CBSE Class 12 Chemistry: 2026 board paper, set 56/5/1

1. In Wolff-Kishner reduction the ketone is treated with hydrazine ($N{H}_{2}N{H}_2$) and then heated with KOH in ethylene glycol. This reduces $>C=O$ to $>C{H}_2$, so the ketone becomes a hydrocarbon. Option (B).

   1 mark

Answer: (B) Wolff-Kishner reduction

1. $RCON{H}_2+B{r}_2+4NaOH\to RN{H}_2+N{a}_{2}C{O}_3+2NaBr+2H_{2}O$. The reagents are the amide, NaOH and $B{r}_2$; $CHC{l}_3$ is not used. Option (C).

   1 mark

Answer: (C) (i), (ii) and (iii)

1. $RN{H}_2+CHC{l}_3+3KOH\to RNC+3KCl+3H_{2}O$. The product is an isocyanide (carbylamine), recognised by its foul smell. Option (D).

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Answer: (D) Isocyanide

1. The 5f, 6d and 7s orbitals of actinoids have comparable energies. Electrons from all three can therefore take part in bonding, which gives many oxidation states. Option (D).

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Answer: (D)

1. This is the Finkelstein reaction: $C{H}_{3}Cl+NaI\to C{H}_{3}I+NaCl$. NaCl is insoluble in dry acetone and precipitates, which drives the reaction forward. So A = $C{H}_{3}I$ and B = NaCl. Option (A).

   1 mark

Answer: (A) A = CH3I, B = NaCl

1. Hinsberg's reagent is benzenesulphonyl chloride, $C_6{H}_{5}S{O}_{2}Cl$. Option (B).

   1 mark

Answer: (B) C6H5SO2Cl

1. For a first order reaction, $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{1386s}=5.0\times 10^{-4}{s}^{-1}$. Option (B).

   1 mark

Answer: (B) 5.0 × 10⁻⁴ s⁻¹

1. The oxalate ion $C_2{O}_4^{2-}$ is a bidentate ligand. It binds the metal through two O atoms and forms a ring, which is a chelate. $N{H}_3$, $H_{2}O$ and $N{O}_2^{-}$ are all monodentate. Option (D).

   1 mark

Answer: (D) Oxalate ion

1. Lucas reagent is conc. HCl with anhydrous $ZnC{l}_2$. With it, 3° alcohols turn turbid at once, 2° alcohols in about 5 minutes, and 1° alcohols give no turbidity at room temperature. Option (A).

   1 mark

Answer: (A) Lucas test

1. In aniline the lone pair is delocalised into the ring, so aniline is the weakest base, weaker than $N{H}_3$.

2. The benzyl group sits between $N{H}_3$ and ethylamine. Diethylamine has two +I alkyl groups and is the strongest. Order: $C_6{H}_{5}N{H}_2<N{H}_3<C_6{H}_{5}C{H}_{2}N{H}_2<C_2{H}_{5}N{H}_2<(C_2{H}_5{)}_{2}NH$. Option (A).

   1 mark

Answer: (A)

1. Cellulose is a polymer of β-D-glucose units, so it is a polysaccharide. Glucose and fructose are monosaccharides and maltose is a disaccharide. Option (D).

   1 mark

Answer: (D) Cellulose

1. The specific sequence of amino acids in a polypeptide chain is called the primary structure of the protein. Option (A).

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Answer: (A) Primary structure

1. The (+) sign shows the compound is dextrorotatory, while D shows its configuration (relative to glyceraldehyde). Both statements are true, and R explains why the name tells us glucose is dextrorotatory. Option (A).

   1 mark

Answer: (A)

1. Mn shows +7 (as in $Mn{O}_4^{-}$), the highest oxidation state in the 3d series, so A is true. Transition metals do show variable oxidation states, so R is true.

2. However, R is only a general fact and does not explain why Mn reaches +7. That is due to the use of all its 4s and 3d electrons ($3d^{5}4s^2$). Option (B).

   1 mark

Answer: (B)

1. The –NO₂ group withdraws electrons by –I and –R effects. This disperses the negative charge of the p-nitrophenoxide ion and stabilises it, making p-nitrophenol more acidic than phenol. R explains A. Option (A).

   1 mark

Answer: (A)

1. Aliphatic aldehydes reduce Fehling's solution, giving a red precipitate of $C{u}_{2}O$, so A is true.

2. In this reaction the aldehyde is oxidised, not reduced, so R is false. Option (C).

   1 mark

Answer: (C)

1. Observed elevation: $\Delta T_b(\text{obs})=100.18-100=0.18\text{K}$

2. Calculated elevation for no association or dissociation: $\Delta T_b(\text{cal})=K_b\times m=0.512\times 1.00$

   ½ mark

3. $\Delta T_b(\text{cal})=0.512\text{K}$

   ½ mark

4. $i=\frac{\text{observed colligative property}}{\text{calculated colligative property}}=\frac{0.18}{0.512}=0.3516\approx 0.35$

   1 mark

Answer: $i\approx 0.35$ (0.3516)

1. Henry's law: at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Equivalently, $p=K_{H}x$, where $x$ is the mole fraction of the gas in solution.

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2. $x_{C{O}_2}=\frac{p}{K_H}=\frac{760\text{mm Hg}}{1.25\times 10^6\text{mm Hg}}$

   ½ mark

3. $x_{C{O}_2}=6.08\times 10^{-4}$

   ½ mark

Answer: $x_{C{O}_2}=6.08\times 10^{-4}$

1. (a) The ${\mathrm{H}}_2$–${\mathrm{O}}_2$ fuel cell was used in the Apollo space programme to provide electrical power.

   1 mark

2. (b) Limiting molar conductivity (${\Lambda }_m^{\circ }$) is the molar conductivity of an electrolyte solution when its concentration approaches zero, i.e. at infinite dilution.

   1 mark

Answer: (a) H₂–O₂ fuel cell; (b) molar conductivity as concentration → 0 (infinite dilution)

1. (a) Coupling takes place at the para position of phenol in alkaline medium:
   ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}+{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\stackrel{\mathrm{O}{\mathrm{H}}^{-}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\text{–}\mathrm{N}=\mathrm{N}\text{–}{\mathrm{C}}_6{\mathrm{H}}_4\text{–}\mathrm{OH}(\mathrm{p})+\mathrm{C}{\mathrm{l}}^{-}+{\mathrm{H}}_2\mathrm{O}$
   The product is p-hydroxyazobenzene, an orange dye.

   1 mark

2. (b) First reduce nitromethane to methanamine, then carry out the carbylamine reaction:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{O}}_2\stackrel{Sn/HCl(\mathrm{or}{\mathrm{H}}_2/Pd)}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$
   $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2+\mathrm{CHC}{\mathrm{l}}_3+3KOH(alc.)\stackrel{\mathrm{\Delta }}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{NC}+3KCl+3{\mathrm{H}}_2\mathrm{O}$

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Answer: (a) p-Hydroxyazobenzene (+ Cl⁻ + H₂O); (b) Reduction to CH₃NH₂, then CHCl₃/alc. KOH gives CH₃NC

1. (a) Hydrolysis of sucrose gives an equimolar mixture of D-(+)-glucose and D-(–)-fructose (invert sugar).

   1 mark

2. (b) Essential amino acids are amino acids that the human body cannot synthesise, so they must be obtained through the diet. Examples are valine, leucine and lysine.

   1 mark

Answer: (a) Glucose and fructose; (b) amino acids not synthesised in the body, which must come from the diet

1. (a) Any two of the fat-soluble vitamins: Vitamin A and Vitamin D (vitamins E and K are also fat soluble).

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2. (b) On acetylation with acetic anhydride, glucose forms a stable glucose pentaacetate. This shows that it contains five –OH groups.

   Since the pentaacetate is stable, the five –OH groups must be on different carbon atoms; two –OH groups on the same carbon would be unstable.

   1 mark

Answer: (a) Vitamins A and D (or E, K); (b) acetylation with acetic anhydride gives stable glucose pentaacetate

1. Cell reaction: $2Cr(\mathrm{s})+3F{\mathrm{e}}^{2+}(\mathrm{aq})\to 2C{\mathrm{r}}^{3+}(\mathrm{aq})+3Fe(\mathrm{s})$, so $n=6$.

   ½ mark

2. $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=(-0.44)-(-0.74)$

   ½ mark

3. $E_{cell}^{\circ }=0.30\text{V}$

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4. Nernst equation: $E_{cell}=E_{cell}^{\circ }-\frac{0.059}{6}\log \frac{[C{r}^{3+}{]}^2}{[F{e}^{2+}{]}^3}=0.30-\frac{0.059}{6}\log \frac{(0.1{)}^2}{(0.01{)}^3}$

5. $=0.30-\frac{0.059}{6}\log 10^4=0.30-\frac{0.236}{6}=0.30-0.0393$

6. $E_{cell}=0.2607\text{V}\approx 0.26\text{V}$

   1 mark

Answer: $E_{cell}\approx 0.26$ V

1. (a) Order of a reaction is the sum of the powers to which the concentrations of the reactants are raised in the experimentally determined rate law.

   1 mark

2. (b)(i) Rate $\propto [B{]}^2$. When [B] is doubled, rate $=k[A](2[B]{)}^2=4k[A][B{]}^2$, so the rate becomes 4 times.

   1 mark

3. (b)(ii) With A in large excess, [A] stays practically constant. Then rate $=k^{\prime }[B{]}^2$ with $k^{\prime }=k[A]$, so the overall order is 2 (pseudo second order).

   1 mark

Answer: (a) Definition; (b)(i) rate becomes 4 times; (ii) order = 2

1. Arrhenius relation: $\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)=\frac{E_a}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$, with $\frac{k_2}{k_1}=2$.

   1 mark

2. $\log 2=\frac{E_a}{2.303\times 8.314}\left(\frac{308-298}{298\times 308}\right)$

   $0.30=\frac{E_a\times 10}{19.147\times 298\times 308}$

   1 mark

3. $E_a=\frac{0.30\times 19.147\times 298\times 308}{10}=52722{\text{J mol}}^{-1}\approx 52.72{\text{kJ mol}}^{-1}$

   1 mark

Answer: $E_a\approx 52.72$ kJ mol⁻¹ (52722 J mol⁻¹)

1. (a) ${\mathrm{K}}_3[\mathrm{Cr}({\mathrm{C}}_2{\mathrm{O}}_4{)}_3]$: Potassium trioxalatochromate(III).

   1 mark

2. (b) In a homoleptic complex the metal is bound to only one kind of donor group, e.g. $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$. In a heteroleptic complex the metal is bound to more than one kind of donor group, e.g. $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_4\mathrm{C}{\mathrm{l}}_2{]}^{+}$.

   1 mark

3. (c) Square planar $[\mathrm{Pt}(\mathrm{N}{\mathrm{H}}_3{)}_2\mathrm{C}{\mathrm{l}}_2]$ (type $M{A}_2{B}_2$) shows geometrical (cis–trans) isomerism.

   1 mark

Answer: (a) Potassium trioxalatochromate(III); (b) one kind vs more than one kind of ligand; (c) geometrical (cis–trans) isomerism

1. (a) Two moles of AgCl per mole means two $\mathrm{C}{\mathrm{l}}^{-}$ are ionisable, i.e. outside the coordination sphere. The third Cl and five ${\mathrm{H}}_2\mathrm{O}$ complete the coordination number 6. Structural formula: $[\mathrm{Cr}({\mathrm{H}}_2\mathrm{O}{)}_5\mathrm{Cl}]\mathrm{C}{\mathrm{l}}_2\cdot {\mathrm{H}}_2\mathrm{O}$.

   1 mark

2. (b) Oxidation state of Fe in $[\mathrm{Fe}({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{3+}$: $x+0=+3$, so +3.

   ½ mark

3. Fe³⁺ is $3d^5$. ${\mathrm{H}}_2\mathrm{O}$ is a weak-field ligand, so the electrons do not pair and the outer 4d orbitals are used. Hybridisation: sp³d² (outer-orbital, high-spin, octahedral).

   ½ mark

4. (c) Ni²⁺ is $3d^8$ with two unpaired electrons. It absorbs light in the visible region, which promotes an electron from $t_{2g}$ to $e_g$ (d–d transition). The complementary colour (green) is seen.

   1 mark

Answer: (a) [Cr(H₂O)₅Cl]Cl₂·H₂O; (b) +3, sp³d²; (c) d–d transition of unpaired 3d electrons (3d⁸)

1. (a) Acetophenone to benzoic acid, by oxidation with alkaline $\mathrm{KMn}{\mathrm{O}}_4$ (or haloform reaction) followed by acidification:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_3\stackrel{(\mathrm{i})\mathrm{KMn}{\mathrm{O}}_4/KOH,\mathrm{\Delta };(\mathrm{ii}){\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$

   Alternatively: ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_3+3NaOI\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+\mathrm{CH}{\mathrm{I}}_3+2NaOH$, then ${\mathrm{H}}_3{\mathrm{O}}^{+}$ gives ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$.

   1 mark

2. (b) Acetonitrile to acetone, by Grignard reagent followed by hydrolysis:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv \mathrm{N}+\mathrm{C}{\mathrm{H}}_3\mathrm{MgBr}\stackrel{\mathrm{dry}\mathrm{ether}}{\to }(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{NMgBr}\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3+\mathrm{N}{\mathrm{H}}_3+\mathrm{Mg}(\mathrm{OH})\mathrm{Br}$

   1 mark

3. (c) Benzoic acid to benzene, by forming the sodium salt and then decarboxylating with soda-lime:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{NaOH}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+\mathrm{NaOH}\stackrel{\mathrm{CaO},\mathrm{\Delta }}{\to }{\mathrm{C}}_6{\mathrm{H}}_6+\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$

   1 mark

Answer: (a) Alk. KMnO₄ (or NaOI), then H₃O⁺; (b) CH₃MgBr, then H₃O⁺; (c) Na salt heated with soda-lime (NaOH/CaO)

1. (i) The –Br group has a –I effect that increases acidity, and it is stronger the closer Br is to –COOH. The –CH₃ group has a +I effect that decreases acidity. Increasing order of acidic strength:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{COOH}<\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{Br})\mathrm{C}{\mathrm{H}}_2\mathrm{COOH}<\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{Br})\mathrm{COOH}$

   1 mark

2. (ii) Acetone has two methyl groups, so its carbonyl carbon is more crowded (steric hindrance) and less electrophilic (+I effect of two CH₃ groups). In ethanal, which has only one CH₃, the nucleophile $\mathrm{C}{\mathrm{N}}^{-}$ attacks the carbonyl carbon more easily. Hence $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$ is more reactive.

   1 mark

3. (iii) Ethanal forms its hydrazone:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}+{\mathrm{H}}_2\mathrm{N}\text{–}\mathrm{N}{\mathrm{H}}_2\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{N}\text{–}\mathrm{N}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i) CH₃CH(CH₃)COOH < CH₃CH(Br)CH₂COOH < CH₃CH₂CH(Br)COOH; (ii) less steric hindrance and greater electrophilicity in CH₃CHO; (iii) CH₃CH=N–NH₂ + H₂O

1. The compound forms a 2,4-DNP derivative, so it has C=O. It reduces Tollens' reagent, so it is an aldehyde. It undergoes the Cannizzaro reaction, so it has no α-H (–CHO on the ring). Vigorous oxidation gives benzene-1,2-dicarboxylic acid, so it has two substituents in ortho positions. With ${\mathrm{C}}_8{\mathrm{H}}_8\mathrm{O}$ the compound is 2-methylbenzaldehyde (o-tolualdehyde), $\mathrm{o}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}$.

   On vigorous oxidation: $\mathrm{o}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}\stackrel{\mathrm{KMn}{\mathrm{O}}_4/KOH,{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\mathrm{o}\text{-}{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{COOH}{)}_2$

   1 mark

2. With 2,4-DNP:
   $\mathrm{o}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}+{\mathrm{H}}_2\mathrm{N}\text{–}\mathrm{NH}\text{–}{\mathrm{C}}_6{\mathrm{H}}_3(\mathrm{N}{\mathrm{O}}_2{)}_2\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{o}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CH}=\mathrm{N}\text{–}\mathrm{NH}\text{–}{\mathrm{C}}_6{\mathrm{H}}_3(\mathrm{N}{\mathrm{O}}_2{)}_2+{\mathrm{H}}_2\mathrm{O}$
   The product is the orange-red 2,4-dinitrophenylhydrazone.

   1 mark

3. Cannizzaro reaction:
   $2\mathrm{o}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}+\mathrm{NaOH}(conc.)\stackrel{\mathrm{\Delta }}{\to }\mathrm{o}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{o}\text{-}\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{COONa}$
   The products are 2-methylbenzyl alcohol and sodium 2-methylbenzoate.

   1 mark

Answer: 2-Methylbenzaldehyde (o-tolualdehyde); gives 2,4-DNP hydrazone; Cannizzaro gives 2-methylbenzyl alcohol + sodium 2-methylbenzoate

1. (a)(i) Friedel–Crafts alkylation at the o- and p-positions:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OC}{\mathrm{H}}_3+\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}\stackrel{anhyd.\mathrm{AlC}{\mathrm{l}}_3,\mathrm{C}{\mathrm{S}}_2}{\to }$ 2-methoxytoluene (minor) + 4-methoxytoluene (major) + HCl

   1 mark

2. (a)(ii) Nitration at the o- and p-positions:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OC}{\mathrm{H}}_3\stackrel{conc.\mathrm{HN}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }$ 2-methoxynitrobenzene (2-nitroanisole, minor) + 4-methoxynitrobenzene (4-nitroanisole, major) + H₂O

   1 mark

3. (b)(i) Use ethyl halide (e.g. bromoethane, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}$) and sodium tert-butoxide, $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}\text{–}\mathrm{ONa}$:
   $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{CONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}\to (\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}\text{–}\mathrm{O}\text{–}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaBr}$
   A tertiary halide is not used because it would undergo elimination.

   1 mark

4. (b)(ii) (Alternative) In anisole, the $\mathrm{O}\text{–}{\mathrm{C}}_6{\mathrm{H}}_5$ bond is stronger than the $\mathrm{O}\text{–}\mathrm{C}{\mathrm{H}}_3$ bond. This is because of resonance and the sp² carbon, which give it partial double-bond character. So after protonation, ${\mathrm{I}}^{-}$ attacks the less hindered $\mathrm{C}{\mathrm{H}}_3$ group by $S_{N}2$ and breaks the $\mathrm{O}\text{–}\mathrm{C}{\mathrm{H}}_3$ bond, giving phenol and $\mathrm{C}{\mathrm{H}}_3\mathrm{I}$, not iodobenzene.

5. (c) The C–O–C bond angle is slightly greater than 109°28′ because of repulsive interaction between the two bulky alkyl groups on oxygen.

   1 mark

Answer: (a)(i) 2- and 4-methoxytoluene; (ii) 2- and 4-nitroanisole; (b)(i) ethyl halide + sodium tert-butoxide; (b)(ii) O–C₆H₅ bond stronger (partial double bond), so O–CH₃ cleaves; (c) repulsion between bulky alkyl groups

1. (a)(i) Aqueous CuCl₂ with Pt electrodes:

   • Anode: chlorine, $2C{\mathrm{l}}^{-}\to \mathrm{C}{\mathrm{l}}_2+2{\mathrm{e}}^{-}$
   • Cathode: copper, $\mathrm{C}{\mathrm{u}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$
   1 mark

2. (a)(ii) Concentrated H₂SO₄ with Pt electrodes:

   • Anode: peroxodisulphate ion, $2S{\mathrm{O}}_4^{2-}\to {\mathrm{S}}_2{\mathrm{O}}_8^{2-}+2{\mathrm{e}}^{-}$
   • Cathode: hydrogen gas, $2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2(\mathrm{g})$
   1 mark

3. (b)(i) $\mathrm{A}{\mathrm{g}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Ag}$. One mole of electrons is needed, so the charge is 1 F (96487 C).

   1 mark

4. (b)(ii) (Alternative) Faraday's second law: when the same quantity of electricity is passed through solutions of different electrolytes, the amounts of the substances liberated at the electrodes are proportional to their chemical equivalent weights.

5. (c) Reaction (I), liberation of Cl₂, actually takes place at the anode. Reaction (II) has the lower E° value, but it is kinetically slow because of the overpotential of oxygen.

   1 mark

Answer: (a)(i) Anode Cl₂, cathode Cu; (ii) anode S₂O₈²⁻, cathode H₂; (b)(i) 1 F; (b)(ii) Faraday's second law; (c) Reaction (I), due to overpotential of O₂

1. (i) $\mathrm{MgB}{\mathrm{r}}_2\to \mathrm{M}{\mathrm{g}}^{2+}+2B{\mathrm{r}}^{-}$, so $i=3$.

   ½ mark

2. $\Delta T_f=i{K}_{f}m=i{K}_f\times \frac{w_B\times 1000}{M_B\times w_A}$

   ½ mark

3. $\Delta T_f=\frac{3\times 1.86\times 10.5\times 1000}{184\times 250}=1.27\text{K}$

   1 mark

4. Freezing point of solution $=T_f^{\circ }-\Delta T_f=273.15-1.27=271.88\text{K}$ (i.e. $-1.27^{\circ }$C)

   1 mark

5. (ii) Difference 1: an ideal solution obeys Raoult's law over the entire range of concentration, whereas a non-ideal solution does not.

   1 mark

6. Difference 2: for an ideal solution ${\Delta }_{mix}H=0$ and ${\Delta }_{mix}V=0$, whereas for a non-ideal solution ${\Delta }_{mix}H\ne 0$ and ${\Delta }_{mix}V\ne 0$. (In ideal solutions A–B interactions equal A–A and B–B interactions; in non-ideal solutions they differ.)

   1 mark

Answer: (i) Freezing point = 271.88 K (−1.27 °C); (ii) ideal obeys Raoult's law, ΔmixH = ΔmixV = 0; non-ideal does not, ΔmixH, ΔmixV ≠ 0

1. (i) ${\mathrm{K}}_2\mathrm{S}{\mathrm{O}}_4\to 2{\mathrm{K}}^{+}+\mathrm{S}{\mathrm{O}}_4^{2-}$, so $i=3$.

   ½ mark

2. $\pi V=i{n}_{B}RT$, so $\pi =\frac{i{w}_{B}RT}{M_{B}V}$

   ½ mark

3. $\pi =\frac{3\times 0.025\times 0.082\times 300}{174\times 2}$ (with $T=27+273=300$ K)

   1 mark

4. $\pi =5.3\times 10^{-3}\text{atm}$

   1 mark

5. (ii) Acetone and chloroform form a maximum boiling azeotrope.

   1 mark

6. Reason: the mixture shows a large negative deviation from Raoult's law. The H of $\mathrm{CHC}{\mathrm{l}}_3$ hydrogen-bonds with the O of acetone, so acetone–chloroform interactions are stronger than those in the pure liquids. The vapour pressure is therefore lowered and the boiling point is raised.

   1 mark

Answer: (i) π ≈ 5.3 × 10⁻³ atm; (ii) maximum boiling azeotrope (negative deviation due to H-bonding)

1. (i)(I) The energies of the $(n-1)d$ and $ns$ orbitals are very close, so electrons from both can take part in bonding. This gives transition metals variable oxidation states.

   1 mark

2. (i)(II) $\mathrm{M}{\mathrm{n}}^{2+}$ is more paramagnetic.

   ½ mark

3. $\mathrm{M}{\mathrm{n}}^{2+}$ is $3d^5$ with 5 unpaired electrons, whereas $\mathrm{T}{\mathrm{i}}^{2+}$ is $3d^2$ with only 2 unpaired electrons.

   ½ mark

4. (i)(III) $\mathrm{M}{\mathrm{n}}^{3+}$ is the strongest oxidising agent.

   ½ mark

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

   ½ mark

6. (ii)(I) $2Mn{\mathrm{O}}_2+4KOH+{\mathrm{O}}_2\to 2{\mathrm{K}}_2\mathrm{Mn}{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}$

   1 mark

7. (ii)(II) $5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+2Mn{\mathrm{O}}_4^{-}+16{\mathrm{H}}^{+}\to 2M{\mathrm{n}}^{2+}+10C{\mathrm{O}}_2+8{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i)(I) similar energies of ns and (n−1)d; (II) Mn²⁺ (5 unpaired e⁻); (III) Mn³⁺ (d⁴ → stable d⁵); (ii)(I) 2K₂MnO₄ + 2H₂O; (II) 2Mn²⁺ + 10CO₂ + 8H₂O

1. (i) Lanthanoid contraction is the steady decrease in the atomic and ionic radii ($M^{3+}$) of the lanthanoids from La to Lu as atomic number increases. It happens because electrons enter the 4f orbitals before 5d. The 4f electrons shield one another poorly from the rising nuclear charge, so the effective nuclear charge increases and the size shrinks.

   1 mark

2. (ii) Most transition metal ions have partly filled d orbitals with unpaired electrons. In a compound the ligands split the d orbitals into sets of slightly different energy. An electron absorbs visible light and jumps from a lower to a higher d level (a d–d transition). The complementary colour of the absorbed light is the colour we see.

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3. (iii) Mn: $M{n}^{2+}$ is $3d^5$, a half-filled configuration with extra stability. This favours forming $M{n}^{2+}$ from Mn, so $E_{M{n}^{2+}/Mn}^o$ ($-1.18$ V) is more negative than the trend predicts.

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4. Zn: $Z{n}^{2+}$ is $3d^{10}$, a completely filled and stable configuration. This also favours oxidation of Zn, so $E_{Z{n}^{2+}/Zn}^o$ ($-0.76$ V) is more negative than expected.

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5. (iv) The most stable oxidation state of copper (in aqueous solution) is +2, i.e. $C{u}^{2+}$.

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6. Reason: $C{u}^{2+}$ is small and highly charged, so its hydration enthalpy (${\Delta }_{hyd}H$) is much more negative than that of $C{u}^{+}$. This more than makes up for the second ionisation enthalpy of Cu. Hence $C{u}^{+}$ disproportionates in water: $2C{u}^{+}\to C{u}^{2+}+Cu$.

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7. (v) The common and most stable oxidation state of cerium (and of all lanthanoids) is +3. $C{e}^{4+}$ therefore readily accepts an electron to revert to $C{e}^{3+}$: $C{e}^{4+}+e^{-}\to C{e}^{3+}$, $E^o=+1.74$ V. Because it is so easily reduced, $C{e}^{4+}$ is a good oxidising agent.

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Answer: (i) Gradual decrease in size of lanthanoids from La to Lu due to poor shielding by 4f electrons; (ii) d–d transitions of unpaired electrons absorbing visible light; (iii) Mn²⁺ is half-filled d⁵ and Zn²⁺ is fully filled d¹⁰, both extra stable; (iv) +2 (Cu²⁺), because of its high hydration enthalpy; (v) Ce⁴⁺ readily changes to the stable +3 state.

1. (i)(I) 2-Bromo-2-methylbutane is more reactive towards $S_{N}1$. It is a tertiary halide, so it ionises to a stable tertiary carbocation $(C{H}_3{)}_2\stackrel{+}{C}C{H}_{2}C{H}_3$. That carbocation is stabilised by +I effect and hyperconjugation. 1-Bromopentane is primary and would form an unstable 1° carbocation.

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2. (i)(II) The Cl atom is bonded directly to an $s{p}^2$ carbon of the C=C double bond ($-C(Cl)=C{H}_2$). It is therefore a vinylic halide.

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3. (i)(III) In the presence of light, air slowly oxidises chloroform to the highly poisonous gas phosgene: $2CHC{l}_3+O_2\stackrel{\text{light}}{\to }2COC{l}_2+2HCl$. Storing it in dark coloured, completely filled bottles keeps out light and air.

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4. (ii)(I) Ambident nucleophiles: nucleophiles that have two nucleophilic centres and can attack through either of them. Example: cyanide ion $[\stackrel{-}{:}C\equiv N:\leftrightarrow :C=\stackrel{-}{N}:]$ attacks through C (giving cyanides) or N (giving isocyanides). Nitrite ion $N{O}_2^{-}$ attacks through O or N.

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5. (ii)(II) Racemic mixture: a mixture containing equal amounts of the dextrorotatory (+) and laevorotatory (–) enantiomers. It shows zero optical rotation because the rotation by one enantiomer is cancelled by the other (external compensation).

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Answer: (i)(I) 2-Bromo-2-methylbutane; (II) vinylic halide; (III) to prevent its oxidation by air in light to poisonous phosgene (COCl₂); (ii)(I) nucleophiles with two nucleophilic sites, e.g. CN⁻, NO₂⁻; (II) equimolar mixture of d- and l-enantiomers, optically inactive.

1. (i)(I) tert-Butyl bromide [$(C{H}_3{)}_{3}C-Br$, 2-bromo-2-methylpropane] is the most reactive isomer of $C_4{H}_{9}Br$ towards $S_{N}1$. It forms the most stable, tertiary carbocation $(C{H}_3{)}_3{C}^{+}$.

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2. (i)(II) Elimination of HBr with alcoholic KOH follows Saytzeff's rule: the more substituted alkene is the major product. The major product is 1-methylcyclohexene (C=C inside the ring at C-1, carrying the $C{H}_3$). Methylenecyclohexane (exocyclic $=C{H}_2$) forms as the minor product.

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3. (i)(III) The –I effect of Cl withdraws electrons, so it deactivates the ring. However, the lone pairs on Cl are donated to the ring by resonance (+R effect), which raises the electron density mainly at the ortho and para positions. As a result, the arenium ion (carbocation intermediate) formed by attack at o/p positions is extra stabilised, since in one structure the positive charge lies on the carbon bearing Cl and is delocalised onto Cl. So substitution is directed to the ortho and para positions.

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4. (ii)(I) Fittig reaction gives biphenyl: $2C_6{H}_{5}Cl+2Na\stackrel{\text{dry ether}}{\to }{C}_6{H}_5-C_6{H}_5+2NaCl$

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5. (ii)(II) On heating, $B{r}_2$ brominates by a free-radical route at the benzylic C–H, which forms a stable tertiary benzylic radical. The major product is 2-bromo-2-(4-nitrophenyl)propane: $p\text{-}{O}_{2}N-C_6{H}_4-CH(C{H}_3{)}_2+B{r}_2\stackrel{\text{heat}}{\to }p\text{-}{O}_{2}N-C_6{H}_4-C(Br)(C{H}_3{)}_2+HBr$

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Answer: (i)(I) tert-Butyl bromide, (CH₃)₃CBr; (II) 1-methylcyclohexene (major); (III) +R effect of Cl stabilises o/p arenium ions; (ii)(I) biphenyl, C₆H₅–C₆H₅; (II) p-O₂N–C₆H₄–C(Br)(CH₃)₂.