Question

${\mathrm{H}}_2\mathrm{O}$ has a higher boiling point than ${\mathrm{H}}_2\mathrm{S}$ due to

• A. Molar Mass
• B. Debye Forces
• C. London Forces
• D. Hydrogen bonding

Answer 1

The correct option is D

Hydrogen bonding

Explanation for correct option:

Option (D);

• In ${\mathrm{H}}_2\mathrm{O}$ there exist strong intermolecular hydrogen bonding, but in the case of ${\mathrm{H}}_2\mathrm{S}$ there is only weak Vander Waals force of attraction exist between the molecule.
• As hydrogen bonding is much stronger than vander Waals force hence it requires more energy to break the bonds between ${\mathrm{H}}_2\mathrm{O}$. Hence, the boiling point in case of ${\mathrm{H}}_2\mathrm{O}$ is greater than${\mathrm{H}}_2\mathrm{S}$.

Explanation for incorrect options:

Option (A);

• The concept of molar mass is not related to increasing the boiling point of ${\mathrm{H}}_2\mathrm{O}$ than ${\mathrm{H}}_2\mathrm{S}$.
• Hence, the option given here doesn't justify the reason for the correct answer.

Option (B);

• Neither ${\mathrm{H}}_2\mathrm{O}$ nor ${\mathrm{H}}_2\mathrm{S}$ has the Debye force of attraction between their molecules.
• So, the option given here doesn't coincide with the correct option.

Option (C);

• There exists no London force of attraction between ${\mathrm{H}}_2\mathrm{O}$ nor ${\mathrm{H}}_2\mathrm{S}$.
• Hence, the correct concept related to the boiling of ${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{H}}_2\mathrm{S}$ can not be understood by using the given option.

Hence, ${\mathrm{H}}_2\mathrm{O}$ has a higher boiling point than ${\mathrm{H}}_2\mathrm{S}$ due to: Hydrogen bonding.