Lesson 2 · 25 min

Molecular polarity

Every bond can be polar while the molecule is not. Add bond dipoles as vectors, then see how lone pairs send NH₃ and NF₃ in opposite directions.

In this lesson

Add bond dipoles as vectors to decide whether a molecule is polar
Recognise the symmetric shapes in which polar bonds give a non-polar molecule
Estimate the dipole moment of a bent molecule from its bond dipoles and bond angle
Use NH₃ versus NF₃ to show what lone pairs contribute and where the bond-dipole model fails

Carbon tetrachloride, $\ce{CCl4}$ , and chloroform, $\ce{CHCl3}$ , differ by a single atom. Both contain clearly polar C–Cl bonds, yet $\ce{CCl4}$ has a dipole moment of exactly zero while $\ce{CHCl3}$ has 1.04 D. Stranger still, more chlorines mean a smaller dipole: $\ce{CH3Cl}$ 1.87 D, $\ce{CH2Cl2}$ 1.60 D, $\ce{CHCl3}$ 1.04 D, $\ce{CCl4}$ 0. Bond polarity and molecular polarity are clearly not the same thing.

Predict first

Which of these molecules is polar (has a non-zero dipole moment)?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

\ce{CO2}

\ce{BF3}

\ce{CH4}

\ce{H2O}

Dipole moments are vectors

Every polar bond has a bond dipole with both size and direction (from δ+ to δ−). A molecule’s dipole moment is the vector sum of all its bond dipoles: opposing components cancel, parallel ones add. Judging polarity therefore takes two steps: find the shape with VSEPR, then lay the bond dipoles along it and add them.

When the central atom has no lone pairs and all the surrounding atoms are identical, the shape is always highly symmetric — linear $\mathrm{AX_2}$ , trigonal planar $\mathrm{AX_3}$ , tetrahedral $\mathrm{AX_4}$ , trigonal bipyramidal $\mathrm{AX_5}$ , octahedral $\mathrm{AX_6}$ — and the bond dipoles cancel completely. A few shapes with lone pairs are symmetric too, such as $\ce{XeF2}$ (linear) and $\ce{XeF4}$ (square planar). Almost everything else is polar.

\vec{\mu} = \sum_i \vec{\mu}_i,\qquad \mu(\mathrm{AX_2}) = 2\,\mu_\mathrm{AX}\cos\frac{\theta}{2}

The molecular dipole is the vector sum of the bond dipoles; for a bent molecule, the two components along the symmetry axis add.

$\vec{\mu}$: the molecule’s dipole moment
$\vec{\mu}_i$: bond dipole of bond i
$\theta$: the X–A–X bond angle
$\mu_\mathrm{AX}$: dipole of one A–X bond (debye)

Measured dipole moments (gas phase, D)
Molecule	Shape	μ / D
$\ce{H2O}$	bent	1.85
$\ce{NH3}$	trigonal pyramidal	1.47
$\ce{NF3}$	trigonal pyramidal	0.235
$\ce{SO2}$	bent	1.63
$\ce{HF}$	linear	1.83
$\ce{HCl}$	linear	1.11
$\ce{CO}$	linear	0.11
$\ce{CH3Cl}$	tetrahedral	1.87
$\ce{CH2Cl2}$	tetrahedral	1.60
$\ce{CHCl3}$	tetrahedral	1.04
$\ce{CCl4}$	tetrahedral	0
$\ce{CO2}$	linear	0

Source: CRC Handbook of Chemistry and Physics, “Dipole Moments” (R. D. Nelson, D. R. Lide, A. A. Maryott, NSRDS-NBS 10), cross-checked with NIST CCCBDB; agreement within about 0.02 D.

NH₃ and NF₃ have the same trigonal pyramidal shape and one lone pair each. The N–F electronegativity difference (0.94) is even larger than N–H’s (0.84). Judged by bond dipoles alone, NF₃ should be the more polar. Experiment says the opposite: NH₃ 1.47 D, NF₃ only 0.235 D.

The key is direction, and the lone pair. In NH₃ nitrogen is more electronegative than hydrogen, so all three bond dipoles point toward N — toward the lone pair side — and the lone pair adds its own outward dipole: they reinforce. In NF₃ fluorine is more electronegative, so the bond dipoles point toward the three F atoms, away from the lone pair, and the two largely cancel. The workbench below first adds the bond dipoles alone, then lets you add the lone pair’s share.

Bond dipoles + lone pairs → molecular dipole

Adding up bond dipoles

Bond dipole (≈ ΔEN debye)
Sum of bond dipoles
Lone pairs
Model total

Side view: arrows point from δ+ to δ−

Contribution per lone pair 0.00 D

Bond dipoles only 1.24 D

Bonds + lone pairs 1.24 D

Measured 0.23 D

The bond dipoles point away from the lone pairs: the two partly cancel.

Challenge: find one lone-pair value that brings both NH₃ and NF₃ close to experiment. You can only get roughly there — the model is too crude.

Start with the lone-pair contribution at 0: the model gives NF₃ 1.24 D and NH₃ 0.95 D — the wrong order. Raise the contribution slowly and watch the two part ways. Finally look at PH₃: its bonds are almost non-polar, and its dipole comes from the lone pair.

Worked example

Estimating water’s dipole from its bond dipoles

If each O–H bond dipole is 1.51 D and the H–O–H angle is 104.5°, what is water’s dipole moment?

Concept check

Which of these chloromethanes has the largest dipole moment?

\ce{CH3Cl}

\ce{CH2Cl2}

\ce{CHCl3}

\ce{CCl4}

Try the calculation

The H–S–H angle of $\ce{H2S}$ is 92.1°. Taking each S–H bond dipole as 0.68 D, what dipole moment does $\mu = 2\mu_\mathrm{AX}\cos(\theta/2)$ give?

Key takeaways

A molecular dipole is the vector sum of the bond dipoles; the shape decides whether they add or cancel.
With no lone pairs on the central atom and identical outer atoms (CO₂, BF₃, CH₄, SF₆) the molecule is non-polar; XeF₂ and XeF₄ are symmetric too.
Lone pairs contribute as well: in NH₃ (1.47 D) they reinforce the bond dipoles, in NF₃ (0.235 D) they oppose them.
The bond-dipole model tells polar from non-polar but cannot give reliable dipole values.

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Data and references

W. M. Haynes (ed.), CRC Handbook of Chemistry and Physics, 97th ed., “Dipole Moments” (from R. D. Nelson, D. R. Lide, A. A. Maryott, NSRDS-NBS 10, 1967).
NIST Computational Chemistry Comparison and Benchmark Database (CCCBDB), Release 22 (2022): experimental dipole moments and geometries.
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed., Topic 14A (electric dipole moments).

Dipole moments are vectors

NH₃ versus NF₃: the lone pair’s share

Estimating water’s dipole from its bond dipoles

Key takeaways

Data and references