Q1 / 90 · KVPY 2015 (SB/SX) · Chemistry
$\mathrm{MnO}_{2}$ when fused with $\mathrm{KOH}$ and oxidized in air gives a dark green compound $\mathrm{X}$. In acidic $\mathrm{X}$. In acidic solution, $\mathrm{X}$ undergoes disproportion to give an intense purple compound $\mathrm{Y}$ and $\mathrm{MnO}_{2}$. The compounds $\mathrm{X}$ and $\mathrm{Y}$. respectively, are
- Correct answerA$\mathrm{K}_{2} \mathrm{MnO}_{4}$ and $\mathrm{KMnO}_{4}$
- B$\mathrm{Mn}_{2} \mathrm{O}_{7}$ and $\mathrm{KMnO}_{4}$
- C$\mathrm{K}_{2} \mathrm{MnO}_{4}$ and $\mathrm{Mn}_{2} \mathrm{O}_{7}$
- D$\mathrm{KMnO}_{4}$ and $\mathrm{K}_{2} \mathrm{MnO}_{4}$
Solution
$2 \mathrm{MnO}_{2}+4 \mathrm{KOH}+\mathrm{O}_{2} \rightarrow 2 \mathrm{~K}_{2} \mathrm{MnO}_{4}+2 \mathrm{H}_{2} \mathrm{O}$
$3 \mathrm{~K}_{2} \mathrm{MnO}_{4}+4 \mathrm{HCl} \rightarrow 2 \mathrm{KMnO}_{4}+\mathrm{MnO}_{2}+4 \mathrm{KCl}+2 \mathrm{H}_{2} \mathrm{O}$
Hence, the correct answer is option: 'A'.