BI3 Lewis Structure in 5 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

BI3 lewis structure has a Boron atom (B) at the center which is surrounded by three Iodine atoms (I). There are 3 single bonds between the Boron atom (B) and each Iodine atom (I).

If you haven’t understood anything from the above image of BI3 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of BI3.

So let’s move to the steps of drawing the lewis structure of BI3.

Steps of drawing BI3 lewis structure

Step 1: Find the total valence electrons in BI3 molecule

In order to find the total valence electrons in a BI3 molecule, first of all you should know the valence electrons present in boron atom as well as iodine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of boron as well as iodine using a periodic table.

Total valence electrons in BI3 molecule

→ Valence electrons given by boron atom:

Boron is a group 13 element on the periodic table. ^[1] Hence the valence electrons present in boron is 3.

You can see the 3 valence electrons present in the boron atom as shown in the above image.

→ Valence electrons given by iodine atom:

Iodine is a group 17 element on the periodic table. ^[2] Hence the valence electrons present in iodine is 7.

You can see the 7 valence electrons present in the iodine atom as shown in the above image.

Hence, 

Total valence electrons in BI3 molecule = valence electrons given by 1 boron atom + valence electrons given by 3 iodine atoms = 3 + 7(3) = 24.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is BI3 and it contains boron atom (B) and iodine atoms (I).

You can see the electronegativity values of boron atom (B) and iodine atom (I) in the above periodic table.

If we compare the electronegativity values of boron (B) and iodine (I) then the boron atom is less electronegative.

So here the boron atom (B) is the center atom and the iodine atoms (I) are the outside atoms.

Step 3: Connect each atoms by putting an electron pair between them

Now in the BI3 molecule, you have to put the electron pairs between the boron atom (B) and iodine atoms (I).

This indicates that the boron (B) and iodine (I) are chemically bonded with each other in a BI3 molecule.

Step 4: Make the outer atoms stable

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of BI3 molecule, you can see that the outer atoms are iodine atoms.

These outer iodine atoms are forming an octet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the BI3 molecule.

The BI3 molecule has a total 24 valence electrons and all these valence electrons are used in the above sketch of BI3.

Hence there are no remaining electron pairs to be kept on the central atom. 

So now let’s proceed to the next step.

Step 5: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of BI3.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on boron (B) atom as well as iodine (I) atoms present in the BI3 molecule.

For calculating the formal charge, you have to use the following formula;

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

You can see the number of bonding electrons and nonbonding electrons for each atom of BI3 molecule in the image given below.

For Boron (B) atom:
Valence electrons = 3 (because boron is in group 13)
Bonding electrons = 6
Nonbonding electrons = 0

For Iodine (I) atom:
Valence electron = 7 (because iodine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
B	=	3	–	6/2	–	0	=	0
I	=	7	–	2/2	–	6	=	0

From the above calculations of formal charge, you can see that the boron (B) atom as well as iodine (I) atom has a “zero” formal charge.

This indicates that the above lewis structure of BI3 is stable and there is no further change in the above structure of BI3.

In the above lewis dot structure of BI3, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of BI3.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

BrO- Lewis Structure	SeOF2 Lewis Structure
SBr6 Lewis Structure	IO3- Lewis Structure
HOFO Lewis Structure	BrF Lewis Structure