Lewis structure of N2O5

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The Lewis structure of N₂O₅ contains four single bonds and two double bonds, with one oxygen in the center, and two nitrogens on either side. Each nitrogen is attached with two other oxygens. The top oxygen atoms and the center oxygen atom have two lone pairs, the bottom oxygen atoms have three 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 bottom oxygen atoms.

Steps

By using the following steps, you can easily draw the Lewis structure of N₂O₅.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Complete octet on atoms by calculating formal charge
#5 Calculate formal charge again 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 N₂O₅ has two nitrogen atoms and five oxygen atoms.

So the total number of valence electrons = (valence electrons of nitrogen atom × 2) + (valence electrons of oxygen atom × 5)

Therefore, the total number of valence electrons = 10 + 30 = 40

• 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, draw the rough skeleton structure for the Lewis structure of N₂O₅ something like this:

Also read: How to draw Lewis structure of CHBr₃ (4 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Use twelve electrons to show six chemical bonds as follows:

#3 Mark lone pairs

As calculated earlier, we have a total of 40 valence electrons. And in the above structure, we have already used twelve valence electrons. Hence, twenty-eight valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty-eight valence electrons = fourteen 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 fourteen lone pairs are used.

So the Lewis structure of N₂O₅ 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 BrCl (4 steps)

#4 Complete octet on atoms by calculating formal charge

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.

Which means that we have to convert two lone pairs. But the question is, which lone pair should we convert to get the best Lewis structure?

There are three possible ways we can convert lone pair: 

1) Convert one lone pair from the top left oxygen atom, and one lone pair from the top right oxygen atom.
2) Convert one lone pair from the bottom left oxygen atom, and one lone pair from the bottom right oxygen atom.
3) Convert two lone pairs from the center oxygen atom.

The best way to predict which lone pair should be converted to get the best Lewis structure is by calculating formal charges.

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 = 6

Formal charge = 5 – 0 – ½ (6) = +2

• For top oxygen and bottom oxygen atoms

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

Formal charge = 6 – 6 – ½ (2) = -1

• For center oxygen atom

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 N₂O₅ looks something like this:

In order to get the best Lewis structure, we have to make the formal charges on atoms closer to zero. And we can do so by choosing the 1) or 2).

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

And then, the Lewis structure of N₂O₅ 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 BrCN (5 steps)

#5 Calculate formal charge again 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 = 4
Bonding electrons = 4

Formal charge = 6 – 4 – ½ (4) = 0

• For bottom oxygen atoms

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

Formal charge = 6 – 6 – ½ (2) = -1

• For center oxygen atom

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 N₂O₅ 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 N₂O₅.

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

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Related

• Lewis structure of CHBr₃
• Lewis structure of BrCl
• Lewis structure of BrCN
• Lewis structure of AsCl₃
• Lewis structure of XeH₄

External links

• Drawing the Lewis Structure for N2O5 – The Geoexchange
• N2O5 Lewis Structure in 5 Steps (With Images) – Pediabay
• N2O5 Lewis Structure, Resonance Structures | Dinitrogen pentoxide – Chemistry School
• Lewis Structure of N2O5 – Reddit
• What is the Lewis structure of “dinitrogen pentoxide”? – Socratic
• N2O5 Lewis Structure & Characteristics (13 Complete Facts) – Lambda Geeks
• Lewis Structure for N2O5 – Chemical Forums
• Draw a Lewis structure for nitrogen pentoxide (N2O5) in which each N is bonded to three O atoms – Chegg
• Write one Lewis structure of N2O5 – OneClass
• Draw a Lewis structure for nitrogen pentoxide (N2O5) in which each N is bonded to three O atoms – Numerade
• N2O5 (Dinitrogen Pentoxide) Oxidation Number – ChemicalAid
• What chemical bonds are found in N2O5? – Quora