# Why is the s-orbital always spherical in shape?

Dec 21, 2017

Because its wave function has no angular dependence.

By definition, an $s$ orbital has zero angular momentum, and $l = 0$. Any nonzero angular momentum leads to atomic orbitals having non-spherical shapes.

Some explicit wave functions for the hydrogen atomic orbitals are:

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

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

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

The main thing you should notice is that all of these $s$ orbital wave functions have no $\theta$ or $\phi$ in them, which are angles in spherical coordinates.

That means there is no way the angles could deviate from a straight integration in spherical coordinates at a constant radius (giving a spherical integration path).