Question

How do you calculate the dipole moment of water?

Answer 1

Let's suppose we put water on the xy-plane like so:

The dipole moment is calculated by looking up the dipole moment contributions from each `"O"-"H"` bond, which are polar, and summing them to get the net dipole. Each contribution is `"1.5 D"` (debyes).

The net dipole points through oxygen down the y-axis in the negative direction.

Note that the dipole projection along the x directions cancel each other out; if the left dipole contribution was pointing to +x, the right contribution points to -x.

As a result, to calculate the net dipole, determine the projection of each dipole in the y direction, and then double it, since both `"OH"` bonds are identical.

The `"H"-"O"-"H"` bond angle of water is pretty much `104.4776^@` (it is okay to use `104.5^@`).

Take the angle used in the projection to be from the vertical until each `"OH"` bond and you will get `104.4776^@"/"2`, then imagine two right triangles.

With that, and the fact that `costheta = cos(-theta)`:

`mu_"y,left contribution" = mu_("OH")xxcos(52.2388^@)`

`= "1.5 D" xx 0.612 = color(green)(0.9187)`

`mu_"y,right contribution" = mu_("OH")xxcos(-52.2388^@)`

`= "1.5 D" xx 0.612 = color(green)(0.9187)`

Finally, we sum them up because they are both in the same y direction:

`color(blue)(mu_"tot") = mu_"y,left contribution" + mu_"y,right contribution"`

`= 0.9187 + 0.9187 = color(blue)("1.837 D")`

which is pretty close to the actual `"1.85 D"`.

Likely the error was either from the referenced `"OH"` dipole moment or the calculated bond angle (via Hartree-Fock theory using a cc-pVQZ basis set, but you don't need to know that).