A) \[2N{{O}_{2}}\to {{N}_{2}}{{O}_{4}}\]
B) \[N{{H}_{4}}OH\to N{{H}_{4}}^{+}+O{{H}^{-}}\]
C) \[{{N}_{2}}{{O}_{5}}+{{H}_{2}}O\to 2HN{{O}_{3}}\]
D) \[2N{{O}_{2}}+{{H}_{2}}O\to HN{{O}_{3}}+HN{{O}_{2}}\]
Correct Answer: D
Solution :
In reaction [d] oxidation number changes from + 4 in \[N{{O}_{2}}\]to + 3 in \[HN{{O}_{2}}\]and + sin \[HN{{O}_{3}}\]
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