B_{2} has **two boron** atoms.

In B_{2} Lewis structure, there is a triple bond between the two boron atoms, and none of the atoms has a lone pair.

## Steps

Use these steps to correctly draw the B_{2} Lewis structure:

#1 First draw a rough sketch

#2 Mark lone pairs on the atoms

#3 Calculate and mark formal charges on the atoms, if required

#4 Convert lone pairs of the atoms, and minimize formal charges

#5 Repeat step 4 if needed, until all charges are minimized, to get a stable Lewis structure

Let’s discuss each step in more detail.

### #1 First draw a rough sketch

- First, determine the total number of valence electrons

In the periodic table, boron lies in group 13. Hence, boron has **three** valence electrons.

Since B_{2} has two boron atoms, so…

Valence electrons of two boron atoms = 3 × 2 = 6

So the **total valence electrons** = 6

Learn how to find: Boron valence electrons

- Second, find the total electron pairs

We have a total of 6 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** = 6 ÷ 2 = 3

- Third, determine the central atom

Here, there are only two atoms and both atoms are boron, so we can assume any one as the central atom.

Let’s assume that the **central atom is left boron**.

- And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 3 electron pairs. And one B — B bond is already marked. So we have to only mark the remaining two 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.

Always start to mark the lone pairs from outside atoms. Here, the outside atom is right boron.

So for right boron, there are **two** lone pairs, and for left boron, there is **zero** lone pair because all two electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Calculate and mark formal charges on the atoms, if required

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

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

For **left boron** atom, formal charge = 3 – 0 – ½ (2) = +2

For **right boron** atom, formal charge = 3 – 4 – ½ (2) = -2

Here, both boron atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both boron atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Convert lone pairs of the atoms, and minimize formal charges

Convert a lone pair of the right boron atom to make a new B — B bond with the left boron atom as follows:

### #5 Repeating step 4 to get a stable Lewis structure

Since there are charges on both boron atoms, again convert a lone pair of the right boron atom to make a new B — B bond with the left boron atom as follows:

In the above structure, you can see that the central atom (left boron) forms an octet. And the outside atom (right boron) also forms an octet. Hence, the octet rule is satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the **stable Lewis structure** of B_{2}.

**Next:** HBrO Lewis structure

## External links

- https://www.answers.com/Q/What_is_the_Lewis_structure_of_B2
- https://quizlet.com/explanations/questions/what-is-the-lewis-structure-of-ceb2-76a0fce5-84c3fd0a-9007-4ee7-afa1-072c96cef4d1

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.