Chemistry (Theory)

For a first order reaction t½ = 0.693/k, which contains no concentration term.
So t½ stays the same whatever the starting concentration → a horizontal line on a t½ vs [R]₀ graph.
(A) (t½ ∝ [R]₀) is zero order; (C) (rate independent of concentration) is also zero order. For first order, rate = k[R] would give a straight rising line, not a horizontal one.

Depression in freezing point ΔTf = i·Kf·m, so the solution with the most particles (largest i) freezes lowest.
Glucose, urea: i = 1; KCl → K⁺ + Cl⁻: i = 2; CaCl₂ → Ca²⁺ + 2Cl⁻: i = 3.
CaCl₂ gives the largest ΔTf, hence the lowest freezing point.

A transition element has an incompletely filled d subshell in the atom or in one of its common oxidation states.
Hg (like Zn, Cd) is 5d¹⁰6s² and Hg²⁺ is 5d¹⁰ – d subshell always full, so it is not a true transition metal.
Sc (3d¹), Cu²⁺ (3d⁹) and Ag²⁺ (4d⁹) have partly filled d orbitals.

In the Arrhenius equation k = A·e^(−Ea/RT), the factor e^(−Ea/RT) is the Boltzmann factor: the fraction of molecules having kinetic energy ≥ Ea.
It is always less than 1 and increases with temperature, which is why rate increases with T. e^(+Ea/RT) would be greater than 1, which is impossible for a fraction.

Cathode: Cu²⁺ (E° = +0.34 V) is reduced more easily than water, so Cu²⁺ + 2e⁻ → Cu deposits on the cathode (not the anode, and no H₂).
Anode: in a reasonably concentrated chloride solution, Cl⁻ is oxidised because of the overpotential of O₂: 2Cl⁻ → Cl₂ + 2e⁻.
So chlorine gas is liberated at the anode.

Aspirin is acetylsalicylic acid (2-acetoxybenzoic acid).
It is made by treating salicylic acid (2-hydroxybenzoic acid) with acetic anhydride (or acetyl chloride) in the presence of an acid: the phenolic –OH becomes –OCOCH₃.

Secondary valency (Werner) = coordination number = number of donor atoms bonded to the metal.
Five NH₃ + one NO₂⁻ (both monodentate) → 6 donor atoms.
So secondary valency = 6. (Primary valency = oxidation state = +3.)

–OH strongly activates the ring and is ortho/para directing, so even dilute HNO₃ nitrates phenol at room temperature.
A mixture of o-nitrophenol and p-nitrophenol is formed (separated by steam distillation).
Picric acid (2,4,6-trinitrophenol) needs concentrated HNO₃.

Enantiomers are non-superimposable mirror images. They have identical physical properties (m.p., b.p., refractive index, solubility) and identical reactivity towards achiral reagents.
They differ only in the direction of rotation of plane polarised light: equal magnitude, opposite sign (+ and −).
So 'same specific rotation' is false.

In strongly acidic nitrating mixture much of aniline is protonated to the anilinium ion C₆H₅NH₃⁺, which is meta directing.
The free aniline that remains gives mainly para product.
Result: about 51% p-nitroaniline, 47% m-nitroaniline and 2% o-nitroaniline. Oxidation products also form – that is why –NH₂ is protected by acetylation first.

Lower pKb = stronger base.
In aryl amines, methyl groups on nitrogen push electron density (+I) onto N, increasing basicity: C₆H₅N(CH₃)₂ (pKb ≈ 8.92) > C₆H₅NHCH₃ (≈ 9.30) > C₆H₅NH₂ (≈ 9.38).
–NO₂ at para position withdraws electrons (−R, −I), making 4-nitroaniline a very weak base (highest pKb).
So N,N-dimethylaniline has the lowest pKb.

Vitamin E (tocopherol) deficiency → increased fragility of RBCs and muscular weakness.
Others: A → night blindness/xerophthalmia; K → delayed blood clotting; D → rickets/osteomalacia.

Fractional distillation works only because the vapour is richer in the more volatile component.
At the azeotropic composition the vapour has exactly the same composition as the liquid, and it boils at a fixed temperature, so no enrichment is possible.
Thus R is the correct explanation of A.

By resonance (+R effect) the –OH group increases electron density at ortho and para positions, so it is ortho/para directing – the assertion (meta) is false.
The same resonance makes the ring more reactive than benzene, so the reason is true.

A is true: actinoids show many oxidation states (+3 to +7) because the 5f, 6d and 7s levels are of comparable energy, so variable numbers of electrons can be used.
R is true: all actinoids are radioactive.
But radioactivity has nothing to do with the range of oxidation states, so R does not explain A.

A is true: the pentaacetate gives no oxime.
This behaviour shows that in glucose (and its pentaacetate) the –CHO group is NOT free – it is tied up in the cyclic hemiacetal structure. So R, which says it indicates a free –CHO group, is false.

Type: negative deviation from Raoult's law.
Reason: the phenolic O–H forms a strong intermolecular hydrogen bond with the lone pair on the nitrogen of aniline (O–H···N). These A–B attractions are stronger than the phenol–phenol and aniline–aniline attractions, so fewer molecules escape → vapour pressure is lower than predicted.
Effect: lower vapour pressure means the boiling point of the solution rises (such mixtures can form maximum-boiling azeotropes).

Any two:
1. Resonance: a lone pair on the halogen is delocalised into the ring, giving the C–X bond partial double-bond character; it is harder to break.
2. The carbon of C–X is sp² hybridised (more s character, more electronegative), so the bond is shorter and stronger than the sp³ C–X bond of haloalkanes.
3. The phenyl cation that SN1 would need is unstable (not resonance stabilised).
4. The electron-rich π cloud of the ring repels the incoming nucleophile; backside attack (SN2) is also blocked by the ring.

(i) Both ions are complex: cation [Ag(NH₃)₂]⁺ (Ag +1) is named first, then anion [Ag(CN)₂]⁻ (Ag +1, anion suffix -ate, Latin 'argent'):
Diamminesilver(I) dicyanidoargentate(I).
(ii) Oxalate C₂O₄²⁻ ×3 = −6, 3K⁺ = +3, so Fe = +3; anionic complex of iron → ferrate:
Potassium trioxalatoferrate(III).

(i) Add AgNO₃ solution to each: [Co(NH₃)₅SO₄]Cl releases Cl⁻ ions and gives a white precipitate of AgCl; [Co(NH₃)₅Cl]SO₄ does not (its Cl is inside the coordination sphere).
Alternatively BaCl₂ gives white BaSO₄ only with [Co(NH₃)₅Cl]SO₄.
(ii) Chelate effect: a di- or polydentate ligand that binds one metal ion through two or more donor atoms forms a ring (chelate); such complexes are more stable than similar complexes with monodentate ligands.
Example: [Co(en)₃]³⁺ is more stable than [Co(NH₃)₆]³⁺; also metal–EDTA complexes.

(i) Peptide linkage: amide bond –CO–NH– formed between the –COOH of one amino acid and the –NH₂ of another (with loss of water); found in proteins.
Glycosidic linkage: oxygen bridge (C–O–C) joining two monosaccharide units; found in di- and polysaccharides.
(ii) Essential amino acids cannot be synthesised by the body and must be supplied in the diet (e.g. valine, leucine, lysine).
Non-essential amino acids can be made in the body (e.g. glycine, alanine).

Elementary step → rate law from stoichiometry: rate = k[A]²[B] (order 3).
Reducing volume to 1/3 makes every concentration 3 times larger.
New rate = k(3[A])²(3[B]) = 27k[A]²[B] → 27 times the original rate.
Order depends on the rate law, not on volume, so the order stays 3.

Let initial pressure P₀ = 0.30 atm and let x atm decompose.
At time t: C₂H₅Cl = P₀ − x, C₂H₄ = x, HCl = x → Pt = P₀ + x, so x = Pt − P₀ = 0.20 atm.
Pressure of reactant left = P₀ − x = 2P₀ − Pt = 0.60 − 0.50 = 0.10 atm.
k = (2.303/t) log[P₀/(2P₀ − Pt)] = (2.303/30) log(0.30/0.10) = (2.303/30) × 0.48
k = 0.0368 s⁻¹ ≈ 3.7 × 10⁻² s⁻¹.
Common mistake: using Pt directly in place of the reactant pressure.

(i) Nernst: Ecell = E°cell − (0.059/n) log Q. At equilibrium Ecell = 0 and Q = Kc, so E°cell = (0.059/n) log Kc. Since Ecell is always zero at equilibrium, Kc can only be related to the constant E°cell.
(ii) A metal with negative E° (below hydrogen, E° = 0) is a better reducing agent than H₂ and reduces H⁺ to H₂. So metal A (−0.24 V) liberates H₂; B (+0.80 V) cannot.
(iii) Charging reverses the discharge reaction (electrolysis):
2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq).

Type: primary cell (cannot be recharged).
Advantage: it gives a constant potential (about 1.35 V) throughout its life, because the overall reaction contains no ions whose concentration changes; a dry cell's voltage falls as it is used.
Overall reaction: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
(anode: Zn(Hg) + 2OH⁻ → ZnO + H₂O + 2e⁻; cathode: HgO + H₂O + 2e⁻ → Hg + 2OH⁻).

Association 2 C₆H₅COOH ⇌ (C₆H₅COOH)₂ with α = 0.88.
i = 1 − α + α/2 = 1 − α/2 = 1 − 0.44 = 0.56.
Molality m = (0.61/122) × (1000/5) = 0.005 × 200 = 1.0 mol kg⁻¹.
ΔTb = i·Kb·m = 0.56 × 2.3 × 1.0 = 1.288 K.
Tb = 46.2 + 1.288 = 47.488 °C ≈ 47.49 °C.
Common mistake: taking i > 1 – association lowers the number of particles, so i < 1.

(a) CHO–(CHOH)₄–CH₂OH + HI, heat → CH₃–CH₂–CH₂–CH₂–CH₂–CH₃ (n-hexane); shows six carbons in a straight chain.
(b) CHO–(CHOH)₄–CH₂OH + [O] (Br₂ water, mild oxidant) → HOOC–(CHOH)₄–CH₂OH (gluconic acid); only –CHO is oxidised, showing an aldehyde group.
(c) CHO–(CHOH)₄–CH₂OH + conc. HNO₃ → HOOC–(CHOH)₄–COOH (saccharic acid); both the –CHO and the primary –CH₂OH are oxidised.

(a) Carboxylic acid molecules are held by stronger and more extensive hydrogen bonding; they exist as cyclic dimers even in the vapour, so more energy is needed to boil them.
(b) The C=O group withdraws electrons, weakening the α C–H bond, and the carbanion (enolate) formed after loss of H⁺ is stabilised by resonance with the carbonyl group.
(c) In strong acid the nitrogen lone pair of NH₃/NH₂–Z is protonated (NH₄⁺, ⁺NH₃–Z), so it is no longer a nucleophile and cannot attack the carbonyl carbon. Hence a weakly acidic medium (pH ≈ 3.5) is used.

(a) Phenol + CHCl₃ + aq. NaOH (≈ 340 K) → intermediate dichloromethyl compound → on acidification (H⁺) gives salicylaldehyde (2-hydroxybenzaldehyde), C₆H₄(OH)CHO.
(b) Phenol + NaOH → sodium phenoxide; heated with CO₂ (4–7 atm, ≈ 400 K) then H⁺ → salicylic acid (2-hydroxybenzoic acid), C₆H₄(OH)COOH.
(c) C₆H₅OCH₃ + CH₃COCl, anhydrous AlCl₃ → 2-methoxyacetophenone (minor) + 4-methoxyacetophenone (major) + HCl. –OCH₃ is o/p directing.

(a) Rate depends only on [X] → SN1 → X must give a stable 3° carbocation: X = (CH₃)₃C–Br (2-bromo-2-methylpropane).
Rate depends on [Y][OH⁻] → SN2; Y is optically active, so it has a chiral carbon: Y = CH₃–CH(Br)–CH₂–CH₃ (2-bromobutane).
(b) X – it reacts by SN1 through a planar carbocation that can be attacked from either face, so an optically active substrate would give a racemic mixture. (X itself is achiral, so its product is not optically active.)
(c) Y – SN2 occurs by backside attack of OH⁻ in one step (Walden inversion), so the product alcohol has the inverted configuration.

(a)(i) In water basicity depends on +I effect, solvation of the conjugate acid by H-bonding, and steric factors together. CH₃NH₃⁺ has three N–H for H-bonding with water and is well stabilised; (CH₃)₃NH⁺ has only one N–H and the three methyl groups also hinder solvation and approach of H⁺. So CH₃NH₂ is the stronger base in water.
(ii) Hoffmann bromamide degradation: CH₃CONH₂ + NaOBr → A = CH₃NH₂ (one carbon less). Benzoylation: CH₃NH₂ + C₆H₅COCl (base) → B = CH₃NHCOC₆H₅ (N-methylbenzamide) + HCl.
(b) A 2° amine gives an N,N-dialkylsulphonamide with no acidic H on N, so it is alkali-insoluble. The only 2° amine of formula C₃H₉N is N-methylethanamine, CH₃CH₂NHCH₃.
(c) –NH₂ makes the ring too reactive (over-nitration, oxidation) and in acid it becomes meta-directing –NH₃⁺. Acetylation (–NHCOCH₃) moderates the activation, prevents anilinium ion formation, and gives mainly p-nitroaniline after hydrolysis.

(a) dx²−y² and dz² (the e_g set). In an octahedral complex the six ligands approach along the x, y, z axes; these two orbitals point directly along the axes, so they suffer greater repulsion and rise in energy (by 0.6Δo), while t₂g (dxy, dyz, dxz) lie between the axes and are lowered.
(b) Co³⁺ = 3d⁶.
(i) F⁻ is a weak-field ligand (Δo < P): t₂g⁴ e_g² (high spin, 4 unpaired).
(ii) NH₃ is a stronger field ligand for Co³⁺ (Δo > P): t₂g⁶ e_g⁰ (low spin, diamagnetic).
(c) [NiCl₄]²⁻: Ni²⁺ is 3d⁸; Cl⁻ is weak field and cannot pair electrons, so two unpaired electrons remain (sp³) → paramagnetic.
[Ni(CO)₄]: Ni(0) is 3d⁸4s²; strong-field CO pushes the 4s electrons into 3d giving 3d¹⁰ with all electrons paired (sp³) → diamagnetic.

(i) (I) No Tollens' → ketone; positive iodoform → CH₃CO– group: pentan-2-one, CH₃COCH₂CH₂CH₃ (3-methylbutan-2-one also fits).
(II) Ketone without CH₃CO– but with α-H: pentan-3-one, CH₃CH₂COCH₂CH₃.
(III) Cannizzaro needs an aldehyde with no α-H: 2,2-dimethylpropanal, (CH₃)₃C–CHO.
(ii) (I) C₆H₅COCH₃ + NaOI (I₂/NaOH) → C₆H₅COONa + CHI₃; then H⁺ → C₆H₅COOH. (Or KMnO₄–KOH, heat, then H₃O⁺.)
(II) C₆H₅CH₂CH₃ + alkaline KMnO₄, heat → C₆H₅COOK (whole side chain oxidised to –COOH); then H₃O⁺ → C₆H₅COOH.

(i) C₆H₅–CH=N–NH–C₆H₃(NO₂)₂, the NO₂ groups at positions 2 and 4 of the second ring (formed from C₆H₅CHO + H₂N–NH–C₆H₃(NO₂)₂ with loss of H₂O).
(ii) Reactivity in nucleophilic addition falls with more alkyl groups (+I, steric hindrance): (t-Bu)₂C=O < CH₃COCH₃ < CH₃CHO.
(iii) Add NaHCO₃ solution: benzoic acid gives brisk effervescence of CO₂; ethyl benzoate (an ester) does not.
(iv) (a) DIBAL-H, then H₂O; or (b) SnCl₂ + HCl, then H₃O⁺ (Stephen reduction): CH₃CN → CH₃CHO.
(v) CrO₃ oxidises the 2° alcohol to a ketone: X = cyclohexanone (C₆H₁₀O, ring with C=O).

(i)(I) E°(M²⁺/M) depends on sublimation enthalpy + (IE₁ + IE₂) − hydration enthalpy. Ionisation and sublimation enthalpies vary irregularly across the series, so E° values are irregular.
(II) Cu has a high enthalpy of atomisation and high IE₂ which are not compensated by its hydration enthalpy; so converting Cu(s) to Cu²⁺(aq) is unfavourable → positive E°.
(III) Mn²⁺ is 3d⁵ (half-filled, extra stable), so Mn is readily oxidised to Mn²⁺ → highly negative E°.
(ii) Acidic: 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O
Neutral/alkaline: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻

(i) (I) Cerium, Ce⁴⁺ (4f⁰). (II) Europium, Eu²⁺ (4f⁷) or ytterbium, Yb²⁺ (4f¹⁴).
(ii) They show variable oxidation states (form unstable intermediates easily), can form complexes, and provide a large surface for adsorption of reactants.
(iii) Cr (3d⁵4s¹) can use more unpaired (n−1)d and ns electrons in metallic bonding than Mn, whose stable half-filled 3d⁵4s² configuration holds its electrons more tightly; stronger metallic bonding → higher m.p.
(iv) Acidified permanganate slowly decomposes, depositing brown MnO₂ and releasing oxygen:
4MnO₄⁻ + 4H⁺ → 4MnO₂ + 3O₂ + 2H₂O
Mn(+7) is reduced and O(−2) is oxidised to O₂ – it is a redox (decomposition) reaction, catalysed by the MnO₂ formed (autocatalysis).

Cell reaction: Mg + 2Ag⁺ → Mg²⁺ + 2Ag, n = 2.
E°cell = E°cathode − E°anode = 0.80 − (−2.37) = 3.17 V.
Nernst: Ecell = E°cell − (0.059/2) log([Mg²⁺]/[Ag⁺]²)
= 3.17 − 0.0295 log(0.01/(0.001)²) = 3.17 − 0.0295 log 10⁴ = 3.17 − 0.0295 × 4
= 3.17 − 0.118 = 3.052 V.
ΔG = −nFEcell = −2 × 96500 × 3.052 = −589 036 J mol⁻¹ ≈ −589 kJ mol⁻¹.
Common mistake: forgetting to square [Ag⁺].

n = 2.
E°cell = −ΔG°/(nF) = 43500/(2 × 96500) = 0.225 V.
Q = [H⁺]²[Cl⁻]²/p(H₂) = (0.1)²(0.2)²/0.4 = (0.01 × 0.04)/0.4 = 10⁻³ (solids omitted).
Ecell = E°cell − (0.059/2) log Q = 0.225 − 0.0295 × (−3) = 0.225 + 0.0885
Ecell = 0.3135 V ≈ 0.314 V.