Lewis Structure of XeI2 (With 5 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of XeI2?

Awesome!

Here, I have explained 5 simple steps to draw the lewis dot structure of XeI2 (along with images).

So, if you are ready to go with these 5 simple steps, then let’s dive right into it!

Lewis structure of XeI2 contains two single bonds between the Xenon (Xe) atom and each Iodine (I) atom. The Xenon atom (Xe) is at the center and it is surrounded by 2 Iodine atoms (I). The Xenon atom has 3 lone pairs and both the Iodine atoms also have 3 lone pairs.

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of XeI2).

5 Steps to Draw the Lewis Structure of XeI2

Step #1: Calculate the total number of valence electrons

Here, the given molecule is XeI2. In order to draw the lewis structure of XeI2, first of all you have to find the total number of valence electrons present in the XeI2 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in XeI2

• For Xenon: 

Xenon is a group 18 element on the periodic table.

Hence, the valence electron present in xenon is also 8 (see below image).

• For Iodine:

Iodine is a group 17 element on the periodic table.

Hence, the valence electrons present in iodine is 7 (see below image).

Hence in a XeI2 molecule, 

Valence electrons given by Xenon (Xe) atom = 8
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in XeI2 molecule = 8 + 7(2) = 22

Step #2: Select the center atom

While selecting the atom, always put the least electronegative atom at the center. 

Here in the XeI2 molecule, if we compare the xenon atom (Xe) and iodine atom (I), then the xenon is less electronegative than iodine.

So, xenon should be placed in the center and the remaining 2 iodine atoms will surround it.

Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of XeI2 molecule, put the two electrons (i.e electron pair) between each xenon atom and iodine atom to represent a chemical bond between them.

These pairs of electrons present between the Xenon (Xe) and Iodine (I) atoms form a chemical bond, which bonds the xenon and iodine atoms with each other in a XeI2 molecule.

Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of XeI2, the outer atoms are iodine atoms.

So now, you have to complete the octet on these iodine atoms (because iodine requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the iodine atoms form an octet.

Also, only 16 valence electrons of XeI2 molecule are used in the above structure.

But there are total 22 valence electrons in XeI2 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 22 – 16 = 6.

So let’s keep these six electrons (i.e 3 electron pairs) on the central atom.

Now, let’s move to the next step.

Step #5: Final step – Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on xenon atom (Xe) as well as each iodine atom (I).

For that, you need to remember the formula of formal charge;

Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2

• For Xenon:
  Valence electron = 8 (as it is in group 18)
  Nonbonding electrons = 6
  Bonding electrons = 4
• For Iodine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
Xe	=	8	–	6	–	4/2	=	0
I	=	7	–	6	–	2/2	=	0

So you can see above that the formal charges on xenon as well as iodine are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of XeI2 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of XeI2 represents the single bond ( | ). So the above lewis dot structure of XeI2 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of PF2Cl3
Lewis Structure of IBr4-
Lewis Structure of SeOBr2
Lewis Structure of HBrO2
Lewis Structure of HBrO3 

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