since i have not much time left, i am begging help from every place i can find.
if there is any physics or chemistry expert here, please help me out!!!
compare and contrast the Bohr-Rutherford model of the atom with the quantum mechanical model. (5pts)
ionization energy trends increase from left to right on the periodic table. There are two exceptions to that rule in the graph below. Explain them using quantum theory. (4pts)
Thank you so much!!!!
I'm moving this to the Help forum. Somebody there can help you out. I know someone… glances at e1n
XD
1st one that's worth 5 points:
in bohr's model of atom, the electron revolves around the nucleus the same way the planets revolve around the sun. however, in quantum mechanics, the electrons are merely probability density and the location of the electrons are entirely probabilistic. this is not the case in bohr's model.
the second one i have no idea. it's a chemistry question.
thank you so much!!!
I don't know what this graph of yours looks like, but I'm guessing it would be boron and oxygen. If those elements are the exceptions, the reasoning has to do with their outer electron shells. With a larger radius, it is easier for these elements to lose electrons from their outer shell. However, both boron and oxygen lose an electron from the 2p orbital when an electron is initially taken from them. In each case, the resulting elements from the removal of the first electron make the element more stable than it was previously. Boron empties its 2p shell, so beryllium's full 2s shell keeps it stable. When oxygen loses an electron, it becomes nitrogen, which has three electrons in its 2p shell. These three electrons occupy different orbitals in the shell. Oxygen has that fourth electron sharing an orbital with one of the other electrons and repulsion is greatly increased. As a result, the ionization energy needed to pull an electron off of oxygen (or boron for that matter) is smaller due to these repulsions (they want to get rid of that electron and reach a stable state). I hope this explanation is coherent enough for you and proves to be useful.
oh.. for my graph it's aluminum and sulfur
so the reasoning for aluminum and sulfur would be the same except for it to be 3p 3s instead of 2p 2s?
You are correct. Aluminum has an outer orbital of 3p with one electron and sulfur has an outer orbital of 3p with four electrons. They are both made more stable with the removal of a single electron for the reasons I mentioned in the last post.
