Two metals X and Y form the salts XSO_(4) and Y_(2)SO_(4) respectively.The solution of salt XSO_(4) is blue in colour whereas that of Y_(2)SO_(4) is formed alongiwth a salt Which turns the solution green. And when barium chloride solution is added to Y_(2)SO_(4) solution, then the same white precipitate Z is formed alongwith colourless common salt solution. (a) What could the metal X and Y be ? (b) Write the name and formula of salt XSO_(4) (c) Write the name and formula of salt Y_(2)SO_(4) (d)What is the name and formula of white precipate Z ? (e) Write the name and formula of the salt which turns the solution green in the first case.

Two metals X and Y form the salts XSO_(4) and Y_(2)SO_(4) respectively

1.	Two metals X and Y form the salts XSO_(4) and Y_(2)SO_(4) respectively.The solution of salt XSO_(4) is blue in colour whereas that of Y_(2)SO_(4) is formed alongiwth a salt Which turns the solution green. And when barium chloride solution is added to Y_(2)SO_(4) solution, then the same white precipitate Z is formed alongwith colourless common salt solution. (a) What could the metal X and Y be ? (b) Write the name and formula of salt XSO_(4) (c) Write the name and formula of salt Y_(2)SO_(4) (d)What is the name and formula of white precipate Z ? (e) Write the name and formula of the salt which turns the solution green in the first case.
Answer» Solution :(a) The available information suggests that metals X and Y are COPPER (Cu) and SODIUM (Na) respectively. (b) The salt X is `CuSO_(4)` and its aqueous solution in BLUE in colour. (c) The salt Y is `Na_(2)SO_(4)` and its aqueous solution is colourless. (d) When `BaCl_(2)` solution is added to `CuSO_(4)` solution, a white precipitate of `BaSO_(4)` (Z) is formed. The solution acquires greenish colour due to formation of `underset((X))(CuSO_(4)) + BaCl_(2) to underset("White PPT (z)")(BaSO_(4)) + underset("Greenish solution")(CuCl_(2))` (e) The green solution as stated above is of cupric chloride.