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#.

Question

Wyatt Adkins · Accepted Answer

I meant to say #Z_(OH) = -0.52672 a.u.#. Here are my calculations:
#mu = qr = Zer# I wrote the following, though perhaps I did it incorrectly: #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)

Luna Chen · Answer

undefined The classical dipole moment for water is different from its experimental result due to several factors including the use of classical approximations, neglect of quantum mechanical effects, and simplifications in the model used for calculation. Additionally, experimental measurements can be influenced by various factors such as temperature, pressure, and molecular interactions, leading to discrepancies between theoretical predictions and observed values. In your specific calculation, the discrepancy may arise from the assumptions and approximations made in the classical model used, such as treating the water molecule as a rigid structure and neglecting the influence of electron distribution and molecular vibrations.