- #1
Pepealej
- 20
- 0
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 :)
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 :)