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

##### 1 Answer

Because the

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

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 (

#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

*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

For **no** angular momentum. Thus, there is **no distortion from a spherical shape**. It is a nonzero

However, for **gives a response** to a magnetic field and produces a magnetic projection in the **That gives your dumbbell shape**.

The only difference with this image is that for