# Why are s orbitals shaped like spheres but p orbitals shaped like dumbbells?

Oct 4, 2016

Because the $s$ orbital is spherical and has neither angular dependence nor angular momentum, whereas the $p$ orbital has angular dependence and an angular momentum.

I'm going to introduce a bit of hard math, but I'll point out the important things. If you have a grasp on some physics and can visualize a little, this should be accessible/approachable.

We can look at example wave functions of the $1 s$ and $2 {p}_{z}$ orbitals for the Hydrogen atom that represent each orbital.

In general, the wave function is notated like this:

${\psi}_{n s \text{/"p"/} e t c} \left(r , \theta , \phi\right) = {R}_{n l} \left(r\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$

where:

• $n$, $l$, and ${m}_{l}$ are the principal, angular momentum, and magnetic quantum numbers, respectively, and we are in spherical coordinates (one radial coordinate and two angular coordinates).
• $R$ is a function of $r$, describing how the radius of the orbital changes, and $Y$ is a function of $\theta$ and $\phi$, describing how the shape of the orbital changes.

The example wave functions are ($1 s , 2 {p}_{z}$):

psi_(1s) = 1/(sqrtpi)(Z/(a_0))^"3/2"e^(-"Zr/a"_0)

psi_(2p_z) = 1/(4sqrt(2pi)) (Z/(a_0))^"5/2" re^(-"Zr/2a"_0)costheta

where $Z$ is the atomic number and ${a}_{0} = 5.29177 \times {10}^{- 11} \text{m}$ is the Bohr radius.

Basically...

• If you examine ${\psi}_{1 s}$, you can see no dependence on $\theta$ or $\phi$, only $r$. That means it's spherically symmetric. If only $r$ is changing, it can only be a sphere, and thus, it has no directionality.
• If you look at ${\psi}_{2 {p}_{z}}$, you can see a dependence on $r$ and a dependence on $\theta$ (in the $\cos \theta$). This is what gives the orbital a non-spherical shape.

It's not obvious however, how the $2 {p}_{z}$ it looks like a dumbbell. That is more obvious if we look at the angular momentum in the $z$ direction, ${L}_{z}$.

The angular momentum is what reacts to a magnetic field, creating a "precessional orbit" about the $z$ axis (a Larmor Procession, as shown below). For the angular momentum operator ${\hat{L}}_{z}$, it corresponds to us observing the "eigenvalue" bb(m_lℏ), which represents the projection of the orbital in the $z$ direction in units of ℏ = (h)/(2pi), where $h$ is Planck's constant.

For $s$ orbitals, $l = 0$, so ${m}_{l} = \left\{0\right\}$ and we have no angular momentum. Thus, there is no distortion from a spherical shape. It is a nonzero ${m}_{l}$ that produces a non-spherical shape! However, for $p$ orbitals, there is $l = 1$, so ${m}_{l} = \left\{- 1 , 0 , + 1\right\}$, which gives a response to a magnetic field and produces a magnetic projection in the $+ z$, $0$, and $- z$ directions. That gives your dumbbell shape. The only difference with this image is that for $2 {p}_{z}$ you only go up to units of 1ℏ, not 2ℏ.