Lewis Structure of BI3 (With 5 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of BI3?

Awesome!

Here, I have explained 5 simple steps to draw the lewis dot structure of BI3 (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 BI3 contains three single bonds between the Boron (B) atom and each Iodine (I) atom. The Boron atom (B) is at the center and it is surrounded by 3 Iodine atoms (I). The Boron atom does not have a lone pair while all three iodine atoms have three lone pairs each.

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 BI3).

5 Steps to Draw the Lewis Structure of BI3

Step #1: Calculate the total number of valence electrons

Here, the given molecule is BI3. In order to draw the lewis structure of BI3, first of all you have to find the total number of valence electrons present in the BI3 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 BI3

• For Boron:

Boron is a group 13 element on the periodic table. ^[1]

Hence, the valence electrons present in boron is 3 (see below image).

• For Iodine:

Iodine is a group 17 element on the periodic table. ^[2]

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

Hence in a BI3 molecule, 

Valence electrons given by Boron (B) atom = 3
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in BI3 molecule = 3 + 7(3) = 24

Step #2: Select the center atom

While selecting the center 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). ^[3]

Here in the BI3 molecule, if we compare the boron atom (B) and iodine atom (I), then boron is less electronegative than iodine.

So, boron 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 BI3 molecule, put the two electrons (i.e electron pair) between each boron atom and iodine atom to represent a chemical bond between them.

These pairs of electrons present between the Boron (B) and Iodine (I) atoms form a chemical bond, which bonds the boron and iodine atoms with each other in a BI3 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 BI3, 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, all the 24 valence electrons of BI3 molecule (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.

Hence there is no change in the above sketch of BI3.

Let’s move to the next step.

Step #5: 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 boron atom (B) 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 Boron:
  Valence electrons = 3 (as it is in group 13)
  Nonbonding electrons = 0
  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
B	=	3	–	0	–	6/2	=	0
I	=	7	–	6	–	2/2	=	0

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

Hence, there will not be any change in the above structure and the above lewis structure of BI3 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of BI3 represents the single bond ( | ). So the above lewis dot structure of BI3 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of CH3I
Lewis Structure of BrO-
Lewis Structure of SeOF2
Lewis Structure of SBr6
Lewis Structure of IO3- 

---

Article by;