What is the VSPER for XeF4 ? (Electron pairs and molecule shape)?

The central atom of XeF₄ is Xe.

Xe has eight valence electrons. Four of them are used to form four Xe-F bonds, and the other four form two lone pairs. Hence, the central atom has 6 valence electron pairs, including four bond pairs and two lone pairs, and thus the 6 valence electron pairs are arranged octahedrally.

A lone pairs exert a stronger repulsion on the other electron pairs than a bond pair. Hence, the two lone pairs are in opposite positions. Hence, the shape of XeF₄ molecule is square planar.

The molecule of XeF₄ is shown below :
https://c2.staticflickr.com/8/7452/27391660204_4265e3dd11_o.png

...
AB4E2
the shape is 'square planar'
http://byjus.com/chemistry/wp-content/uploads/2016/01/VSEPR-Theory.jpg

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