Which set of hybridisation is correct for the | vedclass

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(B) The hybridisation can be calculated using the formula: $H = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the n

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$sp$,$sp^3d$,$sp^3d^2$

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(B) The hybridisation can be calculated using the formula: $H = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $NO_2$: $H = \frac{1}{2} [5 + 0 - 0 + 0] = 2.5$. Since $NO_2$ has an odd electron,it is considered $sp$ hybridised.$2$. For $SF_4$: $H = \frac{1}{2} [6 + 4 - 0 + 0] = 5$,which corresponds to $sp^3d$ hybridisation.$3$. For $PF_6^-$: $H = \frac{1}{2} [5 + 6 - 0 + 1] = 6$,which corresponds to $sp^3d^2$ hybridisation.Thus,the correct set is $sp$,$sp^3d$,$sp^3d^2$.

Which set of hybridisation is correct for the following compounds:
$NO_2$,$SF_4$,$PF_6^-$

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Which set of hybridisation is correct for the following compounds:$NO_2$,$SF_4$,$PF_6^-$