B2 Lewis structure

B₂ has two boron atoms.

In B₂ Lewis structure, there is a triple bond between the two boron atoms, and none of the atoms has a lone pair.

Alternative method: Lewis structure of B₂

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 B₂ 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

Lone pair

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:

Formal charge

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.

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

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:

Final structure

The final structure of B₂ comprises two boron atoms linked together by a triple covalent bond. In this configuration, each boron atom uses all three of its valence electrons to form the triple bond, resulting in a total of six electrons shared between them. Within this layout, neither boron atom reaches a full octet of eight electrons; instead, they exist in an electron-deficient state with six electrons each. Despite the lack of an octet, this arrangement results in a formal charge of zero for both boron atoms. Therefore, this specific electronic distribution serves as the Lewis representation for this triple-bonded state of diboron.

Note: While a triple bond can be drawn to satisfy formal charges, the ground state of B₂ in molecular orbital theory is actually paramagnetic with a single bond and two unpaired electrons, as boron atoms are typically electron-deficient and do not form stable triple bonds under standard conditions.

Next: HBrO Lewis structure

External video

• B2 Lewis Structure: How to Draw the Lewis Structure for B2 – YouTube • Wayne Breslyn

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