Consider the substances hydrogen (h2), fluorine (f2), and hydrogen fluoride (hf). based on their molecular structures, how does the boiling point of hf compare with the boiling points of h2 and n2?

The order of Boiling point is like this:

HF > HI > HBr > HCl

Due to the high electronegativity of flourine, hydrogen bonds can be formed between HF molecules.Therefore more energy is required to break the intermolecular forces in HF than the other hydrogen halides. If the intermolecular forces are strong between atoms then we know that its melting or boiling point is very higher. Since the bond between HF is the strongest therefore its boiling point is high.

The three substances given are

H₂  , F₂  and HF

here in case of HF there will be inter-molecular hydrogen bond [secondary forces] and in H₂ or F₂ there will be only weak london dispersion forces present as inter-molecular forces. Thus H₂ or F₂ will exist as gas at room temperature and thus we can say that they will have very low boiling point.

[The temperature at which the vapour pressure becomes equal to atmospheric pressure] They will easily get converted to vapours while HF molecules will held strong together and will need high temperature to become vapours.

the answer is b. dilution