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I'm reading quantum chemistry. The book says that the orbital angular momentum of a $\pi$ electron along the symmetry axis of a molecule made up of two atoms is $\pm 1$. I think this is a primary question, but I do not know why (I'm a student studying physics)


I currently have a preliminary understanding of this:
$/pi$ orbit
In a molecule inculding two atoms, potential energy axisymmetric about z' axis(the line connecting the two atoms). So, the angular momentum along z axis is quantized. That is to say $m_z$ is good quantum number. Let's consider $\pi$ orbit made up of two $p_z$. The orbital angular momentum along z axis(the symmetry axis of $p_z$) of an electron in $p_z$ is 0. So considering orbital angular momentum along z' axis of this electron, the electron is in $\frac 1 {\sqrt 2}(|+>-|->)$. So the orbital angular momentum along z' axis is either 1 or -1. So the orbital angular momentum along z' axis of a $\pi$ electron is either 1 or -1.

Is my understanding correct?

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  • $\begingroup$ Could you elaborate on your question? Are you asking why the angular momentum l is plus or minus one? Or are you asking why the pi orbitals look like dumbells? And what do you mean by line connecting the atoms? A sigma bond? A pi bond? An exotic bond that was published recently? Are you adking why orbitals have to be oriented a certain way to form bonds? I suspect that you may be using the terms wrongly. $\endgroup$ Mar 21, 2014 at 17:44
  • $\begingroup$ @user97554 I'm asking why the angular momentum l is plus or minus one. Thank you for your advice. I have improved my description. Is it more clear? $\endgroup$
    – Ben
    Mar 22, 2014 at 2:39
  • $\begingroup$ Are you familiar with Laguerre polynomials and spherical Bessel functions? Eigenvalues? $\endgroup$ Mar 22, 2014 at 5:57
  • $\begingroup$ @user97554 Yes. $\endgroup$
    – Ben
    Mar 22, 2014 at 6:40
  • $\begingroup$ I am so sorry. But turns out I misunderstood my knowledge. I thought I knew it but in reality I didn't. I apologize. I think you may have better luck at Physics SE. Again I apologize. $\endgroup$ Mar 22, 2014 at 14:02

1 Answer 1

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Let the molecular axis be the $z$ axis, A and B the two atoms forming the bond. The $2p_z$ AO has atomic quantum number $m_l = 0$ and so has $L_z =0$. The MOs formed from 2pz AOs are therefore $\sigma$ MOs, bonding and antibonding. The $p_x$ AO is a linear combination of $m =+ 1$ and $m = -1$ AOs and has $|m| =1$.

$2p_x \equiv \frac{(2p_1 + 2p_{-1}) } {\sqrt 2}$

$2p_y \equiv \frac{(2p_1 - 2p_{-1}) } {i\sqrt 2}$

Therefore the MOs made from the $2p_x$ and $2p_y$ AOs have $|m| = 1$ and are $\pi$ MOs.

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  • $\begingroup$ +1 for correctly rendering the $m$ quantum number of the $p_x$ and $p_y$ orbitals. Too often I see one of these orbitals specifically assigned $m=+1$ and the other rendered as $m=-1$, causing my eyes to pop out and my face to turn purple (OK, I made up that last part but I do take it badly). $\endgroup$ Feb 25, 2019 at 11:44

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