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Question

What is the geometry of [Ni(CN)4]2−,[NiCl4]2− and [Ni(CO)4]?

A
[NiCl4]2 is square planar and
[Ni(CN)4]2,[Ni(CO)4] are tetrahedral
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B
[Ni(CO)4] is square planar and
[Ni(CN)4]2,[NiCl4]2 are tetrahedral
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C
[Ni(CN)4]2 is square planar and
[NiCl4]2,[Ni(CO)4] are tetrahedral
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D
None of the above
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Solution

The correct option is C [Ni(CN)4]2 is square planar and
[NiCl4]2,[Ni(CO)4] are tetrahedral

[Ni(CN)4]2 is a square planar geometry formed by dsp2 hybridisation and not tetrahedral by sp3.

[NiCl4]2, there is Ni2+ ion, However, in the presence of weak field Cl ligands, NO pairing of d-electrons occurs. Therefore, Ni2+ undergoes sp3 hybridization to make bonds with Cl ligands in tetrahedral geometry

[Ni(CO)4] is also tetrahedral because of sp3

Hence C is correct

1487083_639896_ans_90fbdc38e52d4447a056882c3ce374eb.png

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