Chemistry (55) Board Paper Solution

HSC 12th Board Exam 2025 | Day 08 | Set J-302 (E)

Paper Details

Subject: Chemistry (55)

Date: 2025

Max. Marks: 70

Time: 3 Hours

Given Data:

$ R = 8.314 \text{ J/K/mol} $
Atomic mass Na = 23
$ K_f $ for water = 1.86 K kg mol$^{-1}$
$ 1F = 96500 \text{ C} $
$ N_A = 6.022 \times 10^{23} $

SECTION - A

Q. 1. Select and write the correct answer for the following multiple choice type of questions: [10]

(i) Schottky defect is NOT observed in _____.

(a) NaCl
(b) KCl
(c) AgBr
(d) NiO

Answer: (d) NiO

Explanation: Schottky defect is characteristic of ionic solids with high coordination numbers where cations and anions are of similar size (e.g., NaCl, KCl, AgBr). NiO typically shows metal deficiency defect (non-stoichiometric defect) due to variable oxidation states of Nickel.

(ii) The freezing point of 0.1m aqueous solution of urea, if $ K_f $ for water is 1.86 K kg mol$^{-1}$ is _____.

(a) 1.86 ºC
(b) –1.86 ºC
(c) 0.186 ºC
(d) –0.186 ºC

Answer: (d) –0.186 ºC

Explanation: $$ \Delta T_f = K_f \times m \times i $$ Since urea is non-electrolyte, $ i = 1 $. $$ \Delta T_f = 1.86 \times 0.1 \times 1 = 0.186 \text{ K} $$ $$ T_f = T_f^0 - \Delta T_f = 0 - 0.186 = -0.186^\circ \text{C} $$

(iii) Ozone layer is depleted by _____.

(a) NO
(b) NO$_2$
(c) NO$_3$
(d) N$_2$O$_5$

Answer: (a) NO

Explanation: Nitric oxide (NO) reacts with ozone to deplete it: $ \text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2 $.

(iv) When excess of AgNO$_3$ is added to a complex, one mole of AgCl is precipitated. The formula of complex is _____.

(a) [CoCl$_2$ (NH$_3$)$_4$]Cl
(b) [CoCl (NH$_3$)$_5$]Cl$_2$
(c) [CoCl$_3$ (NH$_3$)$_3$]
(d) [Co (NH$_3$)$_6$]Cl$_3$

Answer: (a) [CoCl$_2$ (NH$_3$)$_4$]Cl

Explanation: Precipitation of 1 mole of AgCl indicates that there is 1 ionizable chloride ion outside the coordination sphere. Formula (a) has one Cl outside the brackets.

(v) The value of $ \Delta n_g $ for the oxidation of 4 mole of sulphur dioxide to sulphur trioxide is _____.

(a) – 2
(b) 2
(c) – 4
(d) 4

Answer: (a) – 2

Explanation: Reaction: $ 4\text{SO}_2(g) + 2\text{O}_2(g) \rightarrow 4\text{SO}_3(g) $ $$ \Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) $$ $$ \Delta n_g = 4 - (4 + 2) = 4 - 6 = -2 $$

(vi) One dimensional nanostructure amongst the following is _____.

(a) Nanoparticles
(b) Nanotubes
(c) Nanofilms
(d) Nanorods

Answer: (b) Nanotubes

Explanation: Nanotubes and Nanowires are classified as 1D nanostructures. Nanoparticles are 0D, and Nanofilms are 2D.

(vii) Which formula co-relates degree of dissociation and concentration of electrolyte?

(a) $ c = \sqrt{\frac{K_a}{\alpha}} $
(b) $ \alpha = \sqrt{\frac{K_a}{c}} $
(c) $ c = \sqrt{K_a \alpha} $
(d) $ c = \sqrt{\frac{\alpha}{K_a}} $

Answer: (b) $ \alpha = \sqrt{\frac{K_a}{c}} $

Explanation: From Ostwald's dilution law for weak electrolytes, $ K_a = \alpha^2 c $, thus $ \alpha = \sqrt{\frac{K_a}{c}} $.

(viii) The highest acidic compound among the following is _____.

(a) m-Hydroxybenzoic acid
(b) o-Aminobenzoic acid
(c) Benzoic acid
(d) p-Methoxybenzoic acid

Answer: (a) m-Hydroxybenzoic acid

Explanation: (a) m-Hydroxybenzoic acid: -OH at meta position exerts only -I effect (Electron Withdrawing), increasing acidity.
(c) Benzoic acid: Standard.
(d) p-Methoxybenzoic acid: -OCH$_3$ exerts strong +R effect (Electron Donating), decreasing acidity.
Therefore, (a) is stronger than benzoic acid.

(ix) The formula used to calculate molar conductivity of an electrolyte is _____.

(a) $ \Lambda = \frac{1000c}{k} $
(b) $ c = \frac{1000\Lambda}{k} $
(c) $ \Lambda = \frac{1000k}{c} $
(d) $ k = \frac{1000}{\Lambda c} $

Answer: (c) $ \Lambda = \frac{1000k}{c} $

(x) Which of the following is a secondary amine?

(a) Cyclohexylamine
(b) Isopropylamine
(c) Diphenylamine
(d) N, N-Dimethylaniline

Answer: (c) Diphenylamine

Explanation: Diphenylamine has the structure $ (C_6H_5)_2NH $, which contains the $ >NH $ group attached to two carbon groups, making it a secondary amine.

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Q. 2. Answer the following questions: [8]

(i) Write the structural formula of N, N-dimethylethanamine.

$$ \text{CH}_3 - \text{CH}_2 - \text{N}(\text{CH}_3)_2 $$

(ii) Write the reagents used for the reduction of carbonyl group in Clemmensen’s reduction.

Zinc amalgam (Zn-Hg) and concentrated Hydrochloric acid (conc. HCl).

(iii) Write the IUPAC name of isoprene.

2-Methylbuta-1,3-diene.

(iv) The rate law equation for A $\to$ Product, is rate = k[A]$^x$. What is the effect of increase in concentration of ‘A’ on rate of reaction, if x < 0?

Since the order of reaction $ x $ is negative, the rate of reaction is inversely proportional to the concentration of A. Therefore, if the concentration of ‘A’ increases, the rate of reaction decreases.

(v) What is the molality of an aqueous solution of KBr having freezing point –3.72ºC (K$_f$ for water is 1.86 K kg mol$^{-1}$)?

KBr is a strong electrolyte dissociating into $ \text{K}^+ $ and $ \text{Br}^- $, so van't Hoff factor $ i = 2 $.
$$ \Delta T_f = 0 - (-3.72) = 3.72^\circ \text{C} $$ $$ \Delta T_f = i \cdot K_f \cdot m $$ $$ 3.72 = 2 \times 1.86 \times m $$ $$ 3.72 = 3.72 \times m $$ $$ m = 1 \text{ mol/kg (or 1 molal)} $$

(vi) Write the balanced chemical equation, when excess of ammonia is treated with chlorine.

When ammonia is in excess:
$$ 8\text{NH}_3 + 3\text{Cl}_2 \rightarrow 6\text{NH}_4\text{Cl} + \text{N}_2 $$

(vii) Write the number of donor atoms present in EDTA, during formation of complex.

EDTA is a hexadentate ligand. It has 6 donor atoms (4 Oxygen atoms and 2 Nitrogen atoms).

(viii) Write the names of the metal elements in brass alloy.

Copper (Cu) and Zinc (Zn).

SECTION - B

Attempt any EIGHT of the following questions: [16]

Q. 3. Derive the relation between half life and rate constant for a first order reaction.

For a first order reaction, the integrated rate law is given by: $$ k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t} $$ where $ [A]_0 $ is initial concentration and $ [A]_t $ is concentration at time $ t $.

At half life ($ t_{1/2} $):
The concentration of reactant becomes half of its initial value. $$ [A]_t = \frac{[A]_0}{2} $$ Substituting these values in the equation: $$ k = \frac{2.303}{t_{1/2}} \log_{10} \frac{[A]_0}{[A]_0 / 2} $$ $$ k = \frac{2.303}{t_{1/2}} \log_{10} (2) $$ Since $ \log_{10}(2) = 0.3010 $: $$ k = \frac{2.303 \times 0.3010}{t_{1/2}} $$ $$ k = \frac{0.693}{t_{1/2}} $$ $$ \therefore t_{1/2} = \frac{0.693}{k} $$ This is the relation between half-life and rate constant for a first order reaction.

Q. 4. (a) State Henry’s law. (b) Define: Osmotic pressure.

(a) Henry's Law:
It states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution. $$ S = K_H P $$ Where $ S $ is solubility, $ P $ is pressure, and $ K_H $ is Henry's constant.

(b) Osmotic Pressure:
It is defined as the excess hydrostatic pressure that must be applied to the solution side to prevent the flow of pure solvent into the solution through a semipermeable membrane.

Q. 5. Write the differences between lanthanoids and actinoids.

Lanthanoids	Actinoids
Differentiating electron enters 4f orbital.	Differentiating electron enters 5f orbital.
Binding energy of 4f orbitals is higher.	Binding energy of 5f orbitals is lower.
They show limited oxidation states (+2, +3, +4). +3 is most common.	They show variable oxidation states (+3, +4, +5, +6, +7).
They are non-radioactive (except Promethium).	All actinoids are radioactive.

Q. 6. Write anomalous behaviour of oxygen with respect to: (i) Atomicity (ii) Oxidation state (iii) Magnetic property (iv) Nature of hydrides.

(i) Atomicity: Oxygen is diatomic ($ \text{O}_2 $), whereas other group 16 elements are polyatomic (e.g., $ \text{S}_8 $).
(ii) Oxidation State: Oxygen typically shows -2 oxidation state. It does not show higher positive oxidation states (+4, +6) due to absence of d-orbitals, unlike other group members.
(iii) Magnetic Property: Molecular oxygen ($ \text{O}_2 $) is paramagnetic, whereas other group 16 elements are diamagnetic.
(iv) Nature of Hydrides: Water ($ \text{H}_2\text{O} $) is a liquid at room temperature due to strong intermolecular hydrogen bonding, while hydrides of other elements (e.g., $ \text{H}_2\text{S} $) are gases.

Q. 7. What is the action of: (i) Liquid bromine in acetic acid on anisole. (ii) Soda-lime on sodium acetate?

(i) Liquid Bromine in acetic acid on Anisole:
Anisole undergoes bromination to give a mixture of o-bromoanisole (minor) and p-bromoanisole (major). $$ \text{Anisole} + \text{Br}_2 \xrightarrow{\text{CH}_3\text{COOH}} \text{p-Bromoanisole (Major)} + \text{o-Bromoanisole (Minor)} $$

(ii) Soda-lime on Sodium Acetate:
Sodium acetate undergoes decarboxylation when heated with soda-lime (mixture of NaOH and CaO) to form methane. $$ \text{CH}_3\text{COONa} + \text{NaOH} \xrightarrow[\Delta]{\text{CaO}} \text{CH}_4 \uparrow + \text{Na}_2\text{CO}_3 $$

Q. 8. Calculate the work done in kJ in a reaction, if volume of the reactant decreases from 8 dm$^3$ to 4 dm$^3$ against 43 bar pressure. [ 1 dm$^3$ bar = 100J ]

Given:
$ V_1 = 8 \text{ dm}^3 $
$ V_2 = 4 \text{ dm}^3 $
$ P_{ext} = 43 \text{ bar} $
Formula: $ W = -P_{ext} (V_2 - V_1) $
Calculation:
$$ W = -43 \text{ bar} \times (4 \text{ dm}^3 - 8 \text{ dm}^3) $$ $$ W = -43 \times (-4) $$ $$ W = +172 \text{ bar dm}^3 $$ Conversion to kJ:
$ 1 \text{ bar dm}^3 = 100 \text{ J} = 0.1 \text{ kJ} $
$$ W = 172 \times 0.1 \text{ kJ} = 17.2 \text{ kJ} $$ Work done is 17.2 kJ.

Q. 9. Explain ionization isomers with suitable example in complexes.

Ionization Isomerism: This type 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. The isomers yield different ions in solution.

Example:
1. $ [Co(NH_3)_5SO_4]Br $ (Red-violet)
Gives precipitate of AgBr with AgNO$_3$ (indicates Br$^-$ outside).
2. $ [Co(NH_3)_5Br]SO_4 $ (Red)
Gives precipitate of BaSO$_4$ with BaCl$_2$ (indicates SO$_4^{2-}$ outside).

Q. 10. Write preparation of glucose from sucrose.

In the laboratory, Glucose is prepared by boiling sucrose (cane sugar) with dilute HCl or H$_2$SO$_4$ in alcoholic solution. Sucrose undergoes hydrolysis to give equimolar mixture of glucose and fructose.

C$_\text{12}$H$_\text{22}$O$_\text{11}$ (Sucrose) + H$_\text{2}$O $\xrightarrow{\text{H}^+}$ C$_\text{6}$H$_\text{12}$O$_\text{6}$ (Glucose) + C$_\text{6}$H$_\text{12}$O$_\text{6}$ (Fructose)

Glucose is separated by adding alcohol as it is less soluble in alcohol than fructose, so it crystallizes out.

Q. 11. How many coulombs of electricity is required to produce 1g of sodium metal by reduction of sodium ion?

Reaction: $ \text{Na}^+ + e^- \rightarrow \text{Na} $
1 mole of electrons (1 Faraday = 96500 C) deposits 1 mole of Na.
Molar mass of Na = 23 g/mol.

To produce 23 g of Na, charge required = 96500 C.
Therefore, to produce 1 g of Na:
$$ Q = \frac{96500}{23} \text{ C} $$ $$ Q \approx 4195.65 \text{ C} $$

Q. 12. Write the structural formula and IUPAC name of the alcohol having molecular formula C$_4$H$_\text{10}$O which does not undergo oxidation under normal condition.

An alcohol that resists oxidation under normal conditions is a tertiary alcohol.
For formula C$_4$H$_\text{10}$O, the tertiary alcohol is tert-butyl alcohol.

Structural Formula:
$$ (\text{CH}_3)_3\text{C-OH} $$ IUPAC Name: 2-Methylpropan-2-ol

Q. 13. Identify ‘A’ and ‘B’ in the following reaction and rewrite the complete reaction :

$ \text{CH}_3 - \text{CH} = \text{CH}_2 \xrightarrow[\text{Peroxide}]{\text{HBr}} \text{A} \xrightarrow[\text{-KBr}]{\text{alcoholic KCN}} \text{B} $

Step 1: Addition of HBr to propene in the presence of peroxide follows the Anti-Markovnikov rule.
A = 1-Bromopropane ($ \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} $).

Step 2: Reaction with alcoholic KCN results in nucleophilic substitution.
B = Butanenitrile ($ \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} $).

Complete Reaction:
$ \text{CH}_3 - \text{CH} = \text{CH}_2 \xrightarrow[\text{Peroxide}]{\text{HBr}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} \xrightarrow{\text{alc. KCN}} \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} $

Q. 14. Write the reaction for the preparation of : (i) acetaldehyde by Rosenmund reaction. (ii) benzaldehyde by Gatterman-Koch formylation.

(i) Acetaldehyde by Rosenmund Reaction:
Reduction of acetyl chloride with hydrogen in presence of Pd/BaSO$_4$. $$ \text{CH}_3\text{COCl} + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4} \text{CH}_3\text{CHO} + \text{HCl} $$

(ii) Benzaldehyde by Gatterman-Koch Formylation:
Benzene reacts with CO and HCl in presence of anhydrous AlCl$_3$ and CuCl. $$ \text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{Anhyd. AlCl}_3 / \text{CuCl}} \text{C}_6\text{H}_5\text{CHO} + \text{HCl} $$

SECTION - C

Attempt any EIGHT of the following questions: [24]

Q. 15. Write the general electronic configuration of 3d series. Draw the structures of sulphuric acid and thiosulphuric acid.

General Electronic Configuration of 3d series:
$ [\text{Ar}] 3d^{1-10} 4s^{1-2} $

Structures:
1. Sulphuric Acid (H$_2$SO$_4$): Tetrahedral arrangement around S. Two S=O bonds and two S-OH bonds.
2. Thiosulphuric Acid (H$_2$S$_2$O$_3$): Similar to sulphuric acid, but one double-bonded Oxygen is replaced by a Sulphur atom (S=S bond).

Q. 16. Define conjugate acid-base pair. The hydroxyl ion concentration in aqueous solution of NaOH is $ 2 \times 10^{-4} $ mol dm$^{-3}$. Calculate pH of the solution.

Conjugate Acid-Base Pair: A pair of acid and base that differs only by a proton (H$^+$) is called a conjugate acid-base pair.

Calculation:
Given $ [\text{OH}^-] = 2 \times 10^{-4} \text{ M} $
$$ \text{pOH} = -\log_{10} [\text{OH}^-] $$ $$ \text{pOH} = -\log_{10} (2 \times 10^{-4}) $$ $$ \text{pOH} = 4 - \log_{10} 2 = 4 - 0.3010 = 3.699 $$ We know, $ \text{pH} + \text{pOH} = 14 $
$$ \text{pH} = 14 - 3.699 $$ $$ \text{pH} = 10.301 $$

Q. 17. What is atom economy? Explain any two applications of nanomaterials.

Atom Economy: It is a measure of the efficiency of a chemical reaction, defined as the ratio of the formula weight of the desired product to the sum of the formula weights of all reactants, expressed as a percentage. It aims to maximize the incorporation of all reactant atoms into the final product.

Applications of Nanomaterials:
1. Medicine: Used in targeted drug delivery systems (e.g., gold nanoparticles) to treat cancer without damaging healthy cells.
2. Electronics: Quantum dots and nanowires are used to manufacture smaller, faster, and more efficient electronic circuits and displays.

Q. 18. What is peptide bond? How is it formed? Write the name and formula of the reagent used to convert alkylhalide to nitroalkane.

Peptide Bond: It is an amide linkage (-CO-NH-) formed between the carboxyl group (-COOH) of one amino acid and the amino group (-NH$_2$) of another amino acid.

Formation: It is formed by a condensation reaction where a water molecule is eliminated.

Reagent for Alkyl Halide to Nitroalkane:
Reagent: Silver Nitrite
Formula: AgNO$_2$
(Note: Using KNO$_2$ gives alkyl nitrite).

Q. 19. (a) Write the reactions for the action of following reagents on phenol: (i) Nitrating mixture (ii) Zinc dust. (b) What is the action of phosphorous pentachloride on ethyl methyl ether?

(a) Reactions on Phenol:
(i) Nitrating Mixture (Conc. HNO$_3$ + Conc. H$_2$SO$_4$): Phenol undergoes nitration to form 2,4,6-trinitrophenol (Picric Acid).
(ii) Zinc Dust: Phenol is reduced to Benzene. $ \text{C}_6\text{H}_5\text{OH} + \text{Zn} \rightarrow \text{C}_6\text{H}_6 + \text{ZnO} $.

(b) PCl$_5$ on Ethyl methyl ether:
The ether linkage is cleaved to form alkyl chlorides.
$$ \text{CH}_3-\text{O}-\text{C}_2\text{H}_5 + \text{PCl}_5 \xrightarrow{\Delta} \text{CH}_3\text{Cl} + \text{C}_2\text{H}_5\text{Cl} + \text{POCl}_3 $$ Products: Methyl chloride, Ethyl chloride, and Phosphoryl chloride.

Q. 20. (a) Write the formula to calculate EAN. (b) Explain formation of [Co(NH$_3$)$_6$]$^{3+}$ complex ion with respect to: (i) Type of hybridisation (ii) Magnetic property.

(a) Formula for EAN (Effective Atomic Number):
$$ \text{EAN} = Z - X + Y $$ Where $ Z $ = Atomic number of metal, $ X $ = Number of electrons lost (oxidation state), $ Y $ = Number of electrons donated by ligands (2 $\times$ Coordination number).

(b) [Co(NH$_3$)$_6$]$^{3+}$:
Cobalt (Z=27), Oxidation state = +3.
Configuration of Co$^{3+}$: [Ar] $ 3d^6 $.
NH$_3$ is a strong field ligand, causing pairing of electrons in 3d orbitals.
(i) Hybridisation: $ d^2sp^3 $ (Inner orbital complex involving two 3d, one 4s, and three 4p orbitals).
(ii) Magnetic Property: All electrons are paired, so it is Diamagnetic.

Q. 21. (a) Calculate spin only magnetic moment of M$^{2+}$ ion. [atomic number of M = 26]. (b) Write condensed electronic configuration of Gadolinium [ Z = 64 ].

(a) Magnetic Moment Calculation:
M = 26 (Iron). M$^{2+}$ means Fe$^{2+}$.
Electronic configuration: [Ar] $ 3d^6 $.
Number of unpaired electrons ($ n $) in d-orbitals (according to Hund's rule) = 4.
$$ \mu = \sqrt{n(n+2)} \text{ B.M.} $$ $$ \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ B.M.} $$

(b) Configuration of Gadolinium (Z=64):
$$ [\text{Xe}] 4f^7 5d^1 6s^2 $$ (Note: 5d gets 1 electron due to stability of half-filled f-subshell).

Q. 22. (a) Write the reducing agents used to convert Fe$_2$O$_3$ to ‘Fe’ in the reduction zone of blast furnace. (b) Write chemical equations involved in : (i) Carbylamine reaction for ethylamine. (ii) Hoffmann Bromamide degradation for acetamide.

(a) Reducing Agents in Blast Furnace:
Carbon monoxide (CO) and Coke (Carbon, C).

(b) Chemical Equations:
(i) Carbylamine Reaction:
$$ \text{C}_2\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc)} \xrightarrow{\Delta} \text{C}_2\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} $$ (Product: Ethyl isocyanide)

(ii) Hoffmann Bromamide Degradation:
$$ \text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{KOH} \rightarrow \text{CH}_3\text{NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O} $$ (Product: Methylamine)

Q. 23. (a) Explain Cannizzaro’s reaction with the help of benzaldehyde. (b) Write the reaction for the conversion of cyclohexene to adipic acid.

(a) Cannizzaro’s Reaction:
Aldehydes having no $\alpha$-hydrogen atom undergo self-oxidation and reduction (disproportionation) when treated with concentrated alkali.
Reaction for Benzaldehyde:
$$ 2\text{C}_6\text{H}_5\text{CHO} + \text{conc. NaOH} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{C}_6\text{H}_5\text{COONa} $$ (Products: Benzyl alcohol and Sodium benzoate).

(b) Cyclohexene to Adipic Acid:
Cyclohexene undergoes vigorous oxidation with hot acidic KMnO$_4$. $$ \text{Cyclohexene} + [O] \xrightarrow{\text{KMnO}_4, \text{H}_2\text{SO}_4, \Delta} \text{HOOC}-(\text{CH}_2)_4-\text{COOH} \text{ (Adipic Acid)} $$

Q. 24. Define zero order reaction. A reaction takes place in two steps : (i) NO(g) + Cl$_2$(g) $\to$ NOCl$_2$(g) (ii) NOCl$_2$(g) + NO(g) $\to$ 2NOCl(g). Write the overall reaction and identify the reaction intermediate.

Zero Order Reaction: A reaction in which the rate of reaction is independent of the concentration of the reactants is called a zero order reaction.

Mechanism Analysis:
Step 1: $ \text{NO} + \text{Cl}_2 \rightarrow \text{NOCl}_2 $
Step 2: $ \text{NOCl}_2 + \text{NO} \rightarrow 2\text{NOCl} $
Overall Reaction: (Add steps) $ 2\text{NO} + \text{Cl}_2 \rightarrow 2\text{NOCl} $
Reaction Intermediate: $ \text{NOCl}_2 $ (It is produced in step 1 and consumed in step 2).

Q. 25. $\Delta$H for formation of ethane gas is – 84.4 kJ at 300 K. Calculate $\Delta$U for the reaction.

Reaction for formation of ethane:
$$ 2\text{C(graphite)} + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g) $$ Calculation of $ \Delta n_g $:
$ \Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) $
$ \Delta n_g = 1 - 3 = -2 $

Formula: $ \Delta H = \Delta U + \Delta n_g RT $
Given: $ \Delta H = -84.4 \text{ kJ} $, $ R = 8.314 \times 10^{-3} \text{ kJ K}^{-1}\text{mol}^{-1} $, $ T = 300 \text{ K} $.

$$ -84.4 = \Delta U + (-2 \times 8.314 \times 10^{-3} \times 300) $$ $$ -84.4 = \Delta U - (2 \times 2.4942) $$ $$ -84.4 = \Delta U - 4.9884 $$ $$ \Delta U = -84.4 + 4.9884 $$ $$ \Delta U = -79.4116 \text{ kJ} $$ $$ \Delta U \approx -79.41 \text{ kJ} $$

Q. 26. Mention the types of polymers formed on the basis of intermolecular forces. Write any two uses of low density polyethylene.

Types of Polymers based on intermolecular forces:
1. Elastomers (Weakest forces, e.g., Rubber).
2. Fibers (Strongest H-bonds/dipole, e.g., Nylon).
3. Thermoplastic Polymers (Intermediate forces, e.g., PVC).
4. Thermosetting Polymers (Cross-linked, rigid, e.g., Bakelite).

Uses of Low Density Polyethylene (LDPE):
1. Manufacturing of squeeze bottles and flexible pipes.
2. Insulation for electrical cables and wires.
3. Packaging films and plastic bags.

SECTION - D

Attempt any THREE of the following questions: [12]

Q. 27. (a) An element with molar mass 27 g/mol forms a cubic unit cell with edge length 405 pm. If density of the crystal is 2.7 g cm$^{-3}$, identify the type of unit cell. (b) Derive the equation of Raoult’s law for binary solution containing non-volatile solute.

(a) Identify Unit Cell:
Given: $ M = 27 \text{ g/mol} $, $ a = 405 \text{ pm} = 4.05 \times 10^{-8} \text{ cm} $, $ \rho = 2.7 \text{ g/cm}^3 $.
Formula: $ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} $
$$ Z = \frac{\rho \cdot a^3 \cdot N_A}{M} $$ $$ a^3 = (4.05 \times 10^{-8})^3 \approx 66.4 \times 10^{-24} \text{ cm}^3 $$ $$ Z = \frac{2.7 \times 66.4 \times 10^{-24} \times 6.022 \times 10^{23}}{27} $$ $$ Z = \frac{2.7}{27} \times (66.4 \times 0.6022) $$ $$ Z = 0.1 \times 40 \approx 4 $$ Since $ Z = 4 $, the unit cell is Face Centered Cubic (FCC).

(b) Raoult's Law Derivation:
Raoult's law states that the vapour pressure of a solvent in a solution is equal to the vapour pressure of pure solvent multiplied by its mole fraction.
$$ P_1 = P_1^0 x_1 $$ For a binary solution, $ x_1 + x_2 = 1 \Rightarrow x_1 = 1 - x_2 $.
Substituting $ x_1 $: $$ P_1 = P_1^0 (1 - x_2) $$ $$ P_1 = P_1^0 - P_1^0 x_2 $$ $$ P_1^0 x_2 = P_1^0 - P_1 $$ $$ x_2 = \frac{P_1^0 - P_1}{P_1^0} $$ This equation relates the relative lowering of vapour pressure to the mole fraction of the solute.

Q. 28. (a) State whether entropy change is positive or negative in the following examples: (i) Melting of ice (ii) Vaporisation of a liquid. (b) Explain ‘common ion effect’ with example.

(a) Entropy Change:
(i) Melting of ice: Positive ($ \Delta S > 0 $). Randomness increases as solid turns to liquid.
(ii) Vaporisation of a liquid: Positive ($ \Delta S > 0 $). Randomness increases significantly as liquid turns to gas.

(b) Common Ion Effect:
It is the suppression of the dissociation of a weak electrolyte by the addition of a strong electrolyte containing a common ion.
Example: Consider the dissociation of weak acid Acetic acid ($ \text{CH}_3\text{COOH} $): $$ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ $$ If Sodium acetate ($ \text{CH}_3\text{COONa} $), a strong electrolyte, is added: $$ \text{CH}_3\text{COONa} \rightarrow \text{CH}_3\text{COO}^- + \text{Na}^+ $$ The concentration of acetate ions ($ \text{CH}_3\text{COO}^- $) increases. According to Le Chatelier's principle, the equilibrium of the acid shifts to the left, suppressing its ionization.

Q. 29. Draw a neat and labelled diagram of a lead accumulator cell. Write the overall reactions taking place at cathode and anode during discharging of the cell.

[Diagram of Lead Accumulator]
Anode: Pb plates
Cathode: PbO$_2$ plates
Electrolyte: Dilute H$_2$SO$_4$ (38%)

Discharging Reactions:
At Anode (Oxidation):
$$ \text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^- $$ At Cathode (Reduction):
$$ \text{PbO}_2(s) + 4\text{H}^+(aq) + \text{SO}_4^{2-}(aq) + 2e^- \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) $$ Overall Cell Reaction:
$$ \text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) $$

Q. 30. (a) Define a unit cell. Which colour is shown by NaCl crystal due to formation of F-centre? (b) Why does fluorine show anomalous behaviour in ‘17 group’ elements?

(a)
Unit Cell: The smallest repeating structural unit of a crystalline solid, which when repeated in different directions, generates the entire crystal lattice.
NaCl Colour: Yellow (due to F-centres).

(b) Anomalous behaviour of Fluorine:
Fluorine differs from other group 17 elements due to:

Smallest atomic size.
Highest electronegativity.
Low F-F bond dissociation enthalpy.
Absence of d-orbitals in its valence shell.

Consequences: It shows only -1 oxidation state, forms H-bonds (HF is liquid), and is highly reactive.

Q. 31. (a) Write salient features of SN$^2$ mechanism. (b) What is the action of following reagents on bromomethane : (i) bromobenzene (ii) mercurous fluoride.

(a) Salient features of SN$^2$ mechanism:
1. It is a one-step concerted mechanism (bond breaking and bond forming occur simultaneously).
2. It is a bimolecular reaction (Rate $\propto [\text{Substrate}][\text{Nucleophile}]$).
3. The nucleophile attacks from the back side of the leaving group.
4. It proceeds through a pentacoordinate transition state.
5. It results in Walden inversion (inversion of configuration).
6. Reactivity order: Methyl > Primary > Secondary > Tertiary halides.

(b) Action of reagents on Bromomethane ($ \text{CH}_3\text{Br} $):
(i) Bromobenzene (Wurtz-Fittig Reaction):
Reacts with bromobenzene and sodium in dry ether to form Toluene. $$ \text{CH}_3\text{Br} + \text{C}_6\text{H}_5\text{Br} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{CH}_3 + 2\text{NaBr} $$ (ii) Mercurous fluoride (Swarts Reaction):
Reacts to form Fluoromethane. $$ 2\text{CH}_3\text{Br} + \text{Hg}_2\text{F}_2 \rightarrow 2\text{CH}_3\text{F} + \text{Hg}_2\text{Br}_2 $$