To determine which halogen acid has the highest bond dissociation enthalpy, we need to analyze the bond strengths of the hydrogen halides (HF, HCl, HBr, HI) and understand the factors affecting these bond strengths.
### Step-by-Step Solution:
1. **Identify the Halogen Acids**:
The halogen acids we are considering are HF (hydrogen fluoride), HCl (hydrogen chloride), HBr (hydrogen bromide), and HI (hydrogen iodide).
2. **Understand Bond Dissociation Enthalpy**:
Bond dissociation enthalpy is the energy required to break a bond in a molecule. A higher bond dissociation enthalpy indicates a stronger bond.
3. **Consider Electronegativity**:
Electronegativity is the tendency of an atom to attract electrons. In the case of halogens, electronegativity decreases from fluorine to iodine:
- Fluorine (F) > Chlorine (Cl) > Bromine (Br) > Iodine (I).
This means that the bond between hydrogen and fluorine (HF) is stronger due to fluorine's high electronegativity, which allows it to attract the shared electrons more effectively.
4. **Analyze Atomic Size**:
As we move down the group from fluorine to iodine, the size of the halogen atoms increases. A smaller atomic size results in a shorter bond length, which typically leads to a stronger bond due to increased overlap of atomic orbitals. Therefore, HF has a shorter bond length compared to HCl, HBr, and HI.
5. **Consider Bond Strength Order**:
Based on the factors discussed:
- HF > HCl > HBr > HI
This order indicates that HF has the strongest bond, followed by HCl, HBr, and finally HI.
6. **Conclusion**:
Since HF has the strongest bond due to its high electronegativity, smaller size, and the presence of hydrogen bonding, it will have the highest bond dissociation enthalpy among the halogen acids.
### Final Answer:
The halogen acid with the highest bond dissociation enthalpy is **HF (hydrogen fluoride)**.
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