
which of the following metals on reacting with sodium hydroxide solution produce hydrogen gas?
(a) \[{\rm{Cu}}\]
(b) \[{\rm{Al}}\]
(c) \[{\rm{Fe}}\]
(d) \[{\rm{Zn}}\].
(A) b and c
(B) b and d
(C) a and d
(D) b only
Answer
555k+ views
Hint:
As we know that the metals have a tendency to donate electrons because the metals are rich in electrons and have very less effect of nuclear charge. The transition metals also have property to exist in a variable oxidation state. As we go from left to right, the tendency to lose electrons also reduces.
Complete step by step solution
The atoms which are given in our question are transition metals except aluminium.
The aluminium element is present at right to the periodic table it exists in compound as \[{\rm{A}}{{\rm{l}}^{3 + }}\] that means it is reducing in nature. The sodium hydroxide when reacted with aluminium it gives sodium aluminate and hydrogen gas.
\[{\rm{2Al + 2NaOH + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{2NaAl}}{{\rm{O}}_{\rm{2}}}{\rm{ + 3}}{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\]
Now come on the transition metal, copper has positive standard electrode potential and it lies below hydrogen in the electrochemical series as it has unpaired electrons, these electrons want to pair up and hence the electrode potential is positive.
When it reacts with sodium hydroxide it gives copper hydroxide but decomposes into copper oxide very quickly.
Iron has the property to react with acids and do not react with base at room temperature because the iron is electron rich element and also has not much effective nuclear charge.
Zinc is a reducing metal and exists as \[{\rm{Z}}{{\rm{n}}^{2 + }}\]. It is an amphoteric metal which means it can react with acid and bases. So, zinc reacts with sodium hydroxide to give sodium zincate and hydrogen gas as-
\[{\rm{2Zn + 2NaOH + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{2NaZn}}{{\rm{O}}_{\rm{2}}}{\rm{ + 3}}{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\]
Therefore, our correct option is option (B).
Note:
The amphoteric nature of elements is found in between groups and in between the periodic table. As we go down the group basic character increases.
As we know that the metals have a tendency to donate electrons because the metals are rich in electrons and have very less effect of nuclear charge. The transition metals also have property to exist in a variable oxidation state. As we go from left to right, the tendency to lose electrons also reduces.
Complete step by step solution
The atoms which are given in our question are transition metals except aluminium.
The aluminium element is present at right to the periodic table it exists in compound as \[{\rm{A}}{{\rm{l}}^{3 + }}\] that means it is reducing in nature. The sodium hydroxide when reacted with aluminium it gives sodium aluminate and hydrogen gas.
\[{\rm{2Al + 2NaOH + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{2NaAl}}{{\rm{O}}_{\rm{2}}}{\rm{ + 3}}{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\]
Now come on the transition metal, copper has positive standard electrode potential and it lies below hydrogen in the electrochemical series as it has unpaired electrons, these electrons want to pair up and hence the electrode potential is positive.
When it reacts with sodium hydroxide it gives copper hydroxide but decomposes into copper oxide very quickly.
Iron has the property to react with acids and do not react with base at room temperature because the iron is electron rich element and also has not much effective nuclear charge.
Zinc is a reducing metal and exists as \[{\rm{Z}}{{\rm{n}}^{2 + }}\]. It is an amphoteric metal which means it can react with acid and bases. So, zinc reacts with sodium hydroxide to give sodium zincate and hydrogen gas as-
\[{\rm{2Zn + 2NaOH + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{2NaZn}}{{\rm{O}}_{\rm{2}}}{\rm{ + 3}}{{\rm{H}}_{\rm{2}}}{\rm{(g)}}\]
Therefore, our correct option is option (B).
Note:
The amphoteric nature of elements is found in between groups and in between the periodic table. As we go down the group basic character increases.
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