PS 7 Sol - CHEM 120A Problem Set 7 Solutions David Homan,...

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CHEM 120A Problem Set 7 Solutions David Hoffman , Doran Bennett , Tara Yacovitch 4/2/2010 1. The solutions can be found in the McQuarrie and Simon Solutions manual. 2. The solutions can be found in the McQuarrie and Simon Solutions manual. 3. The solutions can be found in the McQuarrie and Simon Solutions manual. 4. We will use McQuarrie and Simon’s method of counting microstates where the mi- crostates are listed in a table with the possible M L ’s in the rows and the possible M S ’s in columns. We will consider the following quantum numbers: m l ,l,M L ,L ; m s ,M S and S (a) np 5 n 0 p Here we have two different p orbitals. For the first orbital, l = 1 (p orbital) such that m l = 1 , 0 , - 1 We also find that the spins for paired electrons cancel such that m s = ± 1 2 For the second orbital, l 0 = 1 and m l 0 = 1 , 0 , - 1 m s 0 = ± 1 2 We add the maximum values for the individual orbital quantum numbers to find M L = 2 , 1 , 0 , - 1 , 2 M S = 1 , 0 , - 1 We construct a table and write in the possible microstates, m l m s m l 0 m s 0 . For simplicity, we write α for m s = + 1 2 and β for m s = - 1 2 . Note that we include both 1 α 0 β and 0 β 1 α (and similar states) because the orbitals are non-equivalent, n 6 = n 0 . 1
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P P P P P P P P P M L M S 1 0 -1 2 1 α 1 α 1 α 1 β 1 β 1 β 1 β 1 α 1 1 α 0 α 1 α 0 β , 1 β 0 α 1 β 0 β 0 α 1 α 0 α 1 β , 0 β 1 α 0 β 1 β 0 0 α 0 α 0 α 0 β , 0 β 0 α 0 β 0 β 1 α (-1) α 1 α (-1) β , 1 β (-1) α 1 β (-1) β (-1) α 1 α (-1) α 1 β , (-1) β 1 α (-1) β 1 β -1 (-1) α 0 α (-1) α 0 β , (-1) β 0 α (-1) β 0 β 0 α (-1) α 0 α (-1) β , 0 β (-1) α 0 β (-1) β -2 (-1) α (-1) α (-1) α (-1) β (-1) β (-1) β (-1) β (-1) α From this table we see that there are a total of 36 microstates. The given term symbols 3 P , 3 D , 1 P , 1 D account for 32 of these states. Next, we start collecting microstates for each term symbol. Within a given cell of the table, it doesn’t matter which microstate you cross out. i. Consider microstates with L = 2 and S = 1: M L = 2 , 1 , 0 , - 1 , - 2, a D state. M S = 1 , 0 , - 1, a spin multiplicity of 2 S + 1 = 3 This is a 3 D state. Cross out one microstate for each of the 15 possible com- binations of the above M L and M S . You end up crossing out one microstate in each cell of the table. ii. Consider microstates with L = 2 and S = 0: M L = 2 , 1 , 0 , - 1 , - 2, a D state. M S = 0, a spin multiplicity of 2 S + 1 = 1 This is a 1 D state. Cross out one microstate for each of the 5 possible com- binations of the above M L and M S . iii. Consider microstates with L = 1 and S = 1: M L = 1 , 0 , - 1, a P state.
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PS 7 Sol - CHEM 120A Problem Set 7 Solutions David Homan,...

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