A) \[NO\,\,\,and\,\,\,{{N}_{2}}O\]
B) \[N{{O}_{2}}\,\,\,and\,\,\,{{N}_{2}}O\]
C) \[{{N}_{2}}O\,\,\,and\,\,\,N{{O}_{2}}\]
D) \[N{{O}_{2}}\,\,\,and\,\,\,NO\]
Correct Answer: C
Solution :
| Reaction of Zn with dil. \[HN{{O}_{3}}\left( 20% \right)\] |
| \[4Zn+10HN{{O}_{3}}\left( dil \right)\xrightarrow{{}}\]\[4Zn{{\left( N{{O}_{3}} \right)}_{2}}+5{{H}_{2}}O+{{N}_{2}}O\] |
| Reaction of Zn with dil. \[{{\operatorname{HNO}}_{3}}\left( 10% \right)\] |
| \[5Zn+12HN{{O}_{3}}\xrightarrow{{}}\]\[5Zn{{\left( N{{O}_{3}} \right)}_{2}}+{{N}_{2}}+6{{H}_{2}}O\] |
| Reaction of Zn with dil. \[{{\operatorname{HNO}}_{3}}\left( 30% \right)\] |
| \[3Zn+8HN{{O}_{3}}\xrightarrow{{}}\]\[3Zn{{\left( N{{O}_{3}} \right)}_{2}}+2NO+4{{H}_{2}}O\] |
| Hence, Zn produces different products depending upon percentage of dilute concentration. |
| Reaction of Zn with cone. \[HN{{O}_{3}}\] |
| \[\operatorname{Zn}+4HN{{O}_{3}}\left( conc. \right)\xrightarrow{{}}\]\[\operatorname{Zn}{{\left( N{{O}_{3}} \right)}_{2}}+2{{H}_{2}}O+2N{{O}_{2}}\] |
| Hence, the correct option will be [c]. |
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