Which of the following is the white precipitate for “Two metals X and Y form the salts XSO₄ and Y₂SO₄, respectively. The solution of salt XSO₄ is blue in colour, whereas that of Y₂SO₄ is colourless. When barium chloride solution is added to XSO₄ solution, then a white precipitate Z is formed along with a salt which turns the solution green. And then barium chloride solution is added to Y₂SO₄ solution, and then the same white precipitate Z is formed along with the colourless common salt solution”?

Barium sulphate(BaSO₄) is the white precipitate for “Two metals X and Y form the salts XSO₄ and Y₂SO₄, respectively. The solution of salt XSO₄ is blue in colour, whereas that of Y₂SO₄ is colourless. When barium chloride solution is added to XSO₄ solution, then a white precipitate Z is formed along with a salt which turns the solution green. And then barium chloride solution is added to Y₂SO₄ solution, and then the same white precipitate Z is formed along with the colourless common salt solution”.
When XSO₄ reacts with BaCl₂, it leads to the formation of XCl₂ (green) and BaSO₄ (white precipitate). The green colour solution must be CuCl₂, so metal X would be copper (Cu), and the salt XSO₄ would be CuSO₄. 
The reaction involved is-
CuSO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) + CuCl₂ (aq)
When Y₂SO₄ reacts with BaCl₂, it leads to the formation of NaCl (common salt) and BaSO₄ (white precipitate). So, the metal Y is sodium (Na), and the salt Y₂SO₄ would be Na₂SO₄. The reaction involved is-
Na₂SO₄ (aq) + BaCl₂ (aq) → BaSO₄ (s) + 2NaCl (aq) 
Hence, the white precipitate is barium sulphate (BaSO₄).