Which one of the following molecules will hav | vedclass

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(A) To determine which molecule has all equal $X-F$ bond lengths,we analyze the geometry and hybridization of each:$1$. $SOCl_2F_2$: The molecule has

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$SOCl_2F_2$

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(A) To determine which molecule has all equal $X-F$ bond lengths,we analyze the geometry and hybridization of each:$1$. $SOCl_2F_2$: The molecule has a trigonal bipyramidal geometry with $S$ as the central atom. The $S=O$ bond and the two $Cl$ atoms occupy positions such that the two $S-F$ bonds are equivalent due to symmetry,resulting in equal $S-F$ bond lengths.$2$. $SeF_4$: This molecule has a see-saw geometry ($sp^3d$ hybridization) with one lone pair. The axial and equatorial $Se-F$ bonds have different lengths.$3$. $PBr_2F_3$: This molecule has a trigonal bipyramidal geometry. The $F$ atoms occupy both axial and equatorial positions,which have different bond lengths.$4$. $IF_7$: This molecule has a pentagonal bipyramidal geometry. The axial and equatorial $I-F$ bonds have different lengths.Therefore,$SOCl_2F_2$ is the correct molecule where all $S-F$ bonds are equal.

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	$(D)$ $BF_3$

Which one of the following molecules will have all equal $X-F$ bond lengths? (where $X =$ central atom)

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Match the molecular shapes given in List-$I$ with the examples given in List-$II$.
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 $(D)$ $BF_3$

What is the shape of bromine pentafluoride?

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