Question

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

• A. 1
• B. 5
• C. 4
• D. unpredictable

Answer 1

Answer: (B)

Hexafluoroferrate (III) ion ${\left[Fe{F}_6\right]}^{-3}$
Electronic configuration of $Fe:[Ar]4s^{2}3d^6$
Here $Fe$ shows oxidation state of $+3$.
Thus, its electronic configuration can be given as $[Ar]3d^5$
Thus, it has $5$ unpaired electrons
The electron pairs donate by $F$ enters into $4s,4p$ and two $4d$ subshell to form $s{p}^3{d}^2$ hybrid orbital.