Question

The oxoacid of sulphur which has a lone pair of electrons on the sulphur atom is:

• A. H$_2$SO$_4$
• B. H$_2$SO$_3$
• C. H$_2$S$_2$O$_8$
• D. H$_2$S$_2$O$_7$

Hint:

Sulfur in the +4 oxidation state (as in sulfurous acid, H\(_2\)SO\(_3\)) retains a lone pair of electrons, whereas in the +6 oxidation state (as in sulfuric acid, H\(_2\)SO\(_4\)), no lone pairs are present.

Answer 1

The Correct Option is B

Sulphur in its higher oxidation states (such as in H$_2$SO$_4$, sulfuric acid) typically has no lone pairs of electrons. However, in H$_2$SO$_3$ (sulfurous acid), the sulphur atom is in a lower oxidation state (+4) and retains a lone pair of electrons. This lone pair is a characteristic feature of sulfur in the +4 oxidation state, where sulfur doesn't fully use its available orbitals for bonding with oxygen. Here is the relevant chemical reaction for sulfurous acid: Reduction of sulfurous acid (H\(_2\)SO\(_3\)): \[ \text{H}_2\text{SO}_3 + 2\text{H}^+ + 2e^- \rightarrow \text{H}_2\text{S} + 2\text{H}_2\text{O} \] In this reaction, sulfur is reduced from a +4 oxidation state in H\(_2\)SO\(_3\) to -2 in H\(_2\)S (hydrosulfuric acid). Oxidation of sulfurous acid: \[ \text{H}_2\text{SO}_3 + \text{O}_2 \rightarrow \text{H}_2\text{SO}_4 \] Here, sulfur in H\(_2\)SO\(_3\) is oxidized to H\(_2\)SO\(_4\) (sulfuric acid), where sulfur has a +6 oxidation state.
Thus, H\(_2\)SO\(_3\) (sulfurous acid) is the correct answer as it contains a lone pair on the sulfur atom in the +4 oxidation state.