Which of the following hydrogen halides is the weakest reducing agent?

Hydrogen halides are binary compounds of hydrogen with halogens, including HF, HCl, HBr, and HI. Their reducing ability depends on the ease with which the halide ion (X⁻) can donate electrons and reduce other species. In other words, a stronger reducing agent can more readily lose electrons.

Among the hydrogen halides, HF is the weakest reducing agent. Fluorine is the most electronegative element, which makes the fluoride ion (F⁻) highly stable and reluctant to donate electrons. This high electronegativity and strong H–F bond reduce the reducing power of HF significantly. As a result, HF is much less reactive as a reducing agent compared to other hydrogen halides.

In contrast, HI is the strongest reducing agent among the hydrogen halides. The iodine atom has a larger size and lower electronegativity, making the I⁻ ion less stable and more willing to donate electrons. HBr and HCl have intermediate reducing abilities between HI and HF.

The reducing strength trend for hydrogen halides is:

HF < HCl < HBr < HI

This trend is influenced by bond strength and electronegativity. HF has a very strong H–F bond, which makes it harder to break and release electrons. In HI, the H–I bond is weaker, and the larger iodine atom can more easily provide electrons to reduce other substances.

Understanding the reducing ability of hydrogen halides is important in inorganic chemistry reactions, especially in redox reactions, halogen displacement, and industrial chemical processes. HF’s weak reducing nature also affects its use in chemical reactions, where it often acts as a proton donor rather than a reducing agent.

In summary, HF is the weakest reducing agent among the hydrogen halides due to the strong H–F bond and high electronegativity of fluorine.