Lewis structure of N2O4

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The Lewis structure of N₂O₄ contains three single bonds and two double bonds, with two nitrogens in the center, and each nitrogen is attached with two oxygens. The top oxygen atoms have three lone pairs, the bottom oxygen atoms have two lone pairs, and the nitrogen atoms do not have any lone pair.

Plus, there is a positive (+1) charge on nitrogen atoms, and a negative (-1) charge on top oxygen atoms.

Steps

By using the following steps, you can easily draw the Lewis structure of N₂O₄.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Complete octet on atoms
#5 Calculate formal charge and check stability

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… nitrogen is a group 15 element and oxygen is a group 16 element. Hence, nitrogen has five valence electrons and oxygen has six valence electrons.

Now N₂O₄ has two nitrogen atoms and four oxygen atoms.

So the total number of valence electrons = (valence electrons of nitrogen atom × 2) + (valence electrons of oxygen atom × 4)

Therefore, the total number of valence electrons = 10 + 24 = 34

• 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 nitrogen and oxygen as follows:

Electronegativity value of nitrogen = 3.04
Electronegativity value of oxygen = 3.44

Obviously, nitrogen is less electronegative than oxygen. Hence, assume that left nitrogen is the central atom (as there are two nitrogens).

So now, put two nitrogens in the center and four oxygens on either side. And draw the rough skeleton structure for the Lewis structure of N₂O₄ something like this:

Also read: How to draw Lewis structure of PF₆^– (4 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since two nitrogens are surrounded by four oxygens, use ten electrons to show five chemical bonds as follows:

#3 Mark lone pairs

As calculated earlier, we have a total of 34 valence electrons. And in the above structure, we have already used ten valence electrons. Hence, twenty-four valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty-four valence electrons = twelve lone pairs.

Note that both (nitrogen and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their 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 and right nitrogen, so each oxygen will get three lone pairs. And right nitrogen and central atom (left nitrogen) will not get any lone pair, because all twelve lone pairs are used.

So the Lewis structure of N₂O₄ looks something like this:

In the above structure, you can see that the octet is completed on outside atoms (except right nitrogen). Also, the central atom (left nitrogen) doesn’t form an octet.

So in the next step, we have to complete the octet on left nitrogen and right nitrogen.

Also read: How to draw Lewis structure of XeO₄ (5 steps)

#4 Complete octet on atoms

Remember that nitrogen is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Now both (left nitrogen and right nitrogen) already have six valence electrons. Hence, both nitrogens need two more valence electrons to complete their octet.

So convert one lone pair from the left oxygen atom to make a new bond with the left nitrogen atom, and one lone pair from the right oxygen atom to make a new bond with the right nitrogen atom.

And then, the Lewis structure of N₂O₄ looks something like this:

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

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

Also read: How to draw Lewis structure of CO₃^2- (5 steps)

#5 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 each nitrogen atom

Valence electrons = 5
Nonbonding electrons = 0
Bonding electrons = 8

Formal charge = 5 – 0 – ½ (8) = +1

• For top oxygen atoms

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

Formal charge = 6 – 6 – ½ (2) = -1

• For bottom oxygen atoms

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

Formal charge = 6 – 4 – ½ (4) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of N₂O₄ looks something like this:

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 N₂O₄.

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

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Related

• Lewis structure of PF₆^–
• Lewis structure of XeO₄
• Lewis structure of CO₃^2-
• Lewis structure of SO₄^2-
• Lewis structure of SiS₂

External links

• Chemical Bonding: N2O4 Lewis Structure – The Geoexchange
• N2O4 Lewis Structure, Molecular Geometry, Hybridization, MO Diagram, and Polarity – Techiescientist
• Lewis Structure of N2O4 (Dinitrogen tetroxide) – Drawing Steps – Chemistry School
• N2O4 Lewis Structure in 5 Steps (With Images) – Pediabay
• N2O4 lewis structure, molecular geometry, bond angle, polarity – Topblogtenz
• Draw the Lewis structure for N2O4. What is the geometry of the central atom? Is it polar or nonpolar? – Homework.Study.com
• Draw the Lewis structure of N2O4 and then choose the appropriate pair of hybridization states for the two central atoms – Brainly
• N2O4 Lewis Structure – Laurence Lavelle
• Draw a lewis structure for N2O4 – Chegg
• What is the shape of N2O4? – Quora
• 3.55 Write the Lewis structure of nitrogen tetroxide, N2O4 – TutorBin
• Draw valid Lewis structures for N2O4 in which you vary the connectivity of the six atoms – Numerade