Lewis Structure of AsI3 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of AsI3?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of AsI3 (along with images).

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of AsI3 contains three single bonds between the Arsenic (As) atom and each Iodine (I) atom. The Arsenic atom (As) is at the center and it is surrounded by 3 Iodine atoms (I). The Arsenic atom has 1 lone pair and all the three Iodine atoms 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 AsI3).

6 Steps to Draw the Lewis Structure of AsI3

Step #1: Calculate the total number of valence electrons

Here, the given molecule is AsI3. In order to draw the lewis structure of AsI3, first of all you have to find the total number of valence electrons present in the AsI3 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 AsI3

• For Arsenic:

Arsenic is a group 15 element on the periodic table.

Hence, the valence electrons present in arsenic is 5 (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 AsI3 molecule, 

Valence electrons given by Arsenic (As) atom = 5
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in AsI3 molecule = 5 + 7(3) = 26

Step #2: Select the center atom

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

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table). ^[1]

Here in the AsI3 molecule, if we compare the arsenic atom (As) and iodine atom (I), then the arsenic is less electronegative than iodine.

So, arsenic should be placed in the center and the remaining 3 iodine atoms will surround it.

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

Now in the above sketch of AsI3 molecule, put the two electrons (i.e electron pair) between each arsenic atom and iodine atom to represent a chemical bond between them.

These pairs of electrons present between the Arsenic (As) and Iodine (I) atoms form a chemical bond, which bonds the arsenic and iodine atoms with each other in a AsI3 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 AsI3, the outer atoms are iodine atoms.

So now, you have to complete the octet on these iodine atoms.

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

Also, only 24 valence electrons of AsI3 molecule are used in the above structure.

But there are total 26 valence electrons in AsI3 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 26 – 24 = 2.

So let’s keep these two electrons (i.e electron pair) on the central atom.

Now, let’s move to the next step.

Step #5: Check whether the central atom has octet or not. If it does not have an octet, then move the electron pair from the outer atom to form a double bond or triple bond

In this step, we have to check whether the central atom (i.e arsenic) has an octet or not. 

In simple words, we have to check whether the central Arsenic (As) atom is having 8 electrons or not.

As you can see from the above image, the central atom (i.e arsenic), has 8 electrons. So it fulfills the octet rule and the arsenic atom is stable.

Step #6: 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 arsenic atom (As) 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 Arsenic:
  Valence electron = 5 (as it is in group 15)
  Nonbonding electrons = 2
  Bonding electrons = 6
• 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
As	=	5	–	2	–	6/2	=	0
I	=	7	–	6	–	2/2	=	0

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

Hence, there will not be any change in the above structure and the above lewis structure of AsI3 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of AsI3 represents the single bond ( | ). So the above lewis dot structure of AsI3 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of SbF6-
Lewis Structure of SbCl3
Lewis Structure of C2H4F2
Lewis Structure of Br2O
Lewis Structure of SiH2Cl2 

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