Question

How do you calculate the dipole moment along the `"OH"` bond of water? `r_(O-H) = "0.96 Å"` and `/_HOH = 104.4776^@`, and the charge of the electron is given as `"4.80 A"`.

Answer 1

where:

• `d = "0.96 Å"` is the distance from the nucleus of the `"H"` atom to the nucleus of the `"O"` atom. Its value is the `"OH"` bond length.
• `q = "4.80 D"` is the charge in `"D"` units, debyes (not amperes!).
• `104.4776^@` is the `"H"-"O"-"H"` bond angle, so half of it is `52.2388^@` (not `52.2338^@`, a typo).

From physics, the dipole moment `vecmu` is given by:

`vecmu = sum_i q_ivecr_i`

where `q_i` is the magnitude of the `i`th charge and `vecr_i` is the position of charge `i` from a reference point.

`vecmu_(OH)= "1.5 D"` is a derived quantity that requires you to already have measured `"1.85 D"` for the overall dipole moment of water.

You aren't required to know how to calculate it; you would be given either `vecmu_(t ot)` for the overall molecule, the percent ionic character, or `vecmu` for one bond.

(FYI, the percent ionic character is given by:

`I = (100vecmu_(bond))/(4.803_2vecr)`,

if `vecmu_(bond)` is in `"D"` and `vecr` is in `Å`. So, the % ionic character of the `"O"-"H"` bond of water is 32.5%.)

Then, if you have `vecmu_(bond)`, you can use half the bond angle of water with trigonometry to find the dipole moment of the overall molecule:

https://socratic.org/questions/how-do-you-calculate-the-dipole-moment-of-water

Going backwards though, since `vecmu_(t ot) = nvecmu_(bond) cos theta`, where `theta` is the angle with the vertical, and `n` is the number of identical bonds about a symmetry plane (around which the horizontal dipole moment vectors cancel out), we have that, with TWO `"OH"` bonds:

`vecmu_(OH,L)cos (-theta) + vecmu_(OH,R)cos theta = vecmu_(H_2O)`

`=> color(blue)(vecmu_(OH)) = 1/2("1.85 D")/(cos(52.2388^@))`

`= "1.51 D" ~~ color(blue)("1.5 D")`