What is the spin-only magnetic moment value (B.M.) of a divalent metal ion with atomic number 25, in its aqueous solution?

The element with atomic number 25 is manganese (Mn), which has the electronic configuration $$[Ar]\,3d^5\,4s^2$$. A divalent metal ion $$Mn^{2+}$$ loses the two 4s electrons, giving the configuration $$[Ar]\,3d^5$$ with 5 unpaired electrons.

The spin-only magnetic moment is given by $$\mu = \sqrt{n(n+2)}$$ B.M., where $$n$$ is the number of unpaired electrons. Substituting $$n = 5$$, we get $$\mu = \sqrt{5 \times 7} = \sqrt{35} = 5.92$$ B.M.

In aqueous solution, $$Mn^{2+}$$ forms the high-spin complex $$[Mn(H_2O)_6]^{2+}$$ since water is a weak-field ligand, so all five 3d electrons remain unpaired. The answer is 5.92 B.M., which is option (1).