Lab 11 Introduction

Learning Objectives

By the end of this section, you will be able to:

  • Determine the shape of a molecule using the VSEPR.
  • Draw the Lewis structures of a molecule including bond angles and formal charges.


The three dimensional shape of a molecule gives great insight into its physical and chemical properties. Indeed, modern pharmaceutical companies spend a great deal of time using computer models and manipulating the shapes of prospective drugs in cyberspace to mimic the shape of molecules with proven effectiveness. They can learn a lot about how a chemical will react, even before it is made by studying similarly shaped molecules. This approach potentially saves great deal of effort and expense; if a prospective drug does not function to specifications in a modeling program, the company will not pursue its use.

In this lab, you will build models of several compounds and ions. First, draw Lewis dot structures for the molecules assigned, then construct the molecules out of the kits provided. It is recommended that you follow the general steps outlined below for each new structure. This is a surefire way to systematically obtain the correct geometry.

Steps to Drawing Lewis Structures:

1. Determine the total number of valence electrons that must appear in the structure. All main group atoms have the same number of valence electrons as their group numbers (1-8). A positive charge on an ion removes an electron while if there is a negative charge, it is necessary to add an electron to the structure.

2. Determine the central element. The central element is usually the least electronegative element or the one which can form the most bonds.

3. Draw single bonds between all atoms and the central atom.

4. Fill in the octets on atoms that follow the octet rule by adding electrons in pairs. Most atoms want a total of 8 valence electrons. However, elements in groups 1-3 and metals in group 4 do not require 8 valence electrons, so it is not necessary to add an octet around these elements.

5. Count the valence electrons in the structure. Each bond counts as two electrons.

a. If there are too few electrons, add lone pairs to the central atom until the correct number of electrons are present in the structure. Note that elements that are in groups 5-8 and which are in row 3 or lower may occasionally contain more than 8 electrons in their structures when following this rule. That is acceptable.

b. If there are too many electrons, eliminate a lone pair of electrons from the central atom and a lone pair from a neighboring atom and draw a multiple bond between those atoms.

6. Count the formal charges. The formal charges of all atoms in the correct structure should be minimal (zero). If there must be a formal charge on an atom, a negative formal charge should be attached to the more electronegative atom. Formal charges should add up to be equal to the overall charge of the molecule or ion. The formula for formal charge is

Equation 1

[latex]\text{formal charge}=\text{ # valence electrons}-\left(\frac{1}{2}\text{ number of bond electrons }+\text{ number of lone pair electrons}\right)[/latex]

7. Use the VSEPR to determine the electronic and molecular geometries of the structure.

An example of drawing a Lewis dot structure for NF3 is as follows:

1. Determine the total number of valence electrons that must appear in the structure.

Source (atom, ion or charge) Number Valence Electrons Total
N 1 5 5
F 3 7 21
Total electrons that should appear in the structure: 26

2. Determine the central element. Nitrogen is less electronegative than fluorine so it must be the central element. In addition, N only has 5 valence electrons (and has room for 3 bonds) while each F has 7 valence electrons (and will only make 1 bond). Therefore N must be the central element.

3. Draw single bonds between all atoms and the central atom.

Image showing N in the center with three single lines connecting the N to F. One F is on the left, one on the right, and one below.

4. Fill in the octets on atoms that follow the octet rule by adding electrons in pairs. None of the elements in this structure are from groups 1-3. Neither are they a metal from group 4. Therefore these atoms all want at least 8 valence electrons. Note that N already has 3 bonds or 6 valence electrons around it, so it only needs 1 lone pair to have a total of 8. Each F only has one bond and needs 3 lone pairs to bring the total to 8.

Lewis dot structure for NF3, showing two dots (representing electrongs) over the N and six dots around each of the Fs.

5. Count the valence electrons in the structure. This structure has the 26 electrons originally calculated. Therefore we can move on to looking at formal charge.

6. Count the formal charges. The charge of the molecule is 0, so the formal charges should add up to be 0. The formal charge of N must take into account it is in group 5 (and should have 5 valence electrons) and has 6 bonding electrons and 1 lone pair of electrons:

[latex]\text{FC}_{N}=5-\left(\frac{1}{2}\times{6+ 2}\right)=0[/latex]

The formal charge of N is 0 which means it has exactly what it should have. The formal charge for F can be done for all three atoms. However, in this structure the F are all equivalent (1 bond, 3 lone pairs) and thus the calculation can be done one time.

[latex]\text {FC}_{F} = 7 - \left(\frac{1}{2}\times{2+ 6}\right)= 0[/latex]

All the F also have a formal charge of 0. When the total formal charges for all atoms in the structure are tallied, it is clear the formal charges (0) add up to be the overall charge of the molecule (0). In addition, no atom contains a formal charge other than 0 making it unnecessary to evaluate other ways to draw the structure.

7. Use the VSEPR to determine the electronic and molecular geometries of the structure. The basis of the VSEPR is that like charges repel and try to get as far apart as possible. This structure has 4 regions of electron density around the N. Thus the electronic geometry of N (according to Table 1) is Tetrahedral. Since only 3 of these areas are bonds, while the other is a lone pair, it is possible to determine the molecular geometry of the molecule is Trigonal Pyrimidal. Note that according to Table 1, the areas of electron density will separate by ~ 109.5° to maximize the distance between the electrons in 3 dimensional space.

A further example of this is in comparing CH4 to H2O. Drawing the Lewis structure of each compound gives the following information:

Lewis structure showing C surrounded by four connected H's. Methane contains 4 areas of electron density around the central C atom (all four are bonds). The other image shows water, as an O in with an H on either side, and two dots above and two dots below the O. Water contains 4 areas of electron density around the central O atom (2 are bonds while the other 2 are lone pairs). The image depicts a 109.5 degree angle between the two Hydrogens.

Note that both of these structure also contain four regions of electron density around the central atom (and therefore have the electronic geometry of Tetrahedral) while the number of bonds differs. This difference means that the molecular geometry of methane (which has 4 areas of electron density with all four of those areas bonds) is Tetrahedral. Water has only 2 bonds (the other two areas of electron density around the central oxygen are lone pairs) has the molecular geometry Bent. Table 1 contains a list of specific geometries and bond angles.

Finally, it is necessary to note any polarity in the molecule. A covalent bond is a sharing of electrons. Sometimes, the partners in a bond don’t share electrons equally. This unequal sharing of electrons is called a dipole. The electronegativity scale can be used to see if a dipole exists and which partner wins the greater share of the electrons. The greater the difference between values from the electronegativity scale, the greater the charge separation in the dipole. The most electronegative atom on the periodic table is fluorine. In general, electronegativity decreases as you move left and down away from fluorine. For our purposes, a bond between two different atoms will be considered polar. Dipoles are drawn as arrows along the bond pointing to the more electronegative atom.

In water, the two bonds between O and H atoms are polar. We draw the arrows toward the more electronegative O. This gives an overall polar molecule (consider how vectors are added in mathematics-this is the same concept) with the more electronegative O pulling a partial negative charge (depicted δ-) while the less electronegative H atoms each contain a δ+. However, it is also possible to have polar bonds and overall a nonpolar molecule. Consider the bonds in carbon dioxide. The bonds are polar but since they are equal and head in opposite direction, they cancel.

Carbon dioxide Lewis diagram showing C in the middle, and an O on either side, connected with a double bond. Two dots are above and below each O. Water is also shown with O in the middle (two dots above and below), connected by a single bond to H's on either side.

However, in water, the polar bonds don’t cancel out due to the molecular geometry and therefore water is polar. It may be helpful to remember that an asymmetrical arrangement of atoms will often result in a polar molecule.

Table 1: Electronic and Molecular Geometries

ExampleMolecule Areas of e- Density Electronic Geometry #Bonding Areas #Non-Bonding Areas (Lone Pairs) Molecular Geometry Bond Angles Polar Bond? Polar Molecul?
CO2 2 Linear 2 0 Linear 180 Yes No
BCl3 3 Trigonal Planar 3 0 Trigonal Planar 120 Yes No
SO2 2 1 Bent <120 Yes Yes
CF4 4 Tetrahedral 4 0 Tetrahedral 109.5 Yes No
NH3 3 1 Trigonal Pyrimidal <109.5 Yes Yes
 H2O 2 2 Bent <109.5 Yes Yes
PCl5 5 Trigonal Bipyrimidal 5 0 Trigonal Bipyrimidal 90, 120, 180 Yes No
SCl4 4 1 See-Saw  90, <120, 180 Yes Yes
 IF3 3  2 T-Shaped  90 Yes Yes
 XeCl2 2  3 Linear  180 Yes No
SF6 6 Octahedral 6 0 Octahedral 90 Yes No
BrCl5 5 1 Square Pyrimidal 90, 180 Yes Yes
 XeF4 4 2 Square Planar 90, 180 Yes No