On the catalytic mechanism of Pt 4 +/− in the oxygen transport activation of N2O by CO

Abstract

The two-state reaction mechanism of the Pt 4 +/− with N2O (CO) on the quartet and doublet potential energy surfaces has been investigated at the B3LYP level. The effect of Pt4 − anion assistance is analyzed using the activation strain model in which the activation energy (ΔE ≠) is decomposed into the distortion energies $$ (\Updelta E^{ \ne }_{\text{dist}} ) $$ and the stabilizing transition state (TS) interaction energies $$ (\Updelta E^{ \ne }_{\text{int}} ) $$ , namely $$ \Updelta E^{ \ne } = \Updelta E^{ \ne }_{\text{dist}} + \Updelta E^{ \ne }_{\text{int}} $$ . The lowering of activation barriers through Pt4 − anion assistance is caused by the TS interaction $$ \Updelta E^{ \ne }_{\text{int}} $$ (−90.7 to −95.6 kcal/mol) becoming more stabilizing. This is attributed to the N2O π*-LUMO and Pt d HOMO back-donation interactions. However, the strength of the back-donation interactions has significantly impact on the reaction mechanism. For the Pt4 − anion system, it has very significant back-bonding interaction (N2O negative charge of 0.79e), HOMO has 81.5% π* LUMO(N2O) character, with 3d orbital contributions of 10.7% from Pt(3) and 7.7% from Pt(7) near the 4TS4 transition state. This facilitates the bending of the N2O molecule, the N–O bond weakening, and an O−(2P) dissociation without surface crossing. For the Pt4 + cation system, the strength of the charge transfer is weaker, which leads to the diabatic (spin conserving) dissociation of N2O: N2O(1∑+) → N2(1∑ g + ) + O(1D). The quartet to doublet state transition should occur efficiently near the 4TS1 due to the larger SOC value calculated of 677.9 cm−1. Not only will the reaction overcome spin-change-induced barrier (ca. 7 kcal/mol) but also overcome adiabatic barrier (ca. 40.1 kcal/mol).Therefore, the lack of a thermodynamic driving force is an important factor contributing to the low efficiency of the reaction system.