Question

Among $Ni(CO{)}_4,[Ni\left(CN{)}_4{]}^{2-}and\right[NiC{l}_4^{2-}]$
[IIT-JEE, 1991]

• A. $Ni(CO{)}_4$ and $NiC{l}_4^{2-}$ are diamagnetic and $\left[Ni\right(CN{)}_4{]}^{2-}$ is paramagnetic.
• B. $NiC{l}_4^{2-}$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic and $Ni(CO{)}_4$ is paramagnetic.
• C. $Ni(CO{)}_4$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic and $NiC{l}_4^{2-}$ is paramagnetic.
• D. $Ni(CO{)}_4$ is diamagnetic and $NiCl{4}^{2-}$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are paramagnetic.

Answer 1

The correct option is C $Ni(CO{)}_4$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic and $NiC{l}_4^{2-}$ is paramagnetic.

The concept you need to apply is simple. Figure out which ligand is strong and weak, and hence the electronic configuration of the metal in question.
Let's do this:
(i) $\left[Ni\right(CO{)}_4]$ OS Ni is 0
Ni = (Z=28) $\Rightarrow 3d^{8}4s^2$

CO is a strong field ligand, it causes pairing.

$s{p}^3$-Hybridisation (tetrahedral)
There are no unpaired electrons, so the complex is diamagnetic.
Spin magnetic moment = zero.

(ii) $\left[Ni\right(C{N}_4{)}^{2-},N{i}^{2+}=3d^8$

Cyano is a strong field ligand, it causes pairing.

$ds{p}^2$-hybridisation (square planar)
There are no unpaired electrons so, the complex is diamagnetic.
Spin magnetic moment = zero.

(iii) $[NiC{l}_4{]}^{2-},N{i}^{2+}=3d^8$

Chlorido is a weak field ligand, no pairing
$s{p}^3$-hybridisation (tetrahedral)
There are two unpaired electrons, so the complex is paramagnetic.