Chemistry (Theory)

Third ionisation removes an electron from the M²⁺ ion.
Mn²⁺ is 3d⁵ – a stable half-filled configuration. Removing a third electron breaks this stability, so a very large energy is needed.
Fe²⁺ (3d⁶) actually becomes 3d⁵ after losing an electron, so its third IE is lower; Cr²⁺ (3d⁴) and V²⁺ (3d³) are not specially stable.
Hence Mn has the highest third ionisation enthalpy.

Only a secondary alcohol is oxidised to a ketone.
(A) and (B) are primary alcohols → aldehydes/acids; (C) is tertiary → not oxidised easily.
(D) butan-2-ol is secondary: CH₃CH₂CH(OH)CH₃ → CH₃CH₂COCH₃ (butan-2-one).

A nucleotide = sugar + nitrogenous base + phosphate group.
The phosphate comes from phosphoric acid; it links the sugars to form the sugar-phosphate backbone.
Adenine is itself a base, and uracil is found in RNA, not DNA.

i = (number of particles actually present) / (number of particles expected).
On association, several molecules join into one, so the number of particles decreases.
Therefore i < 1 (e.g. ethanoic acid dimerising in benzene, i ≈ 0.5).

Rate = \( -\frac{d[X]}{dt} = -\frac{1}{2}\frac{d[Y]}{dt} = +\frac{d[P]}{dt} \).
From the last two terms: \( -\frac{d[Y]}{dt} = 2\frac{d[P]}{dt} \), which is option (A).
Common mistake: reactant terms carry a minus sign and each term is divided by its stoichiometric coefficient.

SN1 rate depends on the stability of the carbocation formed in the slow step.
(A) gives an allylic carbocation stabilised by resonance with the C=C → most stable → fastest.
(B) would give a primary carbocation, (C) a simple secondary one.
(D) Aryl halides do not undergo SN1 (C–Br bond has partial double-bond character; phenyl cation is very unstable).

PCl₅ replaces the –OH of a carboxylic acid by –Cl:
CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl.
The α-hydrogens are not substituted (that needs Cl₂/red P – the HVZ reaction).

CN⁻ (nucleophile) attacks the electrophilic carbonyl carbon, the C=O π bond breaks and the alkoxide is protonated: RCHO + HCN → RCH(OH)CN.
Nothing is replaced, so it is addition, started by a nucleophile.

k = A e^(−Eₐ/RT) → log k = log A − \( \frac{E_a}{2.303\,R} \cdot \frac{1}{T} \).
Compare with y = c + mx, where y = log k and x = 1/T.
Slope = −Eₐ/2.303R and intercept = log A.

R–X + R'–O⁻Na⁺ → R–O–R' + NaX is the Williamson ether synthesis (SN2 attack of alkoxide on the alkyl halide; works best with primary halides).

Acid-catalysed dehydration goes through a carbocation.
Carbocation stability: 3° > 2° > 1°, so tertiary alcohols dehydrate most easily.
Order: tert-butyl alcohol > isopropyl alcohol > ethanol.

Glucose is an aldose, so it has a –CHO group (not a ketone).
The –OH on C-5 adds to the C-1 aldehyde, forming a six-membered ring (5 C + 1 O) = pyranose, a cyclic hemiacetal.

A is true: any non-volatile solute lowers the vapour pressure and hence the freezing point (a colligative effect).
R is true: NaCl → Na⁺ + Cl⁻.
But the depression is caused by the presence of solute particles; dissociation only makes it larger (i = 2). A non-electrolyte like glucose also depresses the freezing point. So R does not explain A (official answer: B).

Lanthanoid contraction cancels the expected size increase from 4d to 5d, so Zr (160 pm) and Hf (159 pm) have nearly equal radii.
Same size + same valence configuration → almost identical chemistry → separation is difficult. R correctly explains A.

–I effect of Cl stabilises the carboxylate ion, so chloroacetic acid is the stronger acid (R true).
Stronger acid → lower pKₐ. So chloroacetic acid (pKₐ ≈ 2.9) has a lower pKₐ than ethanoic acid (≈ 4.8). A is false.

R is true: the N lone pair is in conjugation with the benzene ring.
Because of this, aniline is a weaker base than ammonia (pK_b 9.4 vs 4.75). So A is false.

Electrode reaction: Fe²⁺ + 2e⁻ → Fe, n = 2.
Nernst: \( E = E^\circ - \frac{0.059}{n}\log\frac{1}{[Fe^{2+}]} \)
E = −0.45 − (0.059/2) log (1/0.01) = −0.45 − 0.0295 × 2
E = −0.45 − 0.059 = −0.509 V.
Mistake to avoid: [Fe(s)] = 1, so only [Fe²⁺] goes in the denominator.

Molecularity: the number of reacting species (atoms, ions or molecules) that must collide at the same time in an elementary reaction for it to occur. (Always a whole number, never zero.)
Condition: if one of the two reactants is present in large excess (e.g. water in hydrolysis of an ester), its concentration stays almost constant, so the rate depends only on the other reactant → pseudo first-order reaction.
Example: CH₃COOC₂H₅ + H₂O (excess) → CH₃COOH + C₂H₅OH, rate = k'[ester].

(a) On prolonged heating with HI, all –OH groups and the –CHO are reduced: glucose → CH₃(CH₂)₄CH₃ (n-hexane). This shows the six carbons are in a straight chain.
(b) Conc. HNO₃ oxidises both the –CHO and the primary –CH₂OH groups to –COOH: glucose → HOOC–(CHOH)₄–COOH (saccharic acid). This shows glucose has a primary alcohol group.

(i) Electrophilic addition, Markovnikov's rule: H⁺ adds to the terminal CH₂, giving the resonance-stabilised benzylic carbocation C₆H₅–C⁺H–CH₃; Br⁻ then attacks it → (1-bromoethyl)benzene, C₆H₅CH(Br)CH₃.
(ii) In UV light Br₂ forms Br• radicals; at low Br₂ concentration substitution occurs at the allylic position (the allylic radical is resonance-stabilised) → 3-bromocyclohexene (Br on the carbon next to C=C, double bond kept).

(i) Grignard reagents are very reactive; any moisture protonates them: RMgX + H₂O → RH + Mg(OH)X. The reagent is destroyed, so dry ether and dry apparatus are used.
(ii) In water, OH⁻ is strongly solvated and acts mainly as a nucleophile → substitution: R–X + OH⁻ → R–OH.
In alcohol, alkoxide ions (RO⁻) are formed; they are stronger bases and remove a β-hydrogen → β-elimination (dehydrohalogenation) to an alkene.

(a) Clemmensen reduction converts C=O into CH₂:
CH₃COCH₂CH₃ + 4[H] → (Zn-Hg, conc. HCl) → CH₃CH₂CH₂CH₃ + H₂O.
(b) Cannizzaro reaction (benzaldehyde has no α-H): one molecule is oxidised, the other reduced:
2 C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa.

Step 1 – cell constant: G* = κ × R = 1.35 × 10⁻² × 100 = 1.35 cm⁻¹.
Step 2 – conductivity of AgNO₃: κ = G*/R = 1.35/90 = 0.015 S cm⁻¹.
Step 3 – molar conductivity: \( \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{0.015 \times 1000}{0.02} \) = 750 S cm² mol⁻¹.
Keep the units: marks are cut for missing/wrong units.

Let rate = k[NO]ᵖ[Br₂]^q.
(a) Expt 2 ÷ Expt 1: [Br₂] tripled, rate tripled → 3 = 3^q → q = 1.
Expt 3 ÷ Expt 1: [NO] tripled, rate ×9 → 9 = 3ᵖ → p = 2.
Rate = k[NO]²[Br₂] (overall order 3).
(b) From Expt 1: k = 1.0 × 10⁻³ / (0.05² × 0.05) = 1.0 × 10⁻³ / 1.25 × 10⁻⁴ = 8 L² mol⁻² s⁻¹.
(c) Rate = 8 × (0.4)² × 0.2 = 8 × 0.16 × 0.2 = 0.256 mol L⁻¹ s⁻¹ (2.56 × 10⁻¹).

(a) Zn²⁺ + 4 OH⁻ gives charge −2 on the complex, so two K⁺: K₂[Zn(OH)₄].
(b) Conductivity depends on the number of ions produced:
[Cr(NH₃)₃Cl₃] → 0 ions (non-electrolyte);
[Cr(NH₃)₅Cl]Cl₂ → 3 ions;
[Cr(NH₃)₆]Cl₃ → 4 ions.
Order: [Cr(NH₃)₃Cl₃] < [Cr(NH₃)₅Cl]Cl₂ < [Cr(NH₃)₆]Cl₃.
(c)(i) NO₂⁻ is ambidentate (binds through N as nitrito-N or through O as nitrito-O) → linkage isomerism.
(ii) [Co(en)₃]³⁺ is an octahedral complex with three didentate ligands; it has no plane of symmetry and exists as d and l forms → optical isomerism.

(a) In an allylic halide the halogen is on the sp³ carbon next to C=C. In (ii) Br is on C-3, adjacent to the double bond → allylic. In (i) Br is on the sp² carbon itself → vinylic.
(b) 2,4,6-Trinitrochlorobenzene. The three –NO₂ groups at ortho and para positions withdraw electrons, making the ring carbon electron-poor and stabilising the negatively charged intermediate (Meisenheimer complex) by resonance. Chlorobenzene has no such help.
(c) (CH₃)₃C–Cl. Among isomers, more branching → more spherical shape → smaller surface area → weaker van der Waals forces → lowest boiling point.

(a) Step 1 – protonation of alcohol (fast): CH₃CH₂–OH + H⁺ ⇌ CH₃CH₂–OH₂⁺.
Step 2 – loss of water (slow, rate-determining): CH₃CH₂–OH₂⁺ → CH₃CH₂⁺ + H₂O.
Step 3 – loss of a β-proton: CH₃CH₂⁺ → CH₂=CH₂ + H⁺.
(b)(i) Hydroboration–oxidation adds water in an anti-Markovnikov way: CH₃CH₂CH₂CH₂OH (butan-1-ol).
(ii) Kolbe's reaction: phenol → sodium phenoxide; phenoxide + CO₂ → sodium salicylate; acidification → o-hydroxybenzoic acid (salicylic acid).

(a) The amide bond –CO–NH– formed when the –COOH of one amino acid reacts with the –NH₂ of another, with loss of water.
(b) Hydrogen bonds between complementary base pairs (A=T two H-bonds, G≡C three H-bonds).
(c) Starch – a polymer of α-D-glucose. Sucrose is a disaccharide; glucose and fructose are monosaccharides.
(d) Water-soluble: vitamin B group or C. Fat-soluble: vitamin A, D, E or K.

F is a monobasic acid of molar mass 60 → CH₃COOH. So D (oxidised to F) is CH₃CHO and B (oxidised by PCC to D) is CH₃CH₂OH.
D + dil. NaOH → aldol CH₃CH(OH)CH₂CHO; on heating it loses water → E = CH₃CH=CHCHO (but-2-enal, crotonaldehyde).
E + H₂/catalyst (both C=C and C=O reduced) → C = CH₃CH₂CH₂CH₂OH (butan-1-ol).
An ester giving ethanol and butan-1-ol with LiAlH₄ must be ethyl butanoate: CH₃CH₂CH₂COOCH₂CH₃ → CH₃CH₂CH₂CH₂OH + CH₃CH₂OH. Check: C₆H₁₂O₂ ✓.
(Butyl ethanoate, CH₃COOC₄H₉, would also give the same two alcohols and is accepted.)

(a) In a primary battery the reaction occurs only once and the cell cannot be reused (e.g. dry cell, mercury cell). A secondary battery can be recharged by passing current in the opposite direction and reused many times (e.g. lead storage battery).
(b) Overall reaction: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l). It involves no ion in solution whose concentration changes during use, so by the Nernst equation the cell potential stays constant.
(c) During charging the cell works as an electrolytic cell and the discharge reaction is reversed:
At the electrode that becomes Pb: PbSO₄(s) + 2e⁻ → Pb(s) + SO₄²⁻(aq)
At the electrode that becomes PbO₂: PbSO₄(s) + 2H₂O(l) → PbO₂(s) + SO₄²⁻(aq) + 4H⁺(aq) + 2e⁻
Overall: 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq).

(a) When ligands approach in a definite geometry, the five degenerate d-orbitals split into sets of different energy; the energy difference between these sets is the crystal field splitting energy (Δₒ for octahedral).
(b) Ti³⁺ is 3d¹ → t₂g¹e_g⁰. The electron absorbs visible light (around 498 nm, green-yellow) and jumps from t₂g to e_g (d–d transition). The complementary colour, violet, is seen.
(c) Cr³⁺: 3d³ → t₂g³ (d²sp³). Three unpaired electrons → paramagnetic.
Ni²⁺: 3d⁸. CN⁻ is a strong-field ligand; it pairs the electrons so that one 3d orbital is empty → dsp² square planar; no unpaired electrons → diamagnetic.

(i) Henry's law: p = K_H·x, so for a given pressure, solubility x ∝ 1/K_H. O₂ is less soluble, so O₂ has the higher K_H.
(ii) The solution is hypertonic; water moves out of the cells by osmosis and the cells shrink (crenation).
(iii) A₂B₃ → 2A³⁺ + 3B²⁻, so n = 5 ions.
α = (i − 1)/(n − 1) → 0.6 = (i − 1)/4 → i = 3.4.
ΔT_b = i·K_b·m = 3.4 × 0.52 × 1 = 1.768 K.
T_b = 373 + 1.768 = 374.768 K (using 373.15 K gives 374.918 K).

(i) x_A = 0.4, x_B = 0.6.
p_A = 75 × 0.4 = 30 mm Hg; p_B = 25 × 0.6 = 15 mm Hg; P_total = 45 mm Hg.
y_B = p_B/P_total = 15/45 = 1/3 ≈ 0.33 (a mole fraction has no unit).
(ii) A colligative property depends only on the number of solute particles in a given amount of solvent, not on their nature. Osmotic pressure is preferred for macromolecules: it is measured at room temperature (proteins not damaged) and gives a measurable value even for very dilute solutions, using molarity instead of molality.
(iii) π = iCRT. NaCl → Na⁺ + Cl⁻ (i ≈ 2) while glucose does not dissociate (i = 1). So at equal concentration NaCl gives about twice the osmotic pressure; they are not isotonic.

(a) In C₆H₅SO₂N(C₂H₅)₂ the nitrogen carries no hydrogen, so there is no acidic N–H to react with alkali; hence it is insoluble (basis of Hinsberg test for 3° amines' product from 2° amines).
(b) Aniline is a Lewis base and forms a salt with the Lewis acid catalyst AlCl₃. The N becomes positively charged, strongly deactivating the ring, so the reaction fails.
(c) Azo-dye test: with HNO₂ (NaNO₂ + HCl) at 273–278 K aniline gives benzenediazonium chloride, which couples with alkaline β-naphthol to give an orange-red dye. Methylamine gives an unstable aliphatic diazonium salt that releases N₂ and gives no dye.
(d) Phthalimide + ethanolic KOH → potassium phthalimide; + R–X → N-alkylphthalimide; alkaline hydrolysis (NaOH) → R–NH₂ + sodium phthalate.
(e) C₆H₅NH₂ + (CH₃CO)₂O (pyridine) → C₆H₅NHCOCH₃ (acetanilide, activation reduced); + Br₂/CH₃COOH → p-bromoacetanilide; hydrolysis (H⁺ or OH⁻) → p-bromoaniline. Direct bromination would give 2,4,6-tribromoaniline.
(f) C₆H₅N₂⁺Cl⁻ + HBF₄ → C₆H₅N₂⁺BF₄⁻; this with NaNO₂/Cu, heat → C₆H₅NO₂ (nitrobenzene) + N₂ + NaBF₄.
(g) C₆H₅COOH + NH₃ → (heat) C₆H₅CONH₂ (A, benzamide); Hoffmann bromamide degradation: C₆H₅CONH₂ + Br₂ + 4NaOH → C₆H₅NH₂ (B, aniline) + Na₂CO₃ + 2NaBr + 2H₂O.

(i)(1) Zn, Cd, Hg have completely filled d-orbitals (d¹⁰) – no unpaired d-electrons to take part in metallic bonding – so bonding is weak and m.p./b.p. are low.
(2) Cr²⁺ (d⁴) loses an electron to become Cr³⁺ (d³, stable half-filled t₂g³) → good reducing agent. Mn³⁺ (d⁴) gains an electron to become Mn²⁺ (d⁵, stable half-filled) → good oxidising agent.
(3) For Cu, the energy needed (high enthalpy of atomisation + first and second ionisation enthalpies) is not compensated by its hydration enthalpy, so Cu → Cu²⁺ is unfavourable and E° is positive.
(ii)(1) 2KMnO₄ → (513 K) K₂MnO₄ + MnO₂ + O₂.
(2) Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O (check: charge −2 −6 +14 = +6 = 2 × 3; O 7 = 7).

(i) CuCl₂. Cu²⁺(aq) has a much more negative hydration enthalpy, which more than compensates for the second ionisation enthalpy. Cu⁺ in water disproportionates: 2Cu⁺ → Cu²⁺ + Cu.
(ii) (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns².
(iii) Fe³⁺ (3d⁵) has unpaired electrons and shows d–d transitions → coloured. Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) cannot have d–d transitions → colourless.
(iv) Acidify: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O.
Then add KCl: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl (K₂Cr₂O₇ is less soluble and crystallises out).
(v) They can show variable oxidation states (forming unstable intermediates), form complexes with reactants, and provide a large surface area for adsorption of reactants.