Question on d block contraction
If germanium is supposed to be bigger than aluminum, in fact, its empirical values correspond to $\pu{125pm}$ for each one respectively, but germanium being lower in the periodic table is bigger at the same time which is why we say that $\ce{Al &lt; Ga}$. Now yes, the atomic radius of silicon is $\pu{118pm}$ while as we said before, the atomic radius of germanium is $\pu{125pm}$, in the same way the size is deduced based on its position in the periodic table, so $\ce{Si &lt; Al &lt; Ga}$.
This occurs because there is a relationship between the effective nucleus charge and the electronic levels of the atom, since the positive nucleus attracts the electrons, but in turn, the electrons repel each other (except those of the first level), it is by This means that the forces and counter forces end up having an effect on the valence electrons, and these determine the atomic radius of the atom. This difference of attraction by the nucleus and repulsion by the electrons increases as there are more electronic levels, that is, if I have a compound with Z = 11 and e = 11 that we will call &quot;A&quot;, it will be greater than a compound &quot;B&quot; whose Z = 10 e = 10, leaving B &lt;A since although both have the same number of electrons as protons, compound A has an electronic configuration $\ce{1s^2 2s^2 2p^6 3s^1}$ with 11 protons while compound B has an electronic configuration of $\ce{1s^2 2s^2 2p^6}$ with 10 protons, that is, compound A has a higher electronic level than compound B, which gives a larger atomic radius.
Reference
1.https://es.wikipedia.org/wiki/Anexo:Radio_at%C3%B3mico_de_los_elementos_qu%C3%ADmicos
Ah, I just realized that it was Galio and not Germanio, sorry.