The molecular orbital schemes for singlet ($\mathrm{^1\Delta_g}$) and triplet oxygen ($\mathrm{^3\Sigma_g^-}$) are typically given as shown in the image below.

MO schemes of singlet and triplet oxygen
Figure 1: Molecular orbital schemes of two types of singlet oxygen and triplet oxygen with the highest energy electrons highlighted in red.

It is pretty clear that the dioxygen molecule, if just observing the position of the nuclei, must be $D_\mathrm{\infty h}$ symmetric. This implies that both $\unicode{x3c0}^*$ bonds are symmetry equivalent. Therefore, it seems logical that in the $\mathrm{^3\Sigma_g^-}$ triplet state the two orthogonal $\unicode{x3c0}^*$ bonds are each populated by a single electron, leading to an overall rotation-symmetric linear molecule.

In the $\mathrm{^1\Delta_g}$ singlet state, this degeneracy seems to be destroyed, just by looking at the MO scheme. We now seem to have one populated $\unicode{x3c0}^*$ orbital and one unpopulated one, going by my understanding of this scheme.

What does this imply for the electron density distribution of $\mathrm{^1\Delta_g}$ singlet oxygen? Is it valid to assume that two (arbitrary) opposite sides have a higher electron density than the two other orthogonal sides? How can this be interpreted?


1 Answer 1


Probably not a full answer, but hoping to point readers in a useful direction.

If we denote the two $\pi^*$ orbitals as $\pi_x^*$ and $\pi_y^*$ respectively, then the spatial wavefunctions for each state can be written as follows (all lower orbitals are ignored):

$$\begin{align} {}^1\Sigma_\mathrm g^+ &: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) + \pi_x^*(2) \pi_y^*(1)] \\ {}^3\Sigma_\mathrm g^- &: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) - \pi_x^*(2) \pi_y^*(1)] \\ {}^1\!\Delta_\mathrm g &: \begin{cases}\pi_x^*(1) \pi_x^*(2) \\ \pi_y^*(1) \pi_y^*(2) \end{cases} \end{align}$$

For the ${}^1\!\Delta_\mathrm g$ case, one would expect a spatial degeneracy of 2, because the letter $\Delta$ in the term symbol indicates that the quantum number $|\Lambda| = 2$. Therefore, there is one state with $\Lambda = +2$, and one state with $\Lambda = -2$. On the other hand, the $\Sigma$ term has $|\Lambda| = 0$ and hence is spatially non-degenerate (the $^3\Sigma$ term is triply degenerate due to spin).

(Digression: $\Lambda$ represents the projection of the angular momentum along the internuclear axis. As such, there is no further projection quantum number ("$M_\Lambda$") that can take values $-2, -1, 0, +1, +2$. In comparison, in atoms, $\ell$ indicates the total angular momentum and $m_\ell$ the projection of this angular momentum onto the $z$-axis. For more information about term symbols of diatomic molecules, I suggest reading this link. There are many other sources on the Internet, but many are sloppy with notation, which only leads to confusion down the road.)

So, the MO diagram above - which ties the ${}^1\!\Delta_\mathrm g$ state to only one wavefunction - is incomplete. The ${}^1\!\Delta_\mathrm g$ state corresponds to two possible wavefunctions, which are symmetry-equivalent; there is no "preference" for the $x$-axis over the $y$-axis.

If you want to go further than what I've written, then I'd point you to the case of the boron atom which I asked about earlier, which is exactly analogous to this. The point about boron is that it has one electron in the 2p subshell. Does this go into the $\mathrm p_x$, $\mathrm p_y$, or $\mathrm p_z$ orbital? In the case of dioxygen, do the paired electrons go into the $\pi_x^*$ or $\pi_y^*$ orbital?

As you said, it does not make sense for us to assign the electron(s) to any of the orbitals in particular, as that leads to an asymmetric electron density distribution. However, I must admit that I don't fully understand the CASSCF calculation there, and I never quite got round to following up on that question, so you'll have to ask somebody else about that.

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    $\begingroup$ Am I distilling the following take-home message correctly: Assuming we split the space around the bond axis using the planes $x=z$ and $x=-y$ into four equal spaces, it is wrong to imagine that two (arbitrary, but opposite) of these spaces have a higher electron density than the other. $\endgroup$
    – Jan
    Oct 1, 2017 at 14:28
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    $\begingroup$ Yes, that's essentially it. $\endgroup$ Oct 1, 2017 at 14:33
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    $\begingroup$ $$\begin{align} {}^1\Sigma_\mathrm g^+ &: 2^{-1/2}[|\pi_x^*(1) \pi_y^*(2)| \color{red}{-} |\pi_x^*(2) \pi_y^*(1)|] \\ {}^3\Sigma_\mathrm g^- &: 2^{-1/2}[|\pi_x^*(1) \pi_y^*(2)| \color{red}{+} |\pi_x^*(2) \pi_y^*(1)|] \end{align}$$ Acc to doi: 10.1021/cr60272a004 $\endgroup$ Oct 5, 2017 at 8:48
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    $\begingroup$ @Martin-マーチン if you're talking about the equations on p 397, those are the full spatial+spin wavefunctions; what I wrote above is only the spatial part. If you expand the determinants and factorise out the spin part, you get the minus sign for the spatial part of the triplet state (I didn't bother checking the singlet state). (I added the word spatial to hopefully clarify.) $\endgroup$ Oct 5, 2017 at 11:02

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