Abstract

Potential energy surfaces of 10 electronic states of IrH2 and 12 electronic states of IrH+2 are computed. A complete active space multi-configuration self-consistent field (CAS-MCSCF) followed by multireference configuration interaction (MRCI) calculations which included up to 270 000 configurations are employed. In addition spin–orbit effects are studied using the relativistic configuration interaction (RCI) method. It is found that the Ir(2F) state inserts spontaneously into H2 to form a stable IrH2 molecule in a 2A1 ground state which is 28 kcal/mol more stable than Ir(4F)+H2 in the absence of spin–orbit effects. The spin–orbit coupling of the quartet and doublet states provides nonzero transition probability for the insertion of Ir (4F) state into H2. The 3P state of Ir+ was found to insert spontaneously into H2 to form the 3A2 ground state of IrH+2 which is 30 kcal/mol more stable than Ir+(5D)+H2. An excited singlet state of Ir+ also was found to insert into H2 spontaneously. The spin–orbit couplings of quintet and triplet states of IrH+2 at the crossing of these curves provide nonzero transition probability for the insertion of Ir+ (5D) into H2. The bent E ground state of IrH2 in the C22v group was found to be a 63% 2A1, 16% 2B1 and 17% 2A2 mixture. This strong mixing induces a large H–Ir–H bond angle change of 9.5° in the E(III) state of IrH2. The 3A2 (A1) ground state of IrH+2 was found to be a 63% 3A2, 15% 3B2, 12% 3B1 and 7% 1A1 mixture. This strong spin–orbit mixing induces a θe change of almost 9° in the ground state of IrH+2. The adiabatic ionization potential including spin–orbit effects for IrH2 and Ir are calculated as 8.2 and 8.6 eV, respectively. The ground state of IrH2 was found to be ionic (μe=2.2 D) with Ir+H− polarity exhibiting Ir(6s0.756p0.125d8.06) hybridization. The IrH+2 ion in its 3A2 state exhibits Ir+(6s0.626p0.125d7.51) hybridization character.

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