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An inorganic halide (A) reacts with wate...

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)`.

Text Solution

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`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`.
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