Lewis structure of C2Br2

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The Lewis structure of C₂Br₂ contains one triple bond and two single bonds, with two carbons in the center, and each carbon is attached with one bromine. There are three lone pairs on each bromine atom, and the carbon atom does not have any lone pair.

Steps

By using the following steps, you can easily draw the Lewis structure of C₂Br₂.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Complete octet on central atom
#5 Calculate formal charge and check stability

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

• Let’s calculate the total number of valence electrons

We know that… carbon is a group 14 element and bromine is a group 17 element. Hence, carbon has four valence electrons and bromine has seven valence electrons.

Now C₂Br₂ has two carbon atoms and two bromine atoms.

So the total number of valence electrons = (valence electrons of carbon atom × 2) + (valence electrons of bromine atom × 2)

Therefore, the total number of valence electrons = 8 + 14 = 22

• Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for carbon and bromine as follows:

Electronegativity value of carbon = 2.55
Electronegativity value of bromine = 2.96

Obviously, carbon is less electronegative than bromine. Hence, assume that left carbon is the central atom (as there are two carbons).

So now, put two carbons in the center and two bromines on either side. And draw the rough skeleton structure for the Lewis structure of C₂Br₂ something like this:

Also read: How to draw Lewis structure of BrO₄^– (5 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since two carbons are surrounded by two bromines, use six electrons to show three chemical bonds as follows:

#3 Mark lone pairs

As calculated earlier, we have a total of 22 valence electrons. And in the above structure, we have already used six valence electrons. Hence, sixteen valence electrons are remaining.

Two valence electrons represent one lone pair. So sixteen valence electrons = eight lone pairs.

Note that carbon is a period 2 element, so it can not keep more than 8 electrons in its last shell. And bromine is period 4 element, so it can keep more than 8 electrons in its last shell.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are bromines and right carbon, so each bromine will get three lone pairs, and right carbon will get two lone pairs. And the central atom (left carbon) will not get any lone pair, because all eight lone pairs are used.

So the Lewis structure of C₂Br₂ looks something like this:

In the above structure, you can see that the octet is completed on outside atoms. But, the central atom (left carbon) doesn’t form an octet.

So in the next step, we have to complete the octet on the central atom.

Also read: How to draw Lewis structure of HOCN (5 steps)

#4 Complete octet on central atom

Remember that carbon is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Now left carbon already has four valence electrons. Hence, left carbon needs four more valence electrons to complete its octet.

So convert two lone pairs from the right carbon atom to make a new bond with the left carbon atom. And then, the Lewis structure of C₂Br₂ looks something like this:

In the above structure, you can see that the octet is completed on the central atom (left carbon), and also on the outside atoms. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

Also read: How to draw Lewis structure of SF₅^– (4 steps)

#5 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

Collect the data from the above structure and then, write it down below as follows:

• For each carbon atom

Valence electrons = 4
Nonbonding electrons = 0
Bonding electrons = 8

Formal charge = 4 – 0 – ½ (8) = 0

• For each bromine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of C₂Br₂ looks something like this:

In the above structure, you can see that the formal charges of both (carbon and bromine) are zero. Therefore, this is the stable Lewis structure of C₂Br₂.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

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Related

• Lewis structure of BrO₄^–
• Lewis structure of HOCN
• Lewis structure of SF₅^–
• Lewis structure of SeF₂
• Lewis structure of NO₄^3-

External video

• C2Br2 Lewis Structure: How to Draw the Lewis Structure for C2Br2 – Wayne Breslyn

External links

• C2Br2 Lewis Structure in 6 Steps (With Images) – Pediabay
• Dibromoacetylene | C2Br2 | CID 61169 – National Institutes of Health (.gov)
• What is the molecular shape of C2Cl2? – Socratic
• Dibromoacetylene | C2Br2 – ChemSpider