Use the information and data given below to answer the questions $(a)$ to $(c)$. Stronger intermolecular forces result in higher boiling point.
Strength of London forces increases with the number of electrons in the molecule.
Boiling points of $HF, HCl, HBr$ and $HI$ are $293 \ K, 189 \ K, 200 \ K$ and $238 \ K$ respectively.
$(a)$ Which type of intermolecular forces are present in the molecules $HF, HCl, HBr$ and $HI$?
$(b)$ Looking at the trend of boiling points of $HCl, HBr$ and $HI$,explain out of dipole-dipole interaction and London interaction,which one is predominant here.
$(c)$ Why is the boiling point of hydrogen fluoride highest while that of hydrogen chloride lowest?

(N/A) The molecules $HF, HCl, HBr$ and $HI$ are polar,so they exhibit dipole-dipole interactions. Additionally,all molecules possess London dispersion forces.
$(b)$ For $HCl, HBr$ and $HI$,the boiling point increases as the molecular size and number of electrons increase $(HCl < HBr < HI)$. This indicates that London dispersion forces are the predominant factor influencing the boiling point trend in these hydrogen halides.
$(c)$ $HF$ has the highest boiling point due to strong intermolecular hydrogen bonding. $HCl$ has the lowest boiling point because it has the smallest molecular size and the fewest electrons among the group,resulting in the weakest London dispersion forces,and it lacks the strong hydrogen bonding present in $HF$.