A) \[3{{H}_{3}}P{{O}_{2}}\to 2{{H}_{3}}P{{O}_{2}}+P{{H}_{3}}\]
B) \[HCHO+O{{H}^{-}}\to HCO{{O}^{-}}+C{{H}_{3}}OH\]
C) \[N{{H}_{4}}N{{O}_{3}}\to {{N}_{2}}O+2{{H}_{2}}O\]
D) \[3C{{l}_{2}}+6O{{H}^{-}}\to 5C{{l}^{-}}+Cl{{O}^{-}}_{3}+3{{H}_{2}}O\]
Correct Answer: C
Solution :
[c] In \[N{{H}_{4}}N{{O}_{3}}\], there are two different N-atoms \[(NH_{4}^{+},\,\,NO_{3}^{-})\]with different oxidation numbers, thus reaction is not disproportionation.
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