Question

When Zn is treated with an excess of NaOH, the product obtained is:
$\begin{aligned} & a)Zn{{(OH)}_{2}} \\\ & b)ZnOH \\\ & c)N{{a}_{2}}Zn{{O}_{2}} \\\ & d)None\text{ }of\text{ }these. \\\ \end{aligned}$

Answer 1

Hint: Remember that the reactivity of Zinc is lesser than that of Sodium, so no displacement reaction occurs in the given circumstances. With this in mind, try to figure out which reaction really does take place in the given question.

Step-by-Step Solution:
Let us first see if it is electronically feasible for Zinc Hydroxide to exist.

Being a transition element, Zinc exhibits all oxidation states from -2 to 2 as a result of its amphoteric nature. Therefore, we cannot just yet rule out the formation of Zinc Hydroxide as a result of this reaction.

However, let us now recall the order of reactivity of metals. In said order, Sodium ranks among the most reactive of metals while Zinc lies in the middle of the table. This establishes that Zinc is far less reactive than Sodium. As a result of this phenomenon, Zinc cannot displace Sodium from its compounds. Therefore, in the given scenario we can conclude that no displacement reaction takes place and there is no formation of Zinc Hydroxide.

Now that this has been established, let us now look at the formation of dry Sodium Zincate.
Again, according to the metal reactivity order, we observe that the reactivity of Zinc is higher than that of Hydrogen, enabling it to displace the latter from its compounds. Therefore, when Zinc reacts with excess NaOH, it results in the formation of anhydrous Sodium Zincate, making the answer to this question c).

$2Na\left( OH \right)+ZnO\text{ }\to \text{ }N{{a}_{2}}Zn{{O}_{2}}\text{ }+{{H}_{2}}O$
Note:
An effective way of remembering the chemical formula of Sodium Zincate is that it is amalgamation of the chemical formulae of the natural oxides of Sodium ($N{{a}_{2}}O$) and Zinc (ZnO). However, do not ever be under the illusion that Sodium Zincate will be formed by the reaction between Sodium and Zinc Oxide.