JEE Main 2026 — Jan 21, Shift 2 — Chemistry solution

JEE Main2026Jan 21, Shift 2Chemistry

On heating a mixture of common salt and $K_{2} C r_{2} O_{7}$ in equal amount along with concentrated $H_{2} S O_{4}$ in a test tube, a gas is evolved. Formula of the gas evolved and oxidation state of the central metal atom in the gas respectively are :

A
$C r O_{2} C l_{2}$ and +5
B
$C r O_{2} C l_{2}$ and +6
C
$C r_{2} O_{2} C l_{2}$ and +6
D
$C r_{2} O_{2} C l_{2}$ and +3

Solution

This is the classical chromyl chloride test: $4 N a C l + K_{2} C r_{2} O_{7} + 6 H_{2} S O_{4} \to 2 K H S O_{4} + 2 C r O_{2} C l_{2} (↑) + 4 N a H S O_{4} + 3 H_{2} O$ .

The evolved orange-red vapour is chromyl chloride, $C r O_{2} C l_{2}$ . Assigning Cr: $2 (- 2) + 2 (- 1) + C r = 0 \Rightarrow C r = + 6$ .

← Back to full paper

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →