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

1. For a first order reaction, $t_{1/2}=\frac{0.693}{k}$, which does not depend on $[R{]}_0$.

2. So the plot of $t_{1/2}$ against $[R{]}_0$ is a horizontal line. Graph (A) is for zero order, and graph (C) (rate constant with concentration) is also zero order. Correct option: (B).

   1 mark

Answer: Option (B)

1. $\Delta T_f=i{K}_{f}m$. At equal concentration, the solute with the largest van't Hoff factor $i$ gives the largest depression.

2. Values of $i$: glucose 1, urea 1, KCl 2, $\mathrm{CaC}{\mathrm{l}}_2$ 3. So 0.1 M $\mathrm{CaC}{\mathrm{l}}_2$ has the lowest freezing point: option (B).

   1 mark

Answer: Option (B) 0.1 M CaCl₂

1. A transition metal has an incompletely filled d subshell in the atom or in one of its common ions.

2. Hg is $5d^{10}6s^2$, and $\mathrm{H}{\mathrm{g}}^{2+}$ is $5d^{10}$. Its d subshell is always full, so Hg is not a transition metal: option (C).

   1 mark

Answer: Option (C) Hg

1. In the Arrhenius equation $k=A{e}^{-E_a/RT}$, the factor $e^{-E_a/RT}$ is the fraction of molecules with energy $\ge E_a$. Option (B).

   1 mark

Answer: Option (B) $e^{-E_a/RT}$

1. At the cathode, $\mathrm{C}{\mathrm{u}}^{2+}$ is reduced in preference to water, so copper deposits there.

2. At the anode, $\mathrm{C}{\mathrm{l}}^{-}$ is oxidised in preference to water because of the overpotential for ${\mathrm{O}}_2$. So $\mathrm{C}{\mathrm{l}}_2$ is released at the anode: option (D).

   1 mark

Answer: Option (D)

1. Aspirin (acetylsalicylic acid) is made by acetylating salicylic acid, i.e. 2-hydroxybenzoic acid, with acetic anhydride. Option (C).

   1 mark

Answer: Option (C)

1. Secondary valency is the coordination number of the metal.

2. Ligands: 5 $\mathrm{N}{\mathrm{H}}_3$ + 1 $\mathrm{N}{\mathrm{O}}_2^{-}$ = 6. Option (D).

   1 mark

Answer: Option (D) 6

1. The –OH group activates the ring and directs substitution to the ortho and para positions.

2. With dilute $\mathrm{HN}{\mathrm{O}}_3$ at low temperature, phenol gives a mixture of o-nitrophenol and p-nitrophenol. Option (D).

   1 mark

Answer: Option (D)

1. Enantiomers have identical physical properties, such as melting point and refractive index, and the same chemical reactivity.

2. They rotate plane-polarised light by equal amounts but in opposite directions, so their specific rotations are opposite in sign. Statement (B) is not true.

   1 mark

Answer: Option (B)

1. In the strongly acidic nitrating mixture, aniline is protonated to the anilinium ion, which is meta-directing. This gives a large amount of meta product.

2. Direct nitration gives 51% para, 47% meta and 2% ortho products. Option (C).

   1 mark

Answer: Option (C)

1. Lowest $p{K}_b$ means the strongest base.

2. Methyl groups (+I effect) raise the electron density on N, while –$\mathrm{N}{\mathrm{O}}_2$ lowers it. N,N-dimethylaniline is the most basic of these: option (A).

   1 mark

Answer: Option (A)

1. Deficiency of vitamin E (tocopherol) causes increased fragility of red blood cells and muscular weakness. Option (C).

   1 mark

Answer: Option (C) Vitamin E

1. An azeotrope boils at a constant temperature, and its vapour has the same composition as the liquid.

2. Distillation therefore cannot enrich either component. Both A and R are true, and R explains A: option (A).

   1 mark

Answer: Option (A)

1. The –OH group directs incoming groups to the ortho and para positions, not meta, so A is false.

2. –OH does activate the ring by its +R effect, so R is true. Option (D).

   1 mark

Answer: Option (D)

1. Actinoids show a wide range of oxidation states because the 5f, 6d and 7s energy levels are close in energy. A is true.

2. Actinoids are radioactive, so R is true, but radioactivity does not explain the oxidation states. Option (B).

   1 mark

Answer: Option (B)

1. Glucose pentaacetate does not react with hydroxylamine, so A is true.

2. This shows the absence of a free –CHO group (glucose exists in a cyclic hemiacetal form), so R is false. Option (C).

   1 mark

Answer: Option (C)

1. A phenol–aniline mixture shows negative deviation from Raoult's law.

   ½ mark

2. Reason: the acidic phenolic –OH hydrogen forms an intermolecular hydrogen bond with the lone pair on the nitrogen of aniline (${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{O}-\mathrm{H}\cdots \mathrm{N}{\mathrm{H}}_2{\mathrm{C}}_6{\mathrm{H}}_5$). These phenol–aniline (A–B) attractions are stronger than the phenol–phenol (A–A) and aniline–aniline (B–B) attractions, so the escaping tendency of molecules falls and the vapour pressure is lower than Raoult's law predicts.

   1 mark

3. Because the vapour pressure is lowered, the boiling point of the solution increases. It is higher than expected, and such mixtures can form a maximum-boiling azeotrope.

   ½ mark

Answer: Negative deviation, because phenol–aniline H-bonding is stronger than the like–like attractions; the boiling point increases.

1. Resonance effect: a lone pair on the halogen is delocalised into the benzene ring. The C–X bond therefore gets partial double-bond character, becomes stronger and shorter, and is difficult to break.

   1 mark

2. Hybridisation of carbon: the C attached to X is $s{p}^2$ hybridised. With more s-character it is more electronegative, holds the bonding pair more tightly, and the C–X bond is shorter and stronger than in haloalkanes ($s{p}^3$ C). Other acceptable reasons: the phenyl cation is unstable (so $S_{N}1$ is not possible), and the electron-rich ring repels the incoming nucleophile (so backside $S_{N}2$ attack is hindered).

   1 mark

Answer: (1) Resonance gives the C–X bond partial double-bond character; (2) the $s{p}^2$ carbon holds the C–X bond more tightly (also: phenyl cation unstable, nucleophile repelled by the π-electron cloud).

1. (i) The cation is $[\mathrm{Ag}(\mathrm{N}{\mathrm{H}}_3{)}_2{]}^{+}$ (Ag in the +1 state) and the anion is $[\mathrm{Ag}(\mathrm{CN}{)}_2{]}^{-}$ (Ag in the +1 state; an anionic complex takes the suffix -ate, giving argentate). The name is Diamminesilver(I) dicyanidoargentate(I).

   1 mark

2. (ii) Three oxalate (${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$) ligands give a charge of $-6$ and the complex ion is $3-$, so Fe is $+3$. Since the complex is an anion, Fe becomes ferrate. The name is Potassium trioxalatoferrate(III).

   1 mark

Answer: (i) Diamminesilver(I) dicyanidoargentate(I); (ii) Potassium trioxalatoferrate(III)

1. (i) Add $\mathrm{AgN}{\mathrm{O}}_3$ solution to aqueous solutions of each compound:

   • $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{S}{\mathrm{O}}_4]\mathrm{Cl}$ gives free $\mathrm{C}{\mathrm{l}}^{-}$ ions, so a white precipitate of AgCl forms.
   • $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{Cl}]\mathrm{S}{\mathrm{O}}_4$ has its Cl inside the coordination sphere, so there is no precipitate.

   Conversely, $\mathrm{BaC}{\mathrm{l}}_2$ gives a white precipitate of $\mathrm{BaS}{\mathrm{O}}_4$ only with $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5\mathrm{Cl}]\mathrm{S}{\mathrm{O}}_4$. Different ions are released in solution, which proves the two are ionisation isomers.

   1 mark

2. (ii) Chelate effect: a di- or polydentate ligand binding a single metal ion through two or more donor atoms forms a ring (chelate). Such complexes are more stable than similar complexes of monodentate ligands.

   ½ mark

3. Example: $[\mathrm{Co}(\mathrm{en}{)}_3{]}^{3+}$ is more stable than $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$.

   ½ mark

Answer: (i) AgNO₃ gives a white AgCl precipitate only with [Co(NH₃)₅SO₄]Cl (BaCl₂ gives BaSO₄ only with [Co(NH₃)₅Cl]SO₄); (ii) extra stability of complexes formed by chelating ligands, e.g. [Co(en)₃]³⁺.

1. (i) Peptide linkage: an amide bond ($-\mathrm{CO}-\mathrm{NH}-$) formed between the –COOH group of one amino acid and the –NH₂ group of another, with loss of water. It is found in proteins.

   Glycosidic linkage: an oxide (–O–) bridge that joins two monosaccharide units, formed with loss of water. It is found in disaccharides and polysaccharides.

   1 mark

2. (ii) Essential amino acids cannot be synthesised in the body and must be obtained through the diet (e.g. valine, leucine, lysine).

   Non-essential amino acids can be synthesised in the body (e.g. glycine, alanine).

   1 mark

Answer: (i) Peptide: –CONH– amide bond between amino acids; glycosidic: –O– bond between monosaccharides. (ii) Essential amino acids must come from the diet; non-essential ones are made in the body.

1. Since the reaction is elementary (one step), the rate law follows the stoichiometry: ${\text{Rate}}_1=k[A{]}^2[B]$.

2. Reducing the volume to $\frac{1}{3}$ makes each concentration 3 times larger:
   ${\text{Rate}}_2=k(3[A]{)}^2(3[B])=27k[A{]}^2[B]=27\times {\text{Rate}}_1$.

   The rate becomes 27 times the original rate.

   1 mark

3. The order ($2+1=3$) depends only on the rate law, not on the volume, so it remains unchanged.

   1 mark

Answer: The rate increases 27 times; the order remains the same (3).

1. Let $p_0$ be the initial pressure and $x$ the decrease in the pressure of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$ at time $t$:

   	${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$	${\mathrm{C}}_2{\mathrm{H}}_4$	HCl
   $t=0$	$p_0$	0	0
   $t$	$p_0-x$	$x$	$x$

   The total pressure is $p_t=p_0+x$, so $x=p_t-p_0$ and $p_A=p_0-x=2p_0-p_t$.

   1 mark

2. With $p_0=0.30$ atm and $p_t=0.50$ atm:
   $x=0.20$ atm and $p_A=0.60-0.50=0.10$ atm.

   First-order equation: $k=\frac{2.303}{t}\log \frac{p_0}{2p_0-p_t}=\frac{2.303}{30}\log \frac{0.30}{0.10}$

   1 mark

3. $k=\frac{2.303\times 0.48}{30}=\frac{1.105}{30}=3.68\times 10^{-2}{\mathrm{s}}^{-1}$

   1 mark

Answer: $k\approx 3.68\times 10^{-2}{\mathrm{s}}^{-1}$ (0.0368 s⁻¹)

1. (i) At equilibrium the cell reaction stops, so $E_{cell}=0$. The Nernst equation then reduces to $E_{cell}^{\circ }=\frac{0.059}{n}\log K_c$ (at 298 K). Hence $K_c$ is related to $E_{cell}^{\circ }$, while $E_{cell}$ is zero.

   1 mark

2. (ii) Metal A ($E^{\circ }=-0.24$ V). Its reduction potential is below that of the SHE (0.00 V), so it can reduce ${\mathrm{H}}^{+}$ to ${\mathrm{H}}_2$. Metal B (+0.80 V) cannot do this.

   1 mark

3. (iii) On charging, the discharge reaction is reversed:
   $2\mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Pb}(\mathrm{s})+\mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})$

   1 mark

Answer: (i) At equilibrium E_cell = 0, so E°_cell = (0.059/n) log K_c; (ii) Metal A; (iii) 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq)

1. The mercury cell is a primary cell (it cannot be recharged).

   1 mark

2. Advantage over the dry cell: it gives a constant potential of about 1.35 V throughout its life, because the overall reaction involves no ions whose concentration changes. A dry cell's voltage falls as it is used. This makes the mercury cell suitable for hearing aids and watches.

   1 mark

3. Overall reaction: $\mathrm{Zn}(\mathrm{Hg})+\mathrm{HgO}(\mathrm{s})\to \mathrm{ZnO}(\mathrm{s})+\mathrm{Hg}(\mathrm{l})$

   1 mark

Answer: A primary cell; it gives a constant voltage (~1.35 V); Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

1. For dimerisation $2{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}\rightleftharpoons ({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}{)}_2$ with $\alpha =0.88$:
   $i=1-\alpha +\frac{\alpha }{2}=1-\frac{\alpha }{2}=1-0.44=0.56$

   ½ mark

2. Molality: $m=\frac{0.61/122}{5/1000}=\frac{0.005\text{mol}}{0.005\text{kg}}=1\mathrm{mol}\mathrm{k}{\mathrm{g}}^{-1}$

   ½ mark

3. $\Delta T_b=i{K}_{b}m=0.56\times 2.3\times 1=1.288\text{K}$

   1 mark

4. Boiling point of the solution $=46.2+1.288=47.488{}^{\circ }\mathrm{C}\approx 47.49{}^{\circ }\mathrm{C}$

   1 mark

Answer: Boiling point ≈ 47.49 °C (ΔT_b = 1.288 K, i = 0.56)

1. (a) Prolonged heating with HI reduces glucose to n-hexane, showing that its six C atoms are in a straight chain:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{\Delta }{\overset{\mathrm{HI}}{\to }}\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$

   1 mark

2. (b) The mild oxidant bromine water oxidises the –CHO group to give gluconic acid:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\mathrm{B}{\mathrm{r}}_2\mathrm{water}}{\to }\mathrm{COOH}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$

   1 mark

3. (c) Conc. $\mathrm{HN}{\mathrm{O}}_3$ oxidises both the –CHO and the terminal –CH₂OH groups to –COOH, giving saccharic (glucaric) acid:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{conc.\mathrm{HN}{\mathrm{O}}_3}{\to }\mathrm{COOH}-(\mathrm{CHOH}{)}_4-\mathrm{COOH}$

   1 mark

Answer: (a) n-hexane; (b) gluconic acid; (c) saccharic acid

1. (a) Carboxylic acid molecules are held by extensive intermolecular hydrogen bonding. They exist as cyclic dimers even in the vapour phase, so more energy is needed to separate them than for alcohols of similar mass.

   1 mark

2. (b) The carbonyl group is strongly electron-withdrawing, which weakens the α C–H bond. After loss of ${\mathrm{H}}^{+}$, the carbanion (enolate) left behind is resonance-stabilised, with the negative charge shared by the C and O atoms. This makes α-hydrogens acidic.

   1 mark

3. (c) In strongly acidic medium, the lone pair of ammonia or its derivative ($\mathrm{N}{\mathrm{H}}_2-\mathrm{Z}$) is protonated to $\mathrm{N}{\mathrm{H}}_3^{+}-\mathrm{Z}$. The reagent then loses its nucleophilic character and cannot attack the carbonyl carbon, so a weakly acidic medium (pH about 3.5) is used.

   1 mark

Answer: (a) Extensive intermolecular H-bonding and dimer formation; (b) electron-withdrawing C=O and resonance-stabilised enolate; (c) the nucleophile is protonated in strong acid.

1. (a) Reimer–Tiemann reaction: phenol treated with chloroform in aqueous NaOH introduces –CHO at the ortho position. The intermediate (substituted benzal chloride, then sodium salicylaldehyde) is acidified to give salicylaldehyde:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}\underset{\text{aq. NaOH}}{\overset{\mathrm{CHC}{\mathrm{l}}_3}{\to }}o\text{-}{\mathrm{C}}_6{\mathrm{H}}_4({\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+})\mathrm{CHO}\stackrel{{\mathrm{H}}^{+}}{\to }o\text{-}\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CHO}$ (salicylaldehyde)

   1 mark

2. (b) Kolbe's reaction: sodium phenoxide reacts with $\mathrm{C}{\mathrm{O}}_2$ (about 400 K, 4–7 atm) to give sodium salicylate, which is acidified to give salicylic acid:
   ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}\underset{400K,4\text{–}7atm}{\overset{\mathrm{C}{\mathrm{O}}_2}{\to }}o\text{-}\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{CO}{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}\stackrel{{\mathrm{H}}^{+}}{\to }o\text{-}\mathrm{HO}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{COOH}$ (2-hydroxybenzoic acid)

   1 mark

3. (c) Friedel–Crafts acylation of anisole:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OC}{\mathrm{H}}_3+\mathrm{C}{\mathrm{H}}_3\mathrm{COCl}\stackrel{\text{anhyd. }\mathrm{AlC}{\mathrm{l}}_3}{\to }p\text{-}\mathrm{C}{\mathrm{H}}_3\mathrm{O}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{COC}{\mathrm{H}}_3(\text{major})+o\text{-}\mathrm{C}{\mathrm{H}}_3\mathrm{O}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{COC}{\mathrm{H}}_3(\text{minor})+\mathrm{HCl}$

   The products are 4-methoxyacetophenone (major) and 2-methoxyacetophenone (minor).

   1 mark

Answer: (a) Phenol → salicylaldehyde; (b) sodium phenoxide → salicylic acid; (c) anisole → 4-methoxyacetophenone (major) + 2-methoxyacetophenone (minor)

1. (a) The rate depends only on [X], so X reacts by $S_{N}1$; it is a tertiary bromide. **X = $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{Br}$** (2-bromo-2-methylpropane).

   The optically active isomer that reacts by $S_{N}2$ (rate depends on both [Y] and [OH⁻]) must be chiral. **Y = $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}(\mathrm{Br})\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$** (2-bromobutane).

   1 mark

2. (b) X (the $S_{N}1$ substrate) undergoes racemisation. In $S_{N}1$ a planar carbocation forms, and $\mathrm{O}{\mathrm{H}}^{-}$ can attack it equally from either face, giving a racemic mixture.

   1 mark

3. (c) Y gives the product (butan-2-ol) with inversion of configuration. In $S_{N}2$ the nucleophile attacks from the side opposite to the leaving Br⁻ (backside attack), which inverts the configuration (Walden inversion).

   1 mark

Answer: (a) X = (CH₃)₃CBr, Y = CH₃CH(Br)CH₂CH₃; (b) X (S_N1, planar carbocation); (c) Y (S_N2, backside attack)

1. (a)(i) In water, basicity is decided by the combined effect of the +I effect, solvation (H-bonding) of the conjugate acid, and steric hindrance. $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_3^{+}$ has three N–H hydrogens to H-bond with water, so it is well stabilised by solvation. $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{N}{\mathrm{H}}^{+}$ has only one, and its bulky methyl groups also hinder solvation. Hence $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ is the stronger base in aqueous solution.

   1 mark

2. (a)(ii) Hoffmann bromamide degradation gives **A = $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$** (methanamine).

   ½ mark

3. Benzoylation with ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COCl}$ in base gives **B = $\mathrm{C}{\mathrm{H}}_3\mathrm{NHCO}{\mathrm{C}}_6{\mathrm{H}}_5$** (N-methylbenzamide).

   ½ mark

4. (b) [first alternative] A product insoluble in alkali means the amine is secondary. **X = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{NHC}{\mathrm{H}}_3$** (N-methylethanamine).

   (b) [second alternative] Aniline is diazotised with $\mathrm{NaN}{\mathrm{O}}_2+\mathrm{HCl}$ at 273–278 K to form benzenediazonium chloride, which is treated with CuCN/KCN (Sandmeyer reaction):
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2\underset{273\text{–}278K}{\overset{\mathrm{NaN}{\mathrm{O}}_2/HCl}{\to }}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}\stackrel{CuCN/KCN}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CN}+{\mathrm{N}}_2$

   1 mark

5. (c) –NH₂ strongly activates the ring and gets protonated in the nitrating mixture to the meta-directing anilinium ion; it is also oxidised by $\mathrm{HN}{\mathrm{O}}_3$. Acetylation to $-\mathrm{NHCOC}{\mathrm{H}}_3$ lowers the activating effect, prevents anilinium-ion formation and oxidation, and allows controlled nitration giving the p-nitro derivative as the major product.

   1 mark

Answer: (a)(i) Greater solvation (H-bonding) and less steric hindrance of CH₃NH₃⁺; (ii) A = CH₃NH₂, B = CH₃NHCOC₆H₅; (b) X = CH₃CH₂NHCH₃ / aniline → NaNO₂, HCl (0–5 °C) → CuCN/KCN → benzonitrile; (c) to control nitration and prevent anilinium ion formation, giving the p-nitro product.

1. (a) The energies of the $e_g$ orbitals, **$d_{x^2-y^2}$ and $d_{z^2}$**, are raised.

   1 mark

2. Reason: in an octahedral field the six ligands approach along the x, y and z axes. The lobes of $d_{x^2-y^2}$ and $d_{z^2}$ point along these axes, so they are repelled more strongly than $d_{xy}$, $d_{yz}$ and $d_{xz}$, whose lobes lie between the axes.

   1 mark

3. (b)(i) In $[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}$, $\mathrm{C}{\mathrm{o}}^{3+}$ is $3d^6$. ${\mathrm{F}}^{-}$ is a weak-field ligand (${\Delta }_o<P$), giving the high-spin configuration **$t_{2g}^4{e}_g^2$**.

   ½ mark

4. (b)(ii) In $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6{]}^{3+}$, $\mathrm{C}{\mathrm{o}}^{3+}$ is $3d^6$. $\mathrm{N}{\mathrm{H}}_3$ is a strong-field ligand (${\Delta }_o>P$), giving the low-spin configuration **$t_{2g}^6{e}_g^0$**.

   ½ mark

5. (c) [first alternative] In $[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$, $\mathrm{N}{\mathrm{i}}^{2+}$ is $3d^8$. $\mathrm{C}{\mathrm{l}}^{-}$ is a weak-field ligand and does not pair the electrons, so two unpaired electrons remain ($s{p}^3$), making it paramagnetic. In $[\mathrm{Ni}(\mathrm{CO}{)}_4]$, Ni(0) is $3d^{8}4s^2$. CO is a strong-field ligand and pushes the 4s electrons into 3d to give $3d^{10}$ with all electrons paired ($s{p}^3$), making it diamagnetic.

   ½ mark

6. (c) [second alternative] In $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{3-}$, $\mathrm{F}{\mathrm{e}}^{3+}$ is $3d^5$. Strong-field $\mathrm{C}{\mathrm{N}}^{-}$ pairs the electrons into three 3d orbitals, leaving two 3d orbitals empty. Hybridisation is **$d^{2}s{p}^3$ (inner-orbital octahedral) and, with one unpaired electron, the complex is paramagnetic**.

   ½ mark

Answer: (a) $d_{x^2-y^2}$ and $d_{z^2}$, whose lobes lie along the axes of ligand approach; (b)(i) $t_{2g}^4{e}_g^2$, (ii) $t_{2g}^6{e}_g^0$; (c) Cl⁻ weak field leaves 2 unpaired electrons (paramagnetic), CO strong field pairs them all (diamagnetic) / [Fe(CN)₆]³⁻: d²sp³, paramagnetic.

1. (i)(I) No Tollens' test means it is not an aldehyde; a positive iodoform test means it is a methyl ketone ($\mathrm{C}{\mathrm{H}}_3\mathrm{CO}-$). **X = $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$** (pentan-2-one); 3-methylbutan-2-one, $\mathrm{C}{\mathrm{H}}_3\mathrm{COCH}(\mathrm{C}{\mathrm{H}}_3{)}_2$, is also acceptable.

   1 mark

2. (i)(II) It is a ketone that is not a methyl ketone but has α-hydrogens, so it can undergo aldol condensation. **X = $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$** (pentan-3-one).

   1 mark

3. (i)(III) Cannizzaro reaction requires an aldehyde with no α-hydrogen. **X = $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{CHO}$** (2,2-dimethylpropanal).

   1 mark

4. (ii)(I) Acetophenone undergoes the haloform reaction:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_3\stackrel{\mathrm{NaOI}({\mathrm{I}}_2/NaOH)}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+\mathrm{CH}{\mathrm{I}}_3\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$

   1 mark

5. (ii)(II) The side chain of ethylbenzene is oxidised:
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3\underset{\Delta }{\overset{\mathrm{KMn}{\mathrm{O}}_4/KOH}{\to }}{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOK}\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}$

   1 mark

Answer: (i) (I) pentan-2-one; (II) pentan-3-one; (III) 2,2-dimethylpropanal. (ii) (I) NaOI then H₃O⁺; (II) alkaline KMnO₄, heat, then H₃O⁺.

1. (i) The 2,4-DNP derivative of benzaldehyde is ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CH}=\mathrm{N}-\mathrm{NH}-{\mathrm{C}}_6{\mathrm{H}}_3(\mathrm{N}{\mathrm{O}}_2{)}_2$. The benzene ring on NH carries –NO₂ groups at positions 2 and 4.

   1 mark

2. (ii) Reactivity towards HCN falls as steric hindrance and the +I effect at the carbonyl carbon increase:
   Di-tert-butyl ketone < Acetone < Acetaldehyde

   1 mark

3. (iii) Add $\mathrm{NaHC}{\mathrm{O}}_3$ solution to each. Benzoic acid gives brisk effervescence of $\mathrm{C}{\mathrm{O}}_2$; ethyl benzoate gives none.
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COOH}+\mathrm{NaHC}{\mathrm{O}}_3\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2\uparrow$

   1 mark

4. (iv) The reagent is DIBAL-H (diisobutylaluminium hydride) followed by ${\mathrm{H}}_2\mathrm{O}$. Alternatively, use $\mathrm{SnC}{\mathrm{l}}_2/HCl$ followed by ${\mathrm{H}}_3{\mathrm{O}}^{+}$ (Stephen reduction).
   $\mathrm{C}{\mathrm{H}}_3\mathrm{CN}\underset{(ii){\mathrm{H}}_2\mathrm{O}}{\overset{(i)\mathrm{DIBAL}\text{-}\mathrm{H}}{\to }}\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$

   1 mark

5. (v) $\mathrm{Cr}{\mathrm{O}}_3$ oxidises the secondary alcohol to a ketone. X = cyclohexanone, a six-membered ring with a C=O group.

   1 mark

Answer: (i) C₆H₅CH=N–NH–C₆H₃(NO₂)₂-2,4; (ii) Di-tert-butyl ketone < Acetone < Acetaldehyde; (iii) NaHCO₃ test (effervescence with benzoic acid); (iv) DIBAL-H then H₂O (or SnCl₂/HCl, H₃O⁺); (v) X = cyclohexanone

1. (i)(I) $E_{M^{2+}/M}^{\circ }$ depends on the enthalpy of atomisation (sublimation), the sum of the first two ionisation enthalpies (${\Delta }_i{H}_1+{\Delta }_i{H}_2$), and the hydration enthalpy of ${\mathrm{M}}^{2+}$. Ionisation and sublimation enthalpies vary irregularly across the series, so the $E^{\circ }$ values show an irregular trend.

   1 mark

2. (i)(II) Cu has a high enthalpy of atomisation and a high ${\Delta }_i{H}_1+{\Delta }_i{H}_2$. The relatively low hydration enthalpy of $\mathrm{C}{\mathrm{u}}^{2+}$ does not compensate for the energy needed to form $\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})$ from Cu(s), so $E^{\circ }$ is positive (+0.34 V).

   1 mark

3. (i)(III) $\mathrm{M}{\mathrm{n}}^{2+}$ has the extra-stable half-filled $3d^5$ configuration (and Mn has a low atomisation enthalpy). Mn is therefore easily oxidised to $\mathrm{M}{\mathrm{n}}^{2+}$, and $E^{\circ }$ is highly negative.

   1 mark

4. (ii) Acidic solution (I⁻ is oxidised to I₂):
   $2\mathrm{Mn}{\mathrm{O}}_4^{-}+10{\mathrm{I}}^{-}+16{\mathrm{H}}^{+}\to 2\mathrm{M}{\mathrm{n}}^{2+}+5{\mathrm{I}}_2+8{\mathrm{H}}_2\mathrm{O}$

   1 mark

5. Alkaline (or neutral) solution (I⁻ is oxidised to iodate):
   $2\mathrm{Mn}{\mathrm{O}}_4^{-}+{\mathrm{I}}^{-}+{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{Mn}{\mathrm{O}}_2+\mathrm{I}{\mathrm{O}}_3^{-}+2\mathrm{O}{\mathrm{H}}^{-}$

   1 mark

Answer: (i) (I) Irregular variation of ionisation and sublimation enthalpies; (II) high ΔₐH and low hydration enthalpy of Cu²⁺; (III) stable half-filled d⁵ Mn²⁺. (ii) Acidic: 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O; alkaline: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻

1. (i)(I) Cerium (Ce) shows the +4 oxidation state. Ce${}^{4+}$ has the stable noble-gas configuration $4f^0$.

   ½ mark

2. (i)(II) Europium (Eu) shows the +2 oxidation state. Eu${}^{2+}$ has the stable half-filled configuration $4f^7$. Ytterbium (Yb${}^{2+}$, $4f^{14}$) is also accepted.

   ½ mark

3. (ii) Transition metals act as good catalysts for these reasons:

   • They can show variable (multiple) oxidation states, so they easily form and break down unstable intermediates.
   • They can form complexes with the reactants.
   • Their large surface area adsorbs reactant molecules, which weakens the bonds in the reactants and lowers the activation energy.
   1 mark

4. (iii) Cr ($3d^{5}4s^1$) has six unpaired electrons. All of them take part in metallic bonding, so its interatomic metallic bonds are strong. Mn ($3d^{5}4s^2$) has a stable half-filled $3d^5$ subshell that holds its electrons tightly, so fewer electrons are available for metallic bonding. The weaker metallic bonding in Mn gives it a lower melting point than Cr.

   1 mark

5. (iv) An acidified KMnO${}_4$ solution slowly decomposes on standing. It forms a brown precipitate of MnO${}_2$ and gives off oxygen:
   $$4\mathrm{Mn}{\mathrm{O}}_4^{-}(\mathrm{aq})+4{\mathrm{H}}^{+}(\mathrm{aq})\to 4\mathrm{Mn}{\mathrm{O}}_2(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$

   1 mark

6. In this reaction Mn is reduced from +7 to +4, and oxygen is oxidised from −2 to 0. It is therefore a redox reaction (a self-decomposition in which MnO${}_4^{-}$ is reduced).

   1 mark

Answer: (i) (I) Cerium; (II) Europium (or Ytterbium). (ii) Variable oxidation states, complex formation and large surface area. (iii) Greater number of electrons (3d + 4s) in metallic bonding in Cr. (iv) $4\mathrm{Mn}{\mathrm{O}}_4^{-}+4{\mathrm{H}}^{+}\to 4\mathrm{Mn}{\mathrm{O}}_2+3{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$; redox (reduction) reaction.

1. The cell reaction is $\mathrm{Mg}(\mathrm{s})+2\mathrm{A}{\mathrm{g}}^{+}(\mathrm{aq})\to \mathrm{M}{\mathrm{g}}^{2+}(\mathrm{aq})+2\mathrm{Ag}(\mathrm{s})$, so $n=2$.

   $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0.80-(-2.37)=3.17\text{V}$

   1 mark

2. Nernst equation at 298 K:
   $$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{n}\log \frac{[\mathrm{M}{\mathrm{g}}^{2+}]}{[\mathrm{A}{\mathrm{g}}^{+}{]}^2}$$

   1 mark

3. Substitute the concentrations:
   $$E_{cell}=3.17-\frac{0.059}{2}\log \frac{0.01}{(0.001{)}^2}=3.17-0.0295\log 10^4$$
   $$E_{cell}=3.17-0.0295\times 4=3.17-0.118=3.052\text{V}$$

   1 mark

4. Gibbs energy change: $\Delta G=-nF{E}_{cell}$

   ½ mark

5. $\Delta G=-2\times 96500\times 3.052$

   ½ mark

6. $\Delta G=-589036{\text{J mol}}^{-1}\approx -589.04{\text{kJ mol}}^{-1}$

   1 mark

Answer: $E_{cell}=3.052$ V; $\Delta G=-589036$ J mol${}^{-1}$ (≈ −589.04 kJ mol${}^{-1}$)

1. From the reaction, 2 electrons are transferred (${\mathrm{H}}_2\to 2{\mathrm{H}}^{+}+2e^{-}$), so $n=2$. Use $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$.

   ½ mark

2. $$E_{cell}^{\circ }=\frac{-\Delta G^{\circ }}{nF}=\frac{43500}{2\times 96500}=0.2254\text{V}$$

   1 mark

3. Nernst equation at 298 K:
   $$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{2}\log \frac{[{\mathrm{H}}^{+}{]}^2[\mathrm{C}{\mathrm{l}}^{-}{]}^2}{p_{{\mathrm{H}}_2}}$$
   (AgCl and Ag are solids, so they do not appear in Q.)

   ½ mark

4. Calculate the reaction quotient:
   $$Q=\frac{(0.1{)}^2(0.2{)}^2}{0.4}=\frac{0.01\times 0.04}{0.4}=10^{-3}$$

   1 mark

5. $$E_{cell}=0.2254-0.0295\log 10^{-3}=0.2254-0.0295\times (-3)$$

   1 mark

6. $$E_{cell}=0.2254+0.0885=0.3139\text{V}\approx 0.314\text{V}$$

   1 mark

Answer: $E_{cell}\approx 0.314$ V (with $E_{cell}^{\circ }=0.225$ V)