Among `[Ni(CO)_(4)],[Ni(CN)_(4)]^(2-),[NiCl_(2)]^(2-)` species, the hybridisation `""^(**)states` of the Ni atom are, respectively (At. No. of Ni = 28)

Options

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

Answer

c

Explanation

`[Ni(CO)_(4)] OS Ni is 0`
`Ni = (Z =28) implies 3d^(8) 4s^(2)`
`CO` is a strong field ligand it causes pairing
`sp^(3)` Hybridisation(tetrahedral) There are no unpaired electrons, so the complex is diamagnetic Spin magnetic moment = zero
(ii) `[Ni(CN)_(4)]^(2-),Ni^(2+) =3d^(8)`
Cyano is strong field ligand, it causes pairing
`dsp^(2)` hybridisation (squareplanar) There are no unpaired electrons so the complex diamgnetic Spin magnetic moment = zero
(iii) `[NiCI_(4)]^(2-),Ni^(2+) = 3d^(8)`
Chlorido is a weak field no pairing `sp^(3)` hybridisation (tetrahedral) There are two unpaired electrons, so the complex is paramagnetic Spin magnetic moment
`mu =sqrt( n (n +2)) BM = sqrt(2(2 + 2)) BM =sqrt8 BM`