Week 13 Pre Lab Magnetic Behavior Questions PDF

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Assignment For Week 13...

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Magnetic Behavior and Electron Configuration Pre-Lab Questions

Due: Week 13, Friday April 30th BY 11:59 PM.

1. (2 point) Below you are presented with sets of quantum numbers: (n, l, ml , ms ). Only one is a valid set. Which is it? A) (4,3,2,1) B) (3,3,2,1/2) C) (3,1,-1,-1/2) D) (2,0,0,0) E) (5,3,-4,1/2) 2. (2 point) What is the maximum number of electrons can a d subshell can hold? 3. (2 point) One of these subshells is not possible, which is it? A) 3d B) 4d C) 1p D) 4s E) 2s 4. (2 points) Which of the following cations would you expect to be diamagnetic? A. Cu2+ B. Zn+ C. Al2+ D. Ti4+ E. Fe2+ 5. (2 points) Which of the following anions would you expect to be paramagnetic? A. As2− B. S2− C. P3− D. N3− E. Si4− 6. (2 points) How many unpaired electrons would you expect on aluminum in aluminum (III) oxide. 7. (2 points) How many unpaired electrons would you expect on iron in iron II sulfide. 8. (2 points) How many unpaired electrons would you expect on Vanadium in V2 O3

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9. (2 points) How many unpaired electrons would you expect on iron in [Fe(H2 O)6 ]3+ 10. (6 points) Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. These species are called ligands. In the past we have assumed all the d orbitals in some species are degenerate; however, they often are not. Sometimes the ligands bound to a central metal cation can split the d orbitals. That is, some of the d orbitals will be at a lower energy state than others. Ligands that have the ability to cause this splitting are called strong field ligands, CN− is an example of these. If this splitting in the d orbitals is great enough electrons will fill low lying orbitals, pairing with other electrons in a given orbital, before filling higher energy orbitals. In question 7 we had Fe2+ , furthermore we found that there were a certain number (non-zero) of unpaired electrons. Consider now Fe(CN)6 4− : here we also have Fe2+ , but in this case all the electrons are paired, yielding a diamagnetic species. How can you explain this? A) All the d orbitals are degenerate. B) There is 1 low lying d orbital, which will be filled with two electrons before filling the 4, assumed to be degenerate, higher energy orbitals. C) There are 2 low lying d orbitals, which will be filled with 4 electrons before filling the 3, assumed to be degenerate, higher energy orbitals. D) There are 3 low lying d orbitals, which will be filled with 6 electrons before filling the 2, assumed to be degenerate, higher energy orbitals. E) There are 4 low lying d orbitals, which will be filled with 8 electrons before filling the 1 higher energy orbital....