The symmetry of molecule in excited state may not be same as in ground state.

It is possible optimize the molecule, using a multiconfigurational method, to optimize the excited state that you desire. And after you need to examine the orbitals symmetry to classify the symmetry of system. But your output provided an oscilator strength equal 0 indicating that those transitions are forbidden.

As Francisco said, use direct product of the irreps of the occ-virt pairs involved in the definition of the excited state. Gaussian cannot determine the symmetry of the state if more than one result is possible, such as E2UxE2U=A1g+A2g+E2g in D6h symmetry.

Thank you all, I have managed to obtain the symmetry of the excited states (I am calculating only the singlets). My problem now is on optimising it. Whenever I try the optimisation (of the first excited state), I get a transition geometry with four imaginary frequecies

here is the section of the output

Low frequencies --- -2.3563 0.0004 0.0009

****** 4 imaginary frequencies (negative Signs) ******

Harmonic frequencies (cm**-1), IR intensities (KM/Mole), Raman scattering

activities (A**4/AMU), depolarization ratios for plane and unpolarized

incident light, reduced masses (AMU), force constants (mDyne/A),

and normal coordinates:

1 2 3

E" E" E"

Frequencies -- -549.4946 -549.1815 -73.2279

Red. masses -- 7.6720 7.6682 1.1761

Frc consts -- 1.3649 1.3626 0.0037

IR Inten -- 0.0000 0.0000 0.0000

Atom AN X Y Z X Y Z X Y Z

1 6 0.00 0.00 0.55 0.00 0.00 -0.09 0.00 0.00 -0.03

2 6 0.00 0.00 -0.18 0.00 0.00 -0.21 0.00 0.00 0.06

3 6 0.00 0.00 -0.20 0.00 0.00 0.52 0.00 0.00 0.04

4 6 0.00 0.00 0.27 0.00 0.00 -0.05 0.00 0.00 -0.08

5 6 0.00 0.00 -0.35 0.00 0.00 -0.43 0.00 0.00 -0.02

6 6 0.00 0.00 -0.09 0.00 0.00 0.26 0.00 0.00 0.01

7 8 0.00 0.00 0.04 0.00 0.00 0.11 0.00 0.00 -0.01

8 8 0.00 0.00 0.07 0.00 0.00 -0.09 0.00 0.00 0.03

9 8 0.00 0.00 -0.11 0.00 0.00 -0.02 0.00 0.00 -0.02

10 1 0.00 0.00 0.12 0.00 0.00 -0.05 0.00 0.00 0.23

11 1 0.00 0.00 -0.02 0.00 0.00 0.13 0.00 0.00 -0.33

12 1 0.00 0.00 -0.10 0.00 0.00 -0.08 0.00 0.00 0.09

13 1 0.00 0.00 0.32 0.00 0.00 -0.39 0.00 0.00 0.65

14 1 0.00 0.00 -0.49 0.00 0.00 -0.08 0.00 0.00 -0.62

15 1 0.00 0.00 0.18 0.00 0.00 0.47 0.00 0.00 -0.04

Is there any specific keyword which I may use/test to avoid reaching the transition state. The molecule is so planar that the use of dummy atoms does not seem to change the planarity significantly.

Dear Baptista

the fourth excited state is a non-zero oscilator strength. I paste below the results

Excited State 3: Singlet-E" 6.1704 eV 200.93 nm f=0.0000 =0.000

32 -> 34 0.58913

32 -> 35 -0.11201

32 -> 44 -0.20041

33 -> 36 -0.11201

Excited state symmetry could not be determined.

Excited State 4: Singlet-?Sym 6.4152 eV 193.27 nm f=0.0183 =0.000

31 -> 34 -0.14069

32 -> 35 -0.45274

33 -> 36 0.45274

Excited state symmetry could not be determined.

since the first, the second and the third excited states have zero oscillator strength ( and are therefore forbidden states), is it meaningful that when I want to optimise the excited state, I include up to the 4th excited state?

First: are you sure that you are actually following the same state during the optimization procedure? Second: try to turn off the symmetry by the keyword nosym.

Dear Ciro Guido, Thank you so much for your suggestions, I have managed to obtain the excited state optimised geometry (with the CIS method) for the isolated conformer. followed a specific excited state through the use of the Root=N keyword and turned off the symmetry. Thank you all

You're welcome, by the way, as a final suggestion, take care that in the optimization procedure it is also possible that during the nuclei displacement there be a change of the order of the states. Therefore it is usually also a good practice to check that order looking to the orbital components of the vertical excitation energies that are recalculated at each step of the optimization.