A) \[HCl,HBr,HI\]
B) \[HCl,B{{r}_{2}},{{I}_{2}}\]
C) \[C{{l}_{2}},B{{r}_{2}},{{I}_{2}}\]
D) \[C{{l}_{2}},HBr,HI\]
Correct Answer: B
Solution :
\[NaCl+{{H}_{2}}S{{O}_{4}}\xrightarrow{\Delta }NaHS{{O}_{4}}+HCl\uparrow \] \[HCl+{{H}_{2}}S{{O}_{4}}\xrightarrow{\Delta }\text{No}\,\text{reactioin}\] \[NaBr+{{H}_{2}}S{{O}_{4}}\xrightarrow{\Delta }NaHS{{O}_{4}}+HBr\] \[2HBr+{{H}_{2}}S{{O}_{4}}\xrightarrow{\Delta }B{{r}_{2}}\uparrow +S{{O}_{2}}+2{{H}_{2}}O\] \[NaI+{{H}_{2}}S{{O}_{4}}\xrightarrow{\Delta }NaHS{{O}_{4}}+HI\] \[2HI+{{H}_{2}}S{{O}_{4}}\xrightarrow{\Delta }{{I}_{2}}\uparrow +S{{O}_{2}}+2{{H}_{2}}O\] Hence, the gases released are \[\text{HCl,}\,\text{B}{{\text{r}}_{\text{2}}}\]and \[{{\text{I}}_{2}}.\]
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