Let's determine the oxidation states of Mn in the reactants and products:
In MnO$_4^{2-}$: Let $x$ be the oxidation state of Mn.
Then $x + 4(-2) = -2$, so $x = +6$.
In MnO$_4^-$: $x + 4(-2) = -1$, so $x = +7$.
In MnO$_2$: $x + 2(-2) = 0$, so $x = +4$.
The oxidation states of Mn in the reaction are +6, +7, and +4. Therefore, the oxidation states not shown are +2 and +3.
The product (A) formed in the following reaction sequence is:
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