Question

Which atomic orbitals of which subshells have a dumbbell shape?

Answer 1

Well, of the simpler orbitals (i.e. not `(n-2)f` or `(n-3)g` orbitals), the `bb(np_(x,y,z))` and `bb((n-1)d_(z^2))` orbitals are shaped like dumbbells, because their wave functions have...

• for the `np_x,np_y,np_z` case, one nodal plane perpendicular to the principal axis of that orbital.
• for the `(n-1)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.

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The wave function `psi_(nlm_l)(r,theta,phi)` for atomic orbitals, a function of the radial coordinate `r` and two possible angular coordinates `theta,phi`, can be separated into the `R_(nl)(r)Y_(l)^(m_l)(theta,phi)` component functions, where `n,l,m_l` are the familiar quantum numbers.

We'll focus on `Y`.

When `Y_(l)^(m_l)(theta,phi)` contains `trig(theta)trig(phi)`, `trig(theta)`, or `trig(phi)` such that:

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

...we have a dumbbell shape.

Here are examples of such functions for the `2p_x`, `3p_y`, and `3d_(z^2)` orbitals:

`Y_(2p_x)(theta,phi) prop sinthetacosphi`
`Y_(3p_y)(theta,phi) prop sinthetasinphi`
`Y_(3d_(z^2))(theta) prop (3cos^2theta - 1)`

• `Y_(2p_x)(theta,phi) = 0` when `theta = 0,pi, . . .` and/or `phi = pi/2,(3pi)/2, . . .`. By the definition of `theta` and `phi` above, that forms a node along the `z` axis and the `y` axis, giving a nodal `bb(yz)` plane for the `bb(2p_x)` orbital. Indeed, the `yz` plane is perpendicular to the principal, `bb(x)` axis.

So, the `2p_x` orbital is dumbbell-shaped, just like the `2p_y` and `2p_z`.

• `Y_(3p_y)(theta,phi) = 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 `bb(xz)` plane, which is indeed perpendicular to the principal, `bb(y)` axis.

So, the `3p_y` orbital is dumbbell-shaped, just like the `3p_x` and `3p_z`.

• `Y_(3d_(z^2))(theta) = 0` when `3cos^2theta - 1 = 0`. It turns out that solving that gives `theta = arccos(pm1/sqrt3)`, which corresponds to an axis about `bb(54.7^@)` from the `z` axis.

Since `Y_(3d_(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 `~~54.7^@` and `~~234.7^@`, respectively, from the `bb(z)` axis).

One conical node is from `theta = arccos(1/sqrt3)`, and the other is from `theta = arccos(-1/sqrt3)`.

So, the `3d_z^2` orbital is dumbbell-shaped, as shown below with the conical nodes: