A) \[Zn\]act as oxidising agent when react with\[HN{{O}_{3}}\]
B) \[HN{{O}_{3}}\]is weaker acid than \[{{H}_{2}}S{{O}_{4}}\] and \[HCl\]
C) In electrochemical series \[Zn\] is above hydrogen
D) \[NO_{3}^{-}\] is reduced in preference to hydronium ion
Correct Answer: D
Solution :
\[Zn\]is present above \[{{H}_{2}}\] in electrochemical series. So, it liberates hydrogen gas from dilute\[HCl/{{H}_{2}}S{{O}_{4}}\]. But \[HN{{O}_{3}}\] is oxidising agent. The hydrogen obtained in this reaction is converted into\[{{H}_{2}}O\]. In \[HN{{O}_{3}},\,\,NO_{3}^{-}\] ion is reduced and give \[N{{H}_{4}}N{{O}_{3}},\,\,{{N}_{2}}O,\,\,NO\] and \[N{{O}_{2}}\] (based upon the concentration of\[HN{{O}_{3}}\]) \[[Zn+\underset{(nearly\,\,6%)}{\mathop{2HN{{O}_{3}}}}\,\xrightarrow{{}}Zn{{(N{{O}_{3}})}_{2}}+2H]\times 4\] \[HN{{O}_{3}}+8H\xrightarrow{{}}N{{H}_{3}}+3{{H}_{2}}O\] \[N{{H}_{3}}+HN{{O}_{3}}\xrightarrow{{}}N{{H}_{4}}N{{O}_{3}}\] \[4Zn+10HN{{O}_{3}}\xrightarrow{{}}4Zn{{(N{{O}_{3}})}_{2}}\]
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