Atomic radii of transition elements

[Pepealej]

Hi, my question is regarding atomic radii of transition elements.

When we move to the right of the periodic table (in the region of the transition elements) the atomic number increases, thus there is one more proton in the nucleus and on more electron in the atom. The nuclear charge increases due to the extra proton, but the screening increases too due to the extra electron, which is added to a shell with more penetration. Thus the nuclear effective charge remains (aprox.) constant making the atomic radii constant (aprox.) for transition elements.

My question is, how can the extra electron increase screening if it is added to a more energetic shell (3d is more energetic than 4s)? Because a more energetic shell mean farther away from the nucleus, isn't it?

Thanks :)

 

[jbsteele]

The extra electron added to the 3d shell may be further away from the nucleus than the 4s shell, however the 3d shell is more penetrating due to its higher angular momentum. This means that the outer electrons in the 3d shell are able to feel the full effect of the nuclear charge and this helps to screen the nuclear charge better than the 4s shell. Therefore, the extra electron added to the 3d shell increases the screening effect and results in a slightly lower nuclear effective charge, which in turn leads to a smaller atomic radius.