Add time:09/10/2019         Source:sciencedirect.com

The interaction of ClF5 with water in anhydrous hydrogen fluoride solutions was investigated by means of Raman spectroscopy. The reaction between ClF5 and H2O yields only one product (ClO2F) regardless of the temperature of the process, quantity and concentration of chemicals. ClOF3 also did not form during interaction of ClF5 with hydrates of the following metals: Ni, Mn, Al and Si and B2O3. The mechanism will be discussed.The interaction in the system H3OBF4-HF-ClF5 in the temperature range 293–273 K results in practically quantitive yield of ClOF2BF4. The formation of ClOF2BF4 was determined mainly by two successive reactions H3OBF4+ClF5→ClOF2BF4+3HF (1) ClOF2BF4+H3OBF4→ClO2BF4+BF3+3HF (2)Assuming a bimolecular reaction with k1⪢k2, [ClOF2BF4]=k1/k2 [ClF5], the output of the final product is determined by stage (1). Under k2⪢k1, [ClOF2BF4]=0 there is no accumulation of the product.The yield of the product considerably depends on the purity of ClF5; within the stated temperature range the yield is constant.The influence of ClF3 is explained by proceeding of the following concurrent reactions H3OBF4+ClF3→ClOBF4+3HF (3) ClOBF4+ClF5→ClOF2BF4+ClF3 (4) 2ClOBF4→ClO2BF4+ClF+BF3 (5)With ClF5/ClF3>6 the decrease of H3OBF4 concentration is compensated by formation of intermediate ClOBF4 and the output of ClOF2BF4 doesn't decrease.

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