Which reaction involves a change in the elect | vedclass

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(A) To determine the change in electron pair geometry,we analyze the hybridization and geometry of the central atom in each reaction:$A$) $\underline{

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$\underline{B}F_{3} + F^{-} 	o \underline{B}F_{4}^{-}$

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(A) To determine the change in electron pair geometry,we analyze the hybridization and geometry of the central atom in each reaction:$A$) $\underline{B}F_{3} + F^{-} 	o \underline{B}F_{4}^{-}$: $B$ in $BF_{3}$ is $sp^{2}$ hybridized (trigonal planar geometry). $B$ in $BF_{4}^{-}$ is $sp^{3}$ hybridized (tetrahedral geometry). Thus,the geometry changes.$B$) $\underline{N}H_{3} + H^{+} 	o \underline{N}H_{4}^{+}$: $N$ in $NH_{3}$ is $sp^{3}$ hybridized (tetrahedral electron pair geometry). $N$ in $NH_{4}^{+}$ is $sp^{3}$ hybridized (tetrahedral electron pair geometry). No change.$C$) $2\underline{S}O_{2} + O_{2} 	o 2\underline{S}O_{3}$: $S$ in $SO_{2}$ is $sp^{2}$ hybridized (trigonal planar electron pair geometry). $S$ in $SO_{3}$ is $sp^{2}$ hybridized (trigonal planar electron pair geometry). No change.$D$) $H_{2}\underline{O} + H^{+} 	o H_{3}\underline{O}^{+}$: $O$ in $H_{2}O$ is $sp^{3}$ hybridized (tetrahedral electron pair geometry). $O$ in $H_{3}O^{+}$ is $sp^{3}$ hybridized (tetrahedral electron pair geometry). No change.

Which reaction involves a change in the electron pair geometry for the underlined atom?

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