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(B) $P$. $NaBr + 	ext{Conc. } H_2SO_4 ightarrow NaHSO_4 + HBr$. $HBr$ is oxidized by $	ext{conc. } H_2SO_4$ to $Br_2$ (Brown colour) and $SO_2$. $

Vedclass · Accepted Answer

$P-B; Q-C; R-A, D; S-B, D$

Vedclass · Answer

(B) $P$. $NaBr + 	ext{Conc. } H_2SO_4 ightarrow NaHSO_4 + HBr$. $HBr$ is oxidized by $	ext{conc. } H_2SO_4$ to $Br_2$ (Brown colour) and $SO_2$. $Br_2$ is diamagnetic,but $NO_2$ is paramagnetic. Wait,$Br_2$ is brown. $P$ gives $B$.$Q$. $Na_2S + 	ext{dil. } HCl ightarrow 2NaCl + H_2S$. $H_2S$ has a rotten egg smell. $Q$ gives $C$.$R$. $NaNO_2 + 	ext{dil. } HCl ightarrow NaCl + HNO_2$. $HNO_2$ decomposes to $NO$ (colourless gas,paramagnetic). $R$ gives $A, D$.$S$. $NaNO_3 + 	ext{Conc. } H_2SO_4 ightarrow NaHSO_4 + HNO_3$. $HNO_3$ decomposes to $NO_2$ (brown colour,paramagnetic). $S$ gives $B, D$.Thus,the correct matching is $P-B; Q-C; R-A, D; S-B, D$.

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

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