Explain the formation of [Co(NH3)6]3+ ion on the basis of Valence bond theory. Convert the following : (a) acetic acid to ethyl acetate (b) acetic acid to ethyl alcohol.

Answer:

[Co(NH₃)₆]³⁺ by VBT:

1. Oxidation state of cobalt = +3. Co (Z = 27): [Ar] 3d⁷ 4s²; Co³⁺: [Ar] 3d⁶ 4s⁰ 4p⁰, with four unpaired electrons in the free ion.
2. NH₃ is a strong-field ligand. On its approach the 3d electrons pair up: the six electrons occupy three 3d orbitals (all paired), leaving two 3d orbitals empty.
3. Two 3d, one 4s and three 4p orbitals hybridise to give six equivalent d²sp³ hybrid orbitals.
4. Six NH₃ molecules donate a lone pair each into these six empty hybrid orbitals, forming six Co←NH₃ coordinate bonds.

[Orbital diagram: Co³⁺ 3d ↿⇂ ↿⇂ ↿⇂ | then two 3d + 4s + three 4p boxes filled by six electron pairs from 6NH₃]

Result: octahedral geometry, inner orbital (low spin) complex; no unpaired electrons, so it is diamagnetic.

(a) Acetic acid to ethyl acetate (Fischer esterification):

CH₃COOH + C₂H₅OH ⇌(conc. H₂SO₄, Δ) CH₃COOC₂H₅ + H₂O

(b) Acetic acid to ethyl alcohol (reduction with lithium aluminium hydride):

CH₃COOH + 4[H] —(i. LiAlH₄/dry ether, ii. H₃O⁺)→ CH₃CH₂OH + H₂O