BrO- Lewis structure

The information on this page is ✔ fact-checked.

BrO^– (hypobromite) has one bromine atom and one oxygen atom.

In BrO^– Lewis structure, there is a single bond between the bromine and oxygen atom, and on both bromine and oxygen atoms, there are three lone pairs.

Also, there is a negative (-1) charge on the bromine atom.

Steps

Use these steps to correctly draw the BrO^– Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

#1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, bromine lies in group 17, and oxygen lies in group 16.

Hence, bromine has seven valence electrons and oxygen has six valence electrons.

Since BrO^– has one bromine atom and one oxygen atom, so…

Valence electrons of one bromine atom = 7 × 1 = 7
Valence electrons of one oxygen atom = 6 × 1 = 6

Now the BrO^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 6 + 1 = 14

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

• Second, find the total electron pairs

We have a total of 14 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 = 14 ÷ 2 = 7

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since bromine is less electronegative than oxygen, assume that the central atom is bromine.

• And finally, draw the rough sketch

#2 Mark lone pairs on the atoms

Here, we have a total of 7 electron pairs. And one Br — O bond is already marked. So we have to only mark the remaining six electron pairs as lone pairs on the sketch.

Also remember that bromine is a period 4 element, so it can keep more than 8 electrons in its last shell. And oxygen 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 oxygen.

So for each atom, there are three lone pairs.

Mark the lone pairs on the sketch as follows:

#3 Calculate and mark formal charges on the atoms, if required

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

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

For oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

Here, the oxygen atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (bromine) forms an octet. And the outside atom (oxygen) also forms an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on the oxygen atom.

This is okay, because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is oxygen.

Therefore, this structure is the most stable Lewis structure of BrO^–.

And since the BrO^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: HN₃ Lewis structure

📝 Your feedback matters. Visit our contact page.

External video

• BrO- Lewis Structure: How to Draw the Lewis Structure for BrO- – Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/BrO–lewis-structure.html
• https://lambdageeks.com/bro-lewis-structure/
• https://socratic.org/questions/how-do-you-draw-the-lewis-structure-for-the-hypobromite-ion-bro
• https://www.numerade.com/ask/question/draw-the-best-most-probable-lewis-structure-for-bro_-and-determine-the-formal-charge-on-bromine-for-full-marks-show-your-work-1-draw-lewis-structures-for-each-resonance-structure-only-includ-15813/