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Octet Rule

Filtrate obta...

Question

Filtrate obtained after separation of white solid contains_________.

Z n O

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{N a}_{2} {Z n O}_{2}

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M n O

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{N a}_{2} {M n O}_{2}

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Solution

The correct option is B ${N a}_{2} {Z n O}_{2}$
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.
$Z n^{2 +} + 2 N a O H \to N a_{2} Z n O_{2} + 2 H^{+}$
Mn(II) ions react with NaOH to form $M n (O H)_{2}$ which is a white solid and
changes to brown $M n O_{2}$ . This is retained on the filter paper.
$M (I I) + 2 N a O H \to M n (O H)_{2} (w h i t e) + 2 N a^{+}$
$M n (O H)_{2} o x i d a t i o n - ---- \to M n O_{2} (b r o w n) + H_{2}$
HCl reacts with sodium zincate to produce a white ppt which dissolves in excess HCl.
$N a_{2} Z n O_{2} + 2 H C l \to Z n (O H)_{2} ↓ (w h i t e) + 2 N a C l$
$Z n (O H)_{2} + 2 H C l \to Z n C l_{2} + 2 H_{2} O$
When the residue of the filter paper is oxidized with a strong oxidizing agent, reddish violet coloured $M n O_{4}^{-}$ is obtained.
$M n O_{2} 2 [O] - - \to M n O_{4}^{2 -}$

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