Br2O Lewis structure

Br2O (bromoether) has two bromine atoms and one oxygen atom.

In Br2O Lewis structure, there are two single bonds around the oxygen atom, with two bromine atoms attached to it. Each bromine atom has three lone pairs, and the oxygen atom has two lone pairs.

Contents

Steps

Use these steps to correctly draw the Br2O 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 Br2O has two bromine atoms and one oxygen atom, so…

Valence electrons of two bromine atoms = 7 × 2 = 14
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 14 + 6 = 20

• Second, find the total electron pairs

We have a total of 20 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 = 20 ÷ 2 = 10

• Third, determine the central atom

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

Since bromine is less electronegative than oxygen, the central atom should be bromine, right?

But if we place bromine in the center and oxygen outside, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best Lewis structure.

Hence, here we have to assume that the central atom is oxygen.

Therefore, place oxygen in the center and bromines on either side.

• And finally, draw the rough sketch

#2 Mark lone pairs on the atoms

Here, we have a total of 10 electron pairs. And two O — Br bonds are already marked. So we have to only mark the remaining eight 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 atoms are bromines.

So for each bromine, there are three lone pairs, and for oxygen, there are two 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 each bromine atom, formal charge = 7 – 6 – ½ (2) = 0

For oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

Here, both bromine and oxygen atoms do not have charges, so no need to mark the charges.

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

Therefore, this structure is the stable Lewis structure of Br2O.