# IOF5 Lewis structure

IOF5 has one iodine atom, one oxygen atom, and five fluorine atoms.

In IOF5 Lewis structure, there is one double bond and five single bonds around the iodine atom, with one oxygen atom and five fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the oxygen atom has two lone pairs.

Contents

## Steps

Use these steps to correctly draw the IOF5 Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required
#4 Convert lone pairs of the atoms, and minimize formal charges
#5 Repeat step 4 if needed, until all charges are minimized, to get a stable Lewis structure

Let’s discuss each step in more detail.

### #1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, both iodine and fluorine lie in group 17, and oxygen lies in group 16.

Hence, both iodine and fluorine have seven valence electrons, and oxygen has six valence electrons.

Since IOF5 has one iodine atom, one oxygen atom, and five fluorine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of one oxygen atom = 6 × 1 = 6
Valence electrons of five fluorine atoms = 7 × 5 = 35

And the total valence electrons = 7 + 6 + 35 = 48

• Second, find the total electron pairs

We have a total of 48 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 = 48 ÷ 2 = 24

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 24 electron pairs. And six bonds are already marked. So we have to only mark the remaining eighteen 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 both (oxygen and fluorine) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.

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

So for oxygen and each fluorine, there are three lone pairs, and for iodine, there is zero lone pair because all eighteen electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Calculate and mark formal charges on the atoms, if required

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 – 0 – ½ (12) = +1

For oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

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

Here, both iodine and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both iodine and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Convert lone pairs of the atoms, and minimize formal charges

Convert a lone pair of the oxygen atom to make a new I — O bond with the iodine atom as follows:

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

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of IOF5.

Next: CF3 Lewis structure