Lewis structure of NI3

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Lewis structure of NI3
Lewis structure of NI3 | Image: Root Memory

The Lewis structure of NI3 contains three single bonds, with nitrogen in the center, and three iodines on either side. There are three lone pairs on each iodine atom, and one lone pair on the nitrogen atom.

Steps

By using the following steps, you can easily draw the Lewis structure of NI3.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

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 iodine is a group 17 element. Hence, nitrogen has five valence electrons and iodine has seven valence electrons.

Now NI3 has one nitrogen atom and three iodine atoms.

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

Therefore, the total number of valence electrons = 5 + 21 = 26

  • 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 iodine as follows:

Electronegativity value of nitrogen = 3.04
Electronegativity value of iodine = 2.66

Obviously, nitrogen is less electronegative than iodine. Hence, assume that nitrogen is the central atom.

So now, put nitrogen in the center and iodines on either side. And draw the rough skeleton structure for the Lewis structure of NI3 something like this:

Skeleton structure for Lewis structure of NI3 | Image: Root Memory

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since nitrogen is surrounded by three iodines, use six electrons to show three chemical bonds as follows:

Three chemical bonds shown between atoms | Image: Root Memory

#3 Mark lone pairs

As calculated earlier, we have a total of 26 valence electrons. And in the above structure, we have already used six valence electrons. Hence, twenty valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty valence electrons = ten lone pairs.

Note that nitrogen is a period 2 element, so it can not keep more than 8 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its 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 iodines, so each iodine will get three lone pairs. And the central atom (nitrogen) will get one lone pair.

So the Lewis structure of NI3 looks something like this:

Lone pairs marked on Lewis structure of NI3 | 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.

#4 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 nitrogen atom

Valence electrons = 5
Nonbonding electrons = 2
Bonding electrons = 6

Formal charge = 5 – 2 – ½ (6) = 0

  • For each iodine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of NI3 looks something like this:

Formal charges are calculated, and got the stable Lewis structure of NI3 | Image: Root Memory

In the above structure, you can see that the formal charges of both (nitrogen and iodine) are zero. Therefore, this is the stable Lewis structure of NI3.

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

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External links

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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