The eigenstates of OH 2II

The configuration of the outer electrons of OH in the electronic ground state, X2U, is 

This implies that three electrons are in the two pn orbitals with lobes perpendicular to the internuclear O-H axis which serves as the 2-axis of the body-fixed frame of the diatom (not to be confused with the body-fixed system of the triatomic molecule). Following Andresen et al. (1984) we assume that two of the pit electrons are paired and one is unpaired, the latter determining the open-shell character of the OH radical. For a more refined analysis see Alexander and Dagdigian (1984). 

The unpaired pn electron has an orbital angular momentum L with projections A = 1 on the body-fixed 2-axis which — in a classical sense — describe rotation of the electron in opposite directions about the internuclear axis. Furthermore, it possesses a spin S with components = 1/2. Coupling of A and leads to two (2 1 = 2) manifolds with A = 1/2 and 3/2 which are represented by 2nx/2 and 2n3/2, respectively. The splitting is in the range of several hundredths cm-1. 

A and B are the fine-structure and the rotational constants of OH and / is the azimuthal angle out of the rotation plane with p = 0 on the internuclear vector, cos2 / and sin2 / represent, respectively, lobes in the rotation plane and perpendicular to it. The Cj approach the value 2 in the limit of large j which has the consequence that p(A ) is preferentially oriented in the rotation plane while p A ) is preferentially oriented perpendicular to the plane of rotation. Equation (11.15) gives the relative weights of the in-plane and the out-of-plane contributions of the real A-doublet states. They play a vital role in the photolysis of water through the AxBi state (see Section 11.2.2). 

The two most weakly bound electrons in H20(X) occupy a l i orbital which is essentially a pure pn lobe perpendicular to the plane defined by the three nuclei. On excitation to the AlB state one of them is excited to the strongly antibonding 4aJ orbital leaving one unpaired pir electron with its lobe perpendicular to the nuclear plane. Let us assume that H2O does not initially rotate (J = 0) thus that the dissociation plane remains fixed in space for all times. As a consequence of the fragmentation, which proceeds perfectly in-plane, the OH( II) fragment is created with an unpaired electron reminiscent of the A(A//) state. Therefore  

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