Polarity of bonds — dipole moment μ = q·d: Worked examples

The scenario matrix

Every dipole-moment problem is one of these cells. The examples below are labelled with the cell they cover.

#	Case class	What makes it different	Example
A	Single polar bond → find $μ$	Straight $q \cdot d$ , then convert to Debye	Ex 1
B	Back-solve for $q$ or $δ$	You're given $μ$ , must invert the formula	Ex 2
C	% ionic character ( = observed $μ$ ÷ hypothetical $q = e$ dipole, ×100)	Compare real polarity to a fully-transferred electron	Ex 3
D	Symmetric → full cancellation ( $μ = 0$ )	Vectors sum to zero by symmetry	Ex 4
E	Bent / partial cancellation ( $μ \neq = 0$ )	Only the surviving component counts (needs $cos$ )	Ex 5
F	Same geometry, opposite result	Direction of bond arrows flips the answer	Ex 6
G	Real-world word problem	Translate physical words into $q, d$	Ex 7
H	Exam twist (degenerate / trap)	Looks polar but isn't, or vice versa	Ex 8

Percent ionic character means: what fraction of a full electron transfer does the real bond achieve? Quantitatively $% ionic = \frac{μ _{observed}}{μ _{if q = e}} \times 100$ , worked out fully in Example 3.

Prerequisite links you'll lean on: Electronegativity (which end is $δ^{-}$ ), VSEPR Theory (the geometry), Ionic vs Covalent character (% ionic).

Example 1 — Case A: single polar bond

Worked example Find

$μ$ of HF HF has bond length $d = 0.917 \overset{˚}{A} = 0.917 \times 1 0^{- 10} m$ and effective partial charge $q = 0.41 e$ . Find $μ$ in Debye.

Forecast: HF is very polar (fluorine is the greediest atom). We will compute a number in Debye — will it be near $1$ , near $2$ , or bigger? Guess before reading on.

Step 1 — Turn the partial charge into coulombs. $q = 0.41 \times 1.6 \times 1 0^{- 19} = 6.56 \times 1 0^{- 20} C$ Why this step? The formula $μ = q \cdot d$ needs $q$ in coulombs. " $0.41 e$ " is a fraction of the electron charge, so we multiply by $e$ .

Step 2 — Multiply by the distance. $μ = q \cdot d = (6.56 \times 1 0^{- 20}) (0.917 \times 1 0^{- 10}) = 6.02 \times 1 0^{- 30} C\cdotm$ Why this step? This is the definition — charge times separation.

Step 3 — Convert to Debye. $μ = \frac{6.02 \times 1 0 ^{- 30}}{3.336 \times 1 0 ^{- 30}} \approx 1.80 D$ Why this step? Chemists live in Debye; divide by $3.336 \times 1 0^{- 30}$ .

Verify: Experimental $μ_{HF} \approx 1.82 D$ — our answer matches. Units: $C \times m = C\cdotm$ ✔. (For scale, the parent note found $μ_{HCl} \approx 1.03 D$ from $q = 0.17 e$ , so HF being larger fits fluorine being the more electronegative partner.)

Example 2 — Case B: back-solve for the partial charge

Worked example What fraction of an electron is displaced in HBr?

HBr has measured $μ = 0.82 D$ and bond length $d = 1.41 \overset{˚}{A}$ . Find the partial charge $q$ (as a fraction of $e$ , i.e. $δ$ ).

Forecast: HBr is less polar than HCl (Br is less greedy than Cl, so recall the parent note's $δ_{HCl} = 0.17$ ). So expect $δ < 0.17$ . Guess.

Step 1 — Put $μ$ back into SI. $μ = 0.82 \times 3.336 \times 1 0^{- 30} = 2.74 \times 1 0^{- 30} C\cdotm$ Why this step? To invert $μ = q \cdot d$ we need everything in matching SI units.

Step 2 — Solve for $q$ . $q = \frac{μ}{d} = \frac{2.74 \times 1 0 ^{- 30}}{1.41 \times 1 0 ^{- 10}} = 1.94 \times 1 0^{- 20} C$ Why this step? $μ = q \cdot d \Rightarrow q = μ / d$ . Division undoes the multiplication.

Step 3 — Express as a fraction of $e$ . $δ = \frac{q}{e} = \frac{1.94 \times 1 0 ^{- 20}}{1.6 \times 1 0 ^{- 19}} \approx 0.12$ Why this step? " $δ$ " means the charge measured in units of the electron charge (the fraction defined above).

Verify: $δ = 0.12 < 0.17$ (HCl from the parent note) — matches the electronegativity trend. Sanity: $q \cdot d = (1.94 \times 1 0^{- 20}) (1.41 \times 1 0^{- 10}) = 2.74 \times 1 0^{- 30}$ C·m, back to our starting $μ$ ✔.

Example 3 — Case C: percent ionic character

Worked example % ionic character of HBr

Using $μ_{obs} = 0.82 D$ and $d = 1.41 \overset{˚}{A}$ from Ex 2, find the percent ionic character.

Forecast: We already found $δ \approx 0.12$ . Since % ionic is essentially $δ \times 100$ , forecast ≈ 12%.

Step 1 — Compute the 100%-ionic dipole (full charge $q = e$ ). $μ_{ionic} = e \cdot d = (1.6 \times 1 0^{- 19}) (1.41 \times 1 0^{- 10}) = 2.256 \times 1 0^{- 29} C\cdotm$ Convert: $μ_{ionic} = 2.256 \times 1 0^{- 29} /3.336 \times 1 0^{- 30} \approx 6.76 D$ . Why this step? If HBr transferred a whole electron, this is the dipole we'd see over the same distance.

Step 2 — Divide observed by ionic. $% ionic = \frac{μ _{obs}}{μ _{ionic}} \times 100 = \frac{0.82}{6.76} \times 100 \approx 12.1%$ Why this step? The ratio tells us how close the real bond is to a fully transferred electron.

Verify: $12.1% \approx δ \times 100 = 12%$ — the two routes agree, as they must, since $% ionic = \frac{q d}{e d} = \frac{q}{e} = δ$ . See Ionic vs Covalent character.

Example 4 — Case D: full cancellation, $μ = 0$

Worked example Why

$CCl_{4}$ has $μ = 0$ Carbon tetrachloride: four polar C–Cl bonds arranged tetrahedrally (109.5° apart). Find the net $μ$ .

Forecast: Four polar bonds — surely polar? Look at the symmetry first before answering.

Step 0 — Read Figure 1. Figure 1 shows the central carbon (amber dot) with four cyan arrows fanning out to the four chlorine atoms. Each arrow is a bond dipole; by our convention its tail sits on C ( $δ^{+}$ ) and its head on Cl ( $δ^{-}$ ). Look at how symmetrically the four cyan heads are spread around the centre — that even spread is the whole reason they will cancel.

Figure 1 — CCl₄: four equal C→Cl bond-dipole arrows arranged tetrahedrally around carbon.

Step 1 — Draw all four bond dipole arrows from C to each Cl. Why this step? Cl is more electronegative than C, so each arrow points C → Cl ( $δ^{-}$ on Cl).

Step 2 — Group them by symmetry. A regular tetrahedron places the four Cl atoms perfectly symmetrically around C. Pick any three; their vector sum points exactly opposite to the fourth. Why this step? Symmetric arrangements let vectors cancel in pairs/groups.

Step 3 — Add as vectors. $μ_{net} = μ_{1} + μ_{2} + μ_{3} + μ_{4} = 0$ Mathematically, the four direction vectors pointing to the tetrahedron corners sum to zero. Why this step? Dipole moment is a vector — polarity of the molecule is the vector sum, not the count of polar bonds.

Verify: Experimental $μ_{CCl_{4}} = 0 D$ ✔. Numerically the four tetrahedral direction vectors $(1, 1, 1), (1, - 1, - 1), (- 1, 1, - 1), (- 1, - 1, 1)$ — each of length $3$ , not unit length, but all equal, so the equal bond dipoles line up with them — add to $(0, 0, 0)$ , checked below. VSEPR Theory gives this geometry.

Example 5 — Case E: bent molecule, partial cancellation

Worked example Net

$μ$ of $SO_{2}$ from its bond dipoles $SO_{2}$ is bent with an O–S–O angle of $119°$ . Each S–O bond dipole has magnitude $μ_{SO} = 1.62 D$ (pointing S → O). Find the net molecular $μ$ .

Forecast: Bent, so the two arrows don't fully cancel. Net will be less than $2 \times 1.62 = 3.24$ but more than 0. Guess a value.

Step 0 — Read Figure 2. Figure 2 shows sulfur (amber) at the apex; two cyan arrows point down-and-out to the two oxygens (tail on S = $δ^{+}$ , head on O = $δ^{-}$ ). The amber arrow straight down is the net dipole along the angle bisector. Notice how the two cyan arrows' sideways parts point in opposite directions (they will cancel) while their downward parts point the same way (they will add).

Figure 2 — SO₂ bent at 119°: the two cyan bond dipoles leave a net amber dipole along the bisector.

Step 1 — Point both arrows and find the bisector. Both S→O arrows point outward and downward. Their sideways (horizontal) components are equal and opposite; their components along the bisector add. Why this step? We resolve each vector into "along bisector" + "perpendicular to bisector". The perpendicular parts cancel by symmetry.

Step 2 — Keep only the surviving component. Each arrow makes angle $\frac{119°}{2} = 59.5°$ with the bisector, so its useful part is $μ_{SO} cos (59.5°)$ . $μ_{net} = 2 μ_{SO} cos (\frac{119°}{2})$ Why the cosine? $cos$ gives the fraction of a vector lying along the direction we care about (the bisector). That's exactly the projection we need.

Step 3 — Plug in. $μ_{net} = 2 (1.62) cos (59.5°) = 2 (1.62) (0.5075) \approx 1.64 D$ Why this step? Just arithmetic on the surviving components.

Verify: Experimental $μ_{SO_{2}} \approx 1.63 D$ ✔. And $0 < 1.64 < 3.24$ , matching our forecast. (Same $cos$ -of-half-angle trick as H₂O in the parent note.)

Example 6 — Case F: same geometry, opposite result

Worked example

$CO_{2}$ vs $SO_{2}$ — geometry is not the whole story Both look "3 atoms, X in the middle". Yet $CO_{2}$ has $μ = 0$ and $SO_{2}$ has $μ \approx 1.63 D$ . Explain using vectors, and confirm CO₂ = 0.

Forecast: What's different — the bonds or the shape? Decide before reading.

Step 0 — Read Figure 3. In Figure 3, the left panel draws $CO_{2}$ straight (linear): the two cyan C→O arrows point in exactly opposite directions along one line. The right panel draws $SO_{2}$ bent, so its two arrows tilt toward each other, leaving a surviving amber net arrow. Same two-arrow setup — only the angle between the cyan arrows differs.

Figure 3 — Left: linear CO₂ (arrows cancel). Right: bent SO₂ (arrows leave a net amber dipole).

Step 1 — Compare the angles. $CO_{2}$ is linear (180°); $SO_{2}$ is bent (119°). The difference is shape, set by lone pairs (see VSEPR Theory). Why this step? The formula $μ_{net} = 2 μ_{bond} cos (θ /2)$ shows the angle controls everything.

Step 2 — Put $θ = 180°$ into the cosine formula. $μ_{net} = 2 μ_{CO} cos (\frac{180°}{2}) = 2 μ_{CO} cos (90°) = 2 μ_{CO} \cdot 0 = 0$ Why this step? At $180°$ the two arrows point exactly opposite; $cos (90°) = 0$ kills the net.

Step 3 — Contrast with $θ = 119°$ . $cos (59.5°) = 0.51 \neq = 0$ , so $SO_{2}$ survives. Same "two equal bonds" — different angle, opposite outcome. Why this step? Shows one formula covers both the cancelling and non-cancelling case; the limit $θ \to 180°$ smoothly gives 0.

Verify: $cos (90°) = 0$ exactly ⇒ CO₂ net is $0 D$ ✔; $cos (59.5°) \neq = 0$ ⇒ SO₂ nonzero ✔.

Example 7 — Case G: real-world word problem

Worked example A chemist measures a "smeared" charge

A new diatomic molecule XY has a measured dipole moment of $μ = 2.50 D$ . Its bond length is $d = 1.60 \overset{˚}{A}$ . A student claims "a whole electron jumped from X to Y." Test the claim: what fraction of an electron actually moved?

Forecast: If a whole electron moved, $δ = 1$ . Real bonds are partly covalent, so expect $δ < 1$ . Guess.

Step 1 — Convert $μ$ to SI. $μ = 2.50 \times 3.336 \times 1 0^{- 30} = 8.34 \times 1 0^{- 30} C\cdotm$ Why this step? We need SI to divide by a metre length.

Step 2 — Extract $q$ , then $δ$ . $q = \frac{μ}{d} = \frac{8.34 \times 1 0 ^{- 30}}{1.60 \times 1 0 ^{- 10}} = 5.21 \times 1 0^{- 20} C, δ = \frac{q}{e} = \frac{5.21 \times 1 0 ^{- 20}}{1.6 \times 1 0 ^{- 19}} \approx 0.33$ Why this step? Same inversion as Ex 2; $δ$ tells us the actual charge moved.

Step 3 — Answer the claim. Only about $0.33 e$ moved, i.e. the bond is roughly 33% ionic — the student is wrong; no whole electron jumped. Why this step? The word problem asked a yes/no about physical reality; we translate it into $δ$ .

Verify: $δ = 0.33 < 1$ , so not fully ionic ✔. Also $% ionic = 33%$ , a plausibly polar-covalent bond. Round-trip: $0.33 \times 1.6 \times 1 0^{- 19} \times 1.60 \times 1 0^{- 10} /3.336 \times 1 0^{- 30} \approx 2.53 D$ ≈ 2.50 D ✔ (rounding).

Example 8 — Case H: exam twist (degenerate trap)

Worked example "cis vs trans" dichloroethene — the classic trap

Both cis- and trans-1,2-dichloroethene have the same atoms and same two polar C–Cl bonds. One has $μ \approx 1.9 D$ , the other $μ = 0$ . Which is which, and why?

Forecast: Same formula, same bonds — so what could possibly differ? (Answer: arrangement.) Predict which isomer is nonpolar.

Step 0 — Read Figure 4. In Figure 4, both isomers share the same C=C double bond (drawn as a double white line). On the left (trans) the two chlorines sit on opposite corners, so the two cyan C→Cl arrows point in opposite directions and no amber net arrow appears. On the right (cis) both chlorines sit on the same side, so both cyan arrows point the same general way and leave an amber net arrow pointing up.

Figure 4 — Left: trans (dipoles opposite, cancel). Right: cis (dipoles add, net amber dipole survives).

Step 1 — Fix the geometry of each C–Cl arrow. The two carbons are joined by a planar double bond; each carbon carries one Cl. Each C–Cl bond arrow points C → Cl (Cl is the $δ^{-}$ end). Why this step? The molecule's $μ$ is the vector sum of these two arrows, and the geometry fixes the angle between them.

Step 2 — Trans: Cl atoms on opposite sides. By the symmetry of the trans arrangement, the two C–Cl arrows are related by a $180°$ rotation about the molecule's centre — they point in exactly opposite directions. $μ_{trans} = μ_{1} + μ_{2} = 0$ Why this step? Opposite equal vectors cancel — same maths as CO₂, but here forced by isomerism, not by the central atom.

Step 3 — Cis: Cl atoms on the same side. Each $sp^{2}$ carbon has $120°$ bond angles, so each C–Cl bond leans $60°$ off the C=C axis (see VSEPR Theory). In the cis isomer both Cl atoms sit on the same side, so the two arrows' components along the C=C axis point opposite (and partly cancel), while their components perpendicular to the axis both point the same way (upward) and add. The surviving perpendicular components give $μ_{cis} = 2 μ_{CCl} cos (60°) = 2 μ_{CCl} (0.5) = μ_{CCl} \neq = 0$ Why this step? The $120°$ trigonal-planar geometry pins down the $60°$ tilt — we do not guess it — and the projection $cos (60°)$ tells us exactly how much of each bond dipole survives.

Step 4 — State the resolution. Therefore trans-1,2-dichloroethene is nonpolar ( $μ = 0$ ) and cis-1,2-dichloroethene is polar ( $μ \approx 1.9 D$ once the small C–H dipoles are folded in too $)$ . Identical atoms, identical bonds, identical formula — opposite answer, because $μ$ is a vector sum over geometry. Why this step? This is the whole point of Case H: the trap is answered by drawing arrows and summing them with the real geometry, never by counting bonds.

Verify: $cos (60°) = 0.5$ , so $μ_{cis} = μ_{CCl} \neq = 0$ (polar) while the trans sum is $0$ (nonpolar) ✔. This cis/trans polarity split is exactly what Infrared Spectroscopy and Intermolecular forces detect.

Recall

By convention, which way does the

μ

arrow point?

From the positive

δ^{+}

end (tail) to the negative

δ^{-}

end (head).

What is the difference between

δ

and

δ^{+} / δ^{-}

δ

is a number (the fraction of an electron shifted, so

q = δ e

);

δ^{+} / δ^{-}

are labels marking the slightly positive / slightly negative ends.

Given

μ

and

d

, how do you find the partial charge

q

Invert the formula:

q = μ / d

(with both in SI); then

δ = q / e

For a bent AX₂ molecule, what formula gives the net dipole?

μ_{net} = 2 μ_{bond} cos (θ /2)

, where

θ

is the bond angle.

Why does putting

θ = 180°

into that formula give zero?

cos (180°/2) = cos 90° = 0

, so linear symmetric molecules (CO₂) have

μ = 0

Between cis- and trans-1,2-dichloroethene, which is nonpolar?

The trans isomer — its two C–Cl dipoles point

180°

apart and cancel.

Where does the

60°

tilt of each C–Cl bond in cis-dichloroethene come from?

From the trigonal-planar

120°

bond angles at each

sp^{2}

carbon, so each bond leans

60°

off the C=C axis.

How is % ionic character related to the fractional charge

δ

They're equal:

% ionic = \frac{q d}{e d} \times 100 = \frac{q}{e} \times 100 = δ \times 100