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Question:

The hybridization involved in complex [Ni(CN)4]²⁻ is: [Atomic number of Ni=28]

d²sp²

d²sp³

dsp²

sp³

Solution:

[Ni(CN)4]²⁻ is an square planar complex with dsp² hybridization.
The outer electronic configuration of Ni²+ ion is 3d⁸4s⁰. One 3d, one 4s and two 4p orbitals undergo dsp² hybridization four equivalent dsp² hybrid orbitals which are occupied by 4 electron pairs of 4 cyanide ligands. This forms [Ni(CN)4]²⁻. Hence, option C is correct.