Electronic states of atoms, ions, and molecules can be used as reference data. The energies of these states are defined relative to the lowest energy state of the system.
Before proceeding, it is important to point out that atomic states should not be confused with atomic orbital occupancies or electron configurations. To distinguish between them, electron configurations always use lower-case letters, e.g., for carbon, the lowest energy configuration is 1s22s22p2. Its lowest energy state is 3Pg, with higher energy states being 1Dg and 1Sg. Just as electrons in atoms have quantum numbers (n: principal, l: angular, m: magnetic, and s: spin), atomic and excited states also have quantum numbers: N: principal, L: angular, M: magnetic, and S: spin. A state symbol for an atom can be constructed from these letters as N2S+1L. The magnetic quantum number M is not used because, in the absence of a magnetic field, all M states are degenerate.
To help understand the protocols used here, consider a hypothetical atom with a 5Sg ground state and an excited state of symmetry 3Fg, that is, the excited state is a triplet with angular quantum number L = 3. This state is 21 fold degenerate - three from spin and seven from space. For the purpose of parameterization, such degenerate manifolds are regarded as single states.
Each state is defined by three data, with the data being separated by commas.
This is the number of the specific state. If there is only one 3Fg, state, its number will be 1, if there is more than one state with the symmetry 3Fg, the lowest energy state would have the PQN 1, the next, 2, the next 3, and so on.
This defines the spin of the state, and is equal to 2S + 1, where S is the spin angular momentum quantum number. The spin is thus equal to the degeneracy of the spin-state.
For atomic states, the IRs are S, P, D, F, G, H, I, K, L, and M, followed by either (g) or (u) (for gerade or ungerade). Note that there is no J state. For molecular species, the states are the normal IR's.
ROOT=1,2,S(G) This is the 2Sg ground state of an alkali metal atom. Its heat of formation would be zero.
ROOT=1,2,P(U) Excited or 2Pu state of an alkali metal atom. Its heat of formation would be positive.
ROOT=2,5,D(G) Second state of symmetry 5Dg. In this system, there is more than one state of symmetry 5Dg and it is the second energy state of symmetry 5Dg that is being defined, so the PQN is 2.
An easy way to work out the state is to run the system using MOPAC and using keyword MECI. Look at the list of states printed at the bottom of the output file.
A good source for reference data on states is: http://physics.nist.gov/PhysRefData/ASD/levels_form.html. The methods in MOPAC do not include spin-orbit coupling. This means that all spin-orbit split energy levels will be degenerate, i.e., not split. To convert from "exact", i.e., spin-orbit split states to the degenerate state, calculate the barycenter of each manifold of interest.
Spin-orbit splitting will take a degenerate manifold of levels and split it into a set of two or more new degenerate manifolds of levels.
The original manifold, before spin-orbit splitting, will be (2S+1)*(2L+1) fold degenerate, where S is the Spin and L is the spatial or angular quantum number. In atoms and atomic ions, the values of L are 0 for "S" States, 1 for "P", 2 for "D", etc. For molecular systems, the equivalent terms are 0 for non-degenerate, 1/2 for a two-fold degenerate, i.e., an "E" state, 1 for three-fold, i.e., a "T" state, 3/2 for a four fold, i.e., a "G" state, and 2 for a five-fold degenerate molecular state, i.e., a "H" state. For a sodium atom, for example, the first excited State, ROOT=1,2,P(U), is 6-fold degenerate. S = 1/2, so the Spin degeneracy is 2, and IR = P(U), or L = 1, so the spatial degeneracy is 3, for a total degeneracy of 6. For a d1 tetrahedral transition metal complex, the first excited State is 2T, with S=1/2 and L=1. This State is 6 fold degenerate.
Spin-orbit splitting will split the spin-free level according to the triangular conditions. The individual spin-orbit levels can be labeled by J, where J ranges from |S-L| to S+L. For the 2P(U) level of Na, this would be J = 1/2 and 3/2. The J levels would have degeneracy (2J+1), for 2P(U) this would be 2 and 4. The energies of these levels would be E(1/2) and E(3/2).
Converting a set of spin-orbit split levels into a spin-free level involves the weighted average of all the spin-orbit levels. Each spin-orbit level must weighed according to its degeneracy. So for the 2P(U) level of Na, the spin-free level would have an energy of (1/6)(2*E(1/2) + 2*E(3/2)). Because a weighted average is used, this type of averaging is called barycenter averaging.
There are typically several J states in atoms that have high Spin and high angular momentum. Thus for the spin-free 4G(u) atomic state, with S = 3/2 and L = 4, the J states would be 5/2, 7/2, 9/2, and 11/2, and the barycenter would be (1/36)(6*E(5/2) + 8*E(7/2) + 10*E(10/2) + 12*E(11/2)).