# Which d orbital is specified by Y(theta,phi) = (5/(8pi))^(1//2) (3cos^2theta - 1)?

Apr 25, 2016

Your angular wave function looks off. Every wave function ${\psi}_{n l {m}_{l}}$, which describes the state of a quantum mechanical system, must be at LEAST

• orthogonal (${\int}_{\text{allspace" psi_i^"*}} {\psi}_{j} d \tau = 0$ where $i \ne j$)
• normalizable (${\int}_{\text{allspace" psi_i^"*}} {\psi}_{i} d \tau = 1$)

and yours is not. The correct ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$ is:

$\textcolor{b l u e}{{Y}_{2}^{0} \left(\theta , \phi\right) = {\left(\frac{5}{16 \pi}\right)}^{\text{1/2}} \left(3 {\cos}^{2} \theta - 1\right)}$

In this case, what we have here is ONLY the angular component of the wave function for the ${d}_{{z}^{2}}$ orbital using spherical coordinates. Actually, without knowing the radial component, we do not know the principal quantum number of your ${d}_{{z}^{2}}$ orbital.

The full wave function in spherical harmonics is written as the product of the radial and angular components:

$\setminus m a t h b f \left({\psi}_{n l {m}_{l}} \left(\vec{r} , \theta , \phi\right) = {R}_{n l} \left(\vec{r}\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)\right)$

For instance, the full wave function for the $3 {d}_{{z}^{2}}$ orbital is:

${\psi}_{3 {d}_{{z}^{2}}} = {\psi}_{320} \left(\vec{r} , \theta , \phi\right) = {R}_{32} \left(\vec{r}\right) {Y}_{2}^{0} \left(\theta , \phi\right)$

= 1/(81sqrt(6pi))(Z/(a_0))^"3/2" ((Zr)/(a_0))^2e^(-Zr"/"3a_0) (3cos^2theta - 1),

where $Z$ is the atomic number and ${a}_{0}$ is the Bohr radius ($5.29177 \times {10}^{- 11} \text{m}$).

This, you would see is quite different and much more complicated.

You did not include ${R}_{n l} \left(\vec{r}\right)$, so the best I can do is say that you have accidentally specified all ${d}_{{z}^{2}}$ orbitals in existence. There exist the $3 {d}_{{z}^{2}}$, $4 {d}_{{z}^{2}}$, $5 {d}_{{z}^{2}}$, and $6 {d}_{{z}^{2}}$ thus far that have been observed.

Generally they look like dumbbells along the z-axis with a torus (donut) on the xy-plane. The torus has the OPPOSITE sign to the lobes, and the lobes have the SAME sign as each other.

The ${d}_{{z}^{2}}$ have conic nodal surfaces. Specifically:

• The $3 {d}_{{z}^{2}}$ has $n - l - 1 = 3 - 2 - 1 = 0$ radial nodes and $n - 1 = 2$ total nodes. Thus, it has two angular nodes.
• The $4 {d}_{{z}^{2}}$ has $n - l - 1 = 4 - 2 - 1 = 1$ radial node and $n - 1 = 3$ total nodes. Thus, it has two angular nodes as before.
• The $5 {d}_{{z}^{2}}$ has $n - l - 1 = 5 - 2 - 1 = 2$ radial nodes and $n - 1 = 4$ total nodes. Thus, it has two angular nodes as before.
• The $6 {d}_{{z}^{2}}$ has $n - l - 1 = 6 - 2 - 1 = 3$ radial nodes and $n - 1 = 5$ total nodes. Thus, it has two angular nodes as before.

So you can see the pattern that the ${d}_{{z}^{2}}$ orbital simply gains radial nodes as the quantum number $n$ increases.

For instance, here is a picture of the $5 {d}_{{z}^{2}}$ orbital:

Now compare that to the $3 {d}_{{z}^{2}}$ orbital.

You should notice that the $5 {d}_{{z}^{2}}$ is obviously more complex at the torus.