How to calculate dipole moment of #"NH"_3#?
There's no easy way to do it without having a nice molecule whose dipole moment is already known.
We always have to either do it computationally (ab initio) or after already having experimental data available for a classroom exercise...
And in this exercise, we need to calculate the dipole moment along one bond first. Then we can get its
This is because
As a result, we can say that the dipole moment along each
In that case, each
#vecmu = i cdot qvecr#
#q#is the charge of an electron and #vecr#is the separation distance between two charges. Here, we ascribe ionic character to the value of #i#, where #0 < i < 1#.
First, we'll need that its bond length is
Using data from CCCBDB,
Next, we'll need the percent ionic character of an
But this website shows a value of
In that case, the dipole moment along the
#vecmu("N"-"H") = i cdot qvecr#
#= overbrace(0.270 cdot 1.602 xx 10^(-19) cancel"C")^(i cdot q) xx overbrace(1.0124 xx 10^(-10) cancel"m")^(r) xx "1 D"/(3.336 xx 10^(-30) cancel("C"cdot"m"))#
#~~# #"1.313 Debyes"#, or #"D"#
But this is not the net dipole moment. The net dipole moment is what we want, and is the vertical arrow shown below.
Now what we want to do is get the angle relative to the vertical, so that we can get the
This angle is based on a right triangle formed by an
So, this angle is found to be:
#costheta = 0.3816/1.0124#
#=> theta = 67.86^@#
Since the vertical is the
One of these vectors (the
#vecmu_z("N"-"H") = "1.313 D" xx cos(67.86^@) = "0.4948 D"#
The addition of three of these
#color(blue)(vecmu_"net"("NH"_3)) = vecmu_z("N"-"H") + vecmu_z("N"-"H") + vecmu_z("N"-"H")#
#=# #color(blue)("1.484 D")#
for the net dipole moment. The actual value from a casual google search is around
Another reference is from CCCBDB, which says it is