# IF4- Lewis structure

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IF4 (iodine tetrafluoride) has one iodine atom and four fluorine atoms.

In the IF4 Lewis structure, there are four single bonds around the iodine atom, with four fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the iodine atom has two lone pairs.

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

Contents

## Steps

Here’s how you can easily draw the IF4 Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

### #1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, both iodine and fluorine lie in group 17.

Hence, both iodine and fluorine have seven valence electrons.

Since IF4 has one iodine atom and four fluorine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of four fluorine atoms = 7 × 4 = 28

Now the IF4 has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 28 + 1 = 36

• Second, find the total electron pairs

We have a total of 36 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 36 ÷ 2 = 18

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since iodine is less electronegative than fluorine, assume that the central atom is iodine.

Therefore, place iodine in the center and fluorines on either side.

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 18 electron pairs. And four I — F bonds are already marked. So we have to only mark the remaining fourteen electron pairs as lone pairs on the sketch.

Also remember 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.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.

So for each fluorine, there are three lone pairs, and for iodine, there are two lone pairs.

Mark the lone pairs on the sketch as follows:

### #3 If needed, mention formal charges on the atoms

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For iodine atom, formal charge = 7 – 4 – ½ (8) = -1

For each fluorine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, the iodine atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (iodine) forms an octet. And the outside atoms (fluorines) also form an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on the iodine atom.

This is not okay, right? Because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is fluorine.

But if we convert a lone pair of the iodine atom to make a new I — F bond with the fluorine atom, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best Lewis structure.

Therefore, this structure is the most stable Lewis structure of IF4.

And since the IF4 has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: SF3 Lewis structure