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When $$\text{XO}_2$$ is fused with an alkali metal hydroxide in the presence of an oxidizing agent such as $$\text{KNO}_3$$, a dark green product is formed which disproportionate in acid solution to afford a dark purple solution. X is
$$\text{MnO}_2$$ fused with KOH and an oxidizing agent gives dark green potassium manganate, $$\text{K}_2\text{MnO}_4$$. In acid, manganate disproportionates, $$3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$$, giving purple permanganate. So X is Mn.
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