Question

Nickel $(Z=28)$ combines with an uninegative monodentate ligand to form a diamagnetic complex $[Ni{L}_4{]}^{2–}$. The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

• A. $ds{p}^2$, one
• B. $s{p}^3$, two
• C. $ds{p}^2$, zero
• D. $s{p}^3$, zero

Answer 1

The correct option is C $ds{p}^2$, zero

${}_{28}Ni$ has $+2$ oxidation state and since the complex is diamagnetic, all electrons are paired in inner $3d$, orbital to provide the electronic configuration as under.


$L^{–}$ might have been a strong field $C{N}^{–}$. Hybridisation is $ds{p}^2$. Number of unpaired electrons $=0$.