A) \[C{{l}_{2}},HCl\]
B) \[S{{O}_{2}},{{H}_{2}}S{{O}_{4}}\]
C) \[B{{r}_{2}},HBr\]
D) \[{{I}_{2}},HI\]
Correct Answer: D
Solution :
\[X={{I}_{2}},\] \[Y=HF\] \[3{{I}_{2}}+2N{{H}_{3}}\xrightarrow{{}}\underset{(explosive)}{\mathop{N{{H}_{3}}\cdot N{{I}_{3}}}}\,\] \[8N{{I}_{3}}\cdot N{{H}_{3}}\xrightarrow{{}}5{{N}_{2}}+{{I}_{2}}+6N{{H}_{4}}I\] \[{{I}_{2}}+{{H}_{2}}\xrightarrow{{}}\underset{Y}{\mathop{2HI}}\,\] \[3NaI+{{H}_{3}}P{{O}_{4}}\xrightarrow{\Delta }N{{a}_{3}}P{{O}_{4}}+3HI\]You need to login to perform this action.
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