7 Facts On XeO3 Lewis Structure: Detailed Explanations

XeO₃ or xenon trioxide is the oxide of noble gas having a molecular weight of 179.288 g/mol. Let us discuss the details fact of XeO₃.

Xe is a noble gas and it cannot react easily but it can react with most electronegative atoms like O and Cl due to the same threshold energy and form a different molecule. In XeO₃ there are three double bonds present along with one lone pair over Xe, so all eight electrons are being used. The Xe-O bond length is 176 pm.

The O-Xe-O bond angle is not perfect at 109.5⁰ rather it is deviated due to bond pair lone pair repulsion and the bond angle decrease to 103⁰.  Now we discuss the hybridization, lewis structure, shape, and other molecular property of the XeO₃ with a proper explanation in the next part of the article.

1. How to draw the XeO₃ structure?

Lewis structure can help us to find the shape, electrons are involved in the bond formation, and several lone pairs are present. Let us try to draw the lewis structure of XeO₃ in the following steps.

Counting the valence electrons

The total valence electrons for XeO₃ is 26 which can be counted as the valence electrons of each atom in the molecule and just add them together. Xe has 8 valence electrons as it is a noble gas and each O contains 6 valence electrons as it belongs to group 16, so the total valence electrons will be, 8+(6*3) =26.

Choosing the central atom

For drawing the lewis structure we need to choose a central atom and based on the central atoms surrounding atoms are placed accordingly. Based on size and electronegativity, Xe is chosen as the central atom here and all the three O are placed surrounding it and connected with a respective number of bonds.

Satisfying the octet

Here both Xe and O are p-block elements and they need eight electrons in their valence shell Xe is already completed its octet and O formed a double bond and shared two electrons with Xe and completed its octet during the bond formation in XeO₃. The total electrons will be required for octet 8*4 = 32 electrons.

Satisfying the valency

During the satisfaction of their octet, every atom should compatible with their valency. The stable valency of O is 2 and for Xe is 8 . The remaining electrons 32-26 = 6 should be accompanied by their stable valency and those electrons should be shared in 6/2 = 3 bonds and there should minimum 3 bonds are required.

Assign the lone pairs

After the bond formation excess, nonbonded electrons exist as lone pairs over the respective atoms. O has lone pairs after the double bond formation because it has four more electrons in its valence shell and Xe it forms six bonds by using six electrons and the remaining two electrons exist as one lone pair.

2. XeO₃ structure lone pairs

The remaining electrons in the valence orbital after bond formation is known as lone pairs. Let us count the lone pairs of the XeO₃ molecule.

The total lone pairs present over the XeO₃ molecule are 7 pairs which mean 14 lone pairs of electrons. The numbers are contributed from O as well as Xe because both have non-bonded electrons. We just add them together to get a total number of lone pairs over the XeO₃ molecule.

• Now we calculate the lone pairs over the XeO₃ molecule by the formula, non-bonded electrons = valence electrons – bonded electrons
• The lone pairs present over each Cl atom are 8-6= 1
• The lone pairs present over each O atom are 6-2= 4
• So, the total number of lone pairs present over the XeO₃ molecule is 4*3 + 2 = 14 electrons.

3. XeO₃ structure shape

The molecular shape is formed by the proper arrangement of the central and other substituent atoms to avoid repulsion. Let us predict the shape of XeO₃.

The molecular shape of XeO₃ is pyramidal, made by Xe and three O atoms which are confirmed by the following table.

Molecular Formula	No. of bond pairs	No. of lone pairs	Shape	Geometry
AX	1	0	Linear	Linear
AX2	2	0	Linear	Linear
AXE	1	1	Linear	Linear
AX3	3	0	Trigonal planar	Trigonal Planar
AX2E	2	1	Bent	Trigonal Planar
AXE2	1	2	Linear	Trigonal Planar
AX4	4	0	Tetrahedral	Tetrahedral
AX3E	3	1	Trigonal pyramidal	Tetrahedral
AX2E2	2	2	Bent	Tetrahedral
AXE3	1	3	Linear	Tetrahedral
AX5	5	0	trigonal bipyramidal	trigonal bipyramidal
AX4E	4	1	seesaw	trigonal bipyramidal
AX3E2	3	2	t-shaped	trigonal bipyramidal
AX2E3	2	3	linear	trigonal bipyramidal
AX6	6	0	octahedral	octahedral
AX5E	5	1	square pyramidal	octahedral
AX4E2	4	2	square pyramidal	octahedral

So, from the above table, it is confirmed that XeO₃ is an AX₃E type of molecule around the central Xe atom. So, according to the VSEPR (Valence Shell Electrons Pair Repulsion) theory, the AX₃E type of molecule adopted tetrahedral geometry, but the shape changes pyramidal due to the presence of lone pair.

4. XeO₃ hybridization

Mixing of two or more atomic orbitals undergoing hybridization form an equal number of hybrid orbital of equivalent energy. Let us know the hybridization of XeO₃.

The central Xe in the XeO₃ molecule is sp³ hybridized, which can be shown in the following table.

Structure	Hybridization value	State of hybridization of central atom	Bond angle
1.Linear	2	sp /sd / pd	1800
2.Planner trigonal	3	sp2	1200
3.Tetrahedral	4	sd3/ sp3	109.50
4.Trigonal bipyramidal	5	sp3d/dsp3	900 (axial), 1200(equatorial)
5.Octahedral	6	sp3d2/ d2sp3	900
6.Pentagonal bipyramidal	7	sp3d3/ d3sp3	900,720

• We can calculate the hybridization by the convention formula, H = 0.5(V+M-C+A),
• So, the hybridization of central Cl is, ½(8+0+0+0) = 4 (sp³)
• One s orbital and three p orbitals of Xe are involved in the hybridization.
• The lone pairs of Xe are also involved in the hybridization.
• The π bond between Xe and O is not involved in the hybridization.

5. XeO₃ lewis structure formal charge

The formal charge is a hypothetical concept for a molecule to check whether it is charged or neutral or which atom carries the charge. Let us calculate the formal charge of XeO₃.

The formal charge of XeO₃ is not zero. the molecule is neutral and the formal charge posses by the Xe is completely neutralized by the formal charge of each O atom. actually both the atoms show zero formal charges individually and for this reason, the overall formal charge of the molecule is also 0.

• The formula being used for the formal charge is, F.C. = N_v – N_l.p. -1/2 N_b.p.
• The formal charge present over the Xe is 8-2-(6/2) =0
• The formal charge present over the O is 6-4-(4/2) =0
• So, from the above data, we can conclude that O and Xe contain 0 formal charges.
• So, the overall formal charge of the molecule is 0 and the molecule is very stable.

6. XeO₃ lewis structure octet rule

The octet rule is the completion of the valence orbital atom by accepting a suitable le number of electrons to gain noble gas stability. Let us learn about the octet of XeO₃.

Xe and O both obey the octet rule by completing the valence shell with a suitable number of electrons. The valence shell of Xe is already completed as it belongs to the group 18 element and for oxygen, there are only six electrons present in its valence shell, and need two more electrons to complete the octet.

The stable valency of O is two and each O can form a double bond with Xe by sharing their electrons, and this way they complete their octet by accepting two more electrons. Xe forms six bonds with three O atoms and has one lone pair over it. So, via bond formation, both elements complete their octet in XeO_3.

7.  XeO₃ lewis structure resonance

By the concept of resonance, we can predict the delocalization of electron clouds among different skeleton forms. Now we learn more about the resonance of XeO₃.

Resonance is possible in the XeO₃ molecule due to excess electron density present over and it shows different canonical forms. Among the four resonating structures, IV is more stable as it contains more the covalent bond and for this reason, the electronic cloud can delocalize very easily over this molecule.

2^nd most contributing structure of XeO₃ is III because it contains lower number of covalent bonds than IV but higher than the other two and the negative charge is present over the electronegative O atom which is a stabilizing factor. The least contributing structure is I because it has a lower number of covalent bonds.

Conclusion

XeO₃ is the oxide of a noble gas and it can form by the reaction of Xe and O due to the disproportionation of the molecule, It can break into Xe and O with a high explosion, the molecule is not as stable as another oxide. It has an orthorhombic crystal structure with lattice constant a = 6.163 Å, b = 8.115 Å, c = 5.234 Å, and 4 molecules per unit cell.

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