Question: 48. Given below are two statements :
Statement-I : The boiling point of hydrides of Group-16 elements follow the order \mathrm{H}_{2} \mathrm{O}>\mathrm{H}_{2} \mathrm{Te}>\mathrm{H}_{2} \mathrm{Se}>\mathrm{H}_{2} \mathrm{~S}.
Statement-II : On the basis of molecular mass, \mathrm{H}_{2} \mathrm{O} is expected to have lower boiling point than the other members of the group but due to the presence of extensive H -bonding in \mathrm{H}_{2} \mathrm{O}, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both statement-I and Statement-II are true.
(2) Both statement-I and Statement-II are false.
(3) Statement-I is the true but Statement-II is false.
(4) Statement-I is false but Statement-II is true.
Answer: Option (1)
Explanation:
In Group-16 hydrides, boiling point generally increases down the group due to increase in molecular mass and stronger van der Waals forces.
Thus, on the basis of molecular mass alone, the expected order would be
\mathrm{H}_{2} \mathrm{Te}>\mathrm{H}_{2} \mathrm{Se}>\mathrm{H}_{2} \mathrm{S}>\mathrm{H}_{2} \mathrm{O}.
However, \mathrm{H}_{2} \mathrm{O} shows extensive intermolecular hydrogen bonding due to the high electronegativity of oxygen and the small size of the molecule.
This strong hydrogen bonding significantly increases the boiling point of water,
making it higher than all other Group-16 hydrides.
Therefore, the actual boiling point order is
\mathrm{H}_{2} \mathrm{O}>\mathrm{H}_{2} \mathrm{Te}>\mathrm{H}_{2} \mathrm{Se}>\mathrm{H}_{2} \mathrm{S},
which makes Statement-I true.
Statement-II correctly explains why \mathrm{H}_{2} \mathrm{O}
has a higher boiling point despite its lower molecular mass.
Hence, both Statement-I and Statement-II are true.