# How to calculate dipole moment of "NH"_3?

Dec 30, 2017

There's no easy way to do it without having a nice molecule whose dipole moment is already known.

We always have to either do it computationally (ab initio) or after already having experimental data available for a classroom exercise...

And in this exercise, we need to calculate the dipole moment along one bond first. Then we can get its $z$ component, and triple it for the net dipole moment due to the symmetry of ${\text{NH}}_{3}$.

This is because ${\text{NH}}_{3}$ has a three-fold rotational axis. As a result, we can say that the dipole moment along each $\text{N"-"H}$ bond is identical.

In that case, each $\text{N"-"H}$ bond dipole moment is based on:

$\vec{\mu} = i \cdot q \vec{r}$

where $q$ is the charge of an electron and $\vec{r}$ is the separation distance between two charges. Here, we ascribe ionic character to the value of $i$, where $0 < i < 1$.

First, we'll need that its bond length is "1.0124 angstroms, i.e. $1.0124 \times {10}^{- 10} \text{m}$.

Using data from CCCBDB, ${\text{NH}}_{3}$ looks like this:  Next, we'll need the percent ionic character of an $\text{N"-"H}$ bond to estimate the relative sharing of the electrons in the bond. Ironically, we would not know that value unless we already knew the dipole moment....

But this website shows a value of 27.0%, which makes reasonable sense. So, we set $i = 0.270$.

In that case, the dipole moment along the $\text{N"-"H}$ bond might be:

$\vec{\mu} \left(\text{N"-"H}\right) = i \cdot q \vec{r}$

= overbrace(0.270 cdot 1.602 xx 10^(-19) cancel"C")^(i cdot q) xx overbrace(1.0124 xx 10^(-10) cancel"m")^(r) xx "1 D"/(3.336 xx 10^(-30) cancel("C"cdot"m"))

$\approx$ $\text{1.313 Debyes}$, or $\text{D}$

But this is not the net dipole moment. The net dipole moment is what we want, and is the vertical arrow shown below. Now what we want to do is get the angle relative to the vertical, so that we can get the $\boldsymbol{z}$ component of the dipole moment we just found for the bond, since only the vector components in the same direction add.

This angle is based on a right triangle formed by an $\text{N"-"H}$ bond relative to the vertical: So, this angle is found to be:

$\cos \theta = \frac{0.3816}{1.0124}$

$\implies \theta = {67.86}^{\circ}$

Since the vertical is the $z$ axis, the net dipole moment is found from the addition of the $z$ component of all three dipole moment vectors.

One of these vectors (the $z$ component of a diagonal arrow shown above) is:

vecmu_z("N"-"H") = "1.313 D" xx cos(67.86^@) = "0.4948 D"

The addition of three of these $z$-component vectors, one for each $\text{N"-"H}$ bond, gives

$\textcolor{b l u e}{{\vec{\mu}}_{\text{net"("NH"_3)) = vecmu_z("N"-"H") + vecmu_z("N"-"H") + vecmu_z("N"-"H}}}$

$=$ $\textcolor{b l u e}{\text{1.484 D}}$

for the net dipole moment. The actual value from a casual google search is around $\text{1.46 D}$, so that's pretty close!

Another reference is from CCCBDB, which says it is $\text{1.470 D}$. Again, quite close!