# Atomic radii of Sc, Ti, Fe, and Co

The atomic radius of Sc is $\pu{162pm}$, Ti is $\pu{147pm}$, $\ce{Fe}$ is $\pu{126pm}$, and that of $\ce{Co}$ is $\pu{125pm}$.

The electronic configuration of $\ce{Fe}$ is $\ce{[Ar] 3d^6 4s^2}$, and that of $\ce{Co}$ is $\ce{[Ar] 3d^7 4s^2}$. The difference in atomic number, and hence the difference in the number of 3d electrons, is 1. So, due to screening, the extra charge is "canceled" and they have nearly the same radii.

The difference in atomic number and 3d electron count between $\ce{Sc}$ and $\ce{Ti}$ is also the same — 1.

So, I want to know why there is a considerable difference between the radii of Sc and Ti but not between the radii of Fe and Co.

There are different notions of atomic radius; the one you're using seems to be the metallic radius, which is half the distance between nearest neighbors in the metal. This notion is very sensitive to the number of electrons per atom involved in bonding. Scandium has only 3 valence electrons, while $\ce{Ti}$ has 4. These all participate, in some extent, in the "electron soup" that holds metals together. I haven't been able to figure out exactly to what extent, but It's fair to say though that the 4 valence electrons of Ti bind the nuclei together significantly more tightly than the 3 of $\ce{Sc}$. As a result, the $\ce{Ti}$ atoms get significantly closer together. (An analogous situation is the covalent radius of $\ce{F2}$, of around $\pu{70 pm}$, versus that of $\ce{O2}$, of around $\pu{60 pm}$; although the covalent radius tends to decrease across a period, it grows from $\ce{O}$ to $\ce{F}$ because $\ce{F2}$ has a single bond while $\ce{O2}$ has a double one.)
As you progress further along the transition metals the delocalization of d electrons in the metal goes down. That is, although there are more d electrons in $\ce{Co}$ than in $\ce{Fe}$, their effectiveness in binding atoms together isn't really any larger. As a result, the distance between neighbors (hence the metallic radius) is the same for both.