Step 1: Understand polarity and dipole moment.
Dipole moment depends on the magnitude and direction of individual bond dipoles present in a molecule.
If bond dipoles support each other in the same direction, the net dipole moment increases.
If bond dipoles oppose each other, the net dipole moment decreases.
Step 2: Identify the groups present.
The given compounds contain \( -NO_2 \) and \( -NH_2 \) groups.
The \( -NO_2 \) group is a strong electron withdrawing group.
The \( -NH_2 \) group is a strong electron donating group.
Step 3: Compare nitrobenzene.
Nitrobenzene contains only one strong polar group, \( -NO_2 \).
Therefore, it has polarity due to the nitro group, but it does not have an additional donor group to increase charge separation strongly.
Step 4: Compare ortho and meta nitroaniline.
In o-nitroaniline, intramolecular hydrogen bonding may occur between \( -NH_2 \) and \( -NO_2 \), which reduces effective polarity.
In m-nitroaniline, the dipoles of \( -NH_2 \) and \( -NO_2 \) do not reinforce each other maximally.
Step 5: Analyze p-nitroaniline.
In p-nitroaniline, \( -NH_2 \) and \( -NO_2 \) are present opposite to each other on the benzene ring.
The \( -NH_2 \) group donates electron density and the \( -NO_2 \) group withdraws electron density.
This creates strong charge separation across the molecule.
Step 6: Reason for maximum dipole moment.
Due to the para arrangement, the electron donating effect of \( -NH_2 \) and electron withdrawing effect of \( -NO_2 \) act in a conjugated manner.
This increases polarity and gives maximum dipole moment.
Step 7: Conclusion.
Thus, the compound with maximum polarity and dipole moment is p-nitroaniline.
Therefore:
\[
\boxed{\text{p-Nitroaniline}}
\]