Lewis structure of CH2Br2

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Lewis structure of CH2Br2
Lewis structure of CH2Br2 | Image: Root Memory

The Lewis structure of CH2Br2 contains four single bonds, with carbon in the center, and two hydrogens and two bromines on either side. 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 CH2Br2.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

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 CH2Br2 has one carbon atom, two hydrogen atoms, and two bromine atoms.

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

Therefore, the total number of valence electrons = 4 + 2 + 14 = 20

  • 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 carbon is the central atom.

So now, put carbon in the center and hydrogens and bromines on either side. And draw the rough skeleton structure for the Lewis structure of CH2Br2 something like this:

Skeleton structure for Lewis structure of CH2Br2 | Image: Root Memory

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since carbon is surrounded by two hydrogens and two bromines, use eight electrons to show four chemical bonds as follows:

Four chemical bonds shown between atoms | Image: Root Memory

#3 Mark lone pairs

As calculated earlier, we have a total of 20 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, twelve valence electrons are remaining.

Two valence electrons represent one lone pair. So twelve valence electrons = six 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 and bromines. 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 central atom (carbon) will not get any lone pair, because all six lone pairs are used.

Now draw the Lewis structure of CH2Br2 something like this:

Lone pairs marked on Lewis structure of CH2Br2 | Image: Root Memory

In the above structure, you can see that the octet is completed on the central atom (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.

#4 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 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 CH2Br2 looks something like this:

Formal charges are calculated, and got the stable Lewis structure of CH2Br2 | Image: Root Memory

In the above structure, you can see that the formal charges of all atoms are zero. Therefore, this is the stable Lewis structure of CH2Br2.

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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