Question

How many unpaired $e^{-}$ are present in [Ni(CN)${}_4$]${}^{2-}$ ?

• A. 0
• B. 2
• C. 1
• D. 3

Answer 1

The correct option is A 0

Zero unpaired $e^{-}$ are present in [Ni(CN)${}_4$]${}^{2-}$. $Ni$ is in +2 oxidation state. It has $3d^8$ electronic configuration.
It undergoes $ds{p}^2$ hybridization to form 4 $ds{p}^2$ hybrid orbitals which contain 8 electrons from 4 cyanide ligands. 8 electrons of $Ni$ are present in four $3d$ orbitals. Thus, all electrons are paired. Due to this, [Ni(CN)${}_4$]${}^{2-}$ is diamagnetic.