2
$\begingroup$

"If one electron has the quantum numbers n = 4, l = 3, ml = +2, ms =-1/2 and another electron has the quantum numbers n=4, l= 3 ml = -2 ms=+1/2, what are the similarities between the orbitals in which the electron's are found?"

My answer is : They have the same number of energy shells. They have the same shape. Their orientation are different and they have different directions of spin.

I checked the answer and it said that the first one lies on the x and y axis and the other one will lie between the z and y axis. But how do they know that? If m = +2, does that mean it lies on the x and y axis. And if ml = -2, does that mean it lies between the x and y axis?

$\endgroup$
5
$\begingroup$

OP's answer is technically right while additional details about specific orientations of the real $f$-orbitals mentioned somewhere else are basically wrong. At best such details can be thought of the matter of some convention, but it is better to avoid such conventions.

The thing is that for any value of the azimuthal quantum number $l$ an orbital with a non-zero value of the magnetic quantum number $m$ is complex thanks to $\mathrm{e}^{imf}$ factor. Chemists prefer to work with real orbitals formed by taking linear combinations of complex orbitals. However, such linear combinations are not eigenfunctions of $L_z$, or, in other words, $m$ is no longer a good quantum number (but its absolute value is).

In particular, for both $f_{xyz}$ and $f_{z(x^2−y^2)}$ orbitals $|m|=2$, or $m=\pm2$, but the statement that the former corresponds to $m=-2$ and the later to $m=2$ is technically wrong.

$\endgroup$
  • 2
    $\begingroup$ It might be worth looking at McQuarrie & Simon, 'Physical Chemistry' where this is discussed in chapter 6. Examples of complex d orbital equations (unfortunately not f) are given in table 6.5 and of the real d orbitals in table 6.6. $\endgroup$ – porphyrin Sep 19 '16 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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