# C2H2Br2 Lewis structure

C2H2Br2 (1,2-dibromoethylene) has two carbon atoms, two hydrogen atoms, and two bromine atoms.

In the C2H2Br2 Lewis structure, there is a double bond between the two carbon atoms, and each carbon is attached with one hydrogen atom and one bromine atom, and on each bromine atom, there are three lone pairs.

Contents

## Steps

Here’s how you can easily draw the C2H2Br2 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, hydrogen lies in group 1, and bromine lies in group 17.

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

Since C2H2Br2 has two carbon atoms, two hydrogen atoms, and two bromine atoms, so…

Valence electrons of two carbon atoms = 4 × 2 = 8
Valence electrons of two hydrogen atoms = 1 × 2 = 2
Valence electrons of two bromine atoms = 7 × 2 = 14

And the total valence electrons = 8 + 2 + 14 = 24

• Second, find the total electron pairs

We have a total of 24 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 = 24 ÷ 2 = 12

• 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 bromine. 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 hydrogen and bromine on either side.

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 12 electron pairs. And five bonds are already marked. So we have to only mark the remaining seven 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 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, hydrogens, and right carbon. But no need to mark on hydrogen, because each hydrogen has already two electrons.

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 seven 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 – 0 – ½ (6) = +1

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

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

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 formal charges by converting lone pairs of the atoms

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. The outside atoms (right carbon and bromines) also form an octet, and both 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 C2H2Br2.