15
$\begingroup$

Why do the d orbitals have the following notations: $xy, yz, xz, z^2$ and $ x^2-y^2$? What do they represent in their wave-functions?

$\endgroup$
0

1 Answer 1

14
$\begingroup$

The notation is shorthand for the hydrogen atom $\ell = 2$ wavefunctions in real form, in cartesian coordinates. In other words, the Schrodinger equation is solved in spherical coordinates, in terms of spherical harmonics, which involve complex numbers; however, linear combinations of degenerate (equal energy) solutions are also solutions. This permits the wavefunction solutions to be expressed using real numbers only. These real solutions can be transformed to cartesian coordinates.

$z^2$ represents that the wavefunction is proportional to $(3z^2-r^2)/r^2$.

$xy$ represents that the wavefunction is proportional to $xy/r^2$.

$yz$ represents that the wavefunction is proportional to $yz/r^2$.

$xz$ represents that the wavefunction is proportional to $xz/r^2$.

$x^2 - y^2$ represents that the wavefunction is proportional to $(x^2 - y^2)/r^2$.

See this Table of Real Spherical Harmonic Functions for the a list of functions in both spherical coordinates and cartesian coordinates, for s through f orbitals.

$\endgroup$
6
  • 1
    $\begingroup$ Could you comment on the five versus six d-Orbital problem, as in pure versus cartesian orbitals? Compare chemissian.com/ch5 $\endgroup$ Nov 25, 2014 at 2:40
  • 1
    $\begingroup$ @Martin AFAIK, solution of the Schrodinger equation gives 5 orbitals, as in it is enough to consider 5 orbitals to produce all possible wavefunctions. However, calculations using 5 orbitals basis set for some properties require arithmetic on complex numbers, while moving to 6-orbital representation allows to avoid complex numbers in expense of the need to deal with excessive basis set. $\endgroup$
    – permeakra
    Nov 26, 2014 at 10:00
  • $\begingroup$ @permeakra The use of six functions is an older trick because it significantly reduced computational cost and because they are essentially equal to the real solutions of the Schrödinger equation (there is one s-type orbital more). $\endgroup$ Nov 26, 2014 at 10:17
  • $\begingroup$ ccl.net/chemistry/resources/messages/2003/12/12.006-dir/… $\endgroup$ Nov 26, 2014 at 10:27
  • $\begingroup$ other references on the 6-function basis: chem.wayne.edu/schlegel/Pub_folder/174.pdf ; theochem.github.io/horton/tut_gaussian_basis.html $\endgroup$
    – DavePhD
    Nov 26, 2014 at 12:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.