Ch 8 Online Examples Answers Pt2

Ch 8 Online Examples Answers Pt2 - HODSON CHE 1301 Petrucci...

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HODSON CHE 1301 Petrucci Ch8 Example Problems The following problems are taken from old exams . ..... 7. The quantum numbers of the last electron of nickel could be: a)  n   =  3,    =  2,  m    =  0,  m s   =  1/2 b) n   =  3,    =  2,  m    =  1/2,  m s   =  1/2 c) n   =  3,    =  2,  m    =  0,  m s   =  0 d) n   =  4,    =  2,  m    =  0,  m s   =  1/2 e) n   =  3,    =  1,  m    =  0,  m s   =  1/2 Answer: A Explanation: You need to be able to write out the electron configuration of nickel, which is element number 28, and so we have to arrange 28 electrons. They are entered into the lowest energy orbitals possible, the sequence of which we can get from the periodic table. Starting from hydrogen, first period s-block (1s) we work through (like typewriter reading left to right on each period) the 2s, 2p, 3s, 3p, 4s and finally to the 3d energy level. Nickel is the eighth element in from the left, and so the highest energy level present we can represent as 3d 8 . All the levels below this are filled with as many electrons as they can take: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 8 Note that if you add up all the superscripts you get 28 electrons, which is correct for nickel. You may see the same configuration written as (switched 3d and 4s levels): 1s 2 2s 2 2p 6 3s 2 3p 6 3d 8 4s 2 The Petrucci text uses this, and so we must for consistency. Why ? The 4s sublevel is
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This note was uploaded on 12/19/2009 for the course CHE 1301 taught by Professor Klausmeyer during the Spring '08 term at Baylor.

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Ch 8 Online Examples Answers Pt2 - HODSON CHE 1301 Petrucci...

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