Polar molecules — net dipole depends on shape — Chemistry

Before this

This leads to

Two polar bonds, two different answers

Carbon dioxide, CO₂, has two polar bonds. Water, H₂O, also has two polar bonds. Yet water is a polar molecule, with a clear negative end and positive end, and carbon dioxide is not polar at all. Measurements agree: water has a dipole moment of 1.85 debye, and carbon dioxide has none.

Same number of polar bonds, completely different molecules. The difference is shape. The last lecture described each polar bond on its own, and promised to add the bond dipoles together using the shapes from VSEPR. This lecture keeps that promise. You will see that a molecule's polarity depends on two things: how polar its bonds are, and how they are arranged in space.

Bond dipoles are arrows, so they add like arrows

The last lecture drew each polar bond as an arrow, with a cross at the δ+ end, pointing towards the δ− end. That arrow has a size (how polar the bond is) and a direction (which way the electrons have shifted). Quantities like this are called vectors, just like forces and velocities in physics.

Vectors do not add like ordinary numbers. Two forces of 10 N can give a total of 20 N, if they pull the same way, or 0 N, if they pull in opposite directions, or anything in between. Bond dipoles behave in exactly the same way. The molecular dipole is the total of all the bond dipoles, added as arrows.

If the bond dipoles cancel, the molecule has no overall dipole. It is non-polar, even though its bonds are polar.
If they do not cancel, the molecule has a positive end and a negative end. It is polar.

Think of a tug of war. If two equal teams pull in exactly opposite directions, the rope does not move, however hard each team pulls. Carbon dioxide is that tug of war.

Carbon dioxide and water

In CO₂, each C=O bond is polar. The electronegativity difference is $3.44 - 2.55 = 0.89$ , with oxygen δ−. So each bond dipole points from carbon out towards an oxygen. But VSEPR says CO₂ is linear: the two bonds point in exactly opposite directions, at 180°. The two arrows are equal in size and opposite in direction, so they cancel completely. Carbon dioxide is non-polar.

In H₂O, each O–H bond is polar, with a difference of $3.44 - 2.20 = 1.24$ . Each bond dipole points from hydrogen towards oxygen. But water is bent, with an angle of 104.5°, because of the two lone pairs on the oxygen. Now the two arrows do not point in opposite directions. Each one points partly sideways and partly towards the oxygen. The sideways parts cancel, but the parts towards the oxygen add up. So water has an overall dipole, with the oxygen end δ− and the hydrogen end δ+.

Left: CO₂ is linear, so its two bond dipoles point in opposite directions and cancel. Right: H₂O is bent, so both bond dipoles point partly towards oxygen. Their sideways parts cancel, and the rest adds up to a net dipole towards oxygen. Each arrow starts with a cross at the δ+ end.

How much of each bond dipole survives depends on the angle. For two equal bond dipoles at an angle $θ$ , the total is $2 cos (θ /2)$ times one bond dipole. At 180° that is zero. At water's 104.5° it is about 1.22, so the two bonds together give a little more than one bond alone. Change the angle below and watch the total.

Explore — adding two bond dipoles

The two green arrows are equal bond dipoles. Change the angle between them. The thick arrow is their total, the molecular dipole. (In water the arrows point the other way, towards the oxygen, but the size of the total is the same.)

Angle between the bonds: 104.5°

The total of two equal bond dipoles at different angles, as a multiple of one bond dipole (2 × cos of half the angle):

Angle	Total	Example
180°	0	CO₂: non-polar
120°	1.00	the total equals one bond dipole
104.5°	≈ 1.22	H₂O: polar
90°	≈ 1.41	a sharper bend gives a bigger total

Notice what happens near 180°. Even a small bend makes the molecule a little polar. Only a perfectly straight arrangement of two identical bonds cancels exactly.

Real molecules are a little more complicated than two arrows. The lone pairs on water's oxygen also push electron density to that side, and they add to the dipole in the same direction. So the arrow model explains the direction and the main idea, but not every decimal place of the measured 1.85 debye.

Three and four bonds: when symmetry cancels everything

The same idea works with more bonds. In boron trifluoride, BF₃, three B–F bonds point to the corners of a flat triangle, 120° apart. Three equal arrows at 120° cancel exactly, like three equal teams pulling a ring in three evenly spaced directions. BF₃ is non-polar.

In tetrachloromethane, CCl₄, four C–Cl bonds point to the corners of a tetrahedron. Each bond is polar, with a difference of $3.16 - 2.55 = 0.61$ . Here is a neat way to see that they cancel. In a perfect tetrahedron, any three of the bond arrows add up to an arrow exactly as long as the fourth, pointing exactly the opposite way. So the fourth cancels the other three. CCl₄ is non-polar. Methane, CH₄, is non-polar for the same reason.

Now change one atom. In chloromethane, CH₃Cl, one C–Cl bond replaces a C–H bond. The C–Cl bond is far more polar than C–H, which has a difference of only $2.55 - 2.20 = 0.35$ . The symmetry is broken, the bond dipoles no longer cancel, and the molecule is polar, with a measured dipole moment of 1.87 debye.

Is the molecule polar?

1. Draw the shape from VSEPR, including any lone pairs on the central atom.

2. Decide which bonds are polar (difference above about 0.4) and draw their arrows.

3. If no bond is polar, the molecule is non-polar.

4. If bonds are polar, ask whether the arrows cancel. They cancel only when identical bonds are spread evenly around the central atom: linear with two, flat triangle with three, tetrahedral with four, and no lone pairs on the central atom. Otherwise, the molecule is polar.

Stop and think

A student draws dichloromethane, CH₂Cl₂, flat on paper, with the two chlorines opposite each other. "The two C–Cl arrows point in opposite directions, so they cancel. CH₂Cl₂ is non-polar." Is the student right?

Show the answer

No. The flat drawing is not the shape. CH₂Cl₂ is tetrahedral, like methane, and in a tetrahedron no two bonds are opposite each other: every pair of bonds meets at about 109.5°. So the two C–Cl arrows both point partly towards the same side, and they add up instead of cancelling. CH₂Cl₂ is polar. Judge polarity from the three-dimensional shape, never from a flat drawing.

Worked examples

Worked Example 1 — BF₃, very polar bonds in a non-polar molecule

Is BF₃ polar?

Each B–F bond has a difference of $3.98 - 2.04 = 1.94$ . That is above the rough guide of about 1.8 for ionic bonding, yet BF₃ is a gas made of separate molecules. This is a reminder that the guide is rough, not a sharp line. The bonds are very polar, with fluorine δ−.

Boron has three bonding pairs and no lone pair, so BF₃ is trigonal planar, with angles of 120°. Three equal arrows at 120° cancel. So BF₃ is non-polar, even though each bond is more polar than the O–H bond in water.

Worked Example 2 — NH₃, a lone pair breaks the symmetry

Is ammonia, NH₃, polar?

Each N–H bond has a difference of $3.04 - 2.20 = 0.84$ , with nitrogen δ−. Nitrogen has three bonding pairs and one lone pair, so the shape is a trigonal pyramid, with angles of about 107°. The three hydrogens sit below the nitrogen like the legs of a stool.

All three N–H arrows point partly up towards the nitrogen. Their sideways parts cancel, but their upward parts add. The lone pair adds more in the same direction. So NH₃ is polar, with the nitrogen end δ−. Its measured dipole moment is 1.47 debye. Compare BF₃: three bonds in both, but BF₃ is flat and NH₃ is a pyramid, because of the lone pair.

Worked Example 3 — CHCl₃ and CCl₄

Chloroform, CHCl₃, is CCl₄ with one chlorine replaced by hydrogen. Is it polar?

Stand the molecule with the hydrogen on top and the three chlorines below. The three C–Cl arrows point from carbon down towards the chlorines. In CCl₄, the fourth C–Cl arrow pointed straight up and cancelled them. Now there is a C–H bond on top instead.

Which way does the C–H arrow point? Carbon (2.55) is slightly more electronegative than hydrogen (2.20), so carbon is δ− and hydrogen δ+. The arrow points from hydrogen down towards carbon. So it does not oppose the C–Cl arrows at all. It points the same way and adds a little to them. CHCl₃ is polar, with the chlorine end δ− and the hydrogen end δ+.

Worked Example 4 — SO₂ and CO₂ look alike on paper

Sulfur dioxide, SO₂, can be drawn as O=S=O, just like O=C=O. Is it non-polar like CO₂?

No. Drawn with two S=O double bonds, the sulfur also has one lone pair. Carbon in CO₂ has none. So sulfur has three electron regions, and VSEPR makes SO₂ bent, with an angle a little under 120°. Each S=O bond is polar (difference 0.86, oxygen δ−). Bent, like water, so the arrows do not cancel. SO₂ is polar.

This is the classic trap. A lone pair on the central atom does not show in the shape's name, but it bends the molecule. In every shape you have met so far, a lone pair on the central atom means the bond dipoles cannot cancel. (Some larger molecules, with five or six electron regions, are arranged so evenly that they cancel even with lone pairs. You will meet them later.)

Now sort some molecules.

Sorter — polar or non-polar?

Put each molecule in a box. Think of its shape first, then ask whether its bond dipoles cancel.

Ten molecules sorted, each with its reason.

Polar: H₂O (bent); NH₃ (trigonal pyramid); SO₂ (bent, lone pair on S); CHCl₃ (one C–H among three C–Cl); CH₃Cl (one C–Cl among three C–H); HCl (one polar bond, nothing to cancel it).
Non-polar: CO₂ (linear, two equal dipoles cancel); BF₃ (flat triangle, three cancel); CCl₄ (tetrahedral, four cancel); CH₄ (tetrahedral, and C–H is barely polar anyway).

Three classic mistakes

Trap 1 — Polar bonds mean a polar molecule

CO₂, BF₃ and CCl₄ all have polar bonds, and all three are non-polar molecules. Polar bonds are needed for a polar molecule, but they are not enough. The shape decides.

Trap 2 — Judging from a flat drawing

CH₂Cl₂ drawn flat looks as if its C–Cl bonds are opposite. In three dimensions it is tetrahedral, and the bonds are about 109.5° apart. Always picture the VSEPR shape before deciding.

Trap 3 — Forgetting the lone pairs

O=S=O looks like O=C=O, but sulfur has a lone pair and carbon does not. The lone pair bends SO₂, so it is polar. Count the lone pairs on the central atom first.

Try it yourself

Use Pauling electronegativities: H 2.20, C 2.55, N 3.04, O 3.44, F 3.98, Si 1.90, P 2.19, S 2.58, Cl 3.16.

Which of these molecules are polar: CS₂, H₂S, PCl₃, SiF₄? Give the shape of each.

Hint 1
Count bonding pairs and lone pairs on the central atom. Sulfur in H₂S has two lone pairs; phosphorus in PCl₃ has one.

Hint 2
CS₂ is drawn S=C=S, with no lone pair on carbon. SiF₄ has four bonds and no lone pair.

Full answer
CS₂: linear, non-polar. (Its C=S bonds are barely polar anyway: $2.58 - 2.55 = 0.03$ .) H₂S: bent, like water, so polar, though only slightly, because S–H has a difference of only 0.38. PCl₃: trigonal pyramid, like NH₃, so polar (P–Cl difference 0.97). SiF₄: tetrahedral with four identical bonds, so non-polar, even though each Si–F bond is very polar (difference 2.08).
Explain why CCl₄ is non-polar but CH₃Cl is polar, using the idea that in a tetrahedron any three bond arrows add up to the opposite of the fourth.

Hint 1
In CCl₄, all four arrows are the same size.

Hint 2
In CH₃Cl, compare the one C–Cl arrow with the total of the three C–H arrows.

Full answer
In CCl₄, the total of any three C–Cl arrows is exactly as large as the fourth C–Cl arrow and points the opposite way, so everything cancels. In CH₃Cl, the three C–H arrows are small (difference 0.35), and the one C–Cl arrow is large (difference 0.61). The small total cannot cancel the large one, so a net dipole remains, towards the chlorine. CH₃Cl is polar.
Two molecules, XY₂, are made of the same atoms. One is linear and one has a bond angle of 120°. Each X–Y bond dipole is the same size in both. Compare their molecular dipoles.

Hint 1
Use the formula: total = $2 cos (θ /2)$ times one bond dipole.

Hint 2
$cos 9 0^{\circ} = 0$ and $cos 6 0^{\circ} = 0.5$ .

Full answer
Linear: $2 cos 9 0^{\circ} = 0$ , so the molecule is non-polar. At 120°: $2 cos 6 0^{\circ} = 2 \times 0.5 = 1$ , so the molecular dipole is exactly the size of one bond dipole. Same atoms, same bonds, but one molecule is polar and the other is not.
Water's dipole moment is 1.85 debye and its bond angle is 104.5°. If the dipole came only from the two O–H bond dipoles, how large would each bond dipole be? Why is this only an estimate?

Hint 1
Total $= 2 cos (52.2 5^{\circ}) \times b$ , where $b$ is one bond dipole.

Hint 2
$2 cos (52.2 5^{\circ}) \approx 1.224$ .

Full answer
$b = 1.85 \div 1.224 \approx 1.51$ debye for each O–H bond. It is only an estimate. The two lone pairs on oxygen also shift charge towards that side of the molecule, so part of the 1.85 debye comes from them, not from the bonds. The simple arrow model gets the direction right and the size roughly right.

The bigger idea

Water's polarity comes from its bend. Because each water molecule has a δ− end and a δ+ end, water molecules attract each other and attract ions and other polar molecules. That is a large part of why water dissolves salt and sugar but not oil, and why microwave ovens can heat food: the oscillating field keeps turning the polar water molecules back and forth.

It is tempting to go further and say that a straight water molecule would boil far below 0 °C and that life could not exist. That goes beyond what this lecture can show. How strongly molecules hold each other depends on more than the overall dipole, and the next topic, intermolecular forces, takes that apart carefully: first the London forces that act between all molecules, then dipole–dipole forces, then the special case of hydrogen bonding.

Key concepts in this lesson

Each polar bond has a bond dipole, an arrow with a size and a direction; the molecular dipole is the total of these arrows.
A molecule can have polar bonds and still be non-polar, if the bond dipoles cancel, as in CO₂, BF₃ and CCl₄.
Identical bonds spread evenly around a central atom with no lone pairs (linear, trigonal planar, tetrahedral) cancel; bent and pyramidal shapes do not.
For two equal bond dipoles at angle θ, the total is 2cos(θ/2) times one bond dipole: zero at 180°, about 1.22 at water's 104.5°.
Decide polarity from the three-dimensional VSEPR shape, including lone pairs, never from a flat drawing.