# CI4 Lewis structure

CI4 (carbon tetraiodide) has one carbon atom and four iodine atoms.

In CI4 Lewis structure, there are four single bonds around the carbon atom, with four iodine atoms attached to it, and on each iodine atom, there are three lone pairs.

Contents

## Steps

Here’s how you can easily draw the CI4 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, carbon lies in group 14, and iodine lies in group 17.

Hence, carbon has four valence electrons and iodine has seven valence electrons.

Since CI4 has one carbon atom and four iodine atoms, so…

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

And the total valence electrons = 4 + 28 = 32

• Second, find the total electron pairs

We have a total of 32 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 = 32 ÷ 2 = 16

• Third, determine the central atom

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

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

Therefore, place carbon in the center and iodines on either side.

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 16 electron pairs. And four C — I bonds are already marked. So we have to only mark the remaining twelve electron pairs as lone pairs on the sketch.

Also remember that carbon is a period 2 element, so it can not keep more than 8 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

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

So for each iodine, there are three lone pairs, and for carbon, there is zero lone pair because all twelve electron pairs are over.

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 carbon atom, formal charge = 4 – 0 – ½ (8) = 0

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

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

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

Therefore, this structure is the stable Lewis structure of CI4.

Next: BI3 Lewis structure