Observe the following complexes $\begin{aligned} & {\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}…

1. $2$
2. $1$
3. $3$
4. $4$

Solution

$\begin{aligned} & {\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}:-} \\ & \mathrm{Mn}^{3+}=[\mathrm{Ar}]^3 \mathrm{~d}^4\end{aligned}$ $\mathrm{CN}^{-}$is a strong-field ligand so it pairs the electrons. Thus, $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}=$
Thus, it is an inner-orbital but a diamagnetic complex. $\begin{aligned} & {\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}:-} \\ & \mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^5 \end{aligned}$ $\mathrm{CN}^{-}$will pair - up the electrons so :- $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}=$
Thus, it is an inner-orbital and a paramagnetic complex. $\begin{aligned} & {\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}:-} \\ & \mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^6 \end{aligned}$ $\mathrm{C}_2 \mathrm{O}_4{ }^{2-}$ is a weak-field ligand so :-
$\left[\mathrm{Co}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]^{3-}=$
Thus, it is an outer-orbital and paramagnetic complex $\begin{aligned} & {\left[\mathrm{MnCl}_6\right]^{3-}:-} \\ & \mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^4 \end{aligned}$ $\mathrm{Cl}^{-}$is a weak-field ligand so :-
$\left[\mathrm{MnCl}_6\right]^{3-}=$
Thus, it is an inner-orbital and paramagnetic complex $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^4$ :- $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^6$ and $\mathrm{CN}^{-}$is a strong-field ligand. Thus, $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}=$
Thus, it is an inner-orbital and diamagnetic complex. $\left[\mathrm{CoF}_6\right]^{3-}:-$ $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^6$ and $\mathrm{F}^{-}$is a weak-field ligand. Thus,
$\left[\mathrm{CoF}_6\right]^{3-}=$
Thus, it is an outer-orbital and paramagnetic complex. Therefore, there are two inner-orbital and paramagnetic complexes.

Asked in: AP EAMCET 2023 (16 May Shift 2)

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