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

Zinc dissolves in excess of aqueous alkali. <br/><br/> $$\mathrm{Zn}+2 \mathrm{OH}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}+\mathrm{H}_{2} \uparrow$$ (Tetrahydroxozincate(II) ion) <br/><br/> However, this reaction in NCERT is given as <br/><br/> $$\mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2} \uparrow$$ <br/><br/> $\mathrm{ZnO}_{2}^{2-}$ is anhydrous form of $\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$. <br/><br/> $$\mathrm{ZnO}_{2}^{2-}+2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$$ <br/><br/> So in aqueous medium best answer of this question is $\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$.