
Draw the structure of xenon hexafluoride ($Xe{{F}_{6}}$) molecule and state the hybridization of the central atom.
Answer
486.9k+ views
Hint: The oxidation state of Xe in $Xe{{F}_{6}}$ is +6. $Xe{{F}_{6}}$ is a crystalline solid with melting point 49.5$^{o}C$. It is the most volatile of all the fluorides of Xe and its vapour has a greenish yellow colour.
Complete answer:
Electronic combination of Xe (54) in the ground state is given as:
Xe (in ground state) : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}$
Outer shell configuration of Xe is $5{{s}^{2}}5{{p}^{6}}$.
$Xe{{F}_{6}}$ has six fluoride atoms. So, the total number of electrons on Xe is (8+6) =14.
We know that one pair of electrons contains 2 electrons. Therefore, the total number of electrons pairs available is 7.
There are six bonds present in$Xe{{F}_{6}}$. So, 3 electrons from the 5p – orbitals move into 5d orbitals so that six electrons can be made available for bonding. Seven orbitals (one 5s+three 5p+three 5d) hybridize to give $s{{p}^{3}}{{d}^{3}}$ hybridization.
Now, outer shell electronic combination of Xe in +6 oxidation state has become $5{{s}^{2}}5{{p}^{3}}5{{d}^{3}}$.
Six out of the seven orbitals of Xe are used in bonding with six fluorine atoms. One orbital accommodates a pair of electrons.
Valence shell electrons pair theory (VESPER) suggests that the shape of the molecule depends on the total number of electrons including bonding and lone pairs in the valence shell of the Xe atom.
The hybridization of Xe in $Xe{{F}_{6}}$ is $s{{p}^{3}}{{d}^{3}}$. Due to $s{{p}^{3}}{{d}^{3}}$ hybridization, $Xe{{F}_{6}}$ should have pentagonal bipyramidal geometry as shown below. However, no evidence has been provided to prove the existence of this structure.
$Xe{{F}_{6}}$ has been suggested to have a distorted octahedral structure in which the six positions are occupied by fluorine atoms forming an octahedral and one lone pair is present at the center of one of the triangular faces. Distorted octahedral structure of \[Xe{{F}_{6}}\] is given below.
Note: Structure of $Xe{{F}_{6}}$ is controversial. Remember the structure of $Xe{{F}_{6}}$ as an exception. In $Xe{{F}_{6}}$, number of bond pairs (Xe-F) = 6 and number of lone pairs =1.
Complete answer:
Electronic combination of Xe (54) in the ground state is given as:
Xe (in ground state) : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}4{{d}^{10}}5{{s}^{2}}5{{p}^{6}}$
Outer shell configuration of Xe is $5{{s}^{2}}5{{p}^{6}}$.
$Xe{{F}_{6}}$ has six fluoride atoms. So, the total number of electrons on Xe is (8+6) =14.
We know that one pair of electrons contains 2 electrons. Therefore, the total number of electrons pairs available is 7.
There are six bonds present in$Xe{{F}_{6}}$. So, 3 electrons from the 5p – orbitals move into 5d orbitals so that six electrons can be made available for bonding. Seven orbitals (one 5s+three 5p+three 5d) hybridize to give $s{{p}^{3}}{{d}^{3}}$ hybridization.
Now, outer shell electronic combination of Xe in +6 oxidation state has become $5{{s}^{2}}5{{p}^{3}}5{{d}^{3}}$.
Six out of the seven orbitals of Xe are used in bonding with six fluorine atoms. One orbital accommodates a pair of electrons.
Valence shell electrons pair theory (VESPER) suggests that the shape of the molecule depends on the total number of electrons including bonding and lone pairs in the valence shell of the Xe atom.
The hybridization of Xe in $Xe{{F}_{6}}$ is $s{{p}^{3}}{{d}^{3}}$. Due to $s{{p}^{3}}{{d}^{3}}$ hybridization, $Xe{{F}_{6}}$ should have pentagonal bipyramidal geometry as shown below. However, no evidence has been provided to prove the existence of this structure.

$Xe{{F}_{6}}$ has been suggested to have a distorted octahedral structure in which the six positions are occupied by fluorine atoms forming an octahedral and one lone pair is present at the center of one of the triangular faces. Distorted octahedral structure of \[Xe{{F}_{6}}\] is given below.

Note: Structure of $Xe{{F}_{6}}$ is controversial. Remember the structure of $Xe{{F}_{6}}$ as an exception. In $Xe{{F}_{6}}$, number of bond pairs (Xe-F) = 6 and number of lone pairs =1.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
What are the elders in Goa nostalgic about class 11 social science CBSE

Formaldehyde at room temperature is ALiquid BGas CSolid class 11 chemistry CBSE

Define least count of vernier callipers How do you class 11 physics CBSE

Distinguish between Mitosis and Meiosis class 11 biology CBSE

Why are forests affected by wars class 11 social science CBSE

Explain zero factorial class 11 maths CBSE
