The spin only magnetic moment of $$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$$ complexes is ______ B.M. (Nearest integer). (Given atomic number of Mn $$= 25$$)

Manganese has atomic number 25, so its electronic configuration is $$[\text{Ar}]\,3d^5\,4s^2$$. In the complex $$[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$$, Mn is in the +2 oxidation state, giving a $$d^5$$ configuration. Since water is a weak-field ligand, no pairing occurs and all five $$d$$-electrons remain unpaired.

The spin-only magnetic moment is given by $$\mu = \sqrt{n(n+2)}$$ B.M., where $$n$$ is the number of unpaired electrons. Here $$n = 5$$, so $$\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92$$ B.M., which rounds to $$\boxed{6}$$ B.M.