Question

\[HF\] is least volatile and $HCl$ is most volatile amongst hydrogen halides. Explain the given statement with proper reason.

Answer 1

Hint:The boiling point of the substances increases with decrease in the size, as smaller the size higher will be the bond polarity and stronger will be the bond, hence lesser will be the volatility.
The volatility of the hydrogen halides can be compared on the basis of the size of the halides as, on moving down the group, the size increases, and so the electron density decreases.

Complete answer:
In order to answer this question we must first understand the basic differences which are present between the two compounds or acids, hydrochloric acid and hydrofluoric acid. Both contain a hydrogen atom, but if we notice carefully, the halogen groups are different, which is chlorine and fluorine respectively. Now, we will consider the bonding which is present between the hydrogen and the halogens, in their respective molecules.
For hydrofluoric acid, because of the small size of the fluorine atom, the electron density is very high and so the bond polarity between the fluorine and hydrogen is much higher as compared to the hydrochloric acid. So the boiling point of the hydrofluoric acid is higher than that of hydrochloric acid.
The size of chlorine is much larger than fluorine and so the electron density is very low. This results in absence of hydrogen bonding in hydrochloric acid. So the boiling point of the hydrochloric acid is lower hence it is more volatile.
So, the trend of volatility of the hydrogen halides can be represented as,
\[HCl>HBr>HI>HF~\]

Note:The hydrochloric acid has higher volatility value than the hydrofluoric acid because of the larger size of the chlorine.
The hydrofluoric acid has the presence of hydrogen bonding, and so it has a higher boiling point hence less volatile.