Question

Given below are two statements:
Statement I: $\left[Ni\right(CN{)}_4{]}^{2–}$ is square planar and diamagnetic complex, with $ds{p}^2$ hybridization for $Ni$ but $\left[Ni\right(CO{)}_4]$ is tetrahedral, paramagnetic and with $s{p}^3$-hybridization for $Ni$.
Statement II: $[NiC{l}_4{]}^{2–}$ and $\left[Ni\right(CO{)}_4]$ both have same d-electron configuration, have same geometry and are paramagnetic.

In light of the above statements, choose the correct answer from the options given below :

Answer 1

$\left[Ni\right(CN{)}_4{]}^{2-}:d^8$ configuration, $SFL$, $sq.$ planar splitting $\left(ds{p}^2\right),$ diamagnetic.
$\left[Ni\right(CO{)}_4]:d^{10}$ config (after excitation), $SFL$, tetrahedral splitting $\left(s{p}^3\right),$ diamagnetic.
$[NiC{l}_4{]}^{2-}:d^8$ config, $WFL$, tetrahedral splitting $\left(s{p}^3\right),$ paramagnetic (2 unpaired $e^{-}).$
$\left[Ni\right(CN{)}_4{]}^{2–}$ is square planar and diamagnetic with
$\mu =0$
its hybridisation is $ds{p}^2$
$\left[Ni\right(CO{)}_4]$ is tetrahedral but diamagnetic