# Which atomic orbitals of which subshells have a dumbbell shape?

Jan 8, 2017

Well, of the simpler orbitals (i.e. not $\left(n - 2\right) f$ or $\left(n - 3\right) g$ orbitals), the $\boldsymbol{n {p}_{x , y , z}}$ and $\boldsymbol{\left(n - 1\right) {d}_{{z}^{2}}}$ orbitals are shaped like dumbbells, because their wave functions have...

• for the $n {p}_{x} , n {p}_{y} , n {p}_{z}$ case, one nodal plane perpendicular to the principal axis of that orbital.
• for the $\left(n - 1\right) {d}_{{z}^{2}}$ case, two conical nodes (each a special kind of nodal "plane").

If that doesn't make sense, that's OK. Examples that fit the bill for the above orbitals are shown below so you can see.

I go into specific mathematical proofs below (simpler than one might expect) to show why or how the above-described nodal plane or conical nodes are present.

The wave function ${\psi}_{n l {m}_{l}} \left(r , \theta , \phi\right)$ for atomic orbitals, a function of the radial coordinate $r$ and two possible angular coordinates $\theta , \phi$, can be separated into the ${R}_{n l} \left(r\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$ component functions, where $n , l , {m}_{l}$ are the familiar quantum numbers.

We'll focus on $Y$.

When ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$ contains $t r i g \left(\theta\right) t r i g \left(\phi\right)$, $t r i g \left(\theta\right)$, or $t r i g \left(\phi\right)$ such that:

• $Y \left(\theta , \phi\right) = 0$ for a specific plane perpendicular to the principal axis (for a nodal plane), OR
• $Y \left(\theta\right) = 0$ for two different values of $\theta$ (or $\phi$ for $Y \left(\phi\right)$) while the other angular coordinate is constant (for two conical nodes),

...we have a dumbbell shape.

Here are examples of such functions for the $2 {p}_{x}$, $3 {p}_{y}$, and $3 {d}_{{z}^{2}}$ orbitals:

${Y}_{2 {p}_{x}} \left(\theta , \phi\right) \propto \sin \theta \cos \phi$
${Y}_{3 {p}_{y}} \left(\theta , \phi\right) \propto \sin \theta \sin \phi$
${Y}_{3 {d}_{{z}^{2}}} \left(\theta\right) \propto \left(3 {\cos}^{2} \theta - 1\right)$

• ${Y}_{2 {p}_{x}} \left(\theta , \phi\right) = 0$ when $\theta = 0 , \pi , . . .$ and/or $\phi = \frac{\pi}{2} , \frac{3 \pi}{2} , . . .$. By the definition of $\theta$ and $\phi$ above, that forms a node along the $z$ axis and the $y$ axis, giving a nodal $\boldsymbol{y z}$ plane for the $\boldsymbol{2 {p}_{x}}$ orbital. Indeed, the $y z$ plane is perpendicular to the principal, $\boldsymbol{x}$ axis.

So, the $2 {p}_{x}$ orbital is dumbbell-shaped, just like the $2 {p}_{y}$ and $2 {p}_{z}$.

• ${Y}_{3 {p}_{y}} \left(\theta , \phi\right) = 0$ when $\theta = 0 , \pi , . . .$ and/or $\phi = 0 , \pi , . . .$, which marks nodes along the $z$ axis for $\theta = 0 , \pi$ and the $x$ axis for $\phi = 0 , \pi$, giving us a nodal $\boldsymbol{x z}$ plane, which is indeed perpendicular to the principal, $\boldsymbol{y}$ axis.

So, the $3 {p}_{y}$ orbital is dumbbell-shaped, just like the $3 {p}_{x}$ and $3 {p}_{z}$.

• ${Y}_{3 {d}_{{z}^{2}}} \left(\theta\right) = 0$ when $3 {\cos}^{2} \theta - 1 = 0$. It turns out that solving that gives $\theta = \arccos \left(\pm \frac{1}{\sqrt{3}}\right)$, which corresponds to an axis about $\boldsymbol{{54.7}^{\circ}}$ from the $z$ axis.

Since ${Y}_{3 {d}_{{z}^{2}}}$ is not a function of $\phi$ though, $\phi$ is constant, and the angle from the $z$ axis is revolved around the $z$ axis, generating the conical nodes (with a decline of $\approx {54.7}^{\circ}$ and $\approx {234.7}^{\circ}$, respectively, from the $\boldsymbol{z}$ axis).

One conical node is from $\theta = \arccos \left(\frac{1}{\sqrt{3}}\right)$, and the other is from $\theta = \arccos \left(- \frac{1}{\sqrt{3}}\right)$.

So, the $3 {d}_{z}^{2}$ orbital is dumbbell-shaped, as shown below with the conical nodes: