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

Since metal `(X)` reacts with `NaOH` to form `(A)`,which reacts both with an acid and a base to form a soluble complex `(B)`.`(A)` is amphoteric and metal `(X)` is aluminium,`Al`.
`underset((X))(2Al)+NaOHtounderset(("white ppt."))underset("hydroxde")underset("(A),Alumiuium")((Al(OH)_(3))darr)+3Na^(o+)`
`underset(("A"))(Al(OH)_(3))+underset("Excess")(NaOH)rarrunderset(("Sodium complex"))underset("Sodium tetrhydroxoaluminate(III)")(Na^(o+)[Al(OH)_(4)]^(ө))`
`underset(("A"))(Al(OH)_(3))underset("dil")(3HCl)overset(Delta)tounderset("chloride")underset("Aluminium")underset("(C)")(AlCl_(3))+3H_(2)O`
`underset(("A"))(2Al(OH)_(3))overset(Delta)to underset(("D"))(Al_(2)O_(3))+3H_(2)O`
Hence,`(X)` is `Al`
`(A)` is `Al(OH)``
`(B)` is Na^(o+)[Al(OH)_(4)]^(ө)`
`(C) is AlCl_(3)`
`(D) is Al_(2)O_(3)`.