(a) Draw the molecular orbital (MO) diagram of NO molecule. The experimental bond dissociation energy of NO is 626 kJ mol⁻¹ while…

(a) MO Theory of NO and NO⁺

The nitric oxide (NO) molecule possesses 11 valence electrons (5 from N, 6 from O). The molecular orbital configuration is (σ₂ₛ)²(σ^*₂ₛ)²(σ_2p_z)²(π_2pₓ=π_2p_y)⁴(π^*_2pₓ=π^*_2p_y)¹. The bond order is calculated as 1/2(N_b - Nₐ) = 1/2(8 - 3) = 2.5. The presence of a single unpaired electron in the degenerate π^* antibonding orbitals makes NO a paramagnetic radical. This unpaired electron is highly reactive, allowing NO to participate in radical chain reactions, such as auto-oxidation to nitrogen dioxide or reaction with superoxide to form peroxynitrite. It also combines with other radicals and with halogens (2NO + Cl₂ → 2NOCl), binds metal ions such as Fe²⁺ in haem as a ligand, and traps carbon radicals; biologically, as a small, diffusible radical, it acts as a signalling molecule (vasodilation via guanylate cyclase) and in atmospheric chemistry it takes part in ozone and smog reactions.

In contrast, the nitrosonium ion (NO⁺) has 10 valence electrons, with the configuration (σ₂ₛ)²(σ^*₂ₛ)²(σ_2p_z)²(π_2pₓ=π_2p_y)⁴. Here, the antibonding π^* orbitals are empty. The bond order is 1/2(8 - 2) = 3. The removal of an electron from the antibonding π^* orbital in NO⁺ strengthens the bond significantly compared to NO. This increase in bond order from 2.5 to 3 directly explains the higher bond dissociation energy (BDE) of NO⁺ (1047 kJ mol⁻¹) versus NO (626 kJ mol⁻¹), as more energy is required to break a triple bond than a 2.5-order bond.

(b) Diffusion-Limiting Current

For a quiescent solution, the diffusion-limited current density j is governed by Fick’s first law of diffusion. The flux J of the organic compound is given by J = -D dC/dx. Assuming a linear concentration gradient across the diffusion layer thickness δ, where the concentration at the electrode surface is zero (due to rapid oxidation) and in the bulk is C₀, the flux is J = (D C₀)/(δ).

The current density j is related to the flux by Faraday’s law: j = n F J, where n is the number of electrons (6) and F is Faraday’s constant (96485 C mol⁻¹). Given: C₀ = 10⁻² mol L⁻¹ = 10⁻⁵ mol cm⁻³ D = 2 × 10⁻⁵ cm² s⁻¹ δ = 0.05 cm

Calculating the flux: J = ((2 × 10⁻⁵ cm² s⁻¹)(10⁻⁵ mol cm⁻³))/(0.05 cm) = 4 × 10⁻⁹ mol cm⁻² s⁻¹

Calculating the current density: j = 6 × 96485 C mol⁻¹ × 4 × 10⁻⁹ mol cm⁻² s⁻¹ j = 2315.64 × 10⁻⁹ A cm⁻² ≈ 2.32 × 10⁻³ A cm⁻², i.e. 2.32 mA cm⁻². For an electrode of area A (cm²), the limiting current is I_L = nFADC₀/δ = 2.32 × 10⁻³ A × A.

Since the electrode area is not specified, the diffusion-limiting current density is 2.32 mA cm⁻².

(c) Conductivity of Calcium Nitrate

The molar conductivity Λₘ is calculated from the specific conductivity σ and molar concentration C. σ = 26.2 mS cm⁻¹ C = 0.1 mol L⁻¹ (decimolar)

Λₘ (S cm² mol⁻¹) = 1000 κ (S cm⁻¹)/C (mol L⁻¹), with κ = σ = 26.2 mS cm⁻¹ = 0.0262 S cm⁻¹. Λₘ = (0.0262 × 1000)/0.1 = 262 S cm² mol⁻¹. Converting to SI units (S m² mol⁻¹): 262 × 10⁻⁴ = 0.0262 S m² mol⁻¹ or 26.2 mS m² mol⁻¹.

Using Kohlrausch’s law for strong electrolytes at infinite dilution, Λₘ⁰ = ν₊ λ₊⁰ + ν₋ λ₋⁰. For Ca(NO₃)₂, ν₊ = 1, ν₋ = 2. Λₘ⁰ = 1(11.9) + 2(7.14) = 11.9 + 14.28 = 26.18 mS m² mol⁻¹. This equals the measured Λₘ (26.2) within rounding, so, on the data given, the ions can be treated as moving independently with their limiting conductivities (for a real 0.1 M solution Λₘ would be somewhat lower because of interionic attraction).

The molar conductivity of the calcium ions in this solution is therefore λ(Ca²⁺) = t₊Λₘ ≈ 11.9 mS m² mol⁻¹, and that of nitrate is λ(NO₃⁻) ≈ 7.14 mS m² mol⁻¹ per mole of NO₃⁻.

Transport numbers are the fraction of total current carried by each ion: t₊ = (ν₊ λ₊)/(Λₘ) = (1 × 11.9)/26.2 ≈ 0.454 t₋ = (ν₋ λ₋)/(Λₘ) = (2 × 7.14)/26.2 = 14.28/26.2 ≈ 0.545 Check: 0.454 + 0.545 = 0.999 ≈ 1.

(d) Redefining Standard State

The standard Gibbs free energy of formation Δ G^°_f depends on the standard pressure P^°. The relationship between Δ G^° at pressure P₁ and P₂ for a reaction is Δ G^°(P₂) = Δ G^°(P₁) + Δ n_gas RT ln(P₂/P₁). Here, P₁ = 1 atm, P₂ = 2 atm, so the correction term is Δ n_gas RT ln(2).

(i) HCl (g): Formation reaction is 1/2H₂(g) + 1/2Cl₂(g) → HCl(g). Δ n_gas = 1 - (0.5 + 0.5) = 0. Correction = 0 × RT ln(2) = 0. Δ G^°_f(new) = -95.3 kJ mol⁻¹. The value remains unchanged because the number of moles of gas is conserved. The entropy change due to pressure expansion/compression cancels out for reactants and products.

(ii) N₂O (g): Formation reaction is N₂(g) + 1/2O₂(g) → N_2O(g). Δ n_gas = 1 - (1 + 0.5) = -0.5. Correction = -0.5 × (8.314 × 10⁻³ kJ K⁻¹ mol⁻¹) × 298 K × ln(2). RT ln(2) ≈ 1.717 kJ mol⁻¹. Correction = -0.5 × 1.717 = -0.8585 kJ mol⁻¹. Δ G^°_f(new) = 103.7 - 0.86 = 102.84 kJ mol⁻¹.

The result for N₂O becomes less positive (more stable) at higher pressure. This is because the reaction involves a decrease in the number of gas moles (Δ n < 0). According to Le Chatelier’s principle, increasing pressure favors the side with fewer moles (the product). In entropy terms, raising the standard pressure from 1 to 2 atm lowers the standard molar entropy of every gas by R ln 2 (5.76 J K⁻¹ mol⁻¹). The reactants contain 1.5 mol of gas and the product only 1 mol, so the reactants lose more entropy than the product: ΔS°f becomes less negative by 0.5R ln 2, and −TΔS°f makes ΔG°f smaller by 0.86 kJ mol⁻¹ (ΔH° is unchanged for ideal gases). For HCl, reactants and product both contain 1 mol of gas, so the entropy changes cancel, ΔS°f and ΔG°f are unchanged.