The geometry of $X e O F_{4}$ according to VSEPR theory is:

octahedral

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pentagonal planar

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square pyramidal

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trigonal bipyramidal

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Solution

The correct option is D square pyramidal
The geometry of $X e O F_{4}$ by VSEPR theory is square pyramidal.

The central $X e$ atom in $X e O F_{4}$ has one lone pair of electron and 5 bonding domains. Hence, it undergoes $s p^{3} d^{2}$ hybridization.

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