The **Lewis structure of IF**_{4}** ^{–}** contains four single bonds, with iodine in the center, and four fluorines on either side. There are three lone pairs on each fluorine atom, and two lone pairs on the iodine atom.

Plus, there is a negative (-1) charge on the iodine atom.

## Steps

By using the following steps, you can easily draw the Lewis structure of IF_{4}^{–}.

#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… both iodine and fluorine are the group 17 elements. Hence, both iodine and fluorine have **seven** valence electrons.

Now IF_{4}^{–} has one iodine atom and four fluorine atoms.

So the total number of valence electrons = valence electrons of iodine atom + (valence electrons of fluorine atom × 4)

And IF_{4}^{–} has a negative (-1) charge, so we have to add one more electron.

Therefore, the **total number of valence electrons** = 7 + 28 + 1 = 36

- 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 iodine and fluorine as follows:

Electronegativity value of iodine = 2.66

Electronegativity value of fluorine = 3.98

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

So now, put iodine in the center and fluorines on either side. And draw the rough skeleton structure for the Lewis structure of IF_{4}^{–} something like this:

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

### #2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since iodine is surrounded by four fluorines, use eight electrons to show **four chemical bonds** as follows:

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

### #3 Mark lone pairs

As calculated earlier, we have a total of 36 valence electrons. And in the above structure, we have already used eight 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 iodine is a period 5 element, so it can keep more than 8 electrons in its last shell. And fluorine is a period 2 element, so it can not 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 fluorines, so each fluorine will get three lone pairs. And the central atom (iodine) will get two lone pairs.

So the Lewis structure of IF_{4}^{–} looks something like this:

In the above structure, you can see that the octet is completed on the central atom (iodine), 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 SF_{3}^{–} (4 steps)

### #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
**iodine**atom

Valence electrons = 7

Nonbonding electrons = 4

Bonding electrons = 8

Formal charge = 7 – 4 – ½ (8) = -1

- For
**each fluorine**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 IF_{4}^{–} 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 IF**_{4}** ^{–}**.

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

Now IF_{4}^{–} is an ion having a negative (-1) charge, so draw brackets around the above Lewis structure and mention that charge on the top right corner. And then, the Lewis structure of IF_{4}^{–} looks something like this:

## Related

- Lewis structure of ICl
_{3} - Lewis structure of ClF
_{5} - Lewis structure of SF
_{3}^{–} - Lewis structure of CH
_{3}CN - Lewis structure of NOF

## External links

- Drawing the Lewis Structure for IF4 – The Geoexchange
- IF4- Lewis structure: Drawings, Hybridization, Shape, Charges, Pairs – Lambda Geeks
- Lewis structure of IF4 ion – AceOrganicChem
- How can I predict the bond angles for IF4 – Socratic
- IF4- Lewis Structure (Iodine Tetrafluoride) – Pinterest
- What is the molecular shape of IF4? – Quora
- Draw the best Lewis structure of IF4 – Chegg
- What are the Lewis structure and molecular geometry for IF4-? – Pearson
- Is IF4- polar or nonpolar? – Homework.Study.com
- VSEPER shape for IF4- – Laurence Lavelle
- For the anion IF4- How many total valence electrons should the Lewis structure show? – Bartleby

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.