Which dd orbital is specified by Y(theta,phi) = (5/(8pi))^(1//2) (3cos^2theta - 1)Y(θ,ϕ)=(58π)1/2(3cos2θ1)?

1 Answer
Apr 25, 2016

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

  • orthogonal (int_"allspace" psi_i^"*"psi_jd tau = 0allspaceψ*iψjdτ=0 where i ne jij)
  • normalizable (int_"allspace" psi_i^"*"psi_id tau = 1allspaceψ*iψidτ=1)

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

color(blue)(Y_(2)^(0)(theta,phi) = (5/(16pi))^"1/2" (3cos^2theta - 1))Y02(θ,ϕ)=(516π)1/2(3cos2θ1)

In this case, what we have here is ONLY the angular component of the wave function for the d_(z^2)dz2 orbital using spherical coordinates. Actually, without knowing the radial component, we do not know the principal quantum number of your d_(z^2)dz2 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:

http://images.fineartamerica.com/http://images.fineartamerica.com/

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

http://images.fineartamerica.com/http://images.fineartamerica.com/

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