How do you go about guessing the ground-state spin and parity of a nucleus? How do you go about guessing the ground-state spin and parity of a nucleus? Questions of this form seem to be asked frequently here, e.g., for 19F, 23Na, and 87Rb and 40K.
 A: Even-even nuclei
If both N and Z are even, then the ground state is always 0+. The reason for this is that due to the attractive nature of the nuclear force, there is a tendency for nucleons to pair themselves in single-particle states that are related by time-reversal. A pair of states like this has maximum spatial overlap. This is different from what is seen in atomic physics, where the residual interaction (force between electrons) is repulsive.
Odd nuclei
In an odd nucleus, we first need to find out whether the nucleus is spherical or deformed. If one or both of the particle numbers are at or close to a closed shell (magic numbers 2, 8, 20, 28, 50, 82, or 126), then the nucleus will typically be spherical. If both are far from a closed shell, it will typically be a prolate ellipsoid. In between these two cases, we have transitional nuclei, which are usually computationally intractable. The chart below, by Thomas Papenbrock (13th CNS summer school, Oak Ridge, 2014) shows deformed nuclei in red and transitional nuclei as yellow.

Odd and spherical
The single-particle energies are roughly as follows (image from WP). The ground-state spin of the nucleus is determined by the spin of the odd particle, which can be guessed by counting energy levels up to the Fermi level. 

Example: 97Cd has 49 neutrons and 48 protons. Both particle numbers are quite close to the magic number 50, so we can be pretty sure that this nucleus is spherical. The odd neutron hole is in the g9/2 shell, so we predict that this nucleus's ground state is 9/2+. There will also probably be a low-energy 1/2- state, since the p1/2 orbital is nearby. Since the mean-field approximation is only an approximation, and the single-particle levels are not the same for all nuclei, it's quite conceivable that this nucleus actually has a 1/2- ground state. To find out for sure, we would have to look at experimental nuclear data from the ENSDF database.
Odd and deformed
When the nucleus is deformed, the single-particle levels change. For some examples of plots of these energy levels, see a reference such as Hamamoto and Mottelson, "Shape Deformations in Atomic Nuclei," https://arxiv.org/abs/1107.5248 , which has some figures at the end. The simplest, classic model for calculating these energy levels is the Nilsson model. Here is my own open-source implementation of the model, which I used to produce some plots to put in the Wikipedia article.
With prolate deformation, the degeneracy of the energy levels within each subshell is broken. The lowest $j_z$ drops the fastest, and the highest rises the fastest. (This is basically a classical fact, as with a particle orbiting inside an ellipsoidal cavity.) An example is shown below for neutron levels, from the Hamamoto paper.  The x axis is the deformation parameter $\beta$, which is typically about $0.2$ for deformed nuclei. When you couple this odd particle to the rotational degrees of freedom, you get a rotational band in which the lowest spin equals the single particle's value of $j_z$. The energy levels of this band go like $J(J+1)-2j_z^2$ (when $j_z\ne1/2$), so if you're not sure if you have a deformed nucleus, you can check the ENSDF data to see if the excited states have approximately this pattern of energy levels.

Example: 191Yb has $Z=70$ and $N=121$. These are both fairly far from any closed shell, so it's probably deformed. We probably should look in the literature (references found in the bibliography of the ENSDF entry) to get the actual deformation, hopefully a reliable one based on experimental measurements of the quadrupole moment. (Actually this particular nucleus, which is far from the line of stability, has never been studied experimentally.) Let's say for the sake of argument that it's $\beta=0.1$, which is probably wrong but makes it easy to read the diagram. At this deformation, five holes below the $N=126$ shell closure puts us in the middle of the f5/2 shell, which would have $j_z=3/2$. Therefore we would expect this nucleus to have a 3/2- ground state.
Odd-odd nuclei
It is almost never possible to guess the ground-state spin-parity of an odd-odd nucleus without a serious calculation. The two odd particles can be in lots of different states and can have their angular momenta coupled in lots of different ways. An example of this is 40K. The fact that this nucleus has a ground state spin of 4 is claimed to arise in shell model calculations from the configuration with a proton hole in d3/2 and a neutron in f7/2. These two angular momenta can be coupled to make any spin from 2 to 5. The fact that 4 ends up being the ground state is probably because, within the space of all 4- states (there may be hundreds of them in a high-quality calculation), the interactions among them tend to push the lowest one down. The accidents of these interactions may have pushed the 4- down lower than the other possibilities. There is no way of guessing this without diagonalizing a Hamiltonian that's 100x100 or something.
There are some cases of very light odd-odd-nuclei for which the spins can be explained in a simple way. These include ${}^2\text{H}$ (the deuteron) and ${}^6\text{Li}$. In these systems, the crucial point is the strongly attractive nature of the neutron-proton interaction when the two intrinsic spins are opposite.
