Question

How would you account for the following?

[Ni(CO)₄] possesses tetrahedral geometry, while [Ni(CN)₄]²⁻ is square planar.

Answer 1

The difference in geometry between [Ni(CO)₄] and [Ni(CN)₄]²⁻ is due to the nature of the ligands and the type of hybridisation involved in bond formation. In [Ni(CO)₄], nickel is in the zero oxidation state (Ni⁰) with an electron configuration of 3d⁸ 4s². The CO ligand is a neutral, strong field ligand, but since the complex has no overall charge, there is no need for electron pairing in the 3d orbitals. As a result, the metal uses its outer 4s and 4p orbitals to form sp³ hybrid orbitals, leading to a tetrahedral geometry.

In contrast, in [Ni(CN)₄]²⁻, nickel is in the +2 oxidation state (Ni²⁺), with an electron configuration of 3d⁸. The CN⁻ ligand is a strong field ligand that causes the pairing of 3d electrons. This pairing frees up two 3d orbitals, allowing the metal to use two 3d, one 4s, and one 4p orbital for dsp² hybridisation, resulting in a square planar geometry.

Thus, the tetrahedral geometry of [Ni(CO)₄] arises from sp³ hybridisation without d-electron pairing, while the square planar geometry of [Ni(CN)₄]²⁻ is due to dsp² hybridisation after d-electron pairing induced by strong field ligands.