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

Your angular wave function looks off. Every wave function #psi_(nlm_l)#, which describes the state of a quantum mechanical system, must be at LEAST

• orthogonal (#int_"allspace" psi_i^"*"psi_jd tau = 0# where #i ne j#)
• normalizable (#int_"allspace" psi_i^"*"psi_id tau = 1#)

and yours is not. The correct #Y_(l)^(m_l)(theta,phi)# is:

#color(blue)(Y_(2)^(0)(theta,phi) = (5/(16pi))^"1/2" (3cos^2theta - 1))#

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:

#\mathbf(psi_(nlm_l)(vecr,theta,phi) = R_(nl)(vecr)Y_(l)^(m_l)(theta,phi))#

For instance, the full wave function for the #3d_(z^2)# orbital is:

#psi_(3d_(z^2)) = psi_(320)(vecr,theta,phi) = R_(32)(vecr)Y_(2)^(0)(theta,phi)#

#= 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 (#5.29177xx10^(-11) "m"#).

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

You did not include #R_(nl)(vecr)#, so the best I can do is say that you have accidentally specified all #d_(z^2)# orbitals in existence. There exist the #3d_(z^2)#, #4d_(z^2)#, #5d_(z^2)#, and #6d_(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 #3d_(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 #4d_(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 #5d_(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 #6d_(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 #5d_(z^2)# orbital:

Now compare that to the #3d_(z^2)# orbital.

You should notice that the #5d_(z^2)# is obviously more complex at the torus.