Question

Filtrate obtained after separation of white solid contains_________.

• A. $ZnO$
• B. ${Na}_2{ZnO}_2$
• C. $MnO$
• D. ${Na}_2{MnO}_2$

Answer 1

The correct option is B ${Na}_2{ZnO}_2$

The original solution contains Zn(II) and Mn(II) ions.
Zn(II) ions react with NaOH to form soluble sodium zincate. This becomes a part of filtrate.
$Z{n}^{2+}+2NaOH\to N{a}_{2}Zn{O}_2+2H^{+}$
Mn(II) ions react with NaOH to form $Mn(OH{)}_2$ which is a white solid and
changes to brown $Mn{O}_2$. This is retained on the filter paper.
$M\left(II\right)+2NaOH\to Mn\left(OH{)}_2\right(white)+2N{a}^{+}$
$Mn\left(OH{)}_2\stackrel{oxidation}{\to }Mn{O}_2\right(brown)+H_2$
HCl reacts with sodium zincate to produce a white ppt which dissolves in excess HCl.
$N{a}_{2}Zn{O}_2+2HCl\to Zn(OH{)}_2\downarrow (white)+2NaCl$
$Zn(OH{)}_2+2HCl\to ZnC{l}_2+2H_{2}O$
When the residue of the filter paper is oxidized with a strong oxidizing agent, reddish violet coloured $Mn{O}_4^{-}$ is obtained.
$Mn{O}_2\stackrel{2\left[O\right]}{\to }Mn{O}_4^{2-}$