Question

Zinc (II) ion on reaction with $NaOH$ first give a white precipitate which dissolves in excess of $NaOH$ due to the formation of:

• A. $ZnO$
• B. $Zn{\left(OH\right)}_2$
• C. ${\left[Zn{\left(OH\right)}_4\right]}^{2-}$
• D. ${\left[Zn{\left(H_{2}O\right)}_4\right]}^{2+}$

Answer 1

The correct option is C ${\left[Zn{\left(OH\right)}_4\right]}^{2-}$

Zinc (II) ion such as one present in zinc nitrate reacts with sodium hydroxide to form sodium nitrate and a white precipitate of zinc hydroxide.

$2NaOH+Zn(N{O}_3{)}_2\to 2NaN{O}_3+\underset{whiteprecipitate}{Zn(OH{)}_2}$

Zinc hydroxide(white ppt.) reacts with excess sodium hydroxide to form ${\left[Zn{\left(OH\right)}_4\right]}^{2-}$ and gets dissolved.

$Zn(OH{)}_2+\underset{excess}{2NaOH}\to 2N{a}^{+}+{\left[Zn{\left(OH\right)}_4\right]}^{2-}$