A colourless aqueous solution contains nitrates of two metals, X and Y. When it was added to an aqueous solution `NaCl,` a white precipitate was formed. This precipitate was found to be partly soluble in hot water to give a residue P and a solution Q. The residue P was soluble in aq. `NH_(3)` and also in excess sodium thiosulfate. The hot solution Q gave a yellow precipirate with Kl. The metals X and Y, respectively, are :

Options

A. `Ag and Pb`
B. `Ag and Cd`
C. `Cd and Pb`
D. `Cd and Zn`

Explanation

To solve the problem, we need to identify the metals X and Y based on the given information about their reactions and the properties of the precipitates formed.

### Step 1: Identify the Precipitate
When the colorless aqueous solution of nitrates of metals X and Y is mixed with NaCl, a white precipitate is formed. This indicates that one of the metals forms an insoluble chloride. The most common metals that form white precipitates with NaCl are Silver (Ag) and Lead (Pb).

### Step 2: Solubility of the Precipitate
The problem states that the white precipitate is partly soluble in hot water. Silver chloride (AgCl) is known to be sparingly soluble in hot water, while lead(II) chloride (PbCl2) is more soluble in hot water. Therefore, it is likely that the white precipitate is AgCl.

### Step 3: Analyze Residue P and Solution Q
The residue P is said to be soluble in aqueous NH3 and also in excess sodium thiosulfate. Silver chloride (AgCl) dissolves in aqueous ammonia to form a complex ion, [Ag(NH3)2]+, and it is also soluble in sodium thiosulfate, forming a soluble complex. This confirms that one of the metals, X, is Silver (Ag).

### Step 4: Analyze Solution Q
The hot solution Q gives a yellow precipitate with KI. The yellow precipitate formed with KI is likely lead(II) iodide (PbI2), which is known to be yellow. This indicates that the other metal, Y, is Lead (Pb).

### Conclusion
Based on the analysis:
- Metal X is Silver (Ag).
- Metal Y is Lead (Pb).

### Final Answer
The metals X and Y are Silver (Ag) and Lead (Pb), respectively.

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