A sulphate of a metal (A) on heating evolves two gases (B) and (C) and an oxide (D). Gas (B) turns K 2 Cr 2 O 7 paper green while gas (C) forms a trimer in which there is no S-S bond. Compound (

Question

A sulphate of a metal (A) on heating evolves two gases (B) and (C) and an oxide (D).

Gas (B) turns $K_{2} Cr_{2} O_{7}$ paper green while gas (C) forms a trimer in which there is no S-S bond. Compound (D) with conc. HCl , forms a Lewis acid (E) which exists as a dimer.

Compounds (A), (B), (C), (D) and (E) are respectively

TestHub · Accepted Answer

The correct answer is A. The metal sulphate (A) is $F e S O_{4}$ . Upon heating, it decomposes to give $F e_{2} O_{3}$ (D), $S O_{2}$ (B), and $S O_{3}$ (C). $S O_{2}$ turns $K_{2} C r_{2} O_{7}$ paper green, indicating its reducing nature. $S O_{3}$ forms a cyclic trimer $(S_{3} O_{9})$ without S-S bonds. $F e_{2} O_{3}$ reacts with conc. HCl to form $F e C l_{3}$ (E), a Lewis acid that exists as a dimer $(F e C l_{3})_{2}$ .

Concepts used: Thermal decomposition, redox reactions, Lewis acids, structure of sulfur oxides.

$2 F e S O_{4} (s) Δ F e_{2} O_{3} (s) + S O_{2} (g) + S O_{3} (g)$

$F e_{2} O_{3} + 6 H Cl \to 2 F e C l_{3} + 3 H_{2} O$

Chemistry - Salt Analysis Question with Solution | TestHub

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JEE Main (2026)