Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: For which of the following molecule significant $\mu \ne 0$?

To master questions on dipole moments, you must first visualize them as vectors. A dipole moment is not just a number; it has a specific direction pointing from the less electronegative atom to the more electronegative atom.

When multiple polar bonds exist in a single molecule, the overall dipole moment is the vector sum of all individual bond dipoles. If a molecule is perfectly symmetrical and the bond dipoles pull equally in exactly opposite directions, they cancel each other out, resulting in a net dipole moment of zero ($\mu =0$).

Let's analyze the first two molecules: 1,4-dichlorobenzene and 1,4-dicyanobenzene.

In 1,4-dichlorobenzene, the chlorine atoms are attached at the para positions (1 and 4). The $C-Cl$ bond is perfectly linear along the axis of the benzene ring. Because the two $C-Cl$ vectors are equal in magnitude and exactly $180^{\circ }$ apart, they perfectly cancel each other out.

The same logic applies to 1,4-dicyanobenzene. The cyano group ($-C\equiv N$) is linear because the carbon atom is $sp$ hybridized. Therefore, the two $-CN$ groups at opposite ends of the ring pull with equal force in opposite directions, leading to a net dipole moment of zero.

Now, let's look at 1,4-dihydroxybenzene (quinol) and 1,4-dithiobenzene (thioquinol). At first glance, they look just as symmetrical as the previous examples. However, there is a crucial geometric difference.

The oxygen atom in the hydroxyl group ($-OH$) and the sulfur atom in the thiol group ($-SH$) both possess lone pairs of electrons. According to VSEPR theory, these lone pairs repel the bonding pairs, causing the $C-O-H$ and $C-S-H$ bond angles to be bent (approximately $109^{\circ }$), rather than linear.

Because the $C-O-H$ bond is bent, the dipole vector of the $-OH$ group is not aligned with the axis of the benzene ring. Furthermore, there is free rotation around the $C-O$ single bond connecting the hydroxyl group to the ring.

This free rotation means the molecule exists as a dynamic mixture of different spatial arrangements, known as conformers. While there might be one specific conformer where the dipoles happen to cancel (the perfectly anti conformer), the molecule spends time in various other conformers where the dipoles do not align perfectly.

As a result, the time-averaged vector sum of the dipoles is not zero. Therefore, both 1,4-dihydroxybenzene and 1,4-dithiobenzene possess a significant, non-zero net dipole moment ($\mu eq 0$).

Linear substituents at para positions cancel out perfectly, while bent substituents do not due to their geometry and the existence of multiple conformers. Thus, molecules (iii) and (iv) have a significant dipole moment, making option (d) the correct answer.