Polarity and polarizability — dipole moment and induced dipole

A molecule whose charge centers do not coincide carries a permanent dipole, and even one that does not builds an induced dipole $\vec\mu_{\rm ind} = \alpha \vec E$ inside an external field.

Opening

So far we have looked inside a single molecule — its bonds and molecular orbitals. This chapter is about how a molecule looks from the outside: the permanent dipole moment it carries on its own, and the induced dipole it builds in response to an electric field. After this chapter you can explain by geometry why H₂O is polar but CO₂ is not, and say in one line why refractive index, dispersion forces, and Raman scattering all start from the same quantity — polarizability.

Main 1 — The permanent dipole moment

A molecule whose center of positive charge and center of negative charge do not coincide at a single point carries a permanent dipole moment. Its magnitude is

\vec\mu = q \, \vec d

where $q$ is the separated charge and $\vec d$ (vector d) is the displacement vector pointing from the negative charge center to the positive one. In chemistry the direction of $\vec\mu$ (mu, the dipole moment) is taken from negative to positive — the opposite of the physics convention, but we use the chemistry convention throughout this book. The unit is the Debye, with 1 D $= 3.336 \times 10^{-30}$ C·m.

Geometry alone fixes the polarity. Water H₂O is bent, so its two O–H bond dipoles do not cancel and leave $\mu \approx 1.85$ D. Carbon dioxide CO₂, by contrast, is linear O=C=O, so its two C=O dipoles cancel exactly and $\mu = 0$ . HF is $\approx 1.83$ D; ammonia NH₃, with its pyramidal shape, is $\approx 1.47$ D. The key point: even with polar bonds, the molecule as a whole can be nonpolar.

Main 2 — Induced dipole and polarizability

A molecule with no permanent dipole is still not indifferent to an external field. When a field $\vec E$ is applied, the electron cloud is pushed slightly in the direction opposite to the field, and an induced dipole appears. In the linear response regime, where the field is not too strong,

\vec\mu_{\rm ind} = \alpha \, \vec E

so the induced dipole is directly proportional to the field. The proportionality constant $\alpha$ (alpha, the polarizability) is a molecular property measuring how “soft” the electron cloud is. In CGS units it has the dimension of a volume, of order $10^{-30}$ m³ for small molecules. Larger molecules, with more loosely bound electrons, have larger $\alpha$ . Strictly, for an anisotropic molecule $\alpha$ is a rank-2 tensor — a matrix with two indices linking the input direction to the output direction — but here we treat it as a scalar.

Main 3 — Why this matters

Polarizability shows up everywhere in chemistry and optics. The refractive index is tied directly to it through the Clausius–Mossotti relation, $n^2 - 1 \propto N\alpha$ , where $N$ is the number of molecules per unit volume. The dispersion forces between neutral molecules — the subject of chapter 11 — are the effect of each molecule’s instantaneous dipole polarizing its neighbor. Raman scattering selection rules decide whether a vibrational mode is active by the “change in polarizability.” A molecule’s response to a laser field works the same way.

The total dipole of a molecule in a field expands as

\vec\mu_{\rm total} = \vec\mu_0 + \alpha \vec E + O(E^2)

where $\vec\mu_0$ is the permanent dipole, the linear term is the polarizability, and the term of order $E^2$ defines the hyperpolarizability, the basis for nonlinear optics.

In Python

# Plot the induced dipole mu_ind(E) = alpha * E for three polarizabilities.
# In the linear response regime all three curves should be straight lines.
import numpy as np
import matplotlib.pyplot as plt

# Electric field in atomic units (a.u.): 1 a.u. ~ 5.14e11 V/m
E = np.linspace(0, 0.05, 100)
alphas = [1, 5, 20]            # polarizabilities (a.u.)

for alpha in alphas:
    mu_ind = alpha * E         # linear response: mu_ind = alpha * E
    plt.plot(E, mu_ind, label=f"alpha = {alpha} a.u.")

plt.xlabel("electric field E (a.u.)")
plt.ylabel("induced dipole mu_ind (a.u.)")
plt.title("Induced dipole in the linear response regime")
plt.legend()
plt.show()

If all three curves come out as straight lines through the origin, you have checked linear response by hand. Note, though, that for most molecules nonlinear corrections — the hyperpolarizability term — start to matter beyond $E \sim 0.01$ a.u. The linear regime drawn here is a deliberate approximation.

To the next chapter

Chapter 11: Intermolecular forces takes the permanent and induced dipoles of this chapter and shows how they lead to attractions between molecules — dipole–dipole, dipole–induced-dipole, and the purely dispersive force. How a single quantity, polarizability, explains everything from the boiling point of a liquid to the condensation of a gas is the subject of that chapter.