A sodium salt on treatment with $M g C l_{2}$ gives white precipitate only one heating. The anion of the 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_{4}^{2 -}

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Solution

The correct option is A $H C O_{3}^{-}$
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.

$M g C l_{2} + 2 N a H C O_{3} \to M g (H C O_{3})_{2} (S o l u b l e) + 2 N a C l M g (H C O_{3})_{2} H e a t ⟶ M g C O_{3} (w h i t e) ↓ + H_{2} O + C O_{2}$

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