Which one of the following coordination complexes exhibit the lowest value of magnetic moment (in BM)?

1. $\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$
2. $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$
3. $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$
4. $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

Solution

(I) $\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}, \mathrm{Cr}^{3+} \Rightarrow 3 d^3$ $\therefore \mathrm{CN}$ is a strong field ligand.
$\therefore$ Number of unpaired $e^{-}=3$ $ \Rightarrow N=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15} \mathrm{BM} $ (II) $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-} ; \mathrm{Mn}^{3+} \Rightarrow 3 d^4$ $\because \mathrm{CN}$ is a strong field ligand.
$\therefore$ Number of unpaired $e^{-}=2$ $ \Rightarrow \quad N=\sqrt{2(2+2)}=\sqrt{8} \mathrm{BM} $ (III) $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} ; \mathrm{Fe}^{3+} \rightarrow 3 d^5$ $\because \mathrm{CN}$ is a strong field ligand.
$\therefore$ Number of unpaired $e^{-}=2$ $ N=\sqrt{2(2+2)}=\sqrt{8} \mathrm{BM} $ (IV) $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}, \mathrm{Co}^{3+} \longrightarrow 3 d^6$ $\because \mathrm{CN}^{-}$is a strong field ligand.
$\therefore$ Number of unpaired $e^{-}=0$ $ \Rightarrow \quad N=0 $ Hence, minimum value of magnetic moment is for $\left[\mathrm{Co}(\mathrm{CN})_6\right]^{3-}$

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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