Lewis structure of HNO3

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Lewis structure of HNO3
Lewis structure of HNO3 | Image: Root Memory

The Lewis structure of HNO3 contains one double bond and three single bonds, with nitrogen in the center, and hydrogen and three oxygens on either side. The top oxygen atom and left oxygen atom have two lone pairs, the right oxygen atom has three lone pairs, and hydrogen atom and nitrogen atom do not have any lone pair.

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

Steps

By using the following steps, you can easily draw the Lewis structure of HNO3:

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Complete octet on central atom
#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… hydrogen is a group 1 element, nitrogen is a group 15 element, and oxygen is a group 16 element. Hence, hydrogen has one valence electron, nitrogen has five valence electrons, and oxygen has six valence electrons.

Now HNO3 has one hydrogen atom, one nitrogen atom, and three oxygen atoms.

So the total number of valence electrons = valence electrons of hydrogen atom + valence electrons of nitrogen atom + (valence electrons of oxygen atom × 3)

Therefore, the total number of valence electrons = 1 + 5 + 18 = 24

  • Now decide the central atom

We can not assume hydrogen as the central atom, because the central atom is bonded with at least two other atoms. And hydrogen has only one electron in its last shell, so it can not make more than one bond.

Therefore, choose the central atom from nitrogen and oxygen.

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 nitrogen is the central atom.

So now, put nitrogen in the center and hydrogen and oxygen on either side. And draw the rough skeleton structure for the Lewis structure of HNO3 something like this:

Skeleton structure for Lewis structure of HNO3 | Image: Root Memory

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since nitrogen is surrounded by hydrogen and oxygen, use eight electrons to show four chemical bonds as follows:

Four chemical bonds shown between atoms | Image: Root Memory

#3 Mark lone pairs

As calculated earlier, we have a total of 24 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, sixteen valence electrons are remaining.

Two valence electrons represent one lone pair. So sixteen valence electrons = eight lone pairs.

Note that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And 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 hydrogen and oxygen. But hydrogen can not keep more than 2 electrons in its last shell. Hence, don’t mark the lone pairs on hydrogen.

So top oxygen and right oxygen will get three lone pairs, and left oxygen will get two lone pairs. And the central atom (nitrogen) will not get any lone pair, because all eight lone pairs are used.

Now draw the Lewis structure of HNO3 something like this:

Lone pairs marked on Lewis structure of HNO3 | Image: Root Memory

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

So in the next step, we have to complete the octet on the central atom.

#4 Complete octet on central atom

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

Now nitrogen already has six valence electrons. Hence, nitrogen needs two more valence electrons to complete its octet.

So convert one lone pair from the top oxygen atom to make a new bond with the nitrogen atom. And then, the Lewis structure of HNO3 looks something like this:

Lone pair of top oxygen is converted, and octet is completed on atoms | Image: Root Memory

In the above structure, you can see that the octet is completed on the central atom (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.

#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 hydrogen atom

Valence electrons = 1
Nonbonding electrons = 0
Bonding electrons = 2

Formal charge = 1 – 0 – ½ (2) = 0

  • For nitrogen atom

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

Formal charge = 5 – 0 – ½ (8) = +1

  • For top oxygen and 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 HNO3 looks something like this:

Formal charges are calculated, and got the most stable Lewis structure of HNO3 | Image: Root Memory

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 HNO3.

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

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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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