Question

The geometry of $Ni{{\left( CO \right)}_{4}}$ is:
A. Tetrahedral
B. Octahedral
C. Square planar
D. Pyramidal

Answer 1

Hint: The geometry of a complex is defined by the nature of the complex it has which certainly has an effect on the hybridization of the complex.
$CO$ is a strong field ligand in the complex a strong field ligand induces back electron pairing from the valence shell of nickel.
Change in electronic configuration of nickel takes place when $CO$ ligand approaches it.

Complete step by step answer:
The stable state electronic configuration of Nickel is
Argon$\left[ 3{{d}^{8}}4{{s}^{2}}4{{p}^{0}} \right]$.
The valency of Nickel is zero because in the complex, $CO$ is a neutral ligand.
Hence, no electrons will be removed from the above stable state configuration of nickel.
$CO$ is also a strong field ligand and when it approaches nickel, it induces back pairing of electrons from $4s$ to the $3d$ orbital emptying $4s$ orbital.
The new configuration of nickel becomes
argon $\left[ 3{{d}^{10}}4{{s}^{0}}4{{p}^{0}} \right]$
And hence the $3d$ orbital gets filled to 10 electrons making $4s$ orbital and three $4p$orbitals ready for hybridization which decides the geometry and other important properties of the coordinate complex.
All the empty orbitals one $4s$ and three $4p$ mix and hybridize to give $s{{p}^{3}}$ hybridization of the complex.

And as we know this type of hybridization corresponds to the tetrahedral structure of the complex. This is a low spin complex.

Hence the correct option for this question is A.

Note: A square planar complex is generally found to form from a strong ligand and a tetrahedral complex is generally found to be formed from weak field ligand.
The procedure used to solve the above question is done in accordance with valence bond theory.