midterm_ay11_solutions - Chemistry 21b Winter 2011 Midterm...

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Chemistry 21b Winter 2011 Midterm Solution Set T.A.s Scarlett Dong, Jason Crowley 1. (a) This was hopefully a straightforward question. The m l and m s ranges give you 2 L +1 and 2 S +1 values by spanning from (-L,-S ) to ( L,S ) in integer steps, so the independent multiplicities simply multiply, as noted. (b) f electrons mean l =3, so we run the overall value of L from zero to six, or S,P,D,F,G,H,I. Since there are two electrons, there are also singlet and triplet terms associated with each of these. (c) The formula given yields 10!/(8!2!) = 10*9/2=45 states (this would be true for a d 2 configuration as well). The five terms listed give 1*1 + 1*5 + 3*3 + 3*7 + 1*9 = 45 states and so we’ve accounted for all of them. Go with the highest spin multiplicity and then L for the ground state, and since the shell is more than half filled the J levels are inverted. Thus, the ground state should be 3 F 4 , since S =1 and L =3. 2. (a) The microwave region is generally the realm of the rotational spectra of molecules. Very large molecules can sometimes have very low lying vibrational states (so-called ‘torsional’ modes, involving all the atoms in the molecule), but they have to be quite large indeed to have transitions in the microwave region and will definitely not have the regular spacing seen here. Therefore we are clearly seeing a rotational spectrum. For a molecule to have a pure rotational spectrum the first and most important requirement is that it has a permanent dipole moment . Thus, we can immediately eliminate ABA as a candidate. Furthermore, both bent molecules, AAB and ABA, are asymmetric tops, and as such are expected to have a much more congested spectra with less regularity. This leaves linear AAB as the only candidate. Linear molecules have simple spectra with lines separated by two times the rotational constant B. This is indeed what we see, as the lines
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midterm_ay11_solutions - Chemistry 21b Winter 2011 Midterm...

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