# CH2I2 Lewis structure

CH2I2 (diiodomethane) has one carbon atom, two hydrogen atoms, and two iodine atoms.

In CH2I2 Lewis structure, there are four single bonds around the carbon atom, with two hydrogen atoms and two iodine atoms attached to it, and on each iodine atom, there are three lone pairs.

Contents

## Steps

Use these steps to correctly draw the CH2I2 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

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, carbon lies in group 14, hydrogen lies in group 1, and iodine lies in group 17.

Hence, carbon has four valence electrons, hydrogen has one valence electron, and iodine has seven valence electrons.

Since CH2I2 has one carbon atom, two hydrogen atoms, and two iodine atoms, so…

Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of two hydrogen atoms = 1 × 2 = 2
Valence electrons of two iodine atoms = 7 × 2 = 14

And the total valence electrons = 4 + 2 + 14 = 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

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from carbon and iodine. 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 hydrogen and iodine on either side.

• And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 10 electron pairs. And four bonds are already marked. So we have to only mark the remaining six 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. Hydrogen is a period 1 element, so it can not keep more than 2 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 hydrogens and iodines. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for each iodine, there are three lone pairs, and for carbon, there is zero lone pair because all six 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 carbon atom, formal charge = 4 – 0 – ½ (8) = 0

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

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

Here, the 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. The outside atoms (iodines) also form an octet, and each hydrogen forms a duet. Hence, the octet rule and duet rule are satisfied.

Therefore, this structure is the stable Lewis structure of CH2I2.

Next: GaI3 Lewis structure