# BI3 Lewis structure

BI3 (boron triiodide) has one boron atom and three iodine atoms.

In BI3 Lewis structure, there are three single bonds around the boron atom, with three iodine atoms attached to it, and on each iodine atom, there are three lone pairs.

Contents

## Steps

To properly draw the BI3 Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

### #1 Draw a rough sketch of the structure

• First, determine the total number of valence electrons

In the periodic table, boron lies in group 13, and iodine lies in group 17.

Hence, boron has three valence electrons and iodine has seven valence electrons.

Since BI3 has one boron atom and three iodine atoms, so…

Valence electrons of one boron atom = 3 × 1 = 3
Valence electrons of three iodine atoms = 7 × 3 = 21

And the total valence electrons = 3 + 21 = 24

• Second, find the total electron pairs

We have a total of 24 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 = 24 ÷ 2 = 12

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since boron is less electronegative than iodine, assume that the central atom is boron.

Therefore, place boron in the center and iodines on either side.

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 12 electron pairs. And three B — I bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that boron is a period 2 element, so it can not keep more than 8 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are iodines.

So for each iodine, there are three lone pairs, and for boron, there is zero lone pair because all nine electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Indicate formal charges on the atoms, if necessary

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For boron atom, formal charge = 3 – 0 – ½ (6) = 0

For each iodine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both boron and iodine atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (boron) doesn’t form an octet. But, boron has an exception that it does not require eight electrons to form an octet. So no need to worry about the octet rule here.

Therefore, this structure is the stable Lewis structure of BI3.

## FAQs

### Does BI3 follow the octet rule?

BI3 is a molecule that consists of one boron atom and three iodine atoms. The boron atom in BI3 has only six electrons in its valence shell, which is less than the eight electrons required to satisfy the octet rule. Therefore, BI3 does not follow the octet rule.

### Does boron have lone pairs in BI3?

Each iodine atom in BI3 is bonded to the boron atom through a single covalent bond, which means that the boron atom has a total of three bonding pairs of electrons. Since there are no additional electrons around the boron atom, boron has no lone pairs in the BI3 Lewis structure.

### How many valence electrons does BI3 have?

BI3 has one boron and three iodine atoms. Boron, which is in group 3 of the periodic table, has three valence electrons. And each iodine atom, which is in group 17 of the periodic table, has seven valence electrons.

Therefore, in the BI3 molecule, the boron atom contributes three valence electrons, while each iodine atom contributes seven valence electrons, meaning that a total of 3 + 7 + 7 + 7 = 24 valence electrons.