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

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 options are
C $H C O_{3}^{-}$
D $C O_{3}^{2 -}$
$N a_{2} S O_{4} + M g C l_{2} \to M g S O_{4} + 2 N a C l$
$M g C l_{2} + 2 N a H C O_{3} \to M g C O_{3} + 2 N a C l + H_{2} O + C O_{2}$ .( $M g c o_{3}$ is white ppt)
$M g C l_{2} + N a_{2} C O_{3} \to M g C O_{3} + N a C l$ .( $M g c o_{3}$ is white ppt)
$M g C l_{2} + 2 N a N O_{3} \to M g (N O_{3})_{2} + 2 N a C l$
Hence option B & C are correct

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