if one p orbital of oxygen is excluded from participation in the hybridization does that mean that is sp2? (Yes. - GS). I am not sure if i understood this correctly is there any way you could explain how to do this and maybe give another example if possible? (If two p-orbitals are excluded, it is sp - GS). What do you mean by excluded? (Not contributing to the formation of bonding orbitals - GS). and how do you tell? (It is perpendicular to the bond. - GS).

What do we learn from surface of potential energy? is it only the number of reaction steps (No. Also kinetic and thermodynamic parameters. - GS).

^^^so does that mean how fast/slow as well as heat,etc? (Yes. - GS). Also do we learn the same thing from a surface of total energy? (Total energy does not change with reaction coordinates, so that surface would be just a flat plane. - GS).

^^^Does that just mean that the total energy for the reaction is constant and is just measured overall? (I would use the term “total energy of reactants”, which does not change as long as the reactants do not collide with other particles and do not interact with electromagnetic radiation. - GS).

Sorry but could you explain again the purpose of the rate limiting step? (If we want to affect the whole reaction sequence, we can do it most efficiently by affecting the rate limiting step, which is useful for the synthetic design. - GS). I know it takes up the most energy (compared to the other steps in the reaction) and that we determine it from the largest distance from the minima to maxima but I am unclear as what the relevance is? Thanks (The larger that distance, the more difficult it is for the reaction to jump over that “hill”, so the reaction is slower. - GS). Which way was which? If the starting energies are lower than the final states, then it produces heat? or vice versa? (No, you need to supply the missing energy by heat - GS).

Also how can we tell if a reaction is absorbing heat or producing heat from the reaction coordinate diagram? (By comparing energies of the initial and final states. - GS).

Could you briefly repeat what you were saying in class about how molecular shape and molecular geometry? Did you say that molecular shape depends only on the number of electron domains whereas molecular geometry depends on both the number of electron domains as well as how many of those electron domains are lone pairs? (For the molecular geometry, we look at all electron domains. For the molecular shape, we look only at chemical bonds. - GS).

I think I understand hybridization for s and p orbitals but what would the hybridization of a molecule be if if had more than 4 electron domains such as for elements in periods 3,4,5, . . they can exceed the octet rule so how would this affect the molecule’s hybridization overall. so for example what it be for SF6? (We need 6 orbitals from sulfur to account for the 6 bonds of sulfur, so we need to add 2 d-orbitals to the mix of all s- and p-orbitals. The hybridization is sp3d2 - GS).

For the overall formal charge, do we just add up all the formal charges for each atom within the molecule? (Yes, but will be the real charge, not formal charge - GS).

Carbon is more electronegative than Li (because C is more to the right on the periodic table and electronegativity typically increases from left to right and from bottom to top) thus in a compound where Carbon is bonded to Lithium the dipole moment would point to the more electronegative atom which in this case would be the carbon? Is my logic correct for this case? (Yes. - GS).

Do we always assume hybridization is necessary when a compound is formed from two atomic orbitals of differing energy levels? such as a molecule having 2 electrons in 2s and 2 electrons in the 2p orbital? (If orbitals of different energy are contributing to the formation of a chemical bond, and we are interested in the energy of electrons of the bond, we need to look at the relative contribution of those orbitals, and that is what hybridization is. It is not a real process, it is a theoretical tool. - GS). Thanks!

I was just wondering is this reaction correct? CH3OH + HBr = CH3Br + H2O It seems justified to me. (Yes. - GS). Do we put HBr over the reaction line because it is a reagent? (It is acceptable, but it will turn a reaction equation to a reaction scheme. - GS). Okay, thanks! So when would we use a reaction scheme instead of a reaction equation? (When you do not need to balance atoms, for instance when you plan a synthetic strategy or demonstrate chemical properties. - GS).