2XeF2+2H2O→2Xe+4HF+O2
3XeF4+6H2O→2Xe+XeO3+12HF+1.5O2
XeF6+RbF→Rb[XeF7]
XeO3+6HF→XeF6+3H2O
HF is the strong reducing agent and hence, it will definitely reduce xenon's oxidation state. But in the option (A), the oxidation number of xenon is unchanged and hence, it is not feasible.