JEE Main 2024 — Chemistry Inorganic Chemistry

2024mcqmedium

$\mathrm{NaCl}$ reacts with conc. ${H}_{2}{\mathrm{SO}}_{4}$ and ${K}_{2}{\mathrm{Cr}}_{2}{O}_{7}$ to give reddish fumes (B), which react with $\mathrm{NaOH}$ to give yellow solution (C). (B) and (C) respectively are ;

Official previous-year question

Held on 27 Jan 2024 · Verified 6 Jul 2026.

Options

A
${\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2},{\mathrm{Na}}_{2}{\mathrm{CrO}}_{4}$
B
${\mathrm{Na}}_{2}{\mathrm{CrO}}_{4},{\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2}$
C
${\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2},{\mathrm{KHSO}}_{4}$
D
${\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2},{\mathrm{Na}}_{2}{\mathrm{Cr}}_{2}{O}_{7}$

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Solution

Chromyl chloride (B) is produced when a chloride ion reacts with conc. ${H}_{2}{\mathrm{SO}}_{4}$ and ${K}_{2}{\mathrm{Cr}}_{2}{O}_{7}$ .

Passing the chromyl chloride gas through $\mathrm{NaOH}$ turns the solution yellow. This is due to the formation of ${\mathrm{Na}}_{2}{\mathrm{CrO}}_{4}$ (C).

The reaction is as follows,

$\mathrm{NaCl}+\mathrm{conc}.{H}_{2}{\mathrm{SO}}_{4}+{K}_{2}{\mathrm{Cr}}_{2}{O}_{7}\rightarrow \underset{\underset{(\mathrm{Reddish}\mathrm{brown})}{(B)}}{{\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2}}+{\mathrm{KHSO}}_{4}+{\mathrm{NaHSO}}_{4}+{H}_{2}O$

${\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2}+\mathrm{NaOH}\rightarrow \underset{\underset{\mathrm{Yellow}\mathrm{Colour}}{(C)}}{{\mathrm{Na}}_{2}{\mathrm{CrO}}_{4}}+\mathrm{NaCl}+{H}_{2}O$

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Options

${\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2},{\mathrm{Na}}_{2}{\mathrm{CrO}}_{4}$

${\mathrm{Na}}_{2}{\mathrm{CrO}}_{4},{\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2}$

${\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2},{\mathrm{KHSO}}_{4}$

${\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2},{\mathrm{Na}}_{2}{\mathrm{Cr}}_{2}{O}_{7}$

Solution

Chromyl chloride (B) is produced when a chloride ion reacts with conc. ${H}_{2}{\mathrm{SO}}_{4}$ and ${K}_{2}{\mathrm{Cr}}_{2}{O}_{7}$ .

Passing the chromyl chloride gas through $\mathrm{NaOH}$ turns the solution yellow. This is due to the formation of ${\mathrm{Na}}_{2}{\mathrm{CrO}}_{4}$ (C).

The reaction is as follows,

${\mathrm{CrO}}_{2}{\mathrm{Cl}}_{2}+\mathrm{NaOH}\rightarrow \underset{\underset{\mathrm{Yellow}\mathrm{Colour}}{(C)}}{{\mathrm{Na}}_{2}{\mathrm{CrO}}_{4}}+\mathrm{NaCl}+{H}_{2}O$

JEE Main 2024 — Chemistry Inorganic Chemistry

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JEE Main 2024 — Chemistry Inorganic Chemistry

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