Question

Which of the following combinations will produce \[{{\rm{H}}_2}\] gas?
(A) \[{\rm{Au}}\] metal and \[{\rm{NaCN(aq)}}\] in the presence of air
(B) \[{\rm{Cu}}\] metal and conc. \[{\rm{HN}}{{\rm{O}}_{\rm{3}}}\]
(C) \[{\rm{Fe}}\] metal and conc. \[{\rm{HN}}{{\rm{O}}_{\rm{3}}}\]
(D) \[{\rm{Zn}}\] metal and \[{\rm{NaOH}}\,{\rm{(aq)}}\]

Answer 1

Hint: As we know that, the metals of transition series can reduce the number of compounds because these elements are rich in electrons and have d- orbital. Hydrogen gas in any reaction is released when the metal has more reducing power and weaker oxidizing power. If metal has more reducing power, it can give its electrons to hydrogen and then the hydrogen gas will be released.

Complete answer
In the transition rows, the zinc atom is the more reducing element as it has filled d- orbital and increased size among the first transition row. The metal has a tendency to reduce hydrogen as well because its reducing potential in electrochemical series is found below to the hydrogen.
The release of hydrogen gas by zinc occurs as we can see in the following reaction-
\[{\rm{Zn}} + {\rm{ NaOH}} \to {\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{Zn}}{{\rm{O}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}\]
Now let’s move towards the other options
In option (A), gold is a third- row transition metal and when it reacts with sodium cyanide, it forms complex by the coordination bonds as-
\[{\rm{4Au}} + {\rm{8NaCN + }}{{\rm{O}}_{\rm{2}}}{\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{4Na}}\left[ {{\rm{Au}}{{\left( {{\rm{CN}}} \right)}_{\rm{2}}}} \right]{\rm{ + 4NaOH}}\]
In option (B), copper metal is found above than hydrogen in electrochemical series so, it cannot produce hydrogen gas as-
\[{\rm{Cu}} + {\rm{ 4HN}}{{\rm{O}}_3} \to {\rm{Cu}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{2}}} + 2{{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{2N}}{{\rm{O}}_{\rm{2}}}\]
 In option (C), iron metal is also found above than hydrogen in electrochemical series so, it cannot produce hydrogen gas as-
\[{\rm{Fe}} + {\rm{ 6HN}}{{\rm{O}}_3} \to {\rm{Fe}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{3}}} + {\rm{3}}{{\rm{H}}_{\rm{2}}}{\rm{O}} + {\rm{3N}}{{\rm{O}}_{\rm{2}}}\]

Therefore, the correct option is option (D).

Note:
Generally, combination reactions are disproportionation reactions in which reduction and oxidation occurs simultaneously.The same element oxidizes and reduces at the same time.