Choose the correct matching $:-$

Column-$I$	Column-$II$
$P$. $NaBr + \text{Conc. } H_2SO_4 \rightarrow$	$A$. Colourless gas
$Q$. $Na_2S + \text{dil. } HCl \rightarrow$	$B$. Brown colour
$R$. $NaNO_2 + \text{dil. } HCl \rightarrow$	$C$. Rotten egg smell
$S$. $NaNO_3 + \text{Conc. } H_2SO_4 \rightarrow$	$D$. Paramagnetic

During the qualitative analysis of $SO_{3}^{2-}$ using dilute $H_{2}SO_{4}$,$SO_{2}$ gas is evolved which turns $K_{2}Cr_{2}O_{7}$ solution (acidified with dilute $H_{2}SO_{4}$):

$A$ white crystalline salt $A$ reacts with dilute $HCl$ to liberate a suffocating gas $B$ and also forms a yellow precipitate. The gas $B$ turns potassium dichromate acidified with dilute $H_{2}SO_{4}$ to a green coloured solution $C$. $A$,$B$ and $C$ are respectively

In salt analysis,$Cl^{-}$,$Br^{-}$,and $I^{-}$ are identified by adding dilute $HNO_3$ followed by $AgNO_3$ solution to the salt solution. Choose the correct statements from the following:
$A$. $Cl^{-}$ gives pale yellow precipitate.
$B$. $Br^{-}$ gives pale yellow precipitate which is partially soluble in $NH_4OH$.
$C$. $I^{-}$ gives yellow precipitate which is insoluble in $NH_4OH$.
$D$. $Cl^{-}$ gives white precipitate which is soluble in $NH_4OH$.
$E$. $I^{-}$ gives yellow precipitate which is insoluble in $NH_4OH$.
Choose the correct answer from the options given below $:-$

$H_2S$ gas is not evolved when $SO_3^{2-}$ ion reacts with the following reagents.

When $Y$ is treated with $KMnO_4$,then the final precipitate is of:
$'X' \xrightarrow[BaCl_{2} \text{ solution}]{} 'Y' \quad (\text{White ppt., insoluble in dil. } HCl)$
$'X' \xrightarrow[\text{Flame test}]{} \text{Golden yellow flame}$