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: Oxidizing 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 oxidizing agent. (a) \[Li+{{\overset{o}{\mathop{H}}\,}_{2}}\xrightarrow{{}}\overset{+1}{\mathop{2Li}}\,\overset{-1}{\mathop{H}}\,\] Oxidation number of hydrogen is decreasing from 0 to \[-1.\]So \[{{H}_{2}}\] is acting as oxidizing agent in this reaction. (b) \[{{\overset{o}{\mathop{H}}\,}_{2}}+{{I}_{2}}\xrightarrow{{}}\overset{+}{\mathop{2HI}}\,\] \[\because \] Oxidation number of hydrogen is increasing. \[{{H}_{2}}\]is not acting as oxidizing agent in this reaction. (c) \[{{\overset{o}{\mathop{H}}\,}_{2}}+S\xrightarrow{{}}\overset{+\,1}{\mathop{{{H}_{2}}}}\,S\] \[\because \]Oxidation number of hydrogen is increasing \[\therefore \] \[{{H}_{2}}\]is not acting as oxidizing agent in this reaction. \[\therefore \] (a) is correct answer.
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