Based on VSEPR theory,the number of 90^{circ} | vedclass

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(B) The molecule $BrF_5$ has a square pyramidal geometry with one lone pair on the central $Br$ atom.Due to the strong repulsion between the lone pair

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(B) The molecule $BrF_5$ has a square pyramidal geometry with one lone pair on the central $Br$ atom.Due to the strong repulsion between the lone pair and the bonding pairs,the equatorial $F$ atoms are pushed downwards.Consequently,the bond angles between the axial $F$ and the equatorial $F$ atoms are reduced from $90^{\circ}$ to approximately $84.8^{\circ}$.Therefore,there are no $90^{\circ}$ $F-Br-F$ bond angles in the molecule.

List-$I$ (Shape)	List-$II$ (Example)
$(1)$ Linear $(B-A-B)$	$(A)$ $PCl_5$
$(2)$ Trigonal planar	$(B)$ $CH_4$
$(3)$ Tetrahedral	$(C)$ $HgCl_2$
$(4)$ Trigonal bipyramidal	$(D)$ $BF_3$

Based on $VSEPR$ theory,the number of $90^{\circ}$ $F-Br-F$ bond angles in $BrF_5$ is

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