Question

What is the hybridization state and number of unpaired electrons of $Fe$ atom in $[Fe{F}_6{]}^{3-}$ complex?

• A. $d^{2}s{p}^3,4$
• B. $d^{2}s{p}^3,5$
• C. $s{p}^3{d}^2,5$
• D. $s{p}^3{d}^2,3$

Answer 1

The correct option is C $s{p}^3{d}^2,5$

$[Fe{F}_6{]}^{3-}$ is a high spin complex since $F^{-}$ is a weak ligand, it does not cause pairing of electrons in the d-orbital thus, giving 5 unpaired electrons since $Fe$ is in $+3$ oxidation state and its electronic configuration is $\left[Ar\right]3d^5.$ It has 5 unpaired electrons in 3d subshell.