An acidified manganate solution undergoes disproportionation reaction. The spin-only magnetic moment value of the product having manganese in higher oxidation state is ______ B.M. (Nearest integer)

Step 1: Write the disproportionation reaction of manganate ion in acidic medium.

The manganate ion ($$\text{MnO}_4^{2-}$$) has Mn in the +6 oxidation state. In acidic solution, it disproportionates:

$$3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^{-} + \text{MnO}_2 + 2\text{H}_2\text{O}$$

The two products are:

- Permanganate ion ($$\text{MnO}_4^{-}$$): Mn is in +7 oxidation state (higher)

- Manganese dioxide ($$\text{MnO}_2$$): Mn is in +4 oxidation state (lower)

Step 2: Find the spin-only magnetic moment of the product with higher oxidation state.

In $$\text{MnO}_4^{-}$$, Mn is in the +7 state with electronic configuration [Ar] 3d$$^0$$.

Since there are no unpaired electrons ($$n = 0$$):

$$\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0 \text{ B.M.}$$

The spin-only magnetic moment is 0 B.M.