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Question

(C) $PCl_5$: Bond pairs $(bp)$ = $5$,Lone pairs $(lp)$ = $0$. Total = $5$. Hybridization = $sp^3d$. Shape = Trigonal bipyramidal.$BrF_5$: Bond pairs $

Vedclass · Accepted Answer

One of the following is square pyramidal

Vedclass · Answer

(C) $PCl_5$: Bond pairs $(bp)$ = $5$,Lone pairs $(lp)$ = $0$. Total = $5$. Hybridization = $sp^3d$. Shape = Trigonal bipyramidal.$BrF_5$: Bond pairs $(bp)$ = $5$,Lone pairs $(lp)$ = $1$. Total = $6$. Hybridization = $sp^3d^2$. Shape = Square pyramidal.$IF_7$: Bond pairs $(bp)$ = $7$,Lone pairs $(lp)$ = $0$. Total = $7$. Hybridization = $sp^3d^3$. Shape = Pentagonal bipyramidal.Thus,the correct statement is that one of the following $(BrF_5)$ is square pyramidal.

Column $I$	Column $II$
$(a) \ XeO_6^{4-}$	$(P)$ Tetrahedral
$(b) \ ClO_2$	$(Q) \ V\text{-shape}$
$(c) \ NH_4^+$	$(R)$ Trigonal Bipyramidal
$(d) \ XeO_3F_2$	$(S)$ Octahedral

Select the correct statement regarding the shapes of $PCl_5$,$BrF_5$,and $IF_7$.

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Match the column for molecular shape?

Column $I$ Column $II$

$(a) \ XeO_6^{4-}$ $(P)$ Tetrahedral

$(b) \ ClO_2$ $(Q) \ V\text{-shape}$

$(c) \ NH_4^+$ $(R)$ Trigonal Bipyramidal

$(d) \ XeO_3F_2$ $(S)$ Octahedral

The geometry of $I_{3}^{-}$ is

Which of the following is a non-linear molecule?

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