By means of the QCISD(T)/6-311++G(d,p)//MP2/6-311G(d,p) dual-level method, the reaction mechanism between CH2F2 and O(3P) were investigated. Meanwhile, the natural bond orbital (NBO) method was employed to calculate the charge of the transition states occurred in the fluorine elimination and the HF elimination channels. The results indicate that the title reaction exists the hydrogen abstraction (R1), the fluorine abstraction (R2), the fluorine elimination (R3) and the hydrogen elimination (R4) four reaction channels. The reaction energies of each channel are 11.1, 304.0, 78.5 and 31.3 kJ/mol, respectively, and the corresponding barrier heights are 54.6, 351.6, 246.8 and 279.4 kJ/mol, respectively.Being distinctively different from the reactions of CH3F+O(3P) and CHF3+O(3P), the fluorine elimination channel occurs for the title reaction, which is closely related to the asymmetry position of the high electronegativity F2 atom in the transition state TS3. The attraction interaction between F2 and H1 atoms makes the attraction between O and H1 conquer that of F1 and H1, and thus facilitates the fluorine elimination reaction.