# 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".

Jun 10, 2017

where:

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

From physics, the dipole moment $\vec{\mu}$ is given by:

$\vec{\mu} = {\sum}_{i} {q}_{i} {\vec{r}}_{i}$

where ${q}_{i}$ is the magnitude of the $i$th charge and ${\vec{r}}_{i}$ is the position of charge $i$ from a reference point.

${\vec{\mu}}_{O H} = \text{1.5 D}$ is a derived quantity that requires you to already have measured $\text{1.85 D}$ for the overall dipole moment of water.

You aren't required to know how to calculate it; you would be given either ${\vec{\mu}}_{t o t}$ for the overall molecule, the percent ionic character, or $\vec{\mu}$ for one bond.

(FYI, the percent ionic character is given by:

$I = \frac{100 {\vec{\mu}}_{b o n d}}{{4.803}_{2} \vec{r}}$,

if ${\vec{\mu}}_{b o n d}$ is in $\text{D}$ and $\vec{r}$ is in Å. So, the % ionic character of the $\text{O"-"H}$ bond of water is 32.5%.)

Then, if you have ${\vec{\mu}}_{b o n d}$, 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 ${\vec{\mu}}_{t o t} = n {\vec{\mu}}_{b o n d} \cos \theta$, 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 $\text{OH}$ bonds:

${\vec{\mu}}_{O H , L} \cos \left(- \theta\right) + {\vec{\mu}}_{O H , R} \cos \theta = {\vec{\mu}}_{{H}_{2} O}$

$\implies \textcolor{b l u e}{{\vec{\mu}}_{O H}} = \frac{1}{2} \frac{\text{1.85 D}}{\cos \left({52.2388}^{\circ}\right)}$

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