Lewis structure of C2H2Br2

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Lewis structure of C2H2Br2
Lewis structure of C2H2Br2

The Lewis structure of C2H2Br2 contains one double bond and four single bonds, with two carbons in the center, and each carbon is attached with one hydrogen and one bromine. There are three lone pairs on each bromine atom, and carbon atom and hydrogen atom do not have any lone pair.

Steps

By using the following steps, you can easily draw the Lewis structure of C2H2Br2.

#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, hydrogen is a group 1 element, and bromine is a group 17 element. Hence, carbon has four valence electrons, hydrogen has one valence electron, and bromine has seven valence electrons.

Now C2H2Br2 has two carbon atoms, two hydrogen atoms, and two bromine atoms.

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

Therefore, the total number of valence electrons = 8 + 2 + 14 = 24

  • Now decide the central atom

We can not assume hydrogen as 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.

Therefore, choose the central atom from carbon and bromine.

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 hydrogens and two bromines on either side. And draw the rough skeleton structure for the Lewis structure of C2H2Br2 something like this:

Skeleton structure for Lewis structure of C2H2Br2

#2 Show chemical bond

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

Five chemical bonds shown between atoms

#3 Mark lone pairs

As calculated earlier, we have a total of 24 valence electrons. And in the above structure, we have already used ten valence electrons. Hence, fourteen valence electrons are remaining.

Two valence electrons represent one lone pair. So fourteen valence electrons = seven lone pairs.

Note 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.

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

The outside atoms are hydrogens, bromines, and right carbon. But hydrogen can not keep more than 2 electrons in its last shell. Hence, don’t mark the lone pairs on hydrogen.

So each bromine will get three lone pairs, and the right carbon will get one lone pair. And the central atom (left carbon) will not get any lone pair, because all seven lone pairs are used.

Now draw the Lewis structure of C2H2Br2 something like this:

Lone pairs marked on Lewis structure of C2H2Br2

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.

#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 six valence electrons. Hence, left carbon needs two more valence electrons to complete its octet.

So convert one lone pair from the right carbon atom to make a new bond with the left carbon atom. And then, the Lewis structure of C2H2Br2 looks something like this:

Lone pair of right carbon is converted, and octet is completed on atoms

Here, the lone pair of carbon is converted (instead of bromine). This is because carbon is less electronegative than bromine, and so it can give more electrons to share them.

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.

#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 hydrogen atom

Valence electrons = 1
Nonbonding electrons = 0
Bonding electrons = 2

Formal charge = 1 – 0 – ½ (2) = 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 C2H2Br2 looks something like this:

Formal charges are calculated, and got the stable Lewis structure of C2H2Br2

In the above structure, you can see that the formal charges of all atoms are zero. Therefore, this is the stable Lewis structure of C2H2Br2.

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

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Deep

Rootmemory.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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