Analyze Statement I

• In $\mathrm{H_2O}$, the central atom is oxygen.

• Oxygen has $2$ bond pairs and $2$ lone pairs.

• Electron-pair geometry is tetrahedral, but due to two lone pairs, the molecular shape is bent / V-shaped.

• In $\mathrm{SO_2}$, the central atom is sulfur.

• Sulfur has two bonding regions (with two oxygens) and one lone pair.

• The electron-pair geometry is trigonal planar, and with one lone pair the molecular shape is also bent / V-shaped.

Therefore, Statement I is correct.

Analyze Statement II

We compare bond angles:

• In $\mathrm{H_2O}$, the bond angle is approximately $\angle \mathrm{HOH} \approx 104.5^\circ$

• In $\mathrm{SO_2}$, the bond angle is approximately $\angle \mathrm{OSO} \approx 119^\circ$ (slightly less than $120^\circ$ due to lone pair repulsion).

Hence, $\angle \mathrm{OSO} > \angle \mathrm{HOH}$

So the statement “The bond angle of $\mathrm{SO_2}$ is less than that of $\mathrm{H_2O}$” is incorrect.

Therefore, Statement II is incorrect.

Choose the correct option

• Statement I: Correct
• Statement II: Incorrect

So the correct option is: $\boxed{\text{C}}$

Comparison with stored answer

• Derived answer: $\text{C}$
• Stored correct answer: $\text{C}$

They match.