A sodium salt on treatment with $M g C l_{2}$ gives white precipitate only on heating. The anion of sodium salt is:

H C O_{3}^{-}

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C O_{3}^{2 -}

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N O_{3}^{-}

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S O_{3}^{2 -}

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Solution

The correct option is A $H C O_{3}^{-}$
Sodium bicarbonate reacts with $M g C l_{2}$ to give magnesium bicarbonate, $M g (H C O_{3})_{2}$ which on heating produces $M g C O_{3}$ which is white in color.
Reactions:
$M g C l_{2} + 2 N a H C O_{3} \to M g (H C O_{3})_{2} + 2 N a C l$
$M g (H C O_{3})_{2} Δ - \to M g C O_{3} White precipitate + H_{2} O + C O_{2}$
Hence, the sodium salt is sodium bicarbonate and the anion of sodium salt is $H C O_{3}^{-}$ .

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