# IF5 Lewis structure

IF5 (iodine pentafluoride) has one iodine atom and five fluorine atoms.

In the IF5 Lewis structure, there are five single bonds around the iodine atom, with five fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the iodine atom has one lone pair.

Contents

## Steps

To properly draw the IF5 Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

### #1 Draw a rough sketch of the 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 IF5 has one iodine atom and five fluorine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of five fluorine atoms = 7 × 5 = 35

And the total valence electrons = 7 + 35 = 42

• Second, find the total electron pairs

We have a total of 42 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 = 42 ÷ 2 = 21

• 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 Next, indicate lone pairs on the atoms

Here, we have a total of 21 electron pairs. And five I — F bonds are already marked. So we have to only mark the remaining sixteen 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 is one lone pair.

Mark the lone pairs on the sketch as follows:

### #3 Indicate formal charges on the atoms, if necessary

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 – 2 – ½ (10) = 0

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

Here, both iodine and fluorine atoms do not have charges, so no need to mark the charges.

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.

Therefore, this structure is the stable Lewis structure of IF5.

Next: HNO3 Lewis structure