Which of the following is the least volatile hydrogen halide?

\(HF\)

The correct answer is option 2. \(HF\)

The volatility of hydrogen halides generally decreases as the size of the halogen atom increases. This trend is due to the increasing strength of the hydrogen-halogen bond as the size of the halogen increases. Therefore, among the options provided, HF (option 2) is the least volatile hydrogen halide because fluorine is the smallest halogen, resulting in the weakest hydrogen-halogen bond.

Let us look deeper into the concept

The volatility of a compound refers to its tendency to evaporate or vaporize at a given temperature. In the case of hydrogen halides (HX), where X represents a halogen atom (fluorine, chlorine, bromine, or iodine), their volatility is influenced by several factors, with the strength of the hydrogen-halogen bond being one of the most significant.

The hydrogen-halogen bond strength generally follows the trend:

\(HF < HCl < HBr < HI\)
This trend arises due to several factors:

1. Atomic Size:  As you move down the halogen group in the periodic table, the size of the halogen atom increases. For example, fluorine is the smallest halogen, followed by chlorine, bromine, and iodine. This increase in atomic size plays a significant role in determining bond strength.

2. Electronegativity:  Halogens are highly electronegative atoms, meaning they have a strong attraction for electrons. This property influences the polarity of the hydrogen-halogen bond. The smaller the halogen atom, the greater its electronegativity and the more polarizable the bond, resulting in weaker hydrogen-halogen bonds for smaller halogens.

3. Size of the Overlapping Orbitals: In a hydrogen-halogen bond, the overlap between the hydrogen 1s orbital and the halogen's outer p orbital contributes to bond strength. As the size of the halogen atom increases, the overlapping region increases, leading to stronger bonds.

Now, let us apply these factors to understand the volatility of hydrogen halides:

\(HF\): Hydrogen fluoride has the weakest hydrogen-halogen bond among the hydrogen halides because fluorine is the smallest halogen, resulting in the least overlap between the orbitals and the highest electronegativity. Due to its weak intermolecular forces, HF is the least volatile of the hydrogen halides.

\(HCl\): Hydrogen chloride has a stronger hydrogen-halogen bond compared to HF because chlorine is larger than fluorine. However, it is still less volatile than the hydrogen halides with larger halogens.

\(HBr\): Hydrogen bromide has an even stronger hydrogen-halogen bond compared to \(HCl\) because bromine is larger than chlorine. Therefore, \(HBr\) is less volatile than \(HCl\).

\(HI\): Hydrogen iodide has the strongest hydrogen-halogen bond among the hydrogen halides due to the large size of the iodine atom. However, despite having the strongest bond, HI is the most volatile of the hydrogen halides because the larger size of iodine results in weaker intermolecular forces between \(HI\) molecules, allowing it to vaporize more readily.

In summary, hydrogen fluoride \((HF)\) is the least volatile hydrogen halide due to its weaker hydrogen-halogen bond resulting from the small size and high electronegativity of fluorine.