Question

The complex ion $[Ni{F}_4{]}^{2+}$ is tetrahedral. Number of unpaired electrons present in the complex is/are:

• A. 2
• B. 4
• C. 1
• D. 0

Answer 1

The correct option is A 2

$[Ni{F}_4{]}^{2+}$: In this coordination complex, $Ni$ has the electronic configuration of $3d^{8}4s^2$ and $N^{+2}$ has the electronic configuration of $3d^{8}4s^o$.

Also, since fluorine is a weak ligand so, it will form an outer orbital complex with $s{p}^3$ & tetrahedral geometry.

(Refer to Image)

It has $2$ unpaired electrons.