# XeF5+ Lewis structure

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XeF5+ has one xenon atom and five fluorine atoms.

In the XeF5+ Lewis structure, there are five single bonds around the xenon atom, with five fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the xenon atom has one lone pair.

Also, there is a positive (+1) charge on the xenon atom.

Contents

## Steps

To properly draw the XeF5+ Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

### #1 Draw a rough sketch of the structure

• First, determine the total number of valence electrons

In the periodic table, xenon lies in group 18, and fluorine lies in group 17.

Hence, xenon has eight valence electrons and fluorine has seven valence electrons.

Since XeF5+ has one xenon atom and five fluorine atoms, so…

Valence electrons of one xenon atom = 8 × 1 = 8
Valence electrons of five fluorine atoms = 7 × 5 = 35

Now the XeF3+ has a positive (+1) charge, so we have to subtract one electron.

So the total valence electrons = 8 + 35 – 1 = 42

• Second, find the total electron pairs

We have a total of 42 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 42 ÷ 2 = 21

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since xenon is less electronegative than fluorine, assume that the central atom is xenon.

Therefore, place xenon in the center and fluorines on either side.

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 21 electron pairs. And five Xe — F bonds are already marked. So we have to only mark the remaining sixteen electron pairs as lone pairs on the sketch.

Also remember that xenon is a period 5 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.

So for each fluorine, there are three lone pairs, and for xenon, there is one lone pair.

Mark the lone pairs on the sketch as follows:

### #3 Indicate formal charges on the atoms, if necessary

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For xenon atom, formal charge = 8 – 2 – ½ (10) = +1

For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, the xenon atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (xenon) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.

Now there is still a positive (+1) charge on the xenon atom.

This is okay, because the structure with a positive charge on the least electronegative atom is the best Lewis structure. And in this case, the least electronegative element is xenon.

Therefore, this structure is the most stable Lewis structure of XeF5+.

And since the XeF5+ has a positive (+1) charge, mention that charge on the Lewis structure by drawing brackets as follows: