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$^{17}_8 O$ quoted here has a spin of 5/2 and parity of +1 for the ground state, I agree with this, the unpaired neutron is in the $1d_{1/2}$ state so l = 2, spin = 5/2.

Now I want to figure out the spin an parity of the first two excited states, but how do I know which proton and which neutron move up.

I am thinking that the first excited state is where the unpaired neutron moves to the $2s_{1/2}$ state, so that l = 2, and spin = 1/2, thus the parity remains 1.

But how do I figure out the configuration for the second excited state.

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On the same LBL site there is a full set of level drawings. Now all I have to do is recall how to read them. Here's the one for A=17 (PDF link). –  dmckee Feb 21 '13 at 0:32

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A better reference for determining these nuclear states from experiment is NuDat at Brookhaven. Figuring out the interaction energy of a neutron in any particular energy level with a $^{16}O$ core is extremely difficult. There is some model intuition that can help you, as you used for the first excited state. In this case, the second excited state $1/2^-$ must be more complicated. I would guess that it results from a neutron from the closed core joining the extra neutron in the d-shell, leaving a hole in the p-shell that gives the quantum numbers here. As you can see, there's a huge extra energy needed to make this configuration.

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