Mention the number of unpaired electrons and geometry of following complexes :(i) [Ni (Cl)4]2–(ii) [Ni (CN)4]2–Convert the following :(a) Ethanenitrile into ethanal.(b) Cyclohexane into adipic acid.

Answer: [NiCl₄]²⁻: 2 unpaired electrons, tetrahedral; [Ni(CN)₄]²⁻: 0 unpaired electrons, square planar.

In both complexes Ni is in the +2 state: Ni²⁺ = [Ar] 3d⁸.

(i) [NiCl₄]²⁻: Cl⁻ is a weak field ligand, so the 3d electrons do not pair up (3d: ↑↓ ↑↓ ↑↓ ↑ ↑). The four Cl⁻ ligands use the empty 4s and 4p orbitals: sp³ hybridisation, tetrahedral, 2 unpaired electrons (paramagnetic).

(ii) [Ni(CN)₄]²⁻: CN⁻ is a strong field ligand and pairs the two unpaired 3d electrons, leaving one 3d orbital empty (3d: ↑↓ ↑↓ ↑↓ ↑↓ _). One 3d, one 4s and two 4p orbitals hybridise: dsp² hybridisation, square planar, no unpaired electrons (diamagnetic).

(a) Ethanenitrile into ethanal (Stephen reaction):

CH₃–C≡N + SnCl₂ + HCl → (ether) CH₃–CH=NH·HCl (aldimine hydrochloride) → (H₂O, boil) CH₃CHO + NH₄Cl

(b) Cyclohexane into adipic acid: cyclohexane is oxidised by air (cobalt catalyst, heat) to a mixture of cyclohexanol and cyclohexanone, which is then oxidised by concentrated nitric acid with ring opening to adipic acid (hexanedioic acid):

C₆H₁₂ → (O₂/air, Co catalyst, Δ) C₆H₁₁OH + C₆H₁₀O → (conc. HNO₃, Δ) HOOC–(CH₂)₄–COOH

(If one starts from cyclohexene, hot alkaline KMnO₄ followed by acidification gives adipic acid directly.)