BrO2– Lewis structure is made up of two oxygen (O) atoms, and one bromine (Br) atom. The bromine (Br) atom is the central atom in the lewis structure. The lewis structure of BrO2– contains one single bond, one double bond, and a total of 7 lone pairs.
Let’s see how to draw the lewis structure of BrO2– with a simple approach.
Follow some steps for drawing the lewis structure of BrO2–
1. Count total valence electron in BrO2–
To determines the valence electron in BrO2–, look at the periodic group of bromine and oxygen atoms.
By looking at the periodic table, we get to know, that bromine belongs to the 17th periodic group and oxygen to the 16th.
Hence, the valence electron for bromine is 7 and for oxygen, it is 6.
⇒ Total number of the valence electrons in oxygen = 6
⇒ Total number of the valence electrons in bromine= 7
∴ Total number of valence electron available for the BrO2– lewis structure = 7 + 6(2) + 1 = 20 valence electrons [∴ one bromine, two oxygen and one negative ion in BrO2– that also count as a one valence electron]
2. Find the least electronegative atom and placed it at center
In this step, we need to place the less electronegative atom in the BrO2–molecule at the central position, and rests are spaced evenly around it.
A bromine atom(2.96) is less electronegative than an oxygen atom(3.44), hence, put the bromine in the central position of the lewis diagram and oxygen as terminal atoms.
3. Connect outer atoms to central atom with a single bond
Now just attach the single bond for connecting each outer atom(oxygen) to the central atom(bromine).
Count the valence electron we used to draw the above structure. A single bond means two electrons and in the above structure, two single bonds are used.
Therefore, (2 single bonds × 2 electrons) = 4 valence electrons are used in the above structure from a total of 20 valence electrons available for the BrO2– lewis structure.
∴ (20 – 4) = 16 valence electrons
Hence, we are left with 16 valence electrons more.
4. Complete the octet of the outer atom first
Here’s we need to put our remaining valence electron over outer atoms first to complete their octet.
In the case of the BrO2– molecule, oxygen atoms are outer atoms and they need 8 electrons to complete their outer shell.
As you see in the above structure, we put the 6 electrons represented as dots on both oxygens. Now Oxygen atoms completed their octet, since, both have 8 electrons(6 represented as dots + 2 electrons in a single bond).
Again count the total valence electrons used in the above structure.
In the above structure, (2 single bonds mean 4 electrons + 12 electrons represented as dots) = 16 valence electrons are used from the total of 20 electrons available for BrO2- lewis structure.
∴ (20 – 16) = 4 valence electrons
Now we are left with only 4 valence electrons.
5. Complete the octet of the central atom
Bromine is the central atom in BrO2– molecule. And we have 4 remaining valence electrons, hence, put these remaining valence electrons over the Bromine central atom.
If you look at the above structure, we see that all atoms(oxygen and bromine) completed their octet comfortably as each of them has 8 valence electrons(electrons represented as dots + electrons in a single bond) in their outer shell.
Also, we used all 20 valence electrons that are available for BrO2–.
Now we just need to check the stability of the above structure through the formal charge concept.
6. Check the stability with the help of a formal charge concept
The structure with the formal charge close to zero or zero is the best and most stable lewis structure.
To calculate the formal charge on an atom. Use the formula given below-
⇒ Formal charge = (valence electrons – Non bonding electrons – 1/2 bonding electrons)
We will calculate the formal charge on the 5th step structure to verify its stability.
For Bromine atom –
⇒ Valence electron of bromine = 7
⇒ Nonbonding electrons on bromine = 4
⇒ Bonding electrons around bromine = 4 (two single bonds)
∴ Formal charge on bromine atom = (7 – 4 – 4/2) = +1
For oxygen atom –
⇒ Valence electron of oxygen = 6
⇒ Nonbonding electrons on oxygen = 6
⇒ Bonding electrons around oxygen = 2 (one single bond)
∴ Formal charge on oxygen atom = (6 – 6 – 2/2) = -1
The above structure is not stable as it has an uneven formal charge. We have to lower the formal charge of the above structure.
To reduce the formal charge, We have to convert the one lone pair of oxygen atoms to form a double bond with the central atom.
Note: Bromine atomis exceptional to the octet rule as it can hold more than 8 electrons in its outermost shell. It is also called an expanded octet.
Expanded octet: A case where an atom shares more than eight electrons with its bonding partners.
As you see, we reduced the formal charges, and the overall formal charge in the above figure is -1. Since, BrO2– has one negative ion as well, therefore, the entire structure should have a -1 formal charge.
BrO2– lewis structure
As BrO2– molecule contains one negative ion also, so, we need to put the bracket around the BrO2– lewis structure and show a negative ion outside the bracket.
Also check –
- Formal charge calculator
- Lewis structure calculator
- How to draw a lewis structure?
The molecular geometry of BrO2– is Bent. The Bromine (Br) central atom is attached to two oxygen (O) atoms, also, it contains two lone pairs as well. According to VSEPR theory, it will form an AX2N2 type molecule, which implies its molecular geometry will be bent.
A represent the central atom
X represents the number of bonded atom to the central atom
N represent the lone pair on the central atom
The electron geometry for BrO2– will be tetrahedral. Since the Bromine central atom has 4 regions of electron density(2 lone pairs + 2 bonded atoms), it implies, its electron domain geometry will be tetrahedral.
Take a look at the VSEPR chart to understand the molecular and electron geometry of BrO2–.
Hybridization of BrO2–
Apply steric number formula to get the hybridization for BrO2– molecule.
“Steric number is the number of atoms bonded to a central atom of a molecule plus the number of lone pairs on the central atom”
∴ Steric number of Bromine in BrO2– molecule = (Number of bonded atoms attached to bromine + Lone pair on bromine)
∴ Steric number of BrO2– = (2 + 2) = 4
| Steric number | Hybridization |
| 1 | S |
| 2 | Sp |
| 3 | Sp² |
| 4 | Sp³ |
| 5 | Sp³d |
| 6 | Sp³d² |
So, for a steric number of four, we get the Sp3 hybridization for Bromine in BrO2– molecule.
