You can understand this in the following oversimplified model. Let us consider periodic 1D chain of $N$ atoms with lattice constant $a$. Brillouin zone boundaries are located then at $\pm n\pi/a$. Let us consider three configurations:
- Atoms have 1 valence electron on s-type orbital. In this case the
energy level of this atomic orbital is split into $N$-level band.
Only $N/2$ levels are occupied due to spin degeneracy. Fermi
momentum is equal to $k_F = \pm\pi/2a$, hence Fermi surface is fully
inside the 1st Brillouin zone.
- Atoms have 2 valence electrons on s-type orbital. Now all $N$ levels
are occupied and Fermi momentum $k_F = \pm\pi/a$, so that Fermi
surface touches the edges of the 1st BZ.
- Now let us switch to atoms with p-type orbitals occupied. If the
orbitals are oriented like shown in figure below, then two energy
bands of different width would emerge: one is with low mass (from
two orbitals touching each other like $\infty - \infty$) and one is
with large mass (from two orbitals oriented like $8 - 8$), each
having $N$ levels. They have to be occupied up to the same energy
level. Now, if there are 3 valence electrons, then the high mass
band will be fully occupied by 2 valence electrons, whereas the
third electron will be distributed among the low mass band in the
1st BZ and high mass in the 2nd BZ, as shown schematically in the
The general idea is that 2 electrons would fill one band completely and others can stay either in the first, or in the second BZ. It does not mean, that 3 valence electrons must lead to the Fermi surface protracting to the 2nd BZ. In fact, if you rotate p-orbitals in this model 45 degrees, then you will end up with the single band, where each level is degenerate not only in spin, but also in angular momentum. In this case all 3 electrons will stay in the 1st BZ.
Real band structures are quite complicated due to many factors involved, and for complex materials it often occurs that Fermi surfaces touch the BZ boundaries.