XeF5+ Lewis structure

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.

Alternative method: Lewis structure of XeF₅⁺

Rough sketch

• 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

Learn how to find: Fluorine valence electrons

• 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

Lone pair

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:

Formal charge

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:

Final structure

The final structure of XeF₅⁺ consists of a central xenon atom connected to five fluorine atoms through single covalent bonds. In this configuration, the xenon atom utilizes an expanded valence shell to accommodate twelve electrons, which include five bonding pairs and one lone pair. Each fluorine atom fulfills its octet by retaining three lone pairs alongside its single shared bond. This arrangement is the most stable as it results in a +1 formal charge on the central xenon atom, while each fluorine atom maintains a formal charge of zero, aligning with the overall charge of the ion. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of XeF₅⁺.

To properly represent this as a polyatomic ion, the entire Lewis structure is enclosed within square brackets. The overall charge of 1+ is then written as a superscript outside the brackets at the top right, indicating the loss of one electron from the total valence count of the neutral atoms.

Next: XeCl₃^– Lewis structure

External video

• How to Draw the Lewis Dot Structure for XeF5 + – YouTube • Wayne Breslyn

External links

• https://lambdageeks.com/xef5-lewis-structure/