BBr3 lewis structure contains three B-Br bonds with boron in a central position and all three bromine atoms in an outside position. The lewis structure of BBr3 contains a total of 3 bond pairs and 9 lone pairs.
The drawing of the BBr3 lewis’s structure is very easy and simple. Let’s see how to do it.
Follow some steps for drawing the Lewis dot structure for BBr3
1. Count total valence electron in BBr3
First of all, determine the valence electron that is available for drawing the lewis structure of BBr3 because the lewis diagram is all about the representation of valence electrons on atoms.
So, an easy way to find the valence electron of atoms in the BBr3 molecule is, just to look at the periodic group of boron and bromine atoms.
As the boron atom belongs to the 13th group in the periodic table and bromine is situated in the 17th group, hence, the valence electron for the boron is 3, and for the bromine atom, it is 7.
⇒ Total number of the valence electrons in boron = 3
⇒ Total number of the valence electrons in bromine = 7
∴ Total number of valence electron available for the BBr3 Lewis structure = 3 + 7×3 = 24 valence electrons [∴BBr3 molecule has one boron and three bromine atoms]
2. Find the least electronegative atom and place it at center
An atom with a less electronegative value is preferable for the central position in the lewis diagram because they are more prone to share the electrons with surrounding atoms.
In the case of the BBr3 molecule, the boron atom is less electronegative than the bromine atom.
Hence, put the boron atom at the central position of the lewis diagram and all three bromine atoms outside it.
3. Connect outer atoms to the central atom with a single bond
In this step, join all outer atoms to the central atom with the help of a single bond.
In, the BBr3 molecule, bromine is the outer atom, and boron is the central atom. Hence, joined them.
Count the number of valence electrons used in the above structure. There are 3 single bonds used in the above structure, and one single bond means 2 electrons.
Hence, in the above structure, (3 × 2) = 6 valence electrons are used from a total of 24 valence electrons available for drawing the BBr3 Lewis structure.
∴ (24 – 6) = 18 valence electrons
So, we are left with 18 valence electrons more.
4. Place remaining electrons on the outer atom first and complete their octet
Let’s start putting the remaining valence electrons on outer atoms first. In the case of the BBr3 molecule, bromine is the outer atom and each of them needs 8 electrons in its valence shell to complete the octet.
Start putting the remaining electrons on bromine atoms as dots till they complete their octet.
So, all bromine atoms in the above structure completed their octet, because all of them have 8 electrons(electrons represented as dots + 2 electrons in every single bond) in their outermost shell.
Now again count the valence electron in the above structure.
In the above structure, there is 18 electrons are represented as dots + three single bonds that contain 6 electrons means a total of 24 valence electrons is used in the above structure.
So, we have used all the valence electrons available for drawing the lewis structure of BBr3.
5. Complete the octet of the central atom
We don’t have any extra valence electrons left and the central atom boron has only 6 electrons(3 single bonds) in its valence shell.
It should be noted that Boron is exceptional to the octet rule as it can have 8 electrons or less than 8 electrons in the outermost shell to attain stability. Boron is an exception just like aluminum where it can be octet deficient.
Octet deficient molecules are the molecules that can attains the stability by having less than 8 electrons around the atoms. Some examples – Boron, beryllium, aluminium, hydrogen, lithium, helium
But boron and aluminium is two most common element that can fail to complete the octet as they attains stability having only 6 valence electrons.
Therefore, the boron central atom in the BBr3 lewis structure attains stability by just having 6 valence electrons around it.
Let’s check the formal charge for the 4th step structure to verify it’s stable or not.
6. Check the stability with the help of a formal charge concept
The lesser the formal charge on atoms, the better is the stability of the lewis diagram.
To calculate the formal charge on an atom. Use the formula given below-
⇒ Formal charge = (valence electrons – nonbonding electrons – 1/2 bonding electrons)
Let’s count the formal charge for the 4th step structure.
For bromine atom
⇒ Valence electrons of bromine = 7
⇒ Nonbonding electrons on bromine = 6
⇒ Bonding electrons around bromine (1 single bond) = 2
∴ (7 – 6 – 2/2) = 0 formal charge on the bromine atom.
For boron atom
⇒ Valence electrons of boron = 3
⇒ Nonbonding electrons on boron = 0
⇒ Bonding electrons around boron (3 single bonds) = 6
∴ (3 – 0 – 6/2) = 0 formal charge on the boron central atom.
BBr3 lewis structure
Hence, in the above BBr3 lewis structure, all atoms get a formal charge equal to zero. Even the boron central atom has only 6 electrons instead of 8 in the valence shell, it also gets a formal charge equal to zero.
We have no required to form multiple bonds and provide boron atom to 8 electrons in its valence shell.
The boron can achieve stability by just having 6 electrons in the valence shell.
Therefore, the above lewis structure of BBr3 is most stable and appropriate.
Also check –
- Formal charge calculator
- Lewis structure calculator
- How to draw a lewis structure?
