Boron group, such as Al, the outermost layer is 3s2 3p 1, and the fully charged state should be 3s2 3p6, that is, the outermost layer (principal quantum number n=3, including 3s and 3p). S, P are produced due to the difference of angular quantum number l), and the number of electrons in the P orbital is 1, which is not semi-filled at all (semi-filled, then the P orbital should have three electrons and belong to the nitrogen family).
Looking through the electron affinity data, there are two groups with low (negative) values: the second main group and the rare gas. Their explanation is due to the existence of holomorphism (second main family (ns2), zero family (ns2np6)).
As a semi-filled N family and the first main family, the affinity energy is lower than all main families except B family, but higher than B family. I won't explain it in detail here. However, the electron affinity is controlled by two factors, the attraction of the nucleus and the repulsion of the charge outside the nucleus. The smaller the radius, the smaller the distance and the stronger the attraction. However, due to the smaller radius and the higher density of the electron cloud, the repulsive force increases. The two aspects are contradictory, so there is no obvious recursion in the same family and the same period. It can only be said that the electronic structure of group B is just in a good state and has a peak similar to mathematics, so the affinity energy is very small, even less than that of semi-filled group.