# C2Br4 Lewis Structure

C2Br4 (tetrabromoethylene) has two carbon atoms and four bromine atoms. In the lewis structure of C2Br4, there is a double bond between the two carbon atoms, and each carbon is attached with two bromine atoms, and on each bromine atom, there are three lone pairs.

## Steps

Here’s how you can draw the C2Br4 lewis structure step by step.

Step #1: draw sketch
Step #2: mark lone pairs
Step #3: mark charges
Step #4: minimize charges
Step #5: minimize charges again (if there are)

Let’s break down each step in detail.

### #1 Draw Sketch

• First, determine the total number of valence electrons

In the periodic table, carbon lies in group 14, and bromine lies in group 17.

Hence, carbon has four valence electrons and bromine has seven valence electrons.

Since C2Br4 has two carbon atoms and four bromine atoms, so…

Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of four bromine atoms = 7 × 4 = 28

And the total valence electrons = 8 + 28 = 36

• Second, find the total electron pairs

We have a total of 36 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 = 36 ÷ 2 = 18

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since carbon is less electronegative than bromine, 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 left carbon.

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

• And finally, draw the rough sketch

### #2 Mark Lone Pairs

Here, we have a total of 18 electron pairs. And five bonds are already marked. So we have to only mark the remaining thirteen 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 bromine is a period 4 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 bromines and right carbon.

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

Mark the lone pairs on the sketch as follows:

### #3 Mark Charges

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 – 0 – ½ (6) = +1

For right carbon atom, formal charge = 4 – 2 – ½ (6) = -1

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

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

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.

### #4 Minimize Charges

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

In the above structure, you can see that the central atom (left carbon) forms an octet. Hence, the octet rule is satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable lewis structure of C2Br4.