# 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

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 = 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.

**derived quantity that requires you to already have measured**

You aren't required to know how to calculate it; you would be given either

(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

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

Going backwards though, since

#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")#