Question

The $Z{{n}^{+2}}$upon reaction with NaOH first gives out a white precipitate which dissolves in excess NaOH due to the presence of:
$\begin{align}
  & a)ZnO \\
 & b)Zn{{(OH)}_{2}} \\
 & c){{[Zn{{(OH)}_{4}}]}^{2-}} \\
 & d){{[Zn{{({{H}_{2}}O)}_{4}}]}^{2+}} \\
\end{align}$

Answer 1

Hint: Follow the sequence of reactions that leads to the formation of the white precipitate. Then try and point out the complex the formation of which results in the dissolution of the white precipitate formed earlier.

Complete step-by-step answer:
Let us go through this reaction one step at a time.
The initial reaction between the $Z{{n}^{+2}}$ion with NaOH results in the production of $Zn(OH)_2$, a compound that exists as a solid white precipitate.
$Z{{n}^{+2}}+2NaOH\to 2N{{a}^{+}}+Zn{{(OH)}_{2}}\downarrow $
Now, let us very carefully break down the next step of reaction.
When this white precipitate of Zinc Hydroxide is left in excess NaOH, it dissolves due to the production of ${{[Zn{{(OH)}_{4}}]}^{2-}}$.
${{[Zn{{(OH)}_{4}}]}^{2-}}$will dissolve because the ion is normally surrounded by water ligands; when there is excess sodium hydroxide, the hydroxide ions displace the water ligands and the complex will acquire a −2 charge, making it soluble.
The other compounds i.e. ZnO and the complex Zinc forms with 4 water ligands in aqueous medium i.e. ${{[Zn{{({{H}_{2}}O)}_{4}}]}^{2+}}$as not only are neither of them produced as a result of this reaction, but the former is a gas that cannot dissolve in NaOH while the latter is a positive complex which is also insoluble in NaOH.

NOTE: Be extremely cautious of the differences between ${{[Zn{{(OH)}_{4}}]}^{2-}}$and ${{[Zn{{({{H}_{2}}O)}_{4}}]}^{2+}}$ along with their resultant implications. Remember that only when there is a negative charge on the complex does it become soluble in NaOH to ensure that you do not mess up in your solution.