The complex ion $[N i F_{4}]^{2 +}$ is tetrahedral. Number of unpaired electrons present in the complex is/are:

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Solution

The correct option is A 2

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

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