Question

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)

Answer 1

Correct Answer: 0

In the disproportionation of acidified manganate (\( \text{MnO}_4^{2-} \)), manganese (Mn) exists initially in the +6 oxidation state. This reaction will typically result in two products: manganese in lower and higher oxidation states. The key step here is to determine the higher oxidation state among the products.

In an acidified solution, manganate ions (\( \text{MnO}_4^{2-} \)) predominantly disproportionate to permanganate ions (\( \text{MnO}_4^- \)) and manganese dioxide (\( \text{MnO}_2 \)). This can be expressed as:

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

Here, the permanganate ion (\( \text{MnO}_4^- \)) contains manganese in the +7 oxidation state. We need to find the spin-only magnetic moment of Mn in this higher oxidation state.

The formula for calculating the spin-only magnetic moment (\( \mu \)) is given by:

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

Where \( n \) is the number of unpaired electrons. In the permanganate ion (\( \text{MnO}_4^- \)), Mn is in the +7 oxidation state and the electronic configuration is: \( [\text{Ar}] \), which means there are no unpaired electrons (\( n = 0 \)).

Plugging \( n = 0 \) into the formula gives:

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

Therefore, the spin-only magnetic moment for the manganese in the higher oxidation state (\(+7\)) is 0 B.M. This value is indeed within the expected range, confirming its correctness.