December 2016 - January 2017

# VSEPR, the easy way

##

Peter Nassiff and Wendy A. Czerwinski, Burlington High School, Burlington MA

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Textbooks(1-3) and online resources(4-5) teach that it is first necessary to draw a Lewis structure before determining the VSEPR shape. However, there is a quick and easy method to determine VSEPR structures based on the octet rule that does not require drawing a Lewis structure or using complicated equations.6 The method is primarily targeted for chemistry students who already have an understanding of Lewis structures and are focusing on molecular geometry.

The basic idea is that in any Lewis structure, all atoms (except hydrogen), whether single, double or triple bonded require eight valence electrons (VEs). Any valence electrons left over will have to be incorporated as lone pairs around the central atom. So all one has to do is count the number of valence electrons in the structure, subtract the number of valence electrons involved in a bonded atom, eight for all bonded atoms, according to the octet rule, except for H, which requires two. If there are remaining valence electrons, they must be lone pairs (LPs) around the central atom, so the remaining electrons are divided by two to come up with the number of lone pairs. Now determine what the structure is by finding the structure in the VSEPR table that has the correct number of bonding atoms and lone pairs.

Consider CO2 (Fig. 1), which has 16 valence electrons and two bonding atoms on the central C atom. Two atoms bonded to the central atom (2 bonds x 8 VE/atom) means that there are 16 VEs associated with those terminal atoms. (16 – 16 = 0) means there are no VEs left for lone pairs on the central C atom. The structure must be linear since it has two bonding atoms and no lone pairs. Since there are two bonding atoms and no lone pairs around the central carbon atom, there must be multiple bonds, either two double bonds or a single and triple bond, to conform to the octet rule and the correct structure.

CO2: | # VE [(C = 4 + 2O (2 x 6)]=16 VEs
2 <C-O> bonded atoms: | 2 bonds x 8 VE/atom=-16 VEs
# lone pairs: | (16 VE - 16 for 2 bonds)= 0 LPs

Fig. 1: Determining # lone pairs in carbon dioxide

Likewise, SO2 (Fig. 2) has 18 valence electrons and
2 <S-O> bonds around the central atom which require
(2 bonds x 8 VE/atom) 16 VEs; 18 VEs -16 VE for bonding leaves two VEs leftover. These two remaining VE form a lone pair (2 VEs x lone pair/2 VE = 1 LP). SO2 must have one lone pair on the central atom. Two bonding atoms and one lone pair yield a bent geometry.

SO2: | #VE [(S = 6 + 2O (2 x 6)]=18 VEs
2 <S-O> bonds: | 2 bonds x 8 VE/atom=-16 VEs
# lone pairs: | 18 VE - 16 for 2 bonds = 2 2 VEs so 1 LP

Fig. 2: Determining # lone pairs in sulfur dioxide

## The basic procedure involves five steps

1. Count the number of valence electrons.
2. Count the number of atoms that bond to the central atom and multiply by 8 to account for full octets on all atoms involved (except for hydrogen, which requires 2).
3. Find the number of lone pairs on the central atom by subtracting the number of valence electrons on bonded atoms (Step 2) from the total number of valence electrons (Step 1).
4. Divide the number of VEs not in bonds (from Step 3) by 2 to find the number of LPs. For example,

- If the number is 0, there are no lone pairs on the central atom.
- If the number is 2, there is one lone pair on the central atom.
- If the number is 4, there are two lone pairs on the central atom.
- If the number is 6, there are three lone pairs on the central atom.

5. Use the Table of VSEPR electron and molecular geometries (Table 1) to determine the VSEPR geometry

## Examples:

SF6

1. Count valence electrons: for SF6, (6VEs for S, 7VEs for each F) → 6 + 6x7 = 48 valence electrons.
2. Count the number of atoms bonded to the central atom and multiply by 8 to account for full octets on all atoms involved; 6 bonded F atoms requires 6 x 8 VEs, 48 valence electrons.
3. 48 original VEs - 48 involved in bonded atoms = 0
4. No VEs left over so no LPs.
5. SF6 has 6 bonding atoms and no lone pairs; the structure is octahedral.

### SF4

1. Count valence electrons: for SF4, 6+4x7 = 34.
2. Count the number of atoms bonded to the central atom and multiply by 8 to account for full octets on all atoms involved; 4 bonded F atoms accounts for 32 electrons.
3. 34 original VEs - 32 involved in bonded atoms = 2.
4. So there is one lone pair of electrons on the central atom and four bonds.
5. SF4 has 4 bonding atoms and 1 lone pair; the VSEPR structure is trigonal bipyramidal with a seesaw structure.

### I3-

1. Count valence electrons: for I3- , (3 I = 3x7 + 1 for the -1 charge) = 22
2. Count the number of atoms bonded to the central atom and multiply by 8 to account for full octets on all atoms involved; 2 bonded I atoms account for 16 valence electrons.
3. Original 22 VEs - 16 from bonding = 6 remaining VEs
4. There are three lone pairs (6/2) of electrons around the central atom.
5. Two bonding atoms and three lone pairs means that I3- is linear with the 3 LP in a trigonal bipyramidal electron geometry.

### XeF4

1. Count valence electrons: for XeF4, Xe = 8 + 4x7 for
   F = 36
2. Count the number of atoms bonded to the central atom and multiply by 8 to account for full octets on all atoms involved; 4 bonded F atoms account for
   32 electrons.
3. Starting 36 - 32 for bonding = 4.
4. There are two lone pairs (4/2) on the central atom
5. XeF4 has 4 bonding atoms and 2 lone pairs of electrons; The VSEPR electronic structure is octahedral and the molecular structure is square planar.

### BrF3

1. Count valence electrons; for BrF3, Br = 7 + 3x7
   for F = 28
2. Count the number of atoms bonded to the central atom and multiply by 8 to account for full octets on all atoms involved; 3 bonded F atoms account for
   24 VEs.
3. Overall 28 - 24 for bonding = 4.
4. There are two lone pairs (4/2) on the central atom.
5. The structure has 3 bonding atoms and 2 lone pairs: the VSEPR electronic structure trigonal bipyramidal and the molecular structure is T-shaped.

## Conclusion

This technique allows students to quickly determine electronic and molecular geometry without first drawing Lewis structures. They learn that the octet rule is still a useful starting point for finding molecular structures.

## References (websites accessed November 2016)

1. T.L. Brown, H.E. LeMay Jr., B.E. Bursten, C.J. Murphy and P.M. Woodward, Chemistry The Central Science, 12th edition; Prentice Hall: United States, 2012, page 319.

2. J. Nivaldo Tro, Introductory Chemistry Essentials, 5th edition; Prentice Hall, United States, 2015, page 339.

3. S.S. Zumdahl and S.A. Zumdahl, Chemistry, 9th edition, Cengage Learning: United States, 2014, pages 389-402.

4. John Nash, Valence-Shell Electron-Pair Repulsion Theory (VSEPR), Purdue University http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch8/vsepr.html

5. Gérard Dupuis and N. Berland, VSEPR method www.faidherbe.org/site/cours/dupuis/vseprev.htm

6. A.F. Lindmark, “Who Needs Lewis Structures To Get VSEPR Geometries?” Journal of Chemical Education, 2010, 87 (5), pages 487–491.

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