The correct statement with respect to the complexes `Ni(CO)_(4) and |Ni(CN)_(4)|^(2-)` isA. Nickel is in the same oxidation state in bothB. Both have tetrahedral geometryC. Both have square planar geometryD. Have tetrahedral and square planar geometry respectively

The correct statement with respect to the complexes `Ni(CO)_(4) and |N

1.	The correct statement with respect to the complexes `Ni(CO)_(4) and \|Ni(CN)_(4)\|^(2-)` isA. Nickel is in the same oxidation state in bothB. Both have tetrahedral geometryC. Both have square planar geometryD. Have tetrahedral and square planar geometry respectively
Answer» Correct Answer - D Hybridization of nickel in `Ni(CO)_(4)` is `sp^(3)` , therefore geometry of `Ni(CO)_(4)` is tetrahedral whereas hybridization of nickel in `[Ni(CN)_(6)]^(2-)` is `dsp^(2)`, therefore geometry of `[Ni(CN)_(4)]^(2-)` is square planar.

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The correct statement with respect to the complexes `Ni(CO)_(4) and |Ni(CN)_(4)|^(2-)` isA. Nickel is in the same oxidation state in bothB. Both have tetrahedral geometryC. Both have square planar geometryD. Have tetrahedral and square planar geometry respectively

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