Question

Choose the correct statement regarding $\left[Ni\right(CO{)}_4]$:

• A. $\left[Ni\right(CO{)}_4]$ is diamagnetic and tetrahedral
• B. $\left[Ni\right(CO{)}_4]$ is paramagnetic and tetrahedral
• C. $\left[Ni\right(CO{)}_4]$ is diamagnetic and square planar
• D. $\left[Ni\right(CO{)}_4]$ is paramagnetic and square planar

Answer 1

The correct option is A $\left[Ni\right(CO{)}_4]$ is diamagnetic and tetrahedral

$\left[Ni\right(CO{)}_4]$
Oxidation state of $Ni=0$


On approach of $4CO:$
As $CO$ is strong field ligand, it will cause pairing of electrons.


Hence, hybridisation of $Ni$ in $\left[Ni\right(CO{)}_4]$ is $s{p}^3$ and geometry will be tetrahedral. As all electrons are paired in $\left[Ni\right(CO{)}_4]$ it is diamagnetic in nature.