# IO2- Lewis structure

IO2 (iodite) has one iodine atom and two oxygen atoms.

In IO2 Lewis structure, there is one single bond and one double bond around the iodine atom, with two oxygen atoms attached to it. The oxygen atom with a single bond has three lone pairs, the oxygen atom with a double bond has two lone pairs, and the iodine atom also has two lone pairs.

Also, there is a negative (-1) charge on the oxygen atom with a single bond.

Contents

## Steps

To properly draw the IO2 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
#4 Minimize formal charges by converting lone pairs of the atoms
#5 Repeat step 4 if necessary, until all charges are minimized

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, iodine lies in group 17, and oxygen lies in group 16.

Hence, iodine has seven valence electrons and oxygen has six valence electrons.

Since IO2 has one iodine atom and two oxygen atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of two oxygen atoms = 6 × 2 = 12

Now the IO2 has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 12 + 1 = 20

• Second, find the total electron pairs

We have a total of 20 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 = 20 ÷ 2 = 10

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 10 electron pairs. And two I — O bonds are already marked. So we have to only mark the remaining eight 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 oxygen 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 oxygens.

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

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 – 4 – ½ (4) = +1

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

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 Minimize formal charges by converting lone pairs of the atoms

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 (oxygens) also form an octet. Hence, the octet rule is satisfied.

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

This is okay, 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 oxygen.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the most stable Lewis structure of IO2.

And since the IO2 has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows: