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(C) To determine the shapes,we calculate the hybridization and number of lone pairs for each species using the $VSEPR$ theory: $1$. $NO^{+}_2$: $N$ is

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$XeF_4$ and $ICl^{-}_4$

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(C) To determine the shapes,we calculate the hybridization and number of lone pairs for each species using the $VSEPR$ theory: $1$. $NO^{+}_2$: $N$ is central,$sp$ hybridized,linear shape. $NO^{-}_2$: $N$ is central,$sp^2$ hybridized,bent shape. $2$. $PCl_5$: $sp^3d$ hybridized,trigonal bipyramidal. $BrF_5$: $sp^3d^2$ hybridized,square pyramidal. $3$. $XeF_4$: $Xe$ has $8$ valence electrons,$4$ bond pairs,$2$ lone pairs,$sp^3d^2$ hybridization,square planar. $ICl^{-}_4$: $I$ has $7+1=8$ valence electrons,$4$ bond pairs,$2$ lone pairs,$sp^3d^2$ hybridization,square planar. $4$. $TeCl_4$: $sp^3d$ hybridized,see-saw shape. $XeO_4$: $sp^3$ hybridized,tetrahedral shape. Thus,$XeF_4$ and $ICl^{-}_4$ both have a square planar geometry.

Which of the following pairs of species have identical shapes?

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