# CH3CH2Cl Lewis structure

CH3CH2Cl (chloroethane) has two carbon atoms, five hydrogen atoms, and one chlorine atom.

In the CH3CH2Cl Lewis structure, there is a single bond between the two carbon atoms. The left carbon atom is attached with three hydrogen atoms, and the right carbon atom is attached with two hydrogen atoms and one chlorine atom. And on the chlorine atom, there are three lone pairs.

Contents

## Steps

To properly draw the CH3CH2Cl 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, carbon lies in group 14, hydrogen lies in group 1, and chlorine lies in group 17.

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

Since CH3CH2Cl has two carbon atoms, five hydrogen atoms, and one chlorine atom, so…

Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of five hydrogen atoms = 1 × 5 = 5
Valence electrons of one chlorine atom = 7 × 1 = 7

And the total valence electrons = 8 + 5 + 7 = 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 chlorine. Place the least electronegative atom at the center.

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

Here, there are two carbon atoms, so we can assume any one as the central atom.

Let’s assume that the central atom is right carbon.

Therefore, place carbons in the center and hydrogen and chlorine 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 seven bonds are already marked. So we have to only mark the remaining three 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 chlorine is a period 3 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 chlorine. But no need to mark on hydrogen, because each hydrogen has already two electrons.

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

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

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

For chlorine 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 (right carbon) forms an octet. The outside atoms (left carbon and chlorine) also form an octet, and all hydrogens form a duet. Hence, the octet rule and duet rule are satisfied.

Therefore, this structure is the stable Lewis structure of CH3CH2Cl.

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.