CHBr3 Lewis structure

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CHBr₃ (bromoform) has one carbon atom, one hydrogen atom, and three bromine atoms.

In the CHBr₃ Lewis structure , there are four single bonds around the carbon atom, with one hydrogen atom and three bromine atoms attached to it, and on each bromine atom, there are three lone pairs.

Steps

To properly draw the CHBr₃ Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

#1 Draw a rough sketch of the 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 CHBr₃ has one carbon atom, one hydrogen atom, and three bromine atoms, so…

Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of three bromine atoms = 7 × 3 = 21

And the total valence electrons = 4 + 1 + 21 = 26

Learn how to find: Carbon valence electrons, Hydrogen valence electrons, and Bromine valence electrons

• Second, find the total electron pairs

We have a total of 26 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 = 26 ÷ 2 = 13

• 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.

Therefore, place carbon in the center and hydrogen and bromine on either side.

• And finally, draw the rough sketch

#2 Next, indicate lone pairs on the atoms

Here, we have a total of 13 electron pairs. And four bonds are already marked. So we have to only mark the remaining nine 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 hydrogen and bromines. But no need to mark on hydrogen, because hydrogen already has two electrons.

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

Mark the lone pairs on the sketch as follows:

#3 Indicate formal charges on the atoms, if necessary

Use the following formula to calculate the formal charges on atoms:

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

For carbon atom, formal charge = 4 – 0 – ½ (8) = 0

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

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

Here, the atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (carbon) forms an octet. The outside atoms (bromines) also form an octet, and hydrogen forms a duet. Hence, the octet rule and duet rule are satisfied.

Therefore, this structure is the stable Lewis structure of CHBr₃.

Next: XeO₂F₂ Lewis structure

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