Zinc dissolves in excess of aqueous alkali

$\begin{array}{l}Zn+2O{H}^{-}+2H_{2}O\to {\left[Zn{\left(OH\right)}_4\right]}^{2-}+H_2\uparrow \\ Tetrahydorxozincate\left(ii\right)ion\end{array}$

However, this reaction in NCERT is given as

$Zn+2NaOH\text{}\to N_{2}Zn{O}_2+H_2\uparrow$

$Zn{O}_2^{-}$ is anhydrous form of ${\left[Zn{\left(OH\right)}_4\right]}^{2-}$

So in aqueous medium best answer of this question is 

${\left[Zm{\left(OH\right)}_4\right]}^{2-}$