The sum of the spin only magnetic moment values (in B.M.) of [Mn(Br)$$_6$$]$$^{3-}$$ and [Mn(CN)$$_6$$]$$^{3-}$$ is ______.

Manganese has atomic number 25 with ground-state configuration $$[Ar]\,3d^{5}\,4s^{2}$$.

In both complexes the overall charge is $$-3$$ and each ligand carries $$-1$$ charge, so for six ligands:$$x+(-6)=-3\;\Longrightarrow\;x=+3$$Hence Mn is in the $$+3$$ oxidation state, that is $$Mn^{3+}$$ whose electronic configuration is $$3d^{4}$$ (because the two $$4s$$ electrons and one $$3d$$ electron are lost).

For an octahedral $$d^{4}$$ ion two spin arrangements are possible:
• high-spin (weak-field ligand)
• low-spin (strong-field ligand)

Case 1: [Mn(Br)$$\_6$$]$$^{3-}$$
$$Br^-$$ is a weak-field ligand. Thus the complex is high spin:$$t_{2g}^{3}e_{g}^{1}$$The four electrons occupy different orbitals according to Hund’s rule, giving $$n=4$$ unpaired electrons.
Spin-only magnetic moment:
$$\mu_1=\sqrt{n(n+2)}=\sqrt{4(4+2)}=\sqrt{24}\;\text{B.M.}\approx4.90\;\text{B.M.}$$

Case 2: [Mn(CN)$$\_6$$]$$^{3-}$$
$$CN^-$$ is a strong-field ligand. The complex is low spin:$$t_{2g}^{4}e_{g}^{0}$$The occupancy inside $$t_{2g}$$ is $$\uparrow\downarrow\,\uparrow\,\uparrow$$, leaving $$n=2$$ unpaired electrons.
Spin-only magnetic moment:
$$\mu_2=\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}\;\text{B.M.}\approx2.83\;\text{B.M.}$$

Sum of the magnetic moments
$$\mu_{\text{total}}=\mu_1+\mu_2\approx4.90+2.83=7.73\;\text{B.M.}$$

The required sum lies in the range $$7.5\text{-}7.8\;\text{B.M.}$$ (as specified).