Ready to learn how to draw the lewis structure of Br2O?
Here, I have explained 6 simple steps to draw the lewis dot structure of Br2O (along with images).
So, if you are ready to go with these 6 simple steps, then let’s dive right into it!
Lewis structure of Br2O contains two single bonds between the Oxygen (O) atom and each Bromine (Br) atom. The Oxygen atom (O) is at the center and it is surrounded by 2 Bromine atoms (Br). The Oxygen atom has 2 lone pairs and both the Bromine atoms have 3 lone pairs.
Let’s draw and understand this lewis dot structure step by step.
(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of Br2O).
6 Steps to Draw the Lewis Structure of Br2O
Step #1: Calculate the total number of valence electrons
Here, the given molecule is Br2O. In order to draw the lewis structure of Br2O, first of all you have to find the total number of valence electrons present in the Br2O molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).
So, let’s calculate this first.
Calculation of valence electrons in Br2O
- For Bromine:
Bromine is a group 17 element on the periodic table.
Hence, the valence electrons present in bromine is 7 (see below image).
- For Oxygen:
Oxygen is a group 16 element on the periodic table.
Hence, the valence electron present in oxygen is 6 (see below image).
Hence in a Br2O molecule,
Valence electrons given by each Bromine (Br) atom = 7
Valence electrons given by Oxygen (O) atom = 6
So, total number of Valence electrons in Br2O molecule = 7(2) + 6 = 20
Step #2: Select the center atom
While selecting the atom, always put the least electronegative atom at the center.
(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table).
Here in the Br2O molecule, if we compare the bromine atom (Br) and oxygen atom (O), then the oxygen is less electronegative than bromine.
So, oxygen should be placed in the center and the remaining 2 bromine atoms will surround it.
Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of Br2O molecule, put the two electrons (i.e electron pair) between each oxygen atom and bromine atom to represent a chemical bond between them.
These pairs of electrons present between the Oxygen (O) and Bromine (Br) atoms form a chemical bond, which bonds the oxygen and bromine atoms with each other in a Br2O molecule.
Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom
Don’t worry, I’ll explain!
In the Lewis structure of Br2O, the outer atoms are bromine atoms.
So now, you have to complete the octet on these bromine atoms (because bromine requires 8 electrons to have a complete outer shell).
Now, you can see in the above image that all the bromine atoms form an octet.
Also, only 16 valence electrons of Br2O molecule are used in the above structure.
But there are total 20 valence electrons in Br2O molecule (as calculated in step #1).
So the number of electrons left to be kept on the central atom = 20 – 16 = 4.
So let’s keep these four electrons (i.e 2 electron pairs) on the central atom.
Now, let’s move to the next step.
Step #5: Check whether the central atom has octet or not. If it does not have an octet, then move the electron pair from the outer atom to form a double bond or triple bond
In this step, we have to check whether the central atom (i.e oxygen) has an octet or not.
In simple words, we have to check whether the central Oxygen (O) atom is having 8 electrons or not.
As you can see from the above image, the central atom (i.e oxygen), has 8 electrons. So it fulfills the octet rule and the oxygen atom is stable.
Step #6: Final step – Check the stability of lewis structure by calculating the formal charge on each atom
Now, you have come to the final step and here you have to check the formal charge on oxygen atom (O) as well as each bromine atom (Br).
For that, you need to remember the formula of formal charge;
Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2
- For Bromine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2
- For Oxygen:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 4
Bonding electrons = 4
|Formal charge||=||Valence electrons||–||Nonbonding electrons||–||(Bonding electrons)/2|
So you can see above that the formal charges on oxygen as well as bromine are “zero”.
Hence, there will not be any change in the above structure and the above lewis structure of Br2O is the final stable structure only.
Each electron pair (:) in the lewis dot structure of Br2O represents the single bond ( | ). So the above lewis dot structure of Br2O can also be represented as shown below.