Question
When copper is heated with $$conc.$$ $$HN{O_3}$$ it produces
A.
$$Cu{\left( {N{O_3}} \right)_2}\,\,{\text{and}}\,NO$$
B.
$$Cu{\left( {N{O_3}} \right)_2},\,NO\,\,{\text{and}}\,\,N{O_2}$$
C.
$$Cu{\left( {N{O_3}} \right)_2}\,\,{\text{and}}\,\,{N_2}O$$
D.
$$Cu{\left( {N{O_3}} \right)_2}\,\,{\text{and}}\,\,N{O_2}$$
Answer :
$$Cu{\left( {N{O_3}} \right)_2}\,\,{\text{and}}\,\,N{O_2}$$
Solution :
Nitric acid acts as an oxidising agent while reacting with copper. When $$dil.$$ $$HN{O_3}$$ reacts, reaction proceeds as:
$$3Cu + 8HN{O_3}\left( {dil.} \right) \to $$ $$3Cu{\left( {N{O_3}} \right)_2} + 2NO + 4{H_2}O$$
and when conc. $$HN{O_3}$$ is used, reaction proceeds as
$$Cu + 4HN{O_3}\left( {conc.} \right) \to $$ $$Cu{\left( {N{O_3}} \right)_2} + 2N{O_2} + 2{H_2}O$$