The dipole moment ($\mu$) of a molecule is a measure of the polarity of its chemical bonds and is a vector quantity. For a molecule to have a significant non-zero dipole moment, it must possess polar bonds and an asymmetrical molecular geometry such that the individual bond dipoles do not cancel each other out.

Let's analyze each molecule:

1. CO\u2082 (Carbon Dioxide): This molecule has a linear geometry (sp hybridization for carbon). The C=O bonds are polar, but because the molecule is linear, the two bond dipoles are equal in magnitude and opposite in direction, thus they cancel each other out. Hence, $\mu =0$.
2. CH\u2084 (Methane): This molecule has a tetrahedral geometry (sp\u00b3 hybridization for carbon). The C-H bonds are slightly polar, but due to the highly symmetrical tetrahedral arrangement, the individual bond dipoles cancel each other out. Hence, $\mu =0$.
3. H\u2082O (Water): This molecule has a bent or V-shaped geometry (sp\u00b3 hybridization for oxygen with two lone pairs). The O-H bonds are polar. Due to the bent geometry, the bond dipoles do not cancel out, and the lone pair contributions also add to the overall dipole moment. Hence, $\mu \ne 0$.
4. SO\u2082 (Sulfur Dioxide): This molecule has a bent or V-shaped geometry (sp\u00b2 hybridization for sulfur with one lone pair). The S=O bonds are polar. Due to the bent geometry, the bond dipoles do not cancel out, and the lone pair contribution also adds to the overall dipole moment. Hence, $\mu \ne 0$.

Therefore, molecules III (H\u2082O) and IV (SO\u2082) have significant non-zero dipole moments.

• A) I only: Incorrect, CO\u2082 has $\mu =0$.
• B) I and II only: Incorrect, both CO\u2082 and CH\u2084 have $\mu =0$.
• C) III only: Incorrect, SO\u2082 also has $\mu \ne 0$.
• D) III and IV only: Correct, both H\u2082O and SO\u2082 have $\mu \ne 0$.

Correct Answer: (D)