UPHESC Assistant Professor Chemistry 2025 PYQs

Concept: Only the ketone must be reduced; the ester must survive.

1. ${\mathrm{NaBH}}_4$ is a mild hydride donor: reduces aldehydes/ketones but not esters (under normal conditions).
2. ${\mathrm{LiAlH}}_4$ would reduce both (ester → ${\mathrm{CH}}_2\mathrm{OH}$).
3. ${\mathrm{B}}_2{\mathrm{H}}_6$ is an electrophilic reductant used mainly for acids/amides and it also hydroborates C=C; ${\mathrm{NaNH}}_2$ is a base, not a reductant.

Short trick: NaBH₄: C=O of aldehyde/ketone only; LiAlH₄: almost everything.

Concept: In square-planar Pt(II), the ligand trans to the stronger trans-director is replaced; Cl⁻ > NH₃ in the trans-effect series.

1. Step 1: $[{\mathrm{PtCl}}_4{]}^{2-}+{\mathrm{NH}}_3\to [{\mathrm{PtCl}}_3({\mathrm{NH}}_3){]}^{-}$.
2. Step 2: the Cl trans to another Cl (not the one trans to NH₃) is labilised → second NH₃ enters cis to the first.
3. Product: cis-$[{\mathrm{PtCl}}_2({\mathrm{NH}}_3{)}_2]$ (cisplatin). Starting from $[\mathrm{Pt}({\mathrm{NH}}_3{)}_4{]}^{2+}$ + Cl⁻ gives the trans isomer.

Short trick: PtCl₄²⁻ + NH₃ → cis; Pt(NH₃)₄²⁺ + Cl⁻ → trans.

Concept: Strong-field CN⁻ pairs electrons; weak-field F⁻ does not.

1. $[\mathrm{Fe}(\mathrm{CN}{)}_6{]}^{4-}$: Fe(II) $d^6$, low spin $t_{2g}^6$ → n = 0 → 0 BM (diamagnetic).
2. $[{\mathrm{FeF}}_6{]}^{3-}$: Fe(III) $d^5$, high spin $t_{2g}^3{e}_g^2$ → n = 5 → $\sqrt{35}$ = 5.92 BM.

Short trick: Ferrocyanide diamagnetic; d⁵ high spin = 5.92 BM.

Concept: For $\mathrm{M}|{\mathrm{M}}^{\mathrm{n}+}({\mathrm{c}}_1)||{\mathrm{M}}^{\mathrm{n}+}({\mathrm{c}}_2)|\mathrm{M}$, $E=\frac{RT}{nF}\ln \frac{c_2}{c_1}$ ($c_2 > c_1$).

1. As the cell works, the dilute side becomes more concentrated (anode dissolves) and the concentrated side becomes more dilute (cathode deposits).
2. So $c_2/c_1$ falls towards 1 and the EMF decreases continuously.
3. It reaches zero only when the concentrations become equal (cell exhausted), not while it is functioning.

Short trick: Concentration cell runs to equalise concentrations → EMF ↓.

Concept: Find all distributions (configurations) with total energy 3E, then count arrangements $W=\frac{N!}{\prod n_i!}$ for N = 4 atoms.

1. Configuration 1: three atoms at E, one at 0 → $W=\frac{4!}{3!1!}=4$.
2. Configuration 2: one at 2E, one at E, two at 0 → $W=\frac{4!}{1!1!2!}=12$.
3. Total microstates = 4 + 12 = 16 (2 is only the number of configurations).

Short trick: Configurations first, then W = N!/∏n_i! for each, then add.

Concept: Three statistics: Maxwell–Boltzmann (distinguishable), Bose–Einstein (bosons), Fermi–Dirac (fermions).

1. Bosons have integral spin (0, 1, 2 …), symmetric wave functions and any number may occupy one state (photons, ⁴He).
2. Half-integral spin particles (electrons, protons) are fermions → Fermi–Dirac, obey Pauli principle.

Short trick: Bose → integer spin; Fermi → half-integer spin.

Concept: Phosphorescence is radiative emission between states of different multiplicity.

1. $T_1\to S_0+h\nu$: spin-forbidden emission → slow, long-lived glow = phosphorescence.
2. $S_1\to S_0+h\nu$ = fluorescence; $S_1\to T_1$ = intersystem crossing; $T_1\to S_0+\mathrm{\Delta }$ = non-radiative ISC (heat).

Short trick: Fluorescence S→S (fast); phosphorescence T→S (slow).

Concept: All four listed proteins are iron enzymes, so no option fits the question as printed.

1. Cytochrome P₄₅₀ – heme Fe (cysteine-thiolate ligated) monooxygenase.
2. Pyrocatechase (catechol 1,2-dioxygenase) – non-heme Fe(III) intradiol dioxygenase.
3. Tryptophan 2,3-dioxygenase – heme Fe; catalase – heme Fe (Fe(III)/Fe(IV)=O).
4. Typical non-iron examples would be carbonic anhydrase (Zn), tyrosinase (Cu) or nitrogenase's Mo site.

Short trick: P450, catalase, peroxidase, TDO → heme Fe; catechol dioxygenases → non-heme Fe.

Concept: After a small perturbation, the deviation $x$ from equilibrium decays as $\frac{dx}{dt}=-(k_1+k_{-1})x$.

1. So $x=x_0{e}^{-t/\tau }$ with $\frac{1}{\tau }=k_1+k_{-1}$.
2. Hence $\tau =(k_1+k_{-1}{)}^{-1}$; it has units of time, as required (the other forms do not).

Short trick: First-order ⇌ first-order: 1/τ = k_f + k_b.

Concept: An ambident nucleophile has two different donor atoms through which it can attack.

1. ${}^{-}\mathrm{C}\equiv \mathrm{N}\leftrightarrow \mathrm{C}={\mathrm{N}}^{-}$: attacks through C (→ nitriles, e.g. with KCN) or through N (→ isocyanides, e.g. with AgCN).
2. ${\mathrm{CH}}_3^{-}$, ${\mathrm{PH}}_3$ and ${\mathrm{H}}_2\mathrm{O}$ each have only one nucleophilic site.
3. Other examples: ${\mathrm{NO}}_2^{-}$, ${\mathrm{SCN}}^{-}$, ${\mathrm{OCN}}^{-}$, enolates.

Short trick: KCN → R–CN; AgCN → R–NC.

Concept: Typical $\delta$ (ppm): aromatic C 110–160 > alkyne C 65–90 > OCH₃ ~55.

1. z (aromatic C bearing the alkyne, para to OMe) ≈ 114 ppm.
2. y (internal ≡C–Ar) ≈ 84 ppm; x (terminal ≡CH) ≈ 76 ppm – the substituted alkyne carbon is downfield.
3. p (OCH₃) ≈ 55 ppm. Order: z > y > x > p.

Short trick: 13C: C=O > aromatic/alkene > alkyne > O–CH₃ > alkyl.

Concept: N₂ is reduced to NH₃ at the iron–molybdenum cofactor (FeMoco) of nitrogenase.

1. FeMoco = ${\mathrm{MoFe}}_7{\mathrm{S}}_9\mathrm{C}$ cluster with homocitrate (central carbide).
2. The Fe-protein supplies electrons (with ATP hydrolysis) to the MoFe-protein.
3. Alternative nitrogenases use V or only Fe, but the classical cofactor is Fe–Mo; W, Cu, Ni are not involved.

Short trick: Nitrogenase = FeMo; N₂ + 8H⁺ + 8e⁻ + 16 ATP → 2NH₃ + H₂.

Concept: The integrated Clausius–Clapeyron equation assumes the vapour is an ideal gas and the condensed-phase volume is negligible; it applies to liquid ⇌ vapour (and solid ⇌ vapour).

1. (A) and (B) are correct: it relates vapour pressure with temperature, i.e. boiling point with external pressure.
2. (C) is not correct: for solid ⇌ liquid there is no vapour phase, so only the general Clapeyron equation $dP/dT=\mathrm{\Delta }{H}_{fus}/T\mathrm{\Delta }V$ applies.
3. (D) is also not correct: stronger intermolecular forces raise $\mathrm{\Delta }{H}_{vap}$ and raise the boiling point.
4. Since two statements are incorrect, the question has no single answer.

Short trick: Clausius–Clapeyron: liquid/solid ⇌ vapour only; higher ΔH_vap → higher b.p.

Concept: 14 Bravais lattices are distributed among 7 crystal systems.

1. Cubic 3 (P, I, F); tetragonal 2 (P, I); orthorhombic 4 (P, C, I, F); monoclinic 2 (P, C).
2. Triclinic 1 (P); hexagonal 1 (P); rhombohedral/trigonal 1 (R).
3. So a = 1, b = 1, c = 1.

Short trick: 3 + 2 + 4 + 2 + 1 + 1 + 1 = 14.

Concept: At $T\ll {\theta }_D$ the Debye model gives $C_v=\frac{12{\pi }^4}{5}R{\left(\frac{T}{{\theta }_D}\right)}^3$ per mole.

1. Dimensional check: $C_v$ must have units of R and $(T/{\theta }_D)$ must appear as a dimensionless ratio cubed – only (A) satisfies this.
2. Numerically $\frac{12{\pi }^4}{5}R\approx 1944$ J K⁻¹ mol⁻¹ × $(T/{\theta }_D{)}^3$; used to extrapolate $C_p$ to 0 K for third-law entropies.

Short trick: Debye: C_v ∝ T³, constant 12π⁴R/5.

Concept: $\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{[{\mathrm{A}}^{-}]}{[\mathrm{HA}]}$; ${\mathrm{pK}}_{\mathrm{a}}=-\log (1.75\times {10}^{-5})=4.757$.

1. Before: [A⁻] = [HA] = 0.1 → pH = 4.76.
2. NaOH converts 0.01 mol HA to A⁻: [A⁻] = 0.11, [HA] = 0.09.
3. pH = 4.757 + log(0.11/0.09) = 4.757 + 0.087 = 4.84 ≈ 4.847.

Short trick: Equal acid and salt → pH = pK_a; base added shifts the ratio, not much the pH.

Concept: ${\mathrm{PCl}}_5$ (gas) is trigonal bipyramidal ($s{p}^{3}d$, no lone pair).

1. Principal axis $C_3$ through the two axial Cl; three $C_2$ axes along the equatorial P–Cl bonds (⊥ $C_3$) → D group.
2. Horizontal mirror plane containing the three equatorial Cl → ${\sigma }_h$ → $D_{3h}$.
3. Same point group: ${\mathrm{BF}}_3$ (planar), $\mathrm{Fe}(\mathrm{CO}{)}_5$.

Short trick: TBP = D_3h; octahedral = O_h; square pyramid = C_4v.

Concept: Two successive reflections in planes that intersect along a line at angle θ equal a rotation by 2θ about that line.

1. Each reflection reverses handedness; two reflections restore it, so the result is a proper rotation.
2. E.g. in $C_{2v}$: ${\sigma }_v(xz){\sigma }_v^{\prime }(yz)=C_2$ (planes at 90° → rotation by 180°).
3. $S_n$ and i are improper operations (odd number of reflections).

Short trick: σ × σ (intersecting) = C_n; σ_h × C_n = S_n.

Concept: Taube: octahedral complexes are inert when there are no e_g electrons and no vacant t_2g orbital, and CFAE is large.

1. $t_{2g}^6{e}_g^0$ (low-spin $d^6$, e.g. Co(III)) – very high CFSE, no e_g electron → inert.
2. $t_{2g}^3{e}_g^1$ (high-spin $d^4$) and $t_{2g}^6{e}_g^4$ ($d^{10}$) have e_g electrons → labile.
3. $t_{2g}^5$ ($d^5$ low spin) has a vacant t_2g slot → comparatively labile.

Short trick: Inert: d³, low-spin d⁴–d⁶ (best d⁶ LS, e.g. Co³⁺, Cr³⁺).

Concept: The basic unit of all silicates is the ${\mathrm{SiO}}_4$ tetrahedron.

1. Si (+4) with four O (−2 each) → net charge −4: ${\mathrm{SiO}}_4^{4-}$ (orthosilicate, e.g. in olivine, zircon).
2. Sharing of corners gives ${\mathrm{Si}}_2{\mathrm{O}}_7^{6-}$, chains $({\mathrm{SiO}}_3{)}_{\mathrm{n}}^{2\mathrm{n}-}$, sheets $({\mathrm{Si}}_2{\mathrm{O}}_5{)}_{\mathrm{n}}^{2\mathrm{n}-}$, 3-D ${\mathrm{SiO}}_2$.

Short trick: Simple silicate = orthosilicate SiO₄⁴⁻.