In the case of hydrogen bonds between HF molecules, the electronegativity of fluorine is the highest among the halogens, meaning it has the strongest attraction for electrons. This creates a strong partial positive charge on the hydrogen atom of HF, which in turn can form a stronger hydrogen bond with the lone pair of electrons on another fluorine atom.
In comparison, HBr and HI molecules have weaker electronegative bromine and iodine atoms, respectively. This results in a weaker partial positive charge on the hydrogen atom, leading to weaker hydrogen bonds.
Additionally, the smaller size of the fluorine atom in HF allows for closer proximity between the hydrogen and fluorine atoms, further strengthening the hydrogen bond.
Furthermore, the polarity of the H-F bond is higher than that of H-Br and H-I bonds due to the higher electronegativity difference between hydrogen and fluorine. This increased polarity allows for stronger electrostatic interactions between the hydrogen bond donor and acceptor, resulting in a stronger hydrogen bond.
Therefore, the combination of high electronegativity, small molecular size, and high bond polarity contributes to the stronger hydrogen bonds between HF molecules compared to HBr and HI molecules.
