AbstractThe reaction mechanism for the N2H4 homogeneous oxidation in aqueous solutions is more complex than its heterogeneous electrochemical oxidation on the electrode surfaces. Homogenous oxidation of a mixture of non‐labeled (14N2H4) and 15N‐labeled (15N2H4) hydrazine in aqueous solutions produces 14N15N, indicating the intermediate existence of N4H6 or N4H4 intermediates with subsequent hydrogen transfers and splitting lateral N─N bonds. To explain the key part of the hydrazine oxidation reaction, the structures, thermodynamics, and electron characteristics of N4H4 in aqueous solution are investigated. Unlike N4H6, we have not found any spontaneous splitting of the bond between the lateral nitrogens in N4H4. The most probable products of N4H4 disproportionation are H2N─NH2 and N2, which are obtained by splitting the bond between the central nitrogen atoms, and so only 15N2 and 14N2 molecules are formed. Additionally, the formation of H3N─NH and N2 products is also preferred to structures without N─N fissions. The formation of H2N═N and HN═NH is energetically less advantageous. Cyclo‐N4H4 structures are stable, without any N─N fissions, but their energies indicate their vanishing abundance in aqueous solution, so their involvement in hydrazine oxidation is highly improbable. Unlike N4H6, the oxidation of hydrazine to 14N15N molecules cannot be explained by N4H4 intermediates.
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