Dipole moment is shown by

Dipole moment ($$\mu$$) is a vector quantity. It is the vector sum of all individual bond moments in the molecule.

Rule used: If a molecule possesses an element of symmetry (centre of inversion, $$C_2$$ axis perpendicular to the molecular plane, mirror plane cutting all identical polar bonds equally, etc.), the individual bond moments may cancel, giving $$\mu = 0$$. If such complete cancellation does not occur, the molecule shows a net dipole moment ($$\mu \gt 0$$).

Now analyse each option one by one.

Case A:

1,2-dichlorobenzene (ortho-dichlorobenzene)

The two $$\mathrm{C-Cl}$$ bonds make an angle of $$60^{\circ}$$ with each other in the benzene ring. Their bond moments do not lie exactly opposite, so vector cancellation is incomplete. Hence the molecule possesses a finite dipole moment.

Case B:

trans-2,3-dichloro-2-butene : $$\mathrm{CH_3-C(Cl)=C(Cl)-CH_3}$$

The double bond plane contains both $$\mathrm{C-Cl}$$ bonds. In the trans isomer the two Cl atoms are on opposite sides, and the molecule has an internal mirror plane that bisects and opposes the two identical $$\mathrm{C-Cl}$$ bond moments. Vector sum is zero ⇒ $$\mu = 0$$.

Case C:

1,4-chlorobenzene (para-dichlorobenzene)

The two $$\mathrm{C-Cl}$$ bonds are exactly opposite in the ring (180° apart). Their bond moments are equal and collinear but in opposite directions, so they cancel completely ⇒ $$\mu = 0$$.

Case D:

trans-1,2-dinitroethene : $$\mathrm{O_2N-CH=CH-NO_2}$$

Each $$\mathrm{NO_2}$$ group exerts a bond moment vector roughly along the $$\mathrm{C-N}$$ bond. In the trans form these vectors are equal and opposite across the double bond, giving complete cancellation ⇒ $$\mu = 0$$.

Only the ortho isomer in Option A lacks the symmetry necessary for full cancellation, so it alone exhibits a net dipole moment.

Final answer: Option A which is: 1,2-dichlorobenzene