Writing Lewis Structures: Obeying The Octet Rule

A Lewis structure consists of the electron distribution in a compound and the formal charge on each atom. You are expected to be able to draw such structures to represent the electronic structure of compounds. The following examples will be guided by a set of rules. The complete set of rules is provided.

All these examples obey the octet rule: eight electrons in the valence shell is stable.

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Example #1 - Methane CH4

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One carbon has 4 valence electrons
Four hydrogen, each with one valence electron, totals 4
This means there are 8 valence electrons, making 4 pairs, available.

3. Organize the atoms so there is a central atom (usually the least electronegative) surrounded by ligand (outer) atoms. Hydrogen is never the central atom.

4. Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:

5. The formal charge (F) on the central atom is zero. The structure in step 4 is the correct answer.


Example #2 - Ammonia NH3

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One nitrogen has 5 valence electrons
Three hydrogen, each with one valence electron, totals 3
This means there are 8 valence electrons, making 4 pairs, available.

3. Organize the atoms so there is a central atom (usually the least electronegative) surrounded by ligand (outer) atoms. Hydrogen is never the central atom.

It does not matter which of the three sides you use to put hydrogens on.

4. Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:

5. The formal charge (F) on the central atom is zero. The structure in step 4 is the correct answer.


Example #3 - Water H2O

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One oxygen has 6 valence electrons
Two hydrogen, each with one valence electron, totals 2
This means there are 8 valence electrons, making 4 pairs, available.

3. Organize the atoms so there is a central atom (usually the least electronegative) surrounded by ligand (outer) atoms. Hydrogen is never the central atom.

It does matter which of the two sides you use to put hydrogens on. Use sides that are next to each other. DO NOT put the hydrogens 180 degrees apart. There is a reason for this you'll learn later.

4. Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:

5. The formal charge (F) on the central atom is zero. The structure in step 4 is the correct answer.


Example #4 - Carbon tetrachloride CCl4

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One carbon has 4 valence electrons
Four chlorine, each with 7 valence electrons, totals 28
This means there are 32 valence electrons, making 16 pairs, available.

3. Organize the atoms so there is a central atom (usually the least electronegative) surrounded by ligand (outer) atoms. Hydrogen is never the central atom.

4. Determine a provisional electron distribution by arranging the electron pairs (E.P.) until all available pairs have been distributed:

5. The formal charge (F) on the central atom is zero. The structure in step 4 is the correct answer.


Are you ready to try one on your own? Hey, let's do two. Do PF3. then PF5. Go to the answers.

Go to more problems which obey the octet rule.

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