# 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.

Jun 18, 2015

I meant to say ${Z}_{O H} = - 0.52672 a . u .$.

Here are my calculations:
$\mu = q r = Z e r$

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

$\mu = \left[\left(- 0.52672\right) \cancel{e} \cdot - 1.602 \cdot {10}^{- 19} \frac{\cancel{C}}{\cancel{e}}\right] \cdot 0.958 \cdot {10}^{- 10} \cancel{m} \cdot \frac{1 D}{3.33564 \cdot {10}^{- 30} \cancel{C \cdot m}}$

$\approx 2.423419828 D$

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:
$\cos \left({52.2388}^{o}\right) = \frac{\mu}{2.423419828 D}$

$\mu \approx 1.48403 D$

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)