The molecule having sp^3d^2 hybridisation is | vedclass

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(B) is the correct answer.To determine the hybridization,we use the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the num

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$XeOF_4$

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(B) is the correct answer.To determine the hybridization,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.$1. SF_4$: $	ext{Steric Number} = \frac{1}{2} [6 + 4] = 5$ ($sp^3d$ hybridization).$2. XeOF_4$: $	ext{Steric Number} = \frac{1}{2} [8 + 4] = 6$ ($sp^3d^2$ hybridization).$3. PF_5$: $	ext{Steric Number} = \frac{1}{2} [5 + 5] = 5$ ($sp^3d$ hybridization).$4. BrF_3$: $	ext{Steric Number} = \frac{1}{2} [7 + 3] = 5$ ($sp^3d$ hybridization).

The molecule having $sp^3d^2$ hybridisation is

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The overlapping of orbitals in benzene is of the type

Which among the following molecules have $sp^3d$ hybridisation with one lone pair of electrons on the central atom?
$(i)\ SF_4$
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$(iv)\ ClOF_3$

Differentiate between atomic orbitals and hybrid orbitals.

The hybridization of atomic orbitals of nitrogen in $NO_2^+$,$NO_3^-$,and $NH_4^+$ are:

Among the compounds,$BF_3$,$NCl_3$,$H_2S$,$SF_4$ and $BeCl_2$,identify the ones in which the central atom has the same type of hybridization.

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