Consider the following reaction MnO₂ + KOH + O₂ → A + H₂O. Product 'A' in neutral or acidic medium disproportionate 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)

The correct answer is 4.

Step 1: Identifying Product 'A'
When manganese dioxide (MnO₂) is fused with potassium hydroxide (KOH) in the presence of oxygen (O₂), potassium manganate (K₂MnO₄) and water are formed according to the following reaction:
2 MnO₂ + 4 KOH + O₂ → 2 K₂MnO₄ + 2 H₂O
Therefore, product 'A' is potassium manganate, K₂MnO₄ (containing the green manganate ion, MnO₄^2-).

Step 2: Disproportionation of Product 'A'
In neutral or acidic medium, the manganate ion (MnO₄^2-) disproportionates into permanganate ion (MnO₄^-) and manganese dioxide (MnO₂) as shown below:
3 MnO₄^2- + 4 H⁺ → 2 MnO₄^- + MnO₂ + 2 H₂O
Thus, the products 'B' and 'C' are MnO₄^- (permanganate ion) and MnO₂ (manganese dioxide).

Step 3: Finding the Spin-Only Magnetic Moment of B (MnO₄^-)
The atomic number of manganese (Mn) is 25. Its ground-state electronic configuration is:
Mn: [Ar] 3d⁵ 4s²
In the permanganate ion (MnO₄^-), the oxidation state of manganese is +7. The electronic configuration of Mn⁷⁺ is:
Mn⁷⁺: [Ar] 3d⁰ 4s⁰
Since there are no unpaired electrons (n = 0), its spin-only magnetic moment (μ_B) is:

${\mu }_B=\sqrt{0(0+2)}=0\text{BM}$

Step 4: Finding the Spin-Only Magnetic Moment of C (MnO₂)
In manganese dioxide (MnO₂), the oxidation state of manganese is +4. The electronic configuration of Mn⁴⁺ is:
Mn⁴⁺: [Ar] 3d³ 4s⁰
Since there are three unpaired electrons (n = 3), its spin-only magnetic moment (μ_C) is:

${\mu }_C=\sqrt{3(3+2)}=\sqrt{15}\approx 3.87\text{BM}$

Step 5: Calculating the Sum
The sum of the spin-only magnetic moment values of B and C is:

${\mu }_{\text{sum}}={\mu }_B+{\mu }_C=0+3.87=3.87\text{BM}$

Rounding 3.87 to the nearest integer gives 4.