Which of the following is least soluble in water ?

Options

A. AgCl
B. AgBr
C. AgI
D. AgF

Answer

To determine which of the following compounds is least soluble in water, we need to analyze the solubility of silver halides: AgF, AgCl, AgBr, and AgI.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds we are considering are silver fluoride (AgF), silver chloride (AgCl), silver bromide (AgBr), and silver iodide (AgI).

2. **Understand Solubility Trends**: In general, the solubility of silver halides decreases as the size of the halide ion increases. This is due to the increasing covalent character of the bond formed between the silver cation (Ag+) and the halide anions (F-, Cl-, Br-, I-).

3. **Analyze Each Compound**:
- **AgF**: Silver fluoride is relatively soluble in water due to the small size of the fluoride ion (F-).
- **AgCl**: Silver chloride has lower solubility than AgF, but it is still more soluble than AgBr and AgI.
- **AgBr**: Silver bromide has even lower solubility than AgCl.
- **AgI**: Silver iodide is the least soluble of the silver halides due to the large size of the iodide ion (I-), which increases the covalent character of the bond, making it less ionic and thus less soluble in water.

4. **Conclusion**: Based on the analysis, AgI (silver iodide) is the least soluble in water among the given options.

### Final Answer:
The least soluble compound in water is **AgI (silver iodide)**.