A sodium salt of an unknown anion when treated with $M g C l_{2}$ gives white precipitate only on boiling. The anion is :

S O_{4}^{2 -}

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H C O_{3}^{-}

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C O_{3}^{2 -}

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N O_{3}^{-}

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Solution

The correct option is B $H C O_{3}^{-}$
Option (B) is correct.
The anion is $H C O_{3}^{-}$ .
It must be a bicarbonate ion because first magnesium bicarbonate is formed which is soluble, then on heating, magnesium carbonate is formed which is insoluble and forms a precipitate.

$\begin{matrix} M g C l_{2} & + & 2 N a H C O_{3} & ⟶ & M g (H C O_{3})_{2} & + & 2 N a C l s o l u b l e \end{matrix}$
$\begin{matrix} M g (H C O_{3})_{2} & Δ - \to & M g C O_{3} ↓ & + & H_{2} O & + & C O_{2} W h i t e \end{matrix}$

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