# C3H4 Lewis structure

C3H4 (ethylene) has three carbon atoms and four hydrogen atoms.

In C3H4 Lewis structure, there is one triple bond and one single bond between the three carbon atoms. The left carbon is attached with one hydrogen atom, and the right carbon is attached with three hydrogen atoms. And none of the atoms has a lone pair.

Contents

## Steps

Here’s how you can easily draw the C3H4 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
#4 Minimize formal charges by converting lone pairs of the atoms, and try to get a stable Lewis structure
#5 Repeat step 4 again if needed, until all charges are minimized

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 hydrogen lies in group 1.

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

Since C3H4 has three carbon atoms and four hydrogen atoms, so…

Valence electrons of three carbon atoms = 4 × 3 = 12
Valence electrons of four hydrogen atoms = 1 × 4 = 4

And the total valence electrons = 12 + 4 = 16

• Second, find the total electron pairs

We have a total of 16 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 = 16 ÷ 2 = 8

• 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 there are three atoms remaining and all three atoms are carbon, so we can assume any one as the central atom.

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

Therefore, place carbons in the center and hydrogens on either side.

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 8 electron pairs. And six bonds are already marked. So we have to only mark the remaining two 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 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, left carbon, and right carbon. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for left carbon, there are two lone pairs, and for other two carbons, there is zero lone pair because all two 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 left carbon atom, formal charge = 4 – 4 – ½ (4) = -2

For center carbon atom, formal charge = 4 – 0 – ½ (4) = +2

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

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

Here, both left carbon and center carbon atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both left carbon and center carbon 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 left carbon atom to make a new C — C bond with the center carbon atom as follows:

### #5 Since there are charges on atoms, repeat step 4 again

Since there are charges on carbon atoms, again convert a lone pair of the left carbon atom to make a new C — C bond with the center carbon atom as follows:

In the above structure, you can see that the central atom (center carbon) forms an octet. The outside atoms (left carbon and right carbon) also form an octet, and all hydrogens form a duet. Hence, the octet rule and duet rule are satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of C3H4.

Next: PI3 Lewis structure