# N2O3 Lewis structure

N2O3 (dinitrogen trioxide) has two nitrogen atoms and three oxygen atoms.

In the N2O3 Lewis structure, there is a single bond between the two nitrogen atoms. The left nitrogen is attached with two oxygen atoms, and the right nitrogen is attached with one oxygen atom. One oxygen atom with a single bond has three lone pairs, and the two oxygen atoms with double bonds have two lone pairs.

Also, there is a negative (-1) charge on the oxygen atom with a single bond, and a positive (+1) charge on the left nitrogen atom.

Contents

## Steps

Use these steps to correctly draw the N2O3 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
#4 Convert lone pairs of the atoms, and minimize formal charges
#5 Repeat step 4 if needed, until all charges are minimized, to get a stable Lewis structure

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, nitrogen lies in group 15, and oxygen lies in group 16.

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

Since N2O3 has two nitrogen atoms and three oxygen atoms, so…

Valence electrons of two nitrogen atoms = 5 × 2 = 10
Valence electrons of three oxygen atoms = 6 × 3 = 18

And the total valence electrons = 10 + 18 = 28

• Second, find the total electron pairs

We have a total of 28 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 = 28 ÷ 2 = 14

• 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 Mark lone pairs on the atoms

Here, we have a total of 14 electron pairs. And four bonds are already marked. So we have to only mark the remaining ten 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 right nitrogen, there is one lone pair.

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 left nitrogen atom, formal charge = 5 – 0 – ½ (6) = +2

For right 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 Convert lone pairs of the atoms, and minimize formal charges

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

### #5 Repeating step 4 to get a stable Lewis structure

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.

Now there are still charges on the atoms. But we can not convert a lone pair to a bond because nitrogen can not keep more than 8 electrons in its last shell.

The formal charges on atoms are closer to zero. Also, the above structure is more stable than the previous structures. Therefore, this structure is the most stable Lewis structure of N2O3.