A) Fe metal and cone. \[HN{{O}_{3}}\]
B) Cu metal and cone. \[HN{{O}_{3}}\]
C) Zn metal and \[NaOH\text{ }(aq)\]
D) Au metal and \[NaCN\text{ }(aq)\] in the presence of air
Correct Answer: C
Solution :
| Fe becomes passive on reaction with concentrated\[HN{{O}_{3}}\]. However, cold relatively cone.\[{{\operatorname{HNO}}_{3}}\], reacts with Fe as below. |
| \[Fe+6HN{{O}_{3}}\to Fe{{\left( N{{O}_{3}} \right)}_{3}}+3N{{O}_{2}}+3{{H}_{2}}O\] |
| \[Fe+6HN{{O}_{3}}\to Fe{{\left( N{{O}_{3}} \right)}_{3}}+3N{{O}_{2}}+3{{H}_{2}}O\]\[+2{{H}_{2}}O\] |
| \[4Au+8NaCN+{{O}_{2}}+2{{H}_{2}}O\to \]\[4Na[Au {{(CN)}_{2}}]+ 4NaOH\] |
| \[Zn+2NaOH\to N{{a}_{2}}Zn{{O}_{2}}+{{H}_{2}},\] |
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