This situation can be reproduced with CHF_{3}. With `METHOD=HF`

:

```
Standard Nuclear Orientation (Angstroms)
I Atom X Y Z
----------------------------------------------------------------
1 C 0.0000000008 0.0000000013 0.3509718273
2 H 0.0000000013 0.0000000022 1.4457718519
3 F 1.2543992952 -0.0000000000 -0.1315408490
4 F -0.6271996476 1.0863416562 -0.1315408490
5 F -0.6271996476 -1.0863416562 -0.1315408490
----------------------------------------------------------------
Molecular Point Group C3v NOp = 6
Largest Abelian Subgroup Cs NOp = 2
Nuclear Repulsion Energy = 132.95944891 hartrees
```

but with `METHOD=CCSD`

:

```
----------------------------------------------------------------
Standard Nuclear Orientation (Angstroms)
I Atom X Y Z
----------------------------------------------------------------
1 C 0.0000000124 0.3509718273 0.0000000000
2 H 0.0000000202 1.4457718519 0.0000000000
3 F 1.2543992952 -0.1315408490 0.0000000000
4 F -0.6271996885 -0.1315408411 -1.0863416377
5 F -0.6271996885 -0.1315408411 1.0863416377
----------------------------------------------------------------
Molecular Point Group Cs NOp = 2
Largest Abelian Subgroup Cs NOp = 2
Nuclear Repulsion Energy = 132.95944891 hartrees
```

It’s not entirely clear why Q-Chem does not determine that molecular point group symmetry is C3v. However, it is largely irrelevant as Q-Chem always switches to the largest Abelian subgroup in CC/EOM/ADC jobs if the true point group symmetry of the molecule is non-Abelian. Hence the treatment of C3v as Cs in the EOM calculation. Excited state irreps in the Cs point group must be mapped back to the irreps in the C3v point group by the user when interpreting the output.

Using `CC_SYMMETRY=FALSE`

disables symmetry altogether in CC/EOM calculations such that Q-Chem does not even attempt to determine the irreps of the orbitals. The whole thing is treated as if it were C1 symmetry.