Sum of Spin Only Magnetic Moments of Manganese Complexes

To find the sum of the spin-only magnetic moments of $[\text{Mn(Br)}_6]^{3-}$ and $[\text{Mn(CN)}_6]^{3-}$, we determine the oxidation state and $d$-electron configuration of manganese in both complexes.

In both complexes, manganese is in the $+3$ oxidation state, corresponding to a $d^4$ electronic configuration.

1. For $[\text{Mn(Br)}_6]^{3-}$, $\text{Br}^-$ acts as a weak-field ligand, resulting in a high-spin complex with electronic configuration $t_{2g}^3 e_g^1$. The number of unpaired electrons is $n_1 = 4$. Its magnetic moment is: $\mu_1 = \sqrt{n_1(n_1+2)} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.899 \text{ B.M.}$

2. For $[\text{Mn(CN)}_6]^{3-}$, $\text{CN}^-$ acts as a strong-field ligand, resulting in a low-spin complex with electronic configuration $t_{2g}^4 e_g^0$. The number of unpaired electrons is $n_2 = 2$. Its magnetic moment is: $\mu_2 = \sqrt{n_2(n_2+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.828 \text{ B.M.}$

Summing both values yields: $\text{Sum} = \mu_1 + \mu_2 = \sqrt{24} + \sqrt{8} \approx 4.899 + 2.828 = 7.727 \text{ B.M.}$

Thus, the sum of the spin-only magnetic moment values is approximately $7.73 \text{ B.M.}$