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 disproportioates in acidic solution to afforda dark purple solution. X is :

D and F Block Elements PYQ — JEE Main 2018 · 16 Apr · Shift 1 · Q17 with Solution

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Correct answer: D

We are told that $XO_2$ on fusion with alkali metal hydroxide in presence of an oxidizing agent like $KNO_3$ gives a dark green product.
This is a very characteristic reaction of manganese dioxide: $MnO_2 + 2KOH + [O] \rightarrow K_2MnO_4 + H_2O$ Here, $K_2MnO_4$ is potassium manganate, which is dark green in color.
Now, the question says this green product disproportionates in acidic solution to give a dark purple solution.
Acidification of manganate causes disproportionation: $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$
The ion $MnO_4^-$ is permanganate, and its solution is dark purple.
Therefore, the element $X$ must be manganese.
Check options:
- A: Ti — does not fit
- B: V — vanadium gives different colored oxoanions, not this characteristic sequence
- C: Cr — chromate is yellow, dichromate is orange, not dark purple
- D: Mn — correct

Therefore, $X = Mn$ .