Question

Consider following octahedral complexes:

I.$\left[Fe\right(CN{)}_6{]}^{3-}$

II.$\left[Fe\right(H_{2}O{)}_6{]}^{3+}$

Which has fewer unpaired electrons?

• A. $\left[Fe\right(CN{)}_6{]}^{3-}$
• B. $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$
• C. both have equal unpaired electrons.
• D. none of them has unpaired electrons.

Answer 1

The correct option is A $\left[Fe\right(CN{)}_6{]}^{3-}$

Fe is in +3 state in both the complexes. $C{N}^{-}$ is strong field ligand so it will pair the electrons in $t_{2g}$ level.
$\left[Fe\right(CN{)}_6{]}^{3-}$ has only one unpaired electron.
$\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ has 5 unpaired electrons.