The number of molecules/ions from the followi | vedclass

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

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$2$

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(A) To determine the hybridization of the central atom,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.$1$. For $NO_3^{-}$: $N$ has $5$ valence electrons. $SN = \frac{1}{2} (5 + 0 - 0 + 1) = 3$. Hybridization is $sp^2$.$2$. For $BCl_3$: $B$ has $3$ valence electrons. $SN = \frac{1}{2} (3 + 3 - 0 + 0) = 3$. Hybridization is $sp^2$.$3$. For $ClO_2^{-}$: $Cl$ has $7$ valence electrons. $SN = \frac{1}{2} (7 + 0 - 0 + 1) = 4$. Hybridization is $sp^3$.$4$. For $ClO_3$: $Cl$ has $7$ valence electrons. $SN = \frac{1}{2} (7 + 0 - 0 + 0) = 3.5$ (radical species). However,considering the structure with $3$ double bonds and $1$ lone electron,the central $Cl$ atom is $sp^3$ hybridized.Thus,$ClO_2^{-}$ and $ClO_3$ have $sp^3$ hybridization.The total number of such species is $2$.

The number of molecules/ions from the following in which the central atom is involved in $sp^3$ hybridization is: $NO_3^{-}, BCl_3, ClO_2^{-}, ClO_3$

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