Identify, from the following the dimagnetic, tetrahedral complex

Options

A. `[Ni(Cl)_(4)]^(2-)`
B. `[Co(C_(2)O_(4))_(3)]^(3-)`
C. `[Ni(CN)_(4)]^(2-)`
D. `[Ni(CO)_(4)]`

Answer

D

Explanation

Ni in `[Ni(Cl_(4))]^(2-) ` exist as `Ni^(2+)` ion
`Ni^(2) `(configuration)

`:.` cl is a weak field ligand (high spin). It will not go for pairing. `,Ni(Cl)_(4)]^(2-) = `

Tatrehadral with two unpaired electrons (i.e., paramagnetic) with `sp^(3)` hydridisation.
(2) In `[CO(C_(2)O_(4))_(3)]^(3-) , (C_(2)O_(4))_(3)^(2-)` is a badentate ligand thus, give octahedral structure
(3) In `[Ni(CN)_(4)]^(2-)` , Ni exist as `NI^(2+)` ion
`NI^(2+)` (configuration) =

`:. CN^(-)` is a strong field ligand (low spin) the unpaired electrons of 3rd orbital go for pairing to give `[Ni(CN)(4)]^(2-)` as follows :
`[Ni(Cl)_(4)]^(2-) = `

Hence, the structure of `[Ni(CN)_(4)]^(2-)` is square planer with `dsp^(2)` hybridisation having diamagnetic nature.
In `[Ni(CO)_(4)]` Nihas zero oxidation state i.e.

When CO approaches to Ni-meta, electron of Ni, go for rearrangement to give tetrahedral structure with no unpaired electrons (diamagnetic nature) and `sp^(3)` hydridisaton.

Hence, (d) is the correct answer.