Amongst `Ni(CO)_(4),[Ni(CN)_(4)]^(2-) and NiCl_(4)^(2-)`:

Options

A. `Ni(CO)_(4)` and `NiCl_(4)^(2-)` are diamagnetic and `[Ni(CN_(4))]^(2-)` is paramagnetic
B. `NiCl_(4)^(2-)` and `[Ni(CN_(4))]^(2-)` are diamagnetic and `Ni(CO)_(4)` is paramagnetic
C. `Ni(CO)_(4)` and `[Ni(CN_(4))]^(2-)` are diamagnetic and `NiCl_(4)^(2-)` is paramagnetic
D. `Ni(CO)_(4)` is diamagnetic and `NiCl_(4)^(2-)` and `[Ni(CN_(4))]^(2-)` are paramagnetic

Answer

C

Explanation

The electronic configuration in various species are :
`[NiCl_(4)]^(2-)`, the oxidation state of Ni is +2, i.e., `Ni^(2+)`.
The electronic configuration of `Ni^(2+)` is `1s^(2) 2s^(2)2p^(6)3s^(2) 3p^(6)3d^(8)`

(`xx xx` indicates Ione pair from Cl atoms)
Because of presence of 2 unpaired electrons `[NiCl_(4)]^(2-)` is paramagnetic.
In `[Ni(CN_(4))]^(2-) ` the O.S. of Ni is +2, i.e., `Ni^(2+)(d^(8))`

In presence of `CN^(-)` ligand pairing of electrons take place and the hybridisation is `dsp^(2).`

In `[NiCN_(4))]^(2-)` there is no unpaired dlectrons so it is diamagnetic. In `[Ni(CO)_(4)]`, Ni is in zero state. Its configuration is `1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(8)4s^(2). `

In `[Ni(CO)_(4)]`, electrons get shifted from 4s to 3d and then `sp^(3)` hybridisation occurs.

Since there are no unpaired electrons in `[Ni(CO)_(4)]`, so it is diamagnetic.
So (c ) is the correct answer.