The modern preparation entails the reaction of fluorine with a dilute aqueous solution of sodium hydroxide , with sodium fluoride as a side-product : 2 F2 + 2 NaOH → OF2 + 2 NaF + H2O Its powerful oxidizing properties are suggested by the oxidation number of +2 for the oxygen atom , which is unusual .
Sodium fluoride is left as a side-product : 2 F2 + 2 NaOH → OF2 + 2 NaF + H2O It is a very powerful oxidizing agent .