12. Molecular Shapes & Valence Bond Theory
MO Theory: Homonuclear Diatomic Molecules
Homonuclear Diatomic Molecules are composed of 2 identical elements bonded together.
Molecular Orbital Diagrams of Homonuclear Diatomics
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concept
MO Theory: Homonuclear Diatomic Molecules Concept 1
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Now recall that a whole nuclear diatonic molecule is composed of two identical elements bonded together. Good example of this is n. two or F two. Now as a result of increase in electro negativity and a decrease in atomic size. The order of sigma two P and pi two P. Will be reversed for oxygen to neon. So we take a look here at molecular orbital diagrams in this first one. This correlates to only H two to HE two. So hydrogen and di atomic helium which is not a normal structure. And remember we're looking at the valence electrons for these elements in hydrogen and helium. Their valence electrons are found in the one S orbital for in period one of the periodic table. So here we have for instance this could be the atomic orbital of hydrogen or helium and the other hydrogen or helium. We would fill in the atomic orbital's and then distribute those electrons into our molecular orbital's here. Remember sigma one s. Represents our bonding molecular orbital and sigma star one S. Represents our anti bonding molecular orbital. Now if we move to the next one, this correlates to die atomic lithium all the way to die atomic nitrogen here it gets more complex because now we're dealing with period two elements. So we start off with two S and we move our way up to two P. Now here with the two s orbital is atomic orbital's. We fill in first are bonding molecular orbital which is signaled to us and then start filling in our sigma star to us, which represents our anti body molecular orbital. When we get to two P. Gets more complex as the atomic orbital's continue to pull their electrons together to create new molecular orbital's here. The order for it would be pi two P. Sigma two P. Then we'll go pi star two P. Which is an anti body molecular orbital. Two sigma star two P. Another anti body molecular orbital again. We always fill from lowest molecular orbital up now again because of increasing electro negativity and a decrease in atomic size. We're gonna have a slight change when it comes to die atomic oxygen to die atomic neon. If we look the change happens here they flip. So now my pie toupee moves up And my Sigma two p. moved down. So here again, oxygen to neon there's still period two elements. So we're starting out with two S again and we just start pulling together our atomic orbital electrons and dumping them into our molecular orbital's and start filling it up as we move up. So just keep in mind these are the different types of molecular orbital is that can exist based on what type of period element you're dealing with 12 or three. All right, so keep this in mind as we start doing more and more questions dealing with homo nuclear di atomic molecules
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example
MO Theory: Homonuclear Diatomic Molecules Example 1
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using a molecular orbital diagram or M. O diagram. Right? The electron configuration for the dye forming an ion F to minus one. All right. So let's follow the steps step one. We need to determine the number of valence electrons that this particular an ion possesses Flooring is in group seven a. So it has seven valence electrons and there are two of them. So that's times two. So it's 14 valence electrons -1 means we've gained an additional electron. So in total we have 15 valence electrons. Step two, we're going to construct a molecular orbital diagram based on location of the valence electrons. Remember if we're dealing with period one elements, that means the electrons start in one S if we did what period two elements they start with to us and period three elements start with three us. Now we're going to use the molecular orbital diagrams we have here. Now remember we're dealing with two flooring, two foreigners involved because it's F two and it's minus one. So, remember we're going to say it has seven valence electrons. So, I'd be up down up down, Up, up up so that's 5, 6, 7 up up up and then there But remember we have one additional electron, you can put it on either one. So we'll put it here for that additional electron now we would just pull those electrons together into our molecular orbits. So here we have up down up down Up down. So that's 246, 10 11, 12, 13, 14, 15. So this would represent the molecular orbital diagram for the dive flora flooring and I are now here we filled in the total number of valence electrons for each element into the molecular orbital's. And now to fill in our electron configuration is a little bit different. So the way we do it is we show each one of our molecular orbital's as individual things here within parentheses. And up here we put the number of electrons found in each. So if we look we have two electrons within the sigma, two s orbital, two electrons and sigma star to us. Two electrons within sigma two P four total electrons within pi two P four, total electrons within pi star to pee. And then finally one electron within our sigma pop the sigma star to pee. So those are the numbers we place down here. Alright, we're gonna come back down here. We're just gonna fill them in. Right, so what do we say here? We said we had two electrons here. Two electrons here. Two electrons here. Four. And then we had if we come back up here we had four in pi two P four and P I start to pee and one in sigma star to pee. So come back down here make sure we fill them incorrectly. So four. And then finally here we have one. So this would represent the filled in molecular orbital electron configuration for our di flooring. And I'll so again this is the approach you would take in terms of filling out the molecular orbital for your designated diatonic molecule or ion, and then you would set up your molecular orbital electron configuration in this fashion and fill in the number of electrons found within each particular molecular orbital.
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Problem
Determine the number of electrons found in the π2p orbitals for the dioxygen dication, O22+.
A
2
B
0
C
4
D
3
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Problem
Using a MO diagram, write the electron configuration for the P2 molecule
A
(σ2s)2(σ*2s)2(π2p)4(σ2p)2
B
(σ2s)2(σ*2s)2(π2p)2(σ2p)2(π*2p)2
C
(σ3s)2(σ*3s)2(π3p)4(σ3p)2
D
(σ3s)2(σ*3s)2(π3p)2(σ3p)2(π*3p)2
Additional resources for MO Theory: Homonuclear Diatomic Molecules
PRACTICE PROBLEMS AND ACTIVITIES (10)
- Consider the H2+ ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram.
- Consider the H2+ ion. (b) How many electrons are there in the H2+ ion?
- How would we describe a substance that contains only paired electrons and is weakly repelled by a magnetic fi...
- (c) With what neutral homonuclear diatomic molecules are the NO+ and NO- ions isoelectronic (same number of el...
- Consider the molecular orbitals of the P2 molecule. Assume that the MOs of diatomics from the third row of the...
- Consider the molecular orbitals of the P2 molecule. Assume that the MOs of diatomics from the third row of the...
- Look at the MO diagrams of corresponding neutral diatomic species in Figure 8.22, and predict whether each of...
- One of the molecular orbitals of the H2- ion is sketched below: (a) Is the molecular orbital a s or p MO? Is ...
- At high temperatures, sulfur vapor is predominantly in the form of S21g2 molecules. (a) Assuming that the mo...
- Many compounds of the transition-metal elements contain direct bonds between metal atoms. We will assume that ...