Spin only magnetic moment of $$[MnBr_6]^{4-}$$ is ______ B.M. (round off to the closest integer)

We need to find the spin-only magnetic moment of $$[\text{MnBr}_6]^{4-}. Let the oxidation state of Mn be x; then x + 6(-1) = -4 gives x = +2, so Mn is in the +2 oxidation state (Mn^{2+}$$).

Mn has atomic number 25 with configuration [Ar] 3d$$^5 4s^2, so Mn^{2+}$$ is [Ar] 3d$$^5. Since Br^- is a weak field ligand, [\text{MnBr}_6]^{4-}$$ is a high-spin complex.

In a high-spin octahedral d$$^5 complex all five electrons are unpaired, giving n = 5. The spin-only magnetic moment is calculated as \mu = \sqrt{n(n+2)} B.M., which yields \mu = \sqrt{5(5+2)} = \sqrt{5 \times 7} = \sqrt{35} = 5.916 B.M. Rounding to the closest integer: \mu \approx 6$$ B.M.