JEE Main 2026 · 5 Apr · Shift 2 · Chemistry Q70

ChemistrySingle correct+4 / -1Question 70 of 75

A paper dipped in a dil. $\ce{H2SO4}$ solution of 'X' upon treatment with $\ce{SO2}$ gas turns into green. The compound 'X' is :

KI-starch

\ce{KMnO4}

\ce{Pb(CH3COO)2}

\ce{K2Cr2O7}

Answer and worked solution

Official answer: D

The test. $\ce{SO2}$ reduces acidified dichromate (orange) to $\ce{Cr^3+}$ (green):
$\ce{Cr2O7^2- + 3SO2 + 2H+ -> 2Cr^3+ + 3SO4^2- + H2O}$

X: $\ce{K2Cr2O7}$ .

(D)

Topics:Important compounds (KMnO4, K2Cr2O7)Group 16

JEE Main 2026 · 5 Apr · Shift 2, Chemistry question 70