BBr3 Lewis structure

BBr₃ (boron tribromide) has one boron atom and three bromine atoms.

In BBr₃ Lewis structure, there are three single bonds around the boron atom, with three bromine atoms attached to it, and on each bromine atom, there are three lone pairs.

Alternative method: Lewis structure of BBr₃

Rough sketch

• First, determine the total number of valence electrons

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

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

Since BBr₃ has one boron atom and three bromine atoms, so…

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

And the total valence electrons = 3 + 21 = 24

Learn how to find: Boron valence electrons and Bromine valence electrons

• 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 bromine, assume that the central atom is boron.

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

• And finally, draw the rough sketch

Lone pair

Here, we have a total of 12 electron pairs. And three B — Br 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 bromine is a period 3 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 bromines.

So for each bromine, 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:

Formal charge

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 bromine atom, formal charge = 7 – 6 – ½ (2) = 0

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

Final structure

The final structure of BBr₃ features a central boron atom linked to three bromine atoms through single covalent bonds. In this arrangement, the boron atom acts as an electron-deficient center, forming three bonds with no lone pairs and totaling six valence electrons. Within this layout, each of the three bromine atoms successfully satisfies the octet rule by retaining three lone pairs alongside its single shared bond. This configuration represents the most stable state for the molecule because it results in a formal charge of zero for every atom involved. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of boron tribromide.

Next: IF₂^– Lewis structure

External video

• BBr3 Lewis Structure – How to Draw the Lewis Structure for BBr3 – YouTube • Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/BBr3-lewis-structure.html
• https://topblogtenz.com/bbr3-lewis-structure-molecular-geometry-polar-or-non-polar/
• https://lambdageeks.com/bbr3-lewis-structure/
• https://homework.study.com/explanation/draw-lewis-structure-for-bbr3.html
• https://sciedutut.com/bbr3-lewis-structure/