Question

Which of the following is the most volatile compound?
(A) ${\text{HI}}$
(B) ${\text{HCl}}$
(C) ${\text{HBr}}$
(D) ${\text{HF}}$

Answer 1

Hint: Compounds which have very low stability and have very low boiling points are generally volatile in nature or easily evaporate because they will easily break due to many factors like heat, oxidation, reduction, etc.

Complete step by step answer:
The volatile nature of these given compounds will be predicted on the basis of the value of boiling points and the order of boiling point of the above compounds are shown as follow:
${\text{HF > HI > HBr > HCl}}$
- Here ${\text{HF}}$ has the highest boiling point among them due to the presence of strong hydrogen bonding between the most electronegative (${\text{F}}$) and most electropositive (${\text{H}}$) atom.
- After ${\text{HF}}$, ${\text{HI}}$ shows high boiling point then ${\text{HBr}}$ and ${\text{HCl}}$; because of the small size of iodine atom there is a great attraction between hydrogen atom (${\text{H}}$) and iodine atom (${\text{I}}$).
-Boiling point of ${\text{HCl}}$ is very low among all given compounds because of the large size of a chlorine atom, there is not much great force of attraction which doesn’t allow stability to the bonding between the hydrogen atom (${\text{H}}$) and chlorine atom (${\text{Cl}}$).
From the above discussion, it is clear that the boiling point and stability of ${\text{HCl}}$molecule is lower than the others, so ${\text{HCl}}$ is the most volatile compound.
So, the correct answer is “Option B”.

Note: Here some of you may think that the order of boiling point of above-given compounds is like: ${\text{HF > HCl > HBr > HI}}$, on the basis of their electronegativity. And that assumption will be wrong and may lead to giving the wrong answers on the basis of boiling points, so for ignoring this kind of mistake always keep in mind the order which is given in the solution part.