A) a redox reaction
B) a hydrolysis reaction
C) a solvolysis reaction
D) an oscillatory reaction
E) disproportionation
Correct Answer: E
Solution :
\[\underset{+1-1}{\mathop{{{H}_{2}}{{O}_{2}}}}\,+\underset{+1-1}{\mathop{{{H}_{2}}{{O}_{2}}}}\,\xrightarrow{{}}2\underset{+1-2}{\mathop{{{H}_{2}}O}}\,+\underset{0}{\mathop{{{O}_{2}}}}\,\] Here the oxidation number of hydrogen is not changed while the oxidation number of oxygen in\[{{H}_{2}}{{O}_{2}},{{H}_{2}}O\]and\[{{O}_{2}}\]is\[-1,-2\]and 0. Hence, oxygen in undergoing oxidation to\[{{O}_{2}}\](\[-1\]to 0) as well as reduction to\[{{H}_{2}}O(-1\text{ }to-2)\]. Such reactions where a molecule acts as both oxidising and reducing agent are called disproportionation.
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