### Writing Lewis Structures: Expanded and Deficient Octets

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. You might want to examine the steps listed under number 4 of the rules.

All these examples will violate the octet rule: more than eight electrons in the valence shell is stable. Six electrons is a bit of a different story. It will come at the end of the file. Also, all these examples have singles bonds only. Multiple bonds will come later.

Example #1 - SF4

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One sulfur has 6 valence electrons
Four fluorine, each with 7 valence electron, totals 28
This means there are 34 valence electrons, making 17 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.

I put the fluorines like I did because I knew I needed an open space for the unbonded pair on the sulfur. If you put the 4 fluorines around the S like in CH4, that's OK. It gets tiresome making all this little graphic files, but hey, I knew that when I took on this project. OK, enough complaining, back to work.

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 right-hand structure in step 4 is the correct answer.

Example #2 - ClF3

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One chlorine has 7 valence electrons
Three fluorine, each with 7 valence electron, totals 21
This means there are 28 valence electrons, making 14 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 - I3¯

1. This compound is covalent.

2. Determine the total number of valence electrons available:

Three iodine, each with 7 valence electrons, has 21 valence electrons
One additional negative charge gives 1
This means there are 22 valence electrons, making 11 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 right-hand structure in step 4 is the correct answer.

Please note that the iodines are in a straight line. If you placed them at 90 degrees, then this is incorrect. There is a reason for this and it is discussed in the VSEPR section.

Example #4 - XeF4

1. This compound is covalent.

2. Determine the total number of valence electrons available:

One xenon has 8 valence electrons
Four fluorine, each with 7 valence electrons, totals 28
This means there are 36 valence electrons, making 18 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 SF6, then BrF5. Go to the answers.

A separate file on deficient octets is in the works.