Lecture 11: Molecular Orbitals

Lecture 11: Molecular Orbitals Summary Molecular orbital theory is a tool used to predict the shape and behavior of electrons that are shared between atoms. Two or more atomic orbitals are added together to make a linear combination of atomic orbitals—the LCAO method—allowing for quick characterization of the kinds of bonds formed between two atoms. When atomic orbitals are added, they can either constructively interfere, forming a bond, or they can destructively interfere, forming an antibonding state. In a molecule, the bonding state is always lower in energy than the corresponding atomic orbital, while the antibonding state is always higher. Molecular orbital diagrams, or MO diagrams, are a convenient visualization tool to see how electrons are distributed between two atoms: hydrogen is shown to the right as an example. The two hydrogen 1s orbitals combine to form a σ1s bond, which is lower in energy and therefore more stable. If we had started with helium instead, there would have been four electrons to distribute: two would form a σ1s bond, but the other two would go in the σ1s ∗ antibonding state, effectively cancelling out the bond. After placing electrons in the MO diagram, bond order can be calculated: BO = 1 ( of e’ in bonding orbitals − e’ in anti-bonding orbitals) 2 Stronger bonds have higher bond orders, and the bond order must be >0 for a bond to exist at all. For our example above, a hydrogen dimer has BO=1, but a helium dimer has BO=0. This explains why hydrogen gas exists as H2 , but helium gas consists of individual atoms! If an atom has p-orbitals as well as s-orbitals, additional bonds called bonds can form. The three p-orbital domains yield three kinds of bonds, one σ and two : σnpz , πnpx , and πnpy , where n refers to the particular energy level being considered. Note that the σ bond forms for both s orbital overlap as well as one of the p orbital overlaps: the term σbond means that the bond has cylindrical symmetry around the bond axis, something that is not the case 9 Figures above by B.A. Averill and P. Eldridge in General Chemistry. License: CC BY-NC-SA. Source: Open Textbook Library. for the bonds. Anti-bonding σ ∗ and π ∗ states also form from p-orbital MOs. A generic MO diagram for p-orbitals is shown to the left. MOs can also be formed from heterogenous dimers: the same rules apply. Since each of the atoms has distinct energy levels associated with its s- and p-orbitals, the MO diagram formed using two different atoms is usually skewed, and the bonds that result are polar covalent bonds. The atom that is more electronegative is lower in energy. It can be helpful to count the number of electrons in the initial atomic orbitals and make sure that they are all used when filling up the MOs. Remember that if a dimer is positively charged, it has lost an electron, and if it is negatively charged, it has gained an electron: these have to be accounted for as well! If one of the atoms has more electrons than the other, like a bond between H and Cl, the excess electrons form nonbonding pairs—just like the lone pairs in the Lewis diagrams we drew before. If all of the electrons in the MO diagram are paired, then the dimer is diamagnetic, but if there are any electrons left unpaired, it is paramagnetic. For low molecular weight dimers, the 2s and 2p atomic orbitals are very close in energy, and they can interact with each other. For dimers with lower MW than O2 , the π2px,y and σ2pz orbitals switch, and the π2px,y states are filled up first. Why this matters There are two ways to separate pasta from the boiling water once it’s cooked. One way is to pour the mixture through a colander, or a filter, to retain pasta on one side while letting the boiling water pass through. Another is to leave it on the stove and let all of the water boil off. Each of these separation techniques leads to the same outcome, but as you can imagine they are very different. I’m not talking about how the pasta would taste (let’s ignore that part), but rather how much energy it takes to carry out this separation task. Pouring through a filter is easy, fast, and efficient, while leaving the pot on the stove would take longer and require a lot more energy. Now shrink down from pasta to nano-pasta. In other words, the size scale of molecules. We separate molecules from one another all the time for a wide variety of industrial processes. In fact, if we look at the U.S. energy consumption, about 1/3 of it goes into what is vaguely labeled “Industry.” But did you know that 40 percent of this energy for industry serves just one process, separating molecules? The reason that number is so large is that we only use the slow and inefficient thermal approach, not the much more efficient filter approach, to perform the separations. That’s a whopping 12 percent of all energy used in the U.S. that goes into boiling one chemical off of another. So why is it that we don’t just pour these chemicals through a nano-filter, just like we do with pasta and colanders? If we did switch from thermal separation to a filter, we could save up to 90 10 Figure above © source unknown. All rights reserved. This content is excluded from our Creative Commons license. For information, see https://ocw.mit.edu/fairuse. percent of that energy! The reason we don’t, is the filter itself. Current filters aren’t yet up to standard. On the one hand, we’ve got filters made from polymers (materials we’ll learn about later in the semester) that can separate the tiniest of molecules very well, but they’re so delicate that they can’t be used in the harsh chemical and thermal environments of most industrial processes. On the other hand, we’ve got filters made from ceramics (materials with strong ionic and/or covalent bonds) that are super resilient and can handle the conditions, but they can’t get down to the sizes of the small molecules that need to be separated. So it is that we come to our molecular orbitals, orbital filling, and the specific example of N2 vs. O2 covered in class. Because one of the big separations that we need to do on a massive scale involves exactly these two molecules. O2 is plentiful in the air (unless you go into a closed room, light a candle, and wait 12 hours as we learned in lecture 2!), but for many applications we need O2 in much higher concentrations than its naturally occurring 21 percent in air. Take combustion as an example: the 78 percent N2 in the air has a negative impact on combustion processes since the nitrogen molecules get heated up during the reaction to very high temperatures, which is not only inefficient, but breaks down into toxic nitrogen oxide (NOx) gasses. Increasing the amount of oxygen and decreasing the amount of nitrogen leads to much higher combustion efficiencies, lower harmful emissions, and higher processing temperatures. Now, you may be thinking that I’m only talking about fossil fuel processing, and yes that is certainly a prime example of our use of combustion as a society, but getting purified O2 molecules reaches far beyond, to applications in the medical industry to sewage treatment to metal manufacturing, to name only a few. The separation of O2 from N2 is in so much demand globally that it is performed at a quantity of 31 billion kilograms per year. Because the separation is done inefficiently using thermal processes, in this case going cryogenic, which is cooling instead of boiling but for the same purpose, 47 teraBtu (British thermal units, or the amount of thermal energy needed to raise one pound of water by one degree Fahrenheit) of energy are used per year just to do the separation! Just to be clear, that’s 47,000,000,000,000 Btu of energy, or if you prefer 13,774,340,298,094 Watt-hours. A typical household in the U.S. uses about 900,000 Watt-hours of energy per moth, just for reference. How do we save up to 90 percent of the massive energy used to separate O2 from N2 by switching from thermal-based separation to a filter-based approach? The answer lies in those molecular orbitals! They tell us about the bonding of each molecule and the interactions of the molecules with yet-to-be-invented filter materials that could combine the best of both worlds from polymer to ceramic materials. The MOs of N2 and O2 also tell us that these molecules respond differently to externally applied magnetic fields, which may in turn be useful in boosting the separation efficiency. There is a tremendous opportunity for new filters that take these two most abundant molecules in the air and put them into separate compartments, but it all has to start with knowing how the electrons in the molecules behave. And that, of course, we get from molecular orbital theory. Why this employs The problem with fusion energy is that it’s uncontrolled. Way back in the 1950’s, Disney was making feel-good movies (grab some popcorn and check out, “Our Friend the Atom”), about how energy 11 was very soon going to be, “too cheap to meter.” 70 years later, why is this not the case? Fusion is so attractive on so many levels: unlike fission, which is the stuff of current nuclear reactors, fusion has no radioactive waste or byproducts, with the only outcome of a fusion reaction being ridiculous amounts of energy and helium. Fusion is the engine of the stars, so you know this energy is serious. Let’s break it down: remember the battery vs. gasoline comparison I made in Lecture 8? Roughly, the best Li-ion batteries today can store about 1 MJ of energy per kg. Gasoline, on the other hand, can store 45 MJ per kg. But let’s keep going. The explosive TNT stores around 4160 MJ/kg. Uranium when used in nuclear fission stores a whopping 3,456,000 MJ/kg! This incredible energy density is a strong argument for nuclear energy generation and is often invoked. But when we move to fusion, when we move to the stuff of the stars, all of these numbers feel tiny. The energy density of the fuel for fusion, which is a combination of tritium and deuterium, is an incredible 576,000,000 MJ/kg. And that fuel is highly abundant and cheap. That’s why the dream of fusion is still alive even after 70 years of trying, and in fact today there is a huge resurgence in fusion energy. For more information you don’t need to go very far, check out the PSFC (Plasma Science and Fusion Center) right here at MIT. Or the new MIT spin-out called Commonwealth Fusion Systems. If nuclear fusion is to become a reality (in just 10 years according to some, but we’ve also been hearing that since the Disney movie so we need to approach with careful optimism), one of the single most important ingredients to making it work will be magnets. Lots and lots of magnets and very powerful ones. That’s because one of the most likely ways to get fusion to work is to contain the massive energy released (which, by the way, gets to temperatures up to 100 million degrees), is to confine the energy using magnetic fields. Now, in fusion reactor designs the magnetic fields are often generated with superconducting coils, so it’s different from our unpaired electrons in the MO diagram of O2 . But the general idea that a material can be responsive to an external magnetic field comes from its electron filling, and the more electrons that are unpaired like in the O2 molecule, the more responsive it can be. So what are the jobs related to developing new magnets? You could go to work at a nuclear fusion start-up like Commonwealth Fusion, or to the large government-led fusion operations like the ITER program in France. But so many other industries need stronger, cheaper, lighter magnets that the job market extends far beyond just fusion. You could look for jobs in companies that manufacture magnets (there are many), or companies pursuing new ideas for recycling magnets (like Urban Mining Company), or as a scientist at a U.S. lab pushing the frontiers of magnets (like Florida’s National High Magnetic Field Laboratory, or the one at Los Alamos), or at companies trying to make magnets that don’t rely on rare-earth elements (like Toyota for example, among many). 12 Extra practice 1. Consider O1 512 and O2+ a) Draw the MO diagram of each molecule. b) Find the bond order of each. c) Label each one as paramagnetic or diamagnetic. 2. Draw the MO diagram of HCl. 13 Solutions 1. O2 is paramagnetic because it has unpaired electrons. O2+ is also paramagnetic, because it has an unpaired electron. Its bond order is 2.5 14 Solution 1 figure © source unknown. All rights reserved. This content is excluded from our Creative Commons license. For information, see https://ocw.mit.edu/fairuse. Figure © source unknown. All rights reserved. This content is excluded from our Creative Commons license. For information, see https://ocw.mit.edu/fairuse. 2. Figure by B.A. Averill and P. Eldridge in General Chemistry. License: CC BY-NC-SA. Source: Open Textbook Library. 15