The configurations of the complexes of manganese A, B respectively are $t_{2 g}^3 e_g^1, t_{2 g}^4 e_g^0$.…

The configurations of the complexes of manganese A, B respectively are $t_{2 g}^3 e_g^1, t_{2 g}^4 e_g^0$. Then $A, B$ are
  1. $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$
  2. $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$
  3. $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{4-}$
  4. $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$

Solution

The configuration $\mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^1$ represents a total of $3+1=$ 4 electrons in the d-subshell with all electrons unpaired as there is a weak-field ligand. Thus, $\mathrm{Mn}$ is in +3 state and the complex 'A' will be $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$. The configuration $t_{2 g}^4 \mathrm{e}_{\mathrm{g}}^0$ represents the same oxidation state of $\mathrm{Mn}$ but with a strong-field ligand. Thus, the complex ' $\mathrm{B}$ ' will be $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$.

Asked in: AP EAMCET 2023 (18 May Shift 2)

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