Consider PF_{5},BrF_{5},PCl_{3},SF_{6},[ICl_{ | vedclass

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(B) To determine the hybridisation,we use the formula: $	ext{Steric Number} = \frac{1}{2} (V + M - C + A)$,where $V$ is the number of valence electro

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$4$

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(B) To determine the hybridisation,we use the formula: $	ext{Steric Number} = \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.$1$. $PF_{5}$: $	ext{Steric Number} = \frac{1}{2}(5 + 5) = 5$ $(sp^{3}d)$$2$. $BrF_{5}$: $	ext{Steric Number} = \frac{1}{2}(7 + 5) = 6$ $(sp^{3}d^{2})$$3$. $PCl_{3}$: $	ext{Steric Number} = \frac{1}{2}(5 + 3) = 4$ $(sp^{3})$$4$. $SF_{6}$: $	ext{Steric Number} = \frac{1}{2}(6 + 6) = 6$ $(sp^{3}d^{2})$$5$. $[ICl_{4}]^{-}$: $	ext{Steric Number} = \frac{1}{2}(7 + 4 + 1) = 6$ $(sp^{3}d^{2})$$6$. $ClF_{3}$: $	ext{Steric Number} = \frac{1}{2}(7 + 3) = 5$ $(sp^{3}d)$$7$. $IF_{5}$: $	ext{Steric Number} = \frac{1}{2}(7 + 5) = 6$ $(sp^{3}d^{2})$The molecules/ions with $sp^{3}d^{2}$ hybridisation are $BrF_{5}$,$SF_{6}$,$[ICl_{4}]^{-}$,and $IF_{5}$.Total count = $4$.

Consider $PF_{5}$,$BrF_{5}$,$PCl_{3}$,$SF_{6}$,$[ICl_{4}]^{-}$,$ClF_{3}$ and $IF_{5}$.
Amongst the above molecule$(s)$/ion$(s)$,the number of molecule$(s)$/ion$(s)$ having $sp^{3}d^{2}$ hybridisation is $........$

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Consider $PF_{5}$,$BrF_{5}$,$PCl_{3}$,$SF_{6}$,$[ICl_{4}]^{-}$,$ClF_{3}$ and $IF_{5}$.Amongst the above molecule$(s)$/ion$(s)$,the number of molecule$(s)$/ion$(s)$ having $sp^{3}d^{2}$ hybridisation is $........$