Question

$\left[Fe\right(CN{)}_6{]}^{3-}$ ion paramagnetic while $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ ion paramagnetic.

• A. both paramagnetic are equal
• B. $\left[Fe\right(CN{)}_6{]}^{3-}$ > $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$
• C. $\left[Fe\right(CN{)}_6{]}^{3-}$ < $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$
• D. non eof thse

Answer 1

Answer: (C)

$\left[Fe\right(CN{)}_6{]}^{3-}$ < $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$

The given statement is true. $\left[Fe\right(CN{)}_6{]}^{3-}$ ion is weakly paramagnetic while $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ ion strongly paramagnetic.

The magnetic moment of $\left[Fe\right(H_{2}O{)}_6{]}^{3+}$ ion is greater than that of $\left[Fe\right(CN{)}_6{]}^{3-}$ ion.

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

Cyanide ion is strong field ligand. Spin pairing is favoured. Hence, $\left[Fe\right(CN{)}_6{]}^{3-}$ is low spin complex with 1 unpaired electron. Higher is the number of unpaired electrons, higher is the magnetic moment.