Solution:
$\text{Hint: If the dipole moment is zero, then the molecule is nonpolar.}
\text{The observed dipole moment (}\mu_{\text{observed}}\text{) can be greater than the theoretical dipole moment (}\mu_{\text{theoretical}}\text{) due to factors like intramolecular hydrogen bonding.}
\text{1. } \text{CHCl}_3 \text{ (Chloroform): It has a significant dipole moment, but generally, there isn't a large discrepancy between observed and theoretical due to specific interactions causing an increase.}
\text{2. o-dichlorobenzene: The two chlorine atoms are ortho to each other. While it has a dipole moment, it's not typically cited for a significantly higher observed vs. theoretical value due to specific enhancing interactions.}
\text{3. o-dinitrophenol: This structure is not one of the options. Assuming it refers to o-nitrophenol. In o-nitrophenol, there is intramolecular hydrogen bonding between the hydroxyl group and the nitro group. This can reduce the effective dipole moment by creating a pseudo-ring structure, which can make the observed dipole moment lower than what might be theoretically predicted if only considering bond moments without this interaction.}
\text{4. o-chlorophenol: In o-chlorophenol, there is intramolecular hydrogen bonding between the hydroxyl group and the chlorine atom. This intramolecular hydrogen bonding can lead to a conformation that enhances the overall dipole moment or changes the expected vector sum, leading to an observed dipole moment that is greater than the theoretical value calculated without considering such interactions. The electron-withdrawing nature of the chlorine atom along with hydrogen bonding contributes to this effect.}
\text{Therefore, o-chlorophenol is the most likely molecule among the given options to exhibit an observed dipole moment greater than its theoretical dipole moment due to intramolecular hydrogen bonding influencing its conformation and charge distribution.}$