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If NaOH is added to an aqueous solution of ${Zn}^{2 +}$ ions, a white precipitate appears and on adding excess NaOH, the precipitate dissolves. In this solution zinc exists in the :

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both in cationic and anionic parts

there is no zinc left in the solution

cationic part

anionic part.

answer is D.

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Detailed Solution

${Zn}^{+ +} + 2 NaOH ⟶ Zn (OH)_{2} + 2 {Na}^{+} Zn (OH)_{2} + 2 NaOH ⟶ {Na}_{2} {ZnO}_{2} + 2 H_{2} O {Na}_{2} {ZnO}_{2} ⟶ 2 {Na}^{+} + {({ZnO}_{2})}^{2 -}$

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