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

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 `"NH"_3`.

This is because `"NH"_3` has a three-fold rotational axis.

![)

As a result, we can say that the dipole moment along each `"N"-"H"` bond is identical.

In that case, each `"N"-"H"` bond dipole moment is based on:

`vecmu = i cdot qvecr`

where `q` is the charge of an electron and `vecr` 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 xx 10^(-10) "m"`.

Using data from CCCBDB, `"NH"_3` looks like this:

![)

Next, we'll need the percent ionic character of an `"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 `"N"-"H"` bond might be:

`vecmu("N"-"H") = i cdot qvecr`

`= 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"))`

`~~` `"1.313 Debyes"`, or `"D"`

But this is not the net dipole moment. The net dipole moment is what we want, and is the vertical arrow shown below.

![https://chemistry.stackexchange.com/]()

Now what we want to do is get the angle relative to the vertical, so that we can get the `bbz` 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 `"N"-"H"` bond relative to the vertical:

So, this angle is found to be:

`costheta = 0.3816/1.0124`

`=> theta = 67.86^@`

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 `"N"-"H"` bond, gives

`color(blue)(vecmu_"net"("NH"_3)) = vecmu_z("N"-"H") + vecmu_z("N"-"H") + vecmu_z("N"-"H")`

`=` `color(blue)("1.484 D")`

for the net dipole moment. The actual value from a casual google search is around `"1.46 D"`, so that's pretty close!

Another reference is from CCCBDB, which says it is `"1.470 D"`. Again, quite close!