JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year CHEMISTRY Previous Question Paper (June - 2022 - R18)

The nitrogen molecule (N₂) consists of two nitrogen atoms, each with an electronic configuration of 1s² 2s² 2p³. The total number of electrons in N₂ is 14. According to molecular orbital theory, the energy level diagram for N₂ follows the order: σ1s < σ*1s < σ2s < σ*2s < π2p_x = π2p_y < σ2p_z < π*2p_x = π*2p_y < σ*2p_z.

The molecular orbital configuration is: (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (π2p_x)² (π2p_y)² (σ2p_z)².

The bond order is calculated as:

The bond order of 3 indicates a strong triple bond between the nitrogen atoms, making the molecule diamagnetic as all electrons are paired.

In tetrahedral geometry, a central metal ion is surrounded by four ligands positioned at the corners of a tetrahedron. The crystal field splitting energy, denoted as Δ_t, represents the energy gap between the split d-orbitals. Unlike octahedral splitting, the d-orbitals split into two sets: a lower energy doublet (e) consisting of d_z² and d_x²-y² orbitals, and a higher energy triplet (t₂) consisting of d_xy, d_yz, and d_zx orbitals.

The energy separation is defined as:

The e orbitals are stabilized by 0.6Δ_t below the barycenter, while the t₂ orbitals are destabilized by 0.4Δ_t above the barycenter. Because the number of ligands is smaller (4 vs 6) and the approach of ligands does not directly point toward the d-orbitals, Δ_t is always smaller than the pairing energy, resulting in high-spin complexes for tetrahedral geometry.