Question

Excess of NaOH reacts with Zn to form:
(A) $Zn{(OH)_2}$
(B) ZnO
(C) $Zn{H_2}$
(D) $N{a_2}Zn{O_2}$

Answer 1

Hint: Excess of NaOH reacts with Zn to form this oxozincates salt, which is colourless and is a salt of metal and a base. Such salts are very good conductors of electricity in aqueous solution.

Complete step by step answer:
-We all know that zinc is a transition metal and it shows moderate reactivity. Also it is amphoteric in nature and according to the activity series or activity table; the reactivity of zinc is much lower than that of sodium. Hence when we react zinc with excess NaOH it leads to the production of sodium zincate ($N{a_2}Zn{O_2}$) and hydrogen gas. Here the valency of the entire zincate radical as a whole would be 2.
The reaction can be written as:
$Zn + 2NaOH \to N{a_2}Zn{O_2} + {H_2}(g)$
-Sodium zincate is basically a colourless salt formed by the reaction of a metal and a strong base. They can be either anionic zinc oxides or hydroxides (depending on the conditions).
-Mostly the hydroxyl sodium zincates (also known as hydroxyzincates) are prepared by dissolving zinc, zinc oxide or zinc hydroxide in aqueous solution of sodium hydroxide. This can be simplified in the form of following reactions:
$ZnO + {H_2}O + 2NaOH \to N{a_2}Zn{(OH)_4}$
$Zn + 2{H_2}O + 2NaOH \to N{a_2}Zn{(OH)_4} + {H_2}$
The different salts of sodium zincate that can be crystallized out are: $Zn(OH)_4^{2 - }$, $N{a_2}Zn{(OH)_4}$, $Zn(OH)_6^{4 - }$, etc. These salts contain the tetrahedral zincate ions and the octahedral sodium cations.
-The oxide sodium zincates also known as oxozincates, that can be formed are: $N{a_2}Zn{O_2}$, $N{a_2}Z{n_2}{O_3}$, $N{a_{10}}Z{n_4}{O_9}$, etc.
 -When dissolved in water, these salts release ions which conduct electricity. Hence they are good conductors of electricity in aqueous solution.

So, the correct option will be: (D) $N{a_2}Zn{O_2}$

Note: Do not get confused between $N{a_2}Zn{O_2}$ and $Zn{(OH)_2}$. When we react Zn with a normal amount of NaOH then the product formed will be $Zn{(OH)_2}$ (Zinc hydroxide) but if we react it with excess of NaOH then $N{a_2}Zn{O_2}$ (sodium zincate) is formed.