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I am doing some calculations on single oxygen atom (O) using CFOUR. According to a previous answer to this question and according to how CFOUR treats non-Abelian groups, I am considering the (O) atom to belong to the $D_{2h}$ point group. In the ZMAT input file, I should provide the OCCUPATION tag, according to the documentation, this tag the stands for the number of occupied orbitals of each symmetry type.

For the (O) atom, which has an electron configuration of $1s^22s^22p^4$ and according to the character table found in this answer I am assigning the following values OCCUPATION=2-1-1-1-0-0-0-0/2-1-0-0-0-0-0-0. Based on what I understood the first four irreducible representations should correspond to $A_g$,$B_{3u}$,$B_{2u}$,and $B_{1u}$ in the order of $s$,$p_x$,$p_y$,$p_z$ orbitals, I have added (1) in the $\beta$ sets of orbitals since $p_x$ is holding 2 electrons with spin up and spin down. However, I have doubts in my understanding, any guidence would be appreciated.

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Your setting is incorrect for the dominant Slater determinant when calculating the electronic ground state of an O atom:

OCCUPATION=2-1-1-1-0-0-0-0/2-1-0-0-0-0-0-0

Notice that for the N atom, as described here: When orbitals are labeled based on their irreps in D2h, how are the orbitals ordered for an N atom?, we had:

OCCUPATION=2-1-1-0-1-0-0-0/2-0-0-0-0-0-0-0 

This is the hint that the D2h irreps in correspond to the following orbitals:

OCCUPATION=s-p-p-0-p-0-0-0/s-0-0-0-0-0-0-0 

For the element O, you have correctly added one beta electron in a p-orbital, but you have put one of the alpha p-electrons in a different irrep.

A correct setting for an O atom, would be the following:

OCCUPATION=2-1-1-0-1-0-0-0/2-1-0-0-0-0-0-0
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  • $\begingroup$ +1, thank you for the answer, but excuse me that I am still not clear why the representation would be s-p-p-0-p-0-0-0 for the alpha electrons? $\endgroup$ Mar 18 at 5:51
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    $\begingroup$ The authors of CFOUR decided that order. $\endgroup$ Mar 18 at 10:45

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