures (0 – 5°C). The stability of arene diazonium ion is explained on the basis of resonance.

ii) a) Conversion of aniline diazonium chloride to chloro benzene
C₆H₅N₂⁺ C₆H₅Cl + N₂
b) Conversion of aniline diazonium chloride to Iodo benzene
C₆H₅N₂⁺Cl^– + KI → C₆H₅I + N₂ + KCl
c) Conversion of aniline diazonium chloride to Bromo benzene
C₆H₅N₂⁺Cl^– C₆H₅Br + N₂

Question 12.
Complete the following conversions : Aniline to
i) Fluorobenzene
ii) Cyanobenzene
iii) Benzene and
iv) Phenol
Answer:
i) Aniline to Fluorobenzene

ii) Aniline to cyano benzene

iii) Aniline to Benzene

iv) Aniline to phenol

Question 13.
Explain the following name reactions: [A.P. Mar. 18]
i) Sandmeyer reaction
ii) Gatterman reaction
Answer:
i) Sandmeyer reaction: Formation of chioro benzene, Brono benzene (or) cyano benzene from benzene diazonium salts with reagents Cu₂Cl₂/HCl, Cu₂Br₂/HBr, CuCN/KCN is called sandmayers reaction.

ii) Gatterman reaction : Formation of chloro benzene, Bromobenzene from benzene diazonium salts with reagents Cu/HCl, Cu/HBr is referred as gatterman reaction.

Question 14.
Write the steps involved in the coupling of Benzene diazoniumchloride with aniline and phenol.
Answer:
Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form P-hydroxyazobenzene. This type of reactionsis known as coupling reactions. Similarly the reaction of diazonium salt with aniline yields P – amino azobenzene.

Question 15.
Write the equations involved in the conversion of acetamide and propanaldehydeoxime to methyl cyanide and ethyl cyanide respectively.
Answer:

Textual Examples

Question 1.
Write chemical equations for the following reactions:
i) Reaction of ethanolic NH₃ with C₂H₅Cl.
ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of CH₂Cl.
Solution:

Question 2.
Write chemical equations for the following Conversions:
i) CH₃ – CH₂ – Cl into CH₃ – CH₂ – CH₂ – NH₂
ii) C₆H₅ – CH₂ – Cl into C₆H₅ – CH₂ – CH₂ – NH₂
Solution:

Question 3.
Write structures and IUPAC names of
i) the amide which gives propanamine by Hoffmann bromarnide reaction.
ii) the amine produced by the Hoffmann degradation of benzamide.
Solution:
i) Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. Structure and IUPAC name of the starting amide with four carbon atoms are given below:

ii) Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is aromatic primary amine containing six carbon atoms.

Question 4.
Arrange the following in decreasing order of their basic strength:
C₆H₅NH₂, C₂H₅NH₂, (C₂H₅)₂ NH, NH₃. [T.S. Mar.’19]
Solution:
The decreasing order of basic strength of the above amines and ammonia follows the following order:
(C₂H₅)₂NH > C₂H₅NH₂ > NH₃ C₆H₅NH₂
Amines also react with benzoyl chloride (C₆H₅COCl). This reaction is known as benzoylation.
CH₃NH₂ + C₆H₅COCl → CH₃NHCOC₆H₅ + HCl
Methanamine Benzoyl chloride N – Methylbenzamide
What do you think is the product of the reaction of amines with carboxylic acids ? They form salts with amines at room temperature.

Question 5.
How will you convert 4 nitrotoluene to 2 – bromobenzoic acid?
Solution:

Intext Questions

Question 1.
Classify the following amines as primary, secondary or tertiary.

iii) (C₂H₅)₂CHNH₂
iv) (C₂H₅)₂NH
Solution:
i) Primary
ii) Tertiary
iii) Primary
iv) Secondary.

Question 2.
i) Write the structures of different isomeric amines corresponding to the molecular – formula, C₄H₁₁N.
ii) Write IUPAC names of all the isomers.
iii) What type of isomerism is exhibited by different pairs of amines ?
Solution:
i) and ii) Eight isomers of C₄H₁₁N are :

iii) Isomerism exhibited by different amines are:
a) Chain isomers : i.e., have different carbon chains (a) and (b), (c) and (d).
b) Position isomers : i.e., functional group occupy different positions, (b) and (c), (b) and (d), (a) and (d).
e) Metamers: i.e., different alkyl groups are attached to the same functional group, (e) and (0, (g) and (e).
d) Functional isomers : i.e., have different functional groups. All the three categories (1°, 2° and 3°) of amines are the functional isomers of each other.

Question 3.
How will you convert
i) benzene into aniline ?
ii) benzene into N, N-dimethylaniline ?
iii) Cl-(CH₂)₄-Cl into hexan-1, 6-diamine ?
Solution:
i) Benzene into aniline:

ii) benzene into N, N-dimethylaniline

iii) Cl-(CH₂)₄-Cl into hexan-1, 6-diamine

Question 4.
Arrange the following in increasing order of their basic strength.
i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH
ii) C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₆H₅NH₂
iii) CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂
Solution:
i) C₂H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₆H₅NH₂ < (C₂H₅)₂NH
ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N< CH₃NH₂ < (CH₃)₂NH

Question 5.
Complete the following acid-base reactions and name the products.
i) CH₃CH₂CH₂NH₂ + HCl →
ii) (C₂H₅)₃N + HCl →
Solution:

Question 6.
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution.
Solution:
Hofmann’s ammonolysis reaction: In the presence of excess methyl iodide, aniline (primary amine) forms quartemary ammonium salt.

Question 7.
Write the chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Solution:

Question 8.
Write the structures of different isomers corresponding to the molecular formula C₃H₉N. ‘ Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid.
Solution:
C₃H₉N has four isomers :

Question 9.
Convert :
(i) 3-methyl aniline into 3-nitrotoluene.
(ii) anline into 1, 3, 5-tribromobenzene.
Solution:
i) 3-methyl aniline into 3-nitrotoluene.

(ii) Anline into 1, 3, 5-tribromobenzene.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 12(b) Aldehydes, Ketones, and Carboxylic Acids Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 12(b) Aldehydes, Ketones, and Carboxylic Acids

Very Short Answer Questions

Question 1.
Arrange the following compounds in increasing order of their property indicated.

1. Acetaldehyde, Acetone, Methyl t. butyl ketone reactivity towards HCN.
2. Floroacetic acid, monochloroacetic acid, Acetic acid and Dichloroacetic acid (acid strength)

Answer:

1. Due to the presence of groups around the carbonyl group the reactivity of a compound depends on the steric hindrance.
   Greater the steric hindrance, less will be the reactivity of the compound. Reactivity towards HCN is in the following order.
   Methyl tertiary butyl ketone < Acetone < Acetaldehyde
2. Acid strength of given compounds is Dichloro acetic acid > fluoro acetic acid > chloro acetic acid > acetic acid.

Question 2.
Write the reaction showing a-halogenation of carboxylic acid and give its name.
Answer:
Carboxylic acids having a-hydrogens are halogenated at the a-position on treatment with chlorine or bromine in presence of small amount of red phosphorous to give a-halo carboxylic acids.
This reaction is named as Hell-volhard – Zelinsky (HvZ) reactions

Question 3.
Although phenoxide ion has more number of resonating structures thancarboxylate ion carboxylic acid is a stronger acid than phenol. Why ?
Answer:

• Phenoxide ion has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom.
• The negative charge is delocalised over two electronegative oxygen atoms in carboxylate ion whereas in phenoxide ion the negative charge less effectively delocalised over one oxygen atom and less electronegative carbon atoms.
  

Question 4.
How do you distinguish acetophenone and benzophenone?
Answer:
On idoform test Acetophenone gives positive, where as benzophenone (C₆H₅COC₆H₅) does not

Question 5.
Explain the position of electrophilic substitution In benzolc acid.
Answer:
Benzoic acid undergo electrophilic substitution reactions in which carboxyl group acts as a deactivating and meta dirécting group.

Question 6.
Write equations showing the conversion of
i) Acetic acid to Acetyl chloride
ii) Benzoic acid to Benzamide
Answer:
i) acetic acid reacts with PCl₃ (or) PCl₅ (or) SOCl₆ to form acetyl chloride

ii) Benzoic acid reacts with ammonia to form benzamide

Question 7.
An organic acid with molecular formula C₈H₈O₂ on decarboxylation forms Toluene. Identify the organic acid.
Answer:
The organic acid is phenyl acetic acid

Question 8.
List the reagents needed to reduce carboxylic acid to alcohol.
Answer:
The Reagents required to reduce carboxylic acid to alcohol are

1. LiAlH₄/Ether (or) B₂H₆
2. H₄O⁺

Question 9.
Write the mechanism of esterification.
Answer:
Mechanism of esterification of carboxylic acids : The esterification of carboxylic acids with alcohols is a kind of nucleophilic acyl substitution. Protonation of the carbonyl oxygen activates the carbonyl group towards nucleophilic addition of the alcohol. Proton transfer in the tetrahedral intermediate converts the hydroxyl group into – ⁺OH₂ group, which, being a better leaving group, is eliminated as neutral water molecule. The protonated ester so formed finally loses a proton to give the ester.

Question 10.
Compare the acidic strength of acetic acid, Chloroacetic acid, benzoic acid and Phenol. [Mar. 14]
Answer:
Benzoic acid (C₆H₅COOH) > Chloro acetic acid (ClCH₂COOH) > Acetic acid (CH₃COOH) > Phenol (C₆H₅OH)

Short Answer Questions

Question 1.
Write the equations of any aldehyde with Fehlings reagent.
Answer:
Fehling’s reagent is mixture of two solutions Fehling’s A and Fehlings B.
Fehling’s A – aq. CuSO₄ solution
Fehling’s B – Sodium potassium tartarate (Rochelle salt)
Acetaldehyde reacts with Fehlings .eagent and gives a redbrown ppt.
Reaction:
CH₃ – CHO + 2CU⁺² + 5OH^– → RCOO^– + CU₂O + 3H₃O (Red – brown ppt)

Question 2.
What is Tollens reagent? Explain Its reaction with Aldehydes.
Answer:
Tollens Reagent : Freshly prepared ammonicai silver nitrate solution is called Tollens reagent.

On warming an aldehyde with Tollens reagent a bright silver mirror is produced due to formation of silver metal.
R – CHO + 2 [Ag(NH₃)₂]⁺ + 3OH^– → RCOO^– + 2Ag + 2H₂O + 4NH₃

Question 3.
Write the oxidation products of : Acetaldehyc, Acetone and Acetophenone.
Answer:
a) Acetaldehyde under goes oxidation to foim acetic acid.

b) Acetone undergoes oxidation to form acetic acid

c) Acetophenone undergoes oxidation to form benzoicacid and chloroform

Question 4.
Explain why Aldehydes and ketones undergoes nucleophilic addition while alkenes undergoes electrophilic addition though both are unsaturated compounds.
Answer:
The carbon – oxygen double bond in carbonyl compounds is polarised due to higher electronegativity of oxygen relative to carbon. Hence the carbonyl carbon is an electrophilic and carboxyl oxygen is a nucleophilic centre. So aldehydes, ketones undergoes nucleophilic addition reaction.

Alkenes contain which is a source of electron density so electrophiles add no to C = C to give addition products. Hence alkenes undergo electrophilic addition reaction.

Question 5.
Write the IUPAC names of the following:
i) CH₃CH₂CH(Br) CH₃COOH
ii) Ph. CH₃COCH₃COOH
iii) CH₃.CH (CH₃) CH₂COOC₂H₅
Answer:

Question 6.
Arrange the following In the increasing order of their acidic strength:
Benzolc acid, 4 – Methoxybenzoic acid, 4 – Nitrobenzoic acid and 4 – Methylbenzoic acid.
Answer:
Electron donating group (-OCH₃) decreases the acidic strength where as electron withdrawing group (NO₂) increases the same.
Increases order of acidic strength is:
4-Methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3, 4-dinitro benzoic acid.

Question 7.
DescrIbe the following: .
i) Cross aldol condensation
ii) Decarboxylation
Answer:
i) Cross Aldol Condensation : When aldol condensation ¡s earned out between two different aldehydes and (or) ketones, it is called cross aldol condensation.

If both the reactants contain α-hydrogen atoms, it gives a mixture of four products.

Ketones can also be used as one component in the cross aldol reactions

ii) Decarboxylatlon : Carboxylic acids lose carbon dioxide molecule to produce hydrocarbons on heating their sodium salts with sodalizne (a mixutre of NaOH & CaO in ratio 3 : 1)
.-, This reaction is called decarboxylation

Question 8.
ExplaIn the role of electron withdrawing and electron releasing groups on the acidity of carboxylic acids.
Answer:

• Electron with drawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through decocalisation of the negative charge by inductive effect.
• Electron donating groups decreases the äcidity of carboxylic acids by destabilising the conjugate base.
  
  Eg : Cl^– is a electron with drawing group acidic strength order in case of chloro acetic acids
  CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH

Question 9.
Draw the structures of the following derivatives:
i) Acetaldehyde dimethyl acetal
ii) The ethylene ketal of hexan-3-one
iii) The methyl hemiacetal of formaldehyde.
Answer:
The structures of following are
i) Acetaldehyde dimethyl acetal

ii) Ethylene Ketal of hexan-3-one

iii) Methyl hemiacetal of formaldehyde

Question 10.
An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It doesnot reduce Tollens’ reagent but forms sodium hydrogensulphite adduct and gives +ve iodoform test. On vigorous oxidation forms ethanoic and propanoic acids. Write the possible structure of the compound.
Answer:
Step : 1 To determine the molecular formula of the compound.

Emperical formula of the given compound = C₅H₁₀O
Molecular formula = n × (emperical formula)
Where n = [latex]\frac{\text { Molecular mass of the compound }}{\text { Emperical formula mass of compound }}[/latex]
Given, molecular mass = 86
Emperical formula mass of C₅H₁₀O = (12 × 5) + (10 × 1) + 16 = 60 + 10 + 16 = 86
n = [latex]\frac{86}{86}[/latex] = 1
Molecular formula = 1 × (C₅H₁₀O)
∴ Molecular formula = C₅H₁₀O

Long Answer Questions

Question 1.
Explain the following terms. Give an example of the reaction in each case. [A.P. Mar. 18]
i) Cyanohydrin
ii) Acetal
iii) Semicarbazone
iv) Aldol
v) Hemiacetal
vi) Oxime
Answer:
i) Cyanohydrin
Aldehydes and ketones react withk hydrogen cyanide (HCN) forms addition products called (or) known as cyanohydrins.

ii) Acetal
In the presence of dry HCl gas, an aldehyde reacts with two equivalents of a monohydric alchol forms gem-dialkoxy compounds are known as acetals.
—> In acetal two alkoxy groups are present on the terminal C-atom.

iii) Semicarbazone
Aldehydes/ketones react with semicarbazide forms certain compounds called as senilcarbazones.
For example:

iv) Aldol
When an aldehyde ((or) ketone) having at least one a-hydrogen atom undergo a reaction in the presence of dilute alkali as catalyst to form aldol (or) β- hydroxy aldehydes ((or) ketals in case of ketones), the reaction is called aldol condensation.
For example:

v) Hemiacetat : In the presence of dry HCl gas an aldehyde reacts with one molecule of a monohydric alcohol forms gem-alkoxy alcohols. These are known as hemiacetals.
For example:

vi) Oxime: In weak acidic medium, an aldehyde ketone reacts with hydroxylamine forms products which are known as oxims.

Question 2.
Name the following compounds according to IUPAC system of nomenclature:
i) CH₃CH(CH₃)CH₂CH₂CHO
ii) CH₃CH₂COCH (C₂H₅)CH₂CH₂Cl
iii) CH₃CH = CHCHO
iv) CH₃COCH₂COCH₃
Answer:
IUPAC names of following

Question 3.
Draw the structures of the following compounds.
i) 3-Methylbutanal
ii) p-Nitropropiophenone
iii) p-Metylbenzaldehyde
iv) 3-Bromo-4-phenylpentanoic acid
Answer:
i) 3 – Methyl butanal

ii) p-Nitropropiophenone

Question 4.
Write the IUPAC names of the following ketones and Aldehydes. Wherever possible, give also common names.
i) CH₃CO(CH₂)₄ CH₃
ii) CH₃CH₂CHBrCH₂CH (CH₃)CHO
iii) CH₃(CH₂)₅CHO
iv) PhCH = CHCHO
v)
vi) PhCOPh
Answer:
i) CH₃CO(CH₂)₄ CH₃

IUPAC Name: Heptan-2-one
Common Name: Methnyl n-pentyl ketone

ii) CH₃CH₂CHBrCH₂CH (CH₃)CHO

IUPAC Name : 4 – bromo – 2 – methyl hexanal
Common Name : γ – bromo – α – methyl caproaldehyde

iii) CH₃(CH₂)₅CHO

IUPAC Name: Heptanal
Common Name i n – heptyi aldehyde

iv) Ph – CH =CH – CHO

IUPAC Name : 3 – Phenyl Prop-2-en-1-al
Common Name: β – phenyl acrolein

v)
IUPAC Name: Cyclopentane Carbaldehyde

vi) PhCOPh

IUPAC Name : Diphenyl methanone
Common Name : Benzophenone

Question 5.
Draw the structures of the following derivatives.
i) The 2, 4 – dinitrophenylhydrazone of benzaldehyde
ii) Cyclopropanone oxime
iii) Acetaldehyde hemiacetal
iv) The Semicarbazone of cyclobutanone
Answer:
i) The 2, 4 – dinitro phenyl hydrazone of benzaldehyde :

ii) Cyclopropanone oxime

iii) Acetaldehyde hemiacetal

iv) The Semicarbazone of cyclobutanone

Question 6.
Predict the products formed when Cyclohexanecarbaldehyde reacts with following reagents.
i) PhMgBr and then H₃O⁺
ii) Tollens reagent
iii) Semicarbazide and weak acid
iv) Zinc amalgam and dilute HCl
Answer:
The products are formed when cyclohexane carbaldehyde reacts with following.
i) Ph MgBr and the H₃O⁺

ii) Tollens reagent

iii) Semicarbazide and weak acid

iv) Zinc amalgam and dilute HCl

Question 7.
Which of the following compounds would Undergo aldol condensation ? Write the structures of the products expected.
i) 2-Methylpentanal
ii) 1-Phenylpropanone
iii) Phenyl acetaldehyde
iv) 2,2 – Dimethylbutanal
Answer:
Compounds having one (or) more a-H atoms undergo aedol condensation.
So from above only first three compounds having α-H atoms. Therefore they undergo aldol condensation. They are namely

1. 2-methyl Pentanal
2. 1 – Phenyl propanone
3. Phenyl acetaldehyde

i) 2 – Methyl Pentanal


ii) 1 – Phenyl propanone

iii) Phenyl acetaldehyde

Question 8.
An organic compound A(C₉H₁₀O) forms 2, 4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizaro reaction. On vigorous oxidation it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
Answer:

• The compouhd having molecular formula C₉H₁₀O forms a 2, 4 – DNP derivative and reduces Tollen’s reagent. So it is an aldehyde.
• It undergoes cannizaro reaction, so the aldehyde group should be directly attached to the benzene ring.
• On vigorous oxidation it gives 1,2- benzene dicarboxylic acid, so it should be an ortho substituted benzaldehyde. For molecular formula C₉H₁₀O, the possibility is only O-ethyl benzaldehyde.
• The equations for all reactions are given below.
  
  

Question 9.
How do you distinguish the following pairs of compounds ?
i) Propanal and propanone
ii) Acetophenone and benzophenone
iii) Phenol and benzoic acid
iv) Pentan-2-one and Pentan-3-one
Answer:
i) Propanal and Propanone
On idoform test propanone responds, but absence of CH₃CO – group in propanal (CH₃CH₂CHO) it does not respond.

ii) Acetophenone and benzophenone
Acetophenone gives positive idoform test whereas benzophenone (C₆H₅COC₆H₅) does not

iii) Phenol and benzoic acid
Benzoic acid reacts with sodium bicarbonate to produce effervescences of carbon dioxide where as phenol (C₆H₅OH) does not

iv) Pentan-2-one and Pentan – 3 – One
On idoform test Pentan – 2 – One-responds whereas Pentan – 3 – one (C₂H₅ COC₂H₅) does not

Question 10.
How are the following conversions carried in not more than two steps ?
i) Ethanol to 3-hydroxybutanal
ii) Bromobenzene to 1-Phenylethanol
iii) Benzaldehyde to ± Hydroxyphenylacetic acid
iv) Benzaldehyde to benzophenone
Answer:
i) Ethanol to 3-hydroxybutanal

iv) Benzaldehyde to benzophenone

Question 11.
Describe the following. [A.P. & T.S. Mar. 19, 16] [A.P. Mar. 18]
i) Acetylation
ii) Cannizaro reaction
iii) Cross aldol condensation
iv) Decarboxylation
Answer:
i) Acetylation : When active hydrogen atom of alcohol, phenol (or) an amine is replaced by acetyl (CH₃CO^–) group to form corresponding ester (or) amide, the reaction is known as acetylation.

Reagents used are acid chloride (or) acid anhydride in presence of a base like pyridine (or) dimethylaniline.
For example :

ii) Cannizaro reaction: On treating with concentrated alkali, aldehydes which do not have any α – hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction.
This reaction is called cannizaro reaction.

As a result, one molecule of aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.
For example:

iii) Cross aldol condensation: When aldol condensation is carried out between two different aldehydes and (or) ketones, it is called cross aldol condensation.

If both the reactants contain a – hydrogen atoms, it gives a mixture of four products.

Ketones can also be used as one component in the cross aldol reactions

iv) Decarboxylation
Carboxylic acids lose carbon dioxide molecule to produce hydrocarbons on heating their sodium salts with sodalime (a mixutre of NaOH & CaO in ratio 3: 1). This reaction is called decarboxylation.

Question 12.
Complete each synthesis by giving the missing starting material, reagent or product.

Answer:

Question 13.
Explain how methyl ketones are distinguished from other ketones. Write the equations showing it.
Answer:
Oxidation of methyl ketones by haloform reactIon : Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidisedby sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound. The methyl group is converted to haloform. This oxidation does not affect a carbon-carbon double bond, If present in the molecule. lodoform reaction with sodium hypoiodite is also used for detection of CH₃CO group or CH₃CH(OH) group which produces CH₃CO group on oxidation.

Question 14.
Write the equations showing the conversion of the following along with reagents.
i) 1-phenyipropane to Benzoic acid
ii) Benzamide to Benzoic acid
iii) Ethyl butanoate to Butanoic acid.
Answer:
i)

Question 15.
Write the products and reagents needed for the below given conversions
i) 3-Nitrobromobenzene to 3-Nitrobenzoic acid
ii) 4-Methyl’acetophenone to Benzene- 1-4-dicarboxylic acid
Answer:
i) 3-Nitrobromobenzene to 3-Nitrobenzoic acid:

ii) 4-Methyl’acetophenone to Benzene- 1-4-dicarboxylic acid

Textual Examples

Question 1.
Give names of the reagents ot bring about the following transformations:

1. Hexan – 1 – ol to hexanal
2. Cyclohexanol to cyclohexanone
3. p – Fluorotoluene to p – fluorobenzaldehyde
4. Ethanenitrile to ethanal
5. Allyl alcohol to propanal
6. But-2-ene to ethanal

Solution:

1. C₅H₅NH⁺ CrO₃Cl^– (PCC)
2. K₂Cr₂O₇ in acidic medium
3. CrO₃ in the presence of acetic anhydride/1. CrO₂Cl₂₅ 2. HOH
4. (Diisobutyl) aluminiinn hydride (DIBAL-H)
5. PCC
6. O₃/H₂O-Zn dust

Question 2.
Arrange the following compounds in the increasing order of their boiling points:
CH₃CH₂CH₂CHO, CH₃CH₂CH₂CH₂OH, H₅C₂-O-C₂H₅, CH₃CH₂CH₂CH₂CH₃
Solution:
The molecular masses of these compounds are in the range of 72 to 74. Since only butan-1- ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan-l-ol would be the highest. Butanal is more polar than ethoxyethane. Therefore, the intermolecular dipole-dipole attraction is stronger in the former. n-Pentane molecules have only weak vander Waals forces. Hence increasing order of boiling points of the given compounds is as follows :
CH₃CH₂CH₂CH₂CH₃ < H₅C₂-O-C₂H₅ < CH₃CH₂CH₂CHO < CH₃CH₂CH₂CH₂OH

Question 3.
Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal ? Explain your answer.
Solution:
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic them carbon . atom of the carbonyl group present in propanal. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.

Question 4.
An organic compound (A) with molecular formula C₈H₈O forms an orange-red precipitate with 2,4 – DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens or Fehlings reagent, nor does it decolourise bromine water or Baeyer’s reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C₇H₆O₂. Identify the compounds (A) and (B) and explain the reactions involved.
Solution:
(A) forms 2, 4 – DNP derivative. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens or Fehling’s reagent, (A) must be a ketone. (A) responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyers reagent. This indicates the presence of unsaturation due to an aromatic, ring.

Compound (B), being an oxidation product of a ketone should be a carboxylic acid. The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone. The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone). Reactions are as follows:

Question 5.
Write chemical reactions to affect the following transformations :
i) Butan-1-ol to butanoic acid
ii) Benzyl alcohol to phenylethanoic acid
iii) 3-Nitrobromobenzene to 3-nitrobenzoic acid
iv) 4 – Methylacetophenone to benzene – 1, 4 – dicarboxylic acid
v) Cyclohexene to hexane-1, 6 – dioic acid
vi) Butanal to butanoic acid.
Solution:

Intext Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols:


Answer:
Primary alcohols (1), (ii), (iii)
Secondary alcohols (iv) and (v)
Tertiary alcohols (vi)

Question 2.
Identify allylic alcohols In the above examples.
Answer:
Allylic alcohols (ii) and (vi)

Question 3.
Name of the following compounds according to IUPAC system.

Solution:
i) 3-Chloromethyl – 2 isopropylpentan – 1 – ol
ii) 2, 5 – Dimethylhexane – 1, 3 – diol
iii) 3 – Bromocyclohexanol
iv) Hex- 1 -en-3-ol
v) 2-Bromo~3-methylbut-2-en-1-ol

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?

Solution:

Question 5.
Write the structures of the products of the following reactions :

Solution:

Question 6.
Predict the major product of acid catalysed dehydration of
i) 1 – methylcyclohexanol and
ii) butan – 1 – ol
Solution:
i) 1 – Methylcyclohexene
ii) A mixture of but-1-ene and but-2-ene. But-1-ene is the major product formed due to rearrangement to give secondary carbocation.

Question 7.
Write the reactions of Williamson synthesis of 2-ethyoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Solution:

Question 8.
Which of the following Is an appropriate set of reactants for the preparation of 1 – methoxy-4-nltrobenzene?

Solution:
ii)

Question 9.
Predict the products of the following reactions:

Solution:

Question 10.
Write the structures of the following compounds.
i) α-Methoxypropionaldehyde
ii) 3- Hydroxybutanal
iii) 2-Hydroxycyclopentane carbaldehyde
iv) 4-Oxopentanal
v) Di-Sec. butyl ketone
vi) 4 – Fluoroacetophenone
Solution:

Question 11.
Write the structures of products of the following reactions;

Solution:

Question 12.
Arrange the following compounds in increasing order of their boiling points.
CH₃CHO, CH₃CH₂OH, CH₂OCH₃, CH₃CH₂CH₃
Solution:
CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH

Question 13.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions.

1. Ethanal, Porpanal, Propanone, Butanone.
2. Benzaldehyde, p – Toualdehyde, p – Nitrobenzaldehyde, Acetophenone.

Solution:

1. Butanone < Propanone < Propanal < Ethanal
2. Acetophenone < p – Touladehyde, Benzaldehyde < p – Nitgrobenzaldehyde.
   Hint: Consider steric effect and electronic effect.

Question 14.
Predict the products of the following reactions :


Solution:

Question 15.
Give the IUPAC names of the following compounds:
i) Ph CH₂CH₂COOH
ii) (CH₃)₂C = CHCOOH

Solution:
i) 3-Phenylpropanoic acid
ii) 3 – Mehtylbut-2-enoic acid
iii) 2-Metylcyclopentanecarboxylicacid
iv) 2, 4, 6 – Trinitrobenzoic acid

Question 16.
Show how each of the following compounds can be converted to benzoic acid.
i) Ethylbenzene
ii) Acetophenone
iii) Bromobenzene
iv) Phenylethene (Styrene)
Solution:

Question 17.
Which acid of each pair shown here would you expect to be stronger?
i) CH₃CO₂H or CH₂FCO₂H
ii) CH₂FCO₂H or CH₂ClCO₂H
iii) CH₂FCH₂CH₂CO₂H or CH₃CHFCH₂CO₂H

Solution:
i) CH₃COOH
ii) CH₂FCOOH
iii) CH₃CHFCH₂COOH
iv)

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material Lesson 12(a) Alcohols, Phenols, and Ethers

Very Short Answer Questions

Question 1.
Explain why propanol has higher boiling point than that hydrocarbon-butane.
Answer:
Propanol has higher boiling point (391 K) than that hydrocarbon butane (309K).

Reason : In propanol strong intermolecular hydrogen bonding is present between the molecules. But in case of butane weak vander waals force of attractions are present.

Question 2.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses.
Explanation: .

• Alcohols and water are both polar solvents. Alcohol is dissolves in water, due to formation of hydrogen bonding with water molecules.
• Hydro carbons are non polar and these dorv.t form hydrogen bonds with water molecules. So alcohols are soluble in water where as hydrocarbons are not soluble in water.

Question 3.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:
Given molecular formula of monnhydric phenols is C₇H₈O. The no. of possible isomers with molecular formula C₇H₈O are three.

Question 4.
Give the reagents used for the preparation of phenol from chiorobenzene.
Answer:
Phenol is prepared from chiorobenzene as follows. Reagents required are

1. NaOH, 623K, 300 atm
2. HCl.
   Chemical reaction :
   

Question 5.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:
Only 1° – alcohols form Ethers on acid dehydration. But not 2° or 3°-alcohols.

Reason : In case of 2° or 3° alcohols steric hindrance arises. Due to this steric hindrance alkenes are formed but not Ethers.

Question 6.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:
Case – I: When methoxy methane reacts with cold.dil. HI then methyl alcohol and methyl iodide are formed.
Mechanisms:

Case – II : When methoxy methane reacts with hot.conc.HI then only methyl iodide is formed.
Mechanisms:

Question 7.
Name the reagents used in the following reactions.

1. Oxidation of primary alcohol to carboxylic acid
2. Oxidation of primary alcohol to aldehyde.

Answer:

1. The reagents used for the oxidation of 1° – alcohols to carboxylicacid are acidified K₂Cr₂O₇7 (or) Acidic/alkaline KMnO₄ (or) Neutral KMnO₄
2. The reagents used for the oxidation of 1°- alcohols to aldehyde are pyridine chloro chromate (PCC) in CH₂Cl₂.

Question 8.
Write the equations for the following reactions.
i) Bromination of phenol to 2,4, 6-tribromophenol
ii) Benzyl alcohol to benzoic acid.
Answer:
i) Bromination of phenól to 2, 4, 6 tribromophenol.

Question 9.
IdentIfy the reactant needed to form t-.butylalcohol from acetone.
Answer:
When acetone reacts with methyl magnesium bromide followed by the hydrolysis forms t-butyl alcohol.

Question 10.
Write the structures for the following compounds

1. Ethoxyethane
2. Ethoxybutane
3. Phenoxyethane

Answer:

1. Ethoxyethane → CH₃ – CH₂ – O – CH₂ – CH₃
2. Ethoxybutane → CH₃ – CH₂ – O – CH₂ – CH₂ – CH₂ – CH₃
3. Phenoxyethane → C₆H₅ – O – CH₂ – CH₃

Short Answer Questions

Question 1.
Draw the structures of all isomeric alcohols of molecular formula C₅H₁₂O₂ and give their IUFAC names and classify them as primary, secondary and tertiary alcohols.
Answer:

• Given molecular formula of compound is C₅H₁₂O.
• It has eight isomeric alcohols.
  
  

In the above isomeric alcohols (i), (ii), (iii), (iv) and 1°-alcohols; (v), (vi) and (viii) are 2°- alcohols, (vii) is 3°-alcohol.

Question 2.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give rèason.
Answer:
While separating a mixture of ortho and para nitrophenols by steam distillation, ortho nitrophenol is steam volatile.

Reason: In ortho nitrophenol intra molecular hydrogen bonding is present and in case of para nitrophenol inter molecular hydrogen bonding is present. So O-nitrophenol is steam volatile.

Question 3.
Give the equations for the preparation of phenol from Cumene. [Mar. 14]
Answer:
Phenol.is prepared from Cumene as follows.

1. Oxidation of Cumene to Cumene hydroperoxide.
2. Cumene hydroperoxide on acidic hydrolysis to form phenol.
   

Question 4.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Hydration of Ethene to yield ethanol involves 3—step mechanism.
Step – 1: In step – 1 formation of carbocation takes place by the protonation of ethene.

Step – 2 : In step – 2 carbocation formed in the above step attacked by water.

Step – 3 : Ethyl alcohol is formed by he deprotonation in step -3

Question 5.
Explain the acidic nature of phenols and compare with that of alcohols.
Answer:
The reaction of phenol with sodium metal and with aq.NaOH indicates the acidic nature of phenol.
i) Phenol reacts with sodium metal to form sodium phenoxide.

•  In phenol hydroxyl group is attached to the Sp² hydridised carbon of benzene ring which acts as electron with drawing group. The formed phenoxide ion from phenol is more stabilised due to delocalisation of negative charge.

Comparison of acidic character of Phenol and Ethanol:

• The reaction of phenol with aq. NaOH indicates that phenols are stronger acids than alcohols.
• The hydroxyl group attached to an aromatic ring is more acidic than in hydroxyl group is attached to an alkyl group.
• Phenol forms stable phenoxide ion stabilised by resonance but ethoxide ion is not.
  

Question 6.
Write the products formed by the reduction and oxidation of phenol.
Answer:
a) Reduction of phenol: Phenol undergo reduction in presence of zinc dust to form benzene.

b) Oxidation of phenol : Phenol undergo oxidation with chromicacid and forms a conjugated diketone known as benzoquinone.

Question 7.
Ethanol with H₂SO₄, at 443K forms ethene while at 413 K it forms ethoxy ethane. Explain the mechanism.
Answer:
Case – 1: Ethanol reacts with Cone. H₂SO₄ at 443K forms ethene

Mechanism:
Step – 1: Formation of protonated alcohol

Step 2 : Formation of carbo cation

Step 3 : Formation of ethene by elimination of a proton

Case – II : Ethanol reacts with Cone. H₂SO₄ at 413 K to form ethoxy ethane.

Mechanism:
In the above reaction ether formation is a SN reaction. This involve attack of alcohol molecule on a protonated alcohol.

Question 8.
Account for the statement: Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.
Answer:
Alcohols boil at higher temperature than hydrocarbons and ethers of comparable molecular masses.

Explanation : Consider ethanol, propane and methoxy methane which are having comparable molecular masses. The boiling points, molecular masses and structures of the above compounds mentioned below.

The higher boiling points of alcohols are due to the presence of intermolecular hydrogen bonding in them which is lacking in ethers and hydrocarbons.

Question 9.
Explain why in anisole electrophilic substitution takes place at ortho and para positions and not at meta position.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences + R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophihic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m—position. So 0-and P-products are mainly formed during electrophillic substitution reactions.

Question 10.
Write the products of the following reactions :

Answer:

Long Answer Questions

Question 1.
Write the IUPAC name of the following compounds :

Answer:

Question 2.
Write structures of the compounds whose IUPAC names are as follows:
i) 2, Methyl butan—ol
ii) 1-Phenylprpan-2-ol
iii) 3, 5-Dhuethylhexane-1, 3, 5-triol
iv) 2, 3-Diethylphenol
v) 1-Ethoxypropane
vi) 2-Ethoxy-3-methylpentane
vii) Cyclohexylmethanol
viii) 3-Chloromethylpentan-1-ol
Answer:

Question 3.
Write the equations for the preparation of phenol using benzene, conc. H₂SO₄ and NaOH. [Mar. 14]
Answer:
The equations for the preparation of phenol using conc.H₂SO₄ and NaoH as follows

Question 4.
Illustrate hydroboration-oxidation reaction with a suitable example.
Answer:
When alkenes undergo addition reaction with diborane to form tri alkyl boranes. These followed by the oxidation by alkaline H₂O₂ to form alcohols. This reaction is called as hydroboration-oxidation reaction.

Question 5.
Write the IUPAC name of the following compounds:

Answer:

Question 6.
How will you synthesise:
i) 1 – Phenylethanol from a suitable alkene
ii) Cyclohexylmethanol using an alkyl halide by an SN² reaction.
iii) Pentan-1-ol using a suitable alkyl halide.
Answer:
i) Synthesis of 1-phenylethanol from a suitable alkene : When styrene undergo hydrolysis in presence of dil.H₂SO₄ to form 1-phenyl ethanol. It is an example of Marknowni koff’s rule.

ii) Synthesis of cyclohexyl methanol using an alkyl halide by SN² reaction : When cyclohexyl methyl bromide reacts with aq. NaOH to form cyclohexyl methanol.

iii) Synthesis of 1-pentanol using a suitable alkyl halide: When 1-Bromo pentane reacts with aq.KOH to form 1-pentanol.

Question 7.
Explain why-
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
ii) OH group attached to benzene ring activates it towards electrophilic substitution.
Answer:
i) Ortho nitrophenol is more acidic than Ortho methoxyphenol.
Explanation: .

• -NO₂ is an electron withdrawing group and -OCH₃ is electron releasing group.
• By the presence of electron withdrawing group the phenoxide ion is more stabilized. By the presence of electron releasing group the phenoxide ion is less stabilized.
• Due to high stability of phenoxide ion, acidic nature increases.
  

ii) The -OH group attached to benzene ring activates it towards electrophilic substitution.

Explanation : When an electrophile is attacked, the – OH group exerts +R effect on the benzene ring. So electrodensity in the ring increases at ortho and para positions. When an electrophile attacks, substitution takes place at O and p-positions. So benzene ring activates towards electrophilic substitution.

Question 8.
Wth a suitable example write equations for the following:. [T.S. MAr. 19, 18; A.P. Mar. 18, 16] [A.P. Mar. 16]
i) Kolbe’s reaction
ii) Reimer-Tiemann reaction
iii) Williamsons ether synthesis
Answer:
i) Kolbes reaction: Phenol reacts with sodium hydroxide to form sodium phenoxide. This undergoes electrophilhic substitution with CO₂to form salicylic acid.

ii) Relmer-Tlemann reactIon : Phenol reacts with chloroform in presence of NaOH to form salicylaldehyde (O-Hydroxy benzaldehyde). This reaction is known as Reimer-Tiemann reaction. .

iii) Wilhiamsons ether synthesis:

• This method is used for the preparation of symmetrical and unsymmetrical ethers.
• The reaction of an alkyl halide with sodium alkoxide to form ethers is known as Williamsons Synthesis.
  

Question 9.
How are the following conversions carried out?
i) Benzyl chloride to Benzyl alcohol
ii) Ethyl magnesium bromide to Propan-1-ol
iii) 2-Butanone to 2-Butanol
Answer:
i) Conversion of Benzyl chloride to Benzyl alcohol : Ben.zyl chloride reacts with aq. NaOH to form benzyl alcohol.

ii) Conversion of Ethyl magnesium bromide to Propan-1-ol : Ethyl magnesium bromide reacts with form aldehyde followed by hydrolysis to form 1-propanol.

iii) Conversion of 2-Butanone to 2-Butanol: 2-Butanone undergo reduction in presence of LiA/H₄ to form 2-Butanol.

Question 10.
Write the names of the reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
i) 1-Propoxypropane
ii) Ethoxybenzene
iii) 2-Methoxy-2-methylpropane
iv) 1 -Methoxyethane
Answer:
i) Preparation of 1-propoxy propane :

Question 11.
How is 1-propoxypropane synthesized from propan-1-ol ? Write mechanism of this reaction.
Answer:
1 – Proponal reacts with Conc. H₂SO₄ at 413 K to form 1-propoxy propane.

Question 12.
Explain the fact that in aryl alkyl ethers the alkoxy group activates the benzene ring towards electrophilic substitution.
Answer:
Anisole is an aryl alkyl ether. In anisole the group -OCH₃ influences +R effect. This increases the electron density in the benzene ring and it leads to the activation of benzene ring towards electrophillic substitution reactions.

In Anisole eletron density is more at O-and P-Positions but not at m-position. So O-and P-products are mainly formed during electrophillic substitution reactions.

Question 13.
Write equations of the below given reactions:
i) Alkylation of anisole
ii) Nitration of anisole
iii) Friedel-Crafts acetylation of anisole
Answer:
i) Friedel crafts Alkylation of anisole :

ii) Nitration of anisole

Question 14.
Show how you would synthesize the following alcohols from appropriate alkenes?

Answer:

Question 15.
Explain why phenol with bromine water forms 2,4,6-tribromophenol while on reaction with bromine in CS₂ at low temperatures forms para-bromophenol as the major product.
Answer:
a) Phenol under goes Bromination in presence of CS₂ to form p-bromophenol as major product.

b) Phenol undergoes bromination in aqueous medium form 2,4,6 -tribromo phenol (white ppt).

Explanation: In bromination of phenol, the polarisation of Br₂ molecule takes place even in the absence of Lewis acid. This is due to the highly activating effect of -OH group attached to the benzene ring.

Textual Examples

Question 1.
Give IUPAC names of the following compounds:

Solution:
i) 4-Chloro-2, 3-dimethylpentan-1-ol
ii) 2-Ethoxypropane
iii) 2, 6-Dimethyiphenol
iv) 1-Ethoxy-2-nitrocyclohexane

Question 2.
Give the structures and IUPAC names of the products expected from the following reactions :
a) Catalytic reduction of butanal.
b) Hydration of propene in the presence of dilute sulphuric acid.
c) Reaction of propanone with methylmagnesium bromide followed by hydrolysis.
Solution:

Question 3.
Arrange the following sets of compounds in order of their increasing boiling points :
a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.
Solution:
a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
b) n-Butane, ethoxyethane, pentanal and pentan-1-ol.

Question 4.
Arrange the following compounds in increasing order of their acid strength:
Propan-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3, 5-dinitrophenol, phenol, 4-methylphenol.
Solution:
Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2, 4, 6-trinitrophenol.

Question 5.
Write the structures of the major products expected from the following reactions:
a) Mononitratlon of 3-methylphenol .
b) Dinitratlon of 3methylphenol
c) Mononitration of phenyl methanoate.
Solution:
The combined influence of -OH and -CH₃ groups determine the position of the incoming group.

Question 6.
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

i) What would be the major product of this reaction?
ii) Write a suitable reaction for the preparation of t-butylethyl ether.
Solution:
i) The major product of the given reaction is 2-methylprop-1-ene.
It is because sodium ethoxide is a strong nucleophile as well as a strong base. Thus elimination reaction predominates over substitution.

Question 7.
Give the major products that are formed by heating each of the following ethers with HI.

Solution:

Intext Questions

Question 1.
Classify the following as primary, secondary and tertiary alcohols:

Answer:
Primary alcohols (i), (ii), (iii)
Secondary alcohols (iv) and (y)
Tertiary alcohols (vi)

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols are (ii) and (vi)

Question 3.
Name the following compounds according to IUPAC system.


Answer:
i) 3-Chioremethyl 2-isopropylpentan-1-ol
ii) 2, 5-Dimethylhexane-1, 3-dial
iii) 3-Bromocyclohexanol
iv) Hex-1-en-3-ol
v) 2-Bromo-3-methylbut-2-en-1-ol.

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent of methanol?

Answer:

Question 5.
Write structures of the products of the following reactions :


Answer:

Question 6.
Predict the major product of acid catalysed dehydration of

1. 1-methylcyclohexanol and
2. butan-1-ol.

Answer:

1. 1-Methylcyclohexene
2. A mixture of but-1-ene and but-2-ene. But-1-ene is the major product formed due to rearrangement to give,secondary carbocation.

Question 7.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.
Answer:

Question 8.
Predict the products of the following reactions:

Answer:

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 11th Lesson Haloalkanes And Haloarenes Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 11th Lesson Haloalkanes And Haloarenes

Very Short Answer Questions

Question 1.
Write the structures of the following compounds.
i) 2-chloro-3-methylpentane,
ii) 1-Bromo-4-sec-butyl-2-methylbenzene.
Answer:
i) 2-chloro-3-methylpentane
Structure :

ii) 1-Bromo-4-sec-butyl-2-methylbenzene.
Structure :

Question 2.
Which one of the following has highest dipole moment ?
i) CH₂Cl₂
ii) CHCl₃
iii) CCl₄
Answer:
CH₂Cl₂ has high dipole moment (μ = 1.62 D) among the three alkyl halides.
Reason: The total of two C – Cl dipole moments is reinforced by the total of two C-H bonds.
CCl₄ has zero dipole pioment. CHCl₃ has 1.03 D dipole moment due to presence of C-H bond.

Question 3.
What are ambident nucleophiles ?
Answer:
Ambident nucleophiles: The nucleophiles which are able to attack at two (or) more different sites are called ambident nucleophiles.
Eg.: Alkyl halides react with AgCN to form alkyl cyanide and isocyanides.

Question 4.
Write the isomers of the compound having molecular formula C₄H₉Br.
Answer:
Compound having molecular formula C₄H₉Br has five isomers.

Question 5.
Which compound in each of the following pairs will react faster in S_N² reaction with -OH?
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl
Answer:
i) CH₃ – I reacts faster in SN² – reaction with OH than CH₃ – Br.
Reason: The bond dissociation enthalpy of C -1 bond (234 KJ/mole) is less than that of (-Br bond (293 KJ/mole).

ii) CH₃ – Cl reacts faster in S_N2-reaction with -OH than-(CH₃)₃C-Cl.
Reason: The order of reactivity of alkyl halides in S_N2-reactions is l°-alkyl halide > 2°- alkyl halide > 3°-alkyl halide.
Due to high steric hindrance in 3°-alkyl halide i.e., (CH₃C – Cl. It is less reactive towards S_N²-reaction, whereas in case of CH₃-Cl (1°-alkyl halide) less steric hindrance observed. So, it is high reactive towards S_N²-reactions.

Question 6.
Explain why the alkyl halides though polar are immiscible with water.
Answer:
Alkyl halides are polar but these do not dissolve in water i.e., immiscible in water.
Reason : Among water molecules strong inter molecular hydrogen bonding is present. This hydrogen bond is difficult to be broken by alkyl halides.

Question 7.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed aqueous KOH ?
Answer:
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅ the 2^nd one i.e., C₆H₅CHClC₆H₅ gets hydrolysed more easily than C₆H₅CHCl.

This can be explained by considering S_NI reaction mechanism. In case of S_NI reactions reactivity depends upon the stability of carbo cations.
C₆H₅CHClC₆H₅ forms more stable carbo cation than C₆H₅CH₂Cl.

Question 8.
Treatment of alkyl halides with aq.KOH leads to the formation of alcohols, while in presence of alc.KOH what products are formed ?
Answer:

• Treatment of alkyl halides with aq. KOH leads to the formation of alcohols. Here Nucleo- phillic substitution reaction takes place.
  Eg.: C₂H₅Cl + aq.KOH → C₂H₅OH + KCl
• Treatment of alkyl halides with alc.KOH leads to the formation of alkenes. Here elimination reaction takes place.
  E.g.: C₂H₅Cl + alc. KOH → C₂H₄ + KCl + H₂O

Question 9.
What is the stereochemical result of S_N¹ and S_N² reactions ? [T.S. Mar. 17]
Answer:

• The stereochemical result of S_N¹ reaction is reacemisation product.
• The stereochemical result of _N² reaction is inversion product.

Question 10.
What type of isomerism is exhibited by O, m and p-chlorobenzenes ?
Answer:
o, m and p-chloro benzenes exhibits position isomerism.

These are positional isomers.

Question 11.
What are Enantiomers? [T.S. Mar. 19]
Answer:
Enantiomers : The stereo isomers relatëd to each other as non-superimposable mirror images are called enantiomers.

• These have identical physical properties like melting point, boiling points refractive index etc.
• They differ in rotation of plane polarised right.

Short Answer Questions

Questions 1.
Give the IUPAC names of the following compounds.
i) CH₃CH(Cl)CH(I)CH₃
ii) ClCH₂CH = CHCH₂Br
iii) (CCl₃)₃CCl
iv) CH₃C(p-Cl-C₆H₄)₂CH(Br)CH₃
Answer:

Question 2.
Write the structures of the following organic halides.
i) 1-Bromo-4-sec-butyl-2-methylbenzene,
ii) 2-Chioro- 1 -phenylbutane
iii) p-bromochlorobenzene l
iv) 4-t-butyl-3-iodoheptane.
Answer:
i) 1-Bromo-4-sec-butyl-2-methylbenzene.
Structure:

ii) 2-chloro- 1-phenylbutane.
Structure:

iii) p-bromo chlorobenzene Br
Structure:

iv) 4-t-butyl-3-iodoheptane
Structure:

Question 3.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro- compound C₅H₉Cl in bright sunlight. Identitfy the hydrocarbon.
Answer:

• The given compound molecular formula C₅H₁₀. It represents general formula C_nH_2n. It may be an alkene (or) cyclo alkane.
• Given that the alkene does not react with Cl₂ in dark condition it is not an alkene, so it is a cyclo alkene.
  

Question 4.
Which compound in each of the following pairs will react faster in. S_N² reaction with -OH ? [A.P. Mar. 19]
i) CH₃Br or CH₃I
ii) (CH₃)₃CO or CH₃Cl .
Answer:

1. Among CH₃Br and CH₃I, CH₃ – I reacts faster in S_N² reaction with OH^⊗ because bond dissociation energy of C – I is less than the bond dissociation energy of C – Br.
2. Among CH₃Cl and (CH₃)₃CCl, CH₃ – Cl reacts faster in S_N² reaction with OH^⊗ because (CH₃)₃CCl has high steric hindrance than CH₃Cl.

Question 5.
Predict the alkenes that would be formed in the following reactions and identify the major alkene.

Answer:

Question 6.
How will you carry out the following conversions ?
i) Ethane to bromomethene
ii) Toluene to benzyl alcohol
Answer:
i) Conversion of Ethane to Bromo Ethane.

ii) Conversion of toluene to benzyl alcohol:

Question 7.
Explain why the dipole moment of chlorobenzene Is lower than that of cyclohexyichloride.
Answer:
The dipole moment of chloro benzene is lower than that of cylo hexyl chloride.
Explanation:

• The polarity of C – Cl bond in chiorobenzene is less than the polarity of C — Cl bond in cyclo hexyl chloride.
• The above fact is due to the sp² hybridisation of ‘C’ atom in chiorobenzene where as in cyclo hexyl chloride ‘C’ atoms hybridisation is sp³ – hybridisation.
  

Question 8.
Write the mechanism of the following reaction.

Answer:
Given reaction is

CN^⊖ ion is an ambident nucleophile. It has the tendency to attack through C-atom (or) through N-atom. Attack through C-atom results cyanide product and attack through N- atom results isocyanide. But in presence of polar solvent KCN ionises and forms cyanide as major product.

Long Answer Questions

Question 1.
Name the following halides according to IUPAC system and classify them as primary, secondary, tertiary, vinyl or aryl halides,
i) CH₃CH(CH₃)CH(Br)CH₃
ii) CH₃C(Cl)(C₃H₅)CH₂CH₃
iii) m-ClCH₂C₆H₄CH₂C(CH₃)₃
iv) O-Br-C₆H₄CH(CH₃)CH₂CH₃
Answer:

Question 2.
Write the structures of the following organic halogen compounds.
i) 2-Bromo-3-methylhexane
ii) 2-(2-chlorophenyl)-1-iodooctane
iii) 4-tertiary-butyl-3-iodo benzene
iv) 1-Bromo-4-sec-butyl-2-methylbenzene.
Ans:
i) 2-Bromo-3-methyl hexane
Structure:

ii) 2-(2-chloro phenyl) 1-iodo octane
Structure:

iii) 4-tertiary-butyl-3-iodo benzene
Structure:

iv) 1 -Bromo-4-sec-butyl-2 -methyl benzene
Structure:

Question 3.
Discuss the physical properties of haloalkanes.
Answer:
Physical properties of halo alkanes:

1. Pure alkyl halides are colourless, Bromides and iodides exhibits colour when exposed to light.
2. Most of volatile halogen compounds have sweet smell.
3. Lower members of alkyl halides are gases and higher members are liquids (or) solids.
4. The boiling points of chlorides, bromides and iodides are higher than those of hydrocarbons.
5. The boiling points of alkyl halides decrease as follows RI > RBr > RCl > RF.
6. In case of isomeric halo alkanes boiling points decrease with increase in branching.
7. The density of halo alkane, increase with increase in no. of carbon atoms, halogen atoms and atomic mass of the halogen atoms.
8. Halo alkanes, have theTendency to dissolve in organic solvents and are very slightly soluble in water.

Question 4.
Explain the mechanism of Nucleophilic bimolecular substitution (S_N2) reaction with one example. [T.S. Mar. 16] [Mar. 14]
Answer:
Nucleophilic Bimolecular substitution Reaction S_N2:

1. The nucleophilic substitution reaction in which rate depends upon concentration of both reactants is called S_N2 reaction.
2. It follows 2^nd order kinetics. So it is called bimolecular reaction.
   Eg.: Methyl chloride reacts with hydroxide ion and forms methanol and chloride ion.
3. Here the rate of reaction depends upon the concentration of two reactants.
   
4. In the above mechanism the configuration of carbon atom under attack inverts in much the same way as an umbrella is turned inside out when caught in a strong wind. This process is called inversion of configuration.
5. In transition state the carbon atom is simultaneously bonded to the incoming nucleophile and out going group. It is very unstable.
6. The order of reactivity for S_N2 reactions follows : 1°-alkyl halides > 2°-alkyl halides > 3°-alkyl halides.

Question 5.
Explain why allylic and benzylic halides are more reactive towards S_N1 substitution while 1-halo and 2-halobutanes preferentially undergoes S_N2 substitution.
Answer:
1) Allylic and benzylic halides show high reactivity towards the S_N1 reaction.
Reason: The carbocation thus formed gets stabilised through resonance phenomenon as shown below.

2) 1-halo and 2-halo butanes preferentially under goes S_N2 substitute.
Reason : S_N2 reactions involve transition state formation. Higher the steric hindrance lesser the stability of transition state. The given 1-halo and 2-halo butanes have less steric hindrance so these are preferentially undergo S_N2 reaction.

Question 6.
Describe the stereo chemical effect on the hydrolysis of 2-bromobutane.
Answer:
When S-2-Bromobutane is allowed to undergo hydrolysis R-2-butanol is formed with the – OH group occupying the position opposite to what bromide had occupied. This is an example of S_N2 – reaction. S_N2 reactions of optically active halides are accompanied by inversion of configuration.

Question 7.
What is the criteria for optical activity. Give two examples of chiral molecules.
Answer:
Optical activity: The property of rotating the plane polarized light by a chemical substance is called optical activity.

• If the plane polarised light rotates in clock wise direction then it is dextro rotatory [(+) (or) d-forms].
• If the plane polarised light rotates in anti-clock wise direction then it is laevo rotatory [(-) or /-form].

Crieteria for optical activity:

1. Chirality (or) dissymmetry is the necessary and sufficient condition for a molecule to show optical activity.
   Chirality : The objects which are non-superimposable on their mirror images are said to be chiral and this property is known as chirality.
2. Asymmetry (absence of symmetry) of the molecule is responsible for the optical activity of organic compounds.
   Examples of chiral molecules :
   1. 2-Butanol
   2. 2-Chlorobutane
   3. 2-Bromopropanoic acid.

Question 8.
Define the following:    [A.P. Mar. 16]
i) Racemic mixture
ii) Retention of configuration
iii) Enantiomers.
Answer:
i) Racemic mixture: Equal portions of Enantiomers combined to form an optically inactive mixture. This mixture is called racemic mixture.

1. Here rotation due to one isomer will be exactly cancelled by the rotation of due to other isomer.
2. The process of conversion of enantiomer into a racemic mixture is called as racemisation.
   

ii) Retention of configuration: The preservation of the integrity of the spatial arrangement of bonds to an asymmetric centre during a chemical reaction (or) transformation is called Retention of configuration.
General Eg : Conversion of XC_abc chemical species into YC_abc.

Eg : (-) 2 – Methyl 1 – butanol conversion into (+) 1 – chloro 2. Methyl butane

iii) Enantiomers : The stereo isomers related to each other as non-superimposable mirror images are called enantiomers.

• These have identical physical properties like melting point, boiling points refractive index etc.
• They differ in rotation of plane polarised light.

Question 9.
Write the mechanism of dehydrohalogenation of 2-bromobutane.
Answer:
Dehydrohalogenation of 2 – Bromobutane: 2 – Bromobutane reacts with alc.KOH to form 2 – Butene as major product.

Mechanism:

• 2 – Bromobutane heated with alc.KOH elemination of hydrogen atom from β – carbon and bromine atom from α – carbon takes place. This is called β – Elimination.
  
• 2 – Butene is major product formed according to saytzev’s rule. “In dehydrohalogenation reactions the preferred product is that alkene which has the greater no. of alkyl groups attached to double bonded carbon atoms”.

Question 10.
Explain the Grignard reagents preparation and application with suitable example.
Answer:
Alkyl magnesium halides are generally called as Grignard reagenty.
Preparation: These are prepared by the treatment of alkyl halides with magnesium metal in presence of dry ether.

Applications:
Grignard reagents have wide applications in the synthesis of large no. of organic compounds.

1. Preparation of alkanes :
   Grignard reagents reacts with alcohols and forces alkanes.
   
2. Preparation of alcohols : Ethylalcohol is obtained by the action of Methyl magnesium bromide on formal dehyde followed by the hydrolysis.
   
3. Preparation of carboxylicacids:
   Grignard reagent on carboxylation followed by the hydrolysis to form carboxylic acids.
   

Question 11.
A primary alkyl halide C₄H₉Br(A) reacted with alcoholic KOH to give compound B. B on reaction with HBr yields C which is an isomer of A. When A is reacted with sodium metal forms D, C₈H₈ which is different from the compound formed when n-butylbromide is reacted with sodium. Give the structural formulae of A-D and write equations for all the reactions.
Answer:
Given 1° – alkyl halide molecular formula C₄H₉Br
Two isomers possible with molecular formula C₄H₉Br (i- alkyl halides)

Given that compound ‘A’ when reacted with Na does not forms the same product produced by n — Butyl bromide.
∴ isomer I cannot be ‘A’.

Question 12.
Account for the following statements:
i) Aryihalides are extremely less reactive towards Nucleophilic substitution reactions.
ii) p-Nitrochlorobenzene and o, p-dinitrochlorobenzene undergo Nucleophilic substitution readily compared to chlorobenzene.
Answer:
i) Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to following reasons.

• In aryl halides ‘C’ undergoes SP² hybridised and it has greater S – character ånd electro negativity So the C – X bond length is shorter.
• In aryl halides resonance effect plays an important role.
  The electron pairs on halogen atom are in conjugation with it π – electrons of the ring.
  
  In the above C – Br bond acquires apartial double bond nature due to resonance. This bond cleavage is difficult.
• The phenyl cation formed in aryl halides is not stabilised by resonance.

ii) p – nitrochlorobenzene and o, p – dinitro chlorobenzene undergo nucleophillic substitution reachily compared to chlorobenzene due to the following reasons.

• Due to presence of – NO₂ group which is an electron with drawing group at ’O’ and ‘P’ – positions in the ring makes the bond breaking easy.
• As the number of NO₂ groups increases reactivity of aryl halide also increases. This can be evidended by the following reactions.
  
  

Question 13.
Explain how the following conversions are carried out:
i) Propene to Propanol
ii) Ethanol to but-1-yne
iii) 1-Bromopropane to 2-Bromopropane
iv) Aniline to Chlorobenzene.
Answer:
i) Propene to propanol

iv) Aniline to chlorobenzene

Question 14.
What happens when
i) n-butylchloride is treated with alc.KOH.
ii) Bromobenzene is treated with Mg in presence of dry ether.
iii)Methylbromide is treated with sodium in presence of dry ether.
Answer:
i) n-Butylchloride + treated with alc.KOH undergo dehydro halogenation and forms 1-Butene.
CH₃ – CH₂ – CH₂ – CH₂ – Cl + alc. KOH → CH₃ – CH₂ – CH = CH₂ + kCl + H₂O

ii) Bromobenzene is treated with Mg in presence of dry ether forms phenyl magnesium bromide a Grignard reagent.

iii) Methyibromide is treated with sodium in presence of dry ether forms ethane (wurtz reaction)

Question 15.
Write the reactions showing the major and minor products when chlorobenzene is reacted with CH₃Cl and CH₃COCl in presence of AlCl₃.
Answer:
i) Friedel crafts alkylation : When chiorobenzene is treated with CH₃Cl to form 1 – Chloro – 4 – Methyl benzene (major) and 1 – Chloro 2 – Methylbenzene (minor).

ii) Friedel Craft’s acylation: Chlorobenzene reacts with CH₃COCl in presence of Anhydrous AlCl₃ to form 2 – Chloro actophenone, (Minor) and 4 – Chloro acetophenone (major).

Textual Examples

Question 1.
Draw the structures of all the eight structural isomers that have the molecular formula C₅H₁₁Br. Name each isomer according to IUPAC system and classify them as primary, secondary or tertiary bromide.
Solution:
CH₃CH₂CH₂CH₂CH₂Br      1-Bromopentane (1°)
CH₃CH₂CH₂CH(Br)CH₃      2-Bromopentane (2°)
CH₃CH₂CH(Br)CH₂CH₃       3-Bromopentane (2°)
(CH₃)₂CHCH₂CH₂Br            1-Bromo-3-methylbutane (1°)
(CH₃)₂CHCHBrCH₃              2-Bromo-3-methylbutane (2°)
(CH₃)₂CBrCH₂CH₃               2-Bromo-2-methylbutane (3°)
CH₃CH₂CH (CH₃) CH₂Br     1-Bromo-2-methylbutane (1°)
(CH₃)₃CCH₂Br                      1-Bromo-2, 2-dimethyipropane (1°)

Question 2.
Write IUPAC names of the following:

Solution:

1. 4-Bromopent-2-ene
2. 3-Bromo-2-methylbut-l-ene
3. 4-Bromo-3-methylpent-2-ene
4. 1-Bromo-2-methylbut-2-ene
5. 1-Bromobut-2-ene
6. 3-Bromo-2-methylpropene

Question 3.
Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of (CH₃)₂CHCH₂CH₃.
Solution:
In the given molecule, there are four different types of hydrogen atoms. Replacement of these hydrogen atoms will give the following.
(CH₃)₂CHCH₂CH₂Cl
(CH₃)₂CHCH(Cl)CH₃
(CH₃)₂C(Cl)CH₂CH₃
CH₃CH(CH₂Cl)CH₂CH₃

Question 4.
Write the products of the following reactions

Solution:

Question 5.
Haloakanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.
Solution:
KCN is predominantly ionic and provides cyanide ions in solution. Although both carbon and nitrogen atoms are in a position to donate electron pairs, the attack takes place mainly through carbon atom and not through nitrogen atom since C-C bond is more stable than C-N bond. However, AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as the main product.

Question 6.
In the following of halogen compounds, which would undergo S_N2 reaction faster ?

Solution:
It is primary halide and therefore undergoes S_N2 reaction faster.
As iodine is a better leaving group because of its large size, it will be released at a faster rate in the presence of incoming nucleophile.

Question 7.
Predict the order of reactivity of the following compounds in S_N1 and S_N2 reactions.
i) The four isomeric bromobutanes
ii) C₆H₅CH₂Br, C₆H₅CH(C₆H₅)Br, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br
Solution:
i) CH₃CH₂CH₂CH₂Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr (S_N1)
CH₃CH₂CH₂CH₂Br > (CH₃)₂CHCH₂Br > CH₃CH₂CH(Br)CH₃ > (CH₃)₃CBr (S_N2)
Of the two primary bromides, the carbocation intermediate derived from (CH₃)₂CHCH₂Br is more stable than that derived from CH₃CH₂CH₂CH₂Br because of greater electron donating inductive effect of (CH₃)₂CH-group. Therefore, (CH₃)₂CHCH₂Br is more reactive than CH₃CH₂CH₂Br in S_N1 reactions. CH₃CH₂CH(Br)CH₃ is a secondary bromide and (CH₃)₃ CBr is a tertiary bromide. Hence the above order is followed in S_N1. The reactivity in S_N2 reactions follows the reverse order as the steric hinderance around the electrophilic carbon increases in that order.

ii) C₆H₅C(CH₃)(C₆H₅) Br > C₆H₅CH(C₆H₅)Br > C₆H₅CH(CH₃)Br > C₆H₅CH₂Br (S_N1)
C₆H₅C(CH₃) (C₆H₅) Br < C₆H₅CH(C₆H₅)Br < C₆H₅CH(CH₃)Br < C₆H₅CH₂Br (S_N2)
Of the two secondary bromides, the carbocation intermediate obtained from C₆H₅CH(C₆H₅)Br is more stable than obtained from C₆H₅CH(CH₃)Br because it is stabilised by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in S_N1 reactions. A phenyl group is bulkier than a methyl group. Therefore, C₆H₅CH(C₆H₅)Br is less reactive than C₆H₅CH(CH₃)Br in S_N2 reactions.

Question 8.
Identify chiral molecules in each of the following pair of compounds. (Wedge and Dash representations according to Inter 1 yr., fig. 13.1).

Solution:

Question 9.
Although chlorine is an electron withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Why ?
Solution:
Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution.

Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at ortho-and para-positions. The inductive effect is stronger than resonance and causes net electron withdrawl and thus causes net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho-and para-positions and hence makes the deactivation less for ortho-and para-attack. Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.

Intext Questions

Question 1.
Write structures of the following compounds:
i) 2-Chloro-3-methylpentane
ii) 1-Chloro-4-ethylcyclohexane
iii) 4-tert. Butyl-3-iodoheptane
iv) 1, 4-Dibromobut-2-ene
v) 1-Bromo-4-sec. butyl-2-methylbenzene.
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI ?
Answer:
H₂SO₄ cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding acid, HI which is then oxidised by it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:

1. ClCH₂CH₂CH₂Cl
2. ClCH₂ CHClCH₃
3. Cl₂CHCH₂ CH₃
4. CH₃CCl₂ CH₃

Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photo¬chemical chlorination yields.
i) A single monochloride,
ii) Three isomeric monochlorides,
iii) Four isomeric monochlorides.
Answer:
i)

All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.

ii) C^aH₃C^bH₂C^cH₂C^bH₂C^aH₃
The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product.

iii)

Similarly the equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.

Question 5.
Draw the structures of major monohalo products in each of the following reactions.

Answer:

Question 6.
Arrange each set of compounds in order of increasing boiling points.

1. Bromomethane, Bromoform, Chloromethane, Dibromomethane.
2. 1 – Chloropropane, Isopropyl chloride, 1 – Chlorobutane.

Answer:

1. Chloromethane, Bromomethane, Dibromomethane, Bromoform. Boiling point increases with increase in molecular mass.
2. Isoporpylchloride, 1 – Chloropropane, 1 – Chlorobutane. Isopropylchloride being branched has lower b.p. than 1 – Chloropropane.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N2 mechanism ? Explain your answer.

Answer:
i) CH₃CH₂CH₂CH₂Br Being primary halide, there won’t be any steric hindrance.

ii)

Secondary halide reacts faster than tertiary halide.

iii)

The presence of methyl group closer to the halide group will increase the steric hindrance and decrease the rate.

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster S_N1 reaction?

Answer:
i) Tertiary halide reacts faster than secondary halide because of the greater stability of tert-carbocation.

ii) Because of greater stability of secondary carbocation than primary.

Question 9.
Identify A, B, C, D, E, R and R¹ in the following:

Answer:

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 10th Lesson Chemistry In Everyday Life Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 10th Lesson Chemistry In Everyday Life

Very Short Answer Questions

Question 1.
What are drugs?
Answer:
Drug: The chemicals of low molecular masses ranging from 100 to 500 U that react with macromolecular targets to produce biological responses are called drugs.
E.g.: Morphine, Codeine, Heroin etc.,

Question 2.
When are the drugs called medicines?
Answer:
When the biological response of a drug is therapeutic arid useful then the chemical substances (drugs) are called medicines.

Question 3.
Define the term chemotherapy.
Answer:
Chemotherapy : The use of medicines (chemical substances) in the treatment of diseases is called chemotherapy.
In chemotherapy diagnosis, prevention and treatment of diseases involved.

Question 4.
Name the macromolecules that are chosen as drug targets.
Answer:
The macromolecules that are chosen as drug targets are carbohydrates, lipids, proteins, nucleic acids and enzymes.

Question 5.
What are enzymes and receptors ?
Answer:
Enzymes : The proteins which perform the role of biological catalysts in the body are called enzymes.
Receptors : The proteins which are crucial to communication system in the body are called receptors.

Question 6.
Which forces are involved in holding the drug to the active site of enzymes ?
Answer:
The forces involved in holding the drugs to the active site of enzymes are ionic bonds, vander waal’s forces, hydrogen bonds, dipole-dipole interactions etc.,

Question 7.
What are enzyme inhibitors ?
Answer:
The drugs which inhibits the catalytic activity of enzymes and can block the binding site of the enzyme and prevent the binding of substrate are called enzyme inhibitors.

Question 8.
What is allosteric site ?
Answer:
Some drugs do not bind to the enzyme’s active site and bind to a different site of enzyme. This different site of enzyme is called allosteric site.

Question 9.
What are antagonists and agonists ?
Answer:
Antagonists : The drugs that bind to the receptor site and inhibit its natural function are called antagonists.

• These are useful when blocking of message is required.

Agonists : The drugs that mimic the natural messenger by switching on receptors are called agonists.

• These are useful when there is lack of natural chemical messenger.

Question 10.
Why do we need to classify the drugs in different ways ?
Answer:

1. By knowing the pharmacological effect it makes easy for doctors.
2. Antacid can be used in case of excessive acidity in stomach.

Question 11.
What are antacids ? Give example. [Mar. 14]
Answer:
Antacids : Chemicals that remove the excess of acid in the stomach and maintain the pH to normal level are antacids. E.g. : Omeprazole, Lansoprozole etc.,

Question 12.
What are antihistamines ? Give example.
Answer:
Antihistamines : Chemicals that prevent the interaction of histamines with receptors of the stomach wall thus producing less amount of acid are antihistamines.
E.g. : Dimetapp, Terfenadine (Seldane).

Question 13.
While antacids and antiallergic drugs interfere with the function of histamines why do not these interfere with the function of each other ?
Answer:
Antacids and antiallergic drugs donot interfere with the function of each other because they work on different receptors in the body.

Question 14.
What are tranquilizers ? Give example.
Answer:
Tranquilizers : The drugs which are used in the management (or) treatment of psychoes and neuroses are called tranquilizers.
E.g.: Luminal, Seconal, Barbituric acid etc.,

Question 15.
What are barbiturates ?
Answer:
Barbiturates : Derivatives of barbituric acid which functions as important class of tranquilizers are called barbiturates.
E.g.: Veronal, Amytatt etc.,

Question 16.
What are analgesics ? How are they classified ?
Answer:
Analgesics : These are to reduce or totally abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc. Analgesics are classified as

1. Narcotic analgesics : These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
   E.g. : Morphine, Codeine etc.
2. Non-narcotic analgesics : These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
   E.g.: Aspirin, Ibuprofen etc.

Question 17.
What are narcotic analgesics ? Give example.
Answer:
Narcotic analgesics: These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
E.g. : Morphine, Codeine etc.

Question 18.
What are non-narcotic analgesics ? Give example.
Answer:
Non-narcotic analgesics: These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
E.g.: Aspirin, Ibuprofen etc.

Question 19.
What are antimicrobials ?
Answer:
Antimicrobials : The chemical substances which destroy (or) prevent the developement (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials.
E.g.: Lysozyme, Lactic acid etc.,

Question 20.
What are antibiotics ? Give example. [T.S. Mar. 19; A.P. Mar. 16]
Answer:
Antibiotics: The chemical substances produced by micro organisms and inhibit the growth or destroy microorganisms are called antibiotics.
(Or)
The substance produced totally or partly by chemical synthesis which in low concentration inhibits the growth (or) destroy microorganism by intervening in their metabolic process are called antibiotics.
E.g. : Penicillin, Chloramphenicol etc.

Question 21.
What are antiseptics ? Give example. [T.S. Mar. 19; A.P. Mar. 19, 15]
Answer:
Antiseptics : The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
E.g. : Dettol, Bithional etc.

Question 22.
What are disinfectants ? Give example.
Answer:
Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.

• These are applied to inanimate objects like floors, drainage systems etc…
  E.g.:
  • 4% aqueous solution of formaldehyde called formalin is a disinfectant
  • 0.3 ppm chlorine aqueous solution is disinfectant.
  • SO₂ in very low concentration is disinfectant.

Question 23.
Name a substance which can be used as an antiseptic as well as disinfectant.
Answer:
Phenol is used as antiseptic as well as disinfectant.

1. 0.2% phenol is antiseptic.
2. 1% phenol is disinfectant.

Question 24.
What is the difference between antiseptics and disinfectants ?
Answer:
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
Disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.

Question 25.
What are the main constituents of dettol ?
Answer:
Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.

Question 26.
What is tincture of iodine ? What is its use ?
Answer:
Tincture of iodine (antiseptic) is a mixture of 2 – 3% Iodine solution in alcohol-water.

Question 27.
What are antifertility drugs ? Give example.
Answer:
Antifertility drugs: These are birth control pills and contain a mixture of synthetic estrogen and progesterone derivatives.
E.g. : Norethindrone, Ethynylestradiol (novestrol)

Question 28.
Why chemicals are added to food ?
Answer:
Chemicals are added to food for

1. preservation
2. enhancing their appeal
3. adding nutritive value in them.

Question 29.
Name different categories of food additives.
Answer:
The following are the categories of food additives.

1. Food colours
2. Flavours and sweetners
3. Fat Emulsifiers and stabilising agents
4. Anti oxidants
5. Flour improvers – antistaling agents and bleaches
6. Preservatives .
7. Nutritional supplements such as minerals, vitamins and amino acids.

Question 30.
What are artificial sweetening agents ? Give example. [A.P. Mar. 16; 15]
Answer:
The chemical substances which are used instead of sucrose (or) sugar are called artificial sweetening agents.
E.g.: Aspartame, Alitame, saccharin.
These decrease the calorific intake and at the same time several times sweater than sucrose.

Question 31.
Why do we require artificial sweetening agents ?
Answer:

1. Artificial sweetening agents are very useful to diabetic persons.
2. These decrease the calorific in take and at the same time several times sweeter than sucrose.
3. These are harmless.

Question 32.
Why is the use of aspartame limited to cold foods and drinks ?
Answer:
Aspartame is unstable at cooking temperature so it’s use is limited to cold foods and soft drinks. –

Question 33.
Name the sweetening agent used in the preparation of sweets for a diabetic patient.
Answer:
The sweetening agent used in the preparation of sweets for a diabetic patient is saccharin (or) sucralose. It is stable at cooking temperature.

Question 34.
What problem does arise in using alitame as artificial sweetener ?
Answer:
While using alitame as artificial sweetener, the control of sweetness of food is difficult. Alitame is a high potency sweetner.

Question 35.
What are food preservatives ? Give example.
Answer:
The chemical substances which prevent the spoilage of food due to microbial growth are called food preservatives.
E.g.: Sodium benzoate, Salt of sorbic acid etc.

Questions 36.
Name two most familiar antioxidants used as food additives.
Answer:
The most familiar antioxidants are butylated hydroxy toluene (BHT) and butylated hydroxy anisole (BHA)

Question 37.
What is saponification ?
Answer:
The process of formation of soaps Containing sodium salts by heating esters of fatty acid with aq. NaOH solution is called saponification.

Questions 38.
What are soaps chemically ?
Answer:
Chemically soaps are sodium (OF) potassium salts of long chain fatty acids.
E.g.: Sodium stearate.

Question 39.
Why do soaps not work in hard water ?
Answer:
In hard water Ca, Mg-dissolved salts are present. Ca⁺², Mg⁺² ions form insoluble Ca, Mg, soaps respectively when sodium (or) potassium soaps are dissolved in hard water.

1. These insoluble soaps separate as scum in water and are useless as cleansing agent. These are problematic to good washing because the ppt adheres into the fibres of cloth as gummy mass.
2. Hair washed with hard water looks dull.
3. Dye does not absorb evenly on cloth washed with soap using hard water due to this gummy mass.

Question 40.
What are synthetic detergents ?
Answer:
The cleasing agents which are having all the properties of soaps but do not contain any soap are called synthetic detergents.
Synthetic detergents can be used both in soft and hard water as they give foam even in hard water.
E.g.: Sodium dodecyl benzene sulphonate.

Question 41.
What is the difference between a soap and a synthetic detergent ?
Answer:

1. Generally soaps are sodium or potassium salts of long chain fattyacids.
2. Synthetic detergents are cleansing agents having all the properties of soaps and do not contain any soap.
3. Soaps do not work in hard water but synthetic detergents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Question 42.
How are synthetic detergents better than soaps ? [Mar 14]
Answer:
Soaps do not work in hard water but synthetic detergents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Question 43.
Name the different categories of synthetic detergents.
Answer:
Synthetic detergents are classified into three types

1. nionic detergents
2. Cationic detergents
3. Non-ionic detergents.

Question 44.
Can you use soaps and synthetic detergents to check the hardness of water ?
Answer:

1. Soaps are used to check the hardness of water because soaps form insoluble precipitate with hard water.
2. Detergents are soluble in both types of water i.e hardwater and soft water, so detergents are not used to check the hardness of water.

Question 45.
If water contains dissolved calcium hydrogen carbonate, out of soaps and synthetic detergents which one will you use for cleaning clothes and why ?
Answer:
Water contains dissolved calcium hydrogen carbonate is called hard water. This water form insoluble precipitate with soap. Synthetic detergents donot form this type of precipitates, so synthetic detergents are used for cleaning clothes with water containing dissolved Ca(HCO₃)₂.

Short Answer Questions

Question 1.
Explain the term target molecules or drug targets as used in medicinal chemistry.
Answer:
Drug Targets (or) Target molecules: Macro molecules like carbohydrates, lipids, proteins, nucleic acids and enzymes which interact with the drugs are ailed drug targets (or) Target molecules.

Question 2.
Explain the catalytic action of enzymes as drug targets.
Answer:
a) Catalytic action of Enzymes : In the enzyme catalytic activity it performs two functions

1. The first function of an enzyme is to hold the substrate for a chemical reaction. Substrate bind to active site of enzyme through ionic bonding, hydrogen bonding, vander waal force etc.
2. The second function of an enzyme is to provide functional groups that will attack the substrate, and carry out the chemical reaction.

b) Drug- Enzyme interaction :

1. Drugs which inhibit the catalytic activity of the enzymes are called Enzyme inhibitors.
2. Drugs which compete with the substrate for their attachment on active sites of enzymes are called competitive inhibitors.
   

Question 3.
Explain the drug – enzyme interaction.
Answer:
Explanation of drug – Target interaction with Enzymes :

1. The proteins which perform the role of biological catalysts in the body are called enzymes.
2. The binding of inhibitor ai allosteric site changes the shape of the active site.
3. If the bond formed between an enzyme and an inhibitor is a strong covalent bond it cannot be broken easily In this case enzyme is blocked permanently and the body then degrades the enzyme-inhibitoi complex and synthesises the new enzyme.
   

Question 4.
Why are cimetidine and ranitidine better antacids than sodium hydrogen carbonate or magnesium hydroxide or aluminium hydroxide.
Answer:
Sodium hydrogen carbonate or magnesium hydroxide (or) aluminium hydroxide treatment as antacids control only symptoms and not the cause. In advanced stages, ulcers become life threatening and only treatment is removal of affected part of stomach.
The drugs cinetidine, ranitidine prevent the interaction of histamine with the receptors present in the stomach wall and results in release of lesser amount of acid.
So cinetidine, ranitidine are better antacids than NaHCO₃ (or) Mg(OH)₂ (or) Al(OH)₃.

Question 5.
Low level of noradrenaline in the cause of depression. What type of drugs are needed to cure this problem ? Name two drugs.
Answer:
If the level of Noradrenaline is low then the signal sending activity becomes low and the person suffers from depression. In this situations antidepressant drugs (tranquilizers) are required, These drugs inhibit the enzymes which catalyse the degradation of noradrenaline. If the enzyme is inhibited, this important neurotransmitter is slowly metabolised and can activate its receptor for longer periods of time.

Two important antidepressant drugs are IProniazid and phenelzine.

Question 6.
What are analgesics ? How are they classified ? Give examples. [T.S. & A.P. Mar. 19]
Answer:
Analgesics : These are to reduce or totally abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc.
Analgesics are classified as

1. Narcotic analgesics : These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
   E.g. : Morphine, Codeine etc.
2. Non-narcotic analgesics : These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
   E.g.: Aspirin, Ibuprofen etc.

Question 7.
What are different types of microbial drugs ? Give one example for each.
Answer:
Antimicrobials : The chemical substances which destroy or prevent the developement (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials.
E.g.: Lysozyme, Lactic acid etc.,
Different types of antimicrobial drugs are antibiotics, antiseptics, disinfectants.
1) Antibiotics : The chemical substances produced by micro organisms and inhibit the growth or destroy microorganisms are called antibiotics.
(Or)
The substance produced totally or partly by chemical synthesis in low concentration inhibits the growth (or) destroy microorganism by intervening in their metabolic process are called antibiotics.

2) Antiseptics: The chemical compounds that kill (or) prevent the growth of micro organism
are caiied antiseptics. .
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.

3) Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.
These are applied to inanimate objects like floors, drainage systems etc…
E.g.:

• 4% aqueous solution of formaldehyde called formalin is a disinfectant
• 0.3 ppm chlorine aqueous solution is disinfectant.
• SO₂ in very low concentration is disinfectant.

Question 8.
Write the characteristic properties of antibiotics.
Answer:
Characteristic properties of antibiotics :

1. Antibiotic drugs must be the products of metabolism. ,
2. Antibiotic drugs are effective in low concentration.
3. Antibiotic drug retards the growth (or) survival of microorganism.
4. Antibiotic should be a synthetic substance produced as a structural analogue of naturally occurring antibiotic.
5. Antibiotics have either cidal (killing) effect (or) a static (inhibiting) effect on microbes.
   E.g.: Penicillin, Ofloxacin are Bactericidal.
   Erythromycin, Tetracycline are Bacteriostatic.

Question 9.
What is meant by the term broad spectrum antibiotics ? Explain.
Answer:
The range of bacteria (or) other micro organisms that are effected by a certain antibiotic is expressed as its spectrum of action.
Broad spectrum antibiotics : Antibiotics which kill (or) inhibit a wide range of gram-positive and gram-negative bacteria are called broad spectrum antibiotics.

Question 10.
What are broad spectrum and narrow spectrum antibiotics ? Give one example for each.
Answer:
The range of bacteria (or) other micro organisms that are effected by a certain antibiotic is expressed as its spectrum of action.
Broad spectrum antibiotics : Antibiotics which kill (or) inhibit a wide range of gram¬positive and gram-negative bacteria are called broad spectrum antibiotics.

Narrow spectrum antibiotics: Antibiotics which are effective mainly against gram-positive (or) gram-negative bacteria are called narrow spectrum antibiotics,
E.g.: Penicillin – G is a narrow spectrum antibiotic.

Limited spectrum antibiotics : Antibiotics which are effective mainly against a single organism (or) disease are called as limited spectrum antibiotics.

Question 11.
Write notes on antiseptics and disinfectants.
Answer:
Antiseptics: The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
E.g.: Dettol, Bithional etc.
Phenol is used as antiseptic as well as disinfectant. 0.2% phenol is antiseptic.
Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.
Tincture of iodine (antiseptic) is a mixture of 2 – 3% Iodine solution in alcohol-water.

Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.

1. These are applied to inanimate objects like floors, drainage systems etc…
   E.g.: 4% aqueous solution of formaldehyde called formalin is a disinfectant.
2. 0.3 ppm chlorine aqueous solution is disinfectant.
3. SO₂ in very low concentration is disinfectant.

Phenol is used as antiseptic as well as disinfectant 1% phenol is disinfectant.

Question 12.
How do antiseptics differ from disinfectants ? Does the same substance be used as both ? Give one example for each.
Answer:
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.
Disinfectants are applied to inanimate objects such as floors, drainage system, instruments etc.
Phenol is used as antiseptic as well as disinfectant.
i) 0.2% phenol is antiseptic.
ii) 1% phenol is disinfectant.

1. Examples of Antiseptics :
   • Dettol (antiseptic) is a mixture of chloroxylenol and terpineol.
   • Tincture of iodine (antiseptic) is a mixture of 2.3% Iodine solution in alcohol-water.
2. Examples of disinfectants :
   • 4% aqueous solution of formaldehyde called formalin is a disinfectant.
   • 0.3 ppm chlorine aqueous solution is disinfectant.
   • SO₂ in very low concentration is disinfectant.

Phenol is used as anti in very low concentration is disinfectant.

Question 13.
What are the main categories of food additives ?
Answer:
The following are the categories of food additives.

1. Food colours
2. Flavours and sweetners
3. Fat Emulsifiers and stabilising agents
4. Anti oxidants
5. Flour improvers – antistaling agents and bleaches
6. Preservatives
7. Nutritional supplements such as minerals, vitamins and amino acids.

Question 14.
Write notes on antioxidants in food.
Answer:
Antioxidants :

• Antioxidants are important and necessary food additives.
• Antioxidants help in food preservation by retarding the action of oxygen on food. Antioxidants are more reactive towards, oxygen than the food material which they protecting.
• The most familiar antioxidants are Butylated hydroxy toluene (BHT) and Butylated hydroxy anisole (BHA).
• The addition of BHA to butter increases its shelf life from months to years.
• BHT and BHA along with citric acid are added to produce more antioxidant effect.
• SO₂ and sulphites are useful anti Oxidants for wine and beer, sugar syrups and cut, peeled (or) dried fruits and vegetables.

Question 15.
Name different types of soaps.
Answer:
The following are the different types of soaps.

1. Toilet soaps
2. Soaps that float in water
3. Medicated soaps
4. Shaving soaps
5. Laundry soaps
6. Soap powders and scouring soaps etc.

Question 16.
Explain the following terms with suitable examples.
i) Cationic detergents
ii) Anionic detergents
iii) Non-ionic detergents
Answer:
Synthetic detergents are classified into three types.
i) Cationic detergents : These are synthetic detergents.
a) Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides (or) bromides as anions.

b) Cationic part possess a long hydrocarbon chain and a positive charge on nitrogen atom. Hence these are called catiomc detergents.
E.g. : Cetyl trimethyl ammonium bromide
It is used in hair conditioners.

ii) Anionic Detergents: These are synthetic detergents.
a) Anionic detergents are sodium salts of sulphonated long chain alcohols (or) hydrocarbons.
b) Anionic detergents are formed by the treatment of long chain alcohols with conc. H₂SO₄ followed by the neutralisation with alkali.
E.g.: Sodium lauryl sulphate.

These are used for house hold work and in tooth pastes.

iii) Non-ionic detergents : These are synthetic detergents.
a) Non-ionic detergents do not contain any ion in their constitution.
b) The detergent formed by the reaction of stearic acid with poly ethylene glycol is an example of non ionic detergent.

Non-ionic detergents are used in liquid dish washing purpose.

Question 17.
What are biodegradable and non-bio degradable detergents ? Give one example for each.
Answer:

1. Biodegradable detergents :
   The detergents which are degraded (or) decomposed by micro organisms are called biodegradable detergents. Biodegradable detergents have less branching.
   These do not cause water pollution.
   E.g.: n-dodecyle benzene sulphonate, soap (non synthetic),
2. Non Biodegradable detergents :
   The detergents which are not decomposed (or) degraded by microbes (or) micro organisms are called non-biodegradable detergents. These have more branching.
   These cause water pollution.
   E.g.: ABS detergent.

Question 18.
Explain the cleansing action of soaps.
Answer:
Soap is sodium stearate, C₁₇H₃₅COONa, in water soap gives the ions stearate anion and sodium ion.

Cleansing action of soap : Soap anions form a micelle. The grease or dirt of the cloth are absorbed into the interior of the micelle. The tails of the anion are pegged into micelle and these micelle are washed away with the soap solution.

The main function of the soap is therefore, to convert the oily and greasy dirt on the cloth into large colloidal particles. For this a critical concentration of soap solution is required. Soap thus, functions as an emulsifying agent for the water – dirt emulsion. The emulsified grease or dirt is then washed away with soap solution.

Question 19.
Label the hydrophilic and. hydrophobic parts in the following compounds.
i) CH₃ (CH₂)₁₀ CH₂ OSO^–₃ Na⁺
ii) CH₃(CH₂)₁₅N⁺(CH₃)₃ Br^–
iii) CH₃ (CH₂)₁₆ COO (CH₂ CH₂O)_n CH₂ CH₂ OH
Answer:

Question 20.
Draw the structures of following:
i) Serotonin
ii) Bithionol
iii) Chloramphenicol
iv) saccharin
Answer:

Long Answer Questions

Question 1.
Describe the classification of drugs into different classes.
Answer:
Classification of drugs : Drugs are classified in different ways.

1. On the basis of drug action : Based on the actions of drugs on particular biochemical processes they are classified as
   • Antibacterials
   • Antibiotics
   • Hypnotics
   • Sedatives and tranquilizers
   • Cardio vascular drugs
   • Antiseptics etc.
2. On the basis of molecular targets : Biomolecules like Carbohydrates, lipids, proteins, nucleic acids etc. react with drugs. These biomolecules are called drug targets or simply target molecules.
3. On the basis of chemical structure : The drugs with some common structural features generally show similar pharmacological activity.
   E.g.: Sulfonamides have common structural feature.
   So they have same chemotherapeutic action.
4. On the basis of pharmocological effect:
   This type of classification is useful for doctors because it provides them the whole range of drugs available for the treatment.
   E.g.: Analgesics have pair killing effect.

Question 2.
Describe briefly the therapeutic action of different classes of drugs.
Answer:
The therapeutic action of different classes of drugs discussed below.
Antacids : Chemicals that remove the excess of acid in the stomach and maintain the pH to normal level are antacids.
E.g. : Omeprazole, Lansoprazole etc.,

Antihistamines : Chemicals that prevent the interaction of histamines with receptors of the stomach wall thus producing less amount of acid are anti histamines.
E.g.: Dimetapp, Terfenadine (Seldane)

Tranquilizers : The drugs which are used in the management (or) treatment of psychoes and neuroses are called tranquilizers
E.g.: Luminal, Seconal, Barbituric acid etc.,

Barbiturates : Derivatives of barbituric acid which functions as important class of tranquilizers are called barbiturates
E.g.: Veronal, Amytal etc.,

Analgesics : These are to reduce or totally abolish pain without causing impairment of consciousness, mental confusion, incoordination, paralysis, disturbances of nervous system etc.
Analgesics are classified as

1. Narcotic analgesics : These are most potent and clinically useful agents causing depression of central nervous system and at the same time act as strong analgesics.
   E.g.: Morphine, Codeine etc.
2. Non-narcotic analgesics : These drugs are analgesics but they have no addictive properties. Their analgesic use is limited to mild aches and pains.
   E.g. : Aspirin, Ibuprofen etc.

Antimicrobials : The chemical substances which destroy (or) prevent the developement (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials. E.g.: Lysozyme, Lactic acid etc.,

Antibiotics: The chemical substances produced by micro Organisms and inhibit the growth or destroy microorganisms are called antibiotics. .
(Or)
The substance produced totally or partly by chemical synthesis in low concentration inhibits the growth (or) destroy microorganism by intervening in their metabolic process are called antibiotics.
E.g. : Penicillin, Chloramphenicol etc.

Antiseptics : The chemical compounds that kill (or) prevent the growth of micro organism are called antiseptics.
Antiseptics are the chemical substances applied on the living tissues such as wounds, cuts, ulcers and diseased skin surfaces.

Disinfectants : The chemical compounds used for killing (or) preventing the growth of microorganism are called disinfectants.
These are applied to inanimate objects like floors, drainage systems etc.
E.g.:

1. 4% aqueous solution of formaldehyde called formalin is a disinfectant.
2. 0.3 ppm chlorine aqueous solution is disinfectant.
3. SO₂ in very low concentration is disinfectant.

Antifertility drugs: These are birth control pills and contains a mixture of synthetic estrogen and progesterone derivatives.
E.g. : Norethindrone, Ethynylestradiol (novestrol).

Question 3.
Write an essay on antimicrobials.
Answer:
Antimicrobials : The chemical substances which destroy or prevent the developement (or) inhibit the pathogenic action of microbes such as bacteria, fungi, virus are called antimicrobials
E.g. : Lysozyme, Lactic acid etc.,
Different types of anti microbial drugs are antibiotics, antiseptics, disinfectants.

Question 4.
Write notes on the following : [T.S. Mar. 16]
i) Artificial sweetening agents
ii) Food preservatives
iii) Antioxidants in food.
Answer:
i) Artificial sweetening agents: The chemical substances which are used instead of sucrose (or) sugar are called artificial sweetening agents.
E.g.: Aspartame, Alitame, saccharin.

1. These decrease the calorific intake and at the same time several times sweeter than sucrose.
2. Artificial sweetening agents are very useful to diabetic persons.
3. These are harmless.
4. Aspartame is unstable at cooking temperature so it’s use is limited to cold foods and soft drinks.
5. The sweetening agent used in the preparation of sweets for a diabetic patient is saccharin (or) sucralose. It is stable at cooking tenperature.
6. While using alitame as artificial sweetener, the control of sweetness of food is difficult. Alitame is a high potency sweetner.

ii) Food preservatives : The chemical substances which prevent the spoilage of food due to microbial growth are called food preservatives.
Sodium benzoate, C₆H₅COONa is the most important food preservative used in limited quantities and is metabolised by conversion into hippuricacid. Which is finally excreted in the urine. Salts of sorbicacid, propionic acid are also food presevatives.

iii) Antioxidants :

1. Antioxidants are important and necessary food additives.
2. Antioxidants help in food preservation by retarding the action of oxygen on food.
3. Antioxidants are more reactive towards oxygen than the food material which they protecting.
4. The most familiar anti oxidants are Butylated hydroxy toluene (BHT) and Butylated hydroxy anisole (BHA).
5. The addition of BHA to butter increases its shelf life from months to years.
6. BHT and BHA along with citric acid are added to produce more antioxidant effect.
7. SO₂ and sulphits are useful anti oxidants for wine and beer, sugar syrups and cut, peeled (or) dried fruits and vegetables.

Question 5.
Write notes on the following :
i) Soaps
ii) Synthetic detergents.
Answer:
i) Soaps : 1) Soaps are the detergents which improve cleansing properties of water and help in removal of fats which bind other materials to the fabric (or) skin.
Chemically soaps are sodium (or) potassium salts of long chain fatty acids.
E.g.: Sodium stearate.
The process of formation of soaps containing sodium salts by heating esters of fatty acid with aq. NaOH solution is called saponification.
Only sodium and potassium soaps are soluble in water and are used for cleaning purpose. Potassium soaps are soft to the skin and are prepared by using KOH instead of NaOH.

Types of soaps :
The following are the different types of soaps

1. Toilet soaps
2. Soaps that float in water
3. Medicated soaps
4. Shaving soaps
5. Laundry soaps –
6. Soap powders and scouring soaps etc.

In hard water Ca, Mg-dissolved salts are present. Ca⁺², Mg⁺² ions form insoluble Ca, Mg, soaps respectively when sodium (or) potassium soaps are dissolved in hard water.

1. These insoluble soaps separate as scum in water and are useless as cleansing agent. These are problematic to good washing because the ppt adheres into the fibres of cloth as gummy mass.
2. Hair washed with hard water looks dull.
3. Dye does not absorb evenly on cloth washed with soap using hard water due to this gummy mass.

ii) Synthetic detergents : The cleansing agents which are having all the properties of soaps but donot contain any soap are called synthetic detergents.
Synthetic detergents can be used both is soft and hardwater as they give foam even in hard water.
E.g. : Sodium dodecyl benzene sulphonate.
Soaps do not work in hard water but synthetic detegents can be used both in soft and hard water as they give foam even in hard water. Some of the detergents give foam even in ice cold water.

Synthetic Detergents are classified into three types :
i) Cationic detergents : These are synthetic detergents.
a) Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides (or) bromides as anions.

b) Cationic part possess a long hydrocarbon chain and a positive charge on nitrogen atom. Hence these are called cationic detergents.
E.g. : Cetyl trimethyl ammonium bromide.
It is used in hair conditioners.

ii) Anionic Detergents :
These are synthetic detergents.
a) Anionic detergents are sodium salts of sulphonated long chain alcohols (or) hydro-carbons.
b) Anionic detergents are formed by the treatment of long chain alcohols with cone, H2S04 followed by the neutralisation with alkali.
E.g. : Sodium lauryl sulphate.

These are used for house hold work and in tooth pastes.

iii) Non-ionic detergents : .
These are synthetic detergents.
a) Non-ionic detergents donot contain any ion in their constitution.
b) The detergent formed by the reaction of stearic acid with poly ethylene glycol is an example of non ionic detergent.

Non-ionic detergents are used in liquid dish washing purpose.

Intext Questions

Question 1.
Sleeping pills are recommended by doctors to the patients suffering from sleeplessness but it is not advisable to take its doses without consultation with the doctor. Why ?
• Consider the effect of tranquilizers or antidepressants on the nervous system.
Solution:
Sleeping pills contain drugs that may be tranquilizers or antidepressants. They affect the nervous system, relieve anxiety, stress, irritability or excitement. But they should strictly be used under the supervision of a doctor. If not, the uncontrolled and over dosage can cause harm to the body and mind because in higher doses these drugs act as poisons.

Question 2.
With reference to which classification has the statement, “ranitidine is an antacid” been given ?
Solution:
This statement refers to the classification of drugs according to pharmacological effect because in neutralizes the acidity (excess) of stomach.

Question 3.
Why do we require artificial sweetening agents ?
• Mention the function of natural sweeteners, discuss the same for artificial sweeteners.
Solution:
Natural sweeteners (sucrose etc.) provide calories to the body. Taking extra calories is harmful
for diabetic patients. So, artificial sweeteners are used
(a) to control intake of calories and
(b) as a substitute of sugar for diabetics.

Question 4.
Write the chemical equation for preparing sodium soap from glyceryl oleate and glyceryl palmitate. Structural formulae of these compounds are given below.
i) (C₁₅H₃₁COO)₃ C₃H₅ – Glyceryl palmitate
ii) (C₁₇H₃₂COO)₃ C₃H₅ – Glyceryl oleate
• Consider the following method.

Solution:

Question 5.
Following types of non-ionic detergents are present in liquid detergents, emulsifying agents and wetting agents. Label the hydrophilic and hydrophobic parts in the molecule. Identify the functional group (s) present in the molecule.

Solution:

(b) Functional group : Ether and alcohol.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 9th Lesson Biomolecules Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 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.
-H> 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 the 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.
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 4.
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 5.
Write two methods of preparation of glucose.
Answer:
Methods of preparations of glucose :
1) From Sucrose.: Sucrose when boiled with di/.HC/ in alcoholic solution then glucose, fructose are obtained in equal portions.

2) From Starch : Starch undergo hydrolysis, with dil. H2S04 at 393K under pressure to form glucose.

Question 6.
Glucose reacts with bromine water to give gluconic acid. What information do you get from this reaction about the structure of glucose ?
Answer:
Glucose reacts with bromine water to give gluconic acid. This reaction indicates that the carbonyl group present ip,glucose is an aldehyde group.

Question 7.
Glucose and gluconic acid on oxidation with nitric acid give saccharic acid. What Information do you get from this reaction about the structure of glucose ?
Answer:
Glucose and gluconic acid on oxidation with nitric acid give saccharic acid. This indicates the presence of primary alcoholic (-OH) group in glucose.

Question 8.
Glucose reacts with acetic anhydride to form penta acetate. What do you understand about the structure of glucose from this reaction ?
Answer:
Glucose reacts with acetic anhydride to form peta acetate. It confirms the presence of five- OH groups in glucose which are attached to different carbon atoms.

Question 9.
Give any two reasons to understand that glucose molecule has no open chain structure.
Answer:
Open chain structure of glucose could not explain

• why glucose doesnot respond to Schiffs test,
• why it does not react with NaHSO₃ and NH₃,
• why there is mutarotation.

Question 10.
D – glucose means dextro rotatory glucose. Is it true ? Why ?
Answer:
In glucose the carbon atom attached to the lowest CH₂-OH group is found to contain H atom on left side. So it is correlated to glyceraldehyde molecule and gives D-symbol. D-Glucose means not dextro rotatory glucose. D-symbol has nothing to do with optical activity of compound.

Question 11.
What are anomers ?
Answer:
Anomers : The two isomeric structures of a compound which differ in configuration at C-l only are called Anomers.

Question 12.
Write the ring structure of D-glucose. What are their names?
Answer:

Question 13.
Write the ring structure and open chain structure of fructose.
Answer:

Question 14.
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 15.
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 16.
Write the structure of alanine and aspartic acid.
Answer:

Question 17.
What do you mean by essential amino acids ? Give two examples for non essential amino acids? [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 18.
What is zwitter ion ? Give an example.
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 19.
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 20.
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 21.
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.

Question 22.
How are proteins classified with respect to peptide bond ?
Answer:
With respect to peptide bond proteins classified into two types.

1. Fibrous proteins
2. Globular proteins.

Question 23.
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 24.
Write the names of three types of RNA.
Answer:
Three types of RNA are :

1. Messenger RNA (m – RNA)
2. Ribosomal RNA (r – RNA)
3. Transfer RNA (t – RNA).

Question 25.
Write the biological functions of nucleic acids.
Answer:
Biological functions of Nucleic acids :

• DNA is the chemical basis of heredity.
• DNA is the reserve of genetic information.
• DNA is capable of self duplication during cell division and identical DNA strands are transfered to daughter cells.
• RNA molecules are used in the protein synthesis in the cell
• The message for the synthesis of a protien is present in DNA.

Question 26.
Name the vitamin responsible for the coagulation of blood.
Answer:
The vitamin responsible for the coagulation of blood is vitamin K.

Question 27.
What are monosaccharides ?
Answer:
A carbohydrate which donot form any saccharide unit on hydrolysis (or) which cannot be hydrolysed is called monosaccharide.
Eg : glucose, fructose.

Question 28.
What are reducing sugars ?
Answer:
Carbohydrates that reduce Fehling’s reagent, Tollen’s reagent are called reducing sugars.
Eg : glucose.

Question 29.
Write two main functions of carbohydrates in plants.
Answer:
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.

Question 30.
Classify the following into monosaccharides and disaccharides,

1. ribose
2. 2-deoxy ribose
3. maltose
4. fructose.

Answer:

1. Ribose – Monosaccharide
2. 2 – deoxy ribose – Monosaccharide
3. Maltose – Disacch&ride .
4. Fructose – Monosaccharide.

Question 31.
What do you understand by the term glycosidic linkage ?
Answer:
The linkage between two monosacharides of disacharides is called glycosidic linkage.

Question 32.
What is glycogen ? How is it different from starch ?
Answer:
The carbohydrate which is stored in animal body is called glycogen. It is also known as animal starch. Starch is a polysaccharide which is stored in plants where as glycogen is stored in animals (liver, muscles and brain).

• Glycogen mainly resembles amylopectin of starch in structure but glycogen is more branched than amylopectin.

Question 33.
What are the hydrolysis products of
i) sucrose and
ii) lactose ?
Answer:
i) The hydrolysis products of sucrose are glucose and fructose.

ii) The hydrolysis products of lactose are galactose and glucose.

Question 34.
What is the basic structural difference between starch and cellulose ?
Answer:
In cellulose a straight chain polysaccharide composed only of β – D – glucose units which are joined by glycosidic linkage between C – 1 of one glucose unit and C – 4 of the next glucose unit.

Starch consists of two components – Amylose and Amylopectin. Amylose is a long un branched chain with 200 – 1000 α – D – glucose units held by C – 1 to C – 4 glycosidic linkage. Amylopectin is a branched chain polymer of α – D – glucose units in which Chain is formed by C – 1 to C – 4 glycosidic linkage where as branching occurs by C – 1 to C – 6 glycosidic linkage.

Question 35.
What happens when D – glucose is treated with the following reagents :
i) HI
ii) Bromine water
iii) HNOs.
Answer:
i) D – glucose on prolonged heating with HI, glucose forms n – hexane.

Question 36.
Enumerate the reactions of D – glucose which cannot be explained by its open chain structure.
Answer:
The following reactions cannot be explained by the open chain structure of D – glucose.

• Glucose has aldehyde group, but it does not respond to the reagents like Schiff s reagent, NaHSO₃, NH₃ etc., .
• The penta acetyl derivative of glucose doesnot react with hydroxyl amine.
• α and β – methyl glucosides cannot be explained by its open chain structure.

Question 37.
What are essential and non – essential amino acids ? Give one example for each.
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 38.
Define the following as related to proteins :
i) Peptide linkage
ii) Primary structure
iii) Denaturation
Answer:
i) Peptide linkage :
Amide linkages between amino acids are known as peptide linkages. The product obtained from two amino acid molecules through peptide bond is called a dipeptide. Peptide chain may be extended to many amino acids.

Proteins are naturally occuring polypeptides that contain more than 50 amino acid units.

ii) Primary structure : The primary structure of protein is defined as the linear sequence of amino acid residues making up its polypeptide chain. Proteins may be formed of one or more polypeptide chains. The amino acid residues are linked by peptide bonds. The peptide bond is formed between carboxyl group of one amino acid and amino group of adjacent amino acid.
Eg : fibroin of silk.

iii) Protein Denaturation : The phenomenon of disorganization of native protein structure is known as denaturation. Denaturation results in the loss of secondary, tertiary nad 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 agents – Acids, alkalies, organic solvents, urea, salts of heavy metals.

Question 39.
What are the common types of secondary structure of proteins ?
Answer:
Secondary structure: The secondary structure explains the shape of the polypeptide chain. It is derived from the primary structure by the formation of hydrogen bond interactions between amino acid residues.

Disulfide bonds also occur either in same chain or between polypeptide chains. The secondary structure is classified into two types namely helical structure and pleated sheets, e.g.: Keratin of hair.

Question 40.
What type of bonding helps in stabilizing the α – helix structure of proteins ?
Answer:
The α – helix structure of proteins is stabilized by intramolecular hydrogen bonding between = O and – H groups of dlifferent peptide bonds.

Question 41.
Differentiate between globular and fibrous proteins.
Answer:
Globular proteins

• In this proteins the chains of poly peptides coil around to give a spherical shape
• These are soluble in water
• Eg: Insulin

Fibrous proteins

• In this proteins the poly peptides run parallel and are held together by hydrogens and disulphite – bondy
• These are insoluble in water
• Eg: keratin

Question 42.
How do you explain the amphoteric behaviour of amino acids ?
Answer:
Amino acids are the compounds having amino group (-NH₂) i.e., a basic group and carboxyl group (-COOH) i.e., an acidic group

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 is called as zwitter ion.

In zwitter ion form amino acids behave both acidic as well as basic nature. Therefore amino acids are amphoteric in nature.

Question 43.
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, red ness in eyes, xerophthalhia
• 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 44.
What are nucleic acids ? Mention their two important functions.
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).

Biological functions of Nucleic acids :

• DNA is the chemical basis of heredity.
• DNA is the reserve of genetic information.
• DNA is capable of self duplication during cell division and identical DNA strands are transfered to daughter cells.
• RNA molecules are used in the protein synthesis in the,cell.
• The message for the synthesis of a protien is present in DNA.

Question 45.
What is the difference between a nucleoside and a nucleotide ?
Answer:
Nucleosides: Nucleic acid bases combine with pentose sugars to form N – glycosides known as nucleosides.
Eg: Cytidine
Structure of Cytidine

Nucleotides : Nucleotide is a phosphate ester of nucleoside and consists of a purine or pyrimidine base. Nucleotide contains a base moiety, a sugar moiety (i.e.,) a nucleoside moiety and phosphate group.
Nucleotide = base (purine / pyrimidine) + sugar (ribose/deoxyribose) + phosphate
Ex: Adenosine Triphosphate (ATP)

Short Answer Questions

Question 1.
How are the carbohydrates classified on the basis of their
a) Taste
b) Hydrolysis
c) Functional groups
Answer:
a) Taste : On the basis of their taste carbohydrates are classified into
i) Sugars, ii) Non sugars.

1. Sugars : Carbohydrates which are sweet in taste.
   Eg : Sucrose.
2. Non sugars : Carbohydrates which are not sweet in taste.
   Eg : Cellulose.

b) Hydrolysis : On the basis of the 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 monosacchandes.

c) Functional groups: Carbohydrates classified into two types on the basis of functional group present.

1. Aldoses: Carbohydrates having aldehyde group.
   Eg: glucose.
2. Ketoses: Carbohydrates having ketone group.
   Eg: fructose.

Question 2.
Write a brief note on the structure of glucose.
Answer:
1. The molecular formula of glucose is found to be C₆H₁₂O₆. It can be written as CHO – (CHOH)₄ – CH₆OH’.

2. Acylation of glucose with acetic anhydride gives glucose pentacetate. Hence glucose molecule has five -OH groups.

3. Glucose reacts with hydroxylamine and hydrogen cyanide forming corresponding oxime and cyano hydrin respectively. Hence glucose molecule has carbonyl group.

4. Glucose on oxidation with Tollens reagent and Fehling’s solution gives gluconic acid. Hence carbonyl group present in glucose is aldehydic group.

5. Glucose on oxidation with nitric acid gives glucaric (saccharic) acid. This suggests that glucose molecule has primary alcoholic group (-CH₂OH).

6. On reduction with HI, glucose gives n – hexane.

7. Open chain structure of glucose:
Based on above observations, glucose is given the open chain structure.

8. Objections against open chain structure of glucose :
Open chain structure of glucose could not explain

1. Why glucose does not respond to schiffs test.
2. Why it does not react with NaH SO₃, and NH₃.
3. Why there is mutarotation.

Mutarotation :
Change in specific rotation of a compound from certain value to fixed value is known as “Mutarotation”.
The specific rotation of α – glucose is + 110° and that of β – form is + 19.7°. If each of them is separately dissolved in water and allowed to stand, their specific rotation gradually change and reach to a specific constant value + 52.5°.

9. Cyclic structure of glucose :

These two structures have difference in configuration at ‘C₁‘ only and are called ANOMERS.
Cyclic structure of glucose explains all the properties of glucose.

10. Pyranose structure of glucose :

Question 3.
Write short notes on sucrose.
Answer:
Sucrose is a disaccharide. It under go hydrolysis and forms equimolar mixture of D(+) glucose and D(-) fructose.
C₁₂ H₂₂ O₁₁ + H₁₂O → C₆ H₁₂ O₆ (glucose) + C₆ H₁₂ O₆ (fructose).

• In sucrose two monosaccharide units are held together by a glycosidic linkage between C – 1 of α – glucose and C – 2 of β – glucose.
• Sucrose is a non reducing sugar.
  
  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 4.
Write the structure of maltose and lactose. What are the products of hydrolysis of maltose and lactose?
Answer:
Structure of maltose:

• Maltose is formed from two α – glucose units (Disaccharide)
• C – 1 of one glucose linked to C -4 of another glucose
• It is a reducing sugar because C – 1 of 2^nd glucose is solution have free aldehyde group.
  
• maltose on hydrolysis to form two glucose units
  C₁₂ H₂₂ O₁₁ (Maltose) + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (glucose)

Structure of lactose:

• Lactose is milk sugar.
• Lactose is formed by the combination of β – glucose and β – galactose (Disaccharide)
• The linkage between C – 1 of galactose and c – 4 of glucose.
• It is a reducing sugar.
  
• Lactose on hydrolysis to form glucose and galactose
  C₁₂ H₂₂ O₁₁ (Lactose) + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (galactose)

Question 5.
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 of α – 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 :

• ellulose 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 6.
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 hohey is used for a long time as an instant source of energy.

Question 7.
Explain the classification of proteins as primary, secondary, tertiary and quartemary proteins with respect ot their structure.
Answer:
Protein structure’s and shapes are studied at four different levels as
i) Primary ii) Secondary iii) Tertiary iv) Quaternary structures

i) Primary structure : The primary structure of protein is defined as the linear sequence of amino acid residues making up its polypeptide chain. Proteins ‘may be formed of one or more polypeptide chains. The amino acid residues are linked by peptide bonds. The peptide bond is formed between carboxyl group of one amino acid and amino group of adjacent amino acid.
Eg : fibroin of silk.

R₁ – amino acid 1
R₂ – amino acid 2

ii) Secondary structure : The secondary structure explains the shape of the polypeptide chain. It is derived from the primary structure by the formation of hydrogen bond interactions between amino acid residues.

Disulfide bonds also occur either in same chain or between polypeptide chains. The secondary structure is classified into two types namely helical structure and pleated sheets.
Eg. : Keratin of hair

iii) Tertiary structure : Tertiary structure is exhibited by proteins having only one polypeptide chain. The three dimensional structure of a protein is referred to as tertiary structures. Fibrous and globular proteins are formed by tertiary structures. Various bonds are responsible for tertiary structure.

• Hydrophobic interactions
• Hydrogen bonds
• Disulphide bonds (- CH2 – S – S – CH2 -)
• Ionic or electrostatic interactions
• Vander Waal forces
  e.g : Myoglobin, Ribonuclease

iv) Quaternary structure : Two or more poly- peptide chains associate together to protect quaternary structure. A protein with a quaternary structure is referred to as an oligomes.
Eg. : Hemoglobin a chain

Question 8.
Explain the denaturation of proteins. [T.S. Mar. 16]
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 Agents – Acids, alkalies, organic solvents, urea, salts of heavy metals.

Question 9.
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. Practic ally 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) a co enzyme (non-protein part)

Question 10.
Explain the role of sucrose in its hydrolysis.
Answer:

• Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose.
• The laevorotation of fructose (-92.4°) is more than dextrorotation of glucose (+52.5°). This mixture is leavo rotators.
• 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 11.
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 12.
What do you understand by “The two strands of DNA are complementary to each other” ? Explain.
Answer:
Structure of DNA : To understand the structure of DNA, we should know the charagaff s rule, which states that:

• Base composition on DNA for a particular organism is constant throughout all the somatic cells.
• Base composition always varies from one organism to another and is expressed by the dissymmetry ratio [latex]\frac{(A+T)}{(G+C)}[/latex].
• Organisms closely related to each other often have similar base composition and hence – have closed values for their dissymmetry ratios.
• In a given organism the amount of adenine is always equal to the amount of thymine (A = T) and amount of guanine is equal to amount of cytosine (G = C).
• In a given organism the total amount of purine, bases
  are always equal to total number of pyrimidine bases (i.e.,) (A + G = T + C).
  M. Wilkins found that within the crystal there is a repeat distance of 3.4 nm and there are ten subunits per him.
  

Watson – Crick model for DNA : From the above observations Watson and Crick constructed a model for DNA. This model consists of two right handed polynucleotide chains that are complimentary and coiled about the same axis to form double-helix. Some specific base pairs can be spatially accommodated and these are A – T and G – C pairs. The bases are closely associated with each other by hydrogen bonding.

The helix has diameter of about 2.0 nm and contains 10 nucleotide pairs in each turn of helix as shown above :

Question 13.
What are harmones ? Give one example for each. [A.P. & T.S. Mar. 19, 18] [Mar. 14]
i) steroid hormones
ii) Poly peptide hormones and
iii) 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, estrogen.

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

Question 14.
Give the sources of the following vitamins and name the diseases caused by their deficiency (a) A (b) D (c) E and (d) K [A.P. Mar. 16, 15; T.S. Mar. 15]
Answer:

Long Answer Questions

Question 1.
Explain the classification of carbohydrates.
Answer:

Saccharides that reduce Fehling’s reagent, Tollen’s reagent are called reducing sugars.
Eg : glucose.
Saccharides that does not reduce these reagents are called non-reducing sugars.
Eg : sucrose.

Question 2.
Discuss the structure of glucose on the basis of its chemical propeties.
Answer:
1. The molecular formula of glucose is found to be C₆H₁₂O₆. It can be written as ‘CHO – (CHOH)₄ – CH₂OH’.

2. Acylation of glucose with acetic anhydride gives glucose pentacetate. Hence glucose molecule has five ‘-OH’ groups.

3. Glucose reacts with hydroxylamine and hydrogen cyanide forming corresponding oxime and cyano hydrin respectively. Hence glucose molecule has carbonyl group.

4. Glucose on oxidation with Tollen’s reagent and Fehling’s solution gives gluconic acid. Hence carbonyl group present in glucose is aldehydic group.

5. Glucose on oxidation with nitric acid gives glucaric’(saccharic) acid. This suggests that glucose molecule has primary alcoholic group (-CH₂OH)

6. On reduction with HI, glucose gives n – hexane.

This reaction suggests that all six carbons in glucose are linearly linked.

7. Open chain structure of glucose:
Based on above observations, glucose is given the open chain structure.

8. Objections against open chain structure of glucose :
Open chain structure of glucose could not explain

1. Why glucose does not respond to schiff s test.
2. Why it does not react with NaHSO₃ and NH₃.
3. Why there is mutarotation.

Mutarotation :
Change in specific rotation of a compound from certain value to fixed value is known as “Mutarotation”.
The specific rotation of α – glucose is + 110° and that of β – form is + 19.7°. If each of them is separately dissolved in water and allowed to stand, their specific rotation gradually change and reach to a specific constant value + 52.5°.

9. Cyclic structure of glucose :

These two structures have difference in configuration at ‘C₁‘ only and are called
ANOMERS.
Cyclic structure of glucose explains all the properties of glucose.

10. Pyranose structure of glucose :

Question 2.
Write notes on
a) fructose
b) sucrose
c) maltose
d) lactose
Answer:
a) Fructose :

• Fructose is an important ketohexose.
• It is formed by the hydrolysis of sucrose.
  
• It has the ketonic group at C – 2 and six carbons in straight chain as in the glucose.
• It has furanose ring structure.
• Open chain structures:
  
  α – D (-) – Fructo furanose structure
  β – D (-) – Fructo furanose structure
  Haworth structures :
  

b) Sucrose :
Sucrose is a disaccharide. It under go hydrolysis and forms equimolar mixture of D(+) – glucose arid D(-) fructose.
C₁₂ H₂₂ O₁₁ + H₂O → C₆ H₁₂ O₆ (glucose ) + C₆ H₁₂ O₆ (fructose)

In sucrose two monosaccharide units are held together by a glycosidic linkage between C- 1 of α – glucose and C – 2 of β – glucose.

Sucrose is a non reducing sugar.

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.

c) Structure of maltose :
Maltose is formed from two α – glucose units (Disaccharide)
C – 1 of one glucose linked to C – 4 of another glucose
It is a reducing sugar because C – 1 of 2^nd glucose is solution have free aldehyde group.

maltose on hydrolysis to form two glucose units
C₁₂ H₂₂ O₁₁ (Maltose) + H₂O → C₆ H₁₂ O₆ (Glucose ) + C₆ H₁₂ O₆ (Glucose)

d) Structure of lactose :

• Lactose is milk sugar
• Lactose is formed by the combination of β – glucose and β – galactose (Disaccharide).
• The linkage between C – 1 of galactose and c – 4 of glucose >
• It is a reducing sugar
  
• Lactose on hydrolysis to form glucose and galactose
  C₁₂ H₂₂ O₁₁ (Lactose) + H₂O → C₆ H₁₂ O₆ (glucose ) + C₆ H₁₂ O₆ (galactose)

Question 3.
Write notes on
a) starch
b) cellulose
c) importance of carbohydrates.
Answer:
a) Starch:
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 of a – glucose.
• It is constituted by two components Amylose and amylopectin

Amylose :

• It constitutes 15 – 20% of starch -4 Anylose 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.

b) 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.

c) 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 4.
Write notes on amino acids.
Answer:
The organic compounds which contain amino (-NH₂) functional group and carboxyl (-COOH) functional group are called amino acids.
Eg : Glycine, Alanine etc.

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.,

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 is called as zwitter ion.

Peptide linkage :
Amide linkages between amino acids are known as peptide linkages. The product obtained from two amino acid molecules through peptide bond is called a dipeptide. Peptide chain may be extended to many amino acids.
Proteins are naturally occufing polypeptides that contain more than 50 amino acid units.

Question 5.
Write notes on proteins.
Answer:
Proteins : A poly peptide with more than hundred amino acid residues, having molecular mass higher than 10,000 units is called a protein.
g : keratin, myosin, insulin.
With respect to peptide bond proteins classified into two types.

1. Fibrous proteins
2. Globular proteins.

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.

When the poiy 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.

Protein Denaturation: The phenomenon of disorganization of native protein structure is known as denaturation. Denaturation results in the loss of secondary; tertiary,. nad 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, Uy – radiation
Chemical agents – Acids, alkalies, organic solvénts, urea, salts of heavy metals.
Protein strúctures and shapes are studied at four different levels as:
i) Primary
ii) Secondary
iii) Tertiary
iv) Quaternary structures

i) Primary structure: The primary structure of protein is defined as the linear sequence of amino acid residues making up its polypeptide chain. Proteins may be formed of one or more polypeptide chains. The amino acid residues are linked by peptide bonds, The peptide bond is formed between carboxyl group of one amino açid and amino group of adjacent amino acid.
Eg: fibroin of silk.

R₁ – amino acid 1
R₂ – amino acid 2

ii) Secondary structure : The secondary structure explains the shape of the polypeptide chain. It is derived from the primary structure by the formation of hydrogen bond interactions between amino acid residues.

Disulfide bonds also occur either in same chain or between polypeptide chains. The secondary structure is classified into two types namely helical structure and pleated sheets.
Eg. : Keratin of hair

iii) Tertiary structure i Tertiary structure is exhibited by proteins having only one polypeptide chain. The three dimensional structure of a protein is referred to as tertiary structures. Fibrous and globular proteins are formed by tertiary structures. Various bonds are responsible for tertiary structure.

1. Hydrophobic interactions
2. Hydrogen bonds
3. Disulphide bonds (- CH2 – S – S – CH2 -)
4. Ionic or electrostatic interactions
5. Vander Waal forces
   e.g : Myoglobin, Ribonuclease

iv) Quaternary structure : Two or more poly- peptide chains associate together to protect quaternary structure. A protein with a quaternary structure is referred to as an oligomes.
Eg. : Hemoglobin a chain

Question 6.
Write notes on
(a) enzymes and
(b) vitamins.
Answer:
Enzymes :
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)

b) Vitamins :
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 7.
Explain the structures of DNA and RNA.
Answer:
Structure of DNA : To understand the structure of DNA, we should know the charagaff s rule, which states that :

1. Base composition on DNA for a particular organism is constant throughout all the somatic cells.
2. Base composition always varies from one organism to another and is expressed by the dissymmetry ratio [latex]\frac{(\mathrm{A}+\mathrm{T})}{(\mathrm{G}+\mathrm{C})}[/latex]
3. Organisms closely related to each other often have similar base composition and hence have closed values for their dissymmetry ratios.
4. In a given organism the amount of adenine is always equal to the amount of thymine (A = T) and amount of guanine is equal to amount of cytosine (G = C).
5.  In a given organism the total amount of purine bases are always equal to total number of pyrimidine bases (i.e.,) (A + G = T + C).

M. Wilkins found that within the crystal there is a repeat distance of 3.4 nm and there are ten subunits per turn.

Watson – Crick model for DNA : From the above observations Watson and Crick constructed a model for DNA.
This model consists of two right handed polynucleotide chains that are complimentary and coiled about the same axis to form double-helix. Some specific base pairs can be spatially accommodated and these are A – T and G – C pairs. The bases are closely associated with each other by hydrogen bonding.

The helix has diameter of about 2.0 nm and contains 10 nucleotide pairs in each turn of helix as shown above :

Structure of RNA: The native RNA is single stranded rather than a double stranded helical structure characteristic of DNA. However, given the complementary base sequence with opposite polarity, the single strand of RNA may fold back on itself like a hairpin and thus acquire double stranded pattern. In the region of hairpin loops, “A” pairs with ‘U’ and ‘G’ pairs with ‘C’. The base pairing in RNA hairpins is frequently imperfect. The proportion of helical regions in various types of RNA varies over a wide range.

Question 8.
Write notes on the functions of different hormones in the body.
Answer:
Functions of Hormones :

• Hormones help to maintain the balance of biological activities in the body.
• Insulin maintains the blood glucose level with in 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 menustrual cycle.
• Progesterone is responsible for preparing the uterus for implantation of fertilised egg.

Intext Questions

Question 1.
Glucose or sucrose are soluble in water but cyclohexane and benzene (simple six membered ring compounds) are insoluble in water. Explain.
Answer:
In glucose (C₆H₁₂O₆), 5 – OH groups are present and in sucrose (C₁₂H₂₂O₁₁), B – OH groups are present, which are polar in nature. They are involved in intermolecular hydrogen bonding with water molecules, so these compounds are readily soluble in water.

Benzene (C₆H₆) and cyclohexame (C₆H₁₂) are hydrocarbons. They do not have any polar group so they cannot make hydrogen bonds with water. Therefore, they are insoluble in water.

Question 2.
What are the expected products of hydrolysis of lactose ?
Answer:
Lactose forms two molecules of monosaccharides on hydrolysis – One molecule of D-(+) – glucose and one molecule of D – (+) – galactose.

Question 3.
How do you explain the formation of furanose structure for fructose while glucose forms pyranose structure with same molecular formula C₆H₁₂O₆ ?
Answer:
D-glucose, an aldohexose given the characteristic reactions of aldehydic group (e.g., with HCN Tollen’s reagent, Fehling reagent etc.,) but penta acetae of D-glucose does not give these reactions. It means that either – CHO group is absent or it is not available for the chemical reactions in penta acetyl glucose. Infact, the aldehydic group is a part of hemiacetal structure which penta acetyl glucose has. So, it is not available for reaction.

Question 4.
The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Higher the polarity of a group, more is its solubility in water.
Answer:
Amino acids are dipolar in nature (N⁺ H₃ – CHR – COO^–) and have strong dipolar interatctions.
Therefore, these have high melting points. When dissolved in water, these form hydrogen bonds with water molecules, therefore, these are soluble in water, (except at isoelectric point, where they have least solubility) In case of halo acids, these are not dipolar. Only their carboxyl group is involved in hydrogen bonding, not the halogen atom. Therefore, halo acids, have lower melting points and lesser solubility than amino acids.

Question 5.
Why cannot vitamin C be stored in our body ?
Answer:
VitaminC (ascorbic acid) is water soluble. Therefore, it is regularly excreted in urine from the body and cannot be stored.

Question 6.
What products would be formed when a nucleotide from DNA containing thymine is hydrolysed ?
Give the composition of DNA molecule as on hydrolysis it gives its all constituents.
Answer:
When a nucleotide from DNA is hydrolysed, it will give deoxyribose sugar molecules, phosphoric acid, pyrimidine bases, i.e., guanine (G) and adenine (A) and purine bases, i.e., thymine (T) and cytosine (C).

Question 7.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA ?
Single stranded structure of RNA.
Answer:
DNA molecule has a double strand structure and four complementary bases paired with each other. Cytosine (C) pairs with guanine (G) while thymine (T) pairs with adenine (A). Due to such structure, DNA produces the products in definite molar ratio.

The case is not same in RNA, ie., base pairing rule is not followed. Thus, when RNA is hydrolysed, the quantities of different bases are different. This suggests that RNA has single stranded structure.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 8th Lesson Polymers Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 8th Lesson Polymers

Very Short Answer Questions

Question 1.
Define the terms monomer and polymer.
Answer:
Monomer: The repeating structural units of polymer which are derived from simple and reactive molecules linked to one another by covalent bonds are called monomers.
Polymer: A large molecular weight complex compound that is formed by the repeated combination of simple units (monomers) is called a polymer.

Question 2.
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 3.
What is polymerization ? Give an example of polymerization reaction.
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 diamihe and adipic acid leading to the formation of Nylon 6, 6 are examples of two different types of polymerisation reactions.

Question 4.
Give one example each for synthetic and semi-synthetic polymers.
Answer:
Examples of synthetic polymers : Neoprene, Buna-S, Buna-N
Examples of Semi synthetic polymers : Cellulose rayon, cellulose nitrate.

Question 5.
How are the polymers classified on the basis of structure ?
Answer:
On the basis of structure, polymers are classified into three types :

1. Linear polymers : These contains long and straight chains.
   E.g.: PVC, polythene (high density) etc.
2. Branched chain polymers : These contains linear chains having some branches.
   E.g.: low density polythene.
3. Cross linked polymers (or) Network polymers : These are usually formed from bi functional and tri functional monomers and contain strong covalent bonds between various linear polymer chains.
   E.g.: Bakelite, melamine etc. .

Question 6.
Give one example each for linear and branched chain polymers.
Answer:

1. Linear polymers : These contains long and straight chains.
   E.g.: PVC, polythene (high density) etc.
2. Branched chain polymers : These contains linear chains having some branches.
   E.g. : low density polythene.

Question 7.
What are cross linked (or network) polymers ? Give example.
Answer:
Cross linked polymers (or) Network polymers: These are usually formed from bi functional and tri functional monomers and contain strong covalent bonds between various linear polymer chains. E.g. : Bakelite, melamine etc.

Question 8.
What is addition polymer ? Give example. [T.S. Mar. 15]
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 9.
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 10.
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 11.
What are copolymers ? Give example.
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 12.
Is – [-CH₂ – CH (C₆H₅)-]_n – a homopolymer or a copolymer ?
Answer:
-[-CH₂ – CH (C₆H₅-]_n is polystyrene. It is a homopolymer.
It is formed by the polymerisation of single monomer styrene.
(C₆H₅ – CH = CH₂)

Question 13.
Is (NH – CHR- CO)_n a homopolymer or a copolymer ?
Answer:
[NH – CHR – CO]_n is a homopolymer. It is formed by the polymerisation of single monomer a-amino acid (NH₂ – CHR-COOH).

Question 14.
What are the classes of the polymer based on molecular forces ?
Answer:
On the basis of molecular forces polymers are classified into four types.

1. Elastomers: These are rubber like solids with elastic properties.
   E.g.: Buna-S, Buna-N.
2. Fibres : Fibres are the thread forming Solids which possess high tensile strength.
   E.g. : Nylon 6, 6, polyesters
3. 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.
4. 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 15.
What are elastomers ? Give example. .
Answer:
Elastomers: These are rubber like solids with elastic properties. In elastomers the polymer chains are held together by the weak inter molecular forces.
E.g.: Buna – S, Buna – N etc.

Question 16.
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 17.
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 18.
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 19.
Write the name and structure of one of the common initiators used in free radical polymerization reaction.
Answer:
One of the common initiator used in free radical – polymerisation reaction is benzoyl peroxide.
Structure:

Question 20.
How can you differentiate between addition and condensation polymerization?
Answer:

Addition polymerisation

1. Monomers used are unsaturated com­pounds.
2. Polymer formation takes place without loss of atoms (or) groups.
3. It is chain growth polymerisation.
4. These polymers are called addition (or) ‘ chain (or) vinyl polymers.

Condensation polymerisation

1. Monomers are bifunctional, tri-functional compounds.
2. Polymer formation takes place with loss of atoms (or) groups like NH₃, H₂0 etc.
3. It is step growth polymerisation.
4. These are called condensed polymers.

Question 21.
What is Ziegler – Natta catalyst? [T.S. Mar. 19]
Answer:
A mixture of Tri alkyl aluminium and titanium chloride is called Ziegler – Natta catalyst
E.g.: (C₂H₅)₃ Al + TiCl₄

Question 22.
How is Dacron obtained from ethylene glycol and terepthalic acid?
Answer:
Formation of dacron is an example of condensation polymerisation. It is obtained from ethylene glycol and terephthalic acid as follows.

Question 23.
What are the repeating monomeric units of Nylon 6 and Nylon 6, 6?
Answer:
The repeating monomeric units of Nylon – 6 is Capro lactam

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 24.
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.

Question 25.
Arrange the following polymers in increasing order of their molecular forces.

1. Nylon 6, 6 Buna – S, Polythene
2. Nylon 6, Neoprene, Polyvinyl chloride

Answer:

1. Increasing order of inter molecular forces of given polymers is
   Buna – S < polythene < Nylon -6, 6
2. Increasing order of inter molecular forces of given polymers is
   Neoprene < polyvinyl chloride < Nylon – 6.

Question 26.
Identify the monomer in the following polymeric structures

Answer:

1. Monomers present in are decane dioic acid HOOC -CH₂-)₈ COOH and hexamethylene diamine H₂N – (CH₂-)₆ NH₂
2. Monomers present in (-NH – CO – NH – CH₂ -)_n are urea CO (NH₂)₂, formaldehyde HCHO.

Question 27.
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 ([latex]\overline{\mathbf{M}}_{\mathrm{n}}[/latex])
2. Weight average molecular mass ([latex]\overline{\mathbf{M}}_{\mathrm{w}}[/latex])

Question 28.
What is PDI (Poly Dispersity Index) ? [A.P. Mar. 19]
Answer:
Poly Dispersity Index (PDI) : The ratio between weight average molecular mass ([latex]\overline{\mathbf{M}}_{\mathrm{w}}[/latex]) and the number average molecular mass ([latex]\overline{\mathbf{M}}_{\mathrm{n}}[/latex]) of a polymer is called Poly Dispersity Index (PDI).

Question 29.
What is vulcanization of rubber ? [T.S. Mar. 19, 16; A.P. Mar. 17]
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 30.
What is the cross linking agent used in the manufacture of type rubber ?
Answer:
In the manufacture of tyre rubber 5% sulphur is used as cross linking agent. ‘

Question 31.
What is biodegradable polymer ? Give one example of a biodegradable polyester ?
Answer:
Biodegradable poiymers : The polymers degradable by enzymatic hydrolysis and to some extent by oxidation are called biodegradable polymers.
E.g. : Nylon – 2, Nylon – 6, PHBV, Polyglycolic acid, Polylactic acid etc.

Question 32.
What is PHBV ? How is it useful to man ? [A.P. Mar. 19, 18, 16; T.S. Mar. 18, 16, 15]
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.

Question 33.
Give the structure of nylon 2 – nylon 6 ?
Answer:
Nylon 2 – Nylon 6 :
an alternating polyamide copolymer of glycine ( – COOH) and amino caproic acid (H₂N – CH₂)₅ – COOH). It is a biodegradable polymer.
Structure of Nylon 2 – Nylon – 6

Short Answer Questions

Question 1.
Classify the following into addition and condensation polymers [A.P. Mar. 16]
i) Terylene
ii) Bakelite
iii) Polyvinyl chloride
iv) Polythene
Answer:
i) Terylene is a condensation polymer.
ii) Bakelite is a condensation polymer
iii) Polyvinyl chloride is an addition polymer
iv) Polythene is an addition polymer.

Question 2.
How do you explain the functionality of a polymer ?
Answer:
The number of bonding sites present in the monomers of the polymer is called functionality.
E.g.:

1. Functionality of ethene, propene is one.
2. Functionality of ethylene glycol is two.

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.
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-formaldehyd, resin etc….

Question 5.
Explain copolymerization with an example.
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 6.
Explain free radical mechanism for the polymerization of ethene.
Answer:
Polymerisation of ethene to polythene proceeds through free radical mechanism. The initiator used in this reaction is benzoyl peroxide. This mechanism involves three steps.
1) Chain initiation : The addition of phenyl free radical formed by the peroxide to the ethene double bond and forms a larger free radical.

C₆H₅ + CH₂ = CH₂ → C₆H₅ – CH₂ – CH₂

2) Chain propagation : The above formed radical reacts with another molecule of ethene another bigger sized radical is formed. The repetetion of this sequence takes places.

3) Chain Termination : The product radical formed reacts with another radical to form polymerised product.

Question 7.
Write the names and structures of the monomers used for getting the following polymers . [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
Monomers : Phenol, Formal dehyde
Structure :

iv) Polystyrene
Monomer : Styrene
Structure :

Question 8.
Write the names and structures of the monomers of the following polymers.
i) Buna – S
ii) Buna – N
iii) Dacron
iv) Neoprene
Answer:
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, C HOOC COOH

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

Question 9.
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 10.
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 11.
Explain the difference between natural rubber and synthetic rubber.
Answer:
Natural rubber : The rubber which is obtained from natural sources such as plants and animals rubber called natural rubber. E.g.: Starch, cellulose, rubber etc.

Synthetic rubber : The rubber which are artificially prepared i.e. man-made are called synthetic rubbers.
These have wide applications in daily life as well as in industry.

Question 12.
How does the presence of double bonds in rubber molecules influence their structure and reactivity ?
Answer:

1. Natural rubber is cis. poly isoprene. It is obtained by polymerisation of isoprene. The polymerisation takes place at 1, 4 positions. .
2. In rubber molecule double bonds are located between C₂ and C₃ of the monomer
   isoprene.
3. These double bonds (cis) do not allow the polymer chain to come closer. Therefore only weak vander waal’s forces are present.
4. The chains are not linear, they can be stretched just like springs and exhibits elastic properties.

Question 13.
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 addition.
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 14.
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.

Question 15.
Write notes on different types molecular masses of polymers.
Answer:
The molecular mass of the polymer doesnot remain constant as in the case of simple chemical substances. So the molecular weight of a polymer is expressed in terms Of “average value”.
The average molecular masses of polymers expressed in different ways. The following are the important among them.

1. Number average molecular mass ([latex]\overline{\mathrm{M}}_{\mathrm{n}}[/latex]),
2. Weight average molecular mass ([latex]\overline{\mathrm{M}}_{\mathrm{w}}[/latex])

i) Number average molecular mass ([latex]\overline{\mathrm{M}}_{\mathrm{n}}[/latex]) : If a polymer sample contains N₁ particles of mass M₁ each and N₂ particles of Mass M₂ each and so on.
The number average molecular weight ([latex]\overline{\mathrm{M}}_{\mathrm{n}}[/latex]) of the polymer = [latex]\frac{\sum_{\mathrm{n}_{\mathrm{i}}=1}^{\infty} \mathrm{N}_{\mathrm{i}} \mathrm{M}_{\mathrm{i}}}{\sum_{\mathrm{n}_{\mathrm{i}}=1}^{\infty} \mathrm{N}_{\mathrm{i}}}[/latex]
([latex]\overline{\mathrm{M}}_{\mathrm{n}}[/latex]) can be determined by end group analysis method (or) physical methods based on the number of particle present.

ii) Weight average molecular weight ([latex]\overline{\mathrm{M}}_{\mathrm{w}}[/latex]) : Weight average molecular weight ([latex]\overline{\mathrm{M}}_{\mathrm{w}}[/latex]) is calculated by multiplying the molecular weight of each type of particle with contribution of the species to the total weight of the sample.
Weight average molecular weight ([latex]\overline{\mathrm{M}}_{\mathrm{w}}[/latex]) of the polymer = [latex]\frac{\sum_{n_{i}=1}^{\infty} N_{i} M_{i}^{2}}{\sum_{n_{i}=1}^{\infty} N_{i} M_{i}}[/latex]

Long Answer Questions

Question 1.
Write an essay on i) Addition polymerization and ii) Condensation polynerization
Answer:
i) Addition Polymerization : Addition polymerization involves the self addition of unsaturated monomers without loss of any small molecules to form polymer.
The polymer so formed is called addition polymer.
This polymerisation is also called chain growth polymerisation.
In addition polymerisation, monomers should contain double bonds.
This type of polymerisation can be broadly subdivided into 2 groups.
a) ionic polymerisation (cationic & anionic)
b) free radical polymerization
Addition polymerization mechanism involves three steps. They are : i)-Chain initiation step ii) Chain propagation step and iii) Chain termination step.
E.g. : Vinyl Chloride molecules polymerises to PVC.

ii) Condensation polymerization : In this polymerisation monomers combine together with the loss of simple molecules like H₂O, NH₃ etc. The formed polymer is called condensation polymer.
Condensation polymers are generally derived from di (or) polyfunctional monomers. This polymerization is also called step polymerization.
E.g.: i) Nylon – 6,6 is formed from hexamethylene diamine and adipic acid by condensation.

ii) Polyethylene terephthalate (PET) from ethylene glycol and terephtholic acid by condensation reaction.

Question 2.
Explain the classification of polymers based on their source and structure.
Answer:
On the basis of source polymers are classified into three types.

1. Natural polymers
2. Semi synthetic polymers
3. Synthetic polymers

1) 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.

2) Semi synthetic polymers : The polymers which are synthetic derivatives of the natural polymers are called semisynthetic polymers.
E.g. : Cellulose rayon, cellulose nitrate.

3) 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.

Classification of polymers on the basis of structure :
i) Linear polymers : It is one in which the repeating units are similar to the links in a very long chain.

E.g.: Polythene, PVC etc.

ii) Branched chain polymers : It is one in which some of the molecules are attached as side chains to the linear chains.

E.g.: Low density polythene (LDPE) etc.

iii) Cross linked (or) network polymers : More branching at random points connecting many chains give rise to network (or) cross linked polymers.

E.g.: Bakelite, melamine etc.

Question 3.
Explain the classification of polymers based on the mode of polymerization and nature of molecular forces.
Answer:
On the basis of mode of polymerisation polymers are classified into two types :
1) Addition polymers
2) Condensation polymers.
1) 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.
Eg.: Polyethene, poly acrylonitrile.
2) Condensation polymer : The polymer which is formed by the condensation reaction between molecules having more than one functional group is called condensation polymer.
Eg.: Nylon 6, 6, Poly Ethylene terephthalate.

On the basis of molecular forces polymers are classified into four types.

1. Elastomers : These are rubber like solids with elastic properties.
   Eg.: Buna-S, Buna-N.
2. Fibres : Fibres are the thread forming solids which possess high tensile strength.
   Eg.: Nylon 6, 6, polyesters.
3. Thermoplastic Polymers : These are the linear (or) slightly branched long chain molecules capable of softening on heating and hardening on cooling.
   Eg.: Polystyrene, polythene.
4. 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 4.
What are synthetic rubbers ? Explain the preparation and uses of the following
i) Neoprene
ii) Buna-N
iii) Buna-S.
Answer:
The polymer which is capable of getting stretched almost to double the length and which is also vulcanizable as in the case of natural rubber is called a synthetic rubber.
1) These are homo polymers of 1, 3 – butadiene derivatives (or) copolymers in which one of the monomer is 1, 3 – Butadiene (or) its derivatives and the other is any unsaturated monomer.
2) Synthetic rubber return to its original shape and size when the stretched force is removed.
Eg.: Neoprene, Buna – S, Buna – N.

i) Neoprene : Neoprene is formed by the free radical polymerisation of chloroprene (2-Chloro 1, 3 – Butadiene)

Uses:
1) It has superior resistance to vegetable and animal oils.
2) It is used in the manufacture of gaskets, conveyor belts.

ii) Buna – N : Buna – N is the copolymer which is formed by the polymerisation of 1, 3 – Butadiene and acrylonitrile.

Uses:
1) It is resistant to the action of petrol, lubricating oil and organic solvents.
2) It is used in. making oil seals, tank lining.

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

Uses :
1) It is a good substitute for natural rubber.
2) It is used in manufacture of auto tyres.
3) It is used in manufacture of floor tiles.
4) It is used in manufacture of foot wear components, cable insultation

Textual Examples

Question 1.
Is [-CH₂ – CH (C₆H₅-)] _n a homopolymer or a copolymer ?
Solution:
It is a homopolymer and the monomer from which it is obtained is styrene C₆H₅CH = CH₂.

Question 2.
A polymer contains 10 molecules with molecular mass 10,000 and 10 molecules with molecular mass 1,00,000. Calculate number – average molecular mass.
Solution:
[latex]\overline{\mathrm{M}}_{\mathrm{n}}=\frac{\sum \mathrm{N}_{\mathrm{i}} \mathrm{M}_{\mathrm{i}}}{\sum \mathrm{N}_{\mathrm{i}}}=\frac{10 \times 10000+10 \times 100000}{10+10}[/latex] = 55,000

Intext Questions

Question 1.
What are polymers ?
Answer:
Polymers are high molecular mass substances consisting of large -number of repeating structural units. They are also called as macromolecules. Some examples of polymers are polythene, bakelite, rubber, nylon 6, 6 etc.

Question 2.
How are polymers classified on the basis of structure ?
Answer:
On the basis of structure, the polymers are classified as below :

1. Linear polymers such as polythene, polyvinyl chloride, etc.
2. Branched chain polymers such as low density polythene.
3. Cross linked polymers such as bakelite, melamine, etc.

Question 3.
Write the names of monomers of the following polymers.

Answer:

1. Hexamethylene diamine and adipic acid
2. Caprolactam
3. Tetrafluoroethene.

Question 4.
Classify the following as addition and condensation polymers : Terylene, Bakelite, Polyvinyl chloride, polythene.
Answer:
Additon polymers : Polyvinyl chloride, polythene. Condensation polymers : Terylene, Bakelite.

Question 5.
Explain the difference between Buna – N and Buna – S. .
Answer:
Buna – N is a copolymer of 1, 3 – butadiene and acrylonitrile and Buna-S a copolymer of 1, 3—butadiene and styrene.

Question 6.
Arrange the following polymers in increasing order of their intermolecular forces.
i) Nylon 6, 6 Buna — S, Polythene. ii) Nylon 6, Neoprene, Polyvinyl chloride.
Answer:
In order of increasing intermolecular forces.

1. Buna—S, Polythene, Nylon 6, 6.
2. Neoprene, Polyvinyl chloride, Nylon 6.

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material 7th Lesson d and f Block Elements & Coordination Compounds Textbook Questions and Answers.

AP Inter 2nd Year Chemistry Study Material 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.
Which elements of 3d, 4d, and 5d series are not regarded as transition elements and why?
Answer:
Zn (3d-series), Cd (4d-series), Hg (5d-series) are not regarded as transition elements because these contains completely filled d-subshells.

Question 3.
Why are d-block elements called transition elements ?
Answer:
The name transition elements given to d-block is due to their properties which are transitioned between the electropositive s-block metals and electronegative p-block elements.

Question 4.
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 5.
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)d^1-10 ns².

Question 6.
Write the electronic configuration of chromium (Cr) and copper (Cu).
Answer:

• Electronic configuration of chromium (Cr) – [Ar]4s¹d⁵.
• Electronic configuration of copper (Cu) – [Ar]4s¹3d¹⁰.

Question 7.
Why do transition elements exhibits characteristic properties ?
Answer:
Due to the presence of partially filled d-orbitals transition elements exhibit characteristic properties such as variable oxidation states, colour property, magnetic property, complex tendency etc.

Question 8.
Scandium is a transition element. But Zinc is not. Why
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 9.
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 10.
Write the electronic configuration of Co²⁺ and Mn²⁺.
Answer:

• The electronic configuration of Co⁺² is [Ar] 4s⁰ 3d⁷.
• The electronic configuration of Mn⁺² is [Ar] 4s⁰ 3d⁵.

Question 11.
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] 4s¹ 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 12.
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 13.
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 compelete to lose the electrons.

Question 14.
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 15.
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 16.
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 17.
Although Cr, Mo and W belong to the same group (group 6) Cr (VI) is a strong oxidizing agent while Mo (VI) and W (VI) are not. Why ?
Answer:
In group 6, Mo (VI) and W (VI) are more stable than Cr (VI). Thus Cr (VI) in the form of dichromate in acidic medium is a strong oxidising agent. Where as MoO₃ and WO₃ are not.

Question 18.
What do you infer from the fact that M³⁺/M²⁺ standard electrode potential of Mn is comparatively higher and that of Fe is comparitively lower ?
Answer:
M³⁺/M²⁺ standard electrode potential of Mn is comparatively higher than that of Fe is comparatively lower. This is because much larger 3^rd ionisation energy of Mn (d⁵ to d⁴).

Question 19.
Transition elements have high melting points. Why ?
Answer:
Transition elements have high melting points because of involvement of greater number of electrons from (n – 1)d in addition to ns electrons in the interatomic metallic bonding of these metals.

Question 20.
Among the first transition series (3d series) Chromium has highest melting point. Why? Ans. Among first transition series (3d-series) chromium has highest melting point.
Reason : In chromium one unpaired electron per d-orbital is particularly favourable for strong inter atomic interactions.

Question 21.
Compared to s-block elements, the transition elements exhibit higher enthalpy of atomization. Why ?
Answer:
Because of large no. of unpaired electrons in their atoms transition elements have stronger inter atomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

Question 22.
Among the first transition series (3d series) zinc has lowest enthalpy of atomization. Why ?
Answer:
Among the first transition series zinc has lowest enthalpy of atomisation because zinc has no unpaired electrons in their atomic state or in ionic state.

Question 23.
How do you expect the density of transition elements to vary in a given series and why ?
Answer:
In a given series the density of transition elements increases.

Eg. : From Titanium to copper there is a significant increase in the density.
This is due to decrease in metallic radius coupled with increase in atomic mass.

Question 24.
How do the atomic and ionic sizes vary among transition metals in a given series ?
Answer:
In a given series of transition elements atomic and ionic sizes progressively decreased. This is due to entering of a new electron into d-orbital each time.

Question 25.
Why do Mn, Ni and Zn exhibit more negative E^⊖ values than expected ?
Answer:
E⁰ values of Mn, Ni and Zn are more negative than expected from general trend. This is due to the stabilities of half filled d-subshell (d⁵) in Mn⁺² and completely filled d-subshell (d¹⁰) in Zinc. For Nickel E⁰ value is related to the highest negative enthalpy of hydration.

Question 26.
Among the first transition series (3d series) only copper has positive E^⊖M²⁺/M value. Why?
Answer:
Among the first transition series only copper has positive E^⊖M²⁺/M value. This is due to high ∆_aH⁰ and low ∆_hydH⁰ values.

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⁺²oxidises I^– to I₂
2Cu⁺² + 4I^– → Cu₂I₂ + I₂

Question 28.
The highest Mn fluoride is MnF₄ where as the highest oxide is Mn₂O₇ Why ?
Answer:
The ability of oxygen to stabilize the high oxidation states exceeds that of fluoride. Thus the highest Mn fluoride is MnF₄ where as highest oxide is Mn₂O₇.

Question 29.
In its fluoride or Oxide, in which a transition metal exhibits highest oxidation state and why ?
Answer:

• In fluorides highest oxidation numbers are achieved in TiF₄, VF₅ and CrF₆.
• The +7 state of Mn is represented in MnO₃F.
• The highest oxidation number in oxides acheived from Sc₂O₃ to Mn₂O₇.
• In Mn₂O₇, Mn oxidation state is +7.

Question 30.
Why Zn²⁺ is diamagnetic whereas Mn²⁺ is paramagnetic ? [T.S. Mar. 15]
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 31.
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 = [latex]\sqrt{n(n+2)}[/latex] BM
BM = Bohr Magneton.

Question 32.
Calculate the ‘spin only’ magnetic moment of Fe²⁺_(aq) ion.
Answer:
Fe⁺² ion has electronic configuration [Ar] 4s⁰3d⁶
It has four unpaired electrons n = 4
Spin only magnetic moment μ = [latex]\sqrt{n(n+2)}[/latex] BM = [latex]\sqrt{4(4+2)}[/latex] = [latex]\sqrt{24}[/latex] BM = 4.9 BM

Question 33.
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 disproportion 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.
2Cu⁺_(aq) → Cu⁺²_(aq) + Cu_(s)

Question 34.
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. If 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 35.
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 36.
Why do the transition metals form a large number of complex compounds ?
Answer:
Transition elements (metals) forms a large number of complex compounds this is due to

1. Small size of these ions
2. High effective nuclear charge,
3. Possessing in completely filled d-orbitals.

Question 37.
How do transition metals exhibit catalytic activity ?
Answer:
Catalytic properties:

• Transition metals and their compounds form important catalysts in industry and in biological systems.
• The catalytic activity of transition metals depends on their ability to exist in different oxidation states of oxidation (or) to form co-ordination compounds.
  Eg : 1) V₂O₅ is used as catalyst in manufacturing of SO₃ from SO₂.
  
  2) Fe is used as a catalyst in manufacturing of NH₃.
  

Question 38.
Give two reactions in which transition metals or their compounds acts as catalysts.
Answer:
1) V₂O₅ is used as catalyst in manufacturing of SO₃ from SO₂.

2) Fe is used as a catalyst in manufacturing of NH₃.

Question 39.
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 40.
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 41.
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 mainly acidic V₂O₅ reacts with alkali as well as acids to give V0₄^-3 and VO⁺₄.
• CrO is basic
• Mn₂O₇gives HMnO₄ and CrO₃ gives H₂CrO₄ and H₂Cr₂O₇.

Question 42.
What is the effect of increasing pH on a solution of potassium dichromate ?
Answer:
On increasing pH of K₂Cr₂O₇ (orange) it changes into K₂CrO₄ (yellow)

Question 43.
Name the oxometal anions of the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.
Answer:
V₄^-3 ion exhibits ‘+5’ oxidation state which is equal to the V group number
V₄^-3 + 4(-2) = -3, x = +5

Question 44.
Permanganate titrations are carried out in the presence of sulphuric acid but not in presence of hydrochloric acid. Why ?
Answer:
Per manganate titrations are carried out in the presence of sulphuric acid but not in presence of hydrochloric acid because hydrochloric acid is oxidised to chlorine.

Question 45.
What is lanthanoid contraction ? [T.S. Mar. 19]
Answer:
The slow decrease of atomic and ionic radii in lanthanides with increase in atomic number is called lanthanide contraction.

Question 46.
What are the different oxidation states 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 47.
What is mischmetall ? Give its composition and uses. [A.P. Mar. 19, 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 48.
What is actinoid contraction ?
Answer:
The gradual decrease in the size of atoms or M⁺³ ions across, the actinoid series is called actinoid contraction.

Question 49.
What are co-ordination compounds ? Give two examples.
Answer:
Co-ordination 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 coordinate covalent bonds. Such compounds are called co-ordination compounds (or) complex compounds.
Eg.: [Fe(CN)₆]^4-, [Co(NH₃)₆]⁺³.

Question 50.
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 51.
What is a double salt ? Give example.
Answer:
The salts which contains two cations and one anion are called double salts.
These dissociates into simple ions completely when dissolved in water.
Eg : KCl.MgCl₂.6H₃O -Camallite.

Question 52.
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 53.
What is a ligand ?
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 54.
Give one example each for ionic and neutral ligands.
Answer:
Examples for ionic ligands – CN^–, I^–, Cl^–
Examples for Neutral ligands – NH₃, H₂O

Question 55.
How many moles of AgCl is precipitated when 1 mole of CoCl₃ is treated with AgNO₃, solution ?
Answer:
3 moles of AgCl is precipitated by the reaction of 1 mole of CoCl₃ with AgNO₃ solution
3 AgNO₃ + CoCl₃ → CO(NO₃)₃ +3 AgCl ↓

Question 56.
What is a chelate ligand ? Give example.
Answer:
The ligands which can form two co-ordinate covalent bonds through two donar atoms are called bidentate ligands. These bidentate ligands are also called chelate ligands.
Eg.: C₂O₄^-2, CO₃;^-2 etc.

Question 57.
What is an ambidentate ligand ? Give example. [A.P. Mar. 16]
Answer:
A unidentate ligand containing two possible donor atoms can co-ordinate through either of donor atoms. Such ligands are called ambidentate ligands.
Eg : NO₂^–

Question 58.
CUSO₄.5H₂O is blue in colour where as anhydrous CuSO₄ is colourless. Why ?
Answer:
CUSO₄.5H₂O is blue in colour whereas anhydrous CuSO₄ is colourless because in the absence of ligand, crystal field splitting does not occurs.

Question 59.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test for Fe²⁺ ion but CuS04 mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu²⁺ ion. why ?
Answer:

• FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1 : 1 molar ratio gives the test for Fe²⁺ because of formation of double salt (FeSO₄(NH₄)₂ SO₄ 6H₂O Mohr’s salt).
• CuSO₄ mixed with aq.ammonia in 1 : 4 molar ratio does not give the test of Cu⁺² ion because of formation of complex compound [Cu(NH₃)₄]SO₄.

Question 60.
How many geometrical isomers are possible in the following coordination entities ?

1. [Cr(C₂O₄)₃]^3-
2. [CO(NH₃)₃Cl₃]

Answer:

1. [Cr(C₂O₄)₃]^3- : Two geometrical isomers are possible i) cis-isomer, ii) Trans-isomer.
2. [CO(NH₃)₃Cl₃] : Two geometrical isomers are possible i) cis-isomer, ii) Trans-isomer.

Question 61.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution copper sulphate ? Why ?
Answer:
When excess of aq. KCN is mixed with aq.CuSO₄ a complex named potassium tetra cyano cuprate (II) is formal

Question 62.
[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 63.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Why ?
Answer:

• In the complex [Ni(H₂O)₆]²⁺, H₂O molecules are weak ligands they do not cause pairing.
  So the complex has two unpaired electrons, d-d-transitions takes place due to absorption of red light radiation and emission of green colour occurs.
• In the complex [Ni(CN)₄]^2-, CN^– ions are strong ligands they cause pairing. So there is no unpaired electrons so, no d-d transition. Hence it is colourless.

Question 64.
[Fe(CN)₄]^2- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why ?
Answer:
In the given complexes Fe has + 2 oxidation state with 3d⁶ outer electronic configuration. It has four unpaired electrons in presence of weak ligand H₂O. But in presence of strong ligand CN^– the electrons are paired up. Due to the difference in the no. of unpaired electrons both complex have different colours in dilute solutions.

Question 65.
What is the oxidation state of cobalt in (i) K[Co(CO)₄] and (ii) [Co(NH₃)₆]³⁺ ?
Answer:

1. K[CO(CO)₄] : 1 + x + 4(0) = 0, x = -1 .
2. [CO(NH₃)₆]³⁺ : x + 6(0) = + 3, x = + 3

Short Answer Questions

Question 1.
Compared to 3d series the corresponding transition metals of 4d and 5d transition series show high enthalpy of atomization. Explain.
Answer:
Compared to 3d-series the corresponding transition metals of 4d and 5d transition series show high enthalpy of atomisation.

Reason : This is due to the occurrence of much more frequent metal-metal bonding in compounds of heavy transition metals.

Question 2.
Compared to the changes in atomic and ionic sizes of elements of 3d and 4d series, the change in radii of elements of 4d and 5d series is virtually the same Comment.
Answer:
Compared to the change in atomic and ionic sizes of elements 3d and 4d series, the change in radii of elements 4d and 5d series is virtually the same.

This is due to the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbitals results in a regular decrease in atomic radii called lanthanoid contraction. The consequence of lanthanoid contraction is that the 4d and 5d series exhibit same size of radii.

Question 3.
Account for the zero oxidation state of Ni and Fe in [Ni(CO)₄] and [Fe(CO)₅] respectively.
Answer:
In [Ni(CO)₄] and [Fe(CO)₅] the oxidation state of Ni and Fe is zero.
These low oxidation states found when the complex compound has ligands capable of π-acceptor character in addition to the σ-bonding.

Question 4.
Why do the transition metal ions exhibit characteristic colours in aqueous solution. Explain giving examples.
Answer:
Colour property of transition metal ions in aqueous solution is due to the presence of unpaired d-electrons. These unpaired d-electrons from a lower energy level of a metal ion in a complex is excited to a higher energy d-orbital of the same n value. The energy of excitation corresponds to the frequency of light absorbed and this frequency lies in visible region. The colour observed corresponds to the complementary colour of light obsorbed. The frequency of light absorbed is determined by the nature of the ligand. In aqueous solutions water molecules are ligands the colour of the ions observed are listed in the following table.
Colours of Some of the First Row (aquated) Transition Metal Ions

Question 5.
Explain the catalytic action of Iron(DI) in the reaction between I arid S₂O₈^2- ions.
Answer:
Transition metal ions can change their oxidation states and become more effective catalysts. Iron (III) Catalyses the reaction between iodide and per sulphate ions

Catalytic action explained as follows
2Fe⁺³ + 2I^– → 2Fe⁺² + I₂
2Fe⁺² + S₂O₈^-2 → 2Fe⁺³ + 2SO₄^-2

Question 6.
What are interstitial compounds ? How are they formed ? Give two examples.
Answer:
The compounds which are formed when small atoms like H, C or N are trapped inside the crystal lattices of metal are called interstitial compounds.
Eg : TiC, Fe₃H, VH_0.56 etc.

• These are non stoichiometric and are neither typically ionic nor covalent.
• They have high melting points, higher than of pure metals.
• They are very hard, some borides approach diamond in hardness.
• They retain metallic conductivity.
• They are chemically inert.

Question 7.
Write the characteristics of interstitial compounds.
Answer:
Characteristics of interstitial compounds :

• These are non stoichiometric and are neither typically ionic nor covalent.
• They have high melting points, higher than of pure metals.
• They are very hard, some borides approach diamond in hardness.
• They retain metallic conductivity.
• They are chemically inert.

Question 8.
Write the characteristic properties of transition elements. [A.P. Mar. 15]
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 9.
Write down the electronic configuration of ‘

1. Cr³⁺
2. Cu⁺
3. Co²⁺
4. Mn²⁺

Answer:

1. Cr³⁺ electronic configuration is [Ar] 4s⁰3d³
2. Cu⁺ electronic configuration is [Ar] 4s⁰3d¹⁰
3. Co²⁺ electronic configuration is [Ar] 4s⁰3d⁷
4. Mn²⁺ electronic configuration is [Ar] 4s⁰3d⁵

Question 10.
What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : 3d³ 3d⁵ 3d⁸ and 3d⁴ ?
Answer:

1. 3d³-stable oxidation states are +2,+3,+4 and+5 (V)
2. 3d⁵-stable oxidation states are +3, +4 and +6 (Cr)
3. 3d⁵-stable oxidation states are +2, +4, +6 and +7 (Mn)
4. 3d⁸-stable oxidation states are +2, +3 (Co)
5. 3d⁴-This configuration does not exist.

Question 11.
What is lanthanoid contraction ? What are the consequences of lanthanoid contraction?
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 sijnilar properties which makes their separation difficult.

iii) Separation of lathanoids : 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 12.
How is the variability in oxidation states of transition metals different from that of the non transition metals ? Illustrate with examples.
Answer:

• In transition elements the oxidation states vary by unity (due to incomplete filling of d- orbitals)
  Eg : Mn exhibits +2, +3, +4, +5, +6 and +7 all differing by 1.
• In non-transition element, this variation is selective, always differing by 2.
  Eg : S exhibits 2, 4, 6 oxidation states. N exhibits 3, 5 etc.

Question 13.
Describe the preparation of potassium dichromate from iron chromite ore.
Answer:
Preparation of K₂Cr₂O₇ from chromite ore :
i) a) Potassium dichromate is obtained by the fusion of chromite ore (FeCr₂O₄) with sodium (or potassium) carbonate in the presence of excess of air.

b) The solution is filtered and treated with sulphuric acid.

c) Now sodium dichromate is treated with potassium chloride. As a result, potassium dichromate is produced.

ii) Effect of increasing pH on K₂Cr₂O₇ solution :

On increasing pH, K₂Cr₂O₇ changes into K₂CrO₄ (orange to yellow).

Question 14.
Describe the oxidising action of potassium dichromate and write the ionic equations for its.
With (i) iodide (ii) iron (II) solution (iii) H₂S and (iv) Sn(II)
Answer:
Potassium dichromate is a strong oxidising agent. In acidic solution, its oxidising action may be represented as :
Cr₂O₇^2- + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O (E° = 1.33 V)
Ionic reactions :

1. Reaction of K₂Cr₂O₇ with I^–
   Cr₂O₇^2- + 14H⁺ + 6I^– → 2Cr³⁺ + 3I₂ + 7H₂O
2. Reaction of K₂Cr₂O₇ with Fe²⁺ (aq)
   Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
3. Reaction of K₂Cr₂O₇with H₂S
   Cr₂O₇^2- + 8H⁺ + 3H₂S → 2Cr⁺³ + 3S + 7H₂O

Question 15.
Describe the preparation of potassium permanganate.
Answer:
Preparation of KMnO₄ Potassium permanganate is prepared by the fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent like KNO₃. It forms dark green, K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.

Question 16.
How does the acidified permanganate solution react with

1. iron (II) ions
2. SO₂ and
3. oxalic acid.

Write the ionic equations for the reactions.

Answer:
Reactions of KMnO₄ in acidic medium
MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O …………. (i)
1) Iron (II) ions : Ferrous is oxidised to ferric.
[Fe²⁺ → Fe³⁺ + e] × 5 ………………. (ii)
From equation (i) and (ii) 5Fe²⁺ + MnO₄^– + 8H+ → Mn²⁺ + 4H₂O + 5Fe³⁺

2) SO₂ : It is oxidised to SO₄^2- by acidified KMnO₄.
SO₂ + 2H₂O → SO₄^2- + 4H⁺ + 2e^– ……………….(iii)
From equation (i) and (iii) .
5SO₂ + 2MnO₄^– + 2H₂O → 2Mn²⁺ + 4H⁺ + 5SO₄^2-

3) Oxalic acid : C₂O₄^2- → 2CO₂ + 2e^– ……………… (iv)
From equation (1) and (iv)
5C₂O₄^2- + 2MnO₄^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂

Question 17.
Predict which of the ions Cu⁺, Sc³⁺, Mn²⁺, Fe²⁺ are coloured in aqueous solution ? Give reasons.
Answer:
Only those ions will be coloured which have incomplete d-orbitals. Ions which has complete or vacant d-orbitals are colourless.

As Sc³⁺ and Cu⁺ have 3d⁰ and 3d¹⁰ configuration in their valence shell so their aqueous solutions are colourless. All others, i.e., Ti³⁺, V³⁺, Mn²⁺ Fe⁺² and Co²⁺ are coloured in aqueous medium.

Question 18.
Compare the stability of +2 oxidation state of the elements of the first transition series.
Answer:
Element (+2 state) ₂₁Sc²⁺ Ti²⁺ _{22 22}V²⁺ ₂₄Cr²⁺ ₂₅Mn²⁺,
Electronic configuration 3d¹ 3d² 3d³ 3d⁴ 3d⁵
In all the elements listed, the removal of two 4s. Electrons (in Cr²⁺, i.e., from 4s and 1e^– from 3d), the 3d-orbitais get gradually occupied. Since, the number of empty d-orbitais decreases or the number unpaired electrons in 3d-orbitais increases with increase in atomic number of cations, so the stability of the cations (M²⁺) increases from Sc²⁺ to Mn²⁺.

Question 19.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Answer:
Ce(Z = 58) = [Xe] = 4f¹ 5d¹ 6s²
Ce³⁺ = [Xe]4f¹ (only one unpaired electron)
By ‘spin-only’ formula, Magentic moment of Ce³⁺(μ) = [latex]\sqrt{n(n+2)}[/latex]
[∵ n = 1 (unpaired electron)] = [latex]\sqrt{1(1+2)}[/latex] = [latex]\sqrt{3}[/latex] = 1.73 BM

Question 20.
Write down the number of 3d electrons in each of the following ions : Ti²⁺, V²⁺ Cr³⁺ and Mn²⁺ Indicate how would you expect the five 3d orbitais to be occupied for these hydrated ions (octahedral).
Answer:

Question 21.
Explain Werner’s theory of coordination compounds with suitable examples. [A.P. Mar. 18; T.S. Mar. 18, 15] [Mar. 14]
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 numbe (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 SO₄^2- ion 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 22.
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)₈]^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 [Pt Cl₄]^2- is square planar
4. Geometrical shape of [Fe(CN)₈]^4- is octahedral

Question 23.
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) macromolecules, such 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.:
b) Bidentate : Two donor atoms, Eg. : H₂NCH₂CH₂NH₂
(ethane-1, 2-diamine or en), C₂O₄^2- (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(NH_{3/sub>)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/sub>], 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 24.
Explain the terms (i) Unidentate ligand (ii) bidentate ligand (iii) polydentate ligand and (iv) ambidentate ligand giving one example for each.
Answer:
i) Unidentate : The negative ion or neutral molecule having only one donor atom is called unidentate ligand
Eg :
ii) Bidentate (or didentate) : The ions or molecules having two donor atoms are called bidentate ligands.
Eg:
iii) 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 : C₂O₄^-2.
iv) Ambidentate: Ligand which can ligate through two different atoms is called ambidentate ligand. Eg: NO₂^–, SCN^–ions, NO₂^– ion can coordinate either through nitrogen or through oxygen to a central metal atom/ion. Similarly, SCN^– ion can coordinate through the sulphur or nitrogen atom.

Question 25.
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 metal ion, a 5-or 6-membered ring is formed, the effect is known as chelate effect. Example

Question 26.
Give the oxidation numbers of the central metal atoms in the following complex entities
(i) [Ni(CO)₄]
(ii) [CO(NH₃)₆]³⁺
(iii) [Fe(CN)₆]^4- and
(iv) [Fe(C₂O₄)₃]^3-
Answer:
i) [Ni(CO)₄] :
x + 4(0) = 0
x = 0
‘Ni’ oxidation state is zero.

ii) [CO(NH₃)₆]³⁺
x + 6(0) = + 3
x = +3
‘Co’ oxidation state is + 3.

iii) [Fe(CN)₆]^4-
x + 6(-1) = -4
x = + 2
‘Fe’ oxidation state is + 2.

iv) [Fe(C₂O₄)₃]^3-
x + 3(-2) = -3
x = + 3
‘Fe’ oxidation state is + 3.

Question 27.
Using IUPAC norms write the formulas for the following.

1. Tetrahydroxozincate (II)
2. Hexaamminecobalt (III) sulphate
3. Potassium tetrachloropalladate (II) and
4. Potassium tri(oxalato) chromate (III)

Answer:

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

Question 28.
Using IUPAC norms write the systematic names of the following. [A.P. Mar 19]

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

Answer:

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

Question 29.
Explain geometrical isomerism in Co-ordination compounds giving suitable examples. [A.P. Mar 19]
Answer:

• 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 complexes 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 donar atoms of the same ligands occupy adjacent positions at the corners 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.
  

Question 30.
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 ligends. eg.: [Co(NH₃)₄Cl₂]⁺

Long Answer Questions

Question 1.
Explain giving reasons :
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalysts.
Answer:
i) Transition elements have unpaired electrons. Each unpaired electron has a magnetic moment associated with its spin angular momentum and orbital angular momentum. This is the reason of paramagnetism in transition metals.

ii) The reason for the high enthalpy of atomisation is the presence of large number of unpaired electrons in their atoms. These atoms have strong interatomic interaction and hence, stronger bonding between them.

iii) Formation of coloured compounds by transition metals is due to partial adsorption of visible light. The electron absorbs the radiation of a particular frequency (of visible region) and jumps into next orbital.

iv) Catalysts, at the solid surface, involve the formation of bonds between reactants molecules and atoms of the surface of the catalyst (I row transition metals utilise 3d and 48- electrons for bonding). This has the effect of increasing the concentration of the reactants at the catalyst surface and also lowering of the activation energy.
Transition metal ions show variable oxidation states, so they are effective catalysts.

Question 2.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with
(i) iron (II) ions
(ii) SO₂ and
(iii) oxalic acid ?
Write the ionic equations.
Answer:
Preparation of KMnO₄ Potassium permanganate is prepared by the fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent like KNO₃. It forms dark green, K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.

Reactions of KMnO₄ in acidic medium
MnO₄^– + 8H⁺ + 5e^– → Mn²⁺ + 4H₂O …………. (i)
i) Iron (II) ions : Ferrous is oxidised to ferric.
[Fe²⁺ → Fe³⁺ + e] × 5 ………………. (ii)
From equation (i) and (ii) 5Fe²⁺ + MnO₄^– + 8H+ → Mn²⁺ + 4H₂O + 5Fe³⁺

ii) SO₂ : It is oxidised to SO₄^2- by acidified KMnO₄.
SO₂ + 2H₂O → SO₄^2- + 4H⁺ + 2e^– ……………….(iii)
From equation (i) and (iii) .
5SO₂ + 2MnO₄^– + 2H₂O → 2Mn²⁺ + 4H⁺ + 5SO₄^2-

iii) Oxalic acid : C₂O₄^2- → 2CO₂ + 2e^– ……………… (iv)
From equation (1) and (iv)
5C₂O₄^2- + 2MnO₄^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂

Question 3.
Compare the chemistiy of actinoids with that of the lanthanoids with special reference to :
(i) electronic configuration (ii) oxidation state (iii) atomic and ionic sizes and (iv) chemical reactivity.
Answer:

Question 4.
How would you account for the following.
(i) Of the d4 species, Cr²⁺ is strongly reducing while manganese (HI) is strongly oxidising.
(ii) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised,
(iii) The d₁ configuration is very unstable in ions.
Answer:
i) E° value of Cr³⁺/Cr²⁺ is negative (-0.41 V) while that of Mn³⁺/Mn²⁺ is positive (+1.57 V). This means that Cr²⁺ ions can lose electrons to form Cr³⁺ ions and act as a reducing agent while Mn³⁺ ions can accept electrons and can act as an oxidising agent.

ii) Cobalt (III) ion has greater tendency to form complexes than cobalt (II) ion. Therefore, Co (II) ion, being stable in aqueous solution, changes to Co (III) ion, in the presence of complexing reagents and gets oxidised.

iii) Ions of transition metals with d1 configuration tend to lose one electron to acquire d⁰ configuration that is quite stable. Therefore, such ions (with d¹) undergo either oxidation or disproportionation hence unstable.

Question 5.
Give examples and suggest reasons for the following features of the transition medals.
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Answer:
i) Acidic strength of oxides increases with the increase in oxidation state of the element Eg.: MnO(Mn²⁺) is basic whereas Mn₂O₇(Mn⁷⁺) is acidic in nature.

ii) Both oxygen and fluorine being highly electronegative can increase the oxidation state of a particular transition metal. In certain oxides, the element oxygen is involved in multiple bonding with the metal .and this is responsible for the higher oxidation state of the metal.

iii) This is also due to high electronegativity of oxygen eg.: chromium exhibits oxidation states of +6 in oxoanion [CrO₄]^2- and manganese shows oxidation state of +7 in oxoanion [MnO₄]^–

Question 6.
Compare the chemistry of the actinoids with that of lanthanoids with reference to : (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity
Answer:

Question 6.
Explain IUPAC nomenclature of Co-ordination compounds with suitable examples.
Answer:
IUPAC nomenclature : The formula of a compound is an abreviated description of the constitution of the compound. The fpllowing rules are prescribed by IUPAC for naming of Coordination compounds.
i) Positive ions are named first followed by negative ions,
eg.: Potassium hexacyanoferrate (II), K₄[Fe(CN)₆]

ii) Within the Co-ordination sphere ligands are named before the metal atom/ion. However, in formulae, metal ion is written first.
eg.: Tetraammine copper (II) sulphate [Cu(NH₃)₄]SO₄

iii) Prefixes are used to denote the number of same ligands that the present in the Co¬ordination sphere. Complex ions are denoted in paranthesis ( ) and prefixed by bis, tris etc.
Examples:

eg. : [CO(NH₂CH₂CH₂NH₂)Cl₂] Cl is named as dichloro bis (ethylenediamine) cobalt (III) chloride.

iv) Ligands are named in alphabetical order.
eg.: [PtCl₂(NH₃)₂ diammine dichloro platinum (II)

v) Anionic ligands are denoted by a suffix ‘O’ and neutral ligands are denoted by their original names.
eg.: Cl^– – Chloro, CN^– – Cyano
Exception for the above are indicated below.

vi) Oxidation state of the metal ion is indicated by Roman numerical in parenthesis, eg.: [Ag(NH₃)₂] [Ag(CN)₂] is named as diammine silver (I) dicyanoargentate (I).

vii) If the charge of the Co-ordination entity is negative, the name of the metal ends with a suffix-ate.
eg.: [CO(SCN)₄]^2- – tetrahiocyanato cobaltate (II)
Some metal ions are denoted by their names from which their symbols are derived
eg.: Fe – ferrate
Pb – plumbate
Sn – stannate .
Ag-argentate
Au-aurate

viii) Prefixes cis – and trans are used to designate adjacent and opposite geometric locations of the ligands, in a complex.
eg. :

ix) Bridging ligands between two metal ions in a Co-ordination entity are denoted by prefix p(greek letter’mu).
eg.: [(NH₃)₄ CO(OH) (NH₂)Co(NH₃)₄]⁺ is named as μ-amido-μ hydroxo bis (tetraammine) cobalt (IV) .

1. Tetrahydroxozincate (II) – [Zn(OH₄]^-2
2. Hexa ammine cobalt (III) sulphate – [Co(NH₃)₆]₂ (SO₄)₃
3. Potassium tetrachloropalladate – K₂[PdCl₄)
4. Potassium tri(oxalato) chromate (III) – K₃[Cr(C₂O₄)₃]

Question 7.
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) Stereoisomengni: 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.

1. Geomethcal isomerism and
2. optical isomerism

b) Structural isomerism:

1. Linkage isomerism
2. Co-ordination isomerism
3. Ionisation isomerism
4. Hydrate isomerism

a) (1) 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 complexes 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 donar atoms of the same ligands occupy adjacent positions at the corners 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(2) 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 (1) 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(1) 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]

(2) 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]

(3) 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

(4) 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 8.
Discuss the nature of bonding and magnetic behaviour in the following Co-ordination entities on the basis of valence bond theory.
(i) [Fe(CN₆)]^4-
(ii) [FeF₆]^3-
(iii) [Co(C₂O₄)₃]^3- and
(iv) [CoF₆]^3-
Answer:
i) Fe(CN₆)]^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.
Fe³⁺ = 3d⁵4s⁰

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

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 9.
Sketch the spliting of d orbitals in an octahedral crystal field.
Answer:

Question 10.
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 (∆₀) are called weak field ligands. For such ligands ∆₀ < 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 ∆₀ > P.

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, ∆₀ for octahedral field (and ∆_t for tetrahedral field)

If ∆₀ < P (pairing energy), the 4^th electron enters one of the e_g orbitals giving the configuration [latex]t_{2 g}^{3} e_{g}^{1}[/latex] thereby forming high spin complex. Such ligands for which ∆₀ < P are known as weak field ligands.

If ∆₀ > P, the 4^th electron pairs up in one of the t_2g orbitals giving the configuration [latex]\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{0}[/latex], thus forming low spin complexes. Such ligands for which ∆₀ > P are called strong field ligands.

Question 11.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal-carbon bond in metal carbonyls have both sigma (σ) and pi (π) characters. The metal-carbon σ-bond is formed by the donation of lone pair of electrons of the carbonyl carbon to a vacant orbital of the metal. The metal- carbon re-bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant

Question 12.
Explain the applications of Co-ordination compounds in different fields.
Answer:
Applications of Co-ordination compounds : Co-ordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine. These are described below.

Co-ordination compounds find use in many qualitative and quantitative chemical analysis. The familiar colour reactions given by metal ions with a number of ligands (especially chelating ligands), as a result of formation of co-ordination enitites, form the basis for their detection and estimation by classical and instrumental methods of analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), oc-nitroso-p-naphtol, cupron etc.

Hardness of water is estimated by simple titration with Na₂EDTA. The Ca²⁺ and Mg²⁺ ions form stable complexes with EDTA. The selective estimation of these ions can be done due to difference in the stability constants of calcium and magnesium complexes.

Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold for example, combines with cyanide in the presence of oxygen and water to form the co-ordination entity [Au(CN)₂]^– in aqueous solution. Gold can be separated in metallic from this solution by the addition of zinc.

Similarly, purification of metals can be achieved through formation and subsequent decomposition of their coordination compounds. For example, impure nickel is converted to [Ni(CO)₄], which is decomposed to yield pure nickel.

Co-ordination compounds are of great importance in biological systems. The pigment responsible for photosynthesis, chlorophyll, is a co-ordination compound of magnesium. Haemoglobin, the red pigment of blood which acts as oxygen carrier is a co-ordination compound of iron. Vitamin B₁₂, cyanocobalamine, the anti-pernicious anaemia factor, is a coordination compound of cobalt. Among the other compounds of biological importance with coordinated metal ions are the enzymes like, carboxypeptidase A and carbonic anhydrase (catalysts of biological systems).

Co-ordination compounds are used as catalysts for many industrial processes. Examples include rhodium complex, [(Ph₃P)₃RhCl], a Wilkinson catalyst, is used for the hydrogenation of alkenes.

Articles can be electroplated with silver and gold much more smoothly and evenly fr6m solutions of the complexes, [Ag(CN)₂]^– and. [Au(CN)₂]^– than from a solution of simple metal ions.

In black and white photography, the developed film is fixed by washing with hypo solution which dissolves the undecomposed AgBr to form a complex ion, [Ag(S₂O₃)₂]^3-.

There is growing interest in the use of chelate therapy in medicinal chemisty. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. Thus, excess of copper and iro are removed by the chelating ligands D-penicillamine and desferrioxime B via the formation of coordination compounds. EDTA is used in the treatment of lead poisoning. Some co-ordination compounds of platinum effectively inhibit the growth of tumours. Examples are : cis-platin and related compounds.

Textual Examples

Question 1.
On what ground can you say that scandium (Z = 21) is a transition element but zinc (Z = 30) is not ?
Solution:
On the basis of incompletely filled 3d orbitals in case of scandium atom in its atomic state (3d¹), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d¹⁰) in its ground state as well as in its oxidised state, hence it is not regarded as a transition element.

Question 2.
Why do the transition elements exhibit higher enthapies of atomisation ?
Solution:
Because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

Question 3.
Name at transition element which does not exhibit variable oxidation states.
Solution:
Scandium (Z = 21) does not exhibit variable oxidation states.

Question 4.
Why is Cr²⁺ reducing and Mn³⁺ oxidising when both have d4 configuration ?
Solution:
Cr2²⁺ 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 Mn²⁺ to Mn³⁺ results in the half-filled (d<sup5) configuration which has extra stability.

Question 5.
How would you account for the increasing oxidising power in the series VO₂⁺ < Cr₂O₇^2- < MnO₄^–?
Solution:
This is due to the increasing stability of the lower species to which they are reduced.

Question 6.
For the first row transition metals the E^⊖ values are

Explain the non-regularity in the above values.
Solution:
The E^⊖ (M²+/M) values are not regular which can be explained from the non-regular variation of ionisation enthalpies (∆_iH₁ + ∆_iH₂) and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

Question 7.
Why is the E^⊖ value for the Mn³⁺/Mn²⁺ couple much more positive than that for Cr³⁺/ Cr²⁺ or Fe³⁺/Fe²⁺ ? Explain.
Solution:
Much larger third ionisation energy of Mn (where the required change is d⁵ to d⁴) is mainly responsible for this. This also explains why the +3 state of Mn is of little importance.

Question 8.
Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25. [A.P. 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 = [latex]\sqrt{5(5+2)}[/latex] = 5.92 BM

Question 9.
What is meant by ‘disproportionation’ of an oxidation state ? Give an example.
Solution:
When a particular oxidation state becomes less stable relative to other oxidation states, one lower, one higher, it is said to undergo disproportionation. For example, manganese (VI) becomes unstable relative to manganese (VII) and manganese (TV) in acidic solution.
3Mn^VIO₄^2- + 4H⁺ → 2Mn^VIIO₄^– + Mn^IVO₂ + H₂O

Question 10.
Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Solution:
Cerium (Z = 58)

Question 11.
On the basis of the following observations made with aqueous solutions, assign secondary valencies to metals in the following compounds.

Solution:
i) Secondary 4 ii) Secondary 6 iii) Secondary 6 iv) Secondary 6 v) Secondary 4

Question 12.
Write the formulas for the following Co-ordination compounds
a) Tetraammineaquachloro cobalt (III) chloride
b) Potassium tetrahydroxozincate (II)
c) Potassium trioxalatoaluminate (III)
d) Dichlorobis (ethane-1, 2-diamine) cobalt (III)
e) Tetracarbonylnickel (0)
Solution:
a) [Co(NH₃)₄(H₂O)Cl]Cl₂
b) K₂[Zn(OH)₄]
c) K₃[Al(C₂O₄)₃]
d) [CoCl₂(en)₂]⁺
e) [Ni(CO)₄]

Question 13.
Write the IUPAC names of the following coordination compounds.
a) [Pt(NH₃)₂Cl(NO₂)]
b) K₃[Cr(C₂O<sub4)₃]
c) [CoCl₂(en)₂]Cl
d) [Co(NH₃)₅(CO₃)]Cl
e) Hg[Co(SCN)₄]
Solution:
a) Diamminechloronitrito-N-platinum (II)
b) Potassium trioxalatochromate (III)
c) Dichlorobis (ethane-1, 2-diamine) cobalt (III) chloride
d) Pentaairuninecarbonatocobalt (III) chloride
e) Mercury tetrathiocyanocobaltate (III)

Question 14.
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands Co-ordinated with the central metal ion ?
Solution:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the same with respect to each other.

Question 15.
Draw structures of geometrical isomers of [Fe(NH₃)₂(CN)₄]^–
Solution:

Question 16.
Out of the following two Co-ordination entities which is chiral (optically active) ?
a) cis-(CrCl₂(ox)₂]^3-
b) trans-[CrCl₂(ox)₂]^3-
Solution:

Out of the two, (a) cis-(CrCl₂(ox)₂]^3- is chiral (optically active).

Question 17.
The spin only magnetic moment of [MnBr₄]^2- is 5.9 BM. Predict the geometry of the complexion?
Solution:
Since the Co-ordination number of Mn²⁺ ion in the complex ion is 4, it will be either tetrahedral (sp3 hybridisation) or square planar (dsp² hybridisation). But the fact that the magnetic moment of the complex ion is 5.9 BM, it should be tetrahedral in shape rather than square planar because of the presence of five unpaired electrons in the d-orbitals.

Intext Questions

Question 1.
Silver atom has completely filled d-orbitals (4d¹⁰) in its ground State. How can you say that it is a transition element ?
Solution:
Silver (atomic no. = 47), in its +1 oxidation state exhibits 4d¹⁰5s⁰ configuration. But in some compounds it also shows +2 oxidation state, i.e., 4d⁹5s⁰ configuration. Here d-orbital is not completely filled. Therefore, silver is a transition element.

Question 2.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol^-1, why ?
Solution:
In zinc (3d¹⁰4s¹) d-orbital is complete so the electrons of d-orbital are not involved in bonding. So, metallic bond is weaker than the other elements of the series, where the electrons of d-orbitals are involved in metallic bonds. So, enthalpy of atomisation of zinc is lowest in its transition series.

Question 3.
Which of the 3d series of the transition metals exhibits the largest number of oxidation states and why ?
Solution:
Manganese (atomic no. = 25) has electronic configuration [Ar] 3d⁵4s². It shows maximum number of oxidation states, i.e., from +2 to +7 (+2, +3, +4, +5, +6, +7) in its compounds.

Question 4.
The E°(M²⁺/M) value for copper is positive (+0.34 V). What is possibly the reason for this ?
Answer:
E° (M²⁺/M) value for any metal depends on three factors : •
i) ∆_aH (Enthalpy of atomisation); M(s). + ∆_aH → M(g)
ii) ∆_iH (Enthalpy of ionisation); M(g) + ∆_iH → M²⁺(g)
iii) ∆_hydH (Hydration enthalpy); M²⁺(g) + (aq → M²⁺(aq)
Copper has high value of enthalpy of atomisation and low value of enthalpy of hydration. It means that ∆_iH required is not compensated by the energy released. Therefore, E⁰_(Cu^2+/Cu) is positive.

Question 5.
How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements ?
Solution:
In the first transition series, there is irregular variation of ionisation enthalpies because the stability of 3d configuration differs to some extent. Generally the ionisation enthalpy increases with increase in effective nuclear charge, Thus, it is lower for Cr due to absence of any change in the d-configuration while high for Zn as it represents an jonisation from the 4s level. Second ionisation energy, shows the removal of second. The configurations like d⁵ and d¹⁰ are exceptionally stable. So, they have high ionisation enthalpies.

Question 6.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only ?
Solution:
Oxygen as well as fluorine both have high values of electronegativity. So, in their compounds (oxides and fluorides) they can oxidise the metal to their highest oxidation states.

Question 7.
Which is the stronger reducing agent Cr²⁺ or Fe²⁺ and why ?
Solution:
Cr²⁺ is a stronger reducing agent than Fe²⁺. The E° values are
E°_{Cr³⁺/Cr²⁺} = -0-41 V and E⁰Fe³⁺/Fe²⁺ = 0.77 V

Therefore, Cr²⁺ is stronger reducing agent (itself gets oxidised easily) than Fe²⁺.

Question 8.
Calculate the ‘spin only’ magnetic moment of M²⁺(aq) ion (Z = 27).
Solution:
Electronic configuration of M(Z = 27) is [Ar] 3d⁷ 4s²
Electronic configuration of M²⁺ = [Ar]3d⁷ or
Three unpaired electrons are present in M²⁺ (aq) ion. i.e., n = 3. Applying ‘spin only1 formula for magnetic moment.
μ = [latex]\sqrt{n(n+2)}[/latex] = [latex]\sqrt{3(3+2)}[/latex] = [latex]\sqrt{15}[/latex] BM = 3.87 BM
Note : Unit of magnetic moment is Bohr Magnetons (BM).

Question 9.
Explain why Cu⁺ ion is not stablein aqueous solutions ?
Solution:
In aqueous solution, Cu⁺(aq) undergoes disproportionation reaction as follows.
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
The higher stability of Cu²⁺ (aq) in aqueous solution is due to higher negative enthalpy of hydration ∆_hydH° than that of Cu⁺(aq). It compensates the second IE of Cu involved in the formation of Cu²⁺ ion. Hence, Cu⁺ ion changes to more stable Cu²⁺ ion in aqueous medium.

Question 10.
Actinoid contraction is greater from element to element than lanthanoid contraction ?
Solution:
The decrease in atomic (or ionic) radii in actinoid elements (actinoid contraction) is greater than lanthanoid contraction because 5f-electrons have poor shielding effect as compared to 4f- electrons. Therefore, the effect of increased nuclear charge leading to contraction in size is more in case of actinoid elements.

Question 11.
Write the formulae for the following Co-ordination compounds
i) Tetraamminediaquacobalt (III) chloride,
ii) Potassium tetracyanonickelate (II)
iii) Tris-(ethane-l, 2-diamine) chromium (m) chloride
iv) Amminebromidochloridonitrito-N-platinate (II)
v) Dichlorido bis-(ethane-1,2-diamine) platinum (IV) nitrate
Solution:
i)

To find the value of x, we have to find the charge on the complex.
III
[CO(NH₃)₄ (H₂O)₂]^x+ + 3 + 4 × 0 + 2 × 0 = x, × = +3
o, the formula of the complex is [CO(NH₃)₄ (H₂O)₂]Cl₃.

ii)

To find the value of x, find the charge on the complex.
[Ni(CN)₄]^x- (as K+ is positive)
+2 + (-1) x 4 = – x
-x = -2 or x = + 2
So, the formula of the complex is K₂[Ni(CN)₄].
Similarly
iii) [Cr(en)₃]Cl₃
iv) [Pt(NH₃) BrCl(NO₂)]^–
v) [PtCl₂(en)₂] (NO₃)₂
vi) Fe₄[Fe(CN)₆]₃

Question 12.
Write the IUPAC names of the following Co-ordination compounds.
(i) [Co(NH₃)₆]Cl₃
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NH₂CH₃)Cl.
Solution:
i)

Let the oxidation state of Co be x.
x + (0)6 + (-1) × 3 = 0, x + 0 – 3 = 0, x = +3
So, the name of this complex is hexaamminecobalt (III) chloride,

ii) [CO(NH₃)₅Cl]Cl₂
x + (0) × 5 + (-1) × 1 + (-1) × 2 = 0, x + 0 – 3 = 0, x =.+ 3
So, the name of the complex is pentaamminechloridocobalt (III) chloride
[K3[Fe(CN)6]

iii)

Let the oxidation state of Fe is x.
(+ 1) 3 + x + (-1) 6 = 0, + 3 + x – 6 = 0, x = + 3
So, the name of the complex is potassium hexacyanoferrate (III) .

iv)

Let the oxidation state of Fe is x.
(+ 1) 3 + x + (-2) 3 = 0 [∵ Oxalate ion (C₂O₄^2-) bears – 2 charge.]
3 + x- 6 = 0, x = +3
So, the name of the complex is potassium trioxalateferrate (III).

v)

Let the oxidation state of Pd is x.
(+ 1) 2 + x + (-1) 4 = 0, 2 + x – 4 = 0, x = + 2 .
So, the name of the complex is potassium tetrachloridopalladate (II).

vi)

Let the oxidation state of Pt is x.
x + (0) 2 + (-1) × 1 + 0 + (-1) × 1 = 0
x + 0- 1 + 0 -1 = 0, x – 2 = 0, x = + 2
So, the name of the complex is diamminechlorido (methylamine) platinum (II) chloride.

Question 13.
Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers :
(i) K[Cr(H₂O)₂(C₂O₄)₂]
(ii) [Co(en)₃]Cl₃
(iii) [Co(NH₃)₅(NO₂)](NO₃)₂
(iv) [Pt(NH₃)(H₂O)Cl₂]
Solution:
i) [Cr(H₂O)₂(C₂O₄)₂] or K[Cr(H₂O)₂(ox)₂] :
(where, ox = oxalate ion)
a) It exists as geometrical isomers : cis and trans forms.
(In cis form, the same groups occupy adjacent positions while in trans form, they are present at alternate positions.)

b) The cis isomer can also exist as a pair of optical isomers (i.e., d-and 1- forms) (due to absence of plane of symmetry).

ii) [Co(en)₃]Cl₃ : It has two optical isomers (i.e. d-and 1-forms)

iii) [Co(NH₃)₅(NO₂)](NO₃)₂ : It can exist as a pair of ionization isomers as well as linkage iosmers.
Ionization isomers :
[Co(NH₃)₅(NO₂)](NO₃)₂ and [Co(NH₃)₅(NO₃)](NO₂) (NO₃) as they give different ions on ionization.

Linkage isomers : [Co(NH₃)₅(NO₂)](NO₃)₂ and [CO(NH₃)₅ONO](NO₃)₂

iv) [Pt(NH₃)(H₂O)Cl₂] : It can exist as two geometrical iosmers. .

Question 14.
Give evidence that [Co(NH₃)₅Cl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionization isomers.
Solution:
We dissolve both the compounds in water in separate test tubes. To the both test tube
I Step or I Test: Add BaCl₂ solution One compound give white ppt., indicating the presence of SO₄^2- ions.
The other compound does not give white ppt, indicating the absence of SO₄^2- ions.

II Test: Add AgNO₃ solution to both the Compounds in separate test tubes :
Only [II] compound gives white ppt, not the [I] one, due to absence of Cl^– as counter ion.

These two tests prove that the given two compounds are a pair of ionization isomers.

Question 15.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [NiCl₄] ion with tetrahedral geometry is paramagnetlc.
Solution:

(Cl^– being weak field ligand, cannot cause pairing.)

[NiCl₄]^2- ion is paramagnetic in nature as it has two unpaired electrons.

Question 16.
[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral.
Solution:
In the complex [NiCl₄]^2- Ni is in +2 oxidation state and has the configuration 3d⁸4s⁰. The cr ion being a weak field ligand cannot pair the two unpaired electrons present in 3d-orbitals. This means that 3d-orbitals are not involved in hybridization. Thus, the complex is sp3 hybridized (tetrahedral) and.is paramagnetic in nature.

In the complex [Ni(CO)₄], the oxidation state of nickel is zero and electronic configuration is 3d⁸4s². In the presence of the ligand CO, the 4s-electrons shift to the two half-filled 3d-orbitals and make all the electrons paired. The valence 4s and 4p-orbitals are involved in hybridization. Thus, the complex is tetrahedral but diamagnetic in nature.
Outer configuration of Ni atom = 3d⁸ 4s².

Question 17.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic Explain.
Solution:
Outer configuration of ₂₆Fe = 3ds⁶4s². In both the complexes Fe is present as Fes³⁺ ion.
Fes³⁺ ion = 3d⁵4s⁰

In the presence of CN^– (a strong field ligand) the 3d-electrons pair up leaving only one unpaired electron. The hybridization is d²sp³ forming an inner orbital complex.

In the presence of H₂O, (a weak field ligand), 3d-electrons do not pair up. The hybridization is sp³d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 18.
Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Solution:
In [Co(NH₃)₆]³⁺, CO is present as Co³⁺ arid has 3d⁰ configuration.
In the presence of NH₃, the 3d-electrons pair up leaving two d-orbitals empty to the involved in d²sp³ hybridization forming inner orbital complex in the case of [Co(NH₃)₆]³⁺.

Since, (n – 1) d-orbitals are not available but the nd-orbitals are used in bond formation i.e., in hybridization, the complex is called outer orbital complex.

Question 19.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Solution:
₇₈Pt is present in group 10 of the Periodic Table. Its outer configuration is 5d⁹6s¹.

For square plannar shape, the hybridization is dsp². Hence, the unpaired electrons in 5d- orbital pair up to make one d-orbital empty for dsp² hybridization. Thus, there is no unpaired electron.

Question 20.
The hexaquo manganese (II) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using crystal field theory.
Solution:
Mn(II) ion has 3d⁵ configuration. In the presence of H₂O molecules (acting as weak field ligands), the distribution of these five electrons is [latex]\mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^2[/latex] all the electrons remain unpaired to form a high spin complex.

However, in the presence of CN^– (acting as strong field ligands), the distribution of these electrons is [latex]\mathrm{t}_{2 \mathrm{~g}}^5 \mathrm{e}_{\mathrm{g}}^0[/latex], i.e., two t_2g orbitals contain paired electrons while the third t_2g orbital contains one unpaired electron. The complex formed is a low spin complex.

Question 21.
Calculate the overall complex dissociation equilibrium constant for the Cu(NH₃)₄²⁺ ions, given that β₄ for this complex is 2.1 × 10¹³.
Solution:
Overall complex dissociation equilibrium constant
= [latex]\frac{1}{\beta_4}[/latex] = [latex]\frac{1}{2.1 \times 10^{13}}[/latex]
= 4.7 × 10^-14