X and Y are two covalent molecules in which t | vedclass

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(B) The hybridization of the central atom in $XeF_2$ is $sp^3d$ (steric number $5$: $2$ bond pairs + $3$ lone pairs),resulting in a linear shape due t

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$XeF_2, PF_5$

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(B) The hybridization of the central atom in $XeF_2$ is $sp^3d$ (steric number $5$: $2$ bond pairs + $3$ lone pairs),resulting in a linear shape due to the lone pairs occupying equatorial positions. The hybridization of the central atom in $PF_5$ is also $sp^3d$ (steric number $5$: $5$ bond pairs + $0$ lone pairs),resulting in a trigonal bipyramidal shape. Since both have the same hybridization $(sp^3d)$ but different shapes (linear vs. trigonal bipyramidal),$XeF_2$ and $PF_5$ satisfy the given conditions.

$X$ and $Y$ are two covalent molecules in which the hybridization of the central atoms is the same,but their shapes are different. $X$ and $Y$ are:

Similar Questions

What is the hybridisation of each carbon in $H_2C = C = CH_2$?

Which is the correct statement?
As the $s$-character of a hybrid orbital decreases:
$(I)$ The bond angle decreases
$(II)$ The bond strength increases
$(III)$ The bond length increases
$(IV)$ Size of orbitals increases

The hybridisations of $N$,$C$ and $O$ shown in the following compound are:

The correct order regarding the electronegativity of hybrid orbitals of carbon is

In which of the following species is the hybrid state of the central atom the same?

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