4f is a valid orbital.
I would think the s orbital, because it is closer to the nucleus, and because the outer energy level holds more energy than the inner ones.
Lanthanides are in the '4 f ' block (of the so-called 'secundairy transitionals', atom number 58 to 71) filling up 4f orbital (1 to 14 electrons)
They are further away from the nucleus and easier to remove (requires less energy), i.e. first ionization is less.
some atoms have 5, some have none. the max is 5
4f is a valid orbital.
4f orbital
4f
The element cerium has a single electron in the 4f orbital. Its electron configuration is [Xe] 4f1 5d1 6s2.
in 5s it is filled but in 4d or 4s its half
3 nodes in 4f
I would think the s orbital, because it is closer to the nucleus, and because the outer energy level holds more energy than the inner ones.
Lanthanides are in the '4 f ' block (of the so-called 'secundairy transitionals', atom number 58 to 71) filling up 4f orbital (1 to 14 electrons)
They are further away from the nucleus and easier to remove (requires less energy), i.e. first ionization is less.
-20 ℃
This question is tricky because distance and location are not well defined within the electron cloud. So for the sake of explanation, let us assume that the question is "Which has a region maximum probability closest to the nucleus?" I could not find an answer in print for the 4f vs 6s. However, I did find an answer comparing the 3d and 4s. I believe that an understanding of a comparison between the 3d and 4s can be extended to a comparison between the 4f and 6s. The most probable distance of the 3d is less than the 4s. Therefore, we could extend this and say that most likely the 4f sublevel would have a distance of maximum probability closer than the 6s. You did not ask why so I will not go into an explanation.
some atoms have 5, some have none. the max is 5