I performed ORCA SOC CASSCF calculations on the Molecule AsCH3+. Without spin orbit coupling the mullikan term symbol should be 2A'', since due to ionization of an electron the c3v symmetry is lost. However, the change in symmetry is so little that we can talk of an 2E state with 2 nearly degenerate non-bonding p orbitals of arsenic, which have little pi* character. Hence, orbitals with ml = +1, -1.

When spin orbit coupling is switched on, the spin orbit energy difference fits very well with the experiment, but i don't know how to determine which state has the 2E1/2 and which has the 2E3/2 symbol.

Note that in the output Root 0 means the electron is in the px orbital, and in Root 1 in the py orbital.

Here is the important part of the output:

CALCULATED REDUCED SOC MATRIX ELEMENTS

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Block Root

I(Mult) J(Mult) I J cm-1 cm-1 cm-1

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0( 2) 0( 2) 0 0 -0.00 0.00 -0.00

0( 2) 0( 2) 1 0 572.62 809.42 1120.57

0( 2) 0( 2) 1 1 -0.00 0.00 -0.00

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Eigenvectors:

Weight Real Image : Block Root Spin Ms

STATE 0: 0.0000

0.435472 0.633251 -0.185649 : 0 0 1/2 1/2

0.223602 -0.133030 -0.453767 : 0 1 1/2 1/2

0.340926 0.457697 0.362546 : 0 0 1/2 -1/2

STATE 1: 0.0000

0.340926 -0.476667 -0.337217 : 0 0 1/2 1/2

0.435472 0.000000 0.659903 : 0 0 1/2 -1/2

0.223602 -0.472865 -0.000000 : 0 1 1/2 -1/2

STATE 2: 1795.5278

0.125415 0.339840 -0.099621 : 0 0 1/2 1/2

0.776398 0.247866 0.845554 : 0 1 1/2 1/2

0.098186 0.245620 0.194569 : 0 0 1/2 -1/2

STATE 3: 1795.5278

0.098186 0.255806 0.180969 : 0 0 1/2 1/2

0.125415 0.000000 -0.354140 : 0 0 1/2 -1/2

0.776398 -0.881135 0.000000 : 0 1 1/2 -1/2