Zinc reacts with excess of caustic soda to form

Zinc reacts with excess of caustic soda to form
  1. $\mathrm{Zn}(\mathrm{OH})_{2}$
  2. $\mathrm{ZnO}$
  3. $\mathrm{Na}_{2} \mathrm{ZnO}_{2}$
  4. $\mathrm{Zn}(\mathrm{OH})_{2} \mathrm{ZnCO}_{3}$

Solution

$\mathrm{Zn}+\mathrm{NaOH} \longrightarrow \mathrm{Zn}(\mathrm{OH})_{2}$
$\mathrm{Zn}(\mathrm{OH})_{2}+\mathrm{NaOH} \longrightarrow \mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{OH})_{4}ight]$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad$ or $\mathrm{Na}_{2} \mathrm{ZnO}_{2}.2 \mathrm{H}_{2} \mathrm{O}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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