Question

Based on Valence bond theory, explain why the geometry of [Ni(CO)₄] is tetrahedral and [Ni(CN)₄]²- is square planar.

Hint:

To remember: - sp³ Hybridization = Tetrahedral geometry. - dsp² Hybridization = Square Planar geometry. Check the oxidation state and the ligand strength first!

Answer 1

1. [Ni(CO)₄]
Nickel is in zero oxidation state. Electronic configuration of Ni: [Ar] 3d⁸ 4s². CO is a strong field ligand, causing the 4s electrons to shift into the 3d orbital to pair up. • Resulting configuration: [Ar] 3d¹⁰. • The 4s and three 4p orbitals hybridize to form sp³ hybrid orbitals. • Geometry: Tetrahedral. 2. [Ni(CN)₄]²-
Nickel is in +2 oxidation state. Electronic configuration of Ni²⁺: [Ar] 3d⁸ 4s⁰. CN^- is a strong field ligand, forcing the two unpaired electrons in 3d to pair up, leaving one 3d orbital vacant. 2- showing dsp2 hybridization] • One 3d, one 4s, and two 4p orbitals hybridize to form dsp² hybrid orbitals. • Geometry: Square Planar.