p-Block elementsHardBloom L3

Question

A sulphate of a metal (A) on heating evolves two gases (B) and (C) and an oxide (D). Gas (B) turns K2Cr2O7 paper green while gas (C) forms a trimer in which there is no S–S bond. Compound (D) with HCI, forms a Lewis acid (E) which exists as a dimer. Compounds (A), (B), (C), (D) and (E) are respectively

Options

A.FeSO4, SO2, SO3, Fe2O3, FeCl3
B.Al2(SO4)3, SO2, SO3, Al2O3, FeCl3
C.FeS, SO2, SO3, FeSO4, FeCl3
D.FeS, SO2, SO3 , Fe2(PO4)3, FeCl2

Solution

Sol. Only FeSO4​ gives two gases and a oxide

2FeSO4 ​→ SO2​ + SO3 ​+ Fe2​O3​

B is SO2​, C is SO3​ and D is Fe2​O3​.

K2​Cr2​O7​ + 3SO2 ​+ H2​SO4 ​→ K2​SO4 ​+ Cr2​(SO4​)3 ​+ H2​O

Cr2​(SO4​)3​ is green in colour.

Fe2​O3​ + 6HCl → 2FeCl3 ​+ 3H2​O

E is FeCl3​ which is a lewis acid and exists as dimer.

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