When XO$$_2$$ is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO$$_3$$; a dark green product is formed which disproportionate in acidic solution to afford a dark purple solution. X is:

We have an oxide of the type $$\text{XO}_2$$ which is fused with an alkali metal hydroxide, say $$\text{KOH}$$, in the presence of an oxidising agent such as $$\text{KNO}_3$$. Experimentally, this fusion gives a dark green product. Dark green colour, when produced in strongly alkaline medium, immediately suggests the formation of manganate ion $$\text{MnO}_4^{2-}$$, whose potassium salt $$\text{K}_2\text{MnO}_4$$ is characteristically green.

To justify this, let us write the balanced oxidation-reduction reaction that converts $$\text{MnO}_2$$ (manganese dioxide) into potassium manganate under the given conditions:

$$\text{MnO}_2 + 2\,\text{KOH} + \text{KNO}_3 \;\longrightarrow\; \text{K}_2\text{MnO}_4 + \text{KNO}_2 + \text{H}_2\text{O}$$

In the above equation, the oxidising agent $$\text{KNO}_3$$ removes electrons (is reduced to $$\text{KNO}_2$$), while the manganese in $$\text{MnO}_2$$ is oxidised from oxidation state $$+4$$ to $$+6$$ present in $$\text{MnO}_4^{2-}$$. The product $$\text{K}_2\text{MnO}_4$$ is dark green.

Now, when this green manganate solution is acidified, it undergoes a well-known disproportionation reaction. Disproportionation means that the same species is simultaneously oxidised and reduced. In acidic medium, $$\text{MnO}_4^{2-}$$ (oxidation state $$+6$$) converts partly to $$\text{MnO}_4^{-}$$ (oxidation state $$+7$$) and partly to $$\text{Mn}^{4+}$$ (as insoluble $$\text{MnO}_2$$ slick brown precipitate, which may further dissolve depending on conditions). The balanced net reaction in acid can be written as

$$3\,\text{K}_2\text{MnO}_4 + 2\,\text{H}_2\text{O} \;\longrightarrow\; 2\,\text{KMnO}_4 + \text{MnO}_2 + 4\,\text{KOH}$$

Here, the newly formed $$\text{KMnO}_4$$ contains the permanganate ion $$\text{MnO}_4^{-}$$, whose solution is an intense dark purple. This sequence — green manganate turning to purple permanganate on acidification — is the unmistakable signature of compounds of manganese.

Therefore, the element $$\text{X}$$ in $$\text{XO}_2$$ must be manganese, $$\text{Mn}$$.

Hence, the correct answer is Option A.