The hybridization states of [Ni(CO)(4)],...

The hybridization states of `[Ni(CO)_(4)], [Ni(CN)_(4)]^(2-)` and `[NiCl_(4)]^(2-)` species are respectively

`sp^(3), sp^(3), dsp^(2)`

`dsp^(2), sp^(3), sp^(3)`

`sp^(3), dsp^(2), dsp^(2)`

`sp^(3), dsp^(2), sp^(3)`

Text Solution

Verified by Experts

The correct Answer is:

`[Ni(CO)_(4)]` is tetrahedral : `sp^(3)` hybridisation
`[Ni(CN)_(4)]` is square planar : `dsp^(2)` hybridisation
`[NiCl_(4)]^(2-)` is tetrahedral : `sp^(3)` hybridisation

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

`Ni(CO)_(4) and NiCl_(4)^(2-)` are diamagnetic and `[Ni(CN)_(4)]^(2-)` is paramagnetic.

`NiCl_(4)^(2-) and [Ni(CN)_(4)]^(2-)` are diamagnetic and `Ni(CO)_(4)` is paramagnetic.

`Ni(CO)_(4) and [Ni(CN)_(4)]^(2-)` are diamagnetic and `NiCl_(4)^(2-)` is paramagnetic.

`Ni(CO)_(4)` is diamagnetic and `NiCl_(4)^(2-)` and `[Ni(CN)_(4)]^(2-)` are diamagnetic.

Amongst [Ni(CO)_4],[Ni(CN)_4]^2- and [NiCl_4]^2-

`[Ni(CO)_4] and [NiCl_4]^2-` are diamagnetic but `[Ni(CN)_4]^2-` is paramagnetic

`[Ni(Cl)_4] and [NiCN_4]^2-` are diamagnetic but `[Ni(CO)_4]` is paramagnetic

`[Ni(CO)_4] and [NiCN_4]^2-` are diamagnetic but `[Ni(Cl)_4]` is paramagnetic

`[Ni(CO)_4] is diamagnetic but [NiCl_4]^2-` and `[Ni(CN)_4]^2-` are paramagnetic

The oxidation state of nickel in K_(4)[Ni(CN)_(4)] is:

`+4`

`+2`

`-2`