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.

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.

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

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.

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.

For xenon atom

For oxygen atom 

For each fluorine atom 

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

For oxygen atom 

For each fluorine atom 

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).

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