Question

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

Answer 1

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.

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We can look at example wave functions of the `1s` and `2p_z` orbitals for the Hydrogen atom that represent each orbital.

In general, the wave function is notated like this:

`psi_(ns"/"p"/"etc)(r,theta,phi) = R_(nl)(r)Y_(l)^(m_l)(theta,phi)`

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 (`1s, 2p_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 = 5.29177xx10^(-11) "m"` is the Bohr radius.

Basically...

• If you examine `psi_(1s)`, 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_(2p_z)`, you can see a dependence on `r` and a dependence on `theta` (in the `costheta`). This is what gives the orbital a non-spherical shape.

It's not obvious however, how the `2p_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 `hatL_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 = {0}` 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 = {-1,0,+1}`, 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 `2p_z` you only go up to units of `1ℏ`, not `2ℏ`.