Question

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

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

Answer 1

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

Monodentate ligand is uninegative and hence, the oxidation number of Ni in $[Ni{X}_4{]}^{2-}$ is +2.

The electronic configuration of $N{i}^{2+}$ is $\left[Ar\right]3d^8$.

If the compound is diamagnetic it means all the electrons are paired. Hence, there will be one 3d orbital empty and one 4s orbital empty.

The hybridisation involved in the formation of complex $[Ni{L}_4{]}^{2-}$ is $ds{p}^2$.