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Chemistry
XeF4 is square planar where as CCl4 is tetrahedral because
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Q. $XeF_4$ is square planar where as $CCl_4$ is tetrahedral because
AP EAMCET
AP EAMCET 2018
A
In $XeF_4, ‘Xe‘$ is $sp^2$ hybridised and in $CCl_4 \, ‘C'$ is $sp^3$ hybridised
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B
In both $XeF_4$ and $CCl_4$ the central atom is $sp^3$ hybridised
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C
In $XeF_4, ‘Xe'$ is $sp^3d^2$ hybridised but due to the presence of 2 lone pairs of electrons shape is square planar whereas in $CCl_4 \, 'C'$ is $ sp^3$ hybridised
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D
Xe is a noble gas. whereas C is a non-metal
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Solution:
$XeF _{4}$ 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.