$A$ substance reacts with dilute $H_2SO_4$ to release a colorless gas which $(i)$ turns baryta water turbid and $(ii)$ turns acidic dichromate solution green. This reaction indicates the presence of which of the following?
- A$CO_3^{2-}$
- B$S^{2-}$
- C$SO_3^{2-}$
- D$NO_2^-$
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Similar Questions
When $BaCl_2$ is added to a clear solution of a compound $(X)$,a heavy white precipitate is formed which does not dissolve in $dil. HCl$. What is the compound $(X)$?
Easy
View Solution$S^{2-}$ and $SO_3^{2-}$ can be distinguished by using
DifficultAIIMS 2008
View SolutionAmong the following observations, the correct one that differentiates between $SO_{3}^{2-}$ and $SO_{4}^{2-}$ is:
When $Y$ is treated with $KMnO_4$,then the final precipitate is of:
$'X' \xrightarrow[BaCl_{2} \text{ solution}]{} 'Y' \quad (\text{White ppt., insoluble in dil. } HCl)$
$'X' \xrightarrow[\text{Flame test}]{} \text{Golden yellow flame}$
$'X' \xrightarrow[BaCl_{2} \text{ solution}]{} 'Y' \quad (\text{White ppt., insoluble in dil. } HCl)$
$'X' \xrightarrow[\text{Flame test}]{} \text{Golden yellow flame}$
Difficult
View SolutionThe reagent$(s)$ that can selectively precipitate $S^{2-}$ from a mixture of $S^{2-}$ and $SO_4^{2-}$ in aqueous solution is(are):
$(A)$ $CuCl_2$
$(B)$ $BaCl_2$
$(C)$ $Pb(CH_3COO)_2$
$(D)$ $Na_2[Fe(CN)_5NO]$
$(A)$ $CuCl_2$
$(B)$ $BaCl_2$
$(C)$ $Pb(CH_3COO)_2$
$(D)$ $Na_2[Fe(CN)_5NO]$
MediumIIT 2016
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