A 0.3 M HCl solution contains the following ions `Hg^(++),Cd^(++),Sr^(++),Fe^(++)`. The addition of `H_(2)S` to above solution will precipitate

Options

A. Cd, Cu and Hg
B. Cd, Fe and Sr
C. Hg, Cu and Fe
D. Cu, Sr and Fe

Explanation

To determine which ions will precipitate when H₂S is added to a 0.3 M HCl solution containing Hg²⁺, Cd²⁺, Sr²⁺, and Fe²⁺, we need to consider the solubility of the sulfides of these metal ions.

### Step 1: Identify the ions present in the solution
The ions present in the solution are:
- Hg²⁺ (Mercury)
- Cd²⁺ (Cadmium)
- Sr²⁺ (Strontium)
- Fe²⁺ (Iron)

### Step 2: Understand the effect of H₂S
When H₂S is added to the solution, it will react with the metal ions to form metal sulfides. The solubility of these sulfides varies among different metal ions.

### Step 3: Check the solubility of sulfides
- **HgS (Mercury(II) sulfide)**: Insoluble in water, will precipitate.
- **CdS (Cadmium sulfide)**: Insoluble in water, will precipitate.
- **SrS (Strontium sulfide)**: Soluble in water, will not precipitate.
- **FeS (Iron(II) sulfide)**: Insoluble in water, will precipitate.

### Step 4: Conclusion on precipitation
Based on the solubility rules:
- The addition of H₂S will precipitate HgS, CdS, and FeS.
- SrS will remain in solution as it is soluble.

### Final Answer
The addition of H₂S to the 0.3 M HCl solution will precipitate:
- HgS (Mercury(II) sulfide)
- CdS (Cadmium sulfide)
- FeS (Iron(II) sulfide)