Question

Compare the magnetic moment of $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ and $\left[Fe\right(CN{)}_6{]}^{3-}$ ion:

• A. $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ < $\left[Fe\right(CN{)}_6{]}^{3-}$
• B. $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ > $\left[Fe\right(CN{)}_6{]}^{3-}$
• C. $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ = $\left[Fe\right(CN{)}_6{]}^{3-}$
• D. none of these

Answer 1

The correct option is B $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ > $\left[Fe\right(CN{)}_6{]}^{3-}$

In both the complexes, Fe is in +3 oxidation state. $F{e}^{3+}$ has $\left[Ar\right]3d^5$ electronic configuration.

Water is a weak field ligand. So spin pairing is not favored. Hence, $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ is a high spin complex with 5 unpaired electrons.

Cyanide ion is a strong field ligand. So the spin pairing is favored. Hence, $\left[Fe\right(CN{)}_6{]}^{3-}$ is a low spin complex with 1 unpaired electron.

Higher is the number of unpaired electrons, higher is the magnetic moment ( $\sqrt{n(n+1)}$ ), where 'n' is the number of unpaired electrons.

Hence, option $B$ is correct.