# N2O2 Lewis structure

N2O2 (dinitrogen dioxide) has two nitrogen atoms and two oxygen atoms.

In N2O2 Lewis structure, there is a single bond between the two nitrogen atoms, and each nitrogen is attached with one oxygen atom, making a double bond with them. There are two lone pairs on each oxygen atom, and one lone pair on each nitrogen atom.

Contents

## Steps

To properly draw the N2O2 Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary
#4 Minimize formal charges by converting lone pairs of the atoms
#5 Repeat step 4 if necessary, until all charges are minimized

Let’s break down each step in more detail.

### #1 Draw a rough sketch of the structure

• First, determine the total number of valence electrons

In the periodic table, nitrogen lies in group 15, and oxygen lies in group 16.

Hence, nitrogen has five valence electrons and oxygen has six valence electrons.

Since N2O2 has two nitrogen atoms and two oxygen atoms, so…

Valence electrons of two nitrogen atoms = 5 × 2 = 10
Valence electrons of two oxygen atoms = 6 × 2 = 12

And the total valence electrons = 10 + 12 = 22

• Second, find the total electron pairs

We have a total of 22 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 = 22 ÷ 2 = 11

• Third, determine the central atom

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

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

Here, there are two nitrogen atoms, so we can assume any one as the central atom.

Let’s assume that the central atom is left nitrogen.

Therefore, place nitrogens in the center and oxygens on either side.

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 11 electron pairs. And three bonds are already marked. So we have to only mark the remaining eight electron pairs as lone pairs on the sketch.

Also remember that both (nitrogen and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygens and right nitrogen.

So for each oxygen, there are three lone pairs, and for left nitrogen and right nitrogen, there is one lone pair.

Mark the lone pairs on the sketch as follows:

### #3 Indicate formal charges on the atoms, if necessary

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

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

For each nitrogen atom, formal charge = 5 – 2 – ½ (4) = +1

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

Here, both nitrogen and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both nitrogen and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Minimize formal charges by converting lone pairs of the atoms

Convert a lone pair of the left oxygen atom to make a new N — O bond with the left nitrogen atom as follows:

### #5 Repeat step 4 (minimize charges again)

Since there are charges on nitrogen and oxygen atoms, again convert a lone pair of the right oxygen atom to make a new N — O bond with the right nitrogen atom as follows:

In the above structure, you can see that the central atom (left nitrogen) forms an octet. And the outside atoms (right nitrogen and oxygens) also form an octet. Hence, the octet rule is satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of N2O2.

Next: BrF Lewis structure