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(C) The hybridization of the central atom can be determined using the formula $H = \frac{1}{2} [V + X - C + A]$,where $V$ is the number of valence ele

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$SbCl_5^{2-}$

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(C) The hybridization of the central atom can be determined using the formula $H = \frac{1}{2} [V + X - C + A]$,where $V$ is the number of valence electrons,$X$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1. SF_4: H = \frac{1}{2} [6 + 4] = 5 \rightarrow sp^3d$$2. I_3^-: H = \frac{1}{2} [7 + 2 + 1] = 5 \rightarrow sp^3d$$3. PCl_5: H = \frac{1}{2} [5 + 5] = 5 \rightarrow sp^3d$$4. SbCl_5^{2-}: H = \frac{1}{2} [5 + 5 + 2] = 6 \rightarrow sp^3d^2$Thus,$SbCl_5^{2-}$ has a different hybridization $(sp^3d^2)$ compared to the others $(sp^3d)$.

$A. NO_3^-$	$1. sp^3$
$B. NH_2^-$	$2. sp$
$C. SCN^-$	$3. sp^3d$
$D. ICl_2^-$	$4. sp^2$

In which one of the following species does the central atom have a type of hybridization that is not the same as that present in the other three?

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$A. NO_3^-$ $1. sp^3$
$B. NH_2^-$ $2. sp$
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Order of the bond strength of $C-H$ bonds involving $sp, sp^2$ and $sp^3$ hybridized carbon atoms is

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