Xenon tetrafluoride (XeF4) Lewis dot structure, molecular geometry or shape, electron geometry, bond angle, formal charge, hybridization
XeF4 is the chemical formula for xenon tetrafluoride, the first discovered binary compound of a noble gas. It is produced by the chemical reaction of xenon (Xe) with fluorine (F2) and exists as a colorless crystalline solid under normal temperature and pressure conditions.
Let’s find out some interesting facts about XeF4 such as how to draw its Lewis structure, what is its molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, and much more, all through this article.
|Name of Molecule||Xenon tetrafluoride|
|Molecular geometry of XeF4||Square planar|
|Electron geometry of XeF4||Octahedral|
|Bond angle (F-Xe-F)||90° and 180°|
|Total Valence electron in XeF4||36|
|Overall Formal charge in XeF4||Zero|
How to draw lewis structure of XeF4?
The Lewis structure of XeF4 consists of a xenon (Xe) atom present at the center of the molecule. It is bonded to four atoms of fluorine (F) at the sides. There are a total of 6 electron pairs around the central Xe atom in the XeF4 lewis structure. Out of these 6 electron pairs, 4 are bond pairs while there are 2 lone pairs of electrons.
Drawing the Lewis dot structure of XeF4 is quite easy if you follow all the steps given below one by one.
Steps for drawing the Lewis dot structure of XeF4
1. Count the total valence electrons in XeF4
The Lewis dot structure of a molecule is a simplified representation of all the valence electrons present in the molecule. Therefore, the first step whenever we want to draw the Lewis structure of a molecule is to calculate the total valence electrons present in it.
The valence electrons present in different elemental atoms of a molecule can be readily determined from the Periodic Table. As Xe and F are the two different elements present in XeF4, so we will identify them in the Periodic Table.
As Xenon (Xe) is a Noble gas element so it is present in Group VIII A of the Periodic Table which means it has a total of 8 valence electrons. On the other hand, Fluorine (F) is a halogen, located in Group VII A of the Periodic Table thus it has a total of 7 valence electrons.
- Total number of valence electrons in xenon = 8
- Total number of valence electrons in fluorine = 7
∴ The XeF4 molecule is made up of 1 Xe atom and 4 atoms of F. Hence, the total valence electrons available for drawing the Lewis structure of XeF4 = 8 + 4(7) = 36 valence electrons.
2. Find the least electronegative atom and place it at the center
The second step is to identify a central atom that is most likely to share its electrons with the atoms in its surroundings. As electronegativity is defined as the ability of an atom to attract a shared pair of electrons towards itself.
Therefore, the least electronegative or most electropositive atom is the one that shares its electrons.
Fluorine is the most electronegative element in the Periodic Table. Hence the less electronegative Xe is placed at the center of the Lewis structure. All the F atoms are placed in its surroundings.
3. Connect outer atoms with the central atom
Now all the outer atoms are joined to the central atom using straight lines.
As we already identified, the Xe atom is the central atom in XeF4 while F atoms are the outer atoms. So, each F atom is joined to the central Xe using a straight line.
Each straight line represents a single covalent bond i.e., a bond pair containing 2 electrons. There are a total of 4 single bonds in the above structure, so the total valence electrons used so far are 4(2) = 8 valence electrons.
- Total valence electrons available – electrons used till step 3 = 36-8 = 28 valence electrons.
- This means we still have 28 valence electrons available to be accommodated in the Lewis structure of XeF4.
4. Complete the octet of outer atoms
In this step, we complete the octet of outer atoms surrounding the central Xe atom. The F atoms are the outer atoms in the XeF4 molecule. Each F atom requires a total of 8 valence electrons to achieve a stable octet electronic configuration.
All the four F atoms are bonded to the central Xe atom using single bonds (Xe-F) thus each F already has 2 electrons, and it is short of 6 electrons to complete its octet.
Therefore, 6 electrons are placed as 3 lone pairs around each F atom in the Lewis structure of XeF4, as shown in the figure below.
5. Complete the octet of the central atom
- Total valence electrons used till step 4 = 4 single bonds + 4 ( electrons placed around each F atom, shown as dots) = 4(2) + 4 (6) = 32 valence electrons.
- Total valence electrons – electrons used till step 4 = 36-32 = 4 valence electrons.
So, these 4 electrons are placed as lone pairs on the central Xe atom in the XeF4 Lewis structure as shown below.
Xenon (Xe) does not follow the octet rule. It has a total of 12 valence electrons in the Lewis structure of XeF4. During chemical bonding, the incoming valence electrons have access to the 4d subshell of Xe, facilitating the accommodation of more than 8 valence electrons.
In the final step, we need to check the stability of the XeF4 Lewis structure. We can do so by verifying whether there is some formal charge present on XeF4 atoms or not.
6. Check the stability of Lewis’s structure with the help of the formal charge concept
The less the formal charge on the atoms of a molecule, the better the stability of its Lewis structure.
The formal charge can be calculated using the formula given below.
- Formal charge = [ valence electrons – nonbonding electrons- ½ (bonding electrons)]
Now let us count the formal charge present on XeF4 using this formula and the Lewis structure obtained in step 5.
For fluorine atom
- Valence electrons of fluorine = 7
- Bonding electrons = 2
- Non-bonding electrons = 3 lone pairs = 6 electrons
- Formal charge = 7-6-2/2 = 7-6-1 = 7-7 = 0
For xenon atom
- Valence electrons of xenon = 8
- Bonding electrons = 8
- Non-bonding electrons = 2 lone pairs = 4 electrons
- Formal charge = 8-4-8/2 = 8-4-4 = 8-8 = 0
Thus, there are zero formal charges present on all the bonded atoms in the XeF4 molecule which confirms that its Lewis structure is stable and that we have drawn it correctly.
Also check –
What are the electron and molecular geometry of XeF4?
The ideal electron geometry of the xenon tetrafluoride (XeF4) molecule is octahedral. But it is due to the two lone pairs present on the central Xe atom that the XeF4 molecule adopts a different shape or molecular geometry i.e., square planar.
All the four F atoms lie at the four corners of the square arrangement while Xe is present at the center. The two lone pairs present on the central Xe atom are situated at an equal distance above and below the molecule thus giving the molecule a symmetrical shape.
Molecular geometry of XeF4
The XeF4 molecule has a symmetrical molecular geometry or shape i.e., square planar. There are two lone pairs present on the central Xe atom therefore lone pair-bond pair and bond pair-bond pair repulsions do exist in addition to bond pair-bond pair repulsions between the four Xe-F bonds.
The two lone pairs occupy positions farthest away from each other i.e., on opposite sides of the square planar shape while the four F atoms occupy the four corners of the square.
An important point to remember is that the molecular geometry or shape of a molecule depends on the different number of bond pairs and lone pairs present in a molecule.
But the ideal electron geometry only depends on the total number of electron pairs (bond pairs + lone pairs) present in the molecule. Each electron pair is considered a separate region of electron density in the molecule.
Let’s see how this concept applies to the XeF4 molecule.
Electron geometry of XeF4
The valence shell electron pair repulsion (VSEPR) theory of chemical bonding states that the ideal electronic geometry of a molecule containing 6 electron pairs around the central atom is octahedral. As there are 6 electron pairs in total around the central Xe atom in XeF4 (i.e., 4 bond pairs + 2 lone pairs) thus the electron geometry of XeF4 is octahedral.
A more straightforward way of finding the electron and molecular geometry of XeF4 is using the AXN method.
AXN is a simple formula to represent the number of atoms bonded to the central atom in a molecule and the lone pairs present on this central atom.
It is used to predict the geometry and shape of a molecule using the VSEPR concept.
AXN notation for XeF4 molecule
- A in the AXN formula represents the central atom present in a molecule. In the XeF4 molecule, xenon (Xe) acts as the central atom so A=Xe.
- X stands for the number of atoms bonded to the central atom. As 4 F atoms are bonded to the central Xe atom in XeF4 so X=4.
- N denotes the number of lone pairs present on the central atom. As per the Lewis structure of XeF4, there are 2 lone pairs on Xe thus N=2 in the XeF4
This shows that the AXN generic formula for the XeF4 molecule is AX4N2.
Now have a quick look at the VSEPR chart given below and identify the electron and molecular geometry against AX4N2.
The VSEPR chart shown above indicates that an AX4N2 generic formula represents molecules that have an octahedral electron geometry while their molecular geometry or shape is square planar, as we noted down for the XeF4 molecule.
Hybridization of XeF4
The central Xe atom in the XeF4 molecule is sp3d2 hybridized.
The electronic configuration of Xenon (Xe) is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6.
During chemical bond formation, two of the 5p electron pairs of Xe get unpaired and shift to two separate 5d orbitals. Consequently, one 5s orbital, three 5p orbitals, and two 5d orbitals mix to produce six sp3d2 hybrid orbitals. Four of these sp3d2 hybrid orbitals contain a single unpaired electron.
These sp3d2 hybrid orbitals lie in one plane, inclined to one another at 90°, and form a Xe-F sigma bond with the p-orbital of fluorine (F) atom on each side of the square arrangement.
The other two sp3d2 hybrid orbitals contain paired electrons and each electron pair is situated as a lone pair on Xe, directed above and below the XeF4 square planar shape.
A short trick for finding the hybridization present in a molecule is memorizing the table given below. The steric number of the molecule can be used against this table to determine its hybridization.
There are 6 regions of electron density around the central Xe atom in the XeF4 molecule, so its steric number is 6. Thus, it has sp3d2 hybridization.
The XeF4 bond angles
There are two main F-Xe-F bond angles present in the XeF4 molecule. The terminal F atoms lie at a right angle to the central Xe atom while the F-Xe-F atoms that lie on a straight line at the center of the molecule form a 180° bond angle. All the Xe-F bond lengths in the XeF4 molecule are equivalent i.e., 197 pm.
Also check:- How to find bond angle?
Is XeF4 polar or nonpolar?
Fluorine (E.N =3.98) is a highly electronegative atom while Xenon (E.N =2.60) is less electronegative. Thus, each Xe-F bond in the XeF4 molecule is polar due to an electronegativity difference of 3.98 – 2.60 = 1.38 units between the bonded Xe and F atoms.
Each Xe-F bond possesses a specific dipole moment value (symbol μ).
But it is due to the symmetrical shape of the molecule that the Xe-F dipole moments get canceled in opposite directions in the molecule thus XeF4 is a non-polar molecule overall with net μ=0. It has a uniformly distributed electron cloud in its square planar shape.
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How can the Lewis structure of XeF4 be determined?
The Lewis structure of XeF4 can be determined using the following six simple steps:
How many pairs of electrons are there in the Lewis structure of XeF4?
The Lewis structure of XeF4 has a total of 36 valence electrons which means there are 36/2 = 18 electron pairs.
Out of these 18 electron pairs, 4 are bond pairs while the remaining 14 are lone pairs.
Out of the 14 lone pairs, 2 lone pairs are present on the central Xe atom while the remaining 12 electron pairs are equally distributed amongst 4 F atoms (i.e., 3 lone pairs on each F atom).
What is the shape or geometry of H2O and XeF4?
The xenon tetrafluoride (XeF4) molecule has an octahedral electron geometry while its molecular geometry or shape is square planar. There are 6 regions of electron density (4 bond pairs + 2 lone pairs) around the central Xe atom in the molecule.
On the other hand, the water (H2O) molecule has a tetrahedral electron geometry while its molecular geometry or shape is bent. There are a total of 4 electron density regions (2 bond pairs + 2 lone pairs) around the central O atom in the molecule.
Why is the shape of XeF4 not tetrahedral?
The distinction of bond pairs from lone pairs holds primary importance while determining the shape of a molecule.
For a tetrahedral shape, the molecule must have a total of 4 electron pairs around the central atom and each electron pair must be a bond pair.
In XeF4 there are a total of 6 electron pairs around the central Xe atom. Out of which there are 4 bond pairs, but 2 lone pairs are also present. Bond pair-lone pair and lone pair-lone pair repulsions make the molecule attain a square planar shape.
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- The total valence electrons available for drawing the XeF4 Lewis structure are 36.
- The molecular geometry or shape of XeF4 is square planar.
- The ideal electronic geometry of XeF4 is octahedral.
- It is due to the presence of lone pairs on the central Xe atom in the XeF4 molecule that it adopts a different molecular geometry from its electron geometry.
- The central Xe atom in the XeF4 molecule is sp3d2
- The F-Xe-F bonds possess 90° and 180° bond angles in the XeF4.
- XeF4 is a symmetrical molecule overall in which all Xe-F bond lengths are equivalent i.e., 197 pm.
- Each Xe-F bond is polar, but it is due to the symmetrical shape of the molecule that the dipole moments of individually polar Xe-F bonds get canceled in opposite directions thus the molecule overall is non-polar (net μ=0).
- The absence of a formal charge on all the atoms in the XeF4 molecule ensures the stability of its Lewis structure.