Xenon oxytetrafluoride (XeOF4) Lewis dot structure, molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, polar vs nonpolar
XeOF4 is the chemical formula for xenon oxytetrafluoride, a highly reactive chemical compound with powerful oxidative properties.
It exists as a colorless solid at room temperature and is used in the preparation of lubricants, solvents and heat transfer fluids.
In this article, we will teach you how to draw the Lewis structure of xenon oxytetrafluoride (XeOF4), what is its molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, polarity, etc.
So, for all this valuable information, continue reading the article till the end.
The Lewis structure of XeOF4 comprises a carbon (C) atom at the center. It is double-covalently bonded to an oxygen (O) atom on one side and single-covalently bonded to four fluorine (F) atoms on the other sides. There is a lone pair of electrons on the central Xe-atom, while the terminal oxygen and fluorine atoms carry 2 and 3 lone pairs, respectively.
You can easily draw the Lewis structure of XeOF4 by following the simple steps given below.
Steps for drawing the Lewis dot structure of XeOF4
1. Count the total valence electrons present in XeOF4
The three distinct elements present in XeOF4 are xenon, oxygen and fluorine.
Oxygen (O) is located in Group VI A (or 16) of the Periodic Table of Elements containing 6 valence electrons in each atom.
In contrast, Fluorine (F) is a halogen present in Group VII A (or 17), containing a total of 7 valence electrons in each atom.
Meanwhile, xenon (Xe) is a Noble gas element located in Group VIII A (or 18), thus containing 8 valence electrons.
Total number of valence electrons in oxygen = 6
Total number of valence electrons in fluorine = 7
Total number of valence electrons in xenon = 8
The XeOF4 molecule comprises 1 Xe-atom, 1 O-atom and 4 F-atoms.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of XeOF4 = 1(8) + 1(6) + 4(7) = 42 valence electrons.
2. Find the least electronegative atom and place it at the center
By convention, the least electronegative atom out of all those available is chosen as the central atom while drawing the Lewis structure of a molecule.
The least electronegative atom can easily form covalent bonds with other atoms by sharing its electrons.
Out of the three different elements present in XeOF4, fluorine (E.N = 3.98) is the most electronegative element of the Periodic Table. Therefore, it can never be selected as the central atom.
Xenon (E.N = 2.60) is less electronegative than both fluorine and oxygen (E.N = 3.44).
Hence the Xe-atom is placed as the central atom while the 1 O-atom and 4 F-atoms occupy terminal positions, as shown below.
3. Connect the outer atoms with the central atom
In this step, the outer atoms, i.e., 1 O-atom and 4 F-atoms, are joined to the central Xe-atom using single straight lines.
A straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons.
In the above structure, there are 5 single bonds, i.e., 5(2) = 10 valence electrons are already consumed out of the 42 initially available.
Now let’s see in the next steps where to place the remaining 32 valence electrons in the XeOF4 Lewis dot structure.
4. Complete the octet of the outer atoms
Both oxygen and fluorine need a total of 8 valence electrons in order to achieve a stable octet electronic configuration.
In XeOF4, a Xe-O bond represents 2 valence electrons surrounding the O-atom. Therefore, to complete its octet, 6 more electrons are placed around it as 3 lone pairs.
Similarly, a Xe-F bond represents 2 valence electrons already present around each F-atom. Hence 3 lone pairs of electrons are also placed around each F-atom, as shown below.
5. Place the remaining electrons on the central atom
Total valence electrons used till step 4 = 5 single bonds + electrons placed around the O-atom + 4(electrons placed around each F-atom, shown as dots) = 5(2) + 6 + 4(6) = 40 valence electrons.
Total valence electrons – electrons used till step 4 = 42 – 40 = 2 valence electrons.
Thus, these 2 valence electrons are placed as a lone pair on the central Xe-atom in the XeOF4 Lewis structure, as shown below.
In the above Lewis structure, the central Xe-atom has a total of 12 valence electrons surrounding it. This situation falls under the expanded octet rule.
Elements such as sulfur, phosphorus, chlorine, iodine and xenon can accommodate more than 8 valence electrons due to the availability of d-atomic orbitals.
Hence, as a final step, we just need to check the stability of the above Lewis structure by applying the formal charge concept.
6. Check the stability of Lewis’s structure using 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 charges can be calculated using the formula given below.
Formal charge = [valence electrons- nonbonding electrons- ½ (bonding electrons)].
Now let us use this formula and the Lewis structure obtained in step 5 to determine the formal charges present on the XeOF4-bonded atoms.
As per the above calculation, zero or no formal charges are present on either of the F-atoms. However, the central Xe-atom and terminal O-atom carries +1 and -1 formal charges, respectively.
+1 cancels with -1 to give an overall charge of zero on the xenon oxytetrafluoride molecule.
But can we reduce the individual formal charges on the XeOF4 bonded atoms?
Let’s find out in the next step.
7. Minimize formal charges by converting a lone pair into a covalent bond
To reduce the formal charges present on the bonded atoms in XeOF4, a lone pair of electrons from the terminal O-atom is converted into an additional covalent bond between the central Xe-atom and the respective O-atom, as shown below.
In this way, the central Xe-atom has a total of 14 valence electrons in the final XeOF4 Lewis structure, which is quite possible as per the expanded octet rule (discussed above).
Now let us once again check the stability of the above Lewis’s structure by applying the formal charge formula.
For xenon atom
Valence electrons of xenon = 8
Bonding electrons = 4 single bonds + 1 double bond = 4(2) + 4 = 12 electrons
Now zero or no formal charges are present on all the bonded atoms in the XeOF4 Lewis structure, which ensures its incredible stability and that we have drawn it correctly.
So, let’s move ahead and discuss the electron and molecular geometry of xenon oxytetrafluoride (XeOF4).
What are the electron and molecular geometry of XeOF4?
The molecular geometry or shape of XeOF4 is square pyramidal. However, its ideal electronic geometry is octahedral. The presence of a lone pair of electrons on the central Xe-atom in XeOF4 leads to strong lone pair-bond pair electronic repulsions, thus distorting the overall molecular shape.
Molecular geometry of XeOF4
The molecular geometry or shape of XeOF4 w.r.t the central Xe-atom is square pyramidal.
The presence of a lone pair of electrons on the central Xe-atom leads to lone pair-bond pair electronic repulsions in XeOF4. To overcome this strong repulsive effect, the four F-atoms occupy positions such as the four corners of a square.
In contrast, the terminal O-atom is positioned as a pyramid on the top of the molecule, as shown below.
Electron geometry of XeOF4
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule containing a total of 6 electron density regions around the central atom is octahedral.
In XeOF4, the central Xe-atom is directly bonded to 4 F-atoms and an O-atom; it has no lone pair of electrons. This makes a total of 6 electron-density regions surrounding the central Xe-atom.
There is no distinction between bond pairs and lone pairs while determining the electron geometry of a molecule.
As a result, the ideal electron pair geometry of XeOF4 is octahedral.
An easy trick to finding a molecule’s electron and molecular geometry is using the AXN method.
AXN is a simple formula representing the number of bonded atoms and lone pairs on the central atom.
It is used to predict the shape and geometry of a molecule using the VSEPR concept.
AXN notation for XeOF4
A in the AXN formula represents the central atom. In XeOF4, a xenon (Xe) atom is present at the center, so A = Xe.
X denotes the atoms bonded to the central atom. In XeOF4, 1 O-atom and 4 F-atoms are directly bonded to the central Xe-atom, so X= 5.
N stands for the lone pairs present on the central atom. As per the Lewis structure of XeOF4, the central Xe-atom has 1 lone pair of electrons. Thus, N= 1 for XeOF4.
As a result, the AXN generic formula for XeOF4 is AX5N1.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the molecular geometry or shape of a molecule with an AX5N1 generic formula is square pyramidal. In contrast, its ideal electron geometry is octahedral, as we already noted down for xenon oxytetrafluoride (XeOF4).
Hybridization of XeOF4
The central Xe-atom is sp3d2 hybridized in XeOF4.
The ground-state electronic configuration of xenon is [Kr] 4d10 5s2 5p6.
Upon excitation, three of the 5p electrons of xenon shift to its empty 5d atomic orbitals. This results in three half-filled 5p and three half-filled 5d orbitals.
Thus the excited-state electronic configuration of Xenon becomes [Kr] 4d10 5s2 5p3 5d3.
The 5s atomic orbital of xenon thus hybridizes with its three 5p and two 5d orbitals to produce six sp3d2 hybrid orbitals in XeOF4.
Five of the six sp3d2 hybrid orbitals contain a single valence electron only. Thus, xenon uses these sp3d2 hybrid orbitals to form four Xe-F and one Xe-O sigma bond by overlapping with the orbitals of adjacent fluorine and oxygen atoms, respectively.
In contrast, the sixth sp3d2 hybrid orbital containing paired electrons is situated as a lone pair on xenon in XeOF4.
However, the unhybridized d-orbital forms the Xe=O pi bond by overlapping with the unhybridized p-orbital of the adjacent O-atom, as shown below.
Another shortcut to finding the hybridization present in a molecule is using its steric number against the table given below.
The steric number of the Xe-atom in XeOF4 is 6, so it has sp3d2 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The bond angles of XeOF4
The ideal bond angle in a square planar shape is 90°. However, it is due to the asymmetrical square pyramidal shape of XeOF4 that each F-Xe-F bond angle reduces below 90°.
As per Pauling’s electronegativity scale, a polar covalent bond is formed between two dissimilar atoms with an electronegativity difference between 0.4 and 1.6 units.
In XeOF4, a high electronegativity difference of 0.84 units is present between the xenon (E.N = 2.60) and oxygen (E.N = 3.44) atoms in the Xe=O bond. Thus, the Xe=O bond is individually polar and possesses a specific dipole moment value.
In contrast, an even higher electronegativity difference of 1.38 units exists between a xenon and a fluorine (E.N = 3.98) atom in each Xe-F bond. Thus, all four Xe-F bonds present in XeOF4 are strongly polar, possessing even higher dipole moment values.
The more electronegative oxygen and fluorine atoms gain partial negative charges (δ–), while the central Xe-atom obtains a partial positive charge (δ+) in XeOF4.
The unequal Xe=O and Xe-F dipole moments stay uncancelled to yield an overall polar XeOF4 molecule with an asymmetrically distributed electron cloud spread over it (net µ> 0).
The Lewis dot structure of xenon oxytetrafluoride (XeOF4) displays a total of 42 valence electrons, i.e., 42/2 = 21 electron pairs.
Out of the 21 electron pairs, there are 6 bond pairs and 15 lone pairs of electrons.
A xenon (Xe) atom at the center of the XeOF4 Lewis structure is surrounded by 4 F-atoms and 1 O-atom.
There is a lone pair of electrons on the central Xe-atom.
The terminal O-atom carries 2 lone pairs, while 3 lone pairs are present on each F-atom in XeOF4.
How many bond pairs and lone pairs surround the central atom in the XeOF4 Lewis structure?
In XeOF4, the central Xe-atom contains 6 bond pairs and 1 lone pair of electrons. The bond pairs include 4 Xe-F single bonds and 1 Xe=O double bond.
In the best possible Lewis structure of XeOF4, how many double bonds are there on the central xenon atom?
A) 2
B) 1
C) 0
D) 4
E) 3
Option B is the correct answer.
In the Lewis dot structure of XeOF4, xenon forms a double bond with the adjacent oxygen atom only (Xe=O).
What is the molecular geometry of XeOF4?
The molecular geometry or shape of XeOF4 w.r.t the central Xe-atom is square pyramidal. The four F-atoms form a square base, while an O-atom lies as a pyramid at the top.
Why is the molecular shape of XeOF4 different from its electronic geometry?
The molecular shape of XeOF4 is square pyramidal, while its ideal electronic geometry is octahedral.
The molecular geometry or shape of a molecule is controlled by the different number of lone pairs and bond pairs present on the central atom. Contrarily, its electron geometry depends on the total number of electron pairs.
Therefore, the presence of a lone pair of electrons on the central Xe-atom leads to strong lone pair-bond pair electronic repulsions, thus distorting the overall shape of XeOF4.
How is the shape of XeF4 similar to or different from that of XeOF4?
The molecular shape of XeF4is square planar, while that of XeOF4is square pyramidal.
In XeF4, the central Xe-atom is directly bonded to four F-atoms, and there are 2 lone pairs of electrons present on it. The F-atoms form a square base, while the two lone pairs are equally spaced above and below the central atom. Thus, the molecule occupies a symmetrical square planar shape.
In XeOF4, 5 atoms are bonded to the central atom that has 1 lone pair, thus leading to an asymmetrical shape overall.
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In structure of xeof4 in the formation of pie bond between oxygen and xenon is the result of overlapping of P*z_d*yz orbital
In structure of xeof4 in the formation of pie bond between oxygen and xenon is the result of overlapping of P*z_d*yz orbital