Decreasing order of reactivity of HX in the reaction
`ROH + HX to RX + H_(2)O`

Options

A. `HIgt HBr gt HCl gt HF`
B. `HBr gt HCl gt Hl gt HF`
C. `HCl gt HBr gt HI gt HF`
D. `HF gt HBr gt HCl gt HI`

Answer

To determine the decreasing order of reactivity of HX in the reaction \( ROH + HX \rightarrow RX + H_2O \), we need to analyze the halogens (X) involved in the reaction: fluorine (F), chlorine (Cl), bromine (Br), and iodine (I).

### Step-by-Step Solution:

1. **Identify the Halogens**: The halogens in this context are F, Cl, Br, and I. These are all part of Group 17 (the halogens) in the periodic table.

2. **Understand the Reaction**: The reaction involves an alcohol (ROH) reacting with a hydrogen halide (HX) to form an alkyl halide (RX) and water (H2O). The halogen (X) replaces the hydroxyl group (OH) in the alcohol.

3. **Consider Electronegativity and Bond Strength**:
- Fluorine (F) is the most electronegative element, which means it forms a very strong bond with carbon in the alcohol (C-F bond). This strong bond makes it difficult for the hydroxyl group to be replaced.
- Iodine (I), on the other hand, is the least electronegative and has the largest atomic size, resulting in a weaker bond with carbon (C-I bond). This bond is easier to break, allowing for a more reactive substitution.

4. **Evaluate Reactivity Based on Bond Strength**:
- The reactivity of HX in the reaction can be inferred from the bond strength of C-X:
- \( HI \) (weak bond, high reactivity)
- \( HBr \) (moderate bond strength, moderate reactivity)
- \( HCl \) (stronger bond than HBr, lower reactivity)
- \( HF \) (strongest bond, lowest reactivity)

5. **Establish the Decreasing Order of Reactivity**:
- Based on the bond strengths and the ability to replace the hydroxyl group, the order of reactivity from highest to lowest is:
\[ HI > HBr > HCl > HF \]

### Final Answer:
The decreasing order of reactivity of HX in the reaction \( ROH + HX \rightarrow RX + H_2O \) is:
\[ HI > HBr > HCl > HF \]