A sulphate of a metal (A) on heating evolves | vedclass

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(A) The thermal decomposition of ferrous sulphate is given by: $2 FeSO_4 \rightarrow SO_2 + SO_3 + Fe_2O_3$.Here,gas $(B)$ is $SO_2$,gas $(C)$ is $SO_

Vedclass · Accepted Answer

$FeSO_4, SO_2, SO_3, Fe_2O_3, FeCl_3$

Vedclass · Answer

(A) The thermal decomposition of ferrous sulphate is given by: $2 FeSO_4 \rightarrow SO_2 + SO_3 + Fe_2O_3$.Here,gas $(B)$ is $SO_2$,gas $(C)$ is $SO_3$,and oxide $(D)$ is $Fe_2O_3$.Gas $(B)$ $(SO_2)$ is a reducing agent and turns acidified $K_2Cr_2O_7$ paper green due to the formation of $Cr_2(SO_4)_3$.Gas $(C)$ $(SO_3)$ forms a trimer $(SO_3)_3$ which has a cyclic structure with no $S-S$ bonds.Compound $(D)$ $(Fe_2O_3)$ reacts with $HCl$ to form $FeCl_3$ $(E)$,which exists as a dimer $(Fe_2Cl_6)$ and acts as a Lewis acid.Thus,the correct sequence is $FeSO_4, SO_2, SO_3, Fe_2O_3, FeCl_3$.

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