VSEPR Theory Part II

by Janet Gray Coonce, MS

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Transcription begins at 9 minutes 45 seconds into the video.

Now lets use the VSEPR theory to predict the electronic and molecular geometry of carbon dioxide.

The first thing we have to do is to draw a Lewis dot structure.  The first step is to determine which of the atoms is the least electronegative and put that atom in the center.  Looking at the periodic table, the electronegativity increases from left to right.  Carbon is in group IV and oxygen is in group VI, therefore, carbon is less electronegative than oxygen.

 

So to draw the Lewis dot structure, carbon goes in the center with oxygen on either side.  The next step is to determine the number of valence electrons of each atom.  Remember that valence electrons are the electrons available for bonding.  We can determine the number of electrons by looking at the periodic table and the group number of the atoms.  Carbon is in group IV and all members of group IV have 4 valence electrons.  In the Lewis dot structure the 4 electrons are indicated by 4 dots surrounding the carbon atom as illustrated.  Oxygen is a member of group VI and all members of group VI have six valence electrons—6 electrons in the outermost shell.  6 dots are placed around each oxygen atom in the Lewis dot structure as illustrated.  There are a total of 6 + 4 + 6 = 16 valence electrons in the carbon dioxide molecule.  Remember that the valence electrons are illustrated by dots in the Lewis dot structure and we want each of the atoms to have a complete octet after bonding.

Completing the octet:

In order for oxygen to have a complete octet, it needs 2 more electrons.  This is determined by looking a the periodic table.  It is in group VI and has 6 electrons in its valence shell.  Each oxygen atom needs 8 – 6 = 2 electrons to complete its octet. This is convenient because carbon has 4 valence electrons and it needs 8 – 4 = 4 electrons to complete its octet.

 

So if we draw a line between the lone electrons, then by sharing the electrons in a bond, each atom will “feel” like they have a complete octet of valence electrons.  There are 2 pairs of shared electrons connecting each oxygen atom to carbon.  Each pair of shared electrons is a bond so carbon has 2 double bonds, each connecting the carbon to an oxygen atom.

 

The Lewis dot structure is re-drawn showing how to illustrate the 2 double bonds.

 

Each of the double bonds represent a single region of electron density.  The electronic geometry is represented by the purple balloons.  The molecular geometry is represented by the atomic model.  The black ball represents carbon.  The green balls represent oxygen.  The VSEPR theory predicts that the regions of electronegativity will repel each other and assume a geometric shape which will place them as far apart as possible.

The VSEPR theory therefore predicts the geometry to be linear and the bond angle is therefore equal to 180 degrees.

 

Now notice that carbon dioxide and sulfur dioxide have very different molecular geometries.  Carbon dioxide has 2 regions of electron density.  Those regions will repel each other and get as far apart as possible.  This results in a linear electronic and a linear molecular geometry.

Sulfur dioxide has a very different molecular and electronic geometry compared to carbon dioxide.  Sulfur dioxide has a non-bonded electron pair on the central sulfur atom.  This lone pair of electrons is a region of electron density.  There are 2 other regions of electron density, namely the bonds between sulfur and oxygen.  Three resonant structures are possible for sulfur dioxide as illustrated in the Lewis dot structure.  There are 3 regions of electron density in each of the resonant structures.  The lone pair of electrons are more electronegative than the bonds to oxygen.  This increased electronegative force results in a bent molecular and electronic geometry for sulfur dioxide.

This affects the dipole moment of the molecule.  A dipole moment refers to a difference in charge between different poles of the molecule.  Sulfur dioxide has a dipole moment where the end of the molecule closest to oxygen is more electronegative that the end of the molecule closest to sulfur.  This is because oxygen is more electronegative than sulfur.  This can be predicted by noting the position of sulfur and oxygen on the periodic table.  The valence electrons of oxygen are closer to it’s nucleus than the valence electrons of sulfur are from it’s nucleus.  Oxygen will therefore tend to pull the electrons away from sulfur and result in a polar difference as shown in the Lewis dot structure below as a bond dipole symbol ( |—>).

Each of the oxygen molecules have a partial negative charge relative to the central sulfur atom as indicated by the |—> symbol.  Together that gives the lower end of the molecule a net partial negative charge relative to the top of the molecule as indicated by the downward directed |—> symbol.  This predicts that the molecule would have a dipole moment and that sulfur dioxide would be polar.

 

In the case of carbon dioxide there are polar covalent bonds but they are equal and in opposite directions.  There is a partial negative force on each end of the molecule but they are equal and symmetric. There is no difference in charge between ends of the molecule.  Carbon dioxide therefore has no dipole moment.

In this “tug-of-war”model demonstration of carbon dioxide, an equal force is applied in opposite directions.  There is no movement.  There is no dipole moment.

 

In this “tug-of-war” demonstration using a model of sulfur dioxide, there is an equal force applied toward each of the oxygen molecules.  There is a net movement in the direction of the large arrow.  There is a dipole moment because there is a difference in charge between the top and the bottom of the molecule.  The molecule is more positive above and more negative below.  This difference in forces is illustrated by the movement in this demonstration.  The molecular geometry is not linear, it is bent.

So for a molecule consisting of 2 elements, A and B combined in a compound AB₂, and there are no lone pairs of electrons, there will be 2 regions of electron density.  The more electronegative atoms (B) will repel each other equally and get as far apart from each other as possible.  The bond angle will be 180 degrees.  The electronic geometry will be linear.  The molecular geometry will be linear.  The net dipole will be zero in this case since the atoms on each end are the same.  The molecule would not be polar.

 

If the atoms were all different, then the bond dipoles would not be equal and symmetric.  The molecule would be polar.  Using the “tug-of-war” demonstration, there would be movement due to the difference in forces.

 

A molecule with 3 electron densities getting as far apart as possible is illustrated here by the theoretical compound AB₃.  An example would be AlCl₃.  There are no lone pairs of electrons (unshared electrons) so the bond angles will be 120 degrees and equal.  The electronic geometry is trigonal planar.  The molecular geometry is trigonal planar.  Since the 3 bonds are between the same atoms (A-B), the dipole forces would all be equal, and there would be no net dipole.

 

If one of the atoms were different then the forces would not be equal.  In this example, H₂CO, the oxygen end is more electronegative than the hydrogen end.  There is a net dipole and the molecule is polar.

Learning how to predict the polarity of a molecule is a very important concept in chemistry.  This allows us to make predictions about solubility, reactivity, boiling point, melting point, and other properties.

Part III of VSEPR Theory begins at video time 21:18

 

Transcription by James C. Gray MD FACOG