Question

What makes the classical dipole moment for water different than its experimental result? I used `r = 0.958` angstroms, `A_(HOH) = 104.4776^o`, and `q_(OH) = -0.52672 a.u.`, where `a.u. =` electron-charge. My result was `1.484 D`, compared to `1.85 D`.

Answer 1

I meant to say `Z_(OH) = -0.52672 a.u.`.

Here are my calculations:
`mu = qr = Zer`

Maybe I did something wrong, but here's what I wrote:

`mu = [(-0.52672) cancel(e)*-1.602*10^(-19)cancel(C)/cancel(e)]*0.958*10^(-10)cancel(m) * (1 D)/(3.33564*10^(-30) cancel(C*m))`

`~~ 2.423419828 D`

Then, I divided `104.4776^o` by `2` to get `52.2388^o` and treated half of water (primary axis through oxygen, coplanar with the two hydrogens) as a right triangle:
`cos(52.2388^o) = mu/(2.423419828 D)`

`mu ~~ 1.48403 D`

I checked with a quick google search and found the experimental value of `1.85 D`. I also did several calculations on Psi 4, and got, using the following basis sets:

`2.0580 D` (cc-pVDZ)
`2.0262 D` (cc-pVTZ)
`2.0084 D` (cc-pVQZ)
`1.9817 D` (aug-cc-pVQZ)