BrO- Lewis structure

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.

Alternative method: Lewis structure of BrO^–

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

Lone pair

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:

Formal charge

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:

Final structure

The final structure of BrO^– involves a central bromine atom connected to a single oxygen atom through a single covalent bond. Within this layout, both the bromine and oxygen atoms satisfy the octet rule. The bromine atom retains three lone pairs, while the oxygen atom similarly maintains three lone pairs alongside the single shared bonding pair. This arrangement is the most stable because it results in the most favorable formal charge distribution; the oxygen atom carries a formal charge of -1, while the bromine atom carries a formal charge of zero. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of the hypobromite ion.

To complete the representation, draw square brackets around the entire Lewis structure and place a “-” or “-1” sign as a superscript outside the upper right bracket. This notation signifies that the negative charge is a property of the whole ion.

Next: HN₃ Lewis structure

External video

• BrO- Lewis Structure: How to Draw the Lewis Structure for BrO- – YouTube • 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/