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| Q&A
for 20A-2 |
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will the probability density question involve anything other than s orbitals?
No.
is it too late to try and get points back from our midterms?
Yes.
Number three of the practice final exam asks about vibrational frequency. I just wanted to confirm that this material (along with the other chapter 17 material) will not be on the final exam. I think I remember you saying that, but the presence of this question raised some doubt in my mind.
Right. It will not be on the final.
What is the purpose of the negative sign for the equation of Bohr's model? E= - R (Z^2/n^2) What purpose does the - sign play?
The "zero" of energy here is taken to be that of an electron that has no kinetic energy (is at rest) but that is free of interaction with the nucleus (proton), so that its potential energy is zero. The minus sign in the expression for the allowed energies of the bound states makes clear that the energies lie below that of a free electron. The MAGNITUDE of this (negative) energy is the energy that must be provided to free the electron, i.e., ionize the atom.
I just want to see if I understand localize and delocalize, for delocalize, there is resonance for that molecule, where the electrons are moving from one atom to another within the lewist dot structure?. While Localize is a normal atomic behavior? Would we need to know how to list the electron configurations of a triatomic molecule?
A delocalized MO does not necessarily imply resonance, e.g., in CO2 there are two doubly occupied pi delocalized MO's (one from a sum of 2px's on each of the 3 atoms and one from a sum of 2py's); with the localized sigma MO associated with each CO, this implies a double bond for each CO. In NO2-, on the other hand, there is only one delocalized MO (formed from the sum of 3 2pz's in this NONlinear molecule), so the bond order is 3/2 rather than 2, corresponding to the classical situation of resonance. Delocalized "just" means that the MO's enhanced probability density is not associated with a single pair of bonded atoms, but equally with 2 or more pairs. Yes you have to know how to list e configurations for triatomics, but don't memorize anything -- it's an open-book test.
What is the difference(concept wise) between Coluomb's law and Energy dissociation? what is the meaning of (delta)E infinity?
The key is whether one is dissociating an ionic bond into its neutral atoms or into its + and - ions, e.g., whether one is looking at the energy change from NaCl to Na and Cl (in which case the "extra" energy IE - EA is involved) or to Na+ and Cl- (in which case it isn't) -- take a close look at the problem on this from the 2nd midterm, and at the preceding discussion from lectures.
For triatomics, what do you expect us to understand about them? We don't need to draw correlation diagrams for them, but do we have to know how to draw those hibridization models? Like drawing the figure of Sp^2 for a hybridized atom, what specifically is the name of those models? Do we need to know how to draw the destructive and constructive ways that, lets say for example, P orbitals? What exactly is the name of drawing those interferences? Bond overlaps?
For triatomic molecules I expect you to understand the new LCAO and MO ideas involved, e.g., hybridization, and delocalized vs localized bonds, etc. No, you don't need to be able to draw the hybrids, or to draw the positive or destructive interference effects, but rather "only" to understand their consequences.
I just want to make sure that the final exam will NOT be open-book, is it? Also, if not, does that mean we have to remember those mathematical equations? Thank you.
The final exam IS open-book. No need to memorize any equations or particular facts....
Does hybridization occur only for covalent bonds?
For our practical purposes, yes -- that is, in the simplest level of description, hybridization occurs only to maximize overlap of atomic orbitals between bonding atoms and to minimize the repulsions between the resulting bonds (i.e.,to direct them along directions with maximum angle between them). (In reality, ionic bonds are partially covalent (!), and so some hybridization can be involved.)
On question 16.14, the book asks whether the molecule HeH(+) will dissociate to form He + H(+) or He(+) + H. How can we find ou the answer to this question based on the correlation diagram and the magnetism of the molecule (those are the other two pieces of information the question asks you to find)?
You can't really answer this question from the correlation diagram or diamagnetism of the molecule. Rather, the rationalization for He + H(+) is that the two electrons have a lower energy when they can both enjoy the higher charge on the He nucleus.
where can we find the practice test on VOH?
You'll find it as the last item ("practice final and key") on the page...
You mentioned that the final exam will have similar problems to the first and second hour tests, will you be adding extra material in that we learned but was not covered on the 1st and 2nd mid-term? When we use the equation: Enl= (Zeff)^2/n^2 - R with the Absolute value, we get the Ionization energy for the atom. This is true for Hydrogen that R is the IE, but when I tried it on another atom, ex. Oxygen, the IE did not match the ones in the Apendix. Does this only apply to Hydrogen?
Yes, the R(Z^2/n^2) expression is only exact for the ionization energy of hydrogen and other one-electron atoms (He+, Li++, etc.); otherwise, the formula is highly approximate. And, no, I won't be adding any extra material that wasn't covered on the 1st and 2nd hour tests -- rather, only the MO theory we emphasized in lecture since those two tests.
Does the final test include chapter 17 (and today's lecture)? What will be distribution of the problems (if it is allowed to be told)? Which part might be specially emphasized? Sorry I had to leave the end of this morning's lecture so I didn't catch what you've summerized for the final exam.
No, there'll be no questions on Chapter 17 or on Friday's lecture. Rather, I'll be giving you 9 problems to solve, on (but not necessarily in this order): 1. stoichiometry (e.g., limiting reagents) 2. IEs and EAs 3. Franck-Hertz experiment 4. steric number, VSEPR model, hybridization 5. homonuclear diatomic MO theory 6. triatomic MO theory 7. Coulomb's law, ionic bonds 8. effective Z 9. probability densities
Wait, sorry the notes are posted? I only see the lecture 17--20 notes posted?
The notes for 21-25 have been posted since a few days ago....
Can we expect the recent lecture notes been posted online?
They should be posted this morning. (I sent them to voh last night.)
so when forming hybrids for the NO(2) molecule, do you do sp hybridization and put the lone elec into nonbonding or make an sp(2) hybrid and put the lone elec into one of these?? (Is the molec considered linear or trigonal due to odd # of elecs??)
Perhaps it's simplest to start with the 18-valence-electron NO(2)- molecular ion that we discussed in class, and then take away the "extra" electron to make the 17-valence-electron NO(2) neutral molecule. That means taking an electron out of the highest-lying occupied molecular orbital. But it's difficult to figure out (without access to further information from spectroscopic experiments or solutions to the Schrodinger equation) just which one this happens to be in NO(2)-. But it turns out not to matter (whether, say, it's an sp(2) (the one not involved in forming sigma bonding orbitals) or one of the oxygen's non-bonding 2p's...The molecular is bent (distorted trigonal).
I'm doing the assignment of chapter 17, #15 & 16. But I got no idea how to relate them to the lecture. How should we use the given temperature terms?
I'll be discussing this in lecture tomorrow (Wednesday). The equation you need is the second one on p. 605, relating the relative populations of two energy states to the energy difference between them. (In the case of vibrational states of diatomic molecules, it happens that the "degeneracy" factors -- the g's -- are all equal to one.)
so referring to my earlier question, the hybridization would be sp and not sp(2)??
Sorry, but you'll have to remind me of your earlier question!
When you are doing the valence bond theory with NO(2) there is an odd number of electrons, do you make an extra hybrid bond to account for this or put it in a non bonding orbital???
Just put it into the next available MO...(which, as you suggest, happens to be a nonbonding one).
Hi Professor. I am wondering what the next quiz is gonna cover. I think you didn't talk about it in class. Please email me. Thank you very much. Have a good weekend.
The next (last week) quiz will be on MO theory of homonuclear diatomics....
Hi professor, what is meant when they say for CO2, the two electrons on each oxygen atom are nonbonding electrons in 2s atomic orbitals? why are those electrons nonbonding?
Of all the atomic oxygen orbitals that are in principle available for bonding -- the 2s, 2px, 2py, and 2pz -- only the 2pz (pointing along the bond direction of the linear CO2 molecule) have good overlap with the sp-right and sp-left hybrid orbitals of the central carbon. That's because the 2s electrons are held in too close to the oxygen atoms (i.e., the 2pz orbital extends out much closer to the central carbon), while the 2px and 2py give zero overlap (they have the wrong symmetry to interfere constructively with the sp hybrids -- instead they form linear combinations with the 2px and 2py orbitals of the central carbon). This is what it means to say that the 2s electrons on each of the oxygens are "nonbonding" .
Hello, during your lecture you described the pi nonbonding delocalized bond between 3 atoms to be a + c, while the book shows a description that matches a - c, where instead of both positive lobes being on the top, there is only one on top, and one on bottom. I understand that they're both negligible overlap, but will there be a need to follow either a+c or a-c, or will both be acceptable?
Good for you for catching this. And for realizing that there is, in fact, very little difference between the two choices (a+c and a-c), because of how little overlap is involved. I chose a+c so as to simply emphasize the nonbonding character of this MO (and not have you distracted by questions about why the difference combination was chosen). But technically you are right, and of course the two choices correspond to different symmetries ("g" for + and "u" for -).
Hi Professor, I was wondering what the relation between molecular orbitals and the wave function was. On p.571, there is a brief relation of the wave function to homonuclear diatomic molecules. It's a bit confusing and I was wondering if you could clarify.
In the case of the single-electron molecule H2+, the wavefunction depends only on the position of a single electron, and it can be obtained as an exact solution to the Schrodinger equation for a single electron in the presence of two, fixed, protons. The ground state wavefunction, for example, is shown by the bottom picture in Fig. 16.3, p. 569, and it turns out that it can be approximated very well by a sum of H-atom 1s orbitals centered one each of the protons. Similarly, the excited molecular states of H2+, shown by the seven other pictures in Fig. 16.3, can also be approximated by linear combinations of H-atom orbitals. For MANY-electron molecules, however, the H2+ orbitals are only very rough approximations, since the electrons are now interacting directly with eachother. (This situation is similar to the jump we took from H-like (single-electron) atoms to many-electron atoms.) Ands now the molecular orbitals are linear combinations -- not of H-atom orbitals (as was the case, say, with H2+), but rather -- of atomic orbitals for the many-electron atoms that happen to be forming the molecule of interest. Does this help?
It was told in this Monday's lecture (Fig 16.11, Page 579) that for B2, Pi bonding obitals are in the lowest energy level, while for O2, instead, Sigma bonding obitals go to the lowest place. We were given some explanations for the phenomina. Question is: is there rule? Are we sopposed to know the rule and, by use of it, to arrange the electrons by ourselves?
My hope is that you will understand why the sigma orbital comes down in energy (relative to the pi) as the atomic number increases (because of the repulsion from 2s electrons becoming smaller); I don't ask you to memorize just where this happens (i..e., as one moves from N2 to O2), but rather to know how to assign electrons to MOs once one knows their relative ordering. Does that answer your question?
Hi Prof. Gelbart, In the second mid-term, question 1(b), why do we need to calculate the frequency of light emitted from E3-E2? (We only calculated for E2-E1 and E3-E1 in part a though.)
The reason there are THREE frequencies of light emitted is because there are two different excited states, n=3 and n=2, which have been excited in the (Franck-Hertz) experiments. Each of these two states will emit light as the atom undergoes transitions to all possible lower-energy states -- that means 3-to-2 and 3-to-1 transitions for n=3 and 2-to-1 for n=2.
How to understand the concept of "cilindrical symmetry about the internuclear axis" and how are sigma and pi obitals different in it?
Cylindrical syummetry about the internuclear axis means that the wave function has the same value at all points that arise from rotating any given point about this axis. More explicitly, consider some arbitrary point a distance r from the axis. Imagine a circle centered on the axis that passes through this point (and hence has radius r). Sigma orbitals will take on the same value at all points on this circle. Pi orbitals will not. For example, consider a pi orbital formed from a sum of 2px orbitals, one centered on each of the two atoms A and B whose internuclear axis lies along the z-direction. 2px orbitals extend along the x-direction and vanish at points in the yz plane; accordingly, the pi molecular orbital will take on different values at points in the yz plane -- as compared with along the x-axis -- even though they lie the same distance (r, say) from the internuclear (z) axis; accordingly, the pi orbital will not have cylindrical symmetry. We can talk about it further in office hour this afternoon (Tuesday, 4pm) if you like.
ummmm... HI... when we would like to know the average radius of a many electron atom would we use "Z" or "Z[eff]"?
The appropriate (and approximate) equation to use is the following: average radius of outermost orbital = [n^2/Z(eff)]ao, where n is the principle quantum number of the outermost electron, Z(eff) is the effective charge on the nucleus which it sees, and ao is the Bohr radius. Note that the size of an atom is determined by the average radius of its OUTERMOST electron's orbital
What is meant by degenerate and why is that degeneracy removed when it comes to many electron atoms?
Degeneracy just means that there are 2 or more wavefunctions corresponding to the same energy, like the 2s and 2px, 2py, and 2pz orbitals in the H-atom. In many-electron atoms the 2s and 2p orbitals no longer have the same energy, because their electrons see different effective charges on the nuclei.
Hello professor Gelbert, Is there a difference when calculating energy of a Franck-Hertz experiment if the energy is being emitted or absorbed? Would the energy be negative for one because of releasing heat?
In this experiment there is only emission of light, since the excitation of the atoms is achieved by collision with electrons.
specifically how would you approach problem #86a/b??? with the 4(3.14)r^2 or without it
Without it. Only part c involves the volume of the spherial shell of radius r and thickness dr (=4(3.14)(r^2)dr. Can we talk after class (Wednesday, Nov. 16), with paper and pencil? Or at the review session later in the day (5-6 pm)>
can you explain one more time how the probability in a small volume= (wave function)^2dT, goes to the probability of a spherical shell being (wave function)^2 4(3.14)r^2 dr??
Catch me after class tomorrow (Wednesday, Nov. 16), so that we can talk about it -- it's too awkward to do without paper and pencil, or a blackboard....
the first line of page 554 of the textbook says,"the radii of several sets of ions and atoms increase with atomic number in a given group,..." Should "increase" be "decrease?"
If "group" refers to a vertical column, then the radii increase; if it refers to a horizontal row ("period"), then they decrease. Recall that the average radius is DIRECTLY proportional to the square of the principal quantum number (n) of the outermost electron -- which increases down a column (with Z(eff) fixed, and is INVERSELY proportional to Z (eff) of that outermost electron -- which increases across a period (with n fixed).
The last line of page 554, says: (b) Y should be larger because the effective nuclear charge increases through the transition series from Y to Cd. Where do we find the evidence to affirm this point from the book?
A key point is that the effective nuclear charge increases from left to right through a period of the elements. This can be understood, say, in terms of the simple model/theory for Zeff that we discussed in lecture on Wednesday. (Ask questions about this, as you like...) A second key point is that the overall size (radius) of any atom or ion is determined by the average radius of its outermost (i.e., highest-energy, least tightly-bound) electron. Furthermore, the average radius of the outermost electron in any atom is -- to a good approximation -- proportional to the square of that electron's principal quantum nubmer (n) divided by its Zeff. Now, as one moves from left to right from Y to Cd one is basically filling up the five 4d orbitals with from 1 to 10 electrons. n is the same for all of these successive "outermost" electrons, but Zeff is increasing. Therefore atomic size is DEcreasing.
I have asked question about the concept of ms yesterday. Where do I find the posted anwser?
Sorry, I don't remember what the question was! Does "ms" refer to mass spectrometry, or what?
I still don't have clear concept of what is the fourth quantum number, ms.
Indeed, this quantum number is, in many ways, the hardest to undestand, because -- even though it is called the "spin" quantum number, and is often related to the classical spinning motion of a particle -- it describes a property that is still much more abstract than energy (n) or the magnitude (l) or direction (m) of the angular momentum. Strictly speaking, its introduction and explanation require a generalization of the Schrodinger equation to include relativistic (!) effects. But, for our purposes right now, in conjuction with the Pauli Exclusion principle, it "simply" limits (to 2) the number of electrons we can assign to the same (nlm) orbital and tells us whether a many-electron atom will be paramagnetic (un paired spin -- 1 in an orbital) or not (spins paired -- 2 in each orbital).
For 15.12 this still leaves 2 variables: the velocity of the electron in KE and the wavelength of the light. I still don't understand how to solve from this point
Sorry -- I just realized, in looking over your new question, and my answer to your second question from last week, that you had asked about problem 15.12 and I looked at 15.13 instead! In 15.12 you simply need to set hc/l equal to the work function and solve for l, to obtain the maximum wavelenth of light that will eject an electron from a metal with this work function. For 15.13 you want to calculate the energy carried by light of wavelength 2.50x10^-7 m and subtract the work function (7.21x10^-19 J) from it to obtain the maximum kinetic energy of the ejected (emitted) electron. Then set this kinetic energy equal to (1/2)mv^2 and solve for the speed v. (m here is the mass of the electron.)
I also don't understand how to set up 15.34 Do you take the masses of the electron and helium atom into account and if so how??
Calculate the minimum uncertainty in position from eq. 15.16, using for delta-p the product of the electron mass and the speed of light (suggested here as the maximum uncertainty in speed). For helium, every thing is the same except for the mass, just as you have indicated in your question....
How do you set up Question 15.12???
The light's wavelength implies its energy: E=hc/l, where h is Planck's constant, c is the speed of light, and l is the wavelength. The "work function" is the energy needed to free an electron from the metal. So, subtracting the work function from E gives the "left over" energy, i.e., the maximum energy available as kinetic energy (KE). The speed follows from the fact that KE =(1/2)mv^2, where m is the electron mass and v its speed. (v^2 is the square of the speed.) Does this help?
