The **Lewis structure of N _{2}F_{2}** contains one double bond and two single bonds, with two nitrogens in the center, and each nitrogen is attached with one fluorine. There is one lone pair on each nitrogen atom, and three lone pairs on each fluorine atom.

## Steps

By using the following steps, you can easily draw the Lewis structure of N_{2}F_{2}.

#1 Draw skeleton

#2 Show chemical bond

#3 Mark lone pairs

#4 Complete octet on central atom

#5 Calculate formal charge and check stability

Letâ€™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.

- Letâ€™s calculate the total number of valence electrons

We know thatâ€¦ nitrogen is a group 15 element and fluorine is a group 17 element. Hence, nitrogen has **five** valence electrons and fluorine has **seven** valence electrons.

Now N_{2}F_{2} has two nitrogen atoms and two fluorine atoms.

So the total number of valence electrons = (valence electrons of nitrogen atom Ã— 2) + (valence electrons of fluorine atom Ã— 2)

Therefore, the **total number of valence electrons** = 10 + 14 = 24

- 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 fluorine as follows:

Electronegativity value of nitrogen = 3.04

Electronegativity value of fluorine = 3.98

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

So now, put two nitrogens in the center and two fluorines on either side. And draw the rough skeleton structure for the Lewis structure of N_{2}F_{2} something like this:

**Also read:** How to draw Lewis structure of XeOF_{4} (5 steps)

### #2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since two nitrogens are surrounded by two fluorines, use six electrons to show **three chemical bonds** as follows:

### #3 Mark lone pairs

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

Two valence electrons represent one lone pair. So eighteen valence electrons = **nine lone pairs**.

Note that both (nitrogen and fluorine) 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 fluorines and right nitrogen, so each fluorine will get three lone pairs, and right nitrogen will get two lone pairs. And the central atom (left nitrogen) will get one lone pair.

So the Lewis structure of N_{2}F_{2} looks something like this:

In the above structure, you can see that the octet is completed on outside atoms. But, the central atom (left nitrogen) doesnâ€™t form an octet.

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

**Also read:** How to draw Lewis structure of BrCl_{3} (4 steps)

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

So convert one lone pair from the right nitrogen atom to make a new bond with the left nitrogen atom. And then, the Lewis structure of N_{2}F_{2} 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 CSe_{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 – Â½ 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 = 2

Bonding electrons = 6

Formal charge = 5 – 2 – Â½ (6) = 0

- For
**each fluorine**atom

Valence electrons = 7

Nonbonding electrons = 6

Bonding electrons = 2

Formal charge = 7 – 6 – Â½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of N_{2}F_{2} looks something like this:

In the above structure, you can see that the formal charges of both (nitrogen and fluorine) are zero. Therefore, this is the **stable Lewis structure of N _{2}F_{2}**.

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

## Related

- Lewis structure of XeOF
_{4} - Lewis structure of BrCl
_{3} - Lewis structure of CSe
_{2} - Lewis structure of BeBr
_{2} - Lewis structure of XeO
_{2}F_{2}

## External video

- N2F2 Lewis Structure: How to Draw the Lewis Structure for N2F2 – YouTube â€¢ Wayne Breslyn

## External links

- N2F2 Lewis Structure in 6 Steps (With Images) – Pediabay
- N2F2 Lewis Structure, Molecular Geometry, Hybridization & Shape – Geometry of Molecules
- N2F2 Lewis Structure, Geometry, Hybridization, and Polarity – Techiescientist
- Draw the Lewis Structure for N2F2 – Chegg
- N2F2 molecular geometry? – Reddit
- Write the Lewis structure for N2F2. Draw two isomers and determine whether they are polar or non-polar – Brainly
- VSEPR Theory and Predicting Shapes of N2F2 – Chemistry Stack Exchange
- Why does N2F2 have a double bond but N2F4 doesn’t? – Quora
- What is the Lewis structure of N
_{2}F_{2}? – Quizlet

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.