Which of the two ions from the list given bel | vedclass

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(A) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is the number

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$NO_2^-$ and $NO_3^-$

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(A) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1. NO_2^-: SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3 \rightarrow sp^2$ hybridization.$2. NO_3^-: SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3 \rightarrow sp^2$ hybridization.$3. NH_2^-: SN = \frac{1}{2} (5 + 2 - 0 + 1) = 4 \rightarrow sp^3$ hybridization.$4. NH_4^+: SN = \frac{1}{2} (5 + 4 - 1 + 0) = 4 \rightarrow sp^3$ hybridization.$5. SCN^-: SN = \frac{1}{2} (4 + 0 - 0 + 1) = 2 \rightarrow sp$ hybridization.Comparing the results,$NO_2^-$ and $NO_3^-$ both exhibit $sp^2$ hybridization.

Which of the two ions from the list given below have the geometry that is explained by the same hybridization of orbitals: $NO_2^-$,$NO_3^-$,$NH_2^-$,$NH_4^+$,$SCN^-$?

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