Question

\[{{H}_{2}}O\] is a liquid whereas \[{{H}_{2}}S\] is a gas. This is due to______.
(A) The molecular mass of \[{{H}_{2}}S\] is more than that of \[{{H}_{2}}O\]
(B) The molecular mass of \[{{H}_{2}}O\] is more than that of \[{{H}_{2}}S\]
(C) Presence of hydrogen bond in \[{{H}_{2}}S\]
(D) Absence of hydrogen in \[{{H}_{2}}S\]

Answer 1

Hint: We know that this type of liquid and gas difference can occur because of the type of force between the molecules and may be because of the strength difference of those forces.

Step by step solution:
We know that there is hydrogen bonding between water molecules and in \[{{H}_{2}}S\] there is no hydrogen bonding.
* We know that there is a large electronegativity difference between oxygen (O) and hydrogen (H) (from data of electronegativity we can say the difference is about 2.3). This is the reason why the partial positive charged hydrogens are attracted to the lone pairs of the oxygen atoms of other water molecules.
* In other cases there is relatively small electronegativity difference between sulphur and hydrogen (which is about 0.5) is very small to polarise the covalent bond. And, the lone pairs on the sulphur atom are delocalised in the \[s{{p}^{3}}\] orbitals and the d orbitals, this will not happen in the case of oxygen. So, the negative charge of the lone pairs is also distributed and the interaction is lowered.
* So, the hydrogen bonding in water causes the water molecules to be associated with each other due to intermolecular forces of attraction, whereas in the case of \[{{H}_{2}}S\], these forces of attraction are much weaker.
So, from the above explanation we can say that water is liquid and \[{{H}_{2}}S\] is a gaseous form because of absence of hydrogen bonding. Then the correct answer is “D”.

Note: Lesser energy is required to reduce the forces of interaction between the \[{{H}_{2}}S\]molecules than those between water molecules. This energy is available at room temperature and So, \[{{H}_{2}}S\] is a gas and water is still a liquid.