How do you calculate the dipole moment of water?

1 Answer
Feb 2, 2016

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).