Question

: carbonyl(CO) is strong field ligand so [Ni(CO)4] should have square planar shape but by the experiment [Ni(CO)4] is tetrahedral.Explain this statement.

Answer 1

CO forms a coordination bond that has both sigma and pi properties. A non-bonding orbital on the CO will form the primary bond, and an anti-bonding orbital forms a bond as well. Because of this multiple coordination bond, the carbonyl-metal bond is very strong field ligand. The metal carbonyl complex Ni(CO)₄ has Ni present in the Zero Oxidation state. It' s electronic configuration is [Ar] 3d8 4s2. Considering it as a central atom, [CO] is the strong ligand. It Can pair the unpaired electrons. The 4s electrons goes to 3d unpaired electrons so, as to completely fill 3d orbital. Now, 1- 4s and 3- 4p are available. so, these will together form 4 degenerate orbitals therefore, we have sp3 hybridization leading to the tetrahedral shape.

The square planar would have been the case if the Ni had +2 oxidation state. so, Ni²⁺ would be 3d⁸ 4s⁰. As in the case of Ni(CN₄)^2-. Cyanide will cause pairing of the unpaired electrons. So, there the hybridization would be sp2d hence, Square planar