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(A) To determine the geometry,we calculate the hybridization and number of lone pairs using the $VSEPR$ theory:$1$. $BrF_5$: The central atom $Br$ has

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$BrF_5$ & $XeOF_4$

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(A) To determine the geometry,we calculate the hybridization and number of lone pairs using the $VSEPR$ theory:$1$. $BrF_5$: The central atom $Br$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. Total electron pairs = $6$ ($sp^3d^2$ hybridization). Due to $1$ lone pair,the geometry is square pyramidal.$2$. $XeOF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms and $1$ double bond with $O$. It has $1$ lone pair. Total electron pairs = $6$ ($sp^3d^2$ hybridization). Due to $1$ lone pair,the geometry is square pyramidal.$3$. $SbF_5$: The central atom $Sb$ has $5$ valence electrons. It forms $5$ bonds with $F$ atoms and has $0$ lone pairs. Total electron pairs = $5$ ($sp^3d$ hybridization). The geometry is trigonal bipyramidal.$4$. $PCl_5$: The central atom $P$ has $5$ valence electrons. It forms $5$ bonds with $Cl$ atoms and has $0$ lone pairs. Total electron pairs = $5$ ($sp^3d$ hybridization). The geometry is trigonal bipyramidal.Thus,$BrF_5$ and $XeOF_4$ have square pyramidal geometry.

The molecules having square pyramidal geometry are

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