The relative strength of the interionic/ intermolecular forces in a decreasing order is:

We begin by recalling Coulomb’s law, which gives the electrostatic force between two point charges:

$$F \;=\; \dfrac{k\,|q_1 q_2|}{r^{2}}$$

Here $$k$$ is Coulomb’s constant, $$q_1$$ and $$q_2$$ are the magnitudes of the charges and $$r$$ is the separation between them. When both interacting species carry full charges (ions), each particle behaves like a point charge, so the interaction is purely $$1/r^{2}$$ and very strong.

Next, we consider an ion interacting with a polar molecule (a dipole). A permanent dipole may be represented by two equal and opposite charges $$+q$$ and $$-q$$ separated by a small distance $$d$$. The resultant dipole moment is $$\mu = qd$$. When such a dipole approaches an ion of charge $$Q$$, the force varies as

$$F \;\propto\; \dfrac{|Q|\mu}{r^{3}}$$

because the ion feels attractions and repulsions from the two ends of the dipole that do not cancel completely. The extra power of $$r$$ in the denominator ($$r^{3}$$ instead of $$r^{2}$$) shows that the ion-dipole force decays faster with distance and is therefore weaker than the ion-ion force.

Finally, we analyse dipole-dipole interaction. Two permanent dipoles with moments $$\mu_1$$ and $$\mu_2$$ interact with a force that, in its simplest angular average, follows

$$F \;\propto\; \dfrac{\mu_1 \mu_2}{r^{4}}$$

The $$r^{4}$$ dependence means that this force falls off even more rapidly with distance, so it is weaker than both ion-ion and ion-dipole interactions.

Putting these observations together, we obtain the following descending order of strength:

$$\text{ion-ion} \; > \; \text{ion-dipole} \; > \; \text{dipole-dipole}$$

This sequence exactly matches the arrangement given in Option D.

Hence, the correct answer is Option D.