# XeO3 Lewis Structure: Drawings, Hybridization, Shape, Charges, Pair And Detailed Facts

Lewis structure, formal charge calculation, shape, and hybridization of XeO3 molecule are presented here.

There are three oxygen atoms connected with a double bond encircling the central atom Xe in the lewis structure of XeO3. The xenon atom holds one lone pair and three oxygen atoms carry two lone pairs. The hybridization of central atom Xenon is sp3 in XeO3. The molecular geometry is trigonal pyramidal and electron geometry is tetrahedral in XeO3.

• The chemical formula of xenon trioxide XeO3.
• It is a colorless crystalline solid.
•  It is a highly explosive and unstable compound.

This article discusses the shape of XeO3 with some important facts.

## How to draw the lewis structure for XeO3?

Lewis structure gives a clear idea about the presence of bonded and non-bonded electrons. It is also helpful for determining the shape and geometry of the molecule.

One should follow the steps below to draw the lewis structure of XeO3.

• Calculate the total valence electron in a molecule. Xenon(Xe)  comes from group 18  in the periodic table and contain  8 valence electron. Oxygen (O2)  comes from group 16 in the periodic table and contain  6 number of valence electron. Total number of valence electron available for the XeO3 Lewis structure = 8 + 6(3) = 26 valence electrons.
• Keep the least electronegative atom in the center.   Electronegativity for the xenon atom is 2.6 and for the oxygen atom, is 3.44.  Posit xenon atom at the middle and three oxygen atoms at the abutting position.
• Then fix the outer atoms (oxygen) and central atom(xenon) with a single bond. After linking, calculate the valence electrons involved in the structure. i.e 6 valence electrons are used out of 26 valence electrons. Remaining electrons =  26-6=20
• Complete octets on outer side atoms i.e oxygen.

Out of 8 valance electrons of oxygen 6 electrons are denoted as dots and 2 electrons as a single bond.  So there is a total 24 number of electrons used. i.e for 3 oxygen 18 electrons are represented as dots and three single bonds contribute 6 electrons. It can be explained as, (18 + 6) = 24 numbers of electrons are utilized out of 26 electrons for the XeO3 lewis structure. Remaining electrons (26 – 24) = 2 valence electrons.

• Complete central atom octet by moving electrons from the outer atom to form a double bond.
• Now place the 2 valence electron on the xenon atom.
• In Fig-1 XeO3 structure, xenon is bound with the three single bonds means it holds 6 electrons in its valence shell. The remaining two electrons are placed over the xenon central atom.

## XeO3 lewis structure formal charges

• To check the stability of the lewis structure formal charge has to be calculated.
• Formal charge  =  [ number of valence electrons  – number of  nonbonding electrons – ½  number of  bonding electrons]

Calculation of formal charge on oxygen

Oxygen holds six electrons in the valence shell. In fig 1 there are six nonbonding electrons and two bonding electrons (one single bond = 2 electrons). Hence, the formal charge of every oxygen atom (present in fig 1) is (6 – 6 – 2/2) = -1

Similarly, the formal charge on the xenon atom can be calculated.

Similarly, xenon carries eight electrons in the valence shell. In fig 1 there are 2 nonbonding electron and six bonding electron(3 single bond = 6 electrons). Hence, formal charge of central atom xenon (present in fig 1) is (8 – 2 – 6/2) = +3

The above lewis structure of   XeO3 is unstable due to the high formal charge. The formal charge can be reduced by rearrangement of electrons and forming multiple bonds. For that covert the oxygen atom lone pair to bond pair. Lesser is the formal charge more stable is the compound. Conversion of lone pair of an electron to bond pair and formal charge calculation.

Calculation of formal charge of XeO3 after electronic rearrangement i.e after double bond formation is given below.

Formal charge = (valence electrons – nonbonding electrons –  1/2 bonding electrons)

For Oxygen formal charge = 6-4-4/2=0

For Xenon formal charge = 8-4-8/2=0 The most stable and appropriate lewis structure of XeO3 is given below.

## XeO3 Lewis Structure Shape

From Lewis structure, it is observed Xe contains a total of three bond pairs and one lone pair electron present. From VSEPR theory one can predict that it falls under the AX3N category molecule.

A = central atom

X = number of bonding atoms attached to the central atom

N = lone pair on the central atom

According to the VSEPR theory explanation, AX3N kind molecule comes under trigonal pyramidal molecular geometry. The actual geometry of XeO3 should be tetrahedral. Deviation of XeO3 structure is due to the presence of lone pair.  Its shape is trigonal pyramidal

## XeO3 Lewis Structure Lone Pairs

The electrons which are taking part in the formation of bonds are known as bond pair electrons. The electrons which do not involve in bonding are called lone pairs. Lone pairs are denoted as dots in the lewis diagram. Three double bonds present in the XeO3 figure mean 12 bonding electrons are present. It is also shown that 14 nonbonding electrons (denoted as dots) present in XeO3.

In XeO3, all of the atoms (xenon and oxygen) have lone pairs. Lone pair or nonbonded electron = Total number of valance electrons–number of bonded electrons.

In XeO3, number of valence electron (5s2 5p6) in xenon eight. Out of eight, six electrons contribute to bond formation. So the number of the nonbonding electrons is (8 – 6) = 2. The three oxygen has a total of six electrons in its outermost shell (2s2 2p4). Among those six electrons, two electrons participate in double bond formation with xenon. The  nonbonded electrons leftover is (6 – 2) = 4electrons or 2 pair of electrons. This is the same for all of the three oxygen atoms.

## XeO3 Hybridization

Steric number of XeO3 = (Number of bonded atoms attached to xenon + Lone pair on xenon)

Steric number of XeO3 = (3 + 1) = 4. So Xe is sp3 hybridized in XeO3 .

The sp3 hybridization of XeO3 is shown in the Fig-6 below. Xenon uses its one 5s orbital and three 5p orbitals for the sp3 hybridization. The hybrid orbitals form σ bonds and the remaining orbitals i.e 5d orbital form π bonds.

## Xeo3 Lewis Structure Octet Rule

When atoms have 8 electrons in their valence shell, they are said to have an octet.

In XeO3 geometry, xenon holds one lone pair and there are three double bonds fixed with it. So there is a total of 14 electrons in the xenon. That means do not obey the octet rule. It has the potential to expand the octet. It is also known that the Xenon atom has a d- orbital and it can hold more than 8 electrons. Due to the expanded octet, XeO3 obtained a stable structure. But oxygen follows the octet rule. Because it carries six electrons in its valance shell. So, to fill the octet two more electrons are required.

#### Conclusion

XeO3 holds three oxygens which are connected to Xe with a double bond. It is sp3 hybridized and the shape is trigonal pyramidal. It is a polar molecule.