A) \[NO\]
B) \[{{N}_{2}}\]
C) \[N{{O}_{2}}\]
D) \[{{N}_{2}}O\]
Correct Answer: D
Solution :
| Cu produces NO (X) gas when treated with dil.\[HN{{O}_{3}}.\] This gas is reduced to \[N{{H}_{2}}OH\cdot HCI\,(Y)\] by \[SnC{{l}_{2}}/HCl\] and then oxidised by \[HN{{O}_{2}}\] give \[{{N}_{2}}O\,(Z).\]The reaction is as follows: |
| \[Cu+Dil.HN{{O}_{3}}\xrightarrow{{}}\underset{(X)}{\mathop{NO}}\,\xrightarrow{SnC{{l}_{2}}/HCl}\]\[N{{H}_{2}}\underset{(Y)}{\mathop{OH}}\,\cdot HCl\xrightarrow{HN{{O}_{2}}}\underset{(Z)}{\mathop{{{N}_{2}}O}}\,\] |
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