CBSE Class 12 Chemistry: sample paper 2026-27

1. Among isomeric amines, H-bonding decides boiling point: primary (two N–H) > secondary (one N–H) > tertiary (none). So R (350.8 K) is primary, P (329.3 K) is secondary and Q (310.5 K) is tertiary.

2. A primary amine reacts with Hinsberg reagent ($C_6{H}_{5}S{O}_{2}Cl$) to give an N-alkylbenzenesulphonamide. Its N–H is acidic, so the white precipitate dissolves in NaOH. Statement B is therefore correct.

   1 mark

3. The other statements are wrong. A: P has the lower pKb, so P is more basic than R, not less. C: tertiary Q cannot H-bond as well as R, so it is less soluble in water. D: the Gabriel process gives only primary amines.

Answer: Option (B)

1. In $[NiC{l}_4{]}^{2-}$, nickel is Ni²⁺ ($3d^8$). Cl⁻ is a weak field ligand, so the 3d electrons do not pair and the 3d orbitals are not freed for bonding.

2. The metal therefore uses $s{p}^3$ hybridisation and the complex is tetrahedral and paramagnetic. For comparison, $[Ni(CN{)}_4{]}^{2-}$ and $[Cu(N{H}_3{)}_4{]}^{2+}$ are square planar ($ds{p}^2$), and $[Co(en{)}_3{]}^{3+}$ is octahedral.

   1 mark

Answer: Option (C) $[NiC{l}_4{]}^{2-}$

1. In $M_2{O}_3$ the metal is present as $M^{3+}$.

2. The reduction is $M^{3+}+3e^{-}\to M$. One mole of metal needs 3 mol of electrons, which is 3F of charge.

   1 mark

Answer: 3F, option (C)

1. A foul-smelling product with $CHC{l}_3$ and alcoholic KOH is an isocyanide (carbylamine test). Only primary amines give this test, so X must be a primary amine.

2. Reduction of nitrobenzene gives aniline, which is a primary amine. The other options give an isocyanide (A), a nitrile (C) or an amide (D), none of which is a primary amine. So the answer is B.

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Answer: Option (B)

1. By Henry's law, the solubility of a gas is proportional to its partial pressure. Compressing the gas raises the pressure, so more gas dissolves.

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2. Dissolution and escape of gas molecules continue at equal rates once the new state is reached, so the equilibrium is dynamic, not static.

Answer: Option (A)

1. (i) With comparable amounts of both reactants, the rate depends on both: rate $=k[C{H}_{3}COCl][H_{2}O]$. This is second order.

2. (ii) With water in large excess, $[H_{2}O]$ stays practically constant. The rate becomes $k^{\prime }[C{H}_{3}COCl]$, a pseudo first order reaction.

   1 mark

Answer: Option (D)

1. The molecule (glutamic acid) has two –COOH groups and one –NH₂ group, so it is an acidic amino acid.

2. Because the body can synthesise it, it is non-essential. So the answer is A.

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Answer: Option (A)

1. 3-Methylpentan-3-ol is $(C_2{H}_5{)}_{2}C(OH)C{H}_3$, a tertiary alcohol, so it must come from a ketone and a Grignard reagent.

2. (i) $C_2{H}_{5}CO{C}_2{H}_5+C{H}_{3}MgCl$ and (ii) $C_2{H}_{5}COC{H}_3+C_2{H}_{5}MgCl$ both give $(C_2{H}_5{)}_{2}C(OH)C{H}_3$ after hydrolysis. Options (iii) and (iv) use aldehydes, which give secondary alcohols. So the answer is D.

   1 mark

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

1. $M{n}^{3+}$ ($3d^4$) readily gains an electron to form $M{n}^{2+}$, which has the extra-stable half-filled $3d^5$ configuration.

2. Its high $E^{\circ }(M{n}^{3+}/M{n}^{2+})\approx +1.57$ V makes it a strong oxidising agent. So the answer is Mn.

   1 mark

Answer: Option (B) Mn

1. 2-Bromohexane is a secondary halide. Its reaction with OH⁻ here gives a single optically active product.

2. Formation of only one enantiomer, with no racemisation, shows backside attack by OH⁻. This is the $S_{N}2$ mechanism, which gives inversion of configuration. So the answer is A.

   1 mark

Answer: Option (A)

1. For a first order reaction, 75% completion means 25% remains, which is $(1/2{)}^2$ of the initial amount. This takes two half-lives.

2. Time $=2\times t_{1/2}=2\times 16=32$ min.

   1 mark

Answer: 32 min, option (B)

1. HCl is a strong electrolyte and is completely ionised, so at the same concentration it gives the most ions.

2. $C{H}_{3}COOH$ and $N{H}_{4}OH$ are weak electrolytes and only partly ionised. Glucose is a non-electrolyte and gives no ions. So 0.1 M HCl conducts best.

   1 mark

Answer: Option (A) 0.1 M HCl

1. A is true. In the strongly acidic nitrating mixture, aniline is protonated to the anilinium ion, $-N{H}_3^{+}$, which is meta directing. About 47% m-nitroaniline is formed.

2. R is true as a general fact: free –NH₂ is an activating, o/p-directing group. However, it does not explain why the meta product forms. So the answer is B.

   1 mark

Answer: Option (B)

1. A is true: both $Mn{O}_4^{-}$ and $Mn{O}_4^{2-}$ are tetrahedral.

2. R is false: the π bonding in these ions arises from overlap of oxygen p orbitals with manganese d orbitals, not p orbitals. So the answer is C.

   1 mark

Answer: Option (C)

1. A is true: lower ethers such as diethyl ether are about as miscible with water as butan-1-ol, an alcohol of the same molecular mass.

2. R is true: the ether oxygen accepts hydrogen bonds from water molecules, and this is exactly why ethers mix with water. So R correctly explains A.

   1 mark

Answer: Option (A)

1. $C{o}^{3+}$ has a $3d^6$ configuration. NH₃ is a strong field ligand, so all six electrons pair in the $t_{2g}$ set, giving $t_{2g}^6{e}_g^0$ with $d^{2}s{p}^3$ hybridisation.

2. With no unpaired electrons the complex is diamagnetic. The pairing caused by the strong field NH₃ is the reason, so R correctly explains A.

   1 mark

Answer: Option (A)

1. (I) In acidic medium $Mn{O}_4^{-}$ (Mn +7) is reduced to $M{n}^{2+}$ (gain of 5e⁻). $S{O}_3^{2-}$ (S +4) is oxidised to $S{O}_4^{2-}$ (loss of 2e⁻). Equalising electrons and balancing O and H gives:
   $$5S{O}_3^{2-}+2Mn{O}_4^{-}+6H^{+}\to 5S{O}_4^{2-}+2M{n}^{2+}+3H_{2}O$$

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2. (II) Ti ($3d^{2}4s^2$) has few unpaired d electrons. Cr ($3d^{5}4s^1$) has six unpaired electrons. More unpaired electrons take part in metallic bonding, so Cr has stronger interatomic attraction. More energy is needed to break the Cr lattice, so titanium melts at a lower temperature.

   1 mark

Answer: (I) $5S{O}_3^{2-}+2Mn{O}_4^{-}+6H^{+}\to 5S{O}_4^{2-}+2M{n}^{2+}+3H_{2}O$; (II) Cr has more unpaired d electrons, so its metallic bonding is stronger than in Ti.

1. (I) Ti and Sc ions are small and carry a high charge, and they have vacant d orbitals of suitable energy to accept lone pairs from ligands. K and Ca have no vacant d orbitals in use, and their ions are larger with low charge. So Ti and Sc form complexes, but K and Ca do not.

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2. (II) $C{r}_2{O}_7^{2-}$ (Cr +6) gains 6e⁻ to give $2C{r}^{3+}$. Each $F{e}^{2+}$ loses 1e⁻. So:
   $$C{r}_2{O}_7^{2-}+14H^{+}+6F{e}^{2+}\to 2C{r}^{3+}+6F{e}^{3+}+7H_{2}O$$

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Answer: (I) Small size, high charge and available vacant d orbitals of Ti/Sc ions; (II) $C{r}_2{O}_7^{2-}+14H^{+}+6F{e}^{2+}\to 2C{r}^{3+}+6F{e}^{3+}+7H_{2}O$

1. The statement is true only under a certain condition: when the reaction is zero order. For zero order, rate $=k[R{]}^0=k$, so the rate does not depend on reactant concentration and stays constant as the reaction proceeds.

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2. It is not always true. For reactions of order 1, 2, etc., rate $=k[R{]}^n$ with $n>0$. As reactants are used up, [R] falls, so the rate keeps decreasing during the course of the reaction.

   1 mark

Answer: True only for zero order reactions (rate = k). For other orders the rate decreases as concentration falls.

1. (I) In DNA, G pairs with C through three hydrogen bonds and A pairs with T through two hydrogen bonds. So the missing bases are:

   • ?≡C → G (guanine)
   • A=? → T (thymine)
   • ?=A → T (thymine)
   • A=… → T

   The complementary strand runs antiparallel (3'→5').

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2. (II) Lemon juice (citric acid) lowers the pH of milk. This breaks the hydrogen bonds that hold the secondary and tertiary structure of casein. The globular casein unfolds and coagulates, i.e. it is denatured. The primary structure stays intact, and the milk curdles.

   1 mark

Answer: (I) G opposite C; T opposite each A; (II) Casein is denatured: its H-bonds break, the globules unfold and the milk curdles.

1. The nitro group increases the reactivity of chlorobenzene towards nucleophilic substitution.

   1 mark

2. $-N{O}_2$ is strongly electron-withdrawing (–I and –R effects). It lowers the electron density of the ring and favours attack by $O{H}^{-}$. The methoxy group is electron-donating (+R), so it does the opposite.

   ½ mark

3. With $-N{O}_2$ at the ortho or para position, the negative charge of the carbanion (σ-complex) formed after $O{H}^{-}$ attacks is delocalised onto the nitro group by resonance. This stabilises the intermediate, so substitution occurs readily (even at much milder conditions).

   ½ mark

Answer: (i) Nitro, because its –I/–R effect withdraws electrons and stabilises the carbanion intermediate at o/p positions.

1. (I) Mg is more reactive (more negative $E^{\circ }$) than Fe, so it acts as a sacrificial anode and is oxidised in place of iron: $Mg\to M{g}^{2+}+2e^{-}$. The electrons flow to the iron pipe, which becomes the cathode and is not oxidised. This prevents corrosion. The Mg block is replaced from time to time.

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2. (II) Cathode: $M{g}^{2+}+2e^{-}\to Mg(l)$

   ½ mark

3. Anode: $2C{l}^{-}\to C{l}_2(g)+2e^{-}$

   ½ mark

Answer: (I) Sacrificial protection: Mg oxidises instead of Fe; (II) Cathode: $M{g}^{2+}+2e^{-}\to Mg(l)$; Anode: $2C{l}^{-}\to C{l}_2(g)+2e^{-}$

1. (I) Configurations: $T{i}^{2+}$: $3d^2$ (2 unpaired); $C{o}^{2+}$: $3d^7$ (3 unpaired); $F{e}^{2+}$: $3d^6$ (4 unpaired).

   ½ mark

2. $F{e}^{2+}$ has the most unpaired electrons, so it has the highest magnetic moment: $\mu =\sqrt{n(n+2)}=\sqrt{24}=4.90$ BM.

   ½ mark

3. (II)(a) Iron can show variable oxidation states and offers a large surface area for adsorption of $N_2$ and $H_2$. These features help it catalyse ammonia formation in Haber's process.

   1 mark

4. (II)(b) The iron crystal lattice has interstitial voids large enough to trap small non-metal atoms such as O, C or N. This gives interstitial compounds.

   1 mark

Answer: (I) $F{e}^{2+}$ (4 unpaired e⁻, μ ≈ 4.90 BM); (II)(a) variable oxidation states and large surface area; (b) interstitial voids in the lattice that fit small atoms.

1. Arrhenius equation in log form: $\log k=\log A-\frac{E_a}{2.303RT}$.

   ½ mark

2. Comparing with $y=mx+c$ for a plot of $\log k$ vs $1/T$: slope $=-\frac{E_a}{2.303R}$.

   ½ mark

3. Slope from points 1 and 2: $\frac{-4.46-(-3.91)}{0.00333-0.00323}=\frac{-0.55}{0.0001}=-5500$ K

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4. $-\frac{E_a}{2.303R}=-5500$

   $E_a=5500\times 2.303\times 8.314=105307{\text{J mol}}^{-1}\approx 105.3{\text{kJ mol}}^{-1}$

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Answer: $E_a\approx 105.3$ kJ mol⁻¹ (slope = −5500 K)

1. (I) Aldol condensation (aldol addition). Missing reagent: dilute NaOH. Two molecules of ethanal give 3-hydroxybutanal.

   1 mark

2. (II) Etard reaction. Missing reactant: **chromyl chloride, $Cr{O}_{2}C{l}_2$**. It forms the chromium complex, which is hydrolysed to benzaldehyde.

   1 mark

3. (III) Wolff–Kishner reduction. Missing intermediate: the **hydrazone, $>C=N-N{H}_2$**. Heating it with KOH in ethylene glycol gives $>C{H}_2$ and $N_2$.

   1 mark

Answer: (I) Aldol, dil. NaOH; (II) Etard, $Cr{O}_{2}C{l}_2$; (III) Wolff–Kishner, $>C=N–N{H}_2$ (hydrazone)

1. (I)(a) The volume of the solution is slightly more than 20 mL (${\Delta }_{mix}V>0$).

   ½ mark

2. The acetone–CS₂ attractions are weaker than the acetone–acetone and CS₂–CS₂ attractions. Molecules move apart, so volume increases (positive deviation from Raoult's law).

   ½ mark

3. (I)(b) With chloroform, the volume is slightly less than 20 mL (${\Delta }_{mix}V<0$).

   ½ mark

4. Chloroform forms hydrogen bonds with acetone ($C{l}_{3}C-H\cdots O=C(C{H}_3{)}_2$). The stronger A–B attractions pull molecules closer, so volume decreases (negative deviation).

   ½ mark

5. (II) ppm $=\frac{\text{mass of }{O}_2}{\text{mass of solution}}\times 10^6=\frac{6\times 10^{-3}}{1030}\times 10^6$

   ½ mark

6. $=5.8$ ppm

   ½ mark

Answer: (I)(a) Volume > 20 mL (weaker interactions, positive deviation); (b) volume < 20 mL (H-bonding, negative deviation); (II) ≈ 5.8 ppm

1. (I)(a) $C_6{H}_{5}N{O}_2\stackrel{Sn/HCl}{\to }{C}_6{H}_{5}N{H}_2\stackrel{NaN{O}_2/HCl,273-278K}{\to }{C}_6{H}_5{N}_2^{+}C{l}^{-}\stackrel{HB{F}_4}{\to }{C}_6{H}_5{N}_2^{+}B{F}_4^{-}\stackrel{\Delta }{\to }{C}_6{H}_{5}F+B{F}_3+N_2$

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2. (I)(b) $C_6{H}_{5}COOH\stackrel{N{H}_3}{\to }{C}_6{H}_{5}COON{H}_4\stackrel{\Delta ,-H_{2}O}{\to }{C}_6{H}_{5}CON{H}_2\stackrel{(i)LiAl{H}_4,(ii)H_{2}O}{\to }{C}_6{H}_{5}C{H}_{2}N{H}_2$

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3. (II) The $-N{H}_2$ group strongly activates the ring by donating its lone pair, so substitution occurs at all o/p positions (e.g. 2,4,6-tribromoaniline forms).

   ½ mark

4. Acetylation converts $-N{H}_2$ to $-NHCOC{H}_3$. The N lone pair is now also delocalised onto the C=O group, so its activating effect drops. Substitution is controlled and gives mainly the para monosubstituted product, and the amide is hydrolysed back afterwards.

   ½ mark

Answer: (a) Reduction → diazotisation → $HB{F}_4$/heat (Balz–Schiemann); (b) amide formation → $LiAl{H}_4$ reduction; (II) acetylation lowers the activating effect of –NH₂, giving monosubstitution.

1. (I) 1-Bromo-2-(butan-2-yl)benzene, also written as 1-bromo-2-(1-methylpropyl)benzene.

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2. X (with AgCN) = $C{H}_{3}C{H}_{2}C{H}_{2}NC$, propyl isocyanide.

   ½ mark

3. Y (with KCN) = $C{H}_{3}C{H}_{2}C{H}_{2}CN$, butanenitrile (propyl cyanide).

   ½ mark

4. KCN is ionic and gives free $C{N}^{-}$ ions. This ambident nucleophile attacks through carbon, the more nucleophilic site, so a C–C bond forms and the product is the nitrile.

   ½ mark

5. AgCN is mainly covalent (Ag–C bond), so only the nitrogen lone pair is free to attack. A C–N bond forms, giving the isocyanide.

   ½ mark

Answer: (I) 1-Bromo-2-(butan-2-yl)benzene; (II) X = CH₃CH₂CH₂NC, Y = CH₃CH₂CH₂CN

1. (I) Longest chain: $C{H}_3-C(C{H}_3{)}_2-C(Br)(C{H}_3)-C{H}_2-C{H}_3$ (5 C). Numbering from the left gives locants 2,2,3,3, which are lower than the alternative. Name: 3-Bromo-2,2,3-trimethylpentane.

   1 mark

2. (II) $C{H}_2=CH-C{H}_{2}Cl$ (allyl chloride) is hydrolysed more readily than $(C{H}_3{)}_{2}CHCl$.

   1 mark

3. Hydrolysis goes through a carbocation (SN1). Allyl chloride gives the allyl cation, which is resonance-stabilised: $C{H}_2=CH-\stackrel{+}{C}{H}_2\leftrightarrow \stackrel{+}{C}{H}_2-CH=C{H}_2$, with the positive charge spread over two carbons. The 2° isopropyl cation $(C{H}_3{)}_{2}C{H}^{+}$ has no such resonance and is less stable.

   1 mark

Answer: (I) 3-Bromo-2,2,3-trimethylpentane; (II) Allyl chloride, because its carbocation is resonance-stabilised.

1. (I) Relative lowering of vapour pressure: $\frac{p^{\circ }-p_s}{p^{\circ }}=x_B=\frac{n_B}{n_A+n_B}$

   ½ mark

2. $n_A=\frac{1000}{18}=55.56$ mol. A 50% reduction means $\frac{p^{\circ }-p_s}{p^{\circ }}=0.5$.

   ½ mark

3. $0.5=\frac{n_B}{n_B+55.56}$, so $0.5n_B=0.5\times 55.56$ and $n_B=55.56$ mol.

   ½ mark

4. $M_B=\frac{w_B}{n_B}=\frac{100}{55.56}=1.8$ g mol⁻¹

   ½ mark

5. (II) Nitric acid with 32% water (68% HNO₃) forms a maximum-boiling azeotrope (negative deviation). Its liquid and vapour have the same composition, so it boils at constant temperature and cannot be separated by fractional distillation.

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Answer: (I) $M_A=1.8$ g mol⁻¹; (II) it is an azeotrope (same composition in liquid and vapour).

1. (I) Each $-N{O}_2$ group lowers the pKa (p-nitrophenol 7.2, 2,4-dinitrophenol 4.09). A third nitro group lowers it further, so the pKa of 2,4,6-trinitrophenol is less than 4.09. Option D.

   1 mark

2. (II) From the graph: p-bromophenol 9.34 < p-chlorophenol 9.38 < p-fluorophenol 9.9 < phenol 10. Lower pKa means a stronger acid, so p-bromophenol > p-chlorophenol > p-fluorophenol > phenol. Option A.

   1 mark

3. (IIIA) Phenol has the lower pKa (≈10), because it is a stronger acid than cyclohexanol.

   1 mark

4. (IIIA) The phenoxide ion is stabilised by resonance, as its negative charge is delocalised over the ring. In cyclohexoxide the negative charge stays localised on oxygen, so this conjugate base is less stable and cyclohexanol is a weaker acid with higher pKa.

   1 mark

5. (IIIB) 4-Chloro-2,6-dinitrophenol is less acidic than 2,4,6-trinitrophenol.

   1 mark

6. (IIIB) The two compounds differ only in the para group: Cl versus NO₂. The graph shows p-nitrophenol (pKa 7.2) is much more acidic than p-chlorophenol (pKa 9.38). So replacing the para –NO₂ by –Cl raises the pKa, and the trinitro compound with its third strong electron-withdrawing group has the lower pKa (stronger acid).

   1 mark

Answer: (I) D; (II) A; (IIIA) phenol (resonance-stabilised phenoxide); (IIIB) less acidic than picric acid.

1. (I) A brick-red $C{u}_{2}O$ precipitate with Benedict's reagent shows the sugar has a free aldehyde/ketone group, i.e. it is a reducing sugar. Option B.

   1 mark

2. (II) Sucrose (cane sugar) and starch are made by plants. Lactose is milk sugar and glycogen is animal starch. Option B.

   1 mark

3. (IIIA) The sample is maltose. It gives a positive Benedict's test, so it is reducing, which rules out sucrose. It undergoes hydrolysis, so it is not a monosaccharide, which rules out glucose and fructose.

   1 mark

4. (IIIA) Hydrolysis gives only one type of monosaccharide. Lactose gives glucose + galactose, while maltose gives two α-D-glucose units. Maltose also comes from plants (germinating grain). Hence maltose.

   1 mark

5. (IIIB) Ravi's justification: the sample is plant-derived and on hydrolysis gives only one monosaccharide (glucose), which is also true of starch.

   1 mark

6. (IIIB) I do not agree. Starch is insoluble in (cold) water, but the sample dissolved. Starch is also a non-reducing polysaccharide and does not give a brick-red precipitate with Benedict's reagent.

   1 mark

Answer: (I) B; (II) B; (IIIA) Maltose; (IIIB) Ravi is wrong: starch is insoluble and non-reducing.

1. (I)(a) In strongly acidic medium the ammonia derivative ($N{H}_2-Z$) is protonated to ${}^{+}N{H}_3-Z$ and loses its lone pair, so it is no longer nucleophilic. In basic medium the carbonyl oxygen is not protonated, so the carbonyl carbon is not electrophilic enough. A weakly acidic pH of about 3.5 balances both effects and gives the fastest rate.

   1 mark

2. (I)(b) Dry HCl protonates the carbonyl oxygen. This increases the electrophilicity of the carbonyl carbon, so the weak nucleophile (alcohol) can attack to form a hemiacetal and then an acetal. The HCl is kept dry because water would reverse the acetal formation.

   1 mark

3. (I)(c) Aldehydes form crystalline, water-soluble bisulphite addition products that separate from impurities. The adduct is reconverted to the pure aldehyde by treating it with dilute mineral acid or alkali. So the reaction is used for separation and purification.

   1 mark

4. (II) Flowchart:

   Step 1: Add $NaHC{O}_3$ solution to each sample.

   • Brisk effervescence of $C{O}_2$ → acetic acid ($C{H}_{3}COOH+NaHC{O}_3\to C{H}_{3}COONa+H_{2}O+C{O}_2\uparrow$)
   • No effervescence → methanol or acetaldehyde, go to Step 2
   1 mark

5. Step 2: Add 2,4-DNP reagent to the remaining two samples.

   • Yellow/orange precipitate of the 2,4-dinitrophenylhydrazone → acetaldehyde (has a C=O group)
   • No precipitate → methanol
   1 mark

Answer: (I) (a) optimum pH balances protonation of the carbonyl and the nucleophilicity of NH₂–Z; (b) dry HCl activates the C=O; (c) the reversible, separable bisulphite adduct; (II) NaHCO₃ effervescence → acetic acid; then 2,4-DNP precipitate → acetaldehyde; no precipitate → methanol.

1. (I)(a) HCHO > CH₃CHO > C₆H₅CHO. The +I effect of the methyl group and the resonance of the phenyl ring progressively reduce the ease of oxidation.

   1 mark

2. (I)(b) CH₃CONH₂ < CH₃COOH < CH₃COOCH₃ < CH₃COCl. Reactivity rises as the leaving group gets better and as its resonance donation to C=O gets weaker. $C{l}^{-}$ is the best leaving group; $N{H}_2$ donates strongly by resonance and is the poorest leaving group.

   1 mark

3. (I)(c) CH₃COCH₃ < CH₃CHO < CF₃CHO. Two electron-donating methyls lower the electrophilicity, one methyl lowers it less, and the strongly –I $C{F}_3$ group raises it the most.

   1 mark

4. (II) Two isomers of $C_3{H}_{6}O$: propanal, $C{H}_{3}C{H}_{2}CHO$, and propanone, $C{H}_{3}COC{H}_3$.

   1 mark

5. Propanal gives a positive Tollens' test (silver mirror). Its aldehydic C–H allows it to be oxidised to propanoate, reducing $[Ag(N{H}_3{)}_2{]}^{+}$ to metallic Ag. Propanone, a ketone, has no such hydrogen and is not oxidised by this mild reagent.

   1 mark

Answer: (I)(a) HCHO > CH₃CHO > C₆H₅CHO; (b) CH₃CONH₂ < CH₃COOH < CH₃COOCH₃ < CH₃COCl; (c) CH₃COCH₃ < CH₃CHO < CF₃CHO; (II) propanal and propanone; propanal gives Tollens' test.

1. (I) Solution Y, $[Cr(en{)}_3{]}^{3+}$, rotates the plane of polarised light. Its three bidentate en ligands form a chiral octahedral arrangement with no plane of symmetry, so it exists as non-superimposable d and l forms. The fac and mer isomers of $[Cr(H_{2}O{)}_{3}C{l}_3]$ each have a plane of symmetry, so they are optically inactive.

   1 mark

2. (II) $[Ni(CO{)}_4]$ has the higher crystal field splitting.

   ½ mark

3. The determining factor is the nature (field strength) of the ligand. CO is a strong field ligand (it also π-accepts), while $H_{2}O$ is a weak field ligand in the spectrochemical series.

   ½ mark

4. (III) 2 mol AgCl means 2 ionisable $C{l}^{-}$, so the compound is $[Pd(N{H}_3{)}_4]C{l}_2$. Primary valency = oxidation state of Pd = 2.

   ½ mark

5. Secondary valency = coordination number = 4 (four $N{H}_3$ ligands).

   ½ mark

6. (IV) $d^{2}s{p}^3$ (inner-orbital) hybridisation shows that $C{N}^{-}$ is a strong field ligand. It forces the six 3d electrons of $F{e}^{2+}$ to pair up in $t_{2g}$ ($t_{2g}^6{e}_g^0$), leaving two vacant 3d orbitals.

   ½ mark

7. These two 3d orbitals, together with 4s and 4p orbitals, give six $d^{2}s{p}^3$ hybrids. The complex is therefore octahedral, low-spin and diamagnetic.

   ½ mark

8. (V) In $CuS{O}_4\cdot 5H_{2}O$, water molecules coordinate to $C{u}^{2+}$ ($[Cu(H_{2}O{)}_4{]}^{2+}$) and split the d orbitals. A $d$–$d$ transition of the $d^9$ ion absorbs red/orange light, so the compound appears blue. In anhydrous $CuS{O}_4$ there are no water ligands to cause this splitting, so no d–d absorption occurs in the visible region and it appears white.

   1 mark

Answer: (I) Y – [Cr(en)₃]³⁺ (chiral); (II) [Ni(CO)₄], decided by ligand field strength; (III) primary valency 2, secondary valency 4; (IV) CN⁻ is a strong field ligand → low-spin octahedral; (V) H₂O ligands cause d-orbital splitting and d–d transitions (blue); none without water (white).

1. (I) Compound A, $[Co(N{H}_3{)}_6]C{l}_3$, shows the maximum conductivity. In water:

   • A: $[Co(N{H}_3{)}_6]C{l}_3\to [Co(N{H}_3{)}_6{]}^{3+}+3C{l}^{-}$ gives 4 ions
   • B: $[Co(N{H}_3{)}_{5}Cl]C{l}_2\to [Co(N{H}_3{)}_{5}Cl{]}^{2+}+2C{l}^{-}$ gives 3 ions
   • C: $[Co(N{H}_3{)}_{4}C{l}_2]Cl\to [Co(N{H}_3{)}_{4}C{l}_2{]}^{+}+C{l}^{-}$ gives 2 ions

   The more ions there are in solution, the higher the electrolytic conductivity. So A > B > C.

   1 mark

2. (II) When ${\Delta }_0>P$, the ligand is strong field. Pairing in the $t_{2g}$ set costs less energy than promoting an electron to $e_g$. All five electrons therefore enter $t_{2g}$, giving the low-spin configuration $t_{2g}^5{e}_g^0$, with one unpaired electron.

   1 mark

3. (III) Chromium is in the +3 state. The ligands are 2 NH${}_3$ (neutral), 2 H${}_2$O (neutral) and C${}_2$O${}_4^{2-}$ (−2), so the complex ion is $[Cr(N{H}_3{)}_2(H_{2}O{)}_2(C_2{O}_4){]}^{+}$. One Cl${}^{-}$ balances the charge.

   Formula: $[Cr(N{H}_3{)}_2(H_{2}O{)}_2(C_2{O}_4)]Cl$

   1 mark

4. (IV) $[Co(N{H}_3{)}_{2}C{l}_2(en)]Cl$ is optically active and $[Co(N{H}_3{)}_{4}C{l}_2]Cl$ is not.

   • The bidentate en ligand can give an octahedral arrangement with no plane of symmetry. This happens in the isomer with the two Cl cis and the two NH${}_3$ cis. Its mirror image cannot be superimposed on it, so it exists as a pair of enantiomers.
   • $[Co(N{H}_3{)}_{4}C{l}_2{]}^{+}$ is of the type $M{A}_4{B}_2$ with only monodentate ligands. Both its cis and trans forms have a plane of symmetry, so they are achiral.
   1 mark

5. (V) NH${}_3$ is a stronger field ligand than H${}_2$O.

   • It therefore causes a larger crystal field splitting ($\Delta$) in $[Cu(N{H}_3{)}_4{]}^{2+}$.
   • The two complexes absorb light of different wavelengths in their d–d transitions.
   • The complementary colours seen are different: deep blue for the ammine complex and greenish for the aqua complex.

   The factor that decides the colour difference is the nature (field strength) of the ligand, which controls the magnitude of $\Delta$.

   1 mark

Answer: (I) A, $[Co(N{H}_3{)}_6]C{l}_3$ (4 ions); (II) $t_{2g}^5{e}_g^0$; (III) $[Cr(N{H}_3{)}_2(H_{2}O{)}_2(C_2{O}_4)]Cl$; (IV) $[Co(N{H}_3{)}_{2}C{l}_2(en)]Cl$ is optically active (chiral, no plane of symmetry); (V) NH${}_3$ is a stronger ligand, gives larger $\Delta$ and a different absorbed wavelength; ligand strength decides the colour.

1. (I)(a) The cell reaction is $Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$, so $E_{cell}=E_{cell}^0-\frac{0.059}{2}\log \frac{[Z{n}^{2+}]}{[C{u}^{2+}]}$.

   In experiment 4, $[Z{n}^{2+}]=0.01$ M (lowest) and $[C{u}^{2+}]=1.00$ M (highest). This makes $Q_c$ the smallest, at $10^{-2}$. So $\log Q_c$ is most negative, the subtracted term becomes the largest positive addition, and the EMF is the highest.

   1 mark

2. (I)(b) Nernst equation for experiment 3 ($n=2$):
   $E_{cell}=1.10-\frac{0.059}{2}\log \frac{[Z{n}^{2+}]}{[C{u}^{2+}]}=1.10-0.0295\log \frac{1.00}{0.10}$

   ½ mark

3. $E_{cell}=1.10-0.0295\times \log 10=1.10-0.0295=1.0705\approx 1.07$ V

   ½ mark

4. Reason for the difference: The measured value (1.04 V) is lower than the calculated one. When the cell is measured, some current flows, and the internal resistance of the electrolyte, electrodes and salt bridge causes a voltage drop. Non-ideal behaviour (activity is not equal to concentration) also contributes.

   1 mark

5. (II) ${\Lambda }_m=\kappa /c$.

   $1S{m}^{2}mo{l}^{-1}=1S(100cm{)}^{2}mo{l}^{-1}=10^{4}Sc{m}^{2}mo{l}^{-1}$

   So $1S{m}^{2}mo{l}^{-1}=10^{4}Sc{m}^{2}mo{l}^{-1}$, or equivalently $1Sc{m}^{2}mo{l}^{-1}=10^{-4}S{m}^{2}mo{l}^{-1}$.

   1 mark

6. (III) The overall reaction is $Pb+Pb{O}_2+2H_{2}S{O}_4\rightleftharpoons 2PbS{O}_4+2H_{2}O$.

   During charging, $H_{2}S{O}_4$ is regenerated, so the specific gravity of the electrolyte increases.

   ½ mark

7. During discharging, $H_{2}S{O}_4$ is consumed and water is formed, so the specific gravity decreases.

   ½ mark

Answer: (I)(a) Q is smallest (lowest [Zn²⁺], highest [Cu²⁺]), so E is highest; (b) $E=1.07$ V, lower measured value due to internal resistance/non-ideality; (II) $1S{m}^{2}mo{l}^{-1}=10^{4}Sc{m}^{2}mo{l}^{-1}$; (III) specific gravity increases on charging and decreases on discharging.

1. I. Use ${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$.

   For $c_1=0.010$ M, $\sqrt{c_1}=0.100$:
   $123.9={\Lambda }_m^{\circ }-0.100A...(1)$

   ½ mark

2. For $c_2=0.040$ M, $\sqrt{c_2}=0.200$:
   $120.7={\Lambda }_m^{\circ }-0.200A...(2)$

   ½ mark

3. Subtract (2) from (1):
   $123.9-120.7=-0.100A+0.200A$
   $3.2=0.100A$

   ½ mark

4. $A=32Sc{m}^{2}mo{l}^{-1}{M}^{-1/2}$

   ½ mark

5. Substitute $A$ in (1):
   ${\Lambda }_m^{\circ }=123.9+32\times 0.100=123.9+3.2$

   ½ mark

6. ${\Lambda }_m^{\circ }=127.1Sc{m}^{2}mo{l}^{-1}$

   ½ mark

7. II. Electrode reactions (alkaline electrolyte, as in NCERT):

   • Anode: $2H_2(g)+4O{H}^{-}(aq)\to 4H_{2}O(l)+4e^{-}$
   • Cathode: $O_2(g)+2H_{2}O(l)+4e^{-}\to 4O{H}^{-}(aq)$
   • Overall: $2H_2(g)+O_2(g)\to 2H_{2}O(l)$

   (In an acidic medium: anode $2H_2\to 4H^{+}+4e^{-}$; cathode $O_2+4H^{+}+4e^{-}\to 2H_{2}O$.)

   1 mark

8. Advantages over conventional batteries:
   1. They have higher efficiency (about 70%) in converting chemical energy into electrical energy.

   ½ mark

9. 2. They work continuously as long as the reactants H${}_2$ and O${}_2$ are supplied, so they need no recharging. They are also pollution-free, since the only product is water.

   ½ mark

Answer: I. $A=32Sc{m}^{2}mo{l}^{-1}{M}^{-1/2}$, ${\Lambda }_m^{\circ }=127.1Sc{m}^{2}mo{l}^{-1}$; II. Anode: $2H_2+4O{H}^{-}\to 4H_{2}O+4e^{-}$, Cathode: $O_2+2H_{2}O+4e^{-}\to 4O{H}^{-}$; advantages: high efficiency and continuous operation without recharging (pollution-free).