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I am learning excited state calculation. In this paper, the author use "$D_0$","$D_1$" represent excited state of nitrate radical.

I know the symbol such as "$D_1$","$T_2$" means first excited doublet state, second excited triplet state. Can I get specific electron configuration of a molecule with these symbol,like ${}^2\!A_2$ symbol in this answer? Or first and second excited state are different in a range of energy? In other word, Can I write a electron configuration use symbols like "${}^1\!A_2$" for $D_1$ state of nitrate molecule?

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Unfortunately, no. If you don't want to work it out by hand, the best you can hope for is that the authors explicitly specify orbital occupations with something that looks like an electron configuration. This paper doesn't, the previous paper sort of does.

Here is a paper that doesn't have symmetry labels, but does have configurations. For example, on page 2977, the SCF reference for $\ce{CuO}$ is given as

$$ \ce{ (O 2s)^2 (O 2p\sigma)^1 (O 2p\pi)^4 (Cu 3d\sigma)^2 (Cu 3d\pi)^4 (Cu 3d\delta)^4 }, $$

where the wavefunction symmetry is determined by the singly-occupied orbital (everything else is doubly occupied). $\ce{2p\sigma}$ probably lies along the bond, which would be taken as the $z$-axis when looking at a character table. Go to the $C_{\infty v}$ table, check for $z$, and discover that it's $\Sigma^+$, which is even mentioned in the paper. So there are symmetry labels given here, but the linear groups also have Greek characters for their irrep labels...

One interesting point is that non-Abelian groups are not used in practical quantum chemistry. $C_{\infty v}$ would be reduced to $C_{2v}$, and the $^{2}\Sigma^+$ state would become $^{2}\!A_{1}$.

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The ground state radical belongs to the $\ce{D_{3h}}$ point group and has symmetry species $\ce{^2A^'_2}$, there is an excited state also (at $\ce{\approx 15100 cm^{-1}}$) with the same point group with species $\ce{^2E^'}$.
However, to work out species for other excited states you will need to know their point group and which electrons have been excited.
If the excited state has the same point group as the ground state you only need to look at the highest orbital and one other orbital (the one the electron has been excited from) as all others are doubly occupied and so have A (totally symmetric) symmetry species. Multiply the characters in the 'irreps' (characters in rows of point group table) together to get the symmetry species of the excited state.

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    $\begingroup$ Could you "translate" $D_2$ state to wavefunction symmetry irreps as an example? $\endgroup$ – Chao Song Sep 6 '16 at 7:46
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    $\begingroup$ No not without more information. For example to get to the ground state symmetry species, you must know which orbitals the odd electron is in , then work out its symmetry species, (left hand column in point group table) by performing each of the symmetry operations (top row of table) in turn; character is 1 only if the orbital is indistinguishable after any operation compared to that before operation, (else -1) hence work out symmetry species. $\endgroup$ – porphyrin Sep 6 '16 at 8:04

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