Question

xef4 ka akar hota hai​

Answer 1

Answer:

The VSEPR structure of XeF4 is square planar. It is an octahedral but because of the lone pairs, it dictate there 6 Domains around the central atom and the VSEPR theory states any AX4E2 specie with 2 lone pairs is square planar.

Answer 2

$XeF_4$ has square planar geometry

Explanation:

Formula used for calculating molecular weight:

$\text{Number of electrons}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom  = 8

N = number of monovalent atoms bonded to central atom = 4

C = charge of cation  = 0

A = charge of anion  = 0

Now we have to determine the hybridization of the $XeF_4$  molecules.

$\text{Number of electrons}=\frac{1}{2}\times [8+4]=6$

Bond pair electrons = 4

Lone pair electrons = 2

The number of electrons are 6 that means the hybridization will be and the electronic geometry of the molecule will be octahedral.

The molecular geometry will be square planar as two positions will be occupied by lone pair of electrons.

Learn more about hybridization:

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