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Two different atoms from the main group often bond together to make diatomic molecules. In most cases, the molecular orbital picture does not change very much from what we have already seen.
From a Lewis point of view, some very simple compounds include the hydrogen halides. For example, HF contains a hydrogen bonded to a fluorine.
A molecular orbital picture of HF could involve an s orbital on hydrogen interacting with an s orbital on fluorine. It may be that the three p orbitals on fluorine stay out of the way, do not interact with the hydrogen, and remain unchanged in the molecule. In that case, we are dealing with a picture of two s orbitals combining, as in dihydrogen.
The only difference is in the starting energy levels of the atomic electrons. The electrons in fluorine are at lower energy than the electron in hydrogen. Thus, the molecular orbital picture is skewed, with one side of the diagram at lower energy than the other.
- The energetic differences of the two atoms are shown in the molecular orbital interaction diagram.
Together 3 3 7 3 Rule
There is a physical interpretation for this lopsidedness. Notice that the bonding orbital is closer in energy to fluorine than it is to hydrogen. Often, energetic similarity correlates with structural similarity. We think of these bonding electrons as being physically closer to the fluorine as well. In other words, the electrons in this bond are not equally shared, but are held more closely by fluorine.
- More energetic similarity to one atom implies electrons are found closer to that atom.
This idea is consistent with our ideas about bonding between dissimilar atoms. We would think of the H-F bond as a polar, covalent one.
The idea that fluorine's electrons are at lower energy may be confusing. A 1s electron in fluorine is certainly at lower energy than a 2s or 2p electron in fluorine. However, a 1s electron in hydrogen is not necessarily at a lower energy than a 2s or 2p electron in fluorine. All the electrons in fluorine are at lower energy than the electron in hydrogen because fluorine has a much higher nuclear charge, with relatively little shielding from its own 1s electrons. Fluorine has a high Zeff, as evinced by its high electron affinity (high attraction for more electrons) and high ionization potential (large amount of energy needed to strip an electron away).
Exercise (PageIndex{1})
Construct molecular orbital interaction diagrams for the following diatomic molecules. In each case, what is the bond order? Is the compound diamagnetic or paramagnetic?
- carbon monoxide, CO.
- nitrous oxide, NO.
- hypochlorite ion, ClO-.
- Answer a
- Answer b
- Answer c
Attribution
Chris P Schaller, Ph.D., (College of Saint Benedict / Saint John's University)
Diatomic molecules with two non-identical atoms are called heteronuclear diatomic molecules. When atoms are not identical, the molecule forms by combining atomic orbitals of unequal energies. The result is a polar bond in which atomic orbitals contribute unevenly to each molecular orbital.
The application of molecular orbital theory to heteronuclear diatomic molecules is similar to the case of homonuclear diatomics, except that the atomic orbitals from each atom have different energies and contribute unequally to molecular orbitals. Recall that atomic orbitals must have compatible symmetry and similar energy to combine into molecular orbitals. In the case where atomic orbitals of like symmetry have different energies, they combine less favorably than orbitals that are closer to one another in energy. As a general rule, orbitals that have energy differences of greater than 10-14 eV do not combine favorably. In the molecular orbital diagram, the closer a molecular orbital is to an atomic orbital, the more that atomic orbital contributes to the molecular orbital. This last point is helpful for back-of-the napkin estimations of what the molecular orbitals 'look' like.
In this section, you should learn how to generate molecular orbital diagrams of heteronuclear diatomic molecules. To approach such a problem, we must start with a knowledge of the relative energies of electrons in different atomic orbitals. In other words, we need knowledge of the orbital potential energies (or orbital ionization energies).
Orbital Ionization Energies
There are two approaches you can use to 'know' or estimate the atomic orbital energy levels.
1. Use a table of atomic orbital ionization energies, like those found in Table (PageIndex{1}).
2. When you do not have access to a table of values like the one below, use periodic trends in electronegativity and/or ionization energies as your guide to approximate relative values for different atoms.
- 5.3.1: Polar bonds
- 5.3.2: Ionic Compounds and Molecular Orbitals
Sources:
- Gray, Harry. Electrons and Chemical Bonding, Benjamin, 1964.
- Miessler, Gary L, and Donald A. Tarr. Inorganic Chemistry. Upper Saddle River, N.J: Pearson Education, 2014. Print.
Modified or created by Kathryn Haas (khaaslab.com)
Earlier, we talked about the fact that axially symmetric orbitals can mix together to produce new combinations. In HF, we looked at the overlap of the 2s orbital on fluorine with the 1s orbital on hydrogen. However, one of the 2p orbitals can lie along the bond axis and share axial symmetry with the hydrogen 1s orbital.
We will call this fluorine orbital the 2pz orbital. We could just as well label it px or py, but many people refer to the p orbital along the bond axis as the pz. Note that the 2p orbitals that are perpendicular to the bond axis can't interact with the hydrogen 1s orbital. Because the bond axis crosses the p orbital at a node, where the p orbital changes phase, the hydrogen atom can't form either an in-phase or out-of-phase combination with this p orbital. It would actually be in-phase with one part of the orbital and out-of-phase with the other. Instead, these orbitals simply do not interact.
- The hydrogen 1s can combine with the fluorine 2s and the fluorine 2pz.
- Now there are three orbitals that may combine. They will produce three new orbitals.
- We should consider these three orbitals together. What different combinations can they make?
- The orbitals could all be in-phase -- that is, the 1s of hydrogen is in-phase with the 2s of fluorine, and also in-phase with the part of the 2pz that overlaps with it.
- This will be a low-energy, bonding combination.
- The orbitals could be completely out-of-phase. The 1s of hydrogen is out-of-phase with the 2s of fluorine, and also out-of-phase with the part of the 2pz that overlaps with it.
- This will be a high-energy, antibonding combination.
The third combination is a little more subtle. The other possibility is that the 1s orbital of hydrogen is in-phase with one orbital of fluorine, but out-of-phase with the other. However, that means it could be in-phase with the 2s but out-of-phase with the 2pz, or vice versa. That means there are two more combinations, for a total of four. However, since we are using only three atomic orbitals, we can produce only three molecular orbitals. These two possible combinations must somehow be mathematically reduced to one.
Once again, mixing that occurs based on symmetry has some quantitative effects on the energies of the electrons, but in this case the qualitative effect is a subtle change in the ordering of energy levels.
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- A third combination will be partially bonding and partially antibonding.
- This combination will be somewhat like a non-bonding orbital (although its exact energy may be slightly stabilized or destabilized with respect to the original atoms).
Attribution
Chris P Schaller, Ph.D., (College of Saint Benedict / Saint John's University)
We have to start somewhere, so let's start with adding fractions. The most important thing about adding fractions is that they need to be 'like' fractions. That means you need to make sure that both addends (numbers being added) have common denominators. 5/6 + 6/7 may look hard, but when they have the common denominator of 42 it is an easy addition problem. Let's get started.
We'll start with easy ones. In class, you will be given addends with the same denominator. All you have to do is add up the numerators (numbers on top) and then simplify your answer. Let's try some.
1/13 + 6/13 = ?
• Create common denominators. They are already the same at 13, so we do nothing.
• Add the numerators from the two addends. 1 + 6 = 7
• Write the sum of the numerators above the common denominator. 7/13.
•Simplify the fraction. 7/13 cannot be simplified. You are done.
Answer: 1/13 + 6/13 = 7/13
5/9 + 1/9 = ?
• Create common denominators: The denominators are the same. Do nothing.
• Add numerators: 5 + 1 = 6
• Write the sum of the numerators above the common denominator: 6/9.
• Simplify: 6 and 9 have a common factor of 3. When you divide the numerator and denominator by 3 you get 2/3.
Answer: 5/9 + 1/9 = 6/9 = 2/3
Let's get more advanced. What about adding unlike fractions? You don't have common denominators. We've looked at creating equivalent fractions in our earlier pages. You will use that process here.
1/7 + 1/3 = ?
•Create common denominators: We have 7 and 3. They have no common factors, so let's just multiply to create two new equivalent fractions. Remember how we multiplied by equivalents of 1? It went like this...
1/7 = 1/7 * 1 = 1/7 * 3/3 = (1*3)/(7*3) = 3/21
1/3 = 1/3 * 1 = 1/3 * 7/7 = (1*7)/(3*7) = 7/21
You now have the common denominator 21. You can now rewrite the problem as 3/21 + 7/21 = ?
• Add numerators: 3 + 7 = 10
• Write the sum of the numerators above the common denominator: 10/21
• Simplify: 10/21 cannot be simplified. You are done.
Answer: 1/7 + 1/3 = 10/21
You've got common denominators and unlike fractions under control now. Let's look at an example with mixed numbers before we go. Our first example will look at mixed numbers that have common denominators.
2 2/9 + 4 3/9 = ?
• Check for common denominators: They are like fractions with a denominator of 9. Do nothing.
• Add the numerators from each fraction: 2 + 3 = 5
• Write the sum of the numerators above the common denominator: 5/9
• Add the whole numbers: 2 + 4 = 6
• Write out the mixed number: 6 5/9
• Simplify: 5/9 can't be simplified. You are done.
Answer: 2 2/9 + 4 3/9 = 6 5/9
What happens if you wind up with an improper fraction in your answer? You need to simplify that improper fraction and then add the whole numbers. We'll use an example like the last one. We just made the first addend a bit bigger.
2 7/9 + 4 3/9 = ?
• Check for common denominators: They are like fractions with a denominator of 9. Do nothing.
• Add the numerators from each fraction: 7 + 3 = 10
• Write the sum of the numerators above the common denominator: 10/9
• Add the whole numbers: 2 + 4 = 6
• Write out the new mixed number: 6 10/9
• Simplify: This example has an improper fraction that you have to simplify. You need to use division to create a new fraction. 10 ÷ 9 = 1r1. The new mixed number will be 1 1/9. You will need to add that new whole number to your original 6. The entire process goes like this...
6 10/9 = 6 + 10/9 = 6 + 1 1/9 = 6 + 1 + 1/9 = 7 1/9
Answer: 2 7/9 + 4 3/9 = 7 1/9
Let's put it all together with an example that has unlike fractions. You're going to need to make common denominators with this one and simplify an improper fraction.
Common denominators: Start with the fractions. We've got 4 and 8. They have a common factor of four so we only need to fix the 3/4 fraction.
3/4 = 3/4 * 1 = 3/4 * 2/2 = (3*2)/(4*2) = 6/8
Add the numerators: 5 + 6 = 11
Rewrite fraction: Sum of the numerators above the common denominator: 11/8
Add whole numbers: 2 + 5 = 7
New mixed number is 7 11/8
Simplify: We've got an improper fraction. Start dividing... 11 ÷ 8 = 1r3. Your new mixed number is 1 3/8. Add the new mixed number to the original 7.
7 11/8 = 7 + 11/8 = 7 + 1 3/8 = 7 + 1 + 3/8 = 8 3/8
Answer: 2 5/8 + 5 3/4 = 8 3/8
Now it's up to you to practice. If you want to keep going, continue on to subtract fractions. It's very close to addition, so you will do fine.
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