Among `|Ni(CO)_(4)|, |Ni(CN)_(4)|^(2-), |NiCl_(4)|^(2-)` species, the hybridisation state at `Ni` atom are respectively
[ Atomic number of `Ni` = 28 ]

Options

A. `sp^(3),dsp^(2),dsp^(2)`
B. `sp^(3),dsp^(2),sp^(3)`
C. `sp^(3),sp^(3),dsp^(2)`
D. `dsp^(2),sp^(3),sp^(3)`

Answer

B

Explanation

In `Ni(CO)_(4)`, nickel is `sp^(3)` hybridised because in it oxidation state of Ni is zero. So, configuration of
`""_(28)Ni=1s^(2),2s^(2)2p^(6),3s^(2)3p^(6)3d^(8),4s^(2)`

(CO is a strong held ligand, hence pairing of electrons will occur)
In `[Ni(CN)_(4)]^(2-)`, nickel is present as `Ni^(2+)`, so its configuration `=1s^(2),2s^(2)2p^(6),3s^(2)3p^(6)3d^(8)`

`CN^(-)` is strong field ligand, hence it makes `Ni^(2+)` electrons to be paired up.
In `[NiCl_(4)]^(2-)`, nickel is present as `Ni^(2+)`, so its configuration `=1s^(2),2s^(2)2p^(6),3s^(2)3p^(6),3d^(8)`

`Cl^(-)` is weak field ligand, hence in `Ni^(2+)` electrons are not paired.