화학공학소재연구정보센터
Journal of Chemical Physics, Vol.104, No.14, 5558-5571, 1996
Photodissociation of NH2 - 2-Dimensional Potential-Energy Surfaces for the Dissociation into NH and H
Extensive multireference configuration interaction calculations were carried out in order to obtain complete two-dimensional (2D) potential energy surfaces for the amidogen (NH2) radical as functions of both N-H bond lengths keeping the bond angle fixed at its experimental ground state equilibrium value. The eight lowest-lying states (four of each symmetry, A’ and A ") were treated mainly for the purpose of using these surfaces in subsequent studies of the photodissociation dynamics. In analogy with the neighboring dihydrides CH2 and H20 the photodissociation of NH, into NH + H (hydrogen abstraction) takes place preferentially after excitation of the first two Rydberg s states (3(2)A’/2(2)A(1) and 2(2)A "/2(2)B(1)) found closely together at about 7.6 eV. The transition dipole moments connecting the ground state with these two states are large (0.44 a.u, and 0.66 a.u.) in the Franck-Condon region, but the behavior of the potentials in the dissociation channel is quite different. The 3(2)A’/2(2)A(1) State is weakly repulsive whereas the 2(2)A "/2(2)B(1) State is strongly repulsive. This will result in differences in the dissociation dynamics for the two states. The next higher state which should play a role in the NH2 photodissociation is the 4(2)A "/2(2)B(1) Rydberg s state at 9.4 eV, because of its large transition dipole moment with the ground state (0.36 a.u.). Close to this state, many Rydberg p states were found. Due to the high density of states in the region above 9.0 eV, interactions of these states are expected and should lead to complicated dissociation dynamics. Contrary to CH2, the two low-lying valence states for NH2 are found at lower energies [2.2 eV (1(2)A(1)) and 6.5 eV (1(2)B(2))], well separated from the first members of the Rydberg series. These states are less important for the photodissociation of NH2, compared with CH2, because the first state is bound and the transition to the other is dipole-forbidden in C(2v)symmetry. For H2O, the valence states are missing.