Q19 / 150 · MHT CET 2011 · Chemistry

When Zn is treated with excess of $\mathrm{NaOH}$, the product obtained is

  • A
    $\mathrm{Zn}(\mathrm{OH})_{2}$
  • B
    $\mathrm{ZnOH}$
  • Correct answer
    C
    $\mathrm{Na}_{2} \mathrm{ZnO}_{2}$
  • D
    None of the above

Solution

When $\mathrm{Zn}$ is reacted with excess of $\mathrm{NaOH}$, sodium zincate is obtained. $$ \mathrm{Zn}+2 \mathrm{NaOH}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]+\mathrm{H}_{2} $$ soluble Or $$ \mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2} $$

Hence, the correct answer is option: 'C'.

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