The hybridisation of Fe in $[F e F_{6}]^{3 -}$ is $s p^{3} d^{2}$ , the number of unpaired electrons present is

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Solution

The correct option is C 5
Here, $F e^{3 +}$ had $d^{5}$ configuration and it is an outer orbital complex.

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Similar questions

[F e F_{6}]^{3 -}

s p^{3} d^{2}

{[F e F_{6}]}_{6}^{3 -}

What is the hybridization state and number of unpaired electrons of $F e$ atom in $[F e F_{6}]^{3 -}$ complex?

[F e (C O)_{5}]

F e

[F e F_{6}]^{3 -}

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