lanthanide contraction across the period.... help
 

I cant understand lanthanide contraction....as we go along the period if the number of electrons increase in the f-subshell, then the shielding effect should be more pronounced..instead they say that the shielding effect is reduced and hence the nucleus can attract the outer shelll electrons more strongly and ultimately the size decreases...but WHY the addition of the electrons in f-subshell across the period reduce the shielding effect??

Don't try to think of the shielding effect as a complex attribution of an atom. The shielding effect is an attribution of a subshell. The s subshell has the best shielding attributions and the f subshell has the worst.

So try to think of it this way: electrons in the f subshell do not shield the increased nuclear charge as much as an s subshell would. And because of this the nuclear charge can affect electrons on outer layers. This I think is because of the shape of the subshell. You know s subshell is a sphere, p is three dumbell shaped things, etc. So they start out with a sphere covering the nucleus. Three dumbells can't really cover a spheric nucleus that well so the shielding effect is reduces. And as you move along the shape covers less of the still spheric nucleus, therefore allowing the nuclear charge to have its effect on outer electrons.

thanks...that really cleared the concept :)

searching goolge for 15 minutes trying to find a good answer for this lanthanide contraction and this response really cleared up the concept. makes perfect sense now. thanks for the indirect answer!

-Kris

:D

The lanthanide contraction is amazing and the shielding explanation is fascinating.

Without the lanthanide contraction gold would be reactive (and probably worthless) and really dense elements like osmium would have normal densities.