UPSC Chemistry 2026 — Paper I

(a) Two things only: name which van der Waals parameter models intermolecular attraction and which models repulsion (finite molecular size), and give the SI unit of each. No derivation of the equation is wanted. (b) Given surface tension gamma_A > gamma_B, decide for each of the five listed properties (vapour pressure, cohesive force, boiling point, capillary action, wetting) which liquid shows the larger value, with a one-line physical reason each. All five sub-parts must be answered. (c) For two identical blocks (C = 15 J K-1 each) at 300 K and 400 K placed in thermal contact inside an isolated system, calculate (a) the final equilibrium temperature and (b) the total entropy change of the process. (d) For N2O4(g) = 2NO2(g) at 300 K and 1 atm with 20 percent dissociation, compute Kp from the degree of dissociation and then the standard Gibbs free-energy change dG(standard) for the reaction. (e) Apply the phase rule bookkeeping to the NH4Cl(s) = NH3(g) + HCl(g) equilibrium: state the number of components and phases (i) for pure NH4Cl heated in a closed vessel, and (ii) when extra NH3 is also introduced. (f) Recognise the Ag|AgNO3(c1)||AgNO3(c2)|Ag arrangement as a concentration cell without transference and calculate the potential difference dE at 25 degrees C for c1 = 5 x 10^-4 and c2 = 1 x 10^-2 kmol m^-3. (g) Explain why a potentiometric (null / zero-current) measurement is necessary to obtain a meaningful cell potential: the thermodynamic requirement of reversibility, the IR-drop error and the polarisation/composition-change error of a current-drawing measurement. (h) Estimate the activation energy of food-spoilage reactions from the Arrhenius equation, given that the rate is 40 times faster at 30 degrees C (303 K) than at 8 degrees C (281 K). (i) State the apparent colour of a flower that looks red in daylight when it is illuminated only by green light, and justify the answer using selective absorption and the fact that perceived colour is the product of illuminant and reflectance. (j) Explain, using dG = dH - T dS, why spontaneous adsorption of a gas on a solid surface is favoured at low temperature: the sign of the entropy change, the necessarily exothermic enthalpy, and the effect of raising T.

• a = attraction (pressure correction n²a/V²)
• b = repulsion (excluded volume)
• Units: a in Pa m⁶ mol⁻²; b in m³ mol⁻¹
• Vapour pressure: B larger; boiling point: A higher
• Cohesive force: A larger
• Capillary action: A greater rise (h = 2γcosθ/rρg); wetting: B better
• Link to strength of intermolecular attraction
• T_f = 350 K (equal heat capacities)
• ΔS = C ln(T_f/T₁) + C ln(T_f/T₂)
• ΔS ≈ +0.31 J K⁻¹, positive as for an irreversible isolated process
• Equilibrium moles 0.8 and 0.4; total 1.2
• Partial pressures 2/3 and 1/3 atm; K_p = 1/6 ≈ 0.167
• ΔG° = −RT ln K_p ≈ +4.5 kJ mol⁻¹
• (i) C = 1 and P = 2
• (ii) C = 2 and P = 2
• Reasoning by C = S − R − r and the equal-amount restriction
• Concentration cell, E° = 0, n = 1
• ΔE = (RT/F) ln(C₂/C₁) = 0.05916 log(20)
• ΔE ≈ 0.077 V
• No current drawn, so no internal-resistance drop (E − iR)
• Avoids polarisation and concentration change; reversibility
• Gives the equilibrium EMF related to ΔG = −nFE
• Arrhenius equation in the two-temperature form
• T₁ = 281.15 K, T₂ = 303.15 K, ln 40 = 3.689
• E_a ≈ 119 kJ mol⁻¹
• Appears black or very dark
• Red pigment absorbs green and reflects red
• No red component in green light, so nothing is reflected
• ΔG = ΔH − TΔS; ΔS < 0 on adsorption
• ΔH negative (exothermic); adsorption spontaneous when |ΔH| > |TΔS|
• Lower T reduces the −TΔS penalty; Le Chatelier's principle

(a) Sketch the normalised Maxwell speed distribution (1/N)(dN_c/dc) against c for (i) two temperatures with T2 > T1 and (ii) two molar masses with M2 > M1, and justify the shape and the shifts quantitatively using the distribution function. (b(i)) State the interfacial tension between two mutually immiscible liquids A and B whose surface tensions are gamma_A and gamma_B with gamma_A > gamma_B, i.e. apply Antonoff's rule. (b(ii)) Use the Young-Laplace equation for a single spherical liquid surface to compute the excess (differential) pressure across an ethanol droplet of radius 220 nm with gamma = 22.39 mN m^-1 at 20 degrees C. (c(i)) Find the enthalpy change for 1/2 H2(g) + 1/2 I2(g) -> HI(g) at 225 degrees C. This needs two moves: convert the standard formation enthalpy (which refers to I2 in its standard SOLID state) to the gaseous-iodine reaction using the sublimation enthalpy, then transfer the result from 298 K to 498 K by Kirchhoff's law. (c(ii)) For 1/2 N2(g) + 3/2 H2(g) -> NH3(g), compute dS(standard) from the tabulated absolute entropies, back out dH(standard) from dG(standard) at 298 K, and then use the temperature-independent dS assumption to obtain dG(standard) at 400 K.

• Shape: zero at c = 0, single maximum at c_mp = (2RT/M)^½, decays at large c; area = 1
• (i) Higher T: peak shifts right, lower and broader (height ∝ T^-½)
• (ii) Heavier gas: peak shifts left, taller and narrower (height ∝ M^½)
• Labelled axes and both curves on the same axes for each comparison
• γ_AB = γ_A − γ_B (Antonoff's rule)
• Smaller than γ_A; reason from mutual saturation and cross-boundary attraction
• Young-Laplace ΔP = 2γ/r for a single surface
• Unit conversion: 22.39 × 10⁻³ N m⁻¹, 2.20 × 10⁻⁷ m
• ΔP ≈ 2.04 × 10⁵ Pa (about 2 bar)
• ΔH(298 K) = 24.7 − 31.15 = −6.45 kJ mol⁻¹ (using the sublimation enthalpy of I₂)
• ΔC_p = −1.45 J K⁻¹ mol⁻¹
• Kirchhoff: ΔH(498 K) = −6.45 + (−1.45)(200)/1000 ≈ −6.74 kJ mol⁻¹
• ΔS° = 192.45 − 95.805 − 196.02 ≈ −99.4 J K⁻¹ mol⁻¹
• ΔH° = ΔG° + TΔS° ≈ −46.11 kJ mol⁻¹
• ΔG°(400 K) = ΔH° − TΔS° ≈ −6.36 kJ mol⁻¹

(a(i)) Derive the four equivalent definitions of the chemical potential of component i in a mixture as partial derivatives of U, H, A and G, each with its own set of variables held constant, starting from the fundamental equation for an open system. (a(ii)) Sketch the simple eutectic (T versus composition) phase diagram of the Cd-Bi system using the three given data (Cd m.p. 321 degrees C, Bi m.p. 271 degrees C, eutectic at 61 weight percent Bi and 146 degrees C), label every region with the phases present, and explain the shapes of the cooling curves, particularly that of the eutectic mixture. (a(iii)) Explain the working of the acetone-dry ice freezing bath: the role of solid CO2 sublimation, the latent heat absorbed, the function of acetone as heat-transfer medium, and why the bath temperature is self-regulating at about -78 degrees C. (b) Two calculations for Zn2+ moving at 2.74 x 10^-5 m s^-1 in a field of 5 V cm^-1: (i) estimate the hydrated ionic radius from Stokes' law, and (ii) obtain the diffusion coefficient from the Nernst-Einstein (or Stokes-Einstein) relation. (c(i)) Compute the energy of one einstein (one mole of photons) of 800 nm radiation, expressed in SI units. (c(ii)) For the three-step photodecomposition of HI, derive the rate law -d[HI]/dt using the steady-state approximation on the H atom, state the quantum yield phi, and identify the kinetic order with respect to HI. All three quantities must be given.

• Fundamental equation dU = TdS − PdV + Σμᵢdnᵢ giving μᵢ = (∂U/∂nᵢ)_S,V,n
• Legendre transforms giving dH, dA, dG with Σμᵢdnᵢ
• μᵢ = (∂H/∂nᵢ)_S,P = (∂A/∂nᵢ)_T,V = (∂G/∂nᵢ)_T,P
• Only the G form is a partial molar quantity
• Correct diagram: 321 °C, 271 °C, eutectic at 61% Bi and 146 °C, eutectic line
• Four regions labelled: melt; Cd + melt; Bi + melt; Cd + Bi
• Cooling curves: kink at liquidus and halt at 146 °C for off-eutectic melts
• Eutectic melt: single halt only; phase-rule reasoning F = C − P + 1 = 0
• Solid CO₂ sublimes at about −78 °C at 1 atm, absorbing latent heat
• Acetone remains liquid (freezing point about −95 °C) and gives thermal contact
• Temperature stays constant near −78 °C while solid CO₂ remains
• Steady state zeE = 6πηrv with z = 2, E = 500 V m⁻¹
• r ≈ 3.5 × 10⁻¹⁰ m
• D = kT/(6πηr) ≈ 7.0 × 10⁻¹⁰ m² s⁻¹
• Consistency check via Nernst-Einstein
• E per photon = hc/λ ≈ 2.48 × 10⁻¹⁹ J
• One einstein = N_A photons
• ≈ 1.495 × 10⁵ J mol⁻¹ ≈ 149.5 kJ
• Steady-state for H: k₂[H][HI] = I_a
• Steady-state for I: [I] = (I_a/k₃)^½
• −d[HI]/dt = 2I_a
• φ = 2
• Order: zero in HI at constant I_a (first order in absorbed intensity)

(a) Draw the current-voltage (polarographic) wave obtained with a dropping mercury electrode and discuss its three regions - residual current, rising portion with the half-wave potential, and the limiting diffusion current - together with the Ilkovic and Heyrovsky-Ilkovic equations and the experimental precautions. (b(i)) A temperature-jump relaxation problem: derive/quote the relaxation expression for A = B + C, combine it with the equilibrium constant, and calculate BOTH rate constants (the first-order forward constant and the second-order reverse constant) from tau = 3.0 microsecond, K = 2.0 x 10^-16 mol dm^-3 and [B] = [C] = 2.0 x 10^-4 mol dm^-3. (b(ii)) For two parallel first-order reactions D <- B -> D', use the overall decay of B to obtain (k1 + k2) and the branching ratio of the products to split it, thereby calculating k1 and k2 separately. (c) Explain, with a plausible Langmuir-type mechanism, why the kinetic order of a gas-phase decomposition catalysed on a solid surface changes with the pressure of the reactant: derive rate = k theta = k bP/(1 + bP) and examine its low-pressure, high-pressure and intermediate limits with real examples.

• Labelled wave: residual current, rise, plateau (limiting current), E½ at i_d/2
• Diffusion current i_d ∝ C (Ilkovic equation) for quantitative analysis
• E½ characteristic of the species for qualitative analysis; Nernstian log plot gives n
• Role of supporting electrolyte, removal of oxygen, drop oscillations, wide cathodic range of Hg
• Linearised rate law after the T-jump
• 1/τ = k_f + k_r([B]eq + [C]eq)
• k_r ≈ 8.3 × 10⁸ dm³ mol⁻¹ s⁻¹
• k_f = K k_r ≈ 1.7 × 10⁻⁷ s⁻¹
• k₁ + k₂ = ln(10)/40 = 0.0576 min⁻¹
• D = 0.35 kmol m⁻³, [D′]/[D] = k₂/k₁ = 2/7
• k₁ ≈ 0.0448 min⁻¹ and k₂ ≈ 0.0128 min⁻¹ (7.5 × 10⁻⁴ and 2.1 × 10⁻⁴ s⁻¹)
• Langmuir-Hinshelwood mechanism: adsorption, slow surface reaction, desorption
• Rate = kθ = kbP/(1 + bP)
• Low pressure first order; high pressure zero order; intermediate fractional order
• Example or inhibition by strongly adsorbed product (negative order)

(a) Calculate the de Broglie wavelength of a 140 g baseball travelling at 40 m s^-1 and comment on what the magnitude implies about observing wave behaviour for macroscopic objects. (b) Show how dipole-moment measurement distinguishes cis- from trans-1,2-dichloroethylene: the symmetry argument giving mu = 0 for the trans isomer, the vector resultant for the cis isomer, and the experimental method of measuring mu. (c) Draw (not merely describe) all the resonance contributors of the nitrate ion NO3-, and mark the formal charge on every atom in each contributor. A closing statement on the hybrid is expected. (d) Compute the interplanar spacing d(111) for a simple cubic lattice with a = 200 pm using the cubic d-spacing formula, and explain why that formula gives the shortest distance between successive (111) planes. (e) Define ferredoxin as a class of bioinorganic electron-transfer proteins and then give the accepted basis of its classification (by the nuclearity/type of the iron-sulfur cluster, with the plant-type and bacterial-type division). (f) Define the chelate effect (with the thermodynamic reason for it) and then state, with reasons, how the stability of a chelate varies with the size of the ring formed. (g) Justify the description of ferrocene as a sandwich compound and then explain its structure - hapticity, geometry, conformation, metrical data and the 18-electron/bonding rationale. (h) Define ionization isomerism (with an illustrative pair and the test that distinguishes them), then identify the specific type of isomerism shown by each of the three given pairs (i), (ii) and (iii). (i) Identify inorganic benzene as borazine (borazole) B3N3H6 and explain it with a structure - ring geometry, bond lengths, the nature of the pi delocalisation, and how it differs chemically from benzene. (j) Define actinide contraction and explain its cause (poor shielding by 5f electrons plus relativistic effects), supporting it with radius data and stating its chemical consequences.

• λ = h/mv with SI units
• λ ≈ 1.2 × 10⁻³⁴ m
• Comment: immeasurably small, so no observable wave behaviour for macroscopic bodies
• cis: bond dipoles add, μ ≈ 1.9 D
• trans: dipoles cancel, μ = 0
• Identification by measurement of μ
• Three equivalent structures, each with one N=O and two N–O⁻
• Formal charges N +1, =O 0, –O⁻ −1; net −1
• Resonance hybrid: equal N–O bonds, delocalised charge, trigonal planar
• d = a/√(h² + k² + l²)
• d₁₁₁ = 200/√3
• ≈ 115.5 pm
• Small Fe-S electron-transfer protein with cysteine-bound clusters
• Fe(II)/Fe(III) redox cycling
• Classes: [2Fe-2S], [3Fe-4S], [4Fe-4S]
• Definition: enhanced stability with chelating ligands
• Entropy reason: more free molecules released, ΔS > 0
• Five- and six-membered rings most stable; small and large rings less stable
• Example such as [Ni(en)₃]²⁺ versus [Ni(NH₃)₆]²⁺
• Fe²⁺ between two parallel η⁵-C₅H₅ rings
• Each ring donates 6 π electrons; 18-electron configuration
• Equal Fe–C and C–C bond lengths; eclipsed (gas phase), diamagnetic
• Aromatic character or stability
• Definition: same formula, different ions in solution (coordinated versus counter-ion exchange)
• (i) hydrate (solvate) isomerism
• (ii) ionization isomerism, distinguished by BaCl₂ test
• (iii) coordination isomerism
• Borazine B₃N₃H₆, isoelectronic with benzene
• Planar six-membered ring of alternating B and N
• Polar B–N, less aromatic; adds HCl
• Preparation
• Steady decrease of radii from Ac to Lr
• Cause: poor shielding by 5f electrons
• Greater than lanthanide contraction
• Consequence such as similarity of 6d and 5d

(a(i)) For each of the five species SO2, I3(-), SF4, BrF5 and XeF4, draw the Lewis structure and then apply VSEPR: count bond pairs and lone pairs on the central atom, give the electron-pair (parent) geometry and the resulting molecular shape with approximate bond angles. (a(ii)) Draw a correctly ordered molecular orbital energy-level diagram for F2, populate it with the right number of electrons, and from the filling deduce the bond order and state the magnetic behaviour. (b(i)) Explain the very different colour intensities of two Co(II) complexes by invoking the selection rules for electronic spectra - the Laporte rule and its relaxation in a non-centrosymmetric tetrahedral field - and account for the difference in the hue via the ligand-field splitting energy. (b(ii)) State Sidgwick's Effective Atomic Number rule and then calculate the EAN for each of the four given complexes, commenting on which obey and which violate the rule. (c) Describe the metal-metal bonding in the octachlorodirhenate(III) ion: derive the bond order of four, identify the orbital origin of each of the four components (sigma, two pi, delta), and account for the eclipsed geometry and the short Re-Re distance.

• SO₂: three domains, one lone pair, bent
• I₃⁻: five domains, three lone pairs, linear
• SF₄ see-saw; BrF₅ square pyramidal; XeF₄ square planar
• Electron counts and lone-pair placement (equatorial or trans) shown
• Correct order σ2s < σ*2s < σ2p < π2p < π*2p < σ*2p (no s-p mixing)
• Fourteen electrons filled: σ2s² σ*2s² σ2p² π2p⁴ π*2p⁴
• Bond order = 1
• Diamagnetic (all paired)
• Octahedral with centre of symmetry: d-d transition Laporte forbidden (g → g), weak, ε about 10
• Tetrahedral without centre of symmetry: d-p mixing relaxes the rule, ε about 100 times larger
• Δt ≈ 4/9 Δo: absorption shifts to lower energy in the red
• Colour from the transmitted light: pink versus blue
• Rule: metal gains electrons from ligands to reach the next noble-gas configuration
• [Cr(CO)₆] = 36; [Fe(CN)₆]⁴⁻ = 36
• [Co(NH₃)₆]³⁺ = 36
• [Ni(NH₃)₆]²⁺ = 38: rule not obeyed
• Re(III) d⁴, eight d electrons, Re-Re about 224 pm, eclipsed
• σ (dz²), two π (dxz, dyz) and one δ (dxy) bonds
• Configuration σ² π⁴ δ²
• Bond order 4; eclipsed geometry as evidence for the δ bond

(a(i)) Prove by operating with A = d(n)/dx(n) on psi(x) = exp(alpha x) that the result is a constant multiple of psi, thereby showing psi is an eigenfunction, and state the eigenvalue. (a(ii)) Explain why the quantum number n = 0 must be rejected for the particle in a one-dimensional box - i.e. show that it makes the wavefunction vanish identically and therefore violates the requirements of normalisation, physical interpretation and the uncertainty principle. (b(i)) Using the form of the uncertainty relation given in the question, delta-p(x) times delta-x greater than or equal to h, compute the minimum uncertainty in the momentum and hence in the speed of an electron localised to delta-x of about 50 pm, and comment on both results. (b(ii)) Give the three quantum numbers n, l and m(l) for a 2p(z) orbital and explain how each is assigned, including why p(z) specifically corresponds to m(l) = 0. (c(i)) Calculate the packing fraction (packing efficiency) of the simple cubic, body-centred cubic and face-centred cubic lattices from first principles, and on that basis state which is the most economically packed and which is the least. (c(ii)) State Bragg's condition for constructive interference in X-ray diffraction from a crystal and explain it by deriving the path difference between rays reflected from successive lattice planes.

• Repeated differentiation gives αⁿ e^(αx)
• Form Âψ = aψ recognised
• Eigenvalue αⁿ
• ψ = 0 everywhere for n = 0
• No particle: cannot be normalised
• Zero energy would violate the uncertainty principle; minimum energy h²/8mL²
• Δp ≥ h/Δx ≈ 1.3 × 10⁻²³ kg m s⁻¹
• Δv = Δp/mₑ ≈ 1.5 × 10⁷ m s⁻¹
• Comment: about 5% of c, larger than the Bohr-orbit speed 2.2 × 10⁶ m s⁻¹
• Comment: implies no definite orbit; need for orbital or probability description
• n = 2
• l = 1 (p)
• m_l = 0, corresponding to the z orientation
• SC: 1 atom, r = a/2, π/6 = 52.4%
• BCC: 2 atoms, r = √3a/4, π√3/8 = 68.0%
• FCC: 4 atoms, r = √2a/4, π√2/6 = 74.0%
• Most efficient FCC; least SC
• Atoms per cell and touching direction stated for each
• nλ = 2d sin θ
• Path difference 2d sin θ between adjacent planes
• Constructive interference when the path difference is nλ
• Requirement λ ≤ 2d

(a) Differentiate the T (tense, deoxy) and R (relaxed, oxy) quaternary states of haemoglobin - in oxygen affinity, iron stereochemistry and spin state, subunit arrangement and salt bridges, and effector binding - and explain how the T to R switch generates cooperativity. (b(i)) Describe the molecular structure of diborane (bridged, D2h) and account for its bonding, showing why the classical two-centre two-electron picture fails and how three-centre two-electron bonds resolve the electron deficiency. (b(ii)) Set out the classification of silicates on the basis of the number of corner oxygen atoms shared per SiO4 tetrahedron, and then name and describe the structure of the three given anions: SiO4(4-), Si2O7(6-) and Si3O9(6-). (c(i)) Explain why the room-temperature magnetic moments of Eu(3+) and Sm(3+) depart from the Landé/Hund prediction that fits the other lanthanide ions, i.e. why low-lying excited J states and Van Vleck's treatment must be invoked. (c(ii)) Account for the steady fall in metallic radii across the lanthanide series (the lanthanide contraction) and explain the two anomalies at europium and ytterbium, where the radii are abnormally large. (c(iii)) Describe how the lanthanides are separated from one another by methods based on complex formation - chiefly ion-exchange chromatography with a complexing eluent, and solvent extraction - explaining the principle that makes the separation possible and the order of elution. (c(iv)) Explain why Ce(3+) (4f1) and Yb(3+) (4f13) are colourless, i.e. show no absorption in the visible region, yet give sharp absorption bands in the ultraviolet.

• T: deoxy, low O₂ affinity, salt bridges, Fe out of plane; R: oxy, high affinity, salt bridges broken, Fe in plane
• Quaternary rotation of about 15° and change in central cavity
• 2,3-BPG binds T; H⁺ and CO₂ (Bohr effect) favour T
• Cooperativity, sigmoidal curve, MWC concerted model
• Bridged D₂h structure: 4 terminal and 2 bridging H
• 12 valence electrons: 4 terminal 2c–2e bonds and 2 bridging 3c–2e bonds
• Bond lengths or angles (B–Ht 119 pm, B–Hb 133 pm, B–H–B 83°)
• Electron deficiency, sp³ boron, banana bonds
• Classification by number of shared corners: 0, 1, 2, 3, 4
• SiO₄⁴⁻: orthosilicate, isolated tetrahedron
• Si₂O₇⁶⁻: pyrosilicate, two tetrahedra sharing one corner
• Si₃O₉⁶⁻: cyclic trisilicate, six-membered ring, two corners shared per tetrahedron
• Low-lying excited J levels (small spin-orbit splitting)
• Thermally populated at 300 K since separation is comparable with kT
• Observed versus Van Vleck values (Eu³⁺ 0 versus about 3.4 BM; Sm³⁺ 0.84 versus about 1.5 BM)
• Lanthanide contraction: increasing nuclear charge with poor 4f shielding
• Effective nuclear charge rises and outer shells shrink
• Eu and Yb have larger radii: divalent character, only two electrons in metallic bonding
• Stable half-filled 4f⁷ and filled 4f¹⁴
• Cation-exchange resin with Ln³⁺ adsorbed
• Elution with a complexing agent such as citrate or EDTA at controlled pH
• Stability of complexes rises with atomic number; heavier ions elute first
• Fractions collected and processed
• Only one f-f transition ²F₅/₂ ↔ ²F₇/₂, at low (IR) energy
• No visible absorption, so colourless
• UV absorption: 4f → 5d (Ce³⁺), Laporte-allowed and intense
• UV absorption: charge transfer (Yb³⁺)