# How does the length of the z-component of the net dipole moment of ammonia relate to the actual length of the dipole moment vector?

Aug 20, 2017

Well, the length is identical. The dipole moment of ammonia is along its $z$ axis, so the $x$ and $y$ components are zero, and only the $z$ component contributes to the length.

The principal axis of rotation (${C}_{n}$) requires the smallest angle of rotation ${360}^{\circ} / n$ before coinciding with the original molecule.

For instance, the ${C}_{n}$ axis of water requires a ${180}^{\circ}$ rotation to bring it back to an orientation indistinguishable to how it was before. So, we call it ${C}_{2}$ and point it along the $z$ axis on the $y z$ plane:

If you want to visualize this more, click on the image and try out the rotation yourself.

For ammonia, the $z$ axis points vertically through the nitrogen atom, like the handle of an umbrella with the $\text{H}$ atoms as the umbrella head.

The dipole moment of ammonia points exactly along the $z$ axis. So, we write the dipole moment vector $\vec{\mu}$ as:

$\vec{\mu} = \left\langle{\mu}_{x} , {\mu}_{y} , {\mu}_{z}\right\rangle$

$= \left\langle0 , 0 , 1.48\right\rangle$

The length of the vector is given by the vector magnitude:

$\textcolor{b l u e}{| | \vec{\mu} | |} = \sqrt{{\mu}_{x}^{2} + {\mu}_{y}^{2} + {\mu}_{z}^{2}}$

$= \sqrt{{0}^{2} + {0}^{2} + {\left(\text{1.48 D}\right)}^{2}}$

$=$ $\textcolor{b l u e}{\text{1.48 D}}$