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

##### 1 Answer

Your angular wave function looks off. Every **wave function**

*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

#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 **we do not know the principal quantum number** of your

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

#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

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 **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 hastwo angular nodes.- The
#4d_(z^2)# has#n - l - 1 = 4 - 2 - 1 = 1# radial node and#n - 1 = 3# total nodes. Thus, it hastwo angular nodesas before.- The
#5d_(z^2)# has#n - l - 1 = 5 - 2 - 1 = 2# radial nodes and#n - 1 = 4# total nodes. Thus, it hastwo angular nodesas before.- The
#6d_(z^2)# has#n - l - 1 = 6 - 2 - 1 = 3# radial nodes and#n - 1 = 5# total nodes. Thus, it hastwo angular nodesas before.

So you can see the pattern that the

For instance, here is a picture of the

Now compare that to the

You should notice that the