In the adduct compound [XeF5]+[RuF6]−[\text{XeF}_5]^+ [\text{RuF}_6]^-[XeF5]+[RuF6]−, the oxidation number of fluorine is -1.
What are the oxidation numbers of Xe\text{Xe}Xe and Ru\text{Ru}Ru in this compound?
Xe=+4\text{Xe} = +4Xe=+4 and Ru=+6\text{Ru} = +6Ru=+6
Xe=+6\text{Xe} = +6Xe=+6 and Ru=+5\text{Ru} = +5Ru=+5
Xe=+5\text{Xe} = +5Xe=+5 and Ru=+6\text{Ru} = +6Ru=+6
Xe=+6\text{Xe} = +6Xe=+6 and Ru=+6\text{Ru} = +6Ru=+6
