Maria, I am doing my term project on energy bands in solids. I started writing it up explaining the energy spectra in atoms and molecules and cited your page. I fell that it would be good if you show, how the energy spectra in hydrogen atom and H2+ molecule is (in figures) and i can extend it to the solids.
It will be just adding up some figures to your report, as you have already solved the Shroedinger's equation for those problems for the energies.
Thanks Maria, for reminding me about the article "How chemistry and physics meet in the solid state" by Roald Hoffmann. I have that article, i will use that for my write up. Thanks again.
You did a good job in explaining the atomic and molecular orbitals.
Hello Dharma, I believe our topics are similar. I will be covering Energy Bands in Crystalline Solids. I have read your paper and I don't believe we have overlapping materials. I planned on making a link to your page since our topics are similar.
Hi, Maria, thanks for telling us the atomic and molecular orbits. I have one question I still haven't figured out. If the bonding electrons contribute to the numbers of bonds, then what contribution does antibonding electrons do? It is a little hard to imagine a specific picture about the difference of bonding and antibonding electrons in my mind. Hope you can help me. Thank you very much.
Hey, Jialan,your question is a good point to bring up.
It seems I didn't manage to explain it very clearly in the assignment, so, let's try to answer it now:
Let's work for the cases of F2 (single bond), O2 (double bond) and N2 (triple bond) to see how the antibonding orbitals indeed affect the number of bonds.
F2 case
In this case there are four electrons which must be placed in the two 2π* antibonding orbitals. This resulting antibonding character cancels two of the three bonding interactions, leaving a total bond order of 1.
O2
Similarly here, two electrons which must be placed in the two 2π* antibonding orbitals, so, the resulting antibonding character cancels one of the three bonding interactions, leaving a total bond order of 2.
N2
Eventually, for this case, the bond order will be 3, since there are no electrons occupying antibonding orbitals.
I hope I answered your question : )
Hi Maria..you have a great site. I'm glad this question was asked. The concept of bond order shows how both bonding and antibonding sites contribute to the type of bond a molecule will have. The formula: bond order=(1/2)(# of bonding electrons - # of antibonding electrons) provides a nice summary for your reply to the question. Bond order can also answer the question whether a molecule will be stable or not. For example, the order for diatomic helium (4 electrons; 2 bonding, 2 antibonding), which we know is unstable, would be zero.
Great job Maria. Can you explain why Pauli's exclusion principle does not allow two electrons of same spin in same state to be in same state.
Tough question Satyesh ; ). I'll try to explain as well as I can.
Since my background in QM is weak, this is a good chance for me to study these things too.
What first comes to my mind is that each electron in an atom has a unique set of quantum numbers. If this principle did not hold, all of the electrons in an atom would pile up in the lowest energy state.
However, let's use the aspect of symmetric and antisymmetric wave functions in QM to be our starting point for the answer. By symmetric and antisymmetric wave functions we refer to the physical interpretation of the wave function of a system of identical particles. If the sign of Ψ remains unchanged when interchanging the particles occupying any pair of states, the wave function is said to be symmetric with respect to interchange; if the sign changes, the function is antisymmetric. However, the wave function needs to be constructed in such way that the probability density |Ψ|2 always remains the same.
To be more clear: Supposing that electrons a and b are in states r1 and r2 respectively then the two single particle wave functions can be either orNo matter which of the two equations we choose it will always be |ΨI|2=|ΨII|2
Now, if we want to extend this principle to the Pauli's exclusion principle, we see that since we cannot distinguish between the particles, we cannot know which of ΨI or ΨII describes the system. Consequently we have to consider the system as being in some linear combination of ΨI and ΨII. For fermions, there is only one correctly normalized combination possible and this is
Clearly for fermions, if a=b, then Ψ=0, implying that no two fermions can occupy the same state. By considering the form of wave-function for a system of identical particles, we have arrived almost effortlessly at Pauli's exclusion principle !
Eventually, an electron avoids occupied orbitals not due to electrostatic repulsion or some mechanical property but instead due to the anti-symmetry requirement of the wave function of this spin -1/2 particle.
Hey Maria, your webpage is excellent and beautiful. It offers very detail information of atomic and molecular orbitals.
And I have a question about the molecular configuration in your reply to Jialan. Is there any way for us to distinguish (or reasonable guess) whether the energy of $\sigma_{2p}^{*}$ is higer than $\pi_{x}^{*}$ and $\pi_{y}^{*}$or lower than them? I believe we discuss this in class and in personal before, but we did not have a conclusion at that time. After you did such a wonderful work on the atomic and molecular orbitals, do you discover somthing new about the issue??
Hey Bibi, thank you for your comment. Well, some textbooks explain this observation in terms of a concept called s-p mixing. For any atom with an atomic number greater than seven, the π bond is less stable and higher in energy than is the σ bond formed by the two end-on overlapping p orbitals. However, I didn't go further into details for that. ; )
Great Job, Maria. your term paper is well organized.
In addition to ethane example, in polymer science, there is a conjugated conductive system which is a system of atoms covalently bonded with alternating singe and double bonds ( -C=C-C=C-C=C-) in an organic compound. That system results in a general delocalization of the electrons across all of the adjacent parallel p-orbitals of the atoms, which increases stability and thereby lowers the overall energy of the molecules.
your term paper was very helpful. Thanks.