Q: At the end of your lecture today, you mentioned the Bonding, antibonding, and nonbonding properties along with the diagram. When I read the material tuesday, I didn't quite grasp it, and the diagram only confused me a bit more. Is there any way it can be explained without too much pain on your part via e-mail, or would an office visit/help session be better. Thanks for your time.

A: Chapter 1 only mentions that non-bonding molecular orbitals (MO's) are of the same energy as the starting atomic orbitals (AO's) and does not include them in a molecular orbital diagram. I purposely did not want to include them in lecture for that reason. At this point, you should only be aware that non-bonding MO's are of equal energy to the starting AO's. You will encounter the non-bonding molecular orbital in Chem 228 in Chapter 14 during discussion of the allyl cation, anion, and radical.

Another point- Be sure not to confuse "non-bonding MO" with "non-bonding electrons" which refer to "lone pair electrons" which are not involved in bonding to other atoms in a molecule. For example, ammonia has a pair of non-bonding electrons.

Thanks for your question!

Q: (1) Can inorganic molecules bond with organic molecules, and if so, does the whole thing become classified as "organic" simply because it containscarbon?

(2) When an electron moves or shifts into a lower energy level or shell,is the energy that it released emitted as heat or as a photon of light? Or does it depend on the amount of "juice" the electron has?

A: You've raised some interesting questions.

1) There is no definite line between what is considered organic and inorganic but in general I think you could consider an organic molecule to be one that contains carbon. Another term used to describe molecules containing organic and inorganic molecules is organometallic and examples include alkyllithiums, Grignard reagents, and cuprates

2) Molecule which possess an electron in an excited state can decay back to the ground state by two prinicpal modes (there are others) depending on whether the excited atom is in a singlet (S) or triplet state (T). From the triplet state, decay may occur through loss of energy by heat (by so called intersystem crossing) or by phosphoresence (giving up light). The other principal mode is fluorescence which occurs from the singlet state and involves an all at once drop to a lower vibrational state.

If you want a more detailed answer to ques 2, I refer you to the "Bible" of organic chemistry: J. March's "Advanced Organic Chemistry" 3rd Ed, p. 208. It should be in the library.

Q; On Friday, you said that Dipole Moment was the net molecular polarity. Does the Dipole moment only deal with Covalent bonds, or do they also deal with the ionic bonds? Also, does the hybridization affect the Dipole Moment at all? Anyting you can do to clear it up will help; I never really have understood this concept.

A: Ionic substances also have dipole moments and as you might expect these are quite large because they are ionic. Chapter 2 deals with covalent bonds for the most part since of course most organic compounds consist of covalent bonds. As I hope I got across in class, it's important that you are able to predict the polarity of a molecule as that will clue you in as to how it might interact with a nucleophile ("nucleus lover") or an electrophile ("electron lover") later in the course.

Hybridization itself does not necessarily affect the dipole moment directly, more importantly are the atoms in a given molecule and the relative electronegativities of the atoms. For example, a C-Cl bond in a molecule induces a large dipole as does a C-O bond in a molecule. Hopefully this answers your question. If not let me know and I'll try to clarify further.

Thanks for your questions!

Q; What is the answer to 2-44? The Sols. Manual says that this cannot be prepared, but I don't understand. Is it because water is a stronger acid than Ktertbutoxide? What are the rules on this type of problem? I don't think they clarify very well in the book.

A: Concerning your question, you need to consider the difference in pKa values for the two products that would result i.e. water and t-butanol if the following were in fact an equilibrium.

KOt-Bu + H20 -> HOt-Bu + KOH

So in comparing pKa's, you see that water (pka ~15.7) is more acidic by ~ 2.5 pka units relative to t-butanol (pka = 18) so the reaction is hardly an equilibrium and lies far to the right since water is acidic enough to react with the t-butoxide and form t-butanol and potassium hydroxide. So the answer is no, since ktOBu would react with (deprotonate) water.

Thanks for the question.

Q; 1. My notes say that the highest ionization energy is located at the bottom left corner of the periodic table, but the book states that "Alkali metals, at the far left of the periodic table...have low ioniation energy." Which is true?

2. In general, could you say that sp3 hybridied atoms form a single bond, sp2 form a double bond, and sp form a triple bond?

3. Why is the angle in 36c of the first problem set 180 and why is Be sp?

A: 1. I believe I might have done a blunder. Yes, in fact, the lowest ionization energies are found on the left side of the periodic table. Remember, that ionization energies are typcially positive numbers since energy is required to remove an electron whereas electron affinities are negative numbers since energy is typically released when an electron is taken up by an atom.

2. What you may be able to say in general is that a sp3 hybridized atom has no pi bonds, a sp2 hybridized atom has 1 pi bond, and a sp hybridized atom has 2 pi bonds.

3. Since Be has only 2 valence e's in its outermost shell (i.e. 2s), it can only form bonds with two atoms as in H3CBeCH3. It does so by hybridizing its 2s orbital with one 2 p orbital to get two new sp orbitals. The remaining two p orbitals are empty. Be is a rare case where the octet rule cannot be achieved. The Be atom can thus be drawn as having two sigma bonds (consisting of sp orbitals) and two empty p orbitals. Its just like a sp carbon (as in acetylene) except that the p orbitals are empty.

Thanks for the good questions!

I have an addendum to the answer I gave for ques.#2 that you posed. In some rare cases, such as BF3 (which has a empty p orbital and is sp2 hydridized), triplet methylene (which has two half filled p orbitals and is sp hybridized), and singlet methylene (which has one empty p orbital and is sp2 hybridized) and H3CBeCH3 molecule that was in your 3rd question (which has two empty p orbitals and is sp hybridized), the generalization that I made concerning pi orbitals does not hold. In general, the best thing to do when trying to predict the hybridization of a given atom (restricting ourselves to 2nd row elements) is to consider how many electrons are in the valence shell of the atom under consideration and how many bonds it is making to attached atoms and/or how many filled orbitals it has. This will then tell you how many hybrid orbitals are needed to form bonds with the attached atoms. If you need 4 hybrid orbitals then it is sp3 hybridized, if you need 3 hybrid orbitals then it is sp2 hybridized (and then you are left with a p orbital), if you need 2 hybrid orbitals then it is sp hybridized (and then you're left with two p orbitals).

I hope this helps you. I will be making a handout on this.