A sodium salt on treatment with `MgCl_(2)` gives white precipitate on heating. The anion of the sodium salt is:

Options

A. `CO_(3)^(2-)`
B. `HCO_(3)^(Θ)`
C. `SO_(4)^(2+)`
D. `NO_(3)^(Θ)`

Answer

B

Explanation

Given, `NaX+MgCl_(2)overset(Delta)(rarr)`White ppt.
`NaHCO_(3)+MgCl_(2)rarrunderset("Soluble")(Mg(HCO_(3))_(2))+NaCl`
`Mg(HCO_(3))_(2)overset(Delta)(rarr)underset("White ppt".)(MgCO_(3))+H_(2)O+Co_(2)`
Hence, anion of the sodium salt is `HCO_(3)^(Θ)`.