Animated Solution for Chemistry - Chemical Bonding and Molecular Structure: Match the type of interaction in column A with the distance dependence of their interaction energy in column B : $\begin{array}{ll}\text{Column A}& \text{Column B}\\ \text{(I) Ion-ion}& \text{(a) }\frac{1}{r}\\ \text{(II) Dipole-dipole}& \text{(b) }\frac{1}{r^2}\\ \text{(III) London dispersion}& \text{(c) }\frac{1}{r^3}\\ & \text{(d) }\frac{1}{r^6}\end{array}$

Intermolecular forces are the invisible threads that hold matter together. In this problem, we are tasked with matching three fundamental types of interactions—Ion-Ion, Dipole-Dipole, and London Dispersion—with how their potential energy decays as the distance $r$ between the particles increases.

Understanding this distance dependence is crucial because it tells us how "long-range" or "short-range" a force truly is. Let's break them down one by one.

Imagine two point charges, $+q$ and $-q$, separated by a distance $r$. This is the classic electrostatic interaction governed by Coulomb's Law. The force between them is proportional to $1/r^2$. However, the interaction energy (which is the integral of force over distance) is inversely proportional to the first power of the distance.

Therefore, the ion-ion interaction energy $E$ scales as:

This is a very long-range interaction, which is why ionic compounds like $\text{NaCl}$ form strong, extensive crystal lattices. Thus, (I) matches with (a).

Now, let's consider two polar molecules. Each molecule has a permanent dipole moment, meaning it has a partial positive end (${\delta }^{+}$) and a partial negative end (${\delta }^{-}$).

When these dipoles are stationary (like in a solid), their interaction energy drops off much faster than simple point charges because the positive and negative ends partially cancel each other's electric fields at a distance. The energy for stationary dipoles is inversely proportional to the cube of the distance.

This makes it a shorter-range force compared to ion-ion interactions. Thus, (II) matches with (c).

Finally, we arrive at London dispersion forces. These are the weakest and most fleeting of all intermolecular forces, existing even between non-polar atoms like Helium.

They arise from quantum mechanical fluctuations. For a brief instant, the electron cloud of an atom might shift, creating a temporary dipole. This temporary dipole then induces a dipole in a neighboring atom. Because this relies on a chain reaction of temporary fluctuations, the interaction energy decays incredibly fast.

This is a strictly short-range force. Thus, (III) matches with (d).

Bringing it all together: - (I) Ion-ion $\to$ (a) $1/r$ - (II) Dipole-dipole $\to$ (c) $1/r^3$ - (III) London dispersion $\to$ (d) $1/r^6$

This perfectly aligns with option (d). Always trust the physics, even if an answer key has a typo!