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In bent molecular geometry, two neighbours and their central chemical atom form a noncollinear triangle, with an interior bond angle strictly between $0^{\circ }$ and $180^{\circ }$. Water, ${\mathrm{H}}_2\mathrm{O}$, is the standard example. The NIST CCCBDB experimental water entry, internal coordinates, citing Hoy and Bunker (1979) gives the equilibrium H–O–H bond angle as $104.4776^{\circ }$ with quoted uncertainty $0.0019^{\circ }$, usually rounded to $104.5^{\circ }$.

The oxygen in water has two bonding electron domains and two lone pairs. Its ideal electron pair geometry is tetrahedral, while the three atomic nuclei give bent molecular geometry. Three points always lie in a plane; this does not make the four-domain scaffold planar. Nor does the word bent by itself determine an electron domain count: the same atomic shape can arise from other domain arrangements.

In an illustrative model with two equal bond-dipole contributions of magnitude $p$ separated by bond angle $\theta$, their sum of vectors points along the bisector and has magnitude $2p\cos (\theta /2)$. At $180^{\circ }$ it vanishes; at $104.4776^{\circ }$ it is approximately $1.2247p$. This ratio is a geometric calculation, not a prediction of water's measured electric dipole moment. The entire electron density, including contributions represented by lone pairs, matters. Water is a polar molecule; chemical polarity helps explain its interactions with other polar molecules. See McMurry, Organic Chemistry, §2.2: water, lone-pair contributions, and polarity.

Chemical polarity concerns unequal distribution of electron density in a chemical bond or molecule. A polar molecule has a permanent electric dipole moment; the effects of individual bonds must be combined with their spatial arrangement. In a symmetric linear carbon dioxide molecule, opposing bond-dipole contributions cancel. In water, its bent molecular geometry prevents that cancellation, giving a permanent electric dipole moment.^[ref]

The electric dipole moment measures the first spatial moment of signed electric charge. For localized point charges $q_i$ at positions ${\mathbf{r}}_i$, or an integrable charge density $\rho (\mathbf{r})$ with a finite first moment, define^[ref].

The electric dipole moment is a vector with SI units $\mathrm{C}\mathrm{m}$: coulombs times length in metres. Opposite electric charges can cancel in $Q$ while contributing in the same direction to $\mathbf{p}$. For a neutral pair with positive electric charge $q>0$ at ${\mathbf{r}}_{+}$ and negative electric charge $-q$ at ${\mathbf{r}}_{-}$,Thus the electric dipole moment points from negative to positive electric charge, not along the attraction between the two point charges. ^[ref].

Changing the coordinate origin changes the position vectors, not the physical electric charge distribution. If the new origin is at $\mathbf{a}$ relative to the old origin, ${\mathbf{r}}_i^{\prime }={\mathbf{r}}_i-\mathbf{a}$, so direct substitution givesConsequently charge neutrality makes the electric dipole moment independent of the origin. For a system with nonzero total electric charge, always specify the origin: a single point charge at the old origin has $\mathbf{p}=0$ but ${\mathbf{p}}^{\prime }=-Q\mathbf{a}$. This is coordinate bookkeeping, not a newly created electric dipole. The full electric potential stays the same at the same physical point; the terms of its electric multipole expansion and the observation coordinates change together.

As a stipulated numerical example, put $+2\mathrm{nC}$ at $(1,0,0)\mathrm{mm}$ and $-2\mathrm{nC}$ at $(-1,0,0)\mathrm{mm}$. Their total electric charge is zero and $\mathbf{p}=(4\times 10^{-12},0,0)\mathrm{C}\mathrm{m}$. Translating the origin by $(3,-2,0)\mathrm{mm}$ leaves this electric dipole moment unchanged. Adding $+1\mathrm{nC}$ at the old origin instead gives $Q=1\mathrm{nC}$ and ${\mathbf{p}}^{\prime }=(1,2,0)\times 10^{-12}\mathrm{C}\mathrm{m}$. Distinguishing these two cases prevents the mistaken claim that every electric dipole moment is independent of the origin.

The usefulness of the electric dipole moment is compression: many internal electric charges determine one vector governing the leading distant electric potential of a neutral localized object when $\mathbf{p}\ne 0$. ^[ref]. A polar molecule need not literally contain two point charges; its atomic nuclei and electron density collectively give an electric dipole moment. For example, the chemical polarity of water reflects its bent molecular geometry, whereas the symmetric linear molecular geometry of carbon dioxide makes its permanent electric dipole moment vanish. ^[ref]. A collection of such electric dipole moments per unit volume gives electric polarization. ^[ref].

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.

Why do methane, Ammonia, and water have different shapes when each central chemical atom is surrounded by eight valence electrons? The answer separates the positions of atomic nuclei from the electron domains used in a VSEPR model.

Molecular geometry describes the three-dimensional arrangement of the atomic nuclei in a molecule, including bond lengths and bond angles. For a larger molecule, a shape name often describes just the neighbours of one central chemical atom. A Lewis structure records connections and electron bookkeeping; its flat layout does not fix molecular geometry.

In the VSEPR model, electron pair geometry includes both bonding electron domains and lone pairs, whereas molecular geometry counts only the chemical atoms. In methane, four bond pairs and no lone pairs give tetrahedral molecular geometry. In Ammonia, three bond pairs and one lone pair give trigonal pyramidal molecular geometry. In water, two bond pairs and two lone pairs give bent molecular geometry. All three use the same ideal tetrahedral electron pair geometry; this does not make their actual bond angles equal. See Flowers, Theopold, Langley and Robinson, Chemistry 2e, §7.6: electron-pair and molecular geometries.

A static drawing also needs a physical interpretation. The atomic nuclei undergo oscillations; an arrangement at a minimum of potential energy and an experimentally averaged arrangement need not have identical bond lengths or bond angles. This distinction matters when comparing measurements with quantum mechanics calculations; see the IUPAC Gold Book: experimental bond-length conventions. Gas electron diffraction gives a concrete route from a laboratory signal to molecular geometry rather than inferring shape from a drawing alone.

The VSEPR model predicts approximate molecular geometry by arranging the electron domains around a central chemical atom to reduce their mutual repulsion. Start with a Lewis structure, count the domains, and then distinguish the resulting electron pair geometry from the positions of the chemical atoms themselves. A lone pair counts as a domain even though it is not another chemical atom.^[ref]

Three worked counts have the same total of eight valence electrons. In methane, carbon's four plus the contributions of four hydrogen chemical atoms give $4+4=8$: four bond pairs, no lone pairs, and tetrahedral molecular geometry. In Ammonia, nitrogen's five plus three from hydrogen give $5+3=8$: three bond pairs and one lone pair, producing trigonal pyramidal molecular geometry. In water, oxygen's six plus two from hydrogen give $6+2=8$: two bond pairs and two lone pairs, producing bent molecular geometry. All three have four electron domains; their molecular geometries differ.

The regular tetrahedral directions can be represented by the unit vectorsEach pair has dot product $-1/3$, so the ideal bond angle is $\theta =\arccos (-1/3)\approx 109.47^{\circ }$. This is a geometric calculation, rather than a prediction that every four-domain molecule has that exact bond angle. Lone pairs and different kinds of chemical bonds distort the ideal arrangement; water's H–O–H angle is about $104.5^{\circ }$.^[ref]

The model is attractive because small-number bookkeeping produces a spatial prediction without first solving for every molecular orbital. Shape then helps explain chemical polarity and which hydrogen bond donor or hydrogen bond acceptor sites are accessible. It is a qualitative model, not a full calculation of the electronic energy or a universal solution for every chemical compound. Gillespie and Nyholm's 1957 review ^[ref] is a historical anchor for its development.