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

Question

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

1 Answer

Impact of this question

2743 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-18T22:04:43

I meant to say $Z_{O H} = - 0.52672 a . u .$ $Z_(OH) = -0.52672 a.u.$ .

Here are my calculations:
$μ = q r = Z e r$ $mu = qr = Zer$

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)

Science

Math

Humanities

... and beyond

1 Answer

Related questions

Impact of this question