1.(a) Compare average and most probable values of position of an electron in the ground state of hydrogen atom. Explain with the…

(a) Average and most probable position of the 1s electron

For the hydrogen 1s electron the most probable distance from the nucleus is r = a₀ = 52.9 pm, while the average distance is ⟨r⟩ = 3a₀/2 = 79.4 pm. The two differ because the radial distribution is not symmetric about its peak.

Wavefunction and radial distribution. ψ₁ₛ = (1/√(πa₀³)) e^(−r/a₀), so the probability density |ψ|² = e^(−2r/a₀)/(πa₀³) is greatest at the nucleus itself. The chance of finding the electron in a thin spherical shell of radius r and thickness dr is P(r) dr = 4πr²|ψ|² dr:

P(r) = (4r²/a₀³) e^(−2r/a₀)

• Most probable value: dP/dr = (8r/a₀³) e^(−2r/a₀) (1 − r/a₀) = 0 gives r_mp = a₀. The factor r² (shell volume growing with r) competes with the falling exponential, and the product peaks at a₀.
• Average value: ⟨r⟩ = ∫₀^∞ r P(r) dr = (4/a₀³) ∫₀^∞ r³ e^(−2r/a₀) dr = (4/a₀³) × 3!(a₀/2)⁴ = 3a₀/2.

Why they differ. P(r) rises steeply from zero at r = 0 to its peak at a₀. It then falls slowly, with a long tail extending to infinity. Only about 32% of the probability lies inside a₀, and the median is about 1.34 a₀. The mean weights each distance by r, so the far tail pulls ⟨r⟩ beyond the peak, to 1.5 a₀. For a skewed distribution: mode (a₀) < median < mean (1.5 a₀).

(b) Nonbonding interactions in AB

Take the internuclear axis as z (the usual convention). Two atomic orbitals interact, giving a bonding and an antibonding MO, only if they have the same symmetry about the bond axis. Their overlap integral S = ∫φ_A φ_B dτ is then non-zero. If S = 0 by symmetry, the orbitals are orthogonal and stay nonbonding.

• A(2s) with B(2p_y): σ-type with π-type. The + and − lobes of 2p_y overlap the 2s orbital equally, so S = 0 and the pair is nonbonding.
• A(2p_x) with B(2p_y): π(xz) with π(yz). The nodal planes are perpendicular, so S = 0 and the pair is nonbonding.
• A(2p_x) with B(2p_z): π-type with σ-type. The 2p_z lobe meets the + and − lobes of 2p_x equally, so S = 0 and the pair is nonbonding.
• A(2s) with B(2p_z): both σ-type. Head-on overlap gives S ≠ 0, so they form σ bonding and σ∗ MOs.

The nonbonding interactions are therefore A(2s)–B(2p_y), A(2p_x)–B(2p_y) and A(2p_x)–B(2p_z). Only A(2s)–B(2p_z) combine. In each nonbonding case, reflection through the plane containing the bond axis changes the sign of one orbital but not of the other. The in-phase and out-of-phase regions cancel exactly, so these AOs remain nonbonding. (If the axis were chosen as x instead, A's 2s and 2p_x would be σ-type and B's 2p_y and 2p_z π-type, so all four pairs would be nonbonding.)

(c) F-centres: colour and density

(i) Why the colour is characteristic of the crystal. An F-centre (Farbenzentrum) is an anion vacancy occupied by an unpaired electron. It forms in several ways:

• heating an alkali halide in the vapour of an alkali metal (metal-excess defect);
• irradiation with X-rays, γ-rays or neutrons;
• electrolysis (additive colouring).

Whatever the route, the end product is the same entity: an electron trapped in a halide-ion site and held by the surrounding cations. The trapped electron behaves roughly like a particle in a box whose size is fixed by the lattice. It has a ground level and excited levels whose spacing depends only on the lattice spacing, the size of the vacancy, and the charges and polarisability of the surrounding ions. These are properties of the host crystal, not of the method used. Absorption of the matching visible photon (the F-band) promotes the electron, and the crystal shows the complementary colour. The energy falls as the lattice constant increases (Mollwo–Ivey relation, roughly E ∝ 1/a²). Hence NaCl is always yellow, KCl violet and LiCl pink. NaCl heated in potassium vapour is still yellow, because the colour belongs to the NaCl lattice, not to the metal that supplied the electron.

(ii) Effect on density. At each F-centre a site that should hold a halide ion (Cl⁻, 35.5 u) holds only an electron of negligible mass. The lattice volume is essentially unchanged: when excess metal is added, new cation sites extend the lattice with vacant anion sites. The mass per unit volume therefore falls, and the density of the crystal decreases (slightly, since F-centre concentrations are small).

(d) ΔG for O₃ formation in smog

Reaction (per mole of O₃): 3/2 O₂(g) → O₃(g). Here ΔG° = ΔG°f(O₃) − 0 = 163 kJ mol⁻¹.

Reaction quotient: Q = p(O₃)/p(O₂)^(3/2) = (5 × 10⁻⁷)/(0.21)^(3/2) = (5 × 10⁻⁷)/0.0962 = 5.20 × 10⁻⁶

Non-standard free energy: ΔG = ΔG° + RT ln Q

• ln Q = −12.17
• RT ln Q = 8.314 J mol⁻¹ K⁻¹ × 298 K × (−12.17) = −30,150 J mol⁻¹ = −30.1 kJ mol⁻¹
• ΔG = 163 − 30.1 = +132.9 kJ mol⁻¹ (≈ 133 kJ per mol O₃)

(Written as 3O₂ → 2O₃, ΔG° = 326 kJ, Q = (5 × 10⁻⁷)²/(0.21)³ = 2.70 × 10⁻¹¹, and ΔG = +265.7 kJ, exactly twice the value above.)

Interpretation: ΔG is still large and positive, so O₃ cannot form from O₂ spontaneously even at these low smog concentrations. Smog ozone is produced by an external energy input: photolysis of NO₂ by sunlight (NO₂ → NO + O, then O + O₂ → O₃). F is not needed for this part.

(e) ΔG° and K for the Zn–Cu cell

Data interpretation. The couple Zn/Zn²⁺ is written in the oxidation direction, so 0.339 V is taken as the oxidation potential at the anode. Cu²⁺/Cu is written as a reduction, so 0.762 V is its reduction potential at the cathode.

• Anode (oxidation): Zn → Zn²⁺ + 2e⁻.
• Cathode (reduction): Cu²⁺ + 2e⁻ → Cu.
• n = 2.

E°cell = E°ox(anode) + E°red(cathode) = 0.339 + 0.762 = 1.101 V

This agrees with the known e.m.f. of the Daniell cell, about 1.10 V.

Standard free-energy change: ΔG° = −nFE° = −2 × 96485 C mol⁻¹ × 1.101 V = −212,460 J mol⁻¹ ≈ −212.5 kJ mol⁻¹

Equilibrium constant: ln K = nFE°/RT = (2 × 96485 × 1.101)/(8.314 × 298) = 85.75, so log K = 37.24 K ≈ 1.7 × 10³⁷

The large negative ΔG° and enormous K show that the displacement of Cu²⁺ by Zn goes essentially to completion.

(If both figures were instead read as standard reduction potentials, E° = 0.762 − 0.339 = 0.423 V. That would give ΔG° = −81.6 kJ mol⁻¹ and K ≈ 2.0 × 10¹⁴; the method is the same.)