(a) Cite one example of an optically active tetracoordinated complex compound where the metal ion and donor atoms lie on a plane. Justify your answer. 10 marks

(b) Consider aqueous solutions of LaCl₃ (Lanthanum trichloride) and LuCl₃ (Lutetium trich

Question

(a) Cite one example of an optically active tetracoordinated complex compound where the metal ion and donor atoms lie on a plane. Justify your answer. 10 marks

(b) Consider aqueous solutions of LaCl₃ (Lanthanum trichloride) and LuCl₃ (Lutetium trichloride). Which solution shows lower pH? Explain. 10 marks

(c) The reaction cis-2-butene ⇄ trans-2-butene is first order in both the direction. At 25°C, the equilibrium constant is 0.406 and the forward reaction rate constant is 4.21×10⁻⁴ sec⁻¹. Starting with a sample of pure cis isomer with [cis]₀ = 0.115 mol dm⁻³, how long it will take to form half of equilibrium amount of the trans isomer from cis isomer? 10 marks

(d)

At 0°C and 1 atm pressure, the volume of nitrogen gas required to cover a sample of an adsorbent is found to be 130 cm³ g⁻¹. Calculate the surface area per gram of adsorbent. Given that area occupied by a nitrogen molecule is 0.162 (nm)².

[Nₐ = 6.022×10²³ mol⁻¹] 10 marks

(e) Compare and comment on the magnetic properties of the following complexes:

(i) [Cu (OAc)₂]₂

(ii) [Cu (CN)₄]³⁻ 10 marks

UPSC Answer Check · Accepted Answer

(a) A suitable example is [Pt(en)(ox)], ethylenediamineoxalato(2−)platinum(II). Pt(II) is a d8 ion and is four-coordinate in a square-planar arrangement. The Pt atom and the four donor atoms, two nitrogen atoms of ethylenediamine and two oxygen atoms of oxalate, lie in one plane. The complex is nevertheless optically active because the two five-membered chelate rings cannot remain in that plane. They are puckered and twisted above and below the donor-atom plane. Since the two chelate ligands are different, the mirror image of one twisted arrangement cannot be superimposed on the original by any proper rotation; the molecule lacks a plane of symmetry, centre of inversion, or improper rotation axis. The two non-superimposable mirror-image forms, conventionally labelled Δ and Λ, are therefore enantiomers. The important point is that optical activity here does not require a tetrahedral array of donor atoms; it arises from the out-of-plane conformation of the chelate rings while the metal and donor atoms remain coplanar.

(b) In aqueous solution both salts give trivalent lanthanide aqua ions, [La(H2O)6]3+ and [Lu(H2O)6]3+. The relevant difference is ionic size. Across the lanthanide series, the 4f electrons shield the nuclear charge poorly, so the ionic radius contracts from La3+ to Lu3+. For six-coordinate ions, La3+ has a radius of about 1.032 Å and Lu3+ about 0.861 Å. Because Lu3+ is smaller while carrying the same +3 charge, its charge density is higher. In simple terms, the acidity of the aqua ion increases with ionic potential, charge divided by radius. Lu3+ therefore polarises the O–H bonds of the coordinated water molecules more strongly, making the protons more acidic. Hydrolysis occurs to a greater extent: [Lu(H2O)6]3+ + H2O ⇌ [Lu(H2O)5OH]2+ + H3O+. The larger La3+ ion polarises water less, so [La(H2O)6]3+ releases fewer hydronium ions. Consequently, the LuCl3 solution has a higher [H3O+] and a lower pH than the LaCl3 solution.

(c) Let x be the concentration of trans-2-butene formed from pure cis. The reversible first-order rate law is dx/dt = k_f([cis]0 − x) − kᵣ x. At equilibrium, k_f[cis]eq = kᵣ[trans]eq. The numerical value 0.406 is used as [cis]eq/[trans]eq, which is consistent with trans-2-butene being the more stable isomer; hence [trans]eq/[cis]eq = 1/0.406 = 2.46. Therefore kᵣ = k_f × 0.406 = 4.21 × 10⁻⁴ × 0.406 = 1.71 × 10⁻⁴ s⁻¹. The sum of the forward and reverse rate constants is k_f + kᵣ = 5.92 × 10⁻⁴ s⁻¹.

The integrated form for approach to equilibrium is x = x_eq(1 − e^−(k_f+kᵣ)t), where x_eq = [trans]eq. Since [cis]eq + [trans]eq = 0.115 mol dm⁻³ and [cis]eq/[trans]eq = 0.406, [trans]eq = 0.115/(1 + 0.406) = 0.0818 mol dm⁻³. Forming half of the equilibrium amount means x = x_eq/2. Substituting gives 1/2 = 1 − e^−(k_f+kᵣ)t, so e^−(k_f+kᵣ)t = 1/2 and t = ln 2/(k_f + kᵣ). Thus t = 0.693/(5.92 × 10⁻⁴) = 1.17 × 10³ s, or about 19.5 min. Because the required point is one-half of x_eq, the initial concentration cancels; only the sum of the first-order constants controls the time.

(d) The calculation assumes the reported nitrogen volume corresponds to a complete monolayer and that each molecule occupies the given cross-sectional area. Convert 130 cm³ g⁻¹ to 0.130 L g⁻¹. At 0°C and 1 atm, n = PV/RT = (1 atm × 0.130 L g⁻¹)/(0.082057 L atm mol⁻¹ K⁻¹ × 273.15 K) = 5.80 × 10⁻³ mol g⁻¹. The number of nitrogen molecules per gram is n N_A = 5.80 × 10⁻³ × 6.022 × 10²³ = 3.49 × 10²¹ molecules g⁻¹. Each molecule occupies 0.162 nm², which is 0.162 × 10⁻¹⁸ m² = 1.62 × 10⁻¹⁹ m². Multiplying the number of molecules by the area per molecule gives the surface area: 3.49 × 10²¹ × 1.62 × 10⁻¹⁹ = 5.66 × 10² m² g⁻¹. Hence the adsorbent has a surface area of about 566 m² g⁻¹.

(e) The two complexes differ in oxidation state, electron configuration, and magnetic coupling. In [Cu(OAc)2]2, copper is Cu(II), a d9 ion. Each Cu(II) centre has one unpaired electron, and the spin-only moment for an isolated Cu(II) ion would be about 1.73 BM. The acetate complex is a paddlewheel dimer in which two Cu(II) centres are held close together by bridging acetate groups. The unpaired d electrons on the two copper centres interact through the bridging acetate ligands by superexchange. This interaction is antiferromagnetic: the two spins prefer to be paired in a singlet ground state. Because the singlet state is diamagnetic, the observed magnetic moment is reduced relative to two independent Cu(II) ions; at room temperature the effective moment is about 1.4 BM per Cu, reflecting partial thermal population of the excited triplet state.

In [Cu(CN)4]3−, the charge balance gives Cu(I), which is d10. All ten d electrons are paired, and the complex is four-coordinate, commonly tetrahedral, with a closed-shell configuration. There is no unpaired electron and no antiferromagnetic coupling to consider, so the complex is diamagnetic with μ = 0. Thus the acetate dimer is paramagnetic but magnetically weakened by antiferromagnetic Cu–Cu coupling, whereas the cyanide complex is diamagnetic because of its d10 closed-shell Cu(I) centre.

(a) Cite one example of an optically active tetracoordinated complex compound where the metal ion and donor atoms lie on a plane…

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