Hexafluoroferrate (lll) ion is an outer orbital complex. The number of unpaired electrons are:

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Solution

The correct option is C 5
Hexafluoroferrate (III) ion $[F e F_{6}]^{- 3}$
Electronic configuration of $F e : [A r] 4 s^{2} 3 d^{6}$
Here $F e$ shows oxidation state of +3.
Thus, its electronic configuration can be given as $[A r] 3 d^{5}$
Thus, it has 5 unpaired electrons
The electron pairs donate by $F$ enters into $4 s$ , $4 p$ and two $4 d$ subshell to form $s p^{3} d^{2}$ hybrid orbital.

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Similar questions

Hexafluoroferrate (III) ion is an outer orbital complex. The number of unpaired electrons present in Fe is

[C o F_{6}]^{3 -}, C o^{3 +}

s p^{3} d^{2}

[C o (N H_{3})_{6}]^{2 +}

[C o (N H_{3})_{6}]^{3 +}

[C o (N H_{3})_{6}]^{2 +}

[C u (N O_{2})_{6}]^{4 -}

[F e (H_{2} O)_{6}]^{3 +}

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