A) With \[Li\] to form \[LiH\]
B) With \[{{I}_{2}}\] to give HI
C) With S to give \[{{H}_{2}}S\]
D) None of the above
Correct Answer: A
Solution :
Key Idea: Oxidising agent is one which undergoes reduction. Decreasing in oxidation number shows reduction. Find oxidation number in all given reactions to decide in which of them hydrogen is oxidising agent. [a] \[Li={{\overset{0}{\mathop{H}}\,}_{2}}\xrightarrow{{}}\overset{+1}{\mathop{2Li}}\,\,\overset{-1}{\mathop{H}}\,\] Oxidation number of hydrogen is decreasing from 0 to -1. So \[{{H}_{2}}\] is acting as oxidising agent in this reaction. [b] \[{{\overset{0}{\mathop{H}}\,}_{2}}+{{I}_{2}}\xrightarrow{{}}\overset{+1}{\mathop{2HI}}\,\] \[\because \] Oxidation number of hydrogen is increasing. Hz is not acting as oxidising agent in this reaction. [c] \[\overset{0}{\mathop{{{H}_{2}}}}\,+S\xrightarrow{{}}\overset{+1}{\mathop{{{H}_{2}}}}\,S\] \[\because \] Oxidation number of hydrogen is increasing \[\therefore \] \[{{H}_{2}}\] is not acting as oxidising agent in this reaction. \[\therefore \] [a] is correct answer.
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