Question

$ {H_2}S < {H_2}Se < {H_2}Te < {H_2}O $ . This is the correct boiling point order. If true enter $ 1 $ else $ 0. $
(A) $ 1 $
(B) $ 0 $

Answer 1

Hint: Boiling point can be defined as the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. Boiling point directly depends upon the intermolecular forces of attraction between the molecules of the liquid. So, based on these forces we will deduce the correct order of boiling points of above given compounds.

Complete answer:
As we know that boiling point can be defined as the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure.
Now, we can determine the correct order of boiling points of all the given compounds on the basis of intermolecular forces of attraction. Greater the forces of attraction between the molecules in a liquid, more will be the energy required to separate them. And hence, greater will be the boiling point of that liquid.
So, in $ {H_2}O $ , there is intermolecular hydrogen bonding between the oxygen and hydrogen atoms and we know that it is one of the strongest forces. Therefore, $ {H_2}O $ will have the highest boiling point among the given compounds.
Now, we know that on moving down the group in the periodic table the size of the atom increases and hence, the van der Waal forces of attraction also increases.
Therefore, the order of boiling point of rest of the compounds is given as:
  $ {H_2}S < {H_2}Se < {H_2}Te $
Therefore, the correct order of boiling point is:
 $ {H_2}S < {H_2}Se < {H_2}Te < {H_2}O $
Hence, the correct option is (A) $ 1. $

Note:
We should remember that volatility is inversely proportional to the boiling point of the liquid. More the boiling point, less the volatility. This happens because there is a stronger force of attraction between the molecules of the liquid. Similarly, lesser the boiling point, greater will be the volatility.