CBSE Class 12 Chemistry: 2025 board paper, set 56/7/1

1. Colour in transition metal ions arises from d–d transitions, which need a partly filled d subshell.

2. $\mathrm{C}{\mathrm{u}}^{+}$ is $3d^{10}$, a completely filled subshell, so no d–d transition is possible and the ion is colourless. $\mathrm{N}{\mathrm{i}}^{2+}$ ($3d^8$), $\mathrm{C}{\mathrm{o}}^{2+}$ ($3d^7$) and ${\mathrm{V}}^{3+}$ ($3d^2$) all have unpaired electrons and are coloured.

   1 mark

Answer: (A) $\mathrm{C}{\mathrm{u}}^{+}$

1. $\Delta T_b=i{K}_{b}m$. For 1% solutions, the number of particles is proportional to $i/M$.

2. The values of $i/M$ are: KCl $2/74.5=0.027$; glucose $1/180=0.0056$; urea $1/60=0.017$; $\mathrm{CaC}{\mathrm{l}}_2$ $3/111=0.027$. $\mathrm{CaC}{\mathrm{l}}_2$ gives slightly more particles (0.0270 vs 0.0268), so it has the highest boiling point.

   1 mark

Answer: (D) 1% $\mathrm{CaC}{\mathrm{l}}_2$

1. At the anode, water is oxidised in preference to $\mathrm{S}{\mathrm{O}}_4^{2-}$: $2H_{2}O(l)\to O_2(g)+4H^{+}(aq)+4e^{-}$. Oxygen gas is therefore evolved.

   1 mark

Answer: (B) ${\mathrm{O}}_2$ gas

1. The Arrhenius equation gives $\ln k=\ln A-\frac{E_a}{R}\cdot \frac{1}{T}$.

2. A plot of $\ln k$ vs $1/T$ is a straight line with slope $-E_a/R$, from which $E_a$ is found.

   1 mark

Answer: (C) $\ln k$ vs $1/T$

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

   1 mark

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

1. In $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_3\mathrm{C}{\mathrm{l}}_3]$ all three $\mathrm{C}{\mathrm{l}}^{-}$ are ligands inside the coordination sphere, so the complex is neutral and gives no free $\mathrm{C}{\mathrm{l}}^{-}$ ions. Hence 0 mol of AgCl is precipitated.

   1 mark

Answer: (A) 0

1. The $S_{N}1$ rate depends on carbocation stability. 2-Bromo-2-methylpropane forms a stable tertiary carbocation, and $\mathrm{B}{\mathrm{r}}^{-}$ is a good leaving group, so it reacts fastest.

   1 mark

Answer: (C) 2-Bromo-2-methylpropane

1. The alcohol gives up a proton, which is released as ${\mathrm{H}}_2$ when it reacts with an active metal. This shows that alcohols behave as (weak) acids.

   1 mark

Answer: (A) Acidic

1. $\mathrm{C}{\mathrm{S}}_2$ is a non-polar solvent and the temperature is low, so bromination is controlled. Monobromination occurs at the ortho and para positions, giving o- and p-bromophenol.

   1 mark

Answer: (C) o- and p-bromophenol

1. With a large excess of $\mathrm{N}{\mathrm{H}}_3$, each alkyl iodide molecule is far more likely to meet ammonia than an already-formed amine. Further alkylation is suppressed, so the primary amine $\mathrm{RN}{\mathrm{H}}_2$ is the major product.

   1 mark

Answer: (D) Primary amine

1. A primary amine forms an N-alkylbenzenesulphonamide with Hinsberg reagent. Its N–H is acidic because of the strong electron-withdrawing sulphonyl group, so the product dissolves in alkali.

   1 mark

2. Secondary amines, including N-methylaniline, give products with no N–H, which are insoluble in alkali. Tertiary amines do not react. Hence X is $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$.

Answer: (A) $\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$

1. The α-helix is a regular coiling of the polypeptide chain held by intramolecular H-bonds between –C=O and –NH groups. This is the secondary structure of a protein.

   1 mark

Answer: (B) Secondary structure

1. The assertion is true: the isomers are separated by steam distillation.

2. The reason is false: o-nitrophenol is steam volatile because of intramolecular H-bonding, while p-nitrophenol is less volatile because of intermolecular H-bonding (association).

   1 mark

Answer: (C)

1. The higher pressure inside the cooker raises the boiling point of water. Food therefore cooks at a higher temperature and in less time, so R correctly explains A.

   1 mark

Answer: (A)

1. Actinoids show irregular configurations because the 5f, 6d and 7s orbitals have comparable energies.

2. Actinoids are also radioactive, but radioactivity does not explain their configurations. Both statements are true and R is not the correct explanation.

   1 mark

Answer: (B)

1. Vitamin K is fat-soluble, so it is stored in adipose tissue and the liver. A is true and R is false.

   1 mark

Answer: (C)

1. Positive deviation: a solution shows positive deviation from Raoult's law when its vapour pressure is higher than the value predicted by Raoult's law. This happens because the A–B (solute–solvent) attractions are weaker than the A–A and B–B attractions, so molecules escape more easily. For such solutions ${\Delta }_{mix}H>0$ and ${\Delta }_{mix}V>0$.

   1 mark

2. Example: a mixture of ethanol and acetone. Carbon disulphide + acetone is another example.

   ½ mark

3. Solutions showing a large positive deviation form a minimum boiling azeotrope. An example is about 95% ethanol–water.

   ½ mark

Answer: Vapour pressure is higher than predicted by Raoult's law; e.g. ethanol + acetone; minimum boiling azeotrope.

1. Condition: a bimolecular reaction behaves as first order (pseudo first order) when one of the two reactants is present in large excess. Its concentration then stays practically constant during the reaction.

   1 mark

2. Example: acid hydrolysis of ethyl acetate, $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{H}}^{+}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$. Water is in large excess, so rate $=k^{\prime }[\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5]$. Inversion of cane sugar is another example.

   ½ mark

3. Order and molecularity have the same value for elementary reactions, i.e. reactions that occur in a single step.

   ½ mark

Answer: One reactant in large excess (pseudo first order), e.g. acid hydrolysis of ethyl acetate; order = molecularity for elementary (single-step) reactions.

1. IUPAC name: Dichloridobis(ethane-1,2-diamine)platinum(IV) ion. The oxidation state of Pt is $x+0+2(-1)=+2$, which gives $x=+4$.

   1 mark

2. Optically inactive isomer: the trans isomer. Pt sits at the centre of an octahedron. The two $\mathrm{C}{\mathrm{l}}^{-}$ ligands occupy opposite (axial) positions at 180° to each other. The two bidentate en ligands span the four equatorial positions, each en joining two adjacent corners of the square plane. This isomer has a plane of symmetry, so it is optically inactive. (The cis isomer is optically active.)

   1 mark

Answer: Dichloridobis(ethane-1,2-diamine)platinum(IV) ion; the trans isomer (Cl–Pt–Cl at 180°, both en in the equatorial plane) is optically inactive.

1. (i) The carbonate ion is a −2 ligand and Co is +3, so the complex ion is $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{C}{\mathrm{O}}_3){]}^{+}$. One chloride balances it: $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{C}{\mathrm{O}}_3)]\mathrm{Cl}$.

   1 mark

2. (ii) In the given complex, $\mathrm{N}{\mathrm{O}}_2^{-}$ is bonded through N (nitrito-N). Its linkage isomer is $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{ONO})]\mathrm{C}{\mathrm{l}}_2$, in which the ligand is bonded through O. Its name is Pentaamminenitrito-O-cobalt(III) chloride.

   1 mark

Answer: (i) $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_5(\mathrm{C}{\mathrm{O}}_3)]\mathrm{Cl}$; (ii) Pentaamminenitrito-O-cobalt(III) chloride

1. Low reactivity: the lone pair on halogen is in resonance with the benzene ring, so the C–X bond gets partial double bond character. In addition, the carbon of C–X is $s{p}^2$ hybridised, so the bond is shorter and stronger. Both effects make the bond hard to break, and haloarenes are less reactive towards nucleophilic substitution.

   1 mark

2. **Effect of $-N{O}_2$ at o/p positions:** the nitro group withdraws electron density from the ring by its $-R$ and $-I$ effects, which makes the ring carbon easier for a nucleophile to attack. It also stabilises the intermediate carbanion, because the negative charge is delocalised onto the oxygen atoms of $-N{O}_2$. This is possible only when $-N{O}_2$ is at the ortho or para position, so reactivity increases.

   1 mark

Answer: C–X has partial double bond character (resonance, $s{p}^2$ C), so it is hard to break; o/p $-N{O}_2$ withdraws electrons and stabilises the carbanion intermediate.

1. Complementary strands: the two strands are held together by hydrogen bonds formed only between specific base pairs. Adenine pairs with thymine (two H-bonds) and guanine pairs with cytosine (three H-bonds). The base sequence of one strand therefore fixes the sequence of the other. The strands are complementary, not identical.

   1 mark

2. Products of hydrolysis of DNA: 2-deoxy-D-ribose (sugar), nitrogenous bases (adenine, guanine, cytosine, thymine) and phosphoric acid.

   1 mark

Answer: Specific base pairing (A–T, G–C) makes the strands complementary; hydrolysis gives 2-deoxyribose, nitrogenous bases (A, G, C, T) and phosphoric acid.

1. Formula used: $\Delta T_f=i{K}_{f}m$, where $m=\frac{w_B\times 1000}{M_B\times w_A}$.

   ½ mark

2. Substitution: $0.45=i\times 5.12\times \frac{0.3\times 1000}{60\times 30}$

   $0.45=i\times 5.12\times 0.1667=i\times 0.8533$

   So $i=0.527$.

   1 mark

3. For dimerisation, $2\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}\rightleftharpoons (\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}{)}_2$, so $n=2$ and $i=1-\alpha +\frac{\alpha }{2}$. This gives $\alpha =\frac{i-1}{\frac{1}{n}-1}$.

   ½ mark

4. $\alpha =\frac{0.527-1}{\frac{1}{2}-1}=\frac{-0.473}{-0.5}$

   ½ mark

5. $\alpha =0.946$, i.e. percentage association = 94.6%.

   ½ mark

Answer: i = 0.527; degree of association = 0.946, i.e. 94.6%

1. The cell generally used in inverters is the lead storage battery.

   1 mark

2. Anode (oxidation): $\mathrm{Pb}(\mathrm{s})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})\to \mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{e}}^{-}$

   1 mark

3. Cathode (reduction): $\mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})+4{\mathrm{H}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   Overall reaction: $\mathrm{Pb}+\mathrm{Pb}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to 2PbS{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: Lead storage battery; anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻; cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O

1. At cathode: two reductions are possible.

   • $\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Na}$, with $E^{\circ }=-2.71$ V
   • ${\mathrm{H}}_2\mathrm{O}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2+\mathrm{O}{\mathrm{H}}^{-}$, with $E^{\circ }=-0.83$ V

   The reaction with the higher $E^{\circ }$ is preferred, so water is reduced and **${\mathrm{H}}_2$** is evolved.

   1 mark

2. At anode: two oxidations are possible.

   • $\mathrm{C}{\mathrm{l}}^{-}\to \frac{1}{2}\mathrm{C}{\mathrm{l}}_2+{\mathrm{e}}^{-}$, with $E^{\circ }=1.36$ V
   • $2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$, with $E^{\circ }=1.23$ V

   On $E^{\circ }$ values alone, water should be oxidised. However, oxygen evolution has a high overpotential, so oxidation of $\mathrm{C}{\mathrm{l}}^{-}$ is preferred and **$\mathrm{C}{\mathrm{l}}_2$** is liberated.

   1 mark

3. Overall reaction: $2NaCl(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2NaOH(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})$

   1 mark

Answer: H₂O (E° = −0.83 V) is reduced in preference to Na⁺ → H₂ at cathode; Cl₂ forms at anode because of the overpotential of O₂; 2NaCl + 2H₂O → 2NaOH + H₂ + Cl₂

1. Let the initial pressure be $p_i$ and the decrease in ${\mathrm{N}}_2{\mathrm{O}}_5$ pressure be $2x$.

   	$2{\mathrm{N}}_2{\mathrm{O}}_5$	$2{\mathrm{N}}_2{\mathrm{O}}_4$	${\mathrm{O}}_2$
   t = 0	$p_i$	0	0
   t	$p_i-2x$	$2x$	$x$

   Total pressure: $p_t=p_i-2x+2x+x=p_i+x$, so $x=p_t-p_i$.

   1 mark

2. Pressure of ${\mathrm{N}}_2{\mathrm{O}}_5$ at time t: $p_A=p_i-2(p_t-p_i)=3p_i-2p_t$

   $p_A=3(0.5)-2(0.625)=1.5-1.25=0.25$ atm

   1 mark

3. For a first order reaction, $k=\frac{2.303}{t}\log \frac{p_i}{p_A}=\frac{2.303}{100}\log \frac{0.5}{0.25}$

   $k=0.02303\times \log 2=0.02303\times 0.3010$

   $k=6.93\times 10^{-3}{\mathrm{s}}^{-1}$

   1 mark

Answer: k = 6.93 × 10⁻³ s⁻¹

1. (A): $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CH}-\mathrm{C}{\mathrm{H}}_2\mathrm{I}$, 1-iodo-2-methylpropane (isobutyl iodide). It is a primary iodide, and its Wurtz product is not n-octane.

   ½ mark

2. (B): $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{C}{\mathrm{H}}_2$, 2-methylpropene.

   ½ mark

3. (C): $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{I}$, 2-iodo-2-methylpropane. It forms by Markovnikov addition of HI and is an isomer of A.

   ½ mark

4. (D): $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CH}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3{)}_2$, 2,5-dimethylhexane, ${\mathrm{C}}_8{\mathrm{H}}_{18}$. It forms by the Wurtz reaction: $2(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CHC}{\mathrm{H}}_2\mathrm{I}+2\mathrm{Na}\stackrel{\text{dry ether}}{\to }\mathrm{D}+2\mathrm{NaI}$.

   ½ mark

5. Reaction of A with alcoholic KOH (dehydrohalogenation):
   $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CH}-\mathrm{C}{\mathrm{H}}_2\mathrm{I}+\mathrm{KOH}(alc.)\stackrel{\mathrm{\Delta }}{\to }(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{C}{\mathrm{H}}_2+\mathrm{KI}+{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: A = (CH₃)₂CHCH₂I; B = (CH₃)₂C=CH₂; C = (CH₃)₃CI; D = (CH₃)₂CHCH₂CH₂CH(CH₃)₂

1. (a) ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OC}{\mathrm{H}}_3+\mathrm{HI}\to {\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{OH}+\mathrm{C}{\mathrm{H}}_3\mathrm{I}$

   The products are phenol and methyl iodide. The $\mathrm{O}-\mathrm{C}{\mathrm{H}}_3$ bond breaks because the $\mathrm{O}-{\mathrm{C}}_6{\mathrm{H}}_5$ bond has partial double bond character.

   1 mark

2. (b) Phenol + conc. $\mathrm{HN}{\mathrm{O}}_3$ gives 2,4,6-trinitrophenol (picric acid). Its structure is a benzene ring with –OH at C-1 and $-N{O}_2$ groups at C-2, C-4 and C-6.

   1 mark

3. (c) ${\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{MgBr}+\mathrm{HCHO}\to {\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{C}{\mathrm{H}}_2\mathrm{OMgBr}\stackrel{{\mathrm{H}}^{+}/{\mathrm{H}}_2\mathrm{O}}{\to }{\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{Mg}(\mathrm{OH})\mathrm{Br}$

   The product is cyclohexylmethanol, a primary alcohol: a cyclohexane ring bearing a $-C{H}_{2}OH$ group.

   1 mark

Answer: (a) Phenol + CH₃I; (b) 2,4,6-trinitrophenol (picric acid); (c) Cyclohexylmethanol, C₆H₁₁CH₂OH

1. (a) The –COOH group is electron-withdrawing, so it deactivates the benzene ring towards electrophilic substitution. Also, the Lewis acid catalyst $\mathrm{AlC}{\mathrm{l}}_3$ bonds to the carboxyl group (forms a salt or complex). The catalyst is then unavailable, so the Friedel-Crafts reaction does not occur.

   1 mark

2. (b) In CH₃CHO, the methyl group's +I effect lowers the positive charge on the carbonyl carbon, and it also adds steric hindrance. HCHO has no alkyl group, so its carbonyl carbon is more electrophilic and less hindered. HCHO is therefore more reactive towards nucleophilic addition of $\mathrm{C}{\mathrm{N}}^{-}$.

   1 mark

3. (c) In $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-\mathrm{COOH}$, the carboxyl carbon is attached to an $s{p}^2$ carbon. An $s{p}^2$ carbon has more s-character, so it is more electronegative than an $s{p}^3$ carbon. It withdraws electrons and stabilises the carboxylate ion. This effect outweighs resonance, so acidity increases.

   1 mark

Answer: (a) –COOH deactivates the ring and AlCl₃ binds to it; (b) HCHO carbonyl carbon is more electrophilic (no +I or steric effect); (c) the electronegative sp² carbon withdraws electrons and stabilises the carboxylate.

1. (a) Glucose forms a cyanohydrin at the –CHO group:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{HCN}\to \mathrm{CH}(\mathrm{OH})\mathrm{CN}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ (glucose cyanohydrin)

   1 mark

2. (b) Bromine water, a mild oxidant, oxidises –CHO to –COOH:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\mathrm{B}{\mathrm{r}}_2\mathrm{water}}{\to }\mathrm{COOH}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ (gluconic acid)

   1 mark

3. (c) Acetylation of all five –OH groups:
   $\mathrm{CHO}-(\mathrm{CHOH}{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{(\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{)}_2\mathrm{O}}{\to }\mathrm{CHO}-(\mathrm{CHOCOC}{\mathrm{H}}_3{)}_4-\mathrm{C}{\mathrm{H}}_2\mathrm{OCOC}{\mathrm{H}}_3$ (glucose pentaacetate)

   1 mark

Answer: (a) Glucose cyanohydrin; (b) Gluconic acid; (c) Glucose pentaacetate

1. (a) Colour: Ti³⁺ has configuration [Ar]3d¹, i.e. $t_{2g}^1{e}_g^0$ in the octahedral field of water. The electron absorbs visible light (green-yellow region) and is promoted from $t_{2g}$ to $e_g$, a d–d transition. The complex therefore appears violet.

   1 mark

2. (a) On heating: the complex loses its water ligands and becomes colourless. Without ligands there is no crystal field splitting, so no d–d transition is possible.

   1 mark

3. (b)(i) Crystal field splitting energy is the energy difference between the two sets of d-orbitals ($t_{2g}$ and $e_g$). These sets form when the degenerate d-orbitals split under the ligand field in a definite geometry (${\Delta }_o$ for octahedral).

   1 mark

4. (b)(ii) If ${\Delta }_o<P$, the fourth electron enters $e_g$ (high spin): this is a weak field ligand. If ${\Delta }_o>P$, the electron pairs in $t_{2g}$ (low spin): this is a strong field ligand.

5. (c) In tetrahedral complexes, ${\Delta }_t=\frac{4}{9}{\Delta }_o$ is small and generally less than the pairing energy. Electrons occupy the higher orbitals singly instead of pairing, so low spin tetrahedral complexes are rarely observed.

   1 mark

Answer: (a) d–d transition $t_{2g}^1\to e_g^1$ gives colour; on heating, water is lost, there is no splitting and it becomes colourless. (b)(i) Energy gap between $t_{2g}$ and $e_g$. (b)(ii) ${\Delta }_o<P$: weak field; ${\Delta }_o>P$: strong field. (c) ${\Delta }_t$ is too small to force pairing.

1. (a)(i) $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}\stackrel{\mathrm{N}{\mathrm{H}}_3,\mathrm{\Delta }}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CON}{\mathrm{H}}_2\stackrel{\mathrm{B}{\mathrm{r}}_2/NaOH}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2$ (Hoffmann bromamide degradation)

   1 mark

2. (a)(ii) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CN}\stackrel{\mathrm{LiAl}{\mathrm{H}}_4(\mathrm{or}{\mathrm{H}}_2/Ni)}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$

   1 mark

3. (b) In aniline, the lone pair on N is delocalised into the benzene ring by resonance, so it is less available to accept a proton. In methylamine, the +I effect of CH₃ increases electron density on N. Aniline is therefore a weaker base, and its $p{K}_b$ is higher.

   1 mark

4. (c)(i) $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{N}<\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2<(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{NH}$

   This order results from the combined effect of +I, solvation (H-bonding) and steric factors in water.

   1 mark

5. (c)(ii) A = ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2$ (aniline); B = ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{N}}_2^{+}\mathrm{C}{\mathrm{l}}^{-}$ (benzenediazonium chloride).

Answer: (a)(i) CH₃COOH → CH₃CONH₂ → (Br₂/NaOH) CH₃NH₂; (ii) reduction with LiAlH₄ gives CH₃CH₂CH₂NH₂. (b) Resonance delocalises the N lone pair in aniline. (c)(i) (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH; (c)(ii) A = C₆H₅NH₂, B = C₆H₅N₂⁺Cl⁻

1. (i)(I) Mn²⁺ has a stable half-filled $3d^5$ configuration. Mn²⁺ is therefore hard to reduce to Mn, so $E_{M{n}^{2+}/Mn}^{\circ }$ is highly negative. Mn³⁺ ($3d^4$) readily gains an electron to become stable $d^5$ Mn²⁺, so $E_{M{n}^{3+}/M{n}^{2+}}^{\circ }$ is highly positive.

   1 mark

2. (i)(II) In actinoids, the 5f, 6d and 7s orbitals have comparable energies. Electrons from all of them can take part in bonding, so actinoids show a wide range of oxidation states.

   1 mark

3. (i)(III) Transition metals have many unpaired electrons. Electrons from the (n−1)d orbitals, along with the ns electrons, take part in metallic bonding. This gives strong interatomic bonding and hence high melting points.

   1 mark

4. (ii)(I) $5S{\mathrm{O}}_3^{2-}+2Mn{\mathrm{O}}_4^{-}+6{\mathrm{H}}^{+}\to 2M{\mathrm{n}}^{2+}+5S{\mathrm{O}}_4^{2-}+3{\mathrm{H}}_2\mathrm{O}$

   1 mark

5. (ii)(II) $2Mn{\mathrm{O}}_4^{-}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}^{-}\to 2Mn{\mathrm{O}}_2+2O{\mathrm{H}}^{-}+\mathrm{I}{\mathrm{O}}_3^{-}$

   1 mark

Answer: (i) (I) Stable d⁵ Mn²⁺; (II) comparable energies of 5f, 6d, 7s; (III) strong metallic bonding involving (n−1)d electrons. (ii) (I) 2Mn²⁺ + 5SO₄²⁻ + 3H₂O; (II) 2MnO₂ + 2OH⁻ + IO₃⁻

1. (i) Mn and Zn. Removing the third electron would break the stable $d^5$ of Mn²⁺ and the stable $d^{10}$ of Zn²⁺. Ni and Cu are also acceptable.

   1 mark

2. (ii) ${\mathrm{K}}_2\mathrm{Mn}{\mathrm{O}}_4$ is paramagnetic. In it, Mn is in the +6 state ($3d^1$) with one unpaired electron. In $\mathrm{KMn}{\mathrm{O}}_4$, Mn(+7) is $3d^0$, so it is diamagnetic.

   1 mark

3. (iii) The 4d and 5d elements of the same group have almost identical atomic radii (e.g. Zr 160 pm, Hf 159 pm), so they have very similar properties. Another consequence is that lanthanoids are difficult to separate.

   1 mark

4. (iv) Pyrolusite ($\mathrm{Mn}{\mathrm{O}}_2$) is fused with KOH in the presence of air or an oxidising agent such as $\mathrm{KN}{\mathrm{O}}_3$:
   $2Mn{\mathrm{O}}_2+4KOH+{\mathrm{O}}_2\to 2{\mathrm{K}}_2\mathrm{Mn}{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}$

   1 mark

5. (v) Oxygen can form multiple bonds (pπ–dπ) with the metal, but fluorine can form only single bonds. For example, Mn forms $\mathrm{M}{\mathrm{n}}_2{\mathrm{O}}_7$ (+7), whereas its highest fluoride is $\mathrm{Mn}{\mathrm{F}}_4$ (+4).

   1 mark

Answer: (i) Mn, Zn; (ii) K₂MnO₄ (Mn⁶⁺, one unpaired electron); (iii) Zr and Hf have similar radii; (iv) fuse MnO₂ with KOH and an oxidant; (v) oxygen forms multiple bonds with metals.

1. (i)(I) To show the Cannizzaro reaction, the aldehyde must have no α-H: $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{CHO}$, 2,2-dimethylpropanal.

   1 mark

2. (i)(II) To reduce Tollens' reagent it must be an aldehyde, and it must also have a chiral carbon: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2-\stackrel{\ast }{\mathrm{C}}\mathrm{H}(\mathrm{C}{\mathrm{H}}_3)-\mathrm{CHO}$, 2-methylbutanal. C-2 carries H, CH₃, C₂H₅ and CHO.

   1 mark

3. (i)(III) To give a positive iodoform test it must be a methyl ketone: $\mathrm{C}{\mathrm{H}}_3-\mathrm{CO}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$, pentan-2-one. 3-Methylbutan-2-one is also acceptable.

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4. (ii)(I) Clemmensen reduction: the carbonyl group is reduced to $\mathrm{C}{\mathrm{H}}_2$.
   $\mathrm{C}{\mathrm{H}}_3-\mathrm{CO}-\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{Zn}-\mathrm{Hg},conc.\mathrm{HCl}}{\to }\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}$

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5. (ii)(II) Etard reaction: toluene is oxidised to benzaldehyde by chromyl chloride.
   ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{Cr}{\mathrm{O}}_2\mathrm{C}{\mathrm{l}}_2,\mathrm{C}{\mathrm{S}}_2}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}(\mathrm{OCrOHC}{\mathrm{l}}_2{)}_2\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$

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Answer: (i) (I) (CH₃)₃C–CHO; (II) CH₃CH₂CH(CH₃)CHO; (III) CH₃COCH₂CH₂CH₃. (ii) Clemmensen: >C=O → >CH₂ with Zn-Hg/HCl; Etard: toluene → benzaldehyde with CrO₂Cl₂.

1. (i) Methanal adds one molecule of methanol across its C=O bond in the presence of dry HCl to form a hemiacetal:
   $$\mathrm{HCHO}+\mathrm{C}{\mathrm{H}}_3\mathrm{OH}\stackrel{\mathrm{dry}\mathrm{HCl}}{\to }{\mathrm{H}}_2\mathrm{C}(\mathrm{OH})(\mathrm{OC}{\mathrm{H}}_3)$$
   Structure: a central $\mathrm{C}{\mathrm{H}}_2$ carbon bonded to one $-\mathrm{OH}$ group and one $-\mathrm{OC}{\mathrm{H}}_3$ group, written $\mathrm{HO}-\mathrm{C}{\mathrm{H}}_2-\mathrm{O}-\mathrm{C}{\mathrm{H}}_3$ (methoxymethanol).

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2. (ii) In semicarbazide, ${\mathrm{H}}_2\mathrm{N}-\mathrm{CO}-\mathrm{NH}-\mathrm{N}{\mathrm{H}}_2$, the $-\mathrm{N}{\mathrm{H}}_2$ group attached directly to the carbonyl carbon has its lone pair delocalised into the C=O group by resonance (${\mathrm{H}}_2\mathrm{N}-\mathrm{C}(=\mathrm{O})-\leftrightarrow {\mathrm{H}}_2{\mathrm{N}}^{+}=\mathrm{C}(-{\mathrm{O}}^{-})-$). That nitrogen is therefore not nucleophilic. The terminal $-\mathrm{N}{\mathrm{H}}_2$ (on the hydrazine nitrogen) takes no part in resonance, so its lone pair stays free and only this group attacks the carbonyl compound to form the semicarbazone.

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3. (iii) Oxidise ethanol to ethanal with PCC (or Cu at 573 K). Then carry out aldol condensation of ethanal with dilute NaOH:
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\mathrm{PCC}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CHO}$$
   $$2C{\mathrm{H}}_3\mathrm{CHO}\stackrel{dil.\mathrm{NaOH}}{\to }\mathrm{C}{\mathrm{H}}_3-\mathrm{CH}(\mathrm{OH})-\mathrm{C}{\mathrm{H}}_2-\mathrm{CHO}$$
   The product is 3-hydroxybutanal.

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4. (iv) $\mathrm{Cr}{\mathrm{O}}_3$ oxidises the secondary alcohol to a ketone:
   $$\text{cyclohexanol }({\mathrm{C}}_6{\mathrm{H}}_{11}\mathrm{OH})\stackrel{Cr{O}_3}{\to }\text{cyclohexanone }({\mathrm{C}}_6{\mathrm{H}}_{10}\mathrm{O})$$
   The product is a six-membered ring with a C=O group in place of the CH–OH group.

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5. (v) Phthalic acid first forms ammonium phthalate with $\mathrm{N}{\mathrm{H}}_3$. On heating this gives phthalamide, and on strong heating a molecule of $\mathrm{N}{\mathrm{H}}_3$ is lost to give the cyclic imide:
   $${\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{COOH}{)}_2\stackrel{\mathrm{N}{\mathrm{H}}_3}{\to }{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{COON}{\mathrm{H}}_4{)}_2\stackrel{\mathrm{\Delta },-2{\mathrm{H}}_2\mathrm{O}}{\to }{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CON}{\mathrm{H}}_2{)}_2\stackrel{\mathrm{strong}\mathrm{\Delta },-\mathrm{N}{\mathrm{H}}_3}{\to }{\mathrm{C}}_6{\mathrm{H}}_4(\mathrm{CO}{)}_2\mathrm{NH}$$
   The final product is phthalimide.

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Answer: (i) $\mathrm{HO}-\mathrm{C}{\mathrm{H}}_2-\mathrm{OC}{\mathrm{H}}_3$; (ii) the $-\mathrm{N}{\mathrm{H}}_2$ next to C=O is tied up in resonance, so only the terminal $-\mathrm{N}{\mathrm{H}}_2$ is nucleophilic; (iii) ethanol → ethanal (PCC) → aldol with dil. NaOH → 3-hydroxybutanal; (iv) cyclohexanone; (v) phthalimide

1. (i) Cell reaction: $\mathrm{Zn}(\mathrm{s})+2{\mathrm{H}}^{+}(\mathrm{aq})\to \mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$, with $n=2$.

   Nernst equation at 298 K:
   $$E_{cell}=(E_{cathode}^{\circ }-E_{anode}^{\circ })-\frac{0.059}{2}\log \frac{[Z{n}^{2+}]}{[H^{+}{]}^2}$$

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2. Substitute the values:
   $$E_{cell}=[0-(-0.76)]-\frac{0.059}{2}\log \frac{0.1}{(0.01{)}^2}=0.76-0.0295\log 10^3$$

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3. $$E_{cell}=0.76-0.0295\times 3=0.76-0.0885=0.6715\text{V}\approx 0.67\text{V}$$

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

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5. Half-reaction: $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2C{\mathrm{r}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$.

   Each Cr goes from +6 to +3, so 1 mol of $\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7^{2-}$ needs 6 mol of electrons. Electricity required = 6 F (= $6\times 96500=579000$ C).

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Answer: (i) $E_{cell}=0.6715$ V ≈ 0.67 V; (ii) amounts liberated ∝ equivalent weights; 6 F

1. (i) Molar conductivity:
   $${\Lambda }_m=\frac{\kappa \times 1000}{C}=\frac{2.48\times 10^{-2}\times 1000}{0.2}$$

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2. $${\Lambda }_m=124{\text{S cm}}^2{\text{ mol}}^{-1}$$

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3. Limiting molar conductivity:
   $${\Lambda }_m^{\circ }={\lambda }_{K^{+}}^{\circ }+{\lambda }_{C{l}^{-}}^{\circ }=73.5+76.5=150{\text{S cm}}^2{\text{ mol}}^{-1}$$

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4. Degree of dissociation:
   $$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}=\frac{124}{150}=0.827$$

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5. (ii) Mg is oxidised (anode) and $\mathrm{C}{\mathrm{u}}^{2+}$ is reduced (cathode):
   $$E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=0.34-(-2.37)$$

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6. $$E_{cell}^{\circ }=2.71\text{V}$$

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7. With $n=2$:
   $${\Delta }_r{G}^{\circ }=-nF{E}_{cell}^{\circ }=-2\times 96500\times 2.71$$

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8. $${\Delta }_r{G}^{\circ }=-523030{\text{J mol}}^{-1}=-523.03{\text{kJ mol}}^{-1}$$

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9. (iii) The mercury cell is a primary cell.

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10. It is better than a dry cell because it gives a constant potential (about 1.35 V) throughout its life. This is because the overall reaction involves no ions whose concentration changes. It also lasts longer.

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Answer: (i) ${\Lambda }_m=124$ S cm² mol⁻¹, α = 0.827; (ii) $E_{cell}^{\circ }=2.71$ V, ${\Delta }_r{G}^{\circ }=-523030$ J mol⁻¹ (−523.03 kJ mol⁻¹); (iii) primary cell; gives constant voltage throughout its life