The shape of \(XeF_6\) is :

Distorted octahedral

 The shape of \(XeF_6\) is (3) distorted octahedral.

Xenon is a noble gas element which belongs to the fifth period.

Let’s write the electronic configuration of Xenon.

\(Xe = [Kr]4d^{10}5s^25p^6\)

There are 7 valence shell electrons in each fluorine and eight valence shell electrons in the xenon atom. So, the number of electrons present is \(8 + (6 × 7) = 50\) and the total number of electron pairs present is \(25\).

The Lewis structure of Xenon hexafluoride is

In xenon, there are eight electrons in its valence shell. It also has an empty \(5d\) orbital. There are six fluorine atoms in xenon hexafluoride, six covalent bonds are formed. Xenon undergoes \(sp^3d^3\) hybridization to form seven \(sp^3d^3\) hybrid orbitals. Six \(sp^3d^3\) orbitals have one electron each and one \(sp^3d^3\) contain a lone pair of electrons. The six unpaired electrons get paired with six electrons from six fluorine atoms to form xenon hexafluoride, \(XeF_6\).

Due to the presence of one lone pair of electrons, the will be lone-pair –bond-pair repulsion in the molecule and the geometry will be distorted octahedral.