An inorganic halide `(A)` reacts with water to form two acids `(B)` and `(c). (A)` also reacts with `NaOH` to form two salts `(D)` and `( E)` which remain in solution. The solution gives white precipitate with both `AgNO_3` and `BaCl_2` solutions respectively. `(A)` is a useful organic reagent. Identify `(A)` to `( E)`.

An inorganic halide `(A)` reacts with water to form two acids `(B)` an

1.	An inorganic halide `(A)` reacts with water to form two acids `(B)` and `(c). (A)` also reacts with `NaOH` to form two salts `(D)` and `( E)` which remain in solution. The solution gives white precipitate with both `AgNO_3` and `BaCl_2` solutions respectively. `(A)` is a useful organic reagent. Identify `(A)` to `( E)`.
Answer» `underset("Inorganic halide") (A + H_(2) O) rarr (B) + (C)` `(A) + NaOH rarr underset(("Salts which are water soluble")) ubrace((D)+(E))` `(D) + AgNO_(3) rarr` White ppt `( E) + BaCl_(2) rarr` White ppt `AgNO_(3)` gives white ppt. on reaching with `Cl^(Ө)` ions, hence (D) is `NaCl`, `AgNO_(3) + NaCl rarr AgCl darr + NaNO_(3)` `Ag^(oplus) + Cl^(Ө) rarr AgCl darr` `BaCl_(2)` gives white ppt. with `SO_(4)^(2-)` ions, hence (D) is `Na_(2) SO_(4)`. `BaCl_(2) + Na_(2) SO_(4) rarr BaSO_(4) darr + 2 NaCl` `Ba^(2+) + SO_(4)^(2-) rarr BaSO_(4) darr` Hence (A) can be `SO_(2) Cl_(2)`, since this on reacting with water gives acids namely `H_(2) SO_(4)` and `HCl (A)` reacts with `NaOH` to form `Na_(2) SO_(4)` and `NaCl` `underset((A)) (SO_(2) CL_(2)) + 2H_(2) O rarr underset((B)) (H_(2)SO_(4)) + underset((C))(2 HCl)` `underset((A)) (SO_(2) Cl_(2)) + NaOH rarr underset((D))(Na_(2) SO_(4)) + underset((E))(NaCl) + 2 H_(2) O` Both (D) and (E) are water soluble. Hence `(A) SP_(2) Cl_(2), (B)` is `H_(2) SO_(4), (C) is HCl, (D)` is `Na_(2) SO_(4)` and (E) is `NaCl`.