Question

Xe is a noble gas then why does it form $Xe{F}_6$?

Answer 1

Firstly we have to take a look at the electronic configuration of Xe.
The electronic configuration of Xe is $[Kr{]}_4{D}_{10}5s^{2}5p^6$
Energy levels of the 5p orbital, 6s and 4f orbitals are very close and hence energetically accessible. Only a small amount of energy is needed to excite the paired electrons from the 5p orbitals to the 6s and 4f electrons where they can then form covalent bonds with other atoms (i.e. fluorine, where $Xe{F}_6$ is a very easily created compound).
Something interesting though, Xe atoms can form ions containing many bond pairs. Xe atoms can gain 2 unpaired electrons and with the 6 valence unpaired electrons it originally has, it can form bonds. One such molecule formed is $[Xe{F}_8{]}^{2-}$.