Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. $XeF_4$ is square planar where as $CCl_4$ is tetrahedral because

AP EAMCETAP EAMCET 2018

Solution:

image

$XeF _{4}$ is $s p^{3} d^{2}$ hybridised due to $4 \sigma+2 l p$ and shape is square planar.

$CCl _{4}$ is $s p^{3}$ hybridised due to $4 \sigma$ -bond.

image