When XeF6 (xenon hexafluoride) is treated with aqueous NaOH (sodium hydroxide), a series of reactions occur resulting in the formation of sodium hexafluoroxenate(VI) and other products. Let's break down the reaction steps to understand the process in detail:


NaOH dissociates in water to form sodium ions (Na+) and hydroxide ions (OH-).

NaOH → Na+ + OH-


XeF6 is a strong oxidizing agent and reacts with water to form xenonate(VI) ion (XeO6^4-) and fluoride ions (F-).

XeF6 + 2H2O → XeO6^4- + 4HF


The hydroxide ions (OH-) from NaOH react with the hydrogen fluoride (HF) produced in the previous step to form water and fluoride ions (F-).

HF + OH- → H2O + F-


Combining the above steps, the overall reaction can be represented as:

XeF6 + 6NaOH → Na4XeO6 + 4NaF + 3H2O


The reaction between XeF6 and aqueous NaOH yields the following products:

1. Sodium hexafluoroxenate(VI) (Na4XeO6): This compound is a salt containing the hexafluoroxenate(VI) ion (XeO6^4-). It is a yellow-orange solid and is highly soluble in water.

2. Sodium fluoride (NaF): This is a white, crystalline compound formed as a byproduct of the reaction.

3. Water (H2O): Water is formed as a result of the acid-base reaction between HF and OH- ions.


In summary, when XeF6 is treated with aqueous NaOH, a series of reactions occur leading to the formation of sodium hexafluoroxenate(VI), sodium fluoride, and water. This reaction demonstrates the oxidizing properties of XeF6 and the acid-base neutralization between NaOH and HF.