Bond dipole (${\mathbf{p}}_b$)

A bond dipole represents the separation of positive and negative electric charge associated with a chemical bond. In a neutral two-point charge model put $+q$ at ${\mathbf{r}}_A$ and $-q$ at ${\mathbf{r}}_B$, with $q>0$. Its physical electric dipole moment isIf $B$ is the negative end, the physical electric dipole moment points from $B$ to $A$. The customary crossed-tail chemical arrow points towards the negative, more electronegative end, from $A$ to $B$. It therefore has the opposite direction. State the convention before doing vector addition; use physical electric dipole moments in the interaction potential energy $U=-\mathbf{p}\cdot \mathbf{E}$, where $\mathbf{E}$ is the electric field.^[ref]

For a fixed neutral molecule, its complete electric dipole moment follows from all atomic nuclei and the electron density, not just point charges assigned to sites. If $n(\mathbf{r})$ is the expected number of Electrons per unit volume, the corresponding charge density is ${\rho }_e(\mathbf{r})=-en(\mathbf{r})$; the nuclear terms contribute separately. Partitioning that distribution into bond dipoles is a model. Lone pairs and electric polarization within a chemical atom can contribute as well. A measured molecular electric dipole moment cannot uniquely recover every individual bond dipole or partial atomic charge.

Molecular dipole cancellation occurs when nonzero contributions to a molecule's electric dipole moment cancel under vector addition. In a bond dipole model, use ${\mathbf{p}}_{\mathrm{mol}}={\sum }_b{\mathbf{p}}_b$; adding magnitudes instead discards the molecular geometry. A symmetric linear molecule with two equivalent opposed bond dipoles has ${\mathbf{p}}_1=p_0\hat{\mathbf{x}}$ and ${\mathbf{p}}_2=-p_0\hat{\mathbf{x}}$, hence zero total electric dipole moment even though each chemical bond is polarized. Carbon dioxide provides the familiar molecular example; water's bent molecular geometry permits a nonzero total.^[ref]

For an original geometry calculation, give a fictitious three-site model $A{B}_2$ a central effective electric charge $+2q$ at the origin and two terminal electric charges $-q$ atHere $r$ is the stipulated bond length, $\theta$ the angle between the two connections, and $q>0$ is independent of any electronegativity scale. Equivalently, split the central $+2q$ into two colocated $+q$ contributions, one for each neutral bond dipole. With $p_0=qr$, the physical electric dipole moments point from the terminal sites towards the central site:The sideways components cancel; the components along the angle bisector add. Thus $|{\mathbf{p}}_{\mathrm{mol}}|=2p_0\cos (\theta /2)$. At $\theta =180^{\circ }$ the result is zero; at the stipulated $\theta =120^{\circ }$ it is exactly $p_0$. This is an angle example for a model, not the measured angle or electric dipole moment of water.

Opposition alone is insufficient: a linear molecule with unequal opposed bond dipoles of magnitudes $p_1$ and $p_2$ retains magnitude $|p_1-p_2|$. Conversely, symmetry can enforce exact cancellation beyond a particular bond dipole partition. If the complete charge density invariant under point reflection about the origin satisfies $\rho (\mathbf{r})=\rho (-\mathbf{r})$, contributions to $\int \mathbf{r}\rho (\mathbf{r})d^{3}r$ cancel in pairs. That proof uses the whole electric charge distribution, including lone pairs and atomic nuclei. It explains why molecular symmetry is stronger evidence for zero permanent electric dipole moment than a rule based only on tabulated electronegativity differences.

A permanent molecular electric dipole moment is specified relative to the molecule's orientation. An ensemble with random orientations can have zero mean laboratory vector while each molecule is polar. That orientational average differs from internal molecular dipole cancellation. The Stark effect and molecular beam electric resonance probe molecular electric dipole moments through state-dependent responses to an electric field.