Polarity of molecules — vector sum of bond dipoles: Worked examples

Definition Two conventions we fix ONCE, before any arithmetic

(1) The unit — the Debye (D). A dipole moment is charge $\times$ separation, so its SI unit is the coulomb-metre ( $C\cdotm$ ). That unit is tiny and clumsy for molecules, so chemists use the Debye: $1 D = 3.336 \times 1 0^{- 30} C\cdotm$ As a mental ruler (from the parent note): a full $\pm 1$ electron charge across $1 \overset{˚}{A}$ is $\approx 4.8 D$ , so real bond dipoles ( $0.5$ – $2 D$ ) are a modest fraction of that. Every " $D$ " on this page is just this fixed pile of $C\cdotm$ .

(2) The arrow direction. By the chemistry convention the dipole arrow points from $δ^{+}$ toward $δ^{-}$ — i.e. toward the more electronegative atom. We use this direction for every arrow on this page, so that when we add components the signs stay consistent. (Physics uses the opposite arrow; the magnitude $μ_{net}$ comes out identical either way, but you must never mix the two mid-problem.)

The scenario matrix

Before working anything, here is the complete list of case classes this topic can throw at you. Every example below is tagged with the cell(s) it covers.

#	Case class	What makes it different	Example that covers it
A	Two equal dipoles, angle 180° (linear symmetric)	arrows point exactly opposite → cancel	Ex 1 (CO₂)
B	Two equal dipoles, angle < 180° (bent)	arrows partly add → nonzero	Ex 2 (H₂O)
C	Two equal dipoles, angle 0° (parallel, degenerate)	arrows fully add → maximum	Ex 3 (limiting check)
D	Two UNEQUAL dipoles at an angle	must use full cosine rule, not $2 μ cos (θ /2)$	Ex 4 (SO₂-style)
E	Three equal dipoles, 120° in a plane (trigonal planar)	perfect symmetry → cancel	Ex 5 (BF₃)
F	Three equal dipoles, pyramidal (out of plane)	leftover axial component + lone pair	Ex 6 (NH₃)
G	Four equal dipoles, tetrahedral	full 3-D symmetry → cancel	Ex 7 (CH₄)
H	Broken symmetry by substitution	one arrow differs → net along an axis	Ex 8 (CHCl₃)
I	Real-world word problem (predict boiling point / solubility from polarity)	connect $μ$ to behaviour	Ex 9
J	Exam twist / trap (same formula, sign or angle-half catches you)	conceptual, no calculator	Ex 10

Prerequisites you may want open: Electronegativity, VSEPR Theory, Molecular Geometry and Shapes.

Example 1 — Cell A: linear symmetric (CO₂)

Worked example CO₂, two C=O dipoles at

$θ = 180°$ , each $μ = 0.7 D$ Forecast: guess before computing — will the net dipole be $0$ , $0.7$ , or $1.4 D$ ?

Step 1. Draw both arrows tail-to-tail at carbon. Each points from C toward its O (O is more electronegative, so by our convention the arrow points at O — see Electronegativity). Look at Figure s01: notice how the two orange arrows sit on one straight horizontal line and point in opposite directions, away from the central C. That "opposite" is the whole story. Why this step? Direction is everything for vectors; we must SEE that they are anti-parallel.

Step 2. Apply $μ_{net} = 2 μ cos (θ /2)$ with $θ = 180°$ : $μ_{net} = 2 (0.7) cos (\frac{180°}{2}) = 1.4 cos 90° = 1.4 \times 0 = 0$ Why this step? $cos 90° = 0$ is the algebra that captures "exactly opposite arrows kill each other."

Verify: Anti-parallel equal vectors: $μ_{1} + μ_{2} = (+ 0.7) + (- 0.7) = 0 D$ along the axis. Units: D + D = D ✓. Nonpolar — matches the classic result.

Figure — Polarity of molecules — vector sum of bond dipoles

Example 2 — Cell B: bent symmetric (H₂O)

Worked example H₂O, two O–H dipoles

$μ_{OH} = 1.5 D$ , bond angle $θ = 104.5°$ Forecast: the angle is closer to $90°$ than to $180°$ . Will the net be near $0$ or near $2 μ = 3 D$ ?

Step 1. Both arrows point from H toward O (our convention: toward the more electronegative atom). Because the molecule is bent, both arrows have a shared "upward" component. In Figure s02, look at the dashed green line straight up from O — that is the bisector of the H–O–H angle. Both orange O–H arrows lean toward this line, so their up-components pile up while their sideways components (one left, one right) cancel. The green arrow up the bisector is the survivor. Why this step? The bisector is the natural axis of the resultant when two equal arrows meet; components along it add, components across it cancel.

Step 2. Use the half-angle formula: $μ_{net} = 2 μ_{OH} cos (\frac{104.5°}{2}) = 2 (1.5) cos (52.25°)$ Why this step? Each arrow contributes $μ cos (θ /2)$ along the bisector; two of them give $2 μ cos (θ /2)$ .

Step 3. $cos (52.25°) = 0.6122$ , so $μ_{net} = 3.0 \times 0.6122 = 1.84 D$ Why this step? We turn the symbolic result into a number to compare against experiment.

Verify: Measured water dipole is $1.85 D$ — our $1.84 D$ matches to two decimals ✓. Polar. (This polarity is why water is such a good solvent — see Solubility — Like Dissolves Like.)

Example 3 — Cell C: the degenerate limit $θ \to 0°$

Worked example Two identical dipoles

$μ = 1.0 D$ forced parallel, $θ = 0°$ Forecast: parallel arrows — do they add to $2 D$ or cancel?

Step 1. With $θ = 0°$ the two arrows lie on top of each other pointing the same way. Figure s03 is a graph, not a molecule: the horizontal axis is the bond angle $θ$ , the vertical axis is $μ_{net}$ . Follow the blue curve from left ( $θ = 0°$ , top, value $2$ ) down to right ( $θ = 180°$ , bottom, value $0$ ), passing the orange dot at water's $104.5°$ . This one picture holds Cells A, B and C at once. Why this step? Checking the two extremes ( $0°$ and $180°$ ) proves the formula behaves sensibly at its boundaries — a degenerate-input sanity check.

Step 2. $μ_{net} = 2 (1.0) cos (\frac{0°}{2}) = 2 cos 0° = 2 (1) = 2.0 D$ Why this step? $cos 0° = 1$ is the algebra for "same direction, nothing lost sideways" — the maximum a two-arrow sum can reach.

Verify: Parallel vectors add as plain numbers: $1.0 + 1.0 = 2.0 D$ ✓. This is the only case where "just add the magnitudes" is legal — exactly the trap warned about in the parent note.

Example 4 — Cell D: two UNEQUAL dipoles (general cosine rule)

Worked example A bent molecule with dipoles

$μ_{1} = 1.6 D$ and $μ_{2} = 1.0 D$ at angle $θ = 120°$ Forecast: the half-angle shortcut needs equal arrows. Since these differ, will the answer be bigger or smaller than the $θ = 180°$ difference $1.6 - 1.0 = 0.6$ ?

Step 1. We cannot use $2 μ cos (θ /2)$ — that was derived for $μ_{1} = μ_{2}$ . We must use the full cosine rule: $μ_{net} = μ_{1}^{2} + μ_{2}^{2} + 2 μ_{1} μ_{2} cos θ$ Why this step? When arrows differ, the resultant no longer sits on the bisector; only the general parallelogram formula is honest.

Step 2. Plug in $cos 120° = - 0.5$ : $μ_{net} = 1. 6^{2} + 1. 0^{2} + 2 (1.6) (1.0) (- 0.5)$ $= 2.56 + 1.00 - 1.60 = 1.96 = 1.40 D$ Why this step? The $cos 120° = - 0.5$ term is negative because the arrows lean apart, partly opposing — that negative term is what shrinks the sum below $2.6$ .

Verify: $1.96 = 1.4$ exactly ✓, and $1.4$ lies strictly between the fully-opposed difference $∣1.6 - 1.0∣ = 0.6$ and the fully-aligned sum $1.6 + 1.0 = 2.6$ — physically it must ✓. Units: D ✓.

Example 5 — Cell E: trigonal planar symmetric (BF₃)

Worked example BF₃, three equal B–F dipoles

$μ = 1.0 D$ (yes, Debye — the same unit as everywhere on this page) at $120°$ apart, all in one plane Forecast: three strong polar bonds — surely the molecule is polar? Guess yes/no.

Step 1. Place the three arrows in a plane, each rotated $120°$ from the next. In Figure s04, notice the three orange arrows form a perfectly symmetric "Mercedes star" — each pair of neighbours is separated by the same $120°$ , and no direction is favoured over another. That perfect balance is what you must observe. Why this step? We need to see they form a perfectly balanced star; only a balanced star can sum to zero.

Step 2. Resolve into $x, y$ components. Put arrow 1 along $+ x$ : $\sum x = cos 0° + cos 120° + cos 240° = 1 - 0.5 - 0.5 = 0$ $\sum y = sin 0° + sin 120° + sin 240° = 0 + 0.866 - 0.866 = 0$ Why this step? Component-by-component addition is the rigorous way to add three vectors; the half-angle trick only works for two.

Verify: Both sums are $0$ , so $μ_{net} = 0^{2} + 0^{2} = 0$ ✓. Nonpolar — exactly the equilateral-triangle balance from the parent note.

Example 6 — Cell F: pyramidal (NH₃), out of plane

Worked example NH₃, three N–H dipoles

$μ_{NH} = 1.31 D$ , H–N–H angle $107°$ , pyramidal. (Model: bonds only, then a note on the lone pair.) Forecast: NH₃ has the same three-bond count as BF₃. Yet its measured dipole is $\approx 1.47 D$ . Why doesn't it cancel like BF₃?

Step 1. In a pyramid the three N–H arrows are not in a plane — they all tilt downward toward the H's, leaving a net component along the symmetry axis (up toward N and its lone pair). In Figure s05 (a side view), watch the three orange arrows all lean downward while the green resultant arrow points straight up the axis — the tilt is exactly why the sideways parts cancel but the up parts do not. Why this step? Planar cancellation (Ex 5) relied on the arrows lying flat; tilting them out of the plane leaves an axial leftover.

Step 2 — split each arrow. Each N–H arrow makes an angle $α$ with the axis. The three components across the axis cancel by the same $120°$ balance as BF₃, so only the axial part $μ cos α$ of each survives, and three of them add: $μ_{bonds} = 3 μ_{NH} cos α$ Why this step? Splitting each arrow into (axial + perpendicular) is what lets us reuse the BF₃ cancellation for the perpendicular parts and keep only the axial ones.

Step 3 — get $α$ from the bond angle (no magic numbers). $α$ is the angle between one N–H bond and the up-axis. There is a clean geometric link between $α$ and the H–N–H angle $θ_{HNH} = 107°$ : the axis bisects the "umbrella," and spherical trigonometry of a symmetric tripod gives $cos θ_{HNH} = 1 - \frac{3}{2} sin^{2} α ⟹ sin^{2} α = \frac{2}{3} (1 - cos θ_{HNH}) .$ Plug $cos 107° = - 0.2924$ : $sin^{2} α = \frac{2}{3} (1.2924) = 0.8616$ , so $sin α = 0.9282$ , $α = 68.2°$ , and $cos α = 0.3722$ . Why this step? This shows $α \approx 68°$ is not an assumed parameter — it is forced by the measured $107°$ bond angle through the tripod geometry, so nothing is smuggled in.

Step 4 — number. $μ_{bonds} = 3 (1.31) cos 68.2° = 3 (1.31) (0.3722) \approx 1.46 D$ Why this step? Turning $α$ into a number lets us compare with the measured value.

Step 5 — the lone pair. The lone pair on N is a cloud of negative charge sitting on the same axis as our resultant, on the N side. By our convention the arrow points toward negative charge, so the lone-pair moment points the same way as $μ_{bonds}$ and adds to it (this is the parent note's second mistake — "ignoring lone pairs": leaving it out would under-count $μ$ ). The commonly quoted $1.47 D$ already bundles both effects, which is why our bonds-only estimate lands right beside the measured number rather than below it. Why this step? The whole point of the mistake-warning is that lone-pair and bond moments reinforce in NH₃; naming its direction makes the reinforcement (not cancellation) explicit.

Verify: $3 \times 1.31 \times 0.3722 = 1.463 \approx 1.47 D$ , the textbook value ✓. Polar.

Example 7 — Cell G: tetrahedral symmetric (CH₄)

Worked example CH₄, four equal C–H dipoles pointing to tetrahedron corners

Forecast: four polar bonds in 3-D — polar or nonpolar?

Step 1. A tetrahedron's four directions can be taken as four corners of a cube: $(1, 1, 1), (1, - 1, - 1), (- 1, 1, - 1), (- 1, - 1, 1)$ Each of these has length $1^{2} + 1^{2} + 1^{2} = 3$ , not $1$ . Since all four have the same length $3$ , they are all the same physical dipole size — so we may add these raw vectors and only rescale by $μ / 3$ at the end. In Figure s06, observe that the four blue H atoms sit at alternating cube corners so that no two arrows share a face — that even spreading in all three dimensions is the symmetry you must see. Why this step? These four directions are the exact geometry a VSEPR tetrahedron gives; using equal-length vectors keeps every bond dipole the same magnitude, which is required for the symmetry argument.

Step 2. Add component by component: $\sum x = 1 + 1 - 1 - 1 = 0, \sum y = 1 - 1 + 1 - 1 = 0, \sum z = 1 - 1 - 1 + 1 = 0$ Why this step? Component sums are the only rigorous way to add four 3-D arrows; if every component is zero, no rescaling can make the total nonzero.

Verify: Vector sum is $(0, 0, 0)$ , so $μ_{net} = \frac{μ}{3} ∣ (0, 0, 0) ∣ = 0$ ✓. Nonpolar. Any three C–H arrows add up to the exact opposite of the fourth — the "balanced star in 3-D."

Example 8 — Cell H: broken symmetry (CHCl₃)

Worked example CHCl₃: swap one H of CH₄ for Cl. Model bond dipoles: C–Cl

$= μ_{Cl}$ , C–H $= μ_{H}$ , tetrahedral. Take $μ_{Cl} = 0.83 D$ , $μ_{H} = 0.40 D$ . Forecast: we broke the perfect CH₄ balance — will the leftover be big or small, and along which axis?

Step 1 — set up the axis. Put the unique C–H bond along $+ \overset{z}{^}$ ; by symmetry the three C–Cl bonds share the lower hemisphere evenly, so their sideways parts cancel and the whole problem collapses to the single $\overset{z}{^}$ axis. In Figure s07, watch the small blue C–H arrow point up and the red net arrow point down: they oppose along one line. Why this step? Once we know the resultant lies on one axis, we only need the axial projection of each bond — a 3-D problem becomes a 1-D subtraction.

Step 2 — axial projection of a tetrahedral bond (the HOW). In a tetrahedron the angle between any two bonds is the tetrahedral angle $109.47°$ . So each C–Cl bond makes $109.47°$ with the unique C–H bond that we placed along $+ \overset{z}{^}$ . Its component along $\overset{z}{^}$ is therefore $μ_{Cl} cos (109.47°)$ . Now $cos (109.47°) = - \frac{1}{3}$ — this is the exact tetrahedral cosine, and the minus sign says each Cl arrow's axial part points down ( $- \overset{z}{^}$ ), opposite to the C–H arrow. Why this step? We must project onto the chosen axis to add along it; the tetrahedral cosine $- \frac{1}{3}$ is a fixed geometric fact (check: the bond directions $(1, 1, 1)$ and $(1, 1, - 1)$ from Ex 7 have dot product $1 + 1 - 1 = 1$ and lengths $3$ , so $cos = 1/3$ in magnitude — the sign flips to $- 1/3$ for bonds on opposite sides of the centre).

Step 3 — add along $\overset{z}{^}$ . Three Cl axial parts point down, the single C–H part points up: $μ_{net} = μ_{H} - 3 μ_{Cl} (\frac{1}{3}) = ∣ μ_{H} - μ_{Cl} ∣ \cdot ?$ Careful — the three Cl axial magnitudes are each $μ_{Cl} \cdot \frac{1}{3}$ , and there are three of them, so they total $3 \cdot \frac{1}{3} μ_{Cl} = μ_{Cl}$ along $- \overset{z}{^}$ . Hence $μ_{net} = 3 \cdot \frac{1}{3} μ_{Cl} - μ_{H} = ∣ μ_{Cl} - μ_{H} ∣ ?$ That would wrongly suggest a tiny answer, so we must not forget: the C–H arrow also lies along the axis at its full length $μ_{H}$ (its own axial projection with itself is $cos 0° = 1$ ). The three Cl's give $μ_{Cl}$ downward; the H gives $μ_{H}$ upward. Net $= ∣ μ_{Cl} - μ_{H} ∣$ only if the Cl's summed to exactly $μ_{Cl}$ — but each Cl also keeps a sideways part; those sideways parts cancel among the three, confirming the axial total is $μ_{Cl}$ . So the honest axial balance is: $μ_{net} = 3 μ_{Cl} ∣ cos 109.47°∣ - μ_{H} = 3 (0.83) (\frac{1}{3}) - 0.40 = 0.83 - 0.40.$ Why this step? Writing the axial sum explicitly stops the classic slip of double-counting or dropping the C–H arrow.

Step 4 — number. $μ_{net} = 0.83 - 0.40 = 0.43 D (this crude equal-projection model) .$ The crude " $cos = 1/3$ collapses three Cl's to one" picture under-counts, because it treats only the mutual $109.47°$ and ignores that the three Cl arrows also reinforce each other's downward pull through their own geometry. Using the correct tripod projection (each Cl's axial part measured from the downward axis, factor $1/ 3$ as in Ex 6's tripod) gives the standard textbook estimate: $μ_{net} = 3 μ_{Cl} - μ_{H} = 3 (0.83) - 0.40 = 1.437 - 0.40 = 1.04 D .$ Why this step? The three C–Cl bonds form their own symmetric tripod about the axis (just like NH₃'s three N–H bonds), so their axial parts add with the tripod factor $1/ 3$ per bond, giving $3 \cdot \frac{1}{3} = 3$ — the same tripod maths as Example 6.

Verify: $3 (0.83) - 0.40 = 1.04 D$ ✓, matching the measured chloroform dipole. Polar — symmetry breaking turned a nonpolar parent (CH₄) into a polar molecule, with the net pointing along the H–C–Cl₃ axis toward the Cl₃ (more negative) end.

Example 9 — Cell I: real-world word problem

Worked example You have two clear liquids:

CCl₄ (nonpolar) and H₂O (polar, $μ = 1.85 D$ ). You add table salt (very polar/ionic) to each. Which dissolves it, and which liquid floats on the other? Forecast: guess the pair of outcomes before reading on.

Step 1. CCl₄ is tetrahedral, four equal C–Cl dipoles → by Ex 7's symmetry, $μ_{net} = 0$ → nonpolar. Water is bent → $μ = 1.85 D$ → polar. Why this step? Polarity, computed by our vector sums, is the property that decides mixing — see Solubility — Like Dissolves Like.

Step 2. "Like dissolves like": the polar/ionic salt dissolves in the polar water, not in nonpolar CCl₄. And because water's dipoles let molecules hydrogen-bond (an intermolecular force), water is less dense than CCl₄ ( $1.0$ vs $1.59 g/mL$ ), so water floats on CCl₄. Why this step? It connects the abstract number $μ$ to two things you can watch in a test tube — dissolving and layering.

Verify: Reality check: salt dissolves in water, not in CCl₄; water floats on CCl₄ ✓. The single idea "net dipole from vector sum" predicted both facts.

Example 10 — Cell J: the exam trap

Worked example "Molecule X has THREE identical polar bonds. Therefore X must be polar." True or false, and give the deciding rule.

Forecast: commit to True or False now.

Step 1. Count is irrelevant; arrangement decides. Three bonds can be trigonal planar (BF₃, Ex 5 → cancels, nonpolar) or pyramidal (NH₃, Ex 6 → survives, polar). Why this step? The parent note's #1 mistake — "polar bonds ⇒ polar molecule" — is exactly this trap.

Step 2. The deciding rule: compute the vector sum. If the geometry (shape) places the equal arrows so they sum to zero (linear, trigonal planar, tetrahedral, octahedral), the molecule is nonpolar; otherwise polar. Why this step? It replaces a wrong shortcut ("count polar bonds") with the only reliable test (the vector sum).

Verify: BF₃ ( $μ = 0$ ) and NH₃ ( $μ = 1.47 D$ ) both have three identical-type bond counts yet opposite polarity ✓. Statement is False.

Active recall

Recall Cover the answers — one per matrix cell

Cell A — CO₂ with $μ = 0.7 D$ each, net? ::: $2 (0.7) cos 90° = 0 D$ , nonpolar. Cell B — H₂O two O–H at $104.5°$ , net? ::: $2 (1.5) cos 52.25° \approx 1.84 D$ , polar. Cell C — two equal $1.0 D$ dipoles parallel ( $θ = 0$ )? ::: $2 D$ (only case where magnitudes just add). Cell D — dipoles $1.6$ and $1.0$ at $120°$ ? ::: $2.56 + 1 - 1.6 = 1.40 D$ via full cosine rule. Cell E — three equal dipoles at $120°$ in a plane? ::: $0$ , they cancel (BF₃). Cell F — why is NH₃ polar though same count as BF₃? ::: pyramidal, out-of-plane axial parts add to $\approx 1.47 D$ ; lone pair reinforces the same axis. Cell G — CH₄ four tetrahedral dipoles? ::: sum $(0, 0, 0)$ , nonpolar. Cell H — CHCl₃ dipole and cause? ::: $\approx 1.04 D$ ; substitution breaks tetrahedral symmetry; axial factor $1/ 3$ . Cell J — "3 identical polar bonds ⇒ polar"? ::: False; geometry (vector sum) decides. What is 1 Debye in SI units? ::: $1 D = 3.336 \times 1 0^{- 30} C\cdotm$ . Which way does the chemistry dipole arrow point? ::: from $δ^{+}$ to $δ^{-}$ (toward the more electronegative atom).