When salt $BA$ is treated with concentrated $H_2SO_4$, a reddish-brown gas is liberated. The aqueous solution of $BA$ gives a pale yellow precipitate with $AgNO_3$ solution. Which of the following anions $(A^-)$ is present in the salt $BA$?

The brown ring test is used for the detection of which radical?

Choose the correct matching:-

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

An aqueous solution of a salt,when treated with $AgNO_3$ solution,gives a white precipitate which dissolves in $NH_4OH$. The radical present in the salt is:

$A$ 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 $:-$