The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives

We are given the reaction of zinc with excess aqueous alkali. Zinc is an amphoteric metal, meaning it can react with both acids and bases.

When zinc reacts with excess aqueous NaOH (alkali), the reaction proceeds as:

$$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2 \uparrow$$

Here, zinc dissolves in the excess alkali solution and reduces water to liberate hydrogen gas. The sodium zincate $$Na_2ZnO_2$$ formed dissociates in solution to give the zincate ion $$[ZnO_2]^{2-}$$.

Now, one might wonder why the product is $$[ZnO_2]^{2-}$$ rather than $$Zn(OH)_2$$ or $$[Zn(OH)_4]^{2-}$$. When the alkali is in excess, any $$Zn(OH)_2$$ initially formed further reacts with NaOH:

$$Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O$$

This converts the hydroxide completely into the zincate ion $$[ZnO_2]^{2-}$$. The species $$ZnO$$ and $$Zn(OH)_2$$ are only intermediate products and do not persist when alkali is in excess.

Hence, the correct answer is Option D: $$[ZnO_2]^{2-}$$.