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 colourless. When barium chloride solution is added to ${XSO}_{4}$ solution, a white precipitate $Z$ is formed along with a salt which turns the solution green. When barium chloride solution is added to $Y_{2} {SO}_{4}$ solution, then the same white precipitate $Z$ is formed along with a colorless common salt solution.
What could the metals $X$ and $Y$ be?

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Solution

Metals $X$ and $Y$
Metal $X$ is copper $(Cu)$ and metal $Y$ is sodium $(Na)$ .
Proof that metal $X$ is copper $(Cu)$
${CuSO}_{4}$ is blue in color and reacts with barium chloride $({BaCl}_{2})$ to form a white precipitate of barium sulfate $({BaSO}_{4})$ and green colored solution of copper(II) chloride $({CuCl}_{2})$ .
${BaCl}_{2} (aq) + {CuSO}_{4} (s) \to {BaSO}_{4} (s) + {CuCl}_{2} (aq)$
Proof that metal $Y$ is sodium $(Na)$ .
${Na}_{2} {SO}_{4}$ is colorless and reacts with barium chloride $({BaCl}_{2})$ to form a white precipitate of barium sulfate $({BaSO}_{4})$ and a colorless solution of sodium sulfate $(NaCl)$ .
${BaCl}_{2} (aq) + {Na}_{2} {SO}_{4} (aq) \to {BaSO}_{4} (s) + 2 NaCl (aq)$
Therefore, metal $X$ is copper $(Cu)$ and metal $Y$ is sodium $(Na)$ .

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