1.

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

Answer»

`[NI(CN)_4]^(2-)` is SQUARE PLANAR and `[NiCl_4]^(2-)` and `Ni(CO)_4` are tetrahedral
`[NiCl_4]^(2-)` is square planar and `[Ni(CN)_4]^(2-)` and `Ni(CO)_4` are tetrahedral
`Ni(CO)_4` is square planar and`[Ni(CN)_4]^(2-)` and `[NiCl_4]^(2-)` are tetrahedral
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Answer :A


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