CBSE Class 12 Chemistry Sample Paper 2026 27 with Solutions PDF Download

Section-A

Question 1 to 16 are multiple choice questions. Only one of the choices is correct. Select and write the correct choice as well as the answer to these questions.

1. The boiling points and Kb values of isomeric amines are given in table 1.
Amine | Boiling point/K | pKb (approx)
P | 329.3 | 3.00
Q | 310.5 | 9.83
R | 350.8 | 3.22
Table 1
Which one of the following statement is correct? [1 Mark]
(A) Amine P is more volatile than amine R, but less basic
(B) Amine R will give white precipitate with Hinsberg reagent, which is soluble in NaOH
(C) Amine Q is more soluble in water as compared to R.
(D) Amine P and Q can be prepared by Gabriel Phthalimide process.

Answer: (B) Amine R will give white precipitate with Hinsberg reagent, which is soluble in NaOH

Teacher's Note:
a) From the data, P is a secondary amine, Q is a tertiary amine (lowest boiling point, no N-H for hydrogen bonding) and R is a primary amine (highest boiling point).
b) A primary amine gives a sulphonamide with Hinsberg reagent that still has an acidic N-H, so it dissolves in NaOH.
c) Option (A) is wrong because P has the lower pKb, so P is more basic than R.

2. Which of the following complex is tetrahedral in shape? [1 Mark]
(A) [Ni(CN)₄]^2-
(B) [Cu(NH₃)₄]²⁺
(C) [NiCl₄]^2-
(D) [Co(en)₃]³⁺

Answer: (C) [NiCl₄]^2-

Teacher's Note:
a) Cl^- is a weak field ligand, so it does not pair the 3d electrons of Ni²⁺; the complex uses sp³ hybridisation and is tetrahedral.
b) [Ni(CN)₄]^2- and [Cu(NH₃)₄]²⁺ are square planar, while [Co(en)₃]³⁺ is octahedral.

3. The quantity of charge required to obtain one mole of metal "M" from M₂O₃ is: [1 Mark]
(A) 1F
(B) 6F
(C) 3F
(D) 2F

Answer: (C) 3F

Teacher's Note:
a) In M₂O₃ the metal is present as M³⁺, and M³⁺ + 3e^- → M.
b) 3 mol of electrons = 3F; do not count both metal atoms, since only one mole of M is asked (that would give 6F).

4. An organic compound "X" on reaction with chloroform and alcoholic KOH gives foul smelling product.
In which of the following case compound X would be produced: [1 Mark]
(A) Chloromethane is treated with AgCN.
(B) Nitrobenzene is reduced.
(C) Benzene diazonium chloride is treated with Cu/KCN
(D) Aniline reacts with Acetyl chloride.

Answer: (B) Nitrobenzene is reduced.

Teacher's Note:
a) The foul smelling isocyanide is formed by the carbylamine reaction, which is given only by primary amines (aliphatic and aromatic).
b) Reduction of nitrobenzene gives aniline, a primary aromatic amine, so X is aniline.

5. Consider a system in a state of dynamic equilibrium, as shown in Fig. 1(a). The lower part is solution and the upper part is gaseous system at pressure p and temperature T. If the pressure is increased over the solution phase by compressing the gas to a smaller volume as shown in Fig 1(b).
Which of the following statements is true about figure 1(b) [1 Mark]
(A) solubility of gas will increase until dynamic equilibrium is reached
(B) solubility of gas will decrease until dynamic equilibrium is reached
(C) solubility of gas will increase until static equilibrium is reached
(D) solubility of gas will decrease until static equilibrium is reached

[Figure: Figure 1 shows two closed containers with a piston. (a) One weight W on the piston; gas above the solution with gas molecules spread out. (b) Three weights on the piston, which is pushed lower; the gas is compressed to a smaller volume and more gas molecules are shown dissolved in the solution.]

Answer: (A) solubility of gas will increase until dynamic equilibrium is reached

Teacher's Note:
a) This is Henry's law: at constant temperature, the solubility of a gas in a liquid increases with the pressure of the gas.
b) Equilibrium between the gas and dissolved gas is always dynamic, never static.

For Visually Impaired Candidates (in lieu of Q. 5)

According to Henry's law partial pressure P of a gas above the liquid at constant temperature is directly proportional to [1 Mark]
(A) volume of the gas
(B) mole fraction of the gas in the solution
(C) atmospheric pressure
(D) vapour pressure of the solution

Answer: (B) mole fraction of the gas in the solution

Teacher's Note:
a) Henry's law: \( p = K_H\,x \), where x is the mole fraction of the gas in the solution.
b) K_H is Henry's law constant, which depends on the nature of the gas and the temperature.

6. Aavya carries out hydrolysis of acetyl chloride. The reaction is represented by the following equation:
CH₃COCl + H₂O → CH₃COOH + HCl
She carries out the reaction with (i) equimolar amounts of acetyl chloride and water (ii) excess of water. The rate law obeyed in the two cases will be: [1 Mark]
(A) (i) First order (ii) Second Order
(B) (i) First order (ii) First order
(C) (i) Second order (ii) Second order
(D) (i) Second order (ii) First order

Answer: (D) (i) Second order (ii) First order

Teacher's Note:
a) With equimolar amounts, the rate depends on the concentrations of both reactants, so the reaction is second order.
b) With excess water, the concentration of water hardly changes, so the rate depends only on acetyl chloride. This is a pseudo first order (pseudounimolecular) reaction.

7. The amino acid HOOC-CH₂-CH₂-CH (NH₂) COOH, can be synthesised within our body. Which of the following is true about this amino acid?
HOOC-CH₂-CH₂-CH(NH₂) COOH is: [1 Mark]
(A) an acidic and non-essential amino acid.
(B) a neutral and non- essential amino acid.
(C) an acidic and essential amino acid
(D) a neutral and essential amino acid

Answer: (A) an acidic and non-essential amino acid.

Teacher's Note:
a) It can be synthesised in the body, so it is non-essential.
b) It has two -COOH groups and only one -NH₂ group, so it is acidic (this is glutamic acid).

8. Identify the Grignard reagent and carbonyl compound which on reaction followed by hydrolysis will give 3-methyl pentan-3-ol.
(i) C₂H₅COC₂H₅ + CH₃MgCl
(ii) C₂H₅COCH₃ + C₂H₅MgCl
(iii) CH₃CHO + C₂H₅-CH(MgCl)-C₂H₅
(iv) C₂H₅CHO + CH₃-CH(MgCl)-C₂H₅ [1 Mark]
(A) (i) and (iii)
(B) (ii) and (iv)
(C) (iii) and (iv)
(D) (i) and (ii)

[Figure: In (iii) and (iv) the Grignard reagent is drawn as a structure: in (iii) C₂H₅-CH-C₂H₅ with MgCl on the middle carbon; in (iv) CH₃-CH-C₂H₅ with MgCl on the middle carbon.]

Answer: (D) (i) and (ii)

Teacher's Note:
a) 3-Methylpentan-3-ol is a tertiary alcohol, (C₂H₅)₂C(OH)CH₃, so it must come from a ketone.
b) (i) C₂H₅COC₂H₅ + CH₃MgCl and (ii) C₂H₅COCH₃ + C₂H₅MgCl both give C₂H₅C(OH)(CH₃)C₂H₅ after hydrolysis.
c) (iii) and (iv) use aldehydes, so they give secondary alcohols.

9. Which one of the following element will be strong oxidising agent in their +3 oxidation state? (Atomic no. Cr=24, Mn=25, Ti=22, Fe=26) [1 Mark]
(A) Cr
(B) Mn
(C) Ti
(D) Fe

Answer: (B) Mn

Teacher's Note:
a) Mn³⁺ (3d⁴) acts as a strong oxidising agent because it readily gains an electron to become Mn²⁺.
b) Mn²⁺ has the extra stable half-filled 3d⁵ configuration, which drives this change.

10. When (-)-2-bromohexane reacts with sodium hydroxide, (+)-hexan-2-ol is formed. Which one of the following observation would be correct? [1 Mark]
(A) The reaction follows S_N2 mechanism and is accompanied by inversion of configuration.
(B) The reaction follows S_N1 mechanism and is accompanied by inversion of configuration.
(C) The reaction follows S_N2 mechanism and is accompanied by retention of configuration.
(D) The reaction follows S_N1 mechanism and is accompanied by retention of configuration.

Answer: (A) The reaction follows S_N2 mechanism and is accompanied by inversion of configuration.

Teacher's Note:
a) In S_N2, OH^- attacks from the side opposite to Br, so the configuration is inverted.
b) A single optically active product (not a racemic mixture) shows that the reaction is S_N2, not S_N1.

11. If half-life of a first order reaction is 16 minutes, how much time does it take to complete 75% of the reaction? [1 Mark]
(A) 16 min
(B) 32 min
(C) 48 min
(D) 64 min

Answer: (B) 32 min

Teacher's Note:
a) In the first 16 min, 50% of the reaction is completed; in the next 16 min, half of the remaining 50% (that is 25%) reacts.
b) So 50% + 25% = 75% takes 2 half-lives = \( 2 \times 16 = 32 \) min.

12. Which solution has the highest conductivity? [1 Mark]
(A) 0.1 M HCl
(B) 0.1 M CH₃COOH
(C) 0.1 M glucose
(D) 0.1 M NH₄OH

Answer: (A) 0.1 M HCl

Teacher's Note:
a) HCl is a strong acid and is fully dissociated into H⁺ and Cl^- ions; more ions give higher conductivity.
b) CH₃COOH and NH₄OH are weak electrolytes (partly ionised), and glucose is a non-electrolyte.

13. Assertion (A): Nitration of aniline gives significant amount of meta derivative.
Reason (R): -NH₂ is ortho and para directing and highly activating group.
Select the most appropriate answer from the options given below: [1 Mark]
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

Answer: (B) Both A and R are true, and R is not the correct explanation of A.

Teacher's Note:
a) In the strongly acidic nitrating mixture, aniline is protonated to the anilinium ion, which is meta directing; this explains the meta product.
b) The reason is a true fact about -NH₂, but it does not explain the meta product.

14. Assertion (A): Permanganate and manganate ions have tetrahedral structure
Reason (R): In permanganate and manganate ions, pi bonding takes place which is due to overlap of p orbitals of oxygen with p orbitals of manganese.
Select the most appropriate answer from the options given below: [1 Mark]
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

Answer: (C) A is true but R is false.

Teacher's Note:
a) MnO₄^- and MnO₄^2- are both tetrahedral.
b) The pi bonding takes place by overlap of p orbitals of oxygen with d orbitals of manganese, not p orbitals, so R is false.

15. Assertion (A): Miscibility of ethers with water resembles those of alcohols of the same molecular mass.
Reason (R): The oxygen of ether can also form hydrogen bonds with water.
Select the most appropriate answer from the options given below: [1 Mark]
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true

Answer: (A) Both A and R are true, and R is the correct explanation of A.

Teacher's Note:
a) The oxygen atom of an ether can accept a hydrogen bond from water, just like an alcohol.
b) So ethers and alcohols of similar molecular mass show similar miscibility with water.

16. Assertion (A): [Co(NH₃)₆]³⁺ is diamagnetic.
Reason (R): NH₃ is a strong field ligand and causes pairing of electrons.
Select the most appropriate answer from the options given below: [1 Mark]
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, and R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true

Answer: (A) Both A and R are true, and R is the correct explanation of A.

Teacher's Note:
a) Co³⁺ is 3d⁶; the strong field NH₃ pairs all six electrons in the t_2g set (\( t_{2g}^{6}\,e_g^{0} \)).
b) With no unpaired electrons, the complex is diamagnetic, so R correctly explains A.

Section-B

Question No. 17 to 21 are very short answer questions carrying 2 marks each.

Attempt either A or B

17(A) (I) Complete and balance the following reaction
SO₃^2- + MnO₄^- + H⁺ → [1 Mark]

Answer: 2MnO₄^- + 5SO₃^2- + 6H⁺ → 2Mn²⁺ + 5SO₄^2- + 3H₂O

Teacher's Note:
a) In acidic medium Mn changes from +7 to +2 (gains 5 e^-) and S changes from +4 to +6 (loses 2 e^-), so the ratio MnO₄^- : SO₃^2- is 2 : 5.
b) Check the charge balance: left = 2(-1) + 5(-2) + 6(+1) = -6 and right = 2(+2) + 5(-2) = -6.

(II) Why does titanium melt at a lower temperature than chromium? [1 Mark]

Answer: Chromium has more unpaired d electrons than titanium. So chromium has stronger interatomic (metallic) bonding holding its atoms together, and more energy is needed to break it. Hence titanium melts at a lower temperature.

Teacher's Note:
a) Keyword for the mark: more unpaired electrons means stronger interatomic attraction.
b) Ti is 3d²4s² while Cr is 3d⁵4s¹, with six unpaired electrons.

OR

17(B) (I) Why do titanium and scandium forms complex compounds, whereas potassium and calcium do not, although they belong to same period? [1 Mark]

Answer: Titanium and scandium are transition metals. Their ions are comparatively small, have high ionic charge and have vacant d orbitals available for bond formation, so they form complexes. Potassium and calcium lack these features, so they do not.

Teacher's Note:
a) Write all three factors: small size, high ionic charge and availability of d orbitals.
b) K and Ca are s-block metals with larger ions, low charge and no d orbitals for bonding.

(II) Complete and balance the following reaction
Fe²⁺ + Cr₂O₇^2- + H⁺ → [1 Mark]

Answer: Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Teacher's Note:
a) Each Cr goes from +6 to +3, so one Cr₂O₇^2- gains 6 e^- and oxidises 6 Fe²⁺ to Fe³⁺.
b) Balance O with 7H₂O and then H with 14H⁺; check charge: +24 on both sides.

18. "The rate of reaction remains constant during the course of reaction." Is this statement always true or is true under certain conditions? Reflect on the statement. [2 Marks]

Answer:
1. The statement is true only for a zero order reaction, because its rate is independent of the concentration of the reactants (rate = k).
2. For reactions of any other order (first order, second order etc.), the rate decreases as the concentration of the reactants decreases with time. So the statement is not always true.

Teacher's Note:
a) One mark for "true for zero order" and one mark for "rate falls with concentration for other orders".
b) For zero order, rate = \( k[R]^0 = k \), which stays constant.

19. (I) Complete the sequence of nitrogenous bases in the given DNA strand (Figure 2) [1 Mark]

[Figure: Figure 2 shows a DNA double helix with ends labelled 5' and 3' on each strand. Four base pairs are shown from top to bottom: ? ≡ C, A = ?, ? = A, A = ?]

Answer: The completed pairs are: G ≡ C, A = T, T = A, A = T.

Teacher's Note:
a) In DNA, guanine pairs with cytosine by three hydrogen bonds and adenine pairs with thymine by two hydrogen bonds.
b) Use the number of lines (≡ or =) to check each pair.

(II) Milk contains a protein called casein. How does the structure of casein change on addition of a few drops of lemon juice to milk? [1 Mark]

Answer: Lemon juice lowers the pH. This breaks the hydrogen bonds in casein, so the globules of the protein unfold and casein gets denatured (the milk curdles).

Teacher's Note:
a) Key words: change in pH, hydrogen bonds broken, protein denatured.
b) Denaturation changes the secondary and tertiary structure; the primary structure stays the same.

For Visually Impaired Candidates (in lieu of Q. 19)

(I) Name the nitrogenous bases in DNA [1 Mark]

Answer: Adenine, Guanine, Cytosine and Thymine.

Teacher's Note:
a) DNA has thymine; RNA has uracil in place of thymine.
b) Write all four names to get the mark.

(II) Milk contains a protein called casein. How does the structure of casein change on addition of a few drops of lemon juice to milk? [1 Mark]

Answer: Lemon juice lowers the pH. This breaks the hydrogen bonds in casein, so the globules of the protein unfold and casein gets denatured.

Teacher's Note:
a) Key words: change in pH, hydrogen bonds broken, protein denatured.
b) This is why milk curdles when lemon juice is added.

20. Chlorobenzene can be converted into phenol by heating in aqueous sodium hydroxide solution at a temperature of 623K and 300 atm, however presence of a substituent group effects its activity towards this nucleophilic substitution reaction.
Which substituent (i) nitro or (ii) methoxy will increase the reactivity of chlorobenzene and why? [2 Marks]

Answer:
1. The nitro group will increase the reactivity.
2. The nitro group is a strong electron withdrawing group due to its -I and -R (-M) effects, whereas the methoxy group is electron donating.
3. When the nitro group is at the ortho or para position, it withdraws electron density from the ring and stabilises the negative charge of the intermediate by resonance. So chlorobenzene becomes more reactive towards nucleophilic substitution.

Teacher's Note:
a) One mark is for naming the nitro group; the other mark is for the reason.
b) Mention "ortho or para position": a meta nitro group cannot stabilise the charge by resonance.

21. Account for the following:

(I) Underground iron pipelines are often protected by connecting them to a more reactive metal like magnesium. [1 Mark]

Answer: Magnesium is more reactive than iron, so it acts as a sacrificial anode and gets oxidised (Mg → Mg²⁺ + 2e^-). It supplies electrons to the iron pipe, which behaves as the cathode, so the iron does not corrode.

Teacher's Note:
a) Keyword: sacrificial anode (sacrificial protection).
b) The magnesium block is used up and must be replaced from time to time.

(II) Write the reactions occurring at cathode and anode during the electrolysis of molten magnesium chloride. [1 Mark]

Answer:
Cathode: Mg²⁺ + 2e^- → Mg (l)
Anode: 2Cl^- → Cl₂ (g) + 2e^-

Teacher's Note:
a) Reduction always takes place at the cathode and oxidation at the anode.
b) Each half reaction carries half a mark; balance the electrons in both.

Section-C

Question No. 22 to 28 are short answer questions, carrying 3 marks each.

22. (I) Which of the following has the highest magnetic moment? Justify your answer.
Ti²⁺ ,Co²⁺ , Fe²⁺ .
[ Atomic No. Ti = 22 Co = 27 , Fe = 26 ] [1 Mark]

Answer: Fe²⁺ has the highest magnetic moment. Fe²⁺ (3d⁶) has 4 unpaired electrons, which is more than Ti²⁺ (3d², 2 unpaired) and Co²⁺ (3d⁷, 3 unpaired).

Teacher's Note:
a) Magnetic moment \( \mu = \sqrt{n(n+2)} \) BM, so more unpaired electrons give a higher value.
b) For Fe²⁺, \( \mu = \sqrt{24} \approx 4.9 \) BM.

(II) Which characteristic of iron make it suitable to:
a) act as catalyst in Haber's process.
b) form interstitial compound with oxygen. [2 Marks]

Answer:
a) Iron provides a large surface area for adsorption of the reactants and shows variable oxidation states, so it acts as a catalyst in Haber's process.
b) Iron has a crystal lattice with interstitial sites that are large enough to hold small non-metal atoms like oxygen, carbon and nitrogen, so it forms interstitial compounds.

Teacher's Note:
a) For the catalyst part, mention both "large surface area" and "variable oxidation states".
b) Interstitial compounds are formed when small atoms are trapped in the holes of the metal lattice.

23. The values of log k and 1/T for a reaction are given-
S.No. | Log k | 1/T (/K)
1. | -4.46 | 0.00333
2. | -3.91 | 0.00323
3. | -3.37 | 0.00313
4. | -2.82 | 0.00303
5. | -2.31 | 0.00294
The plot of log k vs 1/T obtained for the above data is as shown in Figure 3.
Here k is the rate constant and T is the absolute temperature.
Calculate E_a for the reaction. (Given R= 8.314 JK^-1mol^-1) [3 Marks]

[Figure: Figure 3 is a straight line graph of log k (y-axis, 0 to -5) against 1/T (/K) (x-axis marked 0.00333, 0.00323, 0.00313, 0.00303, 0.00294). The points -4.46, -3.91, -3.37, -2.82 and -2.31 lie on a rising straight line.]

Answer:
Arrhenius equation: \( \log k = \log A - \frac{E_a}{2.303RT} \)
So the plot of log k against 1/T is a straight line with slope \( = -\frac{E_a}{2.303R} \)
Using points 1 and 2: \( \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4.46 - (-3.91)}{0.00333 - 0.00323} = \frac{-0.55}{0.0001} = -5500 \) K
\( -\frac{E_a}{2.303R} = -5500 \)
\( E_a = 5500 \times 2.303 \times 8.314 \) J mol^-1
\( E_a = 105309 \) J mol^-1 \( \approx 105.31 \) kJ mol^-1

Teacher's Note:
a) The slope may vary slightly if other pairs of points are chosen; the marking scheme accepts this.
b) The slope has the unit K (log k has no unit and 1/T is in K^-1); the marking scheme writes it as J/K/mol, but its final value 105.31 kJ mol^-1 is correct.
c) Half a mark each is for the correct value and the correct unit of E_a.

For Visually Impaired Candidates (in lieu of Q. 23)

Rate constant k of a reaction varies with temperature 'T' according to the equation
\( \log k = \log A - \frac{E_a}{2.303RT} \)
where E_a is the activation energy. When a graph is plotted for log k vs 1/T, a straight line with a slope of - 4250 K is obtained. Calculate E_a for the reaction. (Given R= 8.314 JK^-1mol^-1). [3 Marks]

Answer:
\( \log k = \log A - \frac{E_a}{2.303RT} \)
Slope of log k vs 1/T \( = -\frac{E_a}{2.303R} \)
\( -\frac{E_a}{2.303R} = -4250 \) K
\( E_a = 4250 \times 2.303 \times 8.314 \) J mol^-1
\( E_a = 81375 \) J mol^-1 \( \approx 81.38 \) kJ mol^-1

Teacher's Note:
a) The marking scheme uses a slope of -4500 K and gets 86.16 kJ mol^-1, but the question gives -4250 K, which gives 81.38 kJ mol^-1.
b) Always change J mol^-1 to kJ mol^-1 by dividing by 1000 and write the unit.

24. Identify each name reaction and complete the missing reactant/ product in the given reaction sequence.

(I) 2CH₃CHO →(?) CH₃CH(OH)CH₂CHO [1 Mark]

Answer: Aldol condensation; the missing reagent is dil. NaOH.

Teacher's Note:
a) Aldehydes with an alpha-hydrogen form a beta-hydroxy aldehyde (aldol) with dilute alkali.
b) Half a mark for the name and half a mark for the reagent.

(II) C₆H₅CH₃ + __?__ →(CS₂) C₆H₅CH(OCrOHCl₂)₂ →(H₃O⁺) C₆H₅CHO [1 Mark]

[Figure: Toluene (benzene ring with CH₃) + __?__ with CS₂ over the arrow gives a benzene ring carrying CH(OCrOHCl₂)₂, which with H₃O⁺ over the arrow gives a benzene ring carrying CHO (benzaldehyde).]

Answer: Etard reaction; the missing reactant is CrO₂Cl₂ (chromyl chloride).

Teacher's Note:
a) Chromyl chloride oxidises the methyl group to a chromium complex, which on hydrolysis gives benzaldehyde.
b) Half a mark for the name and half a mark for CrO₂Cl₂.

(III) \( \gt \)C=O →(NH₂NH₂, -H₂O) __?__ →(KOH/ethylene glycol, Heat) \( \gt \)CH₂ + N₂ [1 Mark]

Answer: Wolff-Kishner reduction; the missing product is the hydrazone, \( \gt \)C=N-NH₂.

Teacher's Note:
a) The carbonyl group is first converted to a hydrazone, which on heating with KOH in ethylene glycol gives \( \gt \)CH₂ and N₂.
b) Do not confuse it with Clemmensen reduction, which uses Zn-Hg and conc. HCl.

25. (I) Aman prepared a solution by dissolving 10 ml of Acetone and 10 ml of Carbon disulphide. The net volume of resulting solution was not found to be 20 mL.
a) What would be the observation in this case? Comment.
b) In case carbon disulphide is replaced by chloroform in the above case, what would be the expected changes in volume of the resulting solution? Support your answer with appropriate reason. [2 Marks]

Answer:
a) The net volume would be slightly more than 20 mL. The interactions between acetone and carbon disulphide are weaker than those in the pure liquids, which leads to an increase in volume (positive deviation).
b) With chloroform, the net volume would be slightly less than 20 mL. Chloroform and acetone have strong interactions due to hydrogen bonding, so the volume decreases (negative deviation).

Teacher's Note:
a) Positive deviation: \( \Delta_{mix}V \gt 0 \); negative deviation: \( \Delta_{mix}V \lt 0 \).
b) Each part has half a mark for the volume change and half a mark for the reason.

(II) Calculate the concentration of oxygen (in ppm) present in 1 litre of sea water. Given that a litre of sea water weighs 1030 g and contains about 6 × 10^-3 g of dissolved oxygen. [1 Mark]

Answer:
\( \text{ppm} = \frac{\text{Mass of component}}{\text{Total mass of solution}} \times 10^{6} \)
\( = \frac{6 \times 10^{-3}}{1030 + 0.006} \times 10^{6} \approx \frac{6 \times 10^{-3}}{1030} \times 10^{6} \)
\( = 5.8 \) ppm

Teacher's Note:
a) Half a mark for the formula and half a mark for the answer.
b) The mass of oxygen (0.006 g) is so small that it can be ignored in the total mass.

26. (I) Convert the following;
a) Nitrobenzene to flourobenzene
b) Benzoic acid to benzylamine [2 Marks]

Answer:
a) C₆H₅NO₂ →(Sn/HCl) C₆H₅NH₂ →(NaNO₂/HCl) C₆H₅N₂⁺Cl^- →(NaBF₄) C₆H₅F
b) C₆H₅COOH + NH₃ →(heat) C₆H₅CONH₂ →(LiAlH₄) C₆H₅CH₂NH₂

Teacher's Note:
a) In (a), the nitro group is reduced to aniline, diazotised in the cold, and the diazonium salt is changed to the fluoride through the fluoroborate on heating (Balz-Schiemann reaction).
b) In (b), benzamide is reduced by LiAlH₄; the carbon count stays the same, so benzylamine forms.

(II) Aniline undergoes electrophilic substitution reactions readily. However, acetylation of aniline is required to obtain monosubstituted products. Give reason for your answer [1 Mark]

Answer: The -NH₂ group of aniline is strongly activating, so substitution gives 2,4,6-trisubstituted products. Acetylation converts -NH₂ into -NHCOCH₃, which is less activating because the lone pair on nitrogen is in resonance with the carbonyl group. This controls the reaction and gives mainly the para monosubstituted product.

Teacher's Note:
a) Example: aniline with bromine water gives 2,4,6-tribromoaniline at once.
b) After substitution, the acetyl group can be removed by hydrolysis to get back the -NH₂ group.

Attempt either A or B

27(A) (I) Write the IUPAC name of the following compound [1 Mark]

[Figure: A benzene ring carrying Br on one carbon and, on the next (ortho) carbon, a CH group bonded to CH₃ and CH₂CH₃, that is o-BrC₆H₄-CH(CH₃)CH₂CH₃.]

Answer: 1-Bromo-2-(1-methylpropyl)benzene

Teacher's Note:
a) The side chain -CH(CH₃)CH₂CH₃ is named 1-methylpropyl (also called butan-2-yl or sec-butyl).
b) Number the ring so that the substituents get the lowest locants (1 and 2) and write them in alphabetical order.

(II) When CH₃CH₂CH₂Cl reacts with AgCN and KCN, Compound "X" and "Y" respectively are formed. Identify "X" and "Y". Support your answer with appropriate reason. [2 Marks]

Answer:
1. X (with AgCN) = CH₃CH₂CH₂NC (propyl isocyanide).
2. Y (with KCN) = CH₃CH₂CH₂CN (propyl cyanide).
3. KCN is mainly ionic and gives free cyanide ions in solution. The cyanide ion is an ambident nucleophile and attacks through carbon, giving the cyanide (nitrile).
4. AgCN is mainly covalent, so the bond is formed through the nitrogen atom of the cyanide group, giving the isocyanide.

Teacher's Note:
a) Remember: KCN gives -CN (nitrile), AgCN gives -NC (isocyanide).
b) Half a mark each for X, Y and the two reasons.

OR

27(B) (I) What is the IUPAC nomenclature of the compound: (CH₃)₃C-C(CH₃)Br-CH₂CH₃ [1 Mark]

Answer: 3-bromo-2,2,3-trimethylpentane

Teacher's Note:
a) The longest chain has 5 carbons: C(CH₃)₃ gives C1 and C2, then C3 carries Br and CH₃, then C4 and C5.
b) Write prefixes in alphabetical order: bromo before methyl.

(II) Which of the following will be hydrolysed more readily?
CH₂=CH-CH₂Cl or (CH₃)₂CHCl
Support your answer with reason. [2 Marks]

Answer:
1. CH₂=CH-CH₂Cl (allyl chloride) will be hydrolysed more readily than (CH₃)₂CHCl.
2. The allyl carbocation (CH₂=CH-CH₂⁺) formed from it is more stable than the secondary carbocation (CH₃)₂CH⁺, because of resonance: the positive charge is delocalised over two carbon atoms.

Teacher's Note:
a) One mark for the correct compound and one mark for the resonance reason.
b) A more stable carbocation forms faster, so the S_N1 hydrolysis is faster.

28. (I) On adding 100 g of non-volatile solute A in 1000 g of water, the vapour gets reduced by 50 percent. Calculate the molar mass of solute A. [2 Marks]

Answer:
Relative lowering of vapour pressure: \( \frac{P^{\circ} - P_s}{P^{\circ}} = \chi_B = \frac{n_B}{n_A + n_B} \)
\( n_A \) (water) \( = \frac{1000}{18} = 55.56 \) mol
\( \frac{P^{\circ} - P_s}{P^{\circ}} = 0.5 \) (50% reduction)
\( 0.5 = \frac{n_B}{n_B + 55.56} \)
\( 0.5\,n_B + 0.5 \times 55.56 = n_B \)
\( 0.5\,n_B = 0.5 \times 55.56 \), so \( n_B = 55.56 \) mol
\( M_B = \frac{w_B}{n_B} = \frac{100}{55.56} = 1.8 \) g mol^-1

Teacher's Note:
a) With a large lowering (50%), do not use the dilute solution short cut \( \frac{n_B}{n_A} \); use \( \frac{n_B}{n_A + n_B} \) as the scheme does.
b) "Vapour" in the question means vapour pressure of water.

(II) It has been observed that nitric acid solution containing 32% of water by mass, cannot be separated into its components through fractional distillation. Why? [1 Mark]

Answer: Nitric acid containing 32% water by mass forms an azeotrope. An azeotrope has the same composition in the liquid and vapour phases, so it boils without change in composition and cannot be separated by fractional distillation.

Teacher's Note:
a) Keyword: azeotrope, with the same composition in liquid and vapour.
b) This nitric acid-water mixture (68% HNO₃) is a maximum boiling azeotrope, as it shows negative deviation.

Section D

Question No. 29 & 30 are case-based/data -based questions carrying 4 marks each.

29. Acidity of substituted phenols

Phenol and substituted phenols are relatively more acidic than aliphatic alcohols. The pKa value is a quantitative measure of the acidic strength. The effect of substituents on acidity of phenols (pKa values) is given in Figure 4.

[Figure: Figure 4 is a bar graph of pKa value (y-axis, 0 to 12) for substituted phenols: o-cresol 10.29, m-cresol 10.09, p-cresol 10.26, o-nitrophenol 7.23, m-nitrophenol 8.36, p-nitrophenol 7.2, 2,4-dinitrophenol 4.09, p-bromophenol 9.34, p-flourophenol 9.9, p-chlorophenol 9.38, phenol 10.]

Study the graph and answer the following questions

(I) The pKa value of 2,4,6- trinitrophenol will be: [1 Mark]
(A) more than 8.36
(B) between 7.2 and 8.3
(C) between 4.09 and 2.3
(D) less than 4.09

Answer: (D) less than 4.09

Teacher's Note:
a) From the graph, pKa falls from phenol (10) to p-nitrophenol (7.2) to 2,4-dinitrophenol (4.09) as nitro groups are added.
b) A third nitro group lowers the pKa further, so it must be less than 4.09.

(II) The order of acidity of p-halophenols is : [1 Mark]
(A) p-bromophenol \( \gt \) p-chlorophenol \( \gt \) p-flourophenol \( \gt \) phenol
(B) phenol \( \gt \) p-flourophenol \( \gt \) p-chlorophenol \( \gt \) p-bromophenol
(C) p-chlorophenol \( \gt \) p-bromophenol \( \gt \) p-flourophenol \( \gt \) phenol
(D) phenol \( \gt \) p-flourophenol \( \gt \) p-bromophenol \( \gt \) p-chlorophenol

Answer: (A) p-bromophenol \( \gt \) p-chlorophenol \( \gt \) p-flourophenol \( \gt \) phenol

Teacher's Note:
a) Lower pKa means stronger acid. From the graph: p-bromophenol 9.34, p-chlorophenol 9.38, p-flourophenol 9.9, phenol 10.
b) Arrange the compounds in increasing order of pKa to get decreasing order of acidity.

Attempt either (IIIA) or (IIIB).

(IIIA) Predict which one of the following two compounds will have lower value of pKa .
C₆H₁₁OH (cyclohexanol) OR C₆H₅OH (phenol)
Give reason in support of your reason. [2 Marks]

[Figure: Two structures: cyclohexanol (a cyclohexane ring with an OH group) and phenol (a benzene ring with an OH group).]

Answer:
1. Phenol will have the lower pKa value, because phenol is a stronger acid than cyclohexanol.
2. Phenol loses a hydrogen ion to form its conjugate base, the phenoxide ion, which is stabilised by resonance (the negative charge is delocalised over the ring).
3. In cyclohexanol, the conjugate base (cyclohexoxide ion) is less stable because the negative charge stays localised on the oxygen atom.

Teacher's Note:
a) One mark for choosing phenol and one mark for the resonance stabilisation of phenoxide ion.
b) Remember: more stable conjugate base means stronger acid and lower pKa.

OR

(IIIB) Predict whether 4 chloro-2, 6- dintirophenol is more or less acidic than 2,4,6-trinitrophenol? Use the values in the graph to support your answer. [2 Marks]

Answer:
1. From the graph, the pKa of p-nitrophenol is 7.2 while that of p-chlorophenol is 9.38. So a nitro group lowers pKa much more than a chloro group, and the more the nitro groups, the lower the pKa.
2. Therefore the pKa of 4-chloro-2,6-dinitrophenol is higher than that of 2,4,6-trinitrophenol. Higher pKa means lower acidic strength, so 4-chloro-2,6-dinitrophenol is less acidic than 2,4,6-trinitrophenol.

Teacher's Note:
a) Quote the graph values (7.2 and 9.38) as the question asks you to use them.
b) The two compounds differ only at the para position: Cl in one and NO₂ in the other.

For Visually Impaired Candidates (in lieu of Q. 29)

Read the passage and answer the questions that follow

The presence of electron withdrawing groups such as nitro group, enhances the acidic strength of phenol. It is due to the effective delocalisation of negative charge in phenoxide ion. On the other hand, electron releasing groups, such as alkyl groups, in general, do not favour the formation of phenoxide ion resulting in decrease in acid strength. Cresols, for example, are less acidic than phenol. The pKa value is a quantitative measure of the acidic strength. Higher the pKa value lower is the acidic strength.

(I) Which of the following will form least stable phenoxide ion? [1 Mark]
(A) phenol
(B) o-cresol
(C) o-nitrophenol
(D) o-chlorophenol

Answer: (B) o-cresol

Teacher's Note:
a) The methyl group is electron releasing, so it increases the negative charge on the phenoxide ion and makes it less stable.
b) Nitro and chloro groups withdraw electrons and stabilise the phenoxide ion.

(II) Which group causes the maximum increase in pKa value of phenol: [1 Mark]
(A) - CH₃
(B) - OCH₃
(C) - NO₂
(D) - Cl

Answer: (B) - OCH₃

Teacher's Note:
a) Electron releasing groups decrease acidity and so increase the pKa of phenol; -NO₂ and -Cl are electron withdrawing and lower the pKa.
b) The marking scheme gives -OCH₃ as the answer, as it is an electron donating group.

Attempt either (IIIA) or (IIIB)

(IIIA). Which of the following will be more acidic?
Nitrophenol or 2,4-Dinitrophenol. Support your answer with appropriate reason? [2 Marks]

Answer:
1. 2,4-Dinitrophenol is more acidic.
2. The nitro group is a strong electron withdrawing group. It shows both -I and -M effects and withdraws electron density from the ring, so the phenoxide ion formed is stabilised.
3. Two nitro groups make the phenoxide ion more stable than one nitro group does, so 2,4-dinitrophenol is the stronger acid.

Teacher's Note:
a) One mark for the correct compound and one mark for the reason based on -I and -M effects.
b) More electron withdrawing groups at ortho and para positions mean greater acidity.

OR

(IIIB). Predict whether pka value of p-cresol or p-nitrophenol will be higher or lower than phenol. Give reason to support your answer. [2 Marks]

Answer:
1. The pKa of p-cresol will be higher than that of phenol, because the methyl group is electron donating and decreases acidity.
2. The pKa of p-nitrophenol will be lower than that of phenol, because the nitro group is electron withdrawing and increases acidity.

Teacher's Note:
a) Each part has half a mark for the prediction and half a mark for the reason.
b) Remember: higher pKa means weaker acid.

30. Identification of carbohydrate

Students of class 12 are given a sample of carbohydrate obtained from a plant extract. They conducted the following tests to identify the carbohydrate.
The students find that the given carbohydrate is soluble in water. They take about 1-2 mL of aqueous solution of the carbohydrate in a test tube and add 2 mL of Benedict's reagent. The test tube is heated in water bath for 3-5 minutes. A brick red precipitate is obtained.
The sample of the given carbohydrate on hydrolysis with dilute acid, gives only one type of monosaccharide.

Read the passage and answer the following questions:

(I) The sample gives a positive Benedict's test. We can conclude that the given carbohydrate is a [1 Mark]
(A) monosaccharide
(B) reducing sugar
(C) non-reducing sugar
(D) polysaccharide

Answer: (B) reducing sugar

Teacher's Note:
a) Reducing sugars give a brick red precipitate (Cu₂O) with Benedict's solution.
b) A positive test does not prove it is a monosaccharide, since some disaccharides are also reducing.

(II) Which of the following is plant based carbohydrate?
(i) sucrose (ii)lactose (iii) starch (iv) glycogen [1 Mark]
(A) (i) and (ii)
(B) (i) and (iii)
(C) (iii) and (iv)
(D) (i) and (iv)

Answer: (B) (i) and (iii)

Teacher's Note:
a) Sucrose and starch are plant based carbohydrates.
b) Lactose is milk sugar and glycogen is animal starch, so both come from animals.

Attempt either (IIIA) or (IIIB)

(IIIA) Considering all observations reported in the passage the sample is most likely (i) glucose (ii) fructose (iii) sucrose (iv) lactose (v) maltose. Justify your answer. [2 Marks]

Answer:
1. The sample is maltose.
2. It gives a positive Benedict's test, so it is a reducing sugar; hence it is not sucrose.
3. It undergoes hydrolysis, so it is not a monosaccharide; hence it is not glucose or fructose.
4. It gives only one type of monosaccharide on hydrolysis. Lactose gives glucose and galactose, while maltose gives two glucose molecules. Therefore it is maltose.

Teacher's Note:
a) One mark for maltose and one mark for ruling out the other sugars using all three observations.
b) Use elimination: reducing (not sucrose), hydrolysable (not glucose or fructose), one monosaccharide (not lactose).

OR

(IIIB) Ravi reports that the given sample is starch. What could be his justification, based on the observations in the passage? Do you agree with Ravi? Why or why not? [2 Marks]

Answer:
1. Ravi's justification: the carbohydrate gives only one monosaccharide on hydrolysis, and starch on hydrolysis gives only glucose.
2. No, I do not agree. The carbohydrate is soluble in water, while starch is insoluble in water.
3. Also, starch is not a reducing sugar, but the sample gives a positive Benedict's test.

Teacher's Note:
a) One mark for the justification and one mark for disagreeing with correct reasons.
b) Polysaccharides like starch are non-reducing and do not dissolve in water.

Section-E

Question No. 31 to 33 are long answer type questions carrying 5 marks each.

Attempt either A or B

31(A) (I) Account for the following
a. The pH should be around 3.5 during the addition of ammonia derivative compounds to aldehyde/ ketone.
b. Alcohols are added to carbonyl compounds in the presence of dry HCl gas.
c. The reaction of aldehydes with sodium hydrogen sulphite is commonly used for the separation and purification of aldehydes. [3 Marks]

Answer:
a. In a strongly acidic medium, the ammonia derivatives get protonated, which reduces their nucleophilicity. In a basic medium, the carbonyl compound is not protonated, so its electrophilicity for nucleophilic attack stays low. Hence a mildly acidic pH of about 3.5 gives the best rate.
b. Dry HCl gas protonates the oxygen of the carbonyl compound. This increases the electrophilicity of the carbonyl carbon and helps the nucleophilic attack of the alcohol.
c. The hydrogen sulphite addition compound is water soluble and can be converted back into the original aldehyde by treating it with dilute mineral acid or alkali. So this reaction is used to separate and purify aldehydes.

Teacher's Note:
a) For part a, explain both extremes: too acidic and basic medium.
b) Key words: protonation increases electrophilicity; the bisulphite compound is reversible and water soluble.

(II) A student has three colourless samples: methanol, acetaldehyde, and acetic acid. They have only NaHCO₃, and 2,4-DNP reagent
Devise a flowchart to identify each compound using only these two reagents. [2 Marks]

Answer:
Flowchart:
Step 1: Methanol, Acetaldehyde and Acetic acid → Add NaHCO₃
(a) Brisk effervescence of CO₂ → the sample is acetic acid.
(b) No brisk effervescence of CO₂ → the sample is either methanol or acetaldehyde.
Step 2: To the samples in (b) → Add 2,4-DNP reagent
(a) Orange yellow precipitate → the sample is acetaldehyde.
(b) No orange yellow precipitate → the sample is methanol.

Teacher's Note:
a) Carboxylic acids react with NaHCO₃ to give CO₂; alcohols and aldehydes do not.
b) 2,4-DNP gives an orange yellow precipitate only with aldehydes and ketones.
c) Draw the chart as two branching steps; each branch carries half a mark.

OR

31(B) (I) Arrange the following according to the property indicated
a) CH₃CHO, HCHO, C₆H₅CHO (decreasing order of the ease of oxidation)
b) CH₃COCl, CH₃COOH, CH₃CONH₂, CH₃COOCH₃ (increasing order of nucleophilic acyl substitution)
c) CH₃CHO, CH₃COCH₃, CF₃CHO (increasing order of electrophilicity of carbonyl carbon) [3 Marks]

Answer:
a) HCHO \( \gt \) CH₃CHO \( \gt \) C₆H₅CHO
b) CH₃CONH₂ \( \lt \) CH₃COOH \( \lt \) CH₃COOCH₃ \( \lt \) CH₃COCl
c) CH₃COCH₃ \( \lt \) CH₃CHO \( \lt \) CF₃CHO

Teacher's Note:
a) Read the order asked carefully: (a) is decreasing, while (b) and (c) are increasing.
b) In (c), the -CF₃ group strongly withdraws electrons and makes the carbonyl carbon most electrophilic, while two alkyl groups in acetone make it least electrophilic.

(II) Write any two possible isomers of the compound having molecular formula C₃H₆O. Identify which of these isomer/s will give a positive Tollen's test and explain the reason. [2 Marks]

Answer:
1. Propanal: CH₃-CH₂-CHO (an aldehyde).
2. Propanone: CH₃-CO-CH₃ (a ketone).
3. Propanal gives a positive Tollens' test (silver mirror).
4. Reason: Tollens' reagent oxidises aldehydes to carboxylate ions and itself is reduced from Ag⁺ to metallic silver. Ketones do not give this test.

Teacher's Note:
a) Half a mark for each isomer, half a mark for naming propanal, and half a mark for the reason.
b) Propanal and propanone are functional isomers of each other.

Attempt either A or B

32(A) Answer the following questions:

(I) Two test tubes, X and Y, contain solutions of the complexes [Cr(H₂O)₃Cl₃] and [Cr(en)₃]³⁺ respectively. If plane-polarized light is passed through these solutions, which solution will rotate the plane of polarization and why? [1 Mark]

Answer: Solution Y, [Cr(en)₃]³⁺, will rotate the plane of polarisation. Its three bidentate ethylenediamine ligands form a chiral octahedral arrangement, so it is optically active. [Cr(H₂O)₃Cl₃] exists as fac and mer isomers, but both have a plane of symmetry, so it is not optically active.

Teacher's Note:
a) Optical activity needs a chiral structure with no plane of symmetry.
b) [M(AA)₃] type complexes exist as d and l forms.

(II) [Ni(CO)₄] and [Ni(H₂O)₆]²⁺ are two complexes of Ni. Which complex will have higher crystal field splitting energy, and which factor determines this difference? [1 Mark]

Answer: [Ni(CO)₄] will have the higher crystal field splitting, because CO is a strong field ligand while H₂O in [Ni(H₂O)₆]²⁺ is a weak field ligand. The nature of the ligand (strong field or weak field) determines this difference.

Teacher's Note:
a) In the spectrochemical series, CO is near the strong field end and H₂O is a weaker ligand.
b) Half a mark for the complex and half a mark for the factor.

(III) A solution of the coordination compound PdCl₂ .4NH₃ is treated with AgNO₃, and 2 mole of AgCl precipitate form. Determine the primary valency and secondary valency of Pd in the compound. [1 Mark]

Answer: The compound is [Pd(NH₃)₄]Cl₂. Primary valency (oxidation number) of Pd = 2. Secondary valency (coordination number) of Pd = 4, satisfied by the four NH₃ ligands.

Teacher's Note:
a) 2 mol AgCl means both Cl^- ions are outside the coordination sphere.
b) Primary valency is ionisable; secondary valency is non-ionisable.

(IV) The complex [Fe(CN)₆]^4- exhibits d²sp³ hybridisation. Based on this, comment on the strength of the ligand CN^- and its effect on the geometry of the complex. [1 Mark]

Answer: CN^- is a strong field ligand. It pairs up the 3d electrons of Fe²⁺, leaving two inner 3d orbitals free for d²sp³ hybridisation, which gives a low spin octahedral complex.

Teacher's Note:
a) d²sp³ (inner orbital) hybridisation always gives octahedral geometry.
b) Fe²⁺ (3d⁶) with CN^- has all electrons paired, so the complex is diamagnetic.

(V) Why is anhydrous CuSO₄ white, but CuSO₄ .5H₂O blue in colour? [1 Mark]

Answer: CuSO₄.5H₂O is the complex [Cu(H₂O)₄]SO₄.H₂O, in which water acts as a ligand and causes crystal field splitting; d-d transitions then give the blue colour. In anhydrous CuSO₄ there is no water ligand, so crystal field splitting does not occur and it is white (colourless).

Teacher's Note:
a) Keyword: in the absence of ligands, crystal field splitting does not occur.
b) Heating blue CuSO₄.5H₂O removes the water and makes it white.

OR

32(B) Answer the following questions:

(I) Three coordination compounds of cobalt(III) are given:
Compound A: [Co(NH₃)₆]Cl₃
Compound B: [Co(NH₃)₅Cl]Cl₂
Compound C: [Co(NH₃)₄Cl₂]Cl
Which of the three compounds will show maximum electrolytic conductivity and why? [1 Mark]

Answer: Compound A shows the maximum conductivity, because it gives the most ions in solution.
A → [Co(NH₃)₆]³⁺ + 3Cl^- (4 ions)
B → [Co(NH₃)₅Cl]²⁺ + 2Cl^- (3 ions)
C → [Co(NH₃)₄Cl₂]⁺ + Cl^- (2 ions)

Teacher's Note:
a) The more the number of ions in solution, the higher the conductivity.
b) Only Cl^- ions outside the square bracket ionise.

(II) On the basis of crystal field theory, write the electronic configuration for d⁵ ion if \( \Delta_0 \gt P \). [1 Mark]

Answer: \( t_{2g}^{5}\,e_g^{0} \). The strong field causes pairing in the t_2g orbitals, and only one unpaired electron remains.

Teacher's Note:
a) When \( \Delta_0 \gt P \), it costs less energy to pair electrons than to put them in e_g.
b) If \( \Delta_0 \lt P \), the configuration would be \( t_{2g}^{3}\,e_g^{2} \) with five unpaired electrons.

(III) Using IUPAC norms, write the formulae of Diaquadiammine (oxalato) chromium(III) chloride. [1 Mark]

Answer: [Cr(NH₃)₂(H₂O)₂(C₂O₄)]Cl

Teacher's Note:
a) Charge check: Cr(+3) + oxalate(-2) + neutral NH₃ and H₂O = +1, so one Cl^- is needed.
b) In the formula, write ligands in alphabetical order of their symbols after the metal.

(IV) Compare the optical activity of [Co(NH₃)₂Cl₂(en)]Cl and [Co(NH₃)₄Cl₂]Cl. Which one is optically active and why? [1 Mark]

Answer: [Co(NH₃)₂Cl₂(en)]Cl is optically active; [Co(NH₃)₄Cl₂]Cl is not. The bidentate ligand en can produce a chiral (non-superimposable) octahedral arrangement with no mirror plane, giving optical isomers. Complexes with only monodentate ligands of the MA₄B₂ type are symmetric and achiral.

Teacher's Note:
a) A complex is optically active only if its mirror image cannot be superimposed on it.
b) MA₄B₂ complexes show only geometrical (cis-trans) isomerism.

(V) Explain why [Cu(NH₃)₄]²⁺ shows a blue colour while [Cu(H₂O)₄]²⁺ shows a greenish colour. Which factor influences the colour difference? [1 Mark]

Answer: The different ligands (NH₃ and H₂O) produce different crystal field splitting (\( \Delta \)), so the complexes absorb different wavelengths of light and show different colours. The nature (strength) of the ligand decides the crystal field splitting and so the energy of the d-d transition, which affects the colour.

Teacher's Note:
a) NH₃ is a stronger field ligand than H₂O, so it gives a larger splitting.
b) The colour seen is the complementary colour of the light absorbed.

Attempt either A or B

33(A) (I) A student investigates the effect of concentration on the cell potential of a Zn-Cu galvanic cell at 298 K.
The cell is: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq)|Cu(s)
standard electrode potentials:
E⁰(Zn²⁺/Zn)=-0.76 V, E⁰Cu²⁺/Cu=+0.34 V, E^ocell=1.10 V
The following experimental data were collected:
Experiment | [Zn²⁺] (M) | [Cu²⁺] (M) | Measured EMF (V)
1 | 1.00 | 1.00 | 1.10
2 | 0.1 | 1.00 | 1.16
3 | 1.00 | 0.10 | 1.04
4 | 0.01 | 1.00 | 1.22
5 | 1.00 | 0.01 | 0.98

a) Why does experiment 4 show the maximum EMF? [1 Mark]

Answer: By the Nernst equation, \( E_{cell} = E^{\circ}_{cell} - \frac{0.059}{2}\log Q \), where \( Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} \). The cell potential increases when Q decreases. In experiment 4, [Zn²⁺] = 0.01 M is very low and [Cu²⁺] = 1.00 M is high, so Q is very small, log Q is very negative, and the term \( -\frac{0.059}{2}\log Q \) is large and positive. This gives the highest EMF.

Teacher's Note:
a) Lower product ion concentration (Zn²⁺) and higher reactant ion concentration (Cu²⁺) push the cell potential up.
b) The marking scheme prints Q as [Zn²⁺][Cu²⁺]; the correct expression is the ratio [Zn²⁺]/[Cu²⁺].

b) For experiment 3, calculate the theoretical EMF using the Nernst equation. Suggest one possible reason why the measured EMF differs slightly from the calculated value. [2 Marks]

Answer:
\( E_{cell} = E^{\circ}_{cell} - \frac{0.059}{2}\log \frac{[Zn^{2+}]}{[Cu^{2+}]} \)
\( = 1.10 - \frac{0.059}{2}\log \frac{1.00}{0.10} \)
\( = 1.10 - 0.02955 \log 10 \)
\( = 1.10 - 0.02955 \)
\( = 1.07045 \) V \( \approx 1.07 \) V
Reason: when current flows through the cell, it faces some internal resistance (of the electrolyte, electrodes or salt bridge), which lowers the observed cell voltage (measured value 1.04 V).

Teacher's Note:
a) n = 2 for the Zn-Cu cell; do not forget to divide 0.059 by 2.
b) Half a mark each for the formula, substitution, final value with unit, and the reason.

(II) If k is expressed in Sm^-1 and the concentration, c in mol m^-3 then the units of molar conductivity are in Sm²mol^-1. However, if we use Scm^-1 as the units for k and molcm^-3, the units of concentration, then the units of molar conductivity are in Scm²mol^-1. Show how 1 Sm²mol^-1 is related to 1S cm²mol^-1 [1 Mark]

Answer:
\( \Lambda_m \) (S cm² mol^-1) \( = \frac{\kappa\ (\text{S cm}^{-1}) \times 1000\ (\text{cm}^3/\text{L})}{\text{molarity}\ (\text{mol/L})} \)
Since 1 m = 100 cm, 1 m² = 10⁴ cm².
So 1 S m² mol^-1 = 10⁴ S cm² mol^-1, or 1 S cm² mol^-1 = 10^-4 S m² mol^-1.

Teacher's Note:
a) Both units are used in books; always check which one a question uses.
b) Converting m² to cm² means multiplying by \( (100)^2 = 10^4 \).

(III) What happens to the specific gravity of electrolyte in lead storage battery during charging and discharging of battery? [1 Mark]

Answer: During charging, the specific gravity of the electrolyte (sulphuric acid) increases, while during discharging it decreases.

Teacher's Note:
a) On discharging, H₂SO₄ is used up to form PbSO₄, so the acid becomes dilute.
b) On charging, PbSO₄ is changed back and H₂SO₄ is formed again.

OR

33(B) I. For aqueous NaCl at 25 °C the molar conductivities at two concentrations are measured as:
\( \Lambda_m(0.010\ M) = 123.9 \) S cm² mol^-1
\( \Lambda_m(0.040\ M) = 120.7 \) S cm² mol^-1
Molar conductivity for strong electrolytes can be represented as \( \Lambda_m = \Lambda^{\circ}_m - A\sqrt{c} \). Determine \( \Lambda^{\circ} \) (molar conductivity at infinite dilution) and the constant A for NaCl solution. [3 Marks]

Answer:
For \( c_1 = 0.010 \) M: \( 123.9 = \Lambda^{\circ}_m - A\sqrt{0.010} = \Lambda^{\circ}_m - 0.100A \) ... (1)
For \( c_2 = 0.040 \) M: \( 120.7 = \Lambda^{\circ}_m - A\sqrt{0.040} = \Lambda^{\circ}_m - 0.200A \) ... (2)
Subtract (2) from (1): \( 123.9 - 120.7 = -0.100A - (-0.200A) \)
\( 3.2 = 0.100A \)
\( A = \frac{3.2}{0.100} = 32.0 \) S cm² mol^-1 M^-1/2
Substitute A in (1): \( 123.9 = \Lambda^{\circ}_m - 32 \times 0.100 \)
\( \Lambda^{\circ}_m = 123.9 + 3.2 = 127.1 \) S cm² mol^-1

Teacher's Note:
a) \( \sqrt{0.010} = 0.1 \) and \( \sqrt{0.040} = 0.2 \); use square roots of concentration, not the concentration itself.
b) Check: \( 127.1 - 32 \times 0.2 = 120.7 \), which matches the second value.

II. Write the electrode reactions that occur in a hydrogen-oxygen fuel cell. Also, state two advantages of fuel cells compared to conventional batteries. [2 Marks]

Answer:
Anode: 2H₂ → 4H⁺ + 4e^-
Cathode: O₂ + 4H⁺ + 4e^- → 2H₂O
Overall reaction: 2H₂ + O₂ → 2H₂O
Advantages:
1. Higher efficiency: fuel cells convert chemical energy into electrical energy more efficiently than many conventional batteries.
2. Continuous operation: as long as hydrogen and oxygen are supplied, fuel cells keep working, unlike batteries which need recharging.

Teacher's Note:
a) Oxidation of H₂ takes place at the anode and reduction of O₂ at the cathode.
b) Half a mark each for the electrode reactions and for each advantage.