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"#.

1 Answer
Jun 10, 2017

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")#