Q. The hybridization involved in complex $[Ni(CN)_4]^{2-}$ is (At.No. Ni = 28)
A) $d^2sp^2$
B) $d^2sp^3$
C) $dsp^2$
D) $sp^3$
Explanation
[Solution] $Ni^{2+}$ ($3d^8$) with strong-field $CN^-$ pairs its electrons, leaving one 3d orbital free: $dsp^2$ (square planar). Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.