# Ethanol Lewis structure

CH3CH2OH or C2H5OH or C2H6O (ethanol) has two carbon atoms, six hydrogen atoms, and one oxygen atom.

In the ethanol Lewis structure, there are five C — H bonds, one C — C bond, one C — O bond, and one O — H bond. And on the oxygen atom, there are two lone pairs.

Contents

## Steps

To properly draw the ethanol 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 oxygen lies in group 16.

Hence, carbon has four valence electrons, hydrogen has one valence electron, and oxygen has six valence electrons.

Since ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom, so…

Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of six hydrogen atoms = 1 × 6 = 6
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 8 + 6 + 6 = 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 oxygen. Place the least electronegative atom at the center.

Since carbon is less electronegative than oxygen, 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 oxygen 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 eight bonds are already marked. So we have to only mark the remaining two electron pairs as lone pairs on the sketch.

Also remember that both (carbon and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens and oxygens. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for oxygen, there are two lone pairs, and for carbon, there is zero lone pair because all two 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 oxygen atom, formal charge = 6 – 4 – ½ (4) = 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 oxygen) 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 ethanol.

Next: BeF2 Lewis structure