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

This question integrates molecular orbital theory, electrochemical kinetics, electrolytic conductivity, and thermodynamic standard states, requiring a cohesive application of physical chemistry principles.

Part (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. Biologically, this radical character is central to its function as a signaling molecule in vasodilation and atmospheric chemistry.

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.

Part (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 or $2.32$ mA cm⁻².

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

Part (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 = (σ × 1000)/C = (26.2 × 1000)/0.1 = 262,000 mS cm² mol⁻¹? No, standard unit conversion: Λₘ = (κ)/C where κ is in S cm⁻¹ and C in mol cm⁻³, or simply Λₘ (S cm² mol⁻¹) = (κ (S cm⁻¹) × 1000)/(C (mol L⁻¹)). 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 is very close to the measured Λₘ (26.2), indicating the solution is near ideal behavior or the measured value is effectively the limiting value for this calculation context.

The ionic molar conductivities are: λ_Ca²⁺ = 11.9 mS m² mol⁻¹ λ_NO₃⁻ = 7.14 mS m² mol⁻¹

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.

Part (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). Thermodynamically, the entropy of the system decreases upon compression, but the entropic penalty is larger for the reactants (1.5 moles) than for the product (1 mole). Thus, the relative stability of the product increases, lowering the free energy of formation. Conversely, for HCl, where Δ n = 0, pressure has no effect on the equilibrium position or standard free energy.

Conclusion These calculations demonstrate the interconnectedness of physical chemistry domains. MO theory explains the electronic basis of bond strength and reactivity (Part a), while electrochemical principles quantify mass transport limitations (Part b) and ionic mobility (Part c). Finally, thermodynamic corrections (Part d) highlight how standard states are arbitrary conventions that must be adjusted for Δ n_gas to reflect true physical conditions, linking microscopic molecular properties to macroscopic thermodynamic stability.