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

1 Answer
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 #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).

http://teaching.shu.ac.uk/

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.

https://upload.wikimedia.org/

The only difference with this image is that for #2p_z# you only go up to units of #1ℏ#, not #2ℏ#.