12th Chemistry Public Compulsory Questions & Problems with Solutions

12th CHEMISTRY - PUBLIC COMPULSORY QUESTIONS AND PROBLEMS

2. P- Block Elements I

1. A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing agent (C). Identify (A), (B) and (C). (Jun-20)

Answer:
$2LiH + B_2H_6 \xrightarrow{\text{Ether}} 2LiBH_4$
(A) + (B) $\rightarrow$ (C)

Result:
(A) $LiH$ - Lithium Hydride
(B) $B_2H_6$ - Diborane
(C) $LiBH_4$ - Lithium Borohydride

2. Aluminium to thallium ionisation enthalpy decreases is only a marginal difference. Why and explain. (Mar-20)

Answer:
Due to the presence of inner d and f-electrons which has poor shielding effect compared to s and p-electrons.

5. Coordination Chemistry

1. Write the ligand, central metal ion and IUPAC name for the $[Ag(NH_3)_2]^+$ complex. (May 22)

Answer:
Ligand: $NH_3$ (ammine)
Central metal ion: $Ag^+$ (silver)
IUPAC name: Diamminesilver (I) ion

2. Write the ligand, central metal ion and IUPAC name for the $[Co(CN)_2Cl_2]Cl$ complex. (May 22)

Answer:
Ligand: $CN^-$ (cyanido), $Cl^-$ (chlorido)
Central metal ion: $Co^{3+}$ (Cobalt)
IUPAC name: Dichloridodicyanidocobalt(III) chloride

3. Write the IUPAC ligand name for the following (Aug 22)
1) $C_2O_4^{2-}$   2) $H_2O$   3) $Cl^-$   4) $NH_3$

Answer:
1) $C_2O_4^{2-}$ - Oxalato
2) $H_2O$ - Aqua
3) $Cl^-$ - Chlorido
4) $NH_3$ - Ammine

4. For the complex, $[Pt(NO_2)(H_2O)(NH_3)_2]Br$, identify the following.
(a) Central metal atom / ion
(b) Coordination number
(c) Ligand
(d) Oxidation number of central metal ion

Answer:
(a) Central metal atom/ion: $Pt^{2+}$
(b) Coordination number: 4
(c) Ligand: $NO_2^-$ (Nitrito-N), $H_2O$ (Aqua), $NH_3$ (Ammine)
(d) Oxidation number of central metal ion: +2

5. Write the IUPAC name for the following compounds
A) $[Ag(NH_3)_2]^+$
B) $[Co(NH_3)_5Cl]^{2+}$

Answer:
A) $[Ag(NH_3)_2]^+$ - Diamminesilver(I) ion
B) $[Co(NH_3)_5Cl]^{2+}$ - Pentaamminechloridocobalt(III) ion

6. Solid State

1. Classify the following solids (Aug 21)
a) Naphthalene b) Brass c) Diamond d) NaCl e) Glucose g) $SiO_2$

Answer:
Covalent solids: Diamond, $SiO_2$
Molecular solids: Naphthalene, Glucose
Ionic solids: NaCl
Metallic solids: Brass

2. Calculate the number of atoms in a FCC unit cell. (Mar 23)

Answer:

The number of atoms in a FCC unit cell is calculated as:
$FCC = \frac{N_c}{8} + \frac{N_f}{2} = \frac{8}{8} + \frac{6}{2} = 1 + 3 = 4$

3. Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm. calculate the edge length of unit cell. (Mar 24)

Answer:
Given: $CCP = FCC$, $r = 125\text{ pm}$
Formula for FCC: $4r = \sqrt{2}a \implies a = \frac{4r}{\sqrt{2}}$
$a = \frac{4 \times 125}{1.414} = 353.5 \text{ pm}$
Result: Edge length of unit cell (a) = 353.5 pm.

7. Chemical Kinetics

1. What is an order of a reaction? (Mar 24)

Answer:
It is the sum of the powers of concentration terms involved in the experimentally determined rate law.

2. Powdered $CaCO_3$ reacts much faster with dilute HCl than with the same mass of $CaCO_3$ as marble. Give reason (Jun 20)

Answer:
When the $CaCO_3$ particles are smaller, the surface area exposed to the acid is higher.
$\rightarrow$ (Rate of a reaction increases when the surface area of a solid reactant is increased).

3. The rate constant for a first order reaction is $1.54 \times 10^{-3} s^{-1}$. Calculate its half life time. (Jun 20, 24 Mar 25)

Solution:
Given: $k = 1.54 \times 10^{-3} s^{-1}$, $t_{1/2} = ?$
$t_{1/2} = \frac{0.693}{k}$
$t_{1/2} = \frac{0.693}{1.54 \times 10^{-3}} = 450 \text{ s}$

4. Show that in case of first order reaction, the time required for 99% completion is twice times the time required for the completion of 90% of the reaction. (Mar 23)

Solution:
$t = \frac{2.303}{k} \log \frac{[A_o]}{[A]}$

For 90% completion: $A_o = 100$, $A = 100 - 90 = 10$
$t_{90\%} = \frac{2.303}{k} \log \frac{100}{10}$ (or) $t_{90\%} = \frac{2.303}{k} \log 10$

For 99% completion: $A_o = 100$, $A = 100 - 99 = 1$
$t_{99\%} = \frac{2.303}{k} \log \frac{100}{1}$ (or) $t_{99\%} = \frac{2.303}{k} \log 100$

Ratio: $\frac{t_{99\%}}{t_{90\%}} = \frac{\log 100}{\log 10} = \frac{2}{1} = 2$
Therefore, $t_{99\%} = 2 \times t_{90\%}$

5. The rate of the reaction $x + 2y \rightarrow$ product is $4 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$ if $[x]=[y]=0.2\text{M}$ and rate constant at 400K is $2 \times 10^{-2}\text{s}^{-1}$. What is the overall order of the reaction. (Sep 20)

Solution:
$\text{Rate} = k[x]^m[y]^n$
$4 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} = 2 \times 10^{-2} \text{s}^{-1} (0.2 \text{ mol L}^{-1})^m (0.2 \text{ mol L}^{-1})^n$
$\frac{4 \times 10^{-3}}{2 \times 10^{-2}} = (0.2)^{m+n}$
$0.2 = (0.2)^{m+n}$
Comparing the powers on both sides: $m+n = 1$
$\rightarrow$ Overall order of the reaction = 1

6. In a first order reaction $x \rightarrow y$ if k is the rate constant and the initial concentration of the reactant x is 0.1 M. What is the value of $t_{1/2}$ in the reaction? (Jun 25)

Solution:
$\rightarrow$ The half-life of a first order reaction is independent of the initial concentration.
$t_{1/2} = \frac{0.693}{k}$

8. Ionic Equilibrium

1. Calculate the concentration of $OH^-$ in a fruit juice which contains $2 \times 10^{-3}\text{M}$, $H_3O^+$ ion. Identify the nature of the solution. (Jun 23)

Solution:
$K_w = [H_3O^+][OH^-] = 1 \times 10^{-14}$
$[OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1 \times 10^{-14}}{2 \times 10^{-3}} = 5 \times 10^{-12} \text{ M}$
Since $[H_3O^+] (2 \times 10^{-3}\text{M}) \gg [OH^-] (5 \times 10^{-12}\text{M})$, the solution is acidic in nature.

2. Calculate the pH of 0.1M $CH_3COONa$ solution. (pKa for $CH_3COOH$ is 4.74). (Sep 20)

Solution:
$pH = 7 + \frac{pK_a}{2} + \frac{\log c}{2}$
$pH = 7 + \frac{4.74}{2} + \frac{\log 0.1}{2}$
$pH = 7 + 2.37 - 0.5 = 8.87$

3. Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. $K_a$ is $1.8 \times 10^{-5}$. (Aug 21)

Solution:
$pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log 1.8 = 5 - 0.26 = 4.74$
Using Henderson-Hasselbalch equation:
$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$
$pH = 4.74 + \log\frac{0.20}{0.18} = 4.74 + \log\frac{10}{9}$
$pH = 4.74 + \log 10 - \log 9 = 4.74 + 1 - 0.95 = 4.79$

4. Calculate the pH of 0.1M $CH_3COOH$ solution. Dissociation constant of acetic acid is $1.8 \times 10^{-5}$. (Mar 25)

Solution:
$[H^+] = \sqrt{K_a \times C} = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ M}$
$pH = -\log[H^+] = -\log(1.34 \times 10^{-3}) = 3 - \log 1.34 = 3 - 0.1271 = 2.87$

9. Electrochemistry

1. A solution of silver nitrate is electrolysed for 30 minutes with a current of 2 amperes. Calculate the mass of silver deposited at the cathode. (Sep 20)

Solution:
Given: $I = 2\text{A}$, $t = 30 \times 60 \text{ s}$, $Z = \frac{108 \text{ g/mol}}{96500 \text{ C/mol}}$
Using Faraday's First Law, $m = Z I t$
$m = \frac{108}{96500} \times 2 \times 30 \times 60 = \frac{108 \times 36}{965} = 4.03 \text{ g}$

2. A solution of silver nitrate is electrolysed for 20 minutes with a current of 2 amperes. Calculate the mass of silver deposited at the cathode. (July 22)

Solution:
Given: $I = 2\text{A}$, $t = 20 \times 60 \text{ s}$
$m = Z I t = \frac{108}{96500} \times 2 \times 20 \times 60 = \frac{108 \times 24}{965} = 2.68 \text{ g}$

3. A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the cross sectional area is 4.5 sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as $15\ \Omega$. Find the specific conductance of the solution. (Mar 20)

Solution:
Given: $l = 1.5 \text{ cm} = 1.5 \times 10^{-2} \text{ m}$, $A = 4.5 \text{ cm}^2 = 4.5 \times 10^{-4} \text{ m}^2$, $R = 15\ \Omega$
Specific conductance $\kappa = \frac{1}{R} \times \frac{l}{A}$
$\kappa = \frac{1}{15} \times \frac{1.5 \times 10^{-2}}{4.5 \times 10^{-4}} = 22.2 \text{ S m}^{-1}$

4. Can $Fe^{3+}$ oxidises bromide to bromine under standard conditions? Given: $E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771 \text{ V}$; $E^\circ_{Br_2/Br^-} = 1.09 \text{ V}$. (Mar 24)

Solution:
Anode (Oxidation): $2Br^- \rightarrow Br_2 + 2e^-$    ($E^\circ_{ox} = -1.09 \text{ V}$)
Cathode (Reduction): $2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$    ($E^\circ_{red} = +0.771 \text{ V}$)
Overall: $2Fe^{3+} + 2Br^- \rightarrow 2Fe^{2+} + Br_2$
$E^\circ_{cell} = E^\circ_{ox} + E^\circ_{red} = -1.09 + 0.771 = -0.319 \text{ V}$
Since $E^\circ_{cell}$ is negative, $\Delta G^\circ$ is positive and the cell reaction is non-spontaneous. Hence, $Fe^{3+}$ cannot oxidise $Br^-$ to $Br_2$.

5. Reduction potential of two metals $M_1$ and $M_2$ are $E^\circ(M_1^{2+}/M_1) = -2.35 \text{ V}$ and $E^\circ(M_2^{2+}/M_2) = 0.2 \text{ V}$. Predict which one is better for coating the surface of iron. Given: $E^\circ(Fe^{2+}/Fe) = -0.44 \text{ V}$. (Jun 23)

Solution:
Oxidation potential of $M_1$ ($+2.35 \text{ V}$) is more positive than the oxidation potential of Fe ($+0.44 \text{ V}$). This indicates $M_1$ acts as a sacrificial anode and will prevent iron from rusting. Thus, $M_1$ is better for coating.

11. Hydroxy compounds and Ethers

1. Why the C-O-C bond angle is slightly greater than the tetrahedral bond angle.

Answer:
➤ Due to the repulsive interaction between the two bulkier alkyl groups.

2. Compound (A) of molecular formula $C_6H_6O$ gives purple colouration with neutral $FeCl_3$. Compound (A) reacts with ammonia to give Compound (B) and it also reacts with Zn dust to give Compound (C). Identify the Compounds A, B, C and write down the equations.

Answer:
$C_6H_5OH + NH_3 \xrightarrow{\text{Anhydrous } ZnCl_2} C_6H_5NH_2 + H_2O$
(A) $\rightarrow$ (B)

$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$
(A) $\rightarrow$ (C)

Result:
(A) $C_6H_5OH$ - Phenol
(B) $C_6H_5NH_2$ - Aniline
(C) $C_6H_6$ - Benzene

3. Find the products X and Y in the following reactions.
Phenol $\xrightarrow{Zn, \Delta}$ (X)
Phenol $\xrightarrow{NH_3, \Delta, \text{Anhydrous } ZnCl_2}$ (Y)

Answer:
$C_6H_5OH \xrightarrow{Zn, \Delta} C_6H_6 + ZnO$
(X) = $C_6H_6$ Benzene

$C_6H_5OH \xrightarrow{NH_3, \Delta, \text{Anhydrous } ZnCl_2} C_6H_5NH_2 + H_2O$
(Y) = $C_6H_5NH_2$ Aniline

4. How is phenol prepared from chlorobenzene?

Answer:
$C_6H_5-Cl \text{ (chlorobenzene)} + NaOH \xrightarrow{623K, 300\text{ bar}} C_6H_5ONa \xrightarrow{HCl} C_6H_5OH \text{ (phenol)} + NaCl$

5. $C_2H_6O_2$ (A) $\xrightarrow{\text{(i) Anhydrous } ZnCl_2 \text{ (ii) Tautomerisation}}$ (B) $\xrightarrow{\text{Zn-Hg/Conc HCl}}$ (C). Identify A, B, and C in reactions.

Answer:
$HO-CH_2-CH_2-OH \text{ (A)} \xrightarrow{\text{(i) Anhydrous } ZnCl_2 \text{ (ii) Tautomerisation}} CH_3-CHO \text{ (B)}$
$CH_3-CHO \text{ (B)} \xrightarrow{\text{Zn-Hg/Conc HCl (Clemmensen Reduction)}} CH_3-CH_3 \text{ (C)}$

Result:
(A) $HO-CH_2-CH_2-OH$ - Ethylene glycol
(B) $CH_3-CHO$ - Acetaldehyde (Ethanal)
(C) $CH_3-CH_3$ - Ethane

11. Hydroxy compounds and Ethers (Part 2)

1. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5-OH \xrightarrow{\text{Zn dust}} A \xrightarrow[\text{anhydrous } AlCl_3]{CH_3Cl} B \xrightarrow{H^+/KMnO_4} C$

Answer:
$C_6H_5-OH \xrightarrow{\text{Zn dust}} C_6H_6 \text{ (A)} \xrightarrow[\text{anhydrous } AlCl_3]{CH_3Cl} C_6H_5-CH_3 \text{ (B)} \xrightarrow{H^+/KMnO_4} C_6H_5-COOH \text{ (C)}$

Result:
(A) $C_6H_6$ - Benzene
(B) $C_6H_5-CH_3$ - Toluene
(C) $C_6H_5-COOH$ - Benzoic acid

2. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5-OH \xrightarrow{PBr_3} A \xrightarrow{\text{Aq.NaOH}} B \rightarrow C$

Answer:
$CH_3CH_2-OH \xrightarrow{PBr_3} CH_3CH_2Br \text{ (A)} \xrightarrow{\text{Aq.NaOH}} CH_3CH_2-OH \text{ (B)} \rightarrow CH_3CH_2-ONa \text{ (C)}$

Result:
(A) $CH_3CH_2-Br$ - Ethyl bromide (Bromo ethane)
(B) $CH_3CH_2-OH$ - Ethyl alcohol (Ethanol)
(C) $CH_3CH_2-ONa$ - Sodium ethoxide

3. Phenol is distilled with Zn dust followed by Friedel - Crafts alkylation with propyl chloride to give a compound A, A on oxidation gives (B) Identify A and B.

Answer:
$C_6H_5-OH \text{ (phenol)} \xrightarrow{\text{Zn dust}} C_6H_6 \text{ (A)} \xrightarrow[\text{anhydrous } AlCl_3]{CH_3CH_2CH_2Cl} C_6H_5CH_2CH_2CH_3 \text{ (B)} \xrightarrow{H^+/KMnO_4} C_6H_5-COOH \text{ (C)}$

Result:
(A) $C_6H_6$ - Benzene
(B) $C_6H_5CH_2CH_2CH_3$ - n-propyl benzene
(C) $C_6H_5-COOH$ - Benzoic acid

4. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5-OH \xrightarrow{\text{NaOH}} A \xrightarrow[\text{ii) 4-7bar}]{\text{i) } CO_2, 400K} B \xrightarrow{H^+/H_2O} C$

Answer:
Phenol + $NaOH \rightarrow$ Sodium phenoxide (A)
Sodium phenoxide (A) + $CO_2 \xrightarrow{400K, 4-7\text{bar}}$ Sodium salicylate (B)
Sodium salicylate (B) $\xrightarrow{H^+/H_2O}$ Salicylic acid (C)

Result:
(A) Sodium phenoxide
(B) Sodium salicylate
(C) Salicylic acid

13. Organic Nitrogen Compounds

1. There are two isomers with the formula $CH_3NO_2$. How will you distinguish between them? (Jun 24, Mar 25)

Answer:

Nitro form	Aci form
1. Less acidic	1. More acidic
2. Dissolve in NaOH slowly	2. Dissolve in NaOH instantly
3. Decolourises $FeCl_3$ solution	3. With $FeCl_3$ gives reddish brown colour
4. Electrical conductivity is low	4. Electrical conductivity is high

2. Identify the compounds A, B, and C in the following sequence of reaction. (Jun 23)
$C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{HNO_2/273K} B \xrightarrow{C_6H_5OH} C$

Answer:
$C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (A)} + 2H_2O$
$C_6H_5NH_2 \xrightarrow{HNO_2/273K} C_6H_5N_2Cl \text{ (B)}$
$C_6H_5N_2Cl + C_6H_5OH \rightarrow p\text{-Hydroxy azobenzene (C)}$

Result:
(A) $C_6H_5NH_2$ - Aniline
(B) $C_6H_5N_2Cl$ - Benzene diazonium chloride
(C) $C_6H_5-N=N-C_6H_4OH$ - p-Hydroxy azobenzene

3. How is aryl halide prepared by using $Cu_2Cl_2/HCl$ (or) $Cu_2Br_2/HBr$? (or) (Write a note on Sandmeyer reaction.) (Sep 20, Mar 20)

Answer:
$C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2$

4. Identify the compounds A, and B in the following sequence of reactions. (May 22)
$CH_3-Br \xrightarrow{NaN_3} A \xrightarrow{LiAlH_4} B$

Answer:
$CH_3-Br \xrightarrow{NaN_3} CH_3N_3 \text{ (A)} \xrightarrow{LiAlH_4} CH_3-NH_2 \text{ (B)} + N_2$

Result:
(A) $CH_3N_3$ - Methyl azide
(B) $CH_3-NH_2$ - Methyl amine

5. Identify the compounds A, and B in the following sequence of reactions. (May 22)
$CH_3CH_2-NO_2 \xrightarrow{Sn/HCl} A \xrightarrow{CH_3COCl} B$

Answer:
$CH_3CH_2-NO_2 \xrightarrow{Sn/HCl} CH_3CH_2NH_2 \text{ (A)}$
$CH_3CH_2NH_2 \xrightarrow{CH_3COCl} CH_3CH_2-NH-COCH_3 \text{ (B)}$

Result:
(A) $CH_3CH_2NH_2$ - Ethyl amine
(B) $CH_3CH_2-NH-COCH_3$ - N-Ethyl acetamide

6. From the following reaction, Identify A and B. (Sep 21)
$CH_3-NO_2 \xrightarrow{Sn/HCl, 6(H)} A$
$CH_3-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} B$

Answer:
$CH_3-NO_2 \xrightarrow{Sn/HCl, 6(H)} CH_3-NH_2 \text{ (A)}$
$CH_3-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} CH_3-NHOH \text{ (B)}$

Result:
(A) $CH_3-NH_2$ - Methyl amine
(B) $CH_3-NHOH$ - Methyl hydroxylamine

7. Identify the compounds A, B, and C in the following reaction. (Jun 22)
$C_6H_5-NO_2 \xrightarrow{Sn/HCl} A$
$C_6H_5-NO_2 \xrightarrow{Zn/NH_4Cl} B$
$C_6H_5-NO_2 \xrightarrow{Zn/NaOH} C$

Answer:
$C_6H_5-NO_2 \xrightarrow{Sn/HCl, 6(H)} C_6H_5-NH_2 \text{ (A)}$
$C_6H_5-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} C_6H_5-NHOH \text{ (B)}$
$C_6H_5-NO_2 \xrightarrow{Zn/NaOH} C_6H_5-NH-NH-C_6H_5 \text{ (C)}$

Result:
(A) $C_6H_5-NH_2$ - Aniline
(B) $C_6H_5-NHOH$ - Phenyl hydroxylamine
(C) $C_6H_5-NH-NH-C_6H_5$ - Hydrazobenzene

8. Identify A and B. (Mar 20)
A $\xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-CH_2-NH_2$
B $\xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-NH-CH_3$

Answer:
$CH_3CN \text{ (A)} \xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-CH_2-NH_2$
$CH_3NC \text{ (B)} \xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-NH-CH_3$

Result:
(A) $CH_3CN$ - Methyl cyanide
(B) $CH_3NC$ - Methyl isocyanide

9. Identify A and B. (Sep 20)
Dicarboxylic acid $\xrightarrow{SOCl_2}$ A $\xrightarrow{+2NH_3}$ B

Answer:
$HOOC-(CH_2)_n-COOH \xrightarrow{SOCl_2} ClOC-(CH_2)_n-COCl \text{ (A)} + 2NH_3 \rightarrow H_2NOC-(CH_2)_n-CONH_2 \text{ (B)}$

Result:
(A) Acid Chloride derivative
(B) Amide derivative

10. Compound 'A' of molecular $C_2H_4O$ reacts with $Na(Hg)/C_2H_5OH$ gives compound 'B' of molecular formula $C_2H_7N$ which undergo carbylamines test. Compound 'B' on reduction with nitrous acid gives compound 'C' of molecular formula $C_2H_6O$ by liberating nitrogen. Identify A, B and C and write the reactions involved. (Aug 21)

Answer:
$CH_3CN \text{ (A)} \xrightarrow{Na/Hg, C_2H_5OH} CH_3-CH_2-NH_2 \text{ (B)}$
$CH_3-CH_2-NH_2 \text{ (B)} + HNO_2 \rightarrow CH_3-CH_2-OH \text{ (C)}$

Result:
(A) $CH_3-CN$ - Methyl cyanide (Note: Formula in question is a typo for $C_2H_3N$)
(B) $CH_3-CH_2-NH_2$ - Ethyl amine
(C) $CH_3-CH_2-OH$ - Ethyl alcohol (Ethanol)

11. An organic compound (A) of molecular $C_2H_3N$ on reacts with Zn - (Hg)/Conc.HCl to give compound (B) which reacts with $HNO_3$ forming compound C (as major product) and compound (D). Compound C reacts with conc.HCl to give compound (E) (Table vinegar) and hydroxylamine. Identify A, B, C, D and E with suitable reactions. (Mar 24)

Answer:
$CH_3-CHO \text{ (A)} + 4(H) \xrightarrow{Zn-Hg/\text{conc.HCl}} CH_3-CH_3 \text{ (B)}$
$CH_3-CH_3 \text{ (B)} + HNO_3 \xrightarrow{675K} CH_3-CH_2-NO_2 \text{ (C)} + CH_3-NO_2 \text{ (D)}$
$CH_3-CH_2-NO_2 \text{ (C)} \xrightarrow{HCl/H_2O, \text{Boil}} CH_3-COOH \text{ (E)} + NH_2-OH$

Result:
(A) $CH_3-CHO$ - Acetaldehyde (Note: Formula in question is a typo for $C_2H_4O$)
(B) $CH_3-CH_3$ - Ethane
(C) $CH_3-CH_2-NO_2$ - Nitroethane
(D) $CH_3-NO_2$ - Nitromethane
(E) $CH_3-COOH$ - Acetic acid

12. Identify the compounds A, B, and C in the following reaction. (Mar 25)
$C_6H_5NO_2 \xrightarrow{Sn/HCl} A \xrightarrow{NaNO_2/HCl, 0^\circ C} B \xrightarrow{CuCN} C$

Answer:
$C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2 \text{ (A)} \xrightarrow{NaNO_2/HCl, 0^\circ C} C_6H_5N_2Cl \text{ (B)} \xrightarrow{CuCN} C_6H_5CN \text{ (C)}$

Result:
(A) $C_6H_5NH_2$ - Aniline
(B) $C_6H_5N_2Cl$ - Benzene diazonium chloride
(C) $C_6H_5CN$ - Phenyl cyanide (cyanobenzene)

13. Organic Nitrogen Compounds (Part 2)

1. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{HNO_2/273K} B \xrightarrow{H_2O/\Delta} C$

Answer:
$C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (A)} \xrightarrow{HNO_2/273K} C_6H_5N_2Cl \text{ (B)} \xrightarrow{H_2O/\Delta} C_6H_5OH \text{ (C)}$

Result:
(A) $C_6H_5NH_2$ - Aniline
(B) $C_6H_5N_2Cl$ - Benzenediazoniumchloride
(C) $C_6H_5OH$ - Phenol

2. Identify the compounds A and B in the following sequence of reactions.
$C_6H_5N_2Cl \xrightarrow{Cu/HCl} A \xrightarrow{NH_3, Cu_2O/200^\circ C} B$

Answer:
$C_6H_5N_2Cl \xrightarrow{Cu/HCl} C_6H_5Cl \text{ (A)} \xrightarrow{NH_3, Cu_2O/200^\circ C} C_6H_5NH_2 \text{ (B)}$

Result:
(A) $C_6H_5Cl$ - Chlorobenzene
(B) $C_6H_5NH_2$ - Aniline

3. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5N_2Cl \xrightarrow{CuCN} A \xrightarrow{H_2O/H^+} B \xrightarrow{NH_3} C$

Answer:
$C_6H_5N_2Cl \xrightarrow{CuCN} C_6H_5CN \text{ (A)} \xrightarrow{H_2O/H^+} C_6H_5COOH \text{ (B)} \xrightarrow{NH_3} C_6H_5CONH_2 \text{ (C)}$

Result:
(A) $C_6H_5CN$ - Phenylcyanide (Cyanobenzene)
(B) $C_6H_5COOH$ - Benzoic acid
(C) $C_6H_5CONH_2$ - Benzamide

4. Identify the compounds A and B in the following sequence of reactions.
$C_6H_5N_2Cl \xrightarrow{H_2O/\Delta} A \xrightarrow{NH_3, \text{anhydrous } ZnCl_2/300^\circ C} B$

Answer:
$C_6H_5N_2Cl \xrightarrow{H_2O/\Delta} C_6H_5OH \text{ (A)} \xrightarrow{NH_3, \text{anhydrous } ZnCl_2/300^\circ C} C_6H_5NH_2 \text{ (B)}$

Result:
(A) $C_6H_5OH$ - Phenol
(B) $C_6H_5NH_2$ - Aniline

5. Identify the compounds A, B and C in the following sequence of reactions.
$CH_3-NO_2 \xrightarrow{LiAlH_4} A \xrightarrow{2CH_3CH_2Br} B \xrightarrow{H_2SO_4} C$

Answer:
$CH_3-NO_2 \xrightarrow{LiAlH_4} CH_3NH_2 \text{ (A)} \xrightarrow{2CH_3CH_2Br} CH_3-N(CH_2CH_3)_2 \text{ (B)} \xrightarrow{H_2SO_4} [CH_3-NH(CH_2CH_3)_2]^+HSO_4^- \text{ (C)}$

Result:
(A) Methylamine (Methanamine)
(B) N-Ethyl-N-methylethane-1-amine
(C) Ethyl methyl ammonium hydrogen sulphate

6. Identify the compounds A, B and C in the following sequence of reactions.
Butanoic acid $\xrightarrow{liq.NH_3} \dots \xrightarrow{Br_2/KOH} A \xrightarrow{NaNO_2/HCl, 0^\circ C} B \xrightarrow{H^+/KMnO_4} C$

Answer:
$CH_3CH_2CH_2COOH \xrightarrow{liq.NH_3} CH_3CH_2CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2CH_2NH_2 \text{ (A)}$
$CH_3CH_2CH_2NH_2 \text{ (A)} \xrightarrow{NaNO_2/HCl, 0^\circ C} CH_3CH_2CH_2OH \text{ (B)}$
$CH_3CH_2CH_2OH \text{ (B)} \xrightarrow{H^+/KMnO_4} CH_3CH_2COOH \text{ (C)}$

Result:
(A) $CH_3CH_2CH_2NH_2$ - Propanamine
(B) $CH_3CH_2CH_2OH$ - Propanol
(C) $CH_3CH_2COOH$ - Propanoic acid

13. Organic Nitrogen Compounds

1. There are two isomers with the formula $CH_3NO_2$. How will you distinguish between them? (Jun 24, Mar 25)

Answer:

Nitro form ($CH_3NO_2$)	Aci form ($CH_2=N^+(O^-)OH$)
1. Less acidic	1. More acidic
2. Dissolve in NaOH slowly	2. Dissolve in NaOH instantly
3. Decolourises $FeCl_3$ solution	3. With $FeCl_3$ gives reddish brown colour
4. Electrical conductivity is low	4. Electrical conductivity is high

3. How is aryl halide prepared by using $Cu_2Cl_2/HCl$ (or) $Cu_2Br_2/HBr$? (or) (Write a note on Sandmeyer reaction.) (Sep 20, Mar 20)

Answer:
Benzene diazonium chloride reacts with cuprous chloride in HCl to form chlorobenzene.
$C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2 \uparrow$

Additional Important Problems (Mixed)

1. A double salt which contains fourth period alkali metal (A) on heating at 500K gives (B). Aqueous solution of (B) gives white precipitate with $BaCl_2$ and gives a red colour compound with alizarin. Identify A and B.

Answer:
$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O \xrightarrow{500 \text{ K}} K_2SO_4 \cdot Al_2(SO_4)_3 + 24H_2O$
(A) $\rightarrow$ (B)

Result:
(A) Potash alum
(B) Burnt alum

2. Barium has a body centered cubic unit cell with a length of 508pm along an edge. What is the density of barium in $\text{g cm}^{-3}$?

Solution:

Given: For BCC, $n = 2$. $M = 137.3 \text{ g mol}^{-1}$. $a = 508 \text{ pm} = 5.08 \times 10^{-8} \text{ cm}$.
$\rho = \frac{n \times M}{a^3 \times N_A}$
$\rho = \frac{2 \times 137.3}{(5.08 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$
$\rho \approx 3.5 \text{ g cm}^{-3}$

HSC Chemistry 2021 Board Question Paper with Complete Solutions (Code J-565)

Chemistry (55) Board Question Paper Solutions

Year: 2021 | Code: J-565 | Max Marks: 70 | Time: 3 Hrs

General Instructions:

1. The question paper is divided into four sections.
2. Section A: Q. No. 1 contains Ten multiple choice type of questions carrying One mark each. Q. No. 2 contains Eight very short answer type of questions carrying One mark each.
3. Section B: Q. No. 3 to Q. No. 14 are Twelve short answer type of questions carrying Two marks each. (Attempt any Eight)
4. Section C: Q. No. 15 to Q. No. 26 are Twelve short answer type of questions carrying Three marks each. (Attempt any Eight)
5. Section D: Q. No. 27 to Q. No. 31 are Five long answer type of questions carrying Four marks each. (Attempt any Three)
6. Use of log table is allowed. Use of calculator is not allowed.
7. Figures to the right indicate full marks.
8. For each multiple choice type of question, it is mandatory to write the correct answer along with its alphabet e.g. (a)......../ (b)......../ (c)......../ (d)........etc. No mark (s) shall be given, if ONLY the correct answer or the alphabet of the correct answer is written.
9. Only the first attempt will be considered for evaluation.

SECTION - A

Q. 1. (i)

The product obtained in the following reaction
$$ \text{CH}_3-\text{CH}=\text{CH}-\text{CH}_2-\text{CHO} \xrightarrow{\text{H}_2/\text{Ni}} ? \text{ is,} $$

• (a) Pent–3–en–1–ol
• (b) Pentan–1–ol
• (c) Pentan–2–ol
• (d) Pentanal

Solution: (b) Pentan–1–ol

Reason: $H_2/Ni$ is a strong reducing agent. It reduces both the carbon-carbon double bond ($C=C$) and the aldehyde group ($-CHO$) to a primary alcohol. The 5-carbon chain remains intact.
$$ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \text{ (Pentan-1-ol)} $$

Q. 1. (ii)

Amongst the following, the solubility of which ionic solid decreases with increase in temperature?

• (a) KNO₃
• (b) NaBr
• (c) Na₂SO₄
• (d) KCl

Solution: (c) Na₂SO₄

Reason: While most salts (like KNO₃, NaBr, KCl) show increased solubility with temperature (endothermic dissolution), sodium sulphate (Na₂SO₄) shows anomalous behavior. Its solubility increases up to 32.4°C and then decreases as the temperature rises further.

Q. 1. (iii)

The correct IUPAC name of Na₃ [AlF₆] is

• (a) Sodium hexafluoroaluminate (III)
• (b) Sodium hexafluoroaluminate (II)
• (c) Sodium hexafluoroaluminium (III)
• (d) Sodium hexafluoroaluminium (II)

Solution: (a) Sodium hexafluoroaluminate (III)

Reason: The cation is Sodium. The anion is a complex ion $[AlF_6]^{3-}$. Since the complex is anionic, the metal ends with suffix '-ate' (Aluminate). The oxidation state of Al is: $x + 6(-1) = -3 \Rightarrow x = +3$.

Q. 1. (iv)

Which of the following acids has highest pKa value?

• (a) Mono chloroacetic acid
• (b) Dichloroacetic acid
• (c) Trichloroacetic acid
• (d) Acetic acid

Solution: (d) Acetic acid

Reason: Higher pKa means weaker acid. Electron withdrawing groups (like Cl) increase acidity (lower pKa). Acetic acid has no electron withdrawing Chlorine atoms, making it the weakest acid among the options, hence it has the highest pKa.

Q. 1. (v)

Number of carbon atoms present in isoprene unit is ____.

• (a) 6
• (b) 5
• (c) 4
• (d) 3

Solution: (b) 5

Reason: Isoprene is 2-methyl-1,3-butadiene ($C_5H_8$). It contains 5 carbon atoms.

Q. 1. (vi)

The colourless transition metal ion amongst the following is _______.

• (a) Cu⁺
• (b) Cu⁺⁺
• (c) Ni⁺⁺
• (d) Co⁺⁺

Solution: (a) Cu⁺

Reason: A transition metal ion is colourless if it has a completely filled ($d^{10}$) or completely empty ($d^0$) d-orbital.
$Cu^+$ ($Z=29$) config: $[Ar] 3d^{10}$. Since d-orbital is full, d-d transition is not possible.

Q. 1. (vii)

[Image Description: Chlorocyclohexane + Mg (dry ether) → A (H₂O) → B]
The product 'B' in the above reaction sequence is

• (a) Cyclohexane (Ring)
• (b) Chlorocyclohexyl Magnesium (Ring-Mg-Cl)
• (c) Cyclohexyl Magnesium (Ring-Mg)
• (d) Cyclohexane Ring

Solution: (a) Cyclohexane

Reason:
Step 1: Chlorocyclohexane reacts with Mg in dry ether to form Cyclohexylmagnesium chloride (Grignard reagent 'A').
Step 2: Grignard reagent ('A') reacts with H₂O to undergo hydrolysis, replacing -MgCl with -H.
Product B is Cyclohexane.

Q. 1. (viii)

Carbylamine test is given by

• (a) aniline
• (b) dimethylamine
• (c) trimethylamine
• (d) both dimethylamine and trimethylamine

Solution: (a) aniline

Reason: Carbylamine test is given only by primary amines (aliphatic or aromatic). Aniline is a primary aromatic amine. Dimethylamine is secondary, and Trimethylamine is tertiary.

Q. 1. (ix)

A weak monobasic acid is 0.05% dissociated in 0.02 M solution, dissociation constant of the acid is ______.

• (a) \(5 \times 10^{-10}\)
• (b) \(5 \times 10^{-9}\)
• (c) \(50 \times 10^{-9}\)
• (d) \(0.5 \times 10^{-9}\)

Solution: (b) \(5 \times 10^{-9}\)

Calculation:
Degree of dissociation, $\alpha = 0.05\% = 0.0005 = 5 \times 10^{-4}$.
Concentration, $C = 0.02 = 2 \times 10^{-2}$ M.
$$ K_a = C\alpha^2 = (2 \times 10^{-2}) \times (5 \times 10^{-4})^2 $$ $$ K_a = (2 \times 10^{-2}) \times (25 \times 10^{-8}) $$ $$ K_a = 50 \times 10^{-10} = 5 \times 10^{-9} $$

Q. 1. (x)

The rate constant for the reaction
$$ 2N_2O_{5(g)} \rightarrow 2N_2O_{4(g)} + O_{2(g)} \text{ is } 4.98 \times 10^{-4} s^{-1}. $$ The order of reaction is _______

• (a) 0
• (b) 1
• (c) 2
• (d) 3

Solution: (b) 1

Reason: The unit of the rate constant ($k$) is $s^{-1}$ (time^-1). This is the characteristic unit for a First Order reaction.

HSC Chemistry

• Chemistry - March 2025 - English Medium View Answer Key
• Chemistry - March 2025 - Marathi Medium View Answer Key
• Chemistry - March 2025 - Hindi Medium View Answer Key
• Chemistry - March 2024 - English Medium View Answer Key
• Chemistry - March 2024 - Marathi Medium View Answer Key
• Chemistry - March 2024 - Hindi Medium View Answer Key
• Chemistry - March 2023 - English Medium View Answer Key
• Chemistry - July 2023 - English Medium View Answer Key
• Chemistry - March 2022 - English Medium View Answer Key
• Chemistry - July 2022 - English Medium View Answer Key
• Chemistry - March 2021 - English Medium View Answer Key
• Chemistry - March 2013 View
• Chemistry - October 2013 View
• Chemistry - March 2014 View
• Chemistry - October 2014 View
• Chemistry - March 2015 View
• Chemistry - July 2015 View
• Chemistry - March 2016 View
• Chemistry - July 2016 View
• Chemistry - March 2017 View
• Chemistry - July 2017 View
• Chemistry - March 2019 View

Q. 2. Answer the following questions:

(i) Write the name of interhalogen compound of chlorine which has square pyramidal structure.

Answer: Chlorine pentafluoride ($ClF_5$).

(ii) Write the name of sugar present in RNA.

Answer: D-Ribose (or simply Ribose).

(iii) Write the value of \( \frac{2.303 RT}{F} \) in Nernst equation.

Answer: At 298 K, the value is approximately 0.0592 V (or 0.059 V).

(iv) What is the coordination number of atoms in simple cubic crystal lattice?

Answer: The coordination number is 6.

(v) Write the name of nanostructural material used in tyres to increase their life.

Answer: Carbon black.

(vi) Write the name of reagent used during conversion of acetaldehyde to acetaldehyde cyanohydrin.

Answer: Hydrogen Cyanide (HCN) in the presence of a base or Sodium Cyanide (NaCN) followed by acid.

(vii) Write the chemical formula of Haematite.

Answer: Fe₂O₃

(viii) In a particular reaction, 2kJ of heat is released by the system and 6 kJ of work is done on the system. Calculate \(\Delta U\).

Answer: Given:
Heat released ($q$) = $-2 \text{ kJ}$ (negative because heat is lost).
Work done on the system ($w$) = $+6 \text{ kJ}$ (positive because work is done on system).
According to the first law of thermodynamics: $\Delta U = q + w$
$\Delta U = -2 \text{ kJ} + 6 \text{ kJ}$
$\Delta U = +4 \text{ kJ}$

SECTION - B

(Attempt any EIGHT of the following questions)

Q. 3. What are bidentate Ligands? Give one example.

Solution:

Definition: The ligands which bind to the central metal atom or ion through two donor atoms are called bidentate ligands.

Example: Ethylenediamine (en) ($H_2N-CH_2-CH_2-NH_2$) or Oxalate ion ($C_2O_4^{2-}$).

Q. 4. Draw the structure of sulphurous acid. Write two uses of helium.

Solution:

Structure of Sulphurous Acid ($H_2SO_3$):
It contains one S=O double bond, two S-OH groups, and one lone pair on Sulfur.

Uses of Helium:

1. It is used in filling observation balloons and weather balloons due to its lightness and non-inflammability.
2. A mixture of He and O₂ is used for respiration by deep-sea divers (to prevent 'bends').

Q. 5. The molar conductivity of 0.01M acetic acid at 25°C is \(18 \Omega^{-1} cm^2 mol^{-1}\). Calculate its degree of dissociation in 0.01M solution and dissociation constant, if molar conductivity of acetic acid at zero concentration is \(400 \Omega^{-1} cm^2 mol^{-1}\).

Solution:

Given:
Concentration $C = 0.01 \text{ M}$
$\Lambda_m = 18 \Omega^{-1} cm^2 mol^{-1}$
$\Lambda_0 = 400 \Omega^{-1} cm^2 mol^{-1}$

1. Degree of dissociation ($\alpha$):
$$ \alpha = \frac{\Lambda_m}{\Lambda_0} = \frac{18}{400} = 0.045 $$

2. Dissociation constant ($K_a$):
$$ K_a = \frac{C\alpha^2}{1-\alpha} $$ Since $\alpha$ is small ($0.045$), $1-\alpha \approx 1$.
$$ K_a \approx C\alpha^2 = 0.01 \times (0.045)^2 $$ $$ K_a = 0.01 \times 0.002025 $$ $$ K_a = 2.025 \times 10^{-5} $$

Q. 6. Write classification of proteins on the basis of molecular shapes with example.

Solution:

Proteins are classified into two types based on molecular shape:

1. Fibrous Proteins: The polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. They are fibre-like, insoluble in water.
   Example: Keratin (hair, wool), Myosin (muscles).
2. Globular Proteins: The chains of polypeptides coil around to give a spherical shape. They are usually soluble in water.
   Example: Insulin, Albumin.

Q. 7. What is pseudo-first order reaction? Explain with suitable example.

Solution:

Definition: A reaction which has higher order true rate law but experimentally follows first order kinetics because one of the reactants is present in large excess is called a pseudo-first order reaction.

Example: Hydrolysis of methyl acetate in acidic medium.
$$ CH_3COOCH_3 + H_2O \xrightarrow{H^+} CH_3COOH + CH_3OH $$ Here, water is present in excess, so its concentration remains practically constant. Rate depends only on [Methyl acetate].

Q. 8. What is the molar mass of a solute if a solution prepared by dissolving 0.822 g of it in 0.3 dm³ of water has an osmotic pressure of 0.196 atm. at 298 K?

Solution:

Given:
Mass of solute ($W_2$) = 0.822 g
Volume ($V$) = 0.3 dm³ = 0.3 L
Osmotic Pressure ($\pi$) = 0.196 atm
Temperature ($T$) = 298 K
Gas Constant ($R$) = 0.0821 L atm K⁻¹ mol⁻¹

Formula: $M_2 = \frac{W_2 R T}{\pi V}$

Calculation:
$$ M_2 = \frac{0.822 \times 0.0821 \times 298}{0.196 \times 0.3} $$ $$ M_2 = \frac{20.108}{0.0588} $$ $$ M_2 \approx 342 \text{ g/mol} $$ (This corresponds to Sucrose).

Q. 9. Write a note on Kolbe reaction.

Solution:

Kolbe Reaction: It is the reaction used to prepare Salicylic acid from phenol.

When sodium phenoxide is treated with carbon dioxide ($CO_2$) at 398 K under a pressure of 6 atm followed by acid hydrolysis, salicylic acid (2-hydroxybenzoic acid) is formed.

Sodium Phenoxide + CO₂ → Sodium Salicylate →(H⁺) Salicylic Acid

Q. 10. Iron exhibits +2 and +3 oxidation states. Write their electronic configuration. Which will be more stable? Why?

Solution:

Iron (Fe), Z = 26. Configuration: $[Ar] 3d^6 4s^2$.

Electronic Configurations:
$Fe^{2+}$: $[Ar] 3d^6$
$Fe^{3+}$: $[Ar] 3d^5$

Stability:
$Fe^{3+}$ is more stable than $Fe^{2+}$.
Reason: $Fe^{3+}$ has a half-filled d-orbital ($3d^5$), which confers extra stability according to Hund's rule compared to the partially filled $3d^6$ of $Fe^{2+}$.

Q. 11. How is benzophenone prepared from benzonitrile?

Solution:

Benzophenone is prepared by treating Benzonitrile with Phenylmagnesium bromide (Grignard reagent) followed by acid hydrolysis.

$C_6H_5-CN + C_6H_5MgBr \xrightarrow{\text{dry ether}} C_6H_5-C(NMgBr)-C_6H_5$ (Complex)
Complex $+ H_3O^+ \rightarrow C_6H_5-CO-C_6H_5$ (Benzophenone) $+ NH_3 + Mg(OH)Br$

Q. 12. Write names and structure of monomers used in the preparation of Nylon 6, 6 polymer.

Solution:

The monomers are:

1. Adipic Acid (Hexanedioic acid):
   $HOOC-(CH_2)_4-COOH$
2. Hexamethylenediamine (Hexane-1,6-diamine):
   $H_2N-(CH_2)_6-NH_2$

Q. 13. Derive the relationship between pH and pOH.

Solution:

The ionic product of water ($K_w$) is given by:
$$ K_w = [H^+][OH^-] $$ Taking negative logarithm to the base 10 on both sides:
$$ -\log_{10} K_w = -\log_{10} ([H^+][OH^-]) $$ $$ -\log_{10} K_w = (-\log_{10} [H^+]) + (-\log_{10} [OH^-]) $$ By definition, $-\log_{10} K_w = pK_w$, $-\log_{10} [H^+] = pH$, and $-\log_{10} [OH^-] = pOH$.
$$ pK_w = pH + pOH $$ At 298 K, $K_w = 1 \times 10^{-14}$, so $pK_w = 14$.
Therefore, $pH + pOH = 14$.

Q. 14. What is action of the following on chlorobenzene?

• (i) Methyl chloride in presence of anhydrous AlCl₃
• (ii) Fuming H₂SO₄

Solution:

(i) Friedel-Crafts Alkylation:
Chlorobenzene reacts with methyl chloride in the presence of anhydrous $AlCl_3$ to give a mixture of 2-chlorotoluene (minor) and 4-chlorotoluene (major).

(ii) Sulphonation:
Chlorobenzene reacts with fuming sulphuric acid to give 2-chlorobenzenesulphonic acid (minor) and 4-chlorobenzenesulphonic acid (major).

SECTION - C

(Attempt any EIGHT of the following questions)

Q. 15. Calculate the standard enthalpy of \( N_2H_{4(g)} + H_{2(g)} \rightarrow 2NH_{3(g)} \)

if \(\Delta H^\circ(N-H) = 389 \text{ kJ mol}^{-1}\)
\(\Delta H^\circ(H-H) = 435 \text{ kJ mol}^{-1}\)
\(\Delta H^\circ(N-N) = 159 \text{ kJ mol}^{-1}\)

Solution:

Reaction structure:
$H_2N-NH_2 (g) + H-H (g) \rightarrow 2(NH_3)$

Bonds Broken (Reactants):
1 N-N bond
4 N-H bonds (in hydrazine)
1 H-H bond

Bonds Formed (Products):
2 moles of $NH_3$ (each has 3 N-H bonds) = 6 N-H bonds.

$\Delta_r H^\circ = \Sigma \text{Bond Enthalpy (Reactants)} - \Sigma \text{Bond Enthalpy (Products)}$
$= [1(N-N) + 4(N-H) + 1(H-H)] - [6(N-H)]$

Simplifying (canceling 4 N-H):
$= [1(N-N) + 1(H-H)] - [2(N-H)]$

Substituting values:
$= [159 + 435] - [2 \times 389]$
$= 594 - 778$
$= \mathbf{-184 \text{ kJ}}$

Q. 16. Write reactions to prepare ethanamine from

• (i) acetonitrile
• (ii) nitroethane
• (iii) propionamide

Solution:

(i) From Acetonitrile ($CH_3CN$): Mendius Reduction
$CH_3-C\equiv N + 4[H] \xrightarrow{Na/C_2H_5OH} CH_3-CH_2-NH_2$ (Ethanamine)

(ii) From Nitroethane ($C_2H_5NO_2$):
$C_2H_5-NO_2 + 6[H] \xrightarrow{Sn/conc. HCl} C_2H_5-NH_2 + 2H_2O$

(iii) From Propionamide ($C_2H_5CONH_2$): Hoffmann Bromamide Degradation
$C_2H_5-CO-NH_2 + Br_2 + 4KOH \rightarrow C_2H_5-NH_2 + K_2CO_3 + 2KBr + 2H_2O$
(This reaction reduces the carbon chain by one carbon).

Q. 17. Explain three principles of green chemistry.

Solution:

1. Prevention of Waste or By-products: It is better to prevent waste than to treat or clean up waste after it is formed.
2. Atom Economy: Synthetic methods should be designed to maximize the incorporation of all materials used in the process into the final product.
3. Less Hazardous Chemical Synthesis: Wherever practicable, synthetic methods should be designed to use and generate substances that possess little or no toxicity to human health and the environment.

Q. 18. Write chemical equations involved during manufacture of sulphuric acid by contact process. Write two uses of sulphur dioxide.

Solution:

Manufacture of $H_2SO_4$ (Contact Process):

1. Burning of Sulphur to form $SO_2$:
   $S + O_2 \rightarrow SO_2$
2. Catalytic oxidation of $SO_2$ to $SO_3$ ($V_2O_5$ catalyst):
   $2SO_2 + O_2 \rightleftharpoons 2SO_3$
3. Absorption of $SO_3$ in conc. $H_2SO_4$ to form Oleum:
   $SO_3 + H_2SO_4 \rightarrow H_2S_2O_7$ (Oleum)
4. Dilution of Oleum with water:
   $H_2S_2O_7 + H_2O \rightarrow 2H_2SO_4$

Uses of $SO_2$:

• It is used in refining petroleum and sugar.
• It is used as a bleaching agent for wool and silk.

Q. 19. Explain SN² reaction mechanism for alkaline hydrolysis of bromomethane.

Solution:

Reaction: $CH_3Br + OH^- \rightarrow CH_3OH + Br^-$

Mechanism (SN² - Substitution Nucleophilic Bimolecular):

• It is a single-step concerted mechanism.
• Backside Attack: The nucleophile ($OH^-$) attacks the carbon atom from the side opposite to the leaving group ($Br$).
• Transition State: A transition state is formed where the C-OH bond is partially formed and the C-Br bond is partially broken. Carbon is pentacoordinate here.
• Inversion of Configuration: As the leaving group departs, the configuration of the carbon atom inverts (Walden Inversion), like an umbrella turning inside out.

Q. 20. Why La(OH)₃ is the strongest base, while Lu(OH)₃ is the weakest base? Write two applications of catalytic properties of transition metals and compounds.

Solution:

Basicity of Lanthanoids:
Due to Lanthanoid Contraction, the atomic/ionic size decreases from $La^{3+}$ to $Lu^{3+}$. As the size decreases, the covalent character of the M-OH bond increases (and ionic character decreases). Since $La^{3+}$ is the largest, the La-O bond is most ionic and breaks easily to release $OH^-$ ions, making $La(OH)_3$ the strongest base. $Lu(OH)_3$ has the most covalent character, making it the weakest base.

Catalytic Applications:

1. Finely divided Iron (Fe) is used in the Haber process for Ammonia synthesis.
2. Nickel (Ni) is used in the hydrogenation of oils to fats.

Q. 21. Convert the following:

• (i) chlorobenzene to phenol
• (ii) ethanal to ethanol
• (iii) iodomethane to methoxy methane

Solution:

(i) Dow Process:
Chlorobenzene heated with NaOH at 623K and 300 atm followed by acid hydrolysis yields Phenol.
$C_6H_5Cl + NaOH \rightarrow C_6H_5ONa \xrightarrow{H^+} C_6H_5OH$

(ii) Reduction:
Ethanal is reduced using $H_2/Ni$ or $LiAlH_4$.
$CH_3CHO + H_2 \xrightarrow{Ni} CH_3CH_2OH$

(iii) Williamson Synthesis:
Iodomethane reacts with sodium methoxide.
$CH_3I + NaOCH_3 \rightarrow CH_3-O-CH_3 + NaI$

Q. 22. Define Cryoscopic constant. Derive the relation between elevation of boiling point and molar mass of solute.

Solution:

Cryoscopic Constant ($K_f$): It is defined as the depression in freezing point produced when 1 mole of a non-volatile solute is dissolved in 1 kg of solvent.

Relation (Elevation of Boiling Point $\Delta T_b$ and Molar Mass $M_2$):
1. $\Delta T_b \propto m$ (molality).
2. $\Delta T_b = K_b \times m$.
3. Molality $m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{W_2/M_2}{W_1/1000} = \frac{W_2 \times 1000}{M_2 \times W_1}$.
4. Substituting $m$:
$$ \Delta T_b = K_b \times \frac{W_2 \times 1000}{M_2 \times W_1} $$ 5. Rearranging for $M_2$:
$$ M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1} $$

Q. 23. Define solubility product. Derive the relationship between solubility and solubility product for PbI₂.

Solution:

Solubility Product ($K_{sp}$): It is the product of the concentrations of the ions of a sparingly soluble salt in its saturated solution at a given temperature, raised to the power equal to the number of ions produced per formula unit.

Derivation for $PbI_2$:
Equilibrium: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$
Let $s$ be the solubility in mol/L.
$[Pb^{2+}] = s$
$[I^-] = 2s$
$K_{sp} = [Pb^{2+}][I^-]^2$
$K_{sp} = (s)(2s)^2$
$K_{sp} = s \times 4s^2$
$K_{sp} = 4s^3$

Q. 24. A compound forms hexagonal close packed (hcp) structure. What is the number of

• (i) octahedral voids
• (ii) tetrahedral voids
• (iii) total voids formed in 0.4 mol of it?

Solution:

Let $N$ be the number of atoms in the packing.
Number of atoms in 0.4 mol = $0.4 \times 6.022 \times 10^{23} = 2.4088 \times 10^{23}$.

(i) Number of Octahedral voids = $N$ = $2.4088 \times 10^{23}$.
(ii) Number of Tetrahedral voids = $2N$ = $2 \times 2.4088 \times 10^{23} = 4.8176 \times 10^{23}$.
(iii) Total voids = $N + 2N = 3N$.
Total = $3 \times 2.4088 \times 10^{23} = \mathbf{7.2264 \times 10^{23}}$

Q. 25. Illustrate with example, the difference between a double salt and coordinate compounds. Write two applications of coordinate compounds.

Solution:

Double Salt: A molecular compound that dissociates completely into simple ions when dissolved in water.
Example: Mohr's Salt $[FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O]$ gives $Fe^{2+}$, $NH_4^+$, and $SO_4^{2-}$ ions.

Coordinate Compound: A compound containing a complex ion that retains its identity in solution and does not dissociate into simple ions.
Example: $K_4[Fe(CN)_6]$ gives $4K^+$ and $[Fe(CN)_6]^{4-}$. It does not give test for $Fe^{2+}$ or $CN^-$.

Applications: 1. Cisplatin $[Pt(NH_3)_2Cl_2]$ is used in cancer treatment.
2. EDTA is used in the treatment of lead poisoning.

Q. 26. Write a note on 'aldol' condensation reaction of ethanal. Write chemical reaction involved when benzaldehyde is treated with concentrated caustic potash.

Solution:

Aldol Condensation of Ethanal:
When ethanal (containing $\alpha$-hydrogen) is treated with dilute alkali (NaOH), two molecules condense to form 3-hydroxybutanal (Aldol), which upon heating loses water to form But-2-enal (Crotonaldehyde).
$2CH_3CHO \xrightarrow{\text{dil. NaOH}} CH_3-CH(OH)-CH_2-CHO \xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO$

Reaction of Benzaldehyde with Conc. KOH (Cannizzaro Reaction):
Benzaldehyde has no $\alpha$-hydrogen. It undergoes self-oxidation and reduction.
$2C_6H_5CHO + \text{conc. } KOH \rightarrow C_6H_5COOK \text{ (Pot. Benzoate)} + C_6H_5CH_2OH \text{ (Benzyl Alcohol)}$

SECTION - D

(Attempt any THREE of the following questions)

Q. 27. Define isomorphism. Derive integrated rate law expression for first order reaction.

Solution:

Isomorphism: Two or more substances having the same crystal structure are said to be isomorphous (e.g., NaF and MgO).

Integrated Rate Law (1st Order):
Consider $A \rightarrow \text{Products}$.
Rate $= -\frac{d[A]}{dt} = k[A]$.
$\frac{d[A]}{[A]} = -k dt$.
Integrating both sides from $t=0$ ($[A]_0$) to $t=t$ ($[A]_t$):
$$ \int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt $$ $$ \ln[A]_t - \ln[A]_0 = -kt $$ $$ \ln \frac{[A]_0}{[A]_t} = kt $$ Converting $\ln$ to $\log_{10}$: $$ k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t} $$ Or if $a$ is initial conc and $x$ is amount reacted: $k = \frac{2.303}{t} \log_{10} \frac{a}{a-x}$.

Q. 28. What is the action of concentrated H₂SO₄ on,

• (i) CaF₂
• (ii) Cane sugar

What is nucleotide? Write reaction for the preparation of polyacrylonitrile (PAN).

Solution:

Action of Conc. H₂SO₄:
(i) CaF₂: Forms Hydrogen Fluoride.
$CaF_2 + H_2SO_4 \rightarrow CaSO_4 + 2HF$
(ii) Cane Sugar: Dehydrating action causes charring (formation of carbon).
$C_{12}H_{22}O_{11} \xrightarrow{H_2SO_4} 12C + 11H_2O$

Nucleotide: It is the repeating unit of nucleic acids (DNA/RNA), consisting of three components: a pentose sugar, a nitrogenous base, and a phosphate group.

Preparation of PAN: Addition polymerization of Acrylonitrile.
$n(CH_2=CH-CN) \xrightarrow{\text{Polymerization/Peroxide}} -[CH_2-CH(CN)]_n-$ (PAN)

Q. 29. State Kohlrausch law of independent migration of ions. Write and explain two applications of electrochemical series. Write unit of cell constant.

Solution:

Kohlrausch Law: At infinite dilution, each ion migrates independently of its co-ion and contributes to the total molar conductivity of an electrolyte a definite share, which depends only on its own nature.
$\Lambda_0 = \nu_+ \lambda_+^0 + \nu_- \lambda_-^0$.

Applications of Electrochemical Series:
1. To predict feasibility of reaction: A metal with lower electrode potential can displace a metal with higher electrode potential from its salt solution.
2. To calculate EMF: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$.

Unit of Cell Constant ($b$):
$b = l/A$. Unit is $m^{-1}$ or $cm^{-1}$.

Q. 30. Define:

• (i) Intensive property
• (ii) Enthalpy of sublimation

2 moles of an ideal gas are expanded isothermally and reversibly from 20L to 30L at 300K : Calculate the work done. [R = 8.314 JK⁻¹ mol⁻¹]

Solution:

(i) Intensive Property: A property whose value is independent of the amount of substance present in the system (e.g., Temperature, Pressure).

(ii) Enthalpy of Sublimation: The enthalpy change that accompanies the conversion of one mole of a solid directly into its vapor at constant temperature and pressure.

Work Calculation:
Formula: $W_{max} = -2.303 nRT \log_{10} \frac{V_2}{V_1}$
Given: $n=2$, $T=300K$, $R=8.314$, $V_1=20$, $V_2=30$.
$W = -2.303 \times 2 \times 8.314 \times 300 \times \log_{10} \frac{30}{20}$
$W = -11488.28 \times \log_{10} (1.5)$
$W = -11488.28 \times 0.1761$
$W = -2023 \text{ J}$ or $-2.023 \text{ kJ}$.
(Negative sign indicates work done by the system).

Q. 31. Define mineral. Write IUPAC name of [ Fe (CO)₅ ] complex. How will you convert

• (i) methyl iodide to methyl isocyanide.
• (ii) methyl cyanide to ethanoic acid.

Solution:

Mineral: A naturally occurring substance found in the earth's crust containing a metal in free or combined state.

IUPAC Name of $[Fe(CO)_5]$: Pentacarbonyliron(0).

Conversions:
(i) Methyl iodide to Methyl isocyanide: React with Silver Cyanide ($AgCN$).
$CH_3I + AgCN \rightarrow CH_3NC + AgI$

(ii) Methyl cyanide to Ethanoic acid: Acid hydrolysis.
$CH_3CN + 2H_2O + HCl \rightarrow CH_3COOH + NH_4Cl$