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Chemistry 2131:
Organic Chemistry for the Life Sciences (3)

Orbital Hybridizations, VSEPR and Resonance Structures

• last day we discussed the shells that electrons occupy and the orbitals that these shells are subdivided into. We went on to discuss chemical bonds, both ionic and covalent. The covalent bond is particularly important to organic chemistry and involves the sharing of electrons between two nuclei.
• to look at it in more detail, the orbitals that are involved in bond must overlap for the electrons to be shared. Today we'll spend a little more time looking at the shapes and orientations of these things.
• to look at H₂ as an example of a simple covalent bond, the two 1s orbitals overlap, with the electron density highest between the two nuclei along the axis of the bond. This is the simplest example of what we call a sigma bond. By definition a sigma bond is a covalent bond in which the overlap of atomic orbitals is concentrated along the bond axis.
• that's all well and good for an atom as simple as hydrogen, but what about something more complicated, like atoms from the second or third row of the periodic table. Carbon, as we discussed last day has 4 valence shell electrons. Normally we would allocate these to 2 2s, 1 2p_x, and 1 2p_y. We would think of the bonding possibilities as involving the half full p orbitals, but these have an unlikely geometry, that would mean orbitals perpendicular to one another. Also it is harder to figure how the third and fourth bonds are made once the x and y orbitals are full.
• let's take a step back though and forget for a moment about the orbitals that are available, let's think instead about the shapes of molecules predicted simply for electronic considerations
• there is a model for predicting the bond angles of molecules called the valence-shell electron-pair repulsion (VSEPR) model. In this model, we recognize that each atom is surrounded by a shell of valence electrons. These electrons can be forming single, double, or triple bonds or simply by unshared electron pairs. No matter what there involvement, each pair of electrons occupies a space or region, and causes that region to be negatively charged. Since like charges repel each other, it is predicted that these regions of space be as far apart as possible.
• let's apply this VSEPR model to the simplest of molecules, methane, CH₄. First, let's firgure out what the Lewis structure of methane is. Carbon has four unpaired electrons and wants to "get" four more by sharing them with hydrogen. Hydrogen has one electron would like to get another. So it is pretty clear that the structure of methane is a central carbon bonded to 4 hydrogen atoms.
• so, by VSEPR we would predict that there should be four areas of space that are equally far apart. Geometrically, the way to do this is called a tetrahedral arrangement. In a tetrahedral arrangement of atoms, the bond angles, H-C-H, are 109.5¡, and this is precisely what has been determined experimentally for methane.
• OK, let's go back to orbital structures, with that fresh in our minds. We said earlier that the ground state electronic configuration for carbon is 1s², 2s², 2p_x¹, 2p_y¹. How can we reconcile that with the equal bond angles and lengths observed for methane?. We can't and that leads us to the theory of orbital hybridizations.
• the proposal of orbital hybridization came from Linus Pauling. When you hybridize something you combine them and end up with something halfway in between. The hybrid orbitals that we will talk about are of different shapes than the standard s and p orbitals, but in other respects they are similar. They represents regions of space that can accomadate up to 2 electrons.
• the first of the hybrid orbitals that we will discuss is the sp³ orbital. The combination of one 2s and three 2p orbitals results in the formation of four sp³ orbitals. Simplistically, one can think of them as three quarters p-like and one quarter s-like.
• the shape and geometry of these orbitals are very important. Each one has the same shape, an uneven lobed structure, the lobes pointing in opposite directions. Note the p-like shape. The orientation of these is 109.5¡ from each other. This defines the shape know as a tetrahedron. Thus, we say that atoms with sp³ orbital hybridizations are tetrahedral.
• one of the most common places to find tetrahedral orbitals is in carbon. Methane is CH₄. Each sp³ orbital of the elemental carbon has one electron it. In methane each sp³ orbital overlaps with a 1s orbital containing one electron from a hydrogen atom. These orbital overlaps give sigma bonds.
• the sp³ hybridization state is not limited to carbon. Oxygen and nitrogen are also often sp³ hybridized. In the case of nitrogen, one of the orbitals is occupied by an unshared electron pair. Let's look at the predicted structure of ammonia (NH₃). First of all let's figure out the Lewis structure. We mustn't forget about the lone pair of electrons on nitrogen. So, the Lewis structure indicates the presence of 3 N-H bonds and a lone pair. By VSEPR we have 4 regions of space occupied by pairs of electrons, and we should have a tetrahedral arrangement, with one of the areas being occupied by a lone pair. So, we predict bond angles of 109.5¡. In fact the angles have been measured at 107.3¡, this is very close, and also suggests that the repulsion of the unshared electron pair is greater than the repulsion from a bond electron pair. This geometry of nitrogen is often called trigonal pyramidal, but it's important to remember the presence of the lone pair.
• likewise in oxygen two of the orbitals are occupied by electron pairs and two are free to bond with other atoms. To take the most common example, let's predict the structure of water. The Lewis structure has oxygen with 2 unshared electron pairs and two single bonds to hydrogen. Yet again we have four regions of space occupied by electron pairs, so a predicted tetrahedral arrangement. The actual bond angle for water is 104.5¡, further suggesting that the lone pairs repel more strongly.
• that explains a lot of molecules, but not all. What about molecules that have carbon carbon double bonds? Let's start with ethene, C₂H₄. If we try to figure out a Lewis structure for this molecule we realize that there must be a double bond between the two carbon atoms to get full valence shells. In VSEPR the double bond, although it contains 4 electrons, is treated as a single region of space. So each carbon atom in ethene is surrounded by 3 region of electron density that are mutually repulsive.
• without knowing anything about orbital hybridizations, we would predict by VSEPR that the geometry at the carbon atom would be trigonal planar. That is, all of the bonds in a common plane with 120¡ bond angles. In actual fact the H-C-H bond angles in ethene are 117.2¡ and the H-C-C angle is 121.4¡. These are only slightly off of predicted and suggest that the area occupied by the double bond is slightly more repulsive that the area occupied by a single bond.
• clearly the sp³ orbital hybridization cannot accomadate this geolmetry, so we move onto the next hydrid, sp². This hybrid orbital, as the name suggests, is the result of the combination of 1 2s and 2 2p orbitals. There are 3 sp² orbitals. It is important to remember that this leaves the atom with an unhybridized 2p orbital.
• compared to an sp³ orbital, the sp² is more s-like. Again the shape is one with two uneven lobes. It is very similar in shape to the sp³. The important difference is in the geometry of the orbitals. sp² orbitals are in a trigonal planar orientation. This means that the orbitals all lie in a plane, with 120¡ angles between them. The remaining 2p orbital is perpendicular to the plane in which the sp²orbitals lie.
• why do sp² orbitals exist? They are present when atoms form double bonds. A head to head overlap with sp² orbitals gives rise to a standard sigma bond. But there is a free p orbital left on the atoms. These orbitals overlap in a very different way than the other ones we have seen so far. Rather than overlap in a head to head way, with the electron pair residing directly between the nuclei, the lobes above and below the nuclei overlap. This gives a very different type of bond called a pi bond.
• the pi bond is generally weaker than the sigma bond. It has another important feature, a lack of free rotation about the bond axis. Simple sigma bonds allow the atoms to rotate about the bond axis. Pi bonds do not allow free rotation.
• alright back to VSEPR, and our final scenario. We try to explain ethyne, C₂H₂. Figuring out the Lewis structure we predict a triple bond between the 2 carbon atoms. Again, VSEPR treats the triple bond as a single area of electron density (despite the presence of 6 electrons). So, carbon in ethyne has only 2 areas of electron density, and the geometry in which these are farthest apart is in a linear arrangement. So, we predict 180¡ bond angles, and that is what is observed.
• how can we envisage this shape with hybrid orbitals? The answer comes as an sp orbital. As the name suggests it is the combination of one 2s and one 2p, to give the 2 sp orbitals and leaving 2 2p orbitals. You always get 2 sp orbitals. They are again uneven lobed orbitals, but they are less drastically so (given the larger s contribution. The geometry is linear. The 2 remaining p orbitals are perpendicular to the sp orbital axis.
• sp orbitals are present on atoms that will form triple bonds. The head to head overlap of sp orbitals leads to sigma bond formation. The remaining p orbitals overlap as before to give 2 perpendicular pi bonds. So, a triple bond is sigma bond and 2 pi bonds. Obviously, rotation about this bond is highly restricted.
• you should now be able to take a molecular structure, determine the Lewis structure, predict the bond angles and overall shape of the molecule and tell me what the orbital hybridization state at each atom is.
• let's try doing this for ethanal, C₂H₄O.

2. Resonance Structures:

• most of the molecules that we have looked at so far have been pretty simple, in that only one Lewis structure is possible. For many molecules, particularly ions, no one structure is adequate.
• the textbook example of this is the carbonate ion CO₃^2-. The Lewis structure tells us that there must be one oxygen double bonded to the carbon atom and two singly bonded to it bearing negative charges. This would suggest that there are two different types of C-O bonds in the molecule.
• when the structure was determined experimentally it was found that the three C-O bonds are identical. So, no single Lewis structure is adequate to describe this. This brings in the concept of resonance.
• according to the theory of resonance, many molecules are best described by two or more Lewis structures. And the "real" molecule is thought to be a composite or average of these structures. The individual Lewis structures are contributing structures. We can think of the real molecule as a hybrid of these.
• for the purpose of clarity, contributing structures are always separated by double headed arrows.
• there are three rules to follow when writing resonance structures:
  
  1. all of the contributing structures must have the correct number of valence electrons
  2. all of the structures must obey the rules of covalent bonding
  3. the positions of the nuclei must remain the same
• one very common misconception about resonance is that the electrons are continually moving around (that will become more obvious in a minute), there is only one real structure. We use resonance structures to draw versions of electronic distribution that contribute to the real structure.

3. Curved Arrows and Electron Pushing:

• I would like to briefly introduce the concept of electron pushing now, and we will come back to it again and again as the course proceeds. In the context of resonance structures, how do you get from one to another? This redistribution of electrons can be depicted schematically by curved arrows
• the arrow has a very defined shape and orientation. The arrow originates at the source of the electrons and ends at the destination of the electrons.