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The correct geometry and hybridization of \[{\text{Xe}}{{\text{F}}_{\text{4}}}{\text{\;}}\] are:
A.Octahedral, \[s{p^3}{d^2}\]
B.Trigonal bipyramidal, \[s{p^3}d\]
C.Planar triangle, \[s{p^3}{d^3}\]
D.Square planar, \[s{p^3}{d^2}\]

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Hint: To answer this question, you should recall that xenon is a noble gas and when bonding it will have the tendency to use all the eight electrons in the outermost shell to form bond pairs. We know fluorine is a halogen and can form one sigma bond with the central atom. Now, use this information to answer the question.

Complete step by step answer:
It is evident from the formula \[{\text{Xe}}{{\text{F}}_{\text{4}}}\] that it is a \[{\text{A}}{{\text{B}}_{\text{4}}}{{\text{L}}_{\text{2}}}\] type molecule where ${\text{A}}$ is the central atom and ${\text{L}}$ is the peripheral atom. Here, we can see that xenon is bonded with four fluorine atoms.
As we know that fluorine can form a single sigma bond, it will be bonded to ${\text{Xe}}$ with 4 sigma bonds. This leaves xenon with two lone pairs. To avoid repulsion and provide maximum stability, these lone pairs will be attached at an angle of \[180^\circ \].
So, the hybridization of this molecule will be \[s{p^3}{d^2}\]. This means that geometry is octahedral and arrangement of electrons around the central atom is square planar due to the presence of two lone pair electrons.
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Therefore, we can conclude that the correct answer to this question is option D.

Note:
Even if you are not able to calculate the hybridisation using the above-mentioned, you can find the hybridization \[X\] using the formula: \[\dfrac{1}{2}\left( {V + H - C + A} \right)\] where,
$V$= Number of valence electrons in the central atom
$H$= Number of surrounding monovalent atoms
$C$= Cationic charge
$A$= Anionic charge
 The value of \[X\] will determine the hybridisation of the molecule. If $X = 2$, the hybridisation is $sp$. If $X = 3$, the hybridisation is $s{p^2}$. If $X = 4$, the hybridisation is $s{p^3}$; If $X = 5$, the hybridisation is $s{p^3}d$. If $X = 6$, the hybridisation is $s{p^3}{d^2}$. If $X = 7$, the hybridization is $s{p^3}{d^3}$.
Now, to determine the hybridization of \[{\text{Xe}}{{\text{F}}_{\text{4}}}{\text{\;}}\].
Here, the total valence electrons, $V$ is 8 and the number of surrounding monovalent atoms, $H$ is 4. Cationic charge $C$ and anionic charge $A$ are 0.
Substituting in the formula we get,
 $
  X = \dfrac{1}{2}\left( {8 + 4 - 0 + 0} \right) \\
   = \dfrac{1}{2}\left( {12} \right) \\
   = 6 \\
 $
Thus, hybridization of \[{\text{Xe}}{{\text{F}}_{\text{4}}}\] is \[s{p^3}{d^2}\].