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(B) Let us analyze the hybridization and geometry of each species: $NH_3$: $sp^3$ (Pyramidal) $PCl_4^{\oplus}$: $sp^3$ (Tetrahedral) $NH_2^{-}$: $sp^3

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All species are planar

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(B) Let us analyze the hybridization and geometry of each species: $NH_3$: $sp^3$ (Pyramidal) $PCl_4^{\oplus}$: $sp^3$ (Tetrahedral) $NH_2^{-}$: $sp^3$ (Bent) $PCl_3$: $sp^3$ (Pyramidal) $SO_2$: $sp^2$ (Bent) $SO_3$: $sp^2$ (Planar) $BF_3$: $sp^2$ (Planar) $IF_7$: $sp^3d^3$ (Pentagonal bipyramidal) $1$. Hybridization: $NH_3, PCl_4^{\oplus}, NH_2^{-}, PCl_3$ are all $sp^3$ hybridized. So,option $A$ is correct. $2$. Geometry: $NH_3, PCl_4^{\oplus}, NH_2^{-}, PCl_3, IF_7$ are non-planar. Thus,the statement that all species are planar is incorrect. So,option $B$ is incorrect. $3$. $p \pi-p \pi$ and $p \pi-d \pi$ bonds: $SO_2$ and $SO_3$ contain both types of back bonding. So,option $C$ is correct. $4$. Angles: $IF_7$ has $72^{\circ}$ angles in its pentagonal plane. So,option $D$ is correct.

Consider the following species and select the incorrect statement $:-$ $NH_3, PCl_4^{\oplus}, NH_2^{-}, SO_2, SO_3, BF_3, PCl_3, IF_7$

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