UPSC Civil Services (Main) Examination 2026 — Chemistry Optional Paper I: Questions with Model Answers | Plutus IAS

08 Sep UPSC Civil Services (Main) Examination 2026 — Chemistry Optional Paper I: Questions with Model Answers | Plutus IAS

The questions below are from Chemistry Optional Paper I of UPSC Civil Services (Main) Examination 2026 (held 2026-08-30) — the actual paper, which is public. Each carries a model answer written by Aanya in Plutus IAS teaching style, to the marks and word limit.

Official source: official (upsc.gov.in).

Q1. (a) $$dN_c = 4\pi N \left( \frac{M}{2\pi RT} \right)^{3/2} c^2 e^{-Mc^2/2RT} dc$$ The Maxwell distribution of molecular speed of a gaseous system is $$dN_c = 4\pi N \left( \frac{M}{2\pi RT} \right)^{3/2} c^2 e^{-Mc^2/2RT} dc$$ (Symbols have their usual meanings) Plot $$\left( \frac{1}{N} \times \frac{dN_c}{dc} \right)$$ against c— (i) at two different temperatures $T_1$ and $T_2$; $T_2 > T_1$; (ii) at two molar masses $M_1$ and $M_2$; $M_2 > M_1$. Justify your plots. 8+7=15 (b) Two immiscible liquids A and B have surface tension values $\gamma_A$ and $\gamma_B$, respectively. When both the liquids are taken in a container, then what would be the interfacial tension between the two liquid phases? (Given : $\gamma_A > \gamma_B$) Calculate the pressure differential across the surface of ethanol spherical droplet of radius 220 nm at 20 °C. The surface tension of ethanol at that temperature is 22.39 mN m$^{-1}$. [ P.T.O. (c) $$H_2(g); C_{p,m} = 29.08 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$I_2(g); C_{p,m} = 33.56 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$HI(g); C_{p,m} = 29.87 \text{ J K}^{-1} \text{ mol}^{-1}$$ At 25 °C, the latent heat of sublimation per mole of iodine is 62.3 kJ mol⁻¹ and the standard enthalpy of formation of HI(g) is 24.7 kJ mol⁻¹. Calculate the enthalpy change which occurs when HI(g) is formed from the gaseous elements at 225 °C. The mean molar heat capacities over the temperature range 25 °C to 225 °C are : $$H_2(g); C_{p,m} = 29.08 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$I_2(g); C_{p,m} = 33.56 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$HI(g); C_{p,m} = 29.87 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)$$ $$S^\circ_{NH_3} = 192.45 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$S^\circ_{N_2} = 191.61 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$S^\circ_{H_2} = 130.68 \text{ J K}^{-1} \text{ mol}^{-1})$$ Calculate ΔG° at 400 K temperature for the following reaction : $$\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)$$ Assume that ΔS° is independent of temperature. (Given : ΔG°₂₉₈K = -16.496 kJ mol⁻¹ $$S^\circ_{NH_3} = 192.45 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$S^\circ_{N_2} = 191.61 \text{ J K}^{-1} \text{ mol}^{-1}$$ $$S^\circ_{H_2} = 130.68 \text{ J K}^{-1} \text{ mol}^{-1})$$ (15 marks)

How to approach this question

The directive word “Plot” requires you to sketch two pairs of curves and justify their relative positions. The examiner is testing your grasp of (i) the Maxwell speed distribution function, (ii) how temperature and molar mass shift the distribution, and (iii) the physical meaning of the plotted quantity (fractional population per speed interval). A top answer must (1) define the plotted quantity, (2) sketch the four curves with correct relative positions, and (3) explain each shift using the exponential and quadratic factors. The common mistake is to misplace the peaks or to ignore the 1/N normalization, which compresses the vertical axis.

Model answer

The quantity plotted is the normalized Maxwellian

\frac{1}{N}\frac{dN_c}{dc}=4\pi\left(\frac{M}{2\pi RT}\right)^{3/2}c^2\exp\left(-\frac{Mc^2}{2RT}\right).

Part (i) — Temperature variation (T₂ > T₁)

• Peak position c* ∝ √(RT/M): raising T shifts the peak to higher speeds.
• Peak height ∝ T⁻¹·(M/T)¹·⁵: raising T lowers the maximum because the distribution spreads out.
• Curve for T₂ lies to the right and lower than that for T₁.

Part (ii) — Molar-mass variation (M₂ > M₁)

• Peak position c* ∝ √(RT/M): heavier molecules have lower most-probable speeds.
• Peak height ∝ M³·⁵: heavier molecules give a taller, narrower peak.
• Curve for M₂ is left-shifted and higher than that for M₁.

Conclusion
The plots illustrate how thermal energy broadens and lowers the distribution while molar mass sharpens and raises it. These trends are routinely used in effusion and reaction-rate studies.

Q2. (a) Derive the expressions of chemical potential of a component in a mixture system in terms of extensive thermodynamic properties U, H, A and G. Cadmium (m. pt.=321 °C) and bismuth (m. pt.=271 °C) do not form solid solutions or compounds with one another. Their eutectic point lies at 61 weight percent of bismuth and 146 °C. Sketch their phase diagram and label each region to show what phases are present. Explain the cooling curves of this eutectic mixture. How does acetone-dry ice freezing mixture work? (b) In a homogeneous electric field E = 5 V cm⁻¹, the speed ν of Zn²⁺ ions in aqueous solution at 25 °C is 2.74 × 10⁻⁵ m s⁻¹. (i) Estimate the radius r of the hydrated Zn²⁺ ion. The coefficient of viscosity η of water at the given temperature is 0.890 mPa s. (ii) Calculate the diffusion coefficient D of the ion. (c) Calculate the energy of 800 nm electromagnetic radiation per einstein in SI system. $$\begin{array}{l} \mathrm{HI(g)} + h\nu \longrightarrow \mathrm{H(g)} + \mathrm{I(g)} \\ \mathrm{H(g)} + \mathrm{HI(g)} \xrightarrow{k_2} \mathrm{H_2(g)} + \mathrm{I(g)} \\ \mathrm{I(g)} + \mathrm{I(g)} \xrightarrow{k_3} \mathrm{I_2(g)} \end{array}$$ The photodecomposition of HI(g) takes place following three elementary steps : $$\begin{array}{l} \mathrm{HI(g)} + h\nu \longrightarrow \mathrm{H(g)} + \mathrm{I(g)} \\ \mathrm{H(g)} + \mathrm{HI(g)} \xrightarrow{k_2} \mathrm{H_2(g)} + \mathrm{I(g)} \\ \mathrm{I(g)} + \mathrm{I(g)} \xrightarrow{k_3} \mathrm{I_2(g)} \end{array}$$ Find the expression of $-\frac{d[\mathrm{HI(g)}]}{dt}$, quantum yield (φ) and kinetic order of the reaction. (15 marks)

How to approach this question

The directive word “Derive” signals that the examiner wants rigorous thermodynamic derivations and graphical reasoning. The question tests three distinct skills: (1) deriving chemical potential from fundamental potentials (U, H, A, G), (2) interpreting and sketching a binary eutectic phase diagram with cooling-curve explanation, and (3) explaining a practical cryogenic mixture. The common mistake is to mix up the definitions of the four thermodynamic potentials when relating μ to extensive variables; always start from the Gibbs equation dG = V dP − S dT + Σμ dn.

Model answer

Chemical potential in mixtures

For a component i in a mixture, the chemical potential μ_i is defined as the partial molar Gibbs free energy:

• From internal energy U: μ_i = (∂U/∂n_i)_S,V,nj≠i
• From enthalpy H: μ_i = (∂H/∂n_i)_S,P,nj≠i
• From Helmholtz free energy A: μ_i = (∂A/∂n_i)_T,V,nj≠i
• From Gibbs free energy G: μ_i = (∂G/∂n_i)_T,P,nj≠i

Cd–Bi eutectic phase diagram and cooling curves

A binary eutectic diagram for Cd (m.pt. 321 °C) and Bi (m.pt. 271 °C) with no solid solutions shows:

• Liquid (L) region above the liquidus lines.
• Pure solid Cd and Bi regions on the left and right axes.
• A horizontal eutectic isotherm at 146 °C and 61 wt % Bi where L ⇄ Cd(s) + Bi(s).

Cooling a eutectic mixture from the melt yields a single arrest at 146 °C, producing a sharp horizontal plateau on the cooling curve; off-eutectic compositions show two arrests (primary phase + eutectic).

Acetone–dry ice mixture

Solid CO₂ (dry ice) sublimes at −78 °C, absorbing latent heat; acetone provides a high-surface-area liquid matrix that enhances heat transfer, yielding a bath near −78 °C suitable for low-temperature reactions.

Q3. (a) Draw and discuss the polarographic wave obtained in the case of a dropping mercury electrode in polarography. (b) The equilibrium $A \rightleftharpoons B + C$ at 25 °C is subjected to a sudden temperature increase that slightly increases the concentrations of $B$ and $C$. The measured relaxation time is 3.0 μs. The equilibrium constant for the system is $2.0 \times 10^{-16} \text{ mol dm}^{-3}$ at 25 °C, and the equilibrium concentrations of $B$ and $C$ at 25 °C are both $2.0 \times 10^{-4} \text{ mol dm}^{-3}$. Calculate the rate constants involved in the system. A substance $B$ converts simultaneously to the products $D$ and $D'$ by two parallel elementary reactions $D \xleftarrow{k_1} B \xrightarrow{k_2} D'$. The initial concentration of $B$ is $C_{B,0} = 0.5 \text{ kmol m}^{-3}$. After 40 minutes, the concentration of $B$ has decreased to $0.05 \text{ kmol m}^{-3}$; in the same time, the product $D'$ with a concentration $C_{D'} = 0.1 \text{ kmol m}^{-3}$ is formed. Calculate the rate coefficients $k_1$ and $k_2$. (c) The kinetic order of the solid-surface catalyzed gas-phase decomposition reaction varies with the pressure of the gaseous reactant. Explain with plausible mechanism. (15 marks)

How to approach this question

The directive word “draw and discuss” in part (a) asks for a labelled diagram plus a mechanistic explanation of the polarographic wave at a dropping mercury electrode (DME). Examiners test your grasp of Ilkovič equation, diffusion current, half-wave potential and the role of DME in minimising polarisation. A top answer must (1) sketch the S-shaped polarogram, (2) label diffusion current plateau, half-wave potential and residual current, and (3) explain why the DME gives reproducible, poison-free waves. The common mistake is to omit the Ilkovič equation or to confuse half-wave potential with standard electrode potential.

Model answer

a. Polarographic wave at a dropping mercury electrode

The polarogram obtained with a dropping mercury electrode (DME) is an S-shaped current–potential curve (Figure 1). As the applied potential is made more negative, the cathodic current remains near the residual current until the reduction potential of the analyte is reached; thereafter, the current rises steeply and finally levels off at the diffusion current plateau.

Key features and their origin

• Residual current: Small current due to charging of the double layer and traces of reducible impurities; appears at potentials before the analyte reduction.
• Rising portion: Governed by the Butler–Volmer equation; rate of electron transfer is fast, so the current is limited by diffusion of analyte to the expanding mercury drop (Ilkovič equation).
• Half-wave potential (E½): Potential at which the current is half the diffusion current; characteristic of the electroactive species and independent of its concentration.
• Diffusion current plateau: Current limited solely by diffusion; described by the Ilkovič equation:
  i_d = 607 n D^1/2 m^2/3 t^1/6 C
  where n is electrons per molecule, D diffusion coefficient, m mercury flow rate, t drop time and C bulk concentration.

The DME’s periodically renewed surface prevents electrode poisoning and time-dependent polarisation, giving highly reproducible waves—key to quantitative polarography.

b. Rate constants from relaxation and parallel reactions

For the equilibrium A ⇌ B + C, the relaxation time τ = 1/(k_f + k_r) = 3.0 μs. The equilibrium constant K = k_f/k_r = 2.0 × 10⁻¹⁶ mol dm⁻³. At equilibrium, [B] = [C] = 2.0 × 10⁻⁴ mol dm⁻³, so K = [B][C]/[A] ⇒ [A] = ([B][C])/K = 2.0 × 10⁸ mol dm⁻³. Substituting into K = k_f/k_r and τ = 1/(k_f + k_r) gives k_f = 1.0 × 10⁸ s⁻¹ and k_r = 5.0 × 10²³ dm³ mol⁻¹ s⁻¹.

For parallel reactions B → D (k₁) and B → D′ (k₂), the integrated rate law yields k₁ + k₂ = (1/t) ln(C_B,0/C_B) = 5.76 × 10⁻⁵ s⁻¹. The yield of D′ after 40 min (2400 s) is [D′] = (k₂/k_total) C_B,0 (1 − e^−k_totalt), giving k₂ = 2.4 × 10⁻⁵ s⁻¹ and hence k₁ = 3.36 × 10⁻⁵ s⁻¹.

c. Pressure-dependent kinetic order in surface-catalysed decomposition

In gas-phase decomposition on a solid catalyst, the observed order depends on the relative rates of adsorption, surface reaction and desorption, which change with gas pressure.

Mechanistic rationale

• Low pressure (Henry regime): Adsorption is weak and far from saturation; surface coverage θ ∝ P. The surface reaction (RDS) is bimolecular in adsorbed species, so rate = k θ_A² ∝ P² ⇒ second order.
• Intermediate pressure (Langmuir regime): θ approaches a constant; rate = k θ_A ∝ P⁰ ⇒ zero order.
• High pressure (surface saturation): Strong adsorption blocks sites; desorption becomes rate-limiting and rate ∝ P⁻¹ ⇒ negative order.

Thus, the kinetic order decreases from 2 to 0 to −1 as pressure increases, reflecting the shift from adsorption-limited to reaction-limited to desorption-limited regimes.

Q4. (a) What is the de Broglie wavelength for a baseball (140 g) moving at 40 m s$^{-1}$? Comment on your finding. (b) Distinguish between cis- and trans-dichloroethylene by using dipole moments. (c) Provide the resonance structures with formal charges for the nitrate ion. (d) What is the shortest distance between two successive 111 planes of simple cubic lattice of edge length 200 pm? Explain. (e) What is ferredoxin? How is it classified? What is chelate effect? How is it affected by the ring size? Why is ferrocene termed as a sandwich compound? Explain its structure. (i) $[\text{Cr}(\text{H}_2\text{O})_6\text{Cl}_3]$ तथा $[\text{CrCl}(\text{H}_2\text{O})_5]\text{Cl}_2 \cdot \text{H}_2\text{O}$ (ii) $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{NO}_3$ तथा $[\text{Co}(\text{NH}_3)_5\text{NO}_3]\text{SO}_4$ (iii) $[\text{Pt}(\text{NH}_3)_4][\text{PtCl}_6]$ तथा $[\text{Pt}(\text{NH}_3)_4\text{Cl}_2][\text{PtCl}_4]$ What is ionization isomerism? Mention the type of isomerism in the given pairs of complexes : (i) $[\text{Cr}(\text{H}_2\text{O})_6\text{Cl}_3]$ and $[\text{CrCl}(\text{H}_2\text{O})_5]\text{Cl}_2 \cdot \text{H}_2\text{O}$ (ii) $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{NO}_3$ and $[\text{Co}(\text{NH}_3)_5\text{NO}_3]\text{SO}_4$ (iii) $[\text{Pt}(\text{NH}_3)_4] [\text{PtCl}_6]$ and $[\text{Pt}(\text{NH}_3)_4\text{Cl}_2] [\text{PtCl}_4]$ What is inorganic benzene? Explain it with structure. What is actinide contraction? Explain. (15 marks)

How to approach this question

The directive word “Comment” in part (a) asks for both calculation and interpretation; examiners test grasp of de Broglie’s matter-wave concept and the macroscopic–microscopic divide. A top answer must: (1) state the formula and plug in values, (2) compute the wavelength, and (3) comment on why the result is negligible. The common mistake is to stop after the numerical answer without the physical insight.

Model answer

(a) The de Broglie wavelength is given by λ = h/(m v). Substituting h = 6.626 × 10⁻³⁴ J s, m = 0.140 kg, and v = 40 m s⁻¹:

λ = (6.626 × 10⁻³⁴)/(0.140 × 40) ≈ 1.18 × 10⁻³⁴ m.

This wavelength is ~10¹⁹ times smaller than the atomic scale, rendering matter-wave effects undetectable for macroscopic objects like a baseball.

(b) Cis-1,2-dichloroethylene has both chlorine atoms on the same side; trans has them opposite. The cis isomer has a net dipole moment (~1.9 D) because the bond dipoles do not cancel, whereas the trans isomer’s symmetry cancels dipoles, giving a zero dipole moment.

(c) The nitrate ion (NO₃⁻) exhibits three equivalent resonance structures: one N=O double bond alternates among the three oxygen atoms. Formal charges are +1 on N and −1 on each double-bonded O, or 0 on single-bonded O with an additional lone pair; the resonance hybrid averages these.

(d) In a simple cubic lattice of edge a = 200 pm, the spacing between successive {111} planes is d₁₁₁ = a/√(1² + 1² + 1²) = 200/√3 ≈ 115 pm.

(e) Ferredoxin is an iron–sulfur protein that mediates low-potential electron transfer in photosynthesis and nitrogen fixation; it is classified as an electron-transfer metalloprotein. The chelate effect describes the enhanced stability of a complex when a multidentate ligand replaces several monodentate ligands; it is maximized for five- and six-membered chelate rings. Ferrocene, Fe(C₅H₅)₂, is a metallocene with an iron(II) centre sandwiched between two parallel cyclopentadienyl anions, exhibiting D_5h symmetry and aromatic stabilization.

Ionization isomerism occurs when a counter-ion in the outer coordination sphere exchanges with a ligand in the inner sphere, yielding different compounds in solution. The given pairs exhibit: (i) ionization isomerism, (ii) ionization isomerism, and (iii) coordination isomerism.

Inorganic benzene is borazine (B₃N₃H₆), isoelectronic and isostructural with C₆H₆, featuring alternating B–N bonds with partial π-delocalization.

Actinide contraction is the gradual decrease in ionic radii across the actinide series due to poor shielding by 5f electrons, causing contraction similar to the lanthanide contraction but with greater magnitude because 5f orbitals extend farther from the nucleus.

Q5. (a) (1) $\text{SO}_2$ , (2) $\text{I}_3^-$ , (3) $\text{SF}_4$ , (4) $\text{BrF}_5$ , (5) $\text{XeF}_4$ Predict the Lewis structures and geometry of the following as per VSEPR theory : (1) $\text{SO}_2$ , (2) $\text{I}_3^-$ , (3) $\text{SF}_4$ , (4) $\text{BrF}_5$ , (5) $\text{XeF}_4$ Draw the molecular orbital diagram for $\text{F}_2$ molecule, and mention its bond order and magnetic behaviour. (b) Why is the electronic spectrum of $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ pale pink, whereas the spectrum of $[\text{CoCl}_4]^{2-}$ is deep blue? Explain. (1) $[\text{Cr}(\text{CO})_6]$ (2) $[\text{Fe}(\text{CN})_6]^{4-}$ (3) $[\text{Co}(\text{NH}_3)_6]^{3+}$ (4) $[\text{Ni}(\text{NH}_3)_6]^{2+}$ What is EAN rule? Calculate the EAN value of the following : (1) $[\text{Cr}(\text{CO})_6]$ (2) $[\text{Fe}(\text{CN})_6]^{4-}$ (3) $[\text{Co}(\text{NH}_3)_6]^{3+}$ (4) $[\text{Ni}(\text{NH}_3)_6]^{2+}$ (c) Describe the bond order and quadruple bonding in $[\text{Re}_2\text{Cl}_8]^{2-}$ ion. (15 marks)

How to approach this question

This is a multi-part, application-heavy question testing VSEPR theory, molecular orbital theory, crystal field theory, and EAN rule. The examiner is checking: (1) ability to predict geometry using VSEPR, (2) drawing MO diagrams and interpreting bond order/magnetism, (3) explaining d–d transitions in cobalt complexes, and (4) applying the EAN rule. A common mistake is to confuse hybridization labels or miscount electrons in MO diagrams, leading to wrong bond order or magnetic predictions.

Model answer

Part (a)(i): VSEPR predictions

• SO₂: 3 electron domains (1 lone pair + 2 bonding pairs) → bent geometry (∠O–S–O ≈ 119°).
• I₃⁻: 5 electron domains (3 lone pairs + 2 bonding pairs) → linear geometry (∠I–I–I = 180°).
• SF₄: 5 electron domains (1 lone pair + 4 bonding pairs) → see-saw geometry (equatorial lone pair).
• BrF₅: 6 electron domains (1 lone pair + 5 bonding pairs) → square pyramidal geometry.
• XeF₄: 6 electron domains (2 lone pairs + 4 bonding pairs) → square planar geometry. The lone pairs occupy axial positions, minimizing repulsion.

Part (a)(ii): MO diagram for F₂

• Electronic configuration: (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p_z)²(π2p_x)²(π2p_y)²(π*2p_x)¹(π*2p_y)¹.
• Bond order = (8 bonding – 6 antibonding)/2 = 1.
• Two unpaired electrons → paramagnetic.

Part (b): Electronic spectra of cobalt complexes

The pale pink of [Co(H₂O)₆]²⁺ arises from a weak d–d transition (⁴T₁g → ⁴T₂g) in an octahedral (sp³d²) field, where H₂O is a weak-field ligand causing small Δ₀. The deep blue of [CoCl₄]²⁻ results from a stronger d–d transition in a tetrahedral (sp³) field; Cl⁻ is a weak-field ligand but tetrahedral splitting (Δₜ ≈ 4/9 Δ₀) is smaller, shifting absorption to higher energy (blue region).

Part (c): EAN rule and calculations

The Effective Atomic Number (EAN) rule states that stable complexes satisfy EAN = atomic number of next noble gas. EAN = metal electrons + ligand electrons donated.

• [Cr(CO)₆]: Cr(0) has 6 electrons; 6 CO donate 12 electrons → EAN = 6 + 12 = 36.
• [Fe(CN)₆]⁴⁻: Fe(II) has 6 electrons; 6 CN⁻ donate 12 electrons → EAN = 6 + 12 = 36.
• [Co(NH₃)₆]³⁺: Co(III) has 6 electrons; 6 NH₃ donate 12 electrons → EAN = 6 + 12 = 36.
• [Ni(NH₃)₆]²⁺: Ni(II) has 8 electrons; 6 NH₃ donate 12 electrons → EAN = 8 + 12 = 20.

Part (d): Bonding in [Re₂Cl₈]²⁻

The [Re₂Cl₈]²⁻ ion exhibits a quadruple bond (σ²π⁴δ²) between two Re atoms, giving a bond order of 4. This arises from overlap of d-orbitals: one σ bond (dz²), two π bonds (dxz, dyz), and one δ bond (dxy/dx²–y²). The eclipsed Cl–Cl conformation minimizes steric repulsion and stabilizes the δ bond, enabling the short Re–Re distance (~2.24 Å).

Q6. (a) Show that $\psi(x) = e^{\alpha x}$ is an eigenfunction for the operator $\hat{A} = \frac{d^n}{dx^n}$ . What is the eigenvalue? $$\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi(x) = 0;\ 0 \leq x \leq L$$ The time-independent Schrödinger equation for a free particle (of mass $m$ ) in a box of length $L$ (inside the box potential energy is zero, but outside the box is infinity) is $$\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi(x) = 0;\ 0 \leq x \leq L$$ The solution of the above equation is $\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$ ; where $n = 1, 2, 3, \dots$ . Why is $n = 0$ not allowed? Explain. (b) Find the minimum uncertainty in momentum and speed if we wish to locate an electron within an atom so that $\Delta x \approx 50$ pm, using $\Delta p_x \times \Delta x \geq h$ . Comment on your two findings. What are the $n, l$ and $m_l$ values of $2p_z$ orbital? Explain. (c) Out of three lattice structures SC, BCC and FCC, which one is most economically packed and which one least? Give your answer, calculating packing fraction of each lattice structure. For Bragg X-ray diffraction studies of crystalline solid, what is the Bragg condition for constructive interference? Explain. (15 marks)

How to approach this question

The directive word “Show” tests your ability to verify a mathematical statement rigorously; “Why is n=0 not allowed?” tests conceptual grasp of boundary conditions in quantum mechanics. A top answer must (1) prove the eigenfunction-eigenvalue relation with explicit differentiation, (2) justify the exclusion of n=0 via boundary-value analysis, and (3) link the result to physical observables (energy quantisation). The common mistake is to stop after stating the wavefunction form without explicitly computing the second derivative and applying the boundary conditions.

Model answer

Part (a)

To show that ψ(x)=e^{αx} is an eigenfunction of the operator Â=dⁿ/dxⁿ, apply the operator to ψ:

• For n=2, Âψ = d²/dx² (e^{αx}) = α² e^{αx} = α² ψ.
• Hence ψ is an eigenfunction with eigenvalue α².

In the given time-independent Schrödinger equation for a free particle in a box, the general solution is ψ(x)=A sin(kx)+B cos(kx), with k=√(2mE)/ħ. The boundary conditions ψ(0)=ψ(L)=0 yield B=0 and k=nπ/L, n=1,2,3,… Thus ψₙ(x)=√(2/L) sin(nπx/L).

The exclusion of n=0 follows from ψ₀(x)=√(2/L) sin(0)=0, which is not normalisable and represents zero probability density everywhere—physically meaningless for a confined particle.

Part (b)

Using the uncertainty relation Δx Δp ≥ ħ/2 and Δx≈50 pm=5×10⁻¹¹ m, we get Δp ≥ ħ/(2Δx)≈1.05×10⁻²⁴ kg·m/s.

The minimum uncertainty in speed is Δv=Δp/mₑ≈1.16×10⁶ m/s, revealing that locating an electron at atomic scales necessarily imparts a momentum uncertainty comparable to its typical orbital speed, underscoring the wave-particle duality at microscopic scales.

The 2p_z orbital has n=2, l=1, and m_l=0, corresponding to the second principal shell, p-subshell, and zero magnetic quantum number.

Part (c)

Packing fractions:

• SC: 1×(4/3)πr³/a³=π/6≈0.5236
• BCC: 2×(4/3)πr³/a³=√3 π/8≈0.6802
• FCC: 4×(4/3)πr³/a³=√2 π/6≈0.7405

FCC is most economically packed; SC is least. The Bragg condition for constructive interference is 2d sinθ=nλ, where d is the interplanar spacing, θ the incidence angle, n the order, and λ the X-ray wavelength.

Q7. (a) Differentiate between T and R conformation of hemoglobin. (b) Describe the structure and bonding in diborane. (1) $\text{SiO}_4^{4-}$ आयन, (2) $\text{Si}_2\text{O}_7^{6-}$ आयन, (3) $\text{Si}_3\text{O}_9^{6-}$ आयन How are silicates classified? Predict the names and structures of the following silicates : (1) $\text{SiO}_4^{4-}$ ion, (2) $\text{Si}_2\text{O}_7^{6-}$ ion, (3) $\text{Si}_3\text{O}_9^{6-}$ ion (c) Why do $\text{Eu}^{3+}$ and $\text{Sm}^{3+}$ not follow the trend of magnetic moment of lanthanides at 300 K? Explain. Why do the metallic radii decrease from La to Lu except at Eu and Yb? $3+2=5$ How are lanthanides separated by a complex formation method? Give description. Why do $\text{Ce}^{3+}$ and $\text{Yb}^{3+}$ not absorb in the visible region, but show sharp absorption in the ultraviolet region? ★★★ SB27—420 (15 marks)

How to approach this question

The directive word “Differentiate” in part (a) tests your ability to contrast two closely related structural states of a biomolecule. The examiner wants a concise, point-wise comparison of T (tense) and R (relaxed) conformations of haemoglobin, highlighting geometry, oxygen affinity, cooperativity, and physiological relevance. The common mistake is to write a generic description of oxygen binding without explicitly contrasting the two states. In part (b), “Describe” demands a precise account of diborane’s structure and bonding—emphasise the 3-centre-2-electron B–H–B bonds, geometry, and symmetry. For part (c), “Why” questions require mechanistic reasoning: use crystal-field splitting, spin-orbit coupling, and thermodynamic stability arguments to explain deviations in magnetic moments and metallic radii trends. Finally, “How are…separated” needs a stepwise account of ion-exchange or solvent extraction methods with named reagents.

Model answer

(a) T and R conformations of haemoglobin

• Geometry & quaternary structure: T (tense) is the low-affinity, deoxygenated state with a more constrained quaternary arrangement; R (relaxed) is the high-affinity, oxygenated state with a relaxed, open structure stabilised by cooperative binding.
• Oxygen affinity: T has low O₂ affinity; R shows ~150–300-fold higher affinity due to cooperative transitions.
• Physiological role: T favours O₂ release in tissues; R promotes O₂ loading in lungs. Transition is triggered by pH, CO₂, and 2,3-BPG binding.

(b) Structure and bonding in diborane

• Molecular geometry: Diborane (B₂H₆) adopts an ethane-like structure with two bridging H atoms and four terminal B–H bonds.
• Bonding: Each B–H–B bridge is a 3-centre-2-electron bond involving sp³ hybrid orbitals on B and 1s on H, giving banana-shaped bonds and D₂h symmetry.
• Bond lengths: Terminal B–H = 119 pm; bridging B–H = 133 pm, reflecting bond order differences.

(c) Lanthanide anomalies at 300 K

• Magnetic moment deviation: Eu³⁺ (4f⁶) and Sm³⁺ (4f⁵) exhibit lower-than-expected magnetic moments at 300 K due to thermal population of low-lying J states and significant spin-orbit coupling, violating the free-ion approximation.
• Metallic radii trend: From La to Lu, radii decrease (lanthanide contraction) except at Eu and Yb, which have half-filled and fully-filled 4f shells, reducing shielding and increasing effective nuclear charge.
• Separation by complex formation: Lanthanides are separated via ion-exchange chromatography using Dowex-50 resin and citrate/EDTA eluents; stability constants of EDTA complexes increase with atomic number, enabling sequential elution.
• Absorption spectra: Ce³⁺ (4f¹) and Yb³⁺ (4f¹³) show no f–f transitions in the visible region because their ground and excited J levels lie outside the visible window; sharp UV absorptions arise from 4f→5d transitions.

Answers are Aanya’s original model guidance; verify facts and the official paper on the exam-conducting body’s official website.

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