Which of the following compound are having white ppt ?

Options

A. `K_(2)Fe[Fe(CN)_(6)]`
B. `[Fe(H_(2)O)_(3)(SCN_(1))]^(2+)`
C. `ZnS`
D. `Zn(OH)_(2)`

Answer

To determine which of the given compounds produce a white precipitate, we will analyze each compound based on their central metal ion, oxidation state, and the nature of the ligands involved.

### Step-by-Step Solution:

1. **Identify the Compounds:**
The compounds given are:
- K2[Fe(CN)6]
- [Fe(H2O)3(SCN)2]²⁺
- ZnS
- Zn(OH)2

2. **Analyze K2[Fe(CN)6]:**
- Central Metal: Iron (Fe)
- Oxidation State: +2
- Ligands: Cyanide (CN⁻) is a strong field ligand.
- Effect: Strong field ligands cause pairing of electrons. Since all 6 d-electrons are paired, there will be no d-d transition.
- Result: This complex will produce a white precipitate.

3. **Analyze [Fe(H2O)3(SCN)2]²⁺:**
- Central Metal: Iron (Fe)
- Oxidation State: +3
- Ligands: Water (H2O) and thiocyanate (SCN⁻) are weak field ligands.
- Effect: Weak field ligands do not cause pairing of electrons, leading to unpaired electrons and d-d transitions.
- Result: This complex will produce a red precipitate.

4. **Analyze ZnS:**
- Central Metal: Zinc (Zn)
- Oxidation State: +2
- Effect: Zinc in +2 oxidation state has no unpaired electrons.
- Result: There will be no d-d transition, resulting in a white precipitate.

5. **Analyze Zn(OH)2:**
- Central Metal: Zinc (Zn)
- Oxidation State: +2
- Effect: Similar to ZnS, zinc in +2 oxidation state has no unpaired electrons.
- Result: There will be no d-d transition, resulting in a white precipitate.

6. **Conclusion:**
- The compounds that produce a white precipitate are:
- K2[Fe(CN)6] (1st compound)
- ZnS (3rd compound)
- Zn(OH)2 (4th compound)

### Final Answer:
The compounds that have a white precipitate are K2[Fe(CN)6], ZnS, and Zn(OH)2.