# How do you calculate the dipole moment of larger molecules?

I know that the dipole moment is obtained by:

$$\vec{\mu} = Q \cdot \vec{d}$$

with the condition that charges of a pair of atoms are equal and opposite, i.e. $\ce{NaCl}$.

But, in order to obtain the molecular dipole moment of water, how is the $\ce{-OH}$ group handled? What about even larger systems such as $\ce{NH3}$?

For example, the water molecular dipole moment is 1.85 Debye. How was this value was obtained?

I have read 1, 2, 3, and chemistry textbooks.

I am interested in a more mathematical or physical description.

The total dipole of a molecule can be thought of as the sum of dipoles of individual functional groups:

$$\vec{\mu}_{\text{total}} = \sum_{i}^{N_\text{groups}} \vec{\mu}_{i}$$

Because each dipole is represented as a vector with both magnitude and direction, this amounts to vector addition.

In the case of water, consider the dipole of each $\ce{-OH}$ group. The above reference gives it as 1.53 Debye, and the $\ce{H-O-H}$ angle is $105^{\circ}$. Because coordinates aren't given, the angle has to be used instead:

$$\vec{P} + \vec{Q} = \sqrt{\lVert \vec{P} \rVert^{2} + \lVert \vec{Q} \rVert^{2} + 2\lVert\vec{P}\rVert\lVert\vec{Q}\rVert\cos{\theta_{PQ}}}$$

which gives an estimate of 1.64 Debye. Not very good compared to 1.85 Debye! The group addition approximation completely neglects how the electronic structure of the molecule changes upon going from (using water as an example)

$$\ce{2OH^{.}} \rightarrow \ce{H2O}$$

The issue is probably that the two individual $\ce{-OH}$ groups interact strongly and favorably when combined together in water, compared to when they are apart; that is, the electron density changes quite a bit due to the bonding.

I've also neglected to explain how these groups can be defined; there is no unique definition of how to partition a molecule. For example, take a carboxyl group:

Are all four atoms treated together as a single group, or are the $\ce{C=O}$ and $\ce{C-O}$ and $\ce{-OH}$ treated separately? Maybe this is a simple case but it becomes more unclear for something like a protein or a coordination complex.

The molecular dipole can also be calculated using electronic structure theory, which removes the ambiguity surrounding how to partition a molecule and incorporates those "non-additive" effects.

As Geoff mentioned in the comments, the definition of the dipole moment created by a set of charges can also be used:

$$\vec{\mu} = \sum_{a}^{\text{charges}} \vec{r}_a \times q_a$$

where $q_a$ is the charge, and

$$\vec{r}_a = \vec{R}_a - \vec{O} = (R_{ax} - O_x, R_{ay} - O_y, R_{az} - O_z),$$

where $\vec{R}$ is the position of the point charge, and $\vec{O}$ is a common origin, usually taken to be either $(0,0,0)$, the center of mass, or the center of nuclear charge. It doesn't matter what this point is for uncharged (neutral) systems, but there is an origin dependence for anything with a non-zero total charge. The $\{\vec{R}\}$ are taken to be atomic positions, and the charges $\{q\}$ can come from simpler Mulliken or Löwdin population analyses, or more complicated schemes of which many are designed to reproduce total dipole moments. Care must be taken to deal with units properly, specifically when converting: $0.393430307 \, ea_0 = 1 \, \mathrm{Debye}$.

• +1 I think the only thing I'd add is that you could also, in principle add up atomic partial charges: q * v, where v is the atomic position relative to the center, but you'll need to be careful to convert from e*Å to Debye. Of course this requires that the atomic partial charges reproduce the dipole moment, but that's the goal of many partial charge schemes. Aug 18 '16 at 22:04
• I did neglect to point that out, and it's mentioned here. This is also how one would calculate the dipole in molecular mechanics/from a force field. Aug 19 '16 at 5:00
• in H2O the lone pairs of electrons are considered functional groups ? if yes then we should calculate the dipole moment from hydrogen atoms and lone pairs of electrons of the oxygene atom ? Feb 25 '19 at 21:40