$\mathrm{K_2Cr_2O_7}$ paper acidified with dilute $\mathrm{H_2SO_4}$ turns green when exposed to :
Solution
$\mathrm{SO}_{2}$ gets oxidised in presence of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ and it converts to $\mathrm{Cr}^{+3}$ in presence of dil. $\mathrm{H}_{2} \mathrm{SO}_{4}$.
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Similarly, $\mathrm{H}_{2} \mathrm{~S}$ can also get oxidized to sulphur.
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However, most appropriate is (A) Sulphur dioxide.
About this question
Subject: Chemistry · Chapter: p-Block Elements · Topic: Group 13: Boron Family
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