Why do we say that s orbitals have zero angular momentum?

Sep 8, 2017

The angular momentum of any $s$ orbital is zero, since the wave function for an $s$ orbital has no angular dependence.

In other words, recall that angular momentum gives rise to irregular shapes of a given atomic orbital. Well, all $s$ orbitals are spherically symmetric, so angular momentum has no influence on the shape.

Consider some of the $s$ orbital wave functions:

${\psi}_{1 s} \left(r , \theta , \phi\right) = \frac{1}{\sqrt{\pi}} {\left(\frac{Z}{a} _ 0\right)}^{3 / 2} {e}^{- Z r / {a}_{0}}$

${\psi}_{2 s} \left(r , \theta , \phi\right) = \frac{1}{4 \sqrt{2 \pi}} {\left(\frac{Z}{a} _ 0\right)}^{3 / 2} \left(2 - \frac{Z r}{a} _ 0\right) {e}^{- r / 2 {a}_{0}}$

${\psi}_{3 s} \left(r , \theta , \phi\right) = \frac{1}{81 \sqrt{3 \pi}} {\left(\frac{Z}{a} _ 0\right)}^{3 / 2} \left(27 - 18 \frac{Z r}{a} _ 0 + 2 {\left(\frac{Z r}{{a}_{0}}\right)}^{2}\right) {e}^{- Z r / 3 {a}_{0}}$

If you look closely, there is no $\theta$ or $\phi$ anywhere in these wave functions. As a result, the orbitals they describe are perfect spheres, having zero angular momentum.