The geometry of XeOF4 by VSEPR theory is:

A
trigonal bipyramidal
B
square pyramidal
C
octahedral
D
octahedral

Solution:

The valence electrons in ${XeOF}_{4}$ are $42 (= 8 + 6 + 4 x 7)$ . These are distributed as follows.

There are 6 pairs of electrons around $Xe$ . Their orientations will be square bipyramidal. The geometry of ${XeOF}_{4}$ is square pyramid as the lone pair of electrons occupy apical position

The geometry of ${XeOF}_{4}$ by $VSEPR$ theory is:

Solution:

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