The metallic salt `(XY)` is soluble in water.
(a) When the aqueous soluble of `(XY)` is treated with `NaOH` solution, a white precipitate `(A)` is formed. In excess of `NaOH` solution, a white precipitate `(A)` is formed. In excess of `NaOH` solution, white precipitate `(A)` dissolves to form a compound `(B)`. When this solution is boiled with soild `NH_(4) Cl`, a precipitate of compound `(C)` is formed.
(b) An aqueous solution on treatment with `BaCl_(2)` solution gives a white precipitate `(D)` white is insoluble in conc `HCl`.
( c) The metallic salt `(XY)` forms a double salt `(E)` with potassium sulphate.
Identify `(XY),(A),(B),(C),(D)` and `(E)`.

Explanation

To solve the problem step by step, we will identify each compound based on the reactions described in the question.

### Step 1: Identify the metallic salt (XY)
- The metallic salt (XY) is soluble in water and forms a white precipitate (A) when treated with NaOH.
- The white precipitate is likely aluminum hydroxide, \( Al(OH)_3 \), which suggests that the metallic salt contains aluminum ions.
- Therefore, we can assume that \( X \) is \( Al^{3+} \) and \( Y \) is \( SO_4^{2-} \) (sulfate ion) because aluminum sulfate is soluble in water.

**Conclusion:** \( XY = Al_2(SO_4)_3 \)
...