CHBr3 Lewis Structure in 6 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

CHBr3 lewis structure has a Carbon atom (C) at the center which is surrounded by one Hydrogen atom (H) and three Bromine atoms (Br). There is a single bond between the Carbon (C) & Bromine (Br) atoms as well as between the Carbon (C) and Hydrogen (H) atom. There are 3 lone pairs on all three Bromine atoms (Br). 

If you haven’t understood anything from the above image of CHBr3 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of CHBr3.

So let’s move to the steps of drawing the lewis structure of CHBr3.

Steps of drawing CHBr3 lewis structure

Step 1: Find the total valence electrons in CHBr3 molecule

In order to find the total valence electrons in a CHBr3 molecule, first of all you should know the valence electrons present in carbon atom, hydrogen atom as well as bromine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of carbon, hydrogen as well as bromine using a periodic table.

Total valence electrons in CHBr3 molecule

→ Valence electrons given by carbon atom:

Carbon is group 14 element on the periodic table. ^[1] Hence the valence electrons present in carbon is 4.

You can see the 4 valence electrons present in the carbon atom as shown in the above image.

→ Valence electrons given by hydrogen atom:

Hydrogen is group 1 element on the periodic table. ^[2] Hence the valence electron present in hydrogen is 1.

You can see that only 1 valence electron is present in the hydrogen atom as shown in the above image.

→ Valence electrons given by bromine atom:

Bromine is a group 17 element on the periodic table. ^[3] Hence the valence electrons present in bromine is 7.

You can see the 7 valence electrons present in the bromine atom as shown in the above image.

Hence, 

Total valence electrons in CHBr3 molecule = valence electrons given by 1 carbon atom + valence electrons given by 1 hydrogen atom + valence electrons given by 3 bromine atoms = 4 + 1 + 7(3) = 26.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

(Remember: If hydrogen is present in the given molecule, then always put hydrogen outside.)

Now here the given molecule is CHBr3 and it contains carbon atom (C), hydrogen atom (H) and bromine atoms (Br).

So as per the rule we have to keep hydrogen outside.

Now, you can see the electronegativity values of carbon atom (C) and bromine atom (Br) in the above periodic table.

If we compare the electronegativity values of carbon (C) and bromine (Br) then the carbon atom is less electronegative.

So here the carbon atom (C) is the center atom and the bromine atoms (Br) are the outside atom.

Step 3: Connect each atoms by putting an electron pair between them

Now in the CHBr3 molecule, you have to put the electron pairs between the carbon (C) & bromine (Br) atoms and between the carbon (C) & hydrogen (H) atom.

This indicates that these atoms are chemically bonded with each other in a CHBr3 molecule.

Step 4: Make the outer atoms stable

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of CHBr3 molecule, you can see that the outer atoms are hydrogen and bromine atoms.

These hydrogen and bromine atoms are forming a duplet and octet respectively and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the CHBr3 molecule.

The CHBr3 molecule has a total 26 valence electrons and all these valence electrons are used in the above sketch of CHBr3.

Hence there are no remaining electron pairs to be kept on the central atom. 

So now let’s proceed to the next step.

Step 5: Check the octet on the central atom

In this step, you have to check whether the central carbon atom (C) is stable or not.

In order to check the stability of the central carbon (C) atom, we have to check whether it is forming an octet or not.

You can see from the above picture that the carbon atom is forming an octet. That means it has 8 electrons.

And hence the central carbon atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of CHBr3 is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of CHBr3.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on carbon (C) atom, hydrogen (H) atom as well as bromine (Br) atoms present in the CHBr3 molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons for each atom of CHBr3 molecule in the image given below.

For Carbon (C) atom:
Valence electrons = 4 (because carbon is in group 14)
Bonding electrons = 8
Nonbonding electrons = 0

For Hydrogen (H) atom:
Valence electron = 1 (because hydrogen is in group 1)
Bonding electrons = 2
Nonbonding electrons = 0

For Bromine (Br) atom:
Valence electron = 7 (because bromine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

From the above calculations of formal charge, you can see that the carbon (C) atom, hydrogen (H) atom as well as bromine (Br) atoms have a “zero” formal charge.

This indicates that the above lewis structure of CHBr3 is stable and there is no further change in the above structure of CHBr3.

In the above lewis dot structure of CHBr3, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of CHBr3.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

About author

Jay Rana

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

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