Consider the following reaction: $$MnO_2 + KOH + O_2 \to A + H_2O$$. Product A in neutral or acidic medium disproportionates to give products B and C along with water. The sum of spin-only magnetic moment values of B and C is ______ BM. (nearest integer) [Given atomic number of Mn is 25]

We need to find the sum of spin-only magnetic moment values of products B and C.

We identify product A by considering the reaction $$MnO_2 + KOH + O_2 \rightarrow K_2MnO_4 (A) + H_2O$$; the balanced equation is $$2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O$$. Product A is potassium manganate ($$K_2MnO_4$$), where Mn is in the +6 oxidation state.

In neutral or acidic solution, $$K_2MnO_4$$ (Mn$$^{6+}$$) disproportionates according to $$3K_2MnO_4 + 2H_2O \rightarrow 2KMnO_4 + MnO_2 + 4KOH$$, in which Mn$$^{6+}$$ is oxidised to Mn$$^{7+}$$ in $$KMnO_4$$ and reduced to Mn$$^{4+}$$ in $$MnO_2$$. Thus B = $$KMnO_4$$ and C = $$MnO_2$$.

For $$KMnO_4$$, Mn is in the +7 oxidation state. The Mn atom has the configuration [Ar] 3d$$^5$$ 4s$$^2$$, so Mn$$^{7+}$$ loses all 7 electrons giving [Ar] = 3d$$^0$$, resulting in zero unpaired electrons and a spin-only magnetic moment of $$\mu_B = \sqrt{n(n+2)} = \sqrt{0} = 0 \text{ BM}$$.

In $$MnO_2$$, Mn is in the +4 oxidation state, so Mn$$^{4+}$$ has the configuration [Ar] 3d$$^3$$ (lost 4s$$^2$$ and 1 of 3d$$^5$$, leaving 3 electrons in 3d), resulting in 3 unpaired electrons and a magnetic moment of $$\mu_C = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} = 3.87 \text{ BM}$$.

Adding these values gives $$\mu_B + \mu_C = 0 + 3.87 = 3.87 \approx 4 \text{ BM (nearest integer)}$$.

The answer is 4.