Dipole moments and shape

A bond with two ends

In a bond between two different atoms, the more electronegative atom pulls the shared pair closer. It gets a small negative charge, $\delta^-$. The other atom gets a small positive charge, $\delta^+$. The bond is a bond dipole.

The dipole moment measures it: the charge times the distance between the charges, $\mu = q \times d$. Its unit is the debye:

A dipole moment has a direction as well as a size, so it is a vector. In chemistry the arrow points from the positive end to the negative end.

The molecule adds its bond dipoles

The dipole moment of a molecule is the vector sum of its bond dipoles. A lone pair adds a dipole of its own, pointing away from the atom that holds it. So two molecules with the same kind of bonds can have very different dipole moments. Their shapes decide.

For two equal bond dipoles $\mu_b$ at an angle $\theta$:

At $\theta = 180^\circ$, $\cos 90^\circ = 0$, and the two cancel. At small angles they nearly add up to $2\mu_b$.

Worked example. Water has a dipole moment of $1.85\ \text{D}$, and its bond angle is $104.5^\circ$. Find the dipole of one $\text{O-H}$ bond.

1. Half the angle. $\dfrac{104.5^\circ}{2} = 52.25^\circ$, and $\cos 52.25^\circ = 0.612$.
2. Rearrange. $\mu_b = \dfrac{\mu}{2\cos(\theta/2)} = \dfrac{1.85}{2 \times 0.612} = 1.51\ \text{D}$.

This folds the lone pairs into the bond dipoles, as most exam problems do.

When the bond dipoles cancel

• $\text{CO}_2$ is linear. Its two $\text{C=O}$ dipoles point in opposite directions and cancel: $\mu = 0$.
• $\text{BF}_3$ is trigonal planar. Three equal dipoles at $120^\circ$ add to zero, like three people pulling a ring equally hard from three sides.
• $\text{CCl}_4$ is tetrahedral. Any one $\text{C-Cl}$ dipole is cancelled by the sum of the other three.

Each of these molecules has polar bonds, yet it is a nonpolar molecule. Replace one atom and the balance breaks. Chloroform, $\text{CHCl}_3$, is polar, because its $\text{C-H}$ bond cannot cancel the three $\text{C-Cl}$ dipoles.

When they do not cancel

• $\text{H}_2\text{O}$ is bent. Its two $\text{O-H}$ dipoles add in part, to $1.85\ \text{D}$.
• $\text{NH}_3$ is a pyramid. Its three $\text{N-H}$ dipoles all lean the same way, so they cannot cancel.
• $\text{NF}_3$ has the same shape, but a far smaller dipole moment.

Why NH3 beats NF3

Both molecules have a lone pair on nitrogen. It points away from the three outer atoms, and it adds a dipole in that direction.

• In $\text{NH}_3$, nitrogen is more electronegative than hydrogen. The bond dipoles point towards nitrogen, the same way as the lone pair. They add, and $\mu = 1.47\ \text{D}$.
• In $\text{NF}_3$, fluorine is more electronegative than nitrogen. The bond dipoles point away from nitrogen, against the lone pair. They mostly cancel, and $\mu = 0.24\ \text{D}$.

The $\text{N-F}$ bonds are more polar than the $\text{N-H}$ bonds. Yet $\text{NF}_3$ is the less polar molecule. Direction matters as much as size.

A dipole moment tells the shape

A measured dipole moment can settle a question of shape:

• $\text{BCl}_3$ has no dipole moment, so it must be planar.
• $\text{PCl}_3$ has one, so it cannot be planar. It is a pyramid, with a lone pair on phosphorus.
• A molecule $\text{AB}_2$ with $\mu = 0$ must be linear. One with $\mu > 0$ must be bent.

Percent ionic character

If a bond were fully ionic, charges $+e$ and $-e$ would sit a distance $d$ apart. Its dipole moment would be $e \times d$. Compare the measured dipole with it:

A handy number: a charge $e$ at $100\ \text{pm}$ gives $4.803\ \text{D}$.

Worked example. A molecule of potassium chloride in the gas has a dipole moment of $10.27\ \text{D}$ and a bond length of $267\ \text{pm}$. How ionic is the bond?

1. Full charges. $4.803 \times \dfrac{267}{100} = 12.82\ \text{D}$.
2. The ratio. $\dfrac{10.27}{12.82} \times 100 = 80.1\%$.

Even this salt is not fully ionic. The small cation pulls the electron cloud of the chloride towards itself, as Lesson 3 explains.

Cis and trans isomers

1,2-dichloroethene, $\text{ClHC=CHCl}$, comes in two forms. The double bond cannot rotate, so each chlorine stays where it is.

• In the cis isomer, both chlorines lie on the same side. Their $\text{C-Cl}$ dipoles add: $\mu = 1.90\ \text{D}$.
• In the trans isomer, they lie on opposite sides. The dipoles cancel: $\mu = 0$.

The polar cis isomer boils higher, at $60\ ^\circ\text{C}$, against $48\ ^\circ\text{C}$ for the trans isomer. A dipole measurement tells the two apart at once.

Try it in Python

The program adds bond dipoles as vectors in three dimensions. Each bond gets a size in debye and a direction. The tetrahedral directions of $\text{CCl}_4$ point to four corners of a cube.

Common mistakes

• Adding the sizes of the bond dipoles. They are vectors. Add them with their directions.
• Calling a molecule polar because its bonds are polar. $\text{CO}_2$, $\text{BF}_3$ and $\text{CCl}_4$ have polar bonds but no dipole moment.
• Using the whole angle in the cosine. The formula has $\cos(\theta/2)$.
• Forgetting the lone pair. It explains why $\text{NH}_3$ is far more polar than $\text{NF}_3$.
• Mixing units in $e \times d$. Use the bond length in picometres with the handy $4.803\ \text{D}$ per $100\ \text{pm}$.

Remember these

• $\mu = q \times d$, measured in debye. The dipole of a molecule is the vector sum of its bond dipoles.
• Two equal bond dipoles at angle $\theta$: $\mu = 2\mu_b \cos\dfrac{\theta}{2}$.
• Linear $\text{AB}_2$, planar $\text{AB}_3$ and tetrahedral $\text{AB}_4$ have $\mu = 0$.
• $\text{NH}_3$ ($1.47\ \text{D}$) is more polar than $\text{NF}_3$ ($0.24\ \text{D}$), because of the lone pair.
• Ionic character $= \dfrac{\mu}{e \times d} \times 100\%$.
• A cis isomer with two like groups is polar; the trans isomer has $\mu = 0$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Sulfur trioxide, $\text{SO}_3$, has three polar $\text{S=O}$ bonds. Does it have a dipole moment?

2. Does trans-1-bromo-2-chloroethene have a dipole moment of zero?

3. Two equal bond dipoles of $1.0\ \text{D}$ make an angle of $90^\circ$. What is the total?

4. A bond $200\ \text{pm}$ long has a dipole moment of $4.8\ \text{D}$. What is its percent ionic character?