Grossman, CHE 230 1.1 3. Orbitals and hybridization. 3.1 Atomic and Molecular Orbitals. We can use molecular orbital (MO) theory to describe the structure of molecules in more detail. MO theory also provides a means of predicting the shapes of molecules. In MO theory, our first hypothesis is that bonding occurs by the overlap of singly occupied atomic orbitals. So if we look at H2, we see that two singly occupied s orbitals, each with a spherical shape, come together in space to form a new MO with an oblong shape. Electrons in this new MO have less energy than electrons in either of the two atomic orbitals with which we started. As a result, energy is released, and the molecule that is formed is more stable than the individual atoms form which it was formed. You might think that the atoms could keep coming together until they were merged, but at a certain distance the repulsion between the positively charged nuclei becomes important. The balance point is called the bond distance. At this point the energy of the system is at a minimum. We can draw diagrams like the ones below. The diagram on the right shows how two atomic orbitals come together to form a bonding orbital of lower energy than either constituent orbital. It also shows that an anti-bonding orbital of correspondingly higher energy is formed. If two He atoms come together, we can draw exactly the same picture, except that we would have to put the two extra electrons in the anti-bonding orbital. Then we would have equal amounts of loss and gain in energy. The net result would be no gain in energy for the system, and the two He atoms would be happier to fly apart than they would be to stick together. HEnergy0internuclear distancebond distanceH HH HHH(1s) H(1s)Grossman, CHE 230 1.2 3.2 sp3 Hybridization. Remember that C has four valence electrons in the configuration 2s22px12py1, where px and py are two of the three p atomic orbitals (AOs) in shell 2. Since C has two singly occupied AOs, we might expect that it would form two bonds. In fact, C almost always likes to form four bonds, as in methane, CH4. What can we do? The first thing we can do is promote one electron from 2s to 2pz. Then we have the configuration 2s12px12py12pz1, with four singly occupied AOs. This helps a lot. Now we can say that each AO is used to form one bond to H in CH4. atomic C;can make only 2 bondsexcited atomic C;can make 4 bonds,one different from otherssp There is still a problem, though. The angle between each of the p orbitals is 90°, so we might guess that three of the four C–H bonds in methane are 90° apart. In fact every bit of evidence shows that all four C–H bonds are equivalent. Something must be done. What we do is to excited atomic C;can make 4 bonds,one different from othersC(sp3); can make 4 equivalent bondsHHHCHsp3spGrossman, CHE 230 1.3 average the four s and p AOs mathematically. This gives us four equivalent AOs. We call them sp3 orbitals, because they consists of 1 part s orbital and 3 parts p orbital. Their energy is 3/4 of the distance between s and p. They point at 109.5° angles from each other, that is, to the four corners of a tetrahedron. C can use each hybrid AO to overlap with a H(s) orbital to make two new MOs. [How do we derive 109.5°? Look at the depiction of the CH4 tetrahedron within the octahedron. Consider the midpoint, M, of the line connecting any two H atoms. If the length of a side of the octahedron is 1, the distance from M to either H must be (√2)/2, and the distance from M to C must be ½. The H-M-C angle is 90°, so we can use trigonometry (remember SOHCAHTOA?) to conclude that the H–C–M angle is tan–1(√2), or 54.74°. The H–C–H angle is twice this number, or close to 109.5°.] The concept of hybridization is very different from that of bond-making, even though both involve mixing orbitals to make new orbitals. In hybridization, a single atom takes its AOs and uses mathematics to convert them into the same number of new AOs. No physical change has taken place; the total energy of the AOs is the same, and the amount of space covered by those AOs is the same. In bonding, two different atoms come together, and one AO from each overlap to form new MOs. When two AOs come together in this way, one new MO is lower in energy, and one is higher in energy. If each constituent AO contained one electron, then two electrons go into the bonding MO, and a bond is formed because the two electrons have lower energy than they would if the two atoms were far apart. That is, the new hybrid AOs are used to make bonds just as the individual “pure” AOs might have done. C(sp3)H(s)CCHH! *!Grossman, CHE 230 1.4 3.3 Hybridization in Alkenes. We can hybridize orbitals in other ways, too. If we combine the s orbital with two p orbitals and leave the third one unaltered, we have sp2 hybridization. An sp2-hybridized atom has three equivalent sp2 orbitals, each with 1/3 s and 2/3 p character, and one unadulterated p orbital. The sp2 orbitals are 120° apart. If we combine the s orbital with one p orbital and leave the other two unaltered, we have sp hybridization. An sp-hybridized atom has two equivalent sp orbitals, each with 1/2 s and 1/2 p character, and two unadulterated p orbitals. The sp orbitals are 180° apart. The two p orbitals are perpendicular to each other and to the line containing the sp orbitals. Cartoons of sp3, sp2, and sp orbitals all look like a fishy, although in fact they have slightly different shapes. Under what circumstances do we find C(sp3), C(sp2), and C(sp) hybridization, and how does C "choose" a particular hybridization? Suppose CH4 were sp2-hybridized. Three H atoms would be coplanar, and the fourth, whose s orbital overlapped with the p orbital, would have a 90° angle with respect to the other three. Moreover, half the p orbital's bonding density would be wasted on the side of the C atom opposite the unique H atom. So we can see that sp3 hybridization is best for a C atom that is bound to four different atoms. CHHHHCH4 with C(sp2) However, when a C atom is bound to three different atoms, the situation changes. Consider ethene (ethylene, H2C=CH2). If the C atoms were sp3-hybridized, we would need to use two hybrid orbitals to construct the double bond. They would need to be canted with respect to the C–C axis, thus putting only a small amount of electron density between the two C
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