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

A sodium salt on treatment with MgCl_(2) gives white precipitate on he

1.	A sodium salt on treatment with MgCl_(2) gives white precipitate on heating. The anion of the sodium salt is:
Answer» `CO_(3)^(2-)` `HCO_(3)^(Θ)` `SO_(4)^(2+)` `NO_(3)^(Θ)` Solution :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)`.