Question Details

$In a reaction:$

A stands for:

Options

$\begin{matrix} {Mn}^{3 +} \end{matrix}$

$\begin{matrix} {Mn}^{4 +} \end{matrix}$

$\begin{matrix} {Mn}^{2 +} \end{matrix}$

$\begin{matrix} {Mn}^{+} \end{matrix}$

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Correct Answer :

Option C

$\begin{matrix} {Mn}^{2 +} \end{matrix}$

Solution :

The correct answer is:
${Mn}^{2 +}$

Step-by-Step Explanation:

1. Identify the reaction from the image:
The image shows the following incomplete reduction half-reaction in an acidic medium:
${MnO}_{4}^{-} + 8 H^{+} + 5 e^{-} \to A + H_{2} O$

2. Determine the oxidation state change:
In the reactant, the permanganate ion is ${MnO}_{4}^{-}$ .
Let the oxidation state of Manganese (Mn) be x. Since oxygen has an oxidation state of -2:
$x + 4 (- 2) = - 1$
$x - 8 = - 1$
$x = + 7$
So, Manganese starts in the $+ 7$ oxidation state.

3. Account for the added electrons:
During the reaction, 5 electrons ( $5 e^{-}$ ) are gained by the manganese atom.
This reduction reduces the oxidation state by 5 units:
$New oxidation state = + 7 - 5 = + 2$
Therefore, the manganese species formed in the product must be in the $+ 2$ oxidation state, which is represented by the ion:
${Mn}^{2 +}$

4. Check Charge Balance:
We can verify this by balancing the total charge on both sides:
LHS (Reactants side) charge:
$(- 1) from {MnO}_{4}^{-} + 8 (+ 1) from H^{+} + 5 (- 1) from electrons = - 1 + 8 - 5 = + 2$
RHS (Products side) charge:
Since $H_{2} O$ is neutral (charge = 0), the total charge on the product side is determined solely by the species A.
For the charges to balance, the charge of A must be equal to $+ 2$ . This matches the charge of the ${Mn}^{2 +}$ ion.