where Y_b and Y_a represent bonding and anti-bonding MOs and 1s_A and 1s_B represent the atomic orbitals. S is the overlap integral, S = <1s_A|1s_B >. We can designate the MO wavefunction according to the symmetry of the orbital. For linear combinations of s atomic orbitals we use s. If the linear combination is even with respect to inversion (i.e. does not change sign) we call the MO gerade and write it as s_g. If the linear combination is odd with respect to inversion (i.e. does change sign) we call the MO ungerade and write it as s_u. We can also add an asterisk to indicate that this is an anti-bonding orbital.

For molecular hydrogen we can fill the s_g MO with two electrons. The wavefunction is required to be anti-symmetric with respect to electron exchange. This can be ensured by placing the spin-orbitals in a Slater determinant. The two spin orbitals are s_ga and s_gb. The Slater determinant has rows that consist of electrons and columns that consist of spin-orbitals. For two electrons we have:

The energy level ordering shown is typical of the early periodic table. As the 1p_u and 3s_g levels fill, the 3s_g orbital drops in energy below 1p_u. For N₂ the highest occupied MO is 3s_g but for O₂ and F₂ this orbital drops below 1p_u in energy.

However, the valence orbitals are p orbitals in Br so that a more realistic possibility is:

In order for pure rotational spectra to be observed a molecule must possess a permanent dipole moment. The homonuclear diatomic molecules do not possess a dipole moment so they do not have microwave absorption spectra. Molecules such as HBr and CO do have dipole moments are therefore they can interact with microwave radiation as they rotate. This gives rise to pure microwave spectra.
The selection rules for microwave spectra in diatomics is DJ = ± 1. The spectra do not have evenly spaced energy levels, however, the appearance of rotational spectra does lead to evenly spaced lines. Why is this? Examine the energy difference between two successive rotational levels.

Note that the energy level differences will appear as a set of evenly spaced lines with energy difference (h/2p)²/I. Again, these pure rotation spectra not observed for homonuclear diatomics. In units of wavenumber (cm^-1) we can define the rotational term F(J) = E/hc.

For rotational Raman the selection rule is DJ = 0, ± 2. We treat the rotational Raman effect classically. The electric field F impinging on a molecule produces a resulting dipole moment proportional to the field P = aF. F is a time-varying field F = F₀sin(wt). As the molecule rotates the apparent internuclear distance will change (with respect to the fixed electric field) causing the polarizability to change a = a₀ + a₁sin(2w₀t) where w₀ is the rotational frequency of the molecule. The factor of two arises because of the symmetry of a homonuclear diatomic.

P = aF = a₀F₀sin(wt) + a₁sin(2w₀t)F₀sin(wt)

P = a₀F₀sin(wt) + a₁F₀/2[cos([w-2w₀]t) - cos([w+2w₀]t)

There will be light scattered at w the incident frequency. This is known as Rayleigh scattering. There will also be light scattered at w - 2w₀ and w + 2w₀. This light is rotational Raman scattered light. The frequency shift is twice the rotational frequency and thus the quantum mechanical selection rule is DJ = 0, ± 2.

n_observed = n_vib + n_rot

For heteronuclear diatomics (e.g. HBr from above) we can picture the interaction of the changing dipole (dm/dQ) with radiation leading to a change in the vibrational energy of the molecule (absorption or emission). Within the harmonic approximation this would lead to a very boring spectrum with only one line.

No matter what vibrational state |vñ the molecule is found in it would have only one transition energy. There are three factors that make life more interesting:

The absorption and emission spectra must follow both the selection rules for both vibrations Dv = ± 1 and rotations DJ = ± 1. Since DJ = 0 is forbidden there is not observed spectral line at the exact wavenumber of the vibrational transition. This is exemplified with the spectrum of HBr. Note that there is a progression of lines to negative J (the P branch) and a progression of lines to positive J (the R branch). These a centered about the position in wavenumber space of the vibrational transition energy. We will see below that if the orbital angular momentum is not equal to zero the DJ = 0 becomes a valid selection rule. In this case, one observes the Q branch of a rovibrational spectrum.

P branch J à J-1

Q branch J à J

R branch J à J+1

Because the vibrational amplitude increases with vibrational state we expect that the moment of inertia should increase as well. As the excursions of the nuclei increase in amplitude the effective internuclear distance R_e increases and I = mR_e² also increases. We can make the rotational constant B a function of vibrational quantum number B_v.

If we consider the v = 0 à 1 transition, then the frequencies of the P branch will be given by

Since homonuclear diatomic molecules possess no permanent dipole moment they have no change in dipole moment upon vibration. The selection rules for vibrational Raman are the same as those for vibrational absorption and emission Dv = ± 1. At ordinary temperatures most of the molecules are in their ground vibrational states and thus the intensity of the Stoke's line 0 à 1 is much greater than that of the anti-Stoke's line 1 à 0. The mechanism of Raman spectroscopy depends on the polarizability change of the molecule during vibration (da/dQ). Q is the nuclear coordinate.

n_observed = n_electronic + n_vib + n_rot

The transition moment for any of these observed frequencies is the product of the three transition moments:
where n and m represent the quantum numbers of the initial and final electronic states, v and v' those of the initial and final vibrational states and J and J' those of the rotational states. The transition probability for absorption is given by:

The selection rules for all three types of transitions apply.

The motion of the electrons in an atom takes place in a spherically symmetrical field of force. As a consequence the electronic orbital angular momentum L is a constant of the motion (provided spin effects are small). In a diatomic molecule the symmetry of the field in which the electrons move is reduced; there is axial symmetry. As a consequence only the component of angular momentum about the internuclear axis is a constant of the motion. Therefore, it is appropriate to classify the electronic states of diatomic molecules according to the value of M_L than according to the value of L. The corresponding angular momentum vector L represents the component of the orbital angular momentum about the z axis. Its magnitude is L(h/2p). Accordingly as L = 0, 1, 2, 3…, the corresponding molecular state is designated a S, P, D, F,…state, analogous to the mode of designation for atoms.

P, D, and F are doubly degenerate since M_L can have the two values +L and -L; S states are non-degenerate.

Just as in atoms there as 2S+1 possible spin states. If L ¹ 0 (P, D, states) there is an internal magnetic field along the internuclear axis due to the orbital angular momentum of the electrons. This magnetic field causes a precession of S about the field direction with a constant component M_S(h/2p). For molecules, M_S is denoted by S must not be confused with the symbol S.

The total electronic angular momentum is obtained as W = L + S just as J = L + S for atoms. In diatomics J is still the total angular momentum, but it now includes the rotational and electronic contributions. Thus, J is the vector sum of rotational angular momentum N, orbital angular momentum L, and spin angular momentum S. Because of the angular momentum about the internuclear axis W we can consider diatomic molecules as symmetric tops.

Hund's case (a). If L ¹ 0 in diatomic molecules with reasonably heavy atoms electronic motion is strongly coupled to the line joining the nuclei. The coupling of N and W is similar to that of a symmetric top.

I is the ordinary momentum of inertia I = mR². I_e is the moment of inertia of electrons about an axis perpendicular to the internuclear axis. Note that levels with J < W are absent.

Hund's case (b). If L = 0 or in diatomics with light atoms the spin angular momentum is only weakly coupled to the internuclear axis.

For L = 0 the rotational angular momentum vector N is perpendicular to the internuclear axis. In this case the angular momentum vector is called K. There is very slight coupling between spin S and rotation K.

The term symbols for diatomic molecules are analogous to those for atoms. They have the form ^2S+1L, however with the addition of two symmetry specifications. If inversion symmetry is present in the molecule (i.e. for homonuclear diatomics) then the gerade (g) or ungerade (u) [even or odd] property of the wavefunction upon inversion is specified as ^2S+1L_g or ^2S+1L_u. We can also specify the parity of nondegenerate (i.e. S) states with respect to reflection through any plane passing through both nuclei. If the wavefunctions change sign upon reflection the designation - is used and if they do not change sign the designation + is used. The term symbol is ^2S+1L_g(u)^+(-).

For orbital angular momentum state DL = 0, ±1. This means that

S à S

S à P

P à P

P à D

are allowed but other such as S à D are not.

The selection rule for L corresponds exactly to the selection rule for M_L for atoms in an electric or magnetic field.

For homonuclear diatomics g à u (g to g and u to u are forbidden).

S⁺ à S⁺

S^- à S^-

S⁺ à S^-

however both S⁺ and S^- can lead to P states.

Spin S is defined for both Hund's rule case (a) and case (b). The spin selection rule is DS = 0.

For Hund's rule case (a) the quantum number S of the component of spin about the internuclear axis is defined. For this component the selection rule is DS = 0. The component of spin along the internuclear axis does not alter.

For rotational transitions DJ = 0, ±1 (J = 0 to J' = 0 is forbidden).

Furthermore, DJ = 0 if DW = 0.

There is one ground state term ¹S_g⁺ corresponding to the electron configuration (1s_g)². Experimentally, the observation of bands in H₂ is difficult due to the many lines without pronounced bands. H₂ is a case where theory proved valuable in analyzing the spectra and finally obtaining assignments for the transitions.

(1ss_g)(2ps_u^*) (¹S_u⁺) à (1ss_g)² (¹S_g⁺) [B à X] 91689.9 cm^-1

(1ss_g)(2pp_u^*) (¹P_u) à (1ss_g)² (¹S_g⁺) [C à X] 100043.0 cm^-1

One of the things that complicates the H₂ spectrum is that the transition (1ss_g)(2ps_u^*) (³S_u⁺) leads to dissociation and forms a continuum band to higher energy. Notice that this is a triplet state and is therefore forbidden. It is important to recall that forbidden transitions do occur albeit with a smaller transition probability than allowed transitions.

(1ss_g)(2ss_u^*) (³S_u⁺) à (1ss_g)² (¹S_g⁺) [a à X] 95938 cm^-1

(1ss_g)(2pp_u^*) (³P_u) à (1ss_g)² (¹S_g⁺) [c à X] 95744 cm^-1

nds ß à 1s (¹S_g⁺)

ndp ß à 1s (¹P_g)

There is one ground state term ¹S_g⁺ corresponding to the electron configuration (1p_u)⁴(3s_g)². We can depict the ground state as follows:

The lowest excited electron configurations are (1p_u)⁴(3s_g)(1p_g^*) and (1p_u)³(3s_g)²(1p_g^*). They account for the lowest observed excited states ¹P_g, ³P_g, and ³S_u⁺.

(3s_g)(1p_g^*) (¹P_g) à (3s_g)² (¹S_g⁺) [a à X] 68,956.6 cm^-1

(3s_g)(1p_g^*) (³S_u⁺) à (3s_g)² (¹S_g⁺) [A à X] 49,756.5 cm^-1

(3s_g)(1p_g^*) (³P_g) à (3s_g)² (¹S_g⁺) [B à X] 59,626 cm^-1 

There are three possible ground state terms ³S_g^-, ¹D_g ,¹S_g⁺ corresponding to the electron configuration (3s_g)²(1p_u)⁴(1p_g^*)². We can depict these as follows:

According the Hund's rule the ³S_g^- state is the ground state. The ¹D_g state is 0.98 eV (7,900 cm^-1) higher in energy and the ¹S_g⁺ state is 1.63 eV (13,150 cm^-1) higher in energy. Singlet oxygen is an important molecule because of its ability to react with organic molecules.

The first excited electron configuration is (3s_g)²(1p_u)³(1p_g^*)³. These configurations give rise to the states ¹S_u⁺, ¹S_u^-, ¹D_u ,³S_u⁺, and ³S_u^-. Of these, ³S_u⁺ and ³S_u^- have been observed.

The ground state of iodine and other dihalogens is shown below. The ground state is ¹S_g⁺.

(1p_u ^*)³(3s_u^*) (³P_u) à (1p_u ^*)⁴ (¹S⁺) [A à X] 11,803 cm^-1

(1p_u ^*)³(3s_u^*) (³P_u) à (1p_u ^*)⁴ (¹S⁺) [B à X] 15,598 cm^-1

Carbon monoxide has an identical electronic structure to N₂. It has a bond order of 3 as does N₂ and has, in fact, one of the strongest molecular bonds known. Since the CO molecule has unequal nuclear charges it has ground state dipole moment. Furthermore, the distinction between gerade and ungerade is no longer needed for CO (or any other heteronuclear diatomic molecule). The ground state term is thus ¹S⁺.

(3s)(1p^*) (¹P) à (3s)² (¹S⁺) [A à X] 64,746.5 cm^-1

(3s)(1p^*) (³P) à (3s)² (¹S⁺) [a à X] 48,473.9 cm^-1