Lewis structure of BrO2-

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Lewis structure of BrO2-
Lewis structure of BrO2

The Lewis structure of BrO2 contains one double bond and one single bond, with bromine in the center, and two oxygens on either side. The left oxygen atom has two lone pairs, the right oxygen atom has three lone pairs, and the bromine atom also has two lone pairs.

Plus, there is a negative (-1) charge on the right oxygen atom.

Steps

By using the following steps, you can easily draw the Lewis structure of BrO2.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)
#5 Convert lone pair and calculate formal charge again (if formal charges are not closer to zero)

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

  • Let’s calculate the total number of valence electrons

We know that… bromine is a group 17 element and oxygen is a group 16 element. Hence, bromine has seven valence electrons and oxygen has six valence electrons.

Now BrO2 has one bromine atom and two oxygen atoms.

So the total number of valence electrons = valence electrons of bromine atom + (valence electrons of oxygen atom × 2)

And BrO2 has a negative (-1) charge, so we have to add one more electron.

Therefore, the total number of valence electrons = 7 + 14 + 1 = 22

  • Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for bromine and oxygen as follows:

Electronegativity value of bromine = 2.96
Electronegativity value of oxygen = 3.44

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

So now, put bromine in the center and oxygens on either side. And draw the rough skeleton structure for the Lewis structure of BrO2 something like this:

Skeleton structure for Lewis structure of BrO2

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since bromine is surrounded by two oxygens, use four electrons to show two chemical bonds as follows:

Two chemical bonds shown between atoms

#3 Mark lone pairs

As calculated earlier, we have a total of 22 valence electrons. And in the above structure, we have already used four valence electrons. Hence, eighteen valence electrons are remaining.

Two valence electrons represent one lone pair. So eighteen valence electrons = nine lone pairs.

Note that bromine is 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.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are oxygens, so each oxygen will get three lone pairs. And the central atom (bromine) will get two lone pairs.

So the Lewis structure of BrO2 looks something like this:

Lone pairs marked on Lewis structure of BrO2

In the above structure, you can see that the octet is completed on the central atom (bromine), and also on the outside atoms. Therefore, the octet rule is satisfied.

Now calculate the formal charge and check the stability of the above structure.

#4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

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

Collect the data from the above structure and then, write it down below as follows:

  • For bromine atom

Valence electrons = 7
Nonbonding electrons = 4
Bonding electrons = 4

Formal charge = 7 – 4 – ½ (4) = +1

  • For each oxygen atom

Valence electrons = 6
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 6 – 6 – ½ (2) = -1

Mention the formal charges of atoms on the structure. So the Lewis structure of BrO2 looks something like this:

Formal charges are not closer to zero

In the above structure, you can see that the formal charges of atoms are not closer to zero. Therefore, convert lone pair and calculate formal charge again.

#5 Convert lone pair and calculate formal charge again

As mentioned earlier, bromine is a period 4 element, so it can keep more than 8 electrons in its last shell.

So convert one lone pair from one oxygen atom to make a new bond with the bromine atom. And then, the Lewis structure of BrO2 looks something like this:

Lone pair of left oxygen is converted, and octet is completed on atoms

Now one last thing we need to do is, calculate the formal charge again and check the stability of the above structure.

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

Collect the data from the above structure and then, write it down below as follows:

  • For bromine atom

Valence electrons = 7
Nonbonding electrons = 4
Bonding electrons = 6

Formal charge = 7 – 4 – ½ (6) = 0

  • For left oxygen atom

Valence electrons = 6
Nonbonding electrons = 4
Bonding electrons = 4

Formal charge = 6 – 4 – ½ (4) = 0

  • For right oxygen atom

Valence electrons = 6
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 6 – 6 – ½ (2) = -1

Mention the formal charges of atoms on the structure. So the Lewis structure of BrO2 looks something like this:

Formal charges are calculated, and got the most stable Lewis structure of BrO2

In the above structure, you can see that the formal charges of atoms are closer to zero. Therefore, this is the most stable Lewis structure of BrO2.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

Now BrO2 is an ion having a negative (-1) charge, so draw brackets around the above Lewis structure and mention that charge on the top right corner. And then, the Lewis structure of BrO2 looks something like this:

Lewis structure of BrO2 showing a negative (-1) charge

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Deep

Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

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