C2H4 Lewis structure

C2H4 Lewis Structure
C2H4 Lewis structure | Image: Learnool

C2H4 (ethylene or ethene) has two carbon atoms and four hydrogen atoms.

In the C2H4 Lewis structure, there is a double bond between the two carbon atoms, and each carbon is attached with one hydrogen atom, and none of the atoms has a lone pair.

Rough sketch

  • First, determine the total number of valence electrons
Periodic table | Image: Learnool

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 C2H4 has two carbon atoms and four hydrogen atoms, so…

Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of four hydrogen atoms = 1 × 4 = 4

And the total valence electrons = 8 + 4 = 12

  • Second, find the total electron pairs

We have a total of 12 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 = 12 ÷ 2 = 6

  • 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 only two atoms remaining and both atoms are carbon, 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 hydrogens on either side.

  • And finally, draw the rough sketch
C2H4 Lewis Structure (Step 1)
Rough sketch of C2H4 Lewis structure | Image: Learnool

Lone pair

Here, we have a total of 6 electron pairs. And five bonds are already marked. So we have to only mark the remaining one electron pair as a lone pair 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 and left carbon. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for left carbon, there is one lone pair, and for right carbon, there is zero lone pair because all electron pairs are over.

Mark the lone pairs on the sketch as follows:

C2H4 Lewis Structure (Step 2)
Lone pairs marked on C2H4 Lewis structure | Image: Learnool

Formal charge

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 – 2 – ½ (6) = -1

For right carbon atom, formal charge = 4 – 0 – ½ (6) = +1

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

← Scroll horizontally to view all columns →
AtomValence electronsNon-bonding electronsBonding electronsFormal charge
Carbon (left)426-1
Carbon (right)406+1
Hydrogen (all)1020

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

C2H4 Lewis Structure (Step 3)
Formal charges marked on C2H4 Lewis structure | Image: Learnool

The above structure is not a stable Lewis structure because both carbon atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

Convert a lone pair of the left carbon atom to make a new C — C bond with the right carbon atom as follows:

C2H4 Lewis Structure (Step 4)
Lone pair of left carbon is converted, and got the stable Lewis structure of C2H4 | Image: Learnool

Final structure

The final structure of C2H4 involves two central carbon atoms connected to each other by a double covalent bond. In this arrangement, each carbon atom is also single-bonded to two hydrogen atoms. This configuration allows both carbon atoms to satisfy the octet rule by sharing four pairs of electrons total, while each hydrogen atom reaches its stable duet state. Within this layout, there are no lone pairs present on any of the atoms. This configuration is the most stable because it results in formal charges of zero for every atom in the molecule, representing the most energetically favorable state. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of ethylene (ethene).

Next: SF4 Lewis structure

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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.

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