Analytical ChemistryMedium
Question
Treatment of a gas ' X ' with a freshly prepared ferrous sulphate solution gives a compound ' Y ' as a brown ring. The compounds X and Y are.
Options
A.NO and $\lbrack Fe(NO)\rbrack{SO}_{4}$
B.${NO}_{2}$ and $\left\lbrack Fe\left( {NO}_{2} \right) \right\rbrack{SO}_{4}$
C.$N_{2}O$ and $\left\lbrack Fe\left( N_{2}O \right) \right\rbrack{SO}_{4}$
D.$N_{2}O_{4}$ and $\left\lbrack Fe\left( N_{2}O_{4} \right) \right\rbrack{SO}_{4}$
Solution
Sol. [Key Concept: Brown-ring test: $NO + {FeSO}_{4} \rightarrow \lbrack Fe(NO)\rbrack{SO}_{4}$, so gas X is $NO$ and the brown compound Y is $\lbrack Fe(NO)\rbrack{SO}_{4}$.]
$$NO + {FeSO}_{4}\overset{\phantom{\text{~}\text{dil.}\text{~}H_{2}{SO}_{4}\ }}{\rightarrow}\lbrack Fe(NO)\rbrack{SO}_{4}$$
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