If filling the orbitals according to the Aufbau principle, which of the following elements will have the wrong configuration?
#"Fe"# , #"Zn"# , #"Cd"# , #"Hg"#
1 Answer
Well, first of all, that filling order has been made too specific, so this question is flawed in that
Under a certain ASSUMED filling order that is blindly followed,
ORBITAL FILLING ORDERS
In general, the Aufbau principle actually says that the orbitals are USUALLY filled from
Some exceptions are due to relative orbital size or shape, relative closeness of certain orbital energies (e.g.
If we, for the moment, accept this particular filling order suggested by the question, then we would ASSUME a filling order of
However, when one blindly fills the orbitals using this so-called "rule" without considering the orbital energy variations (i.e. by assuming no exceptions ever exist ever), and suggests filling the
#=> overbrace(color(red)([Ar] 3d^8))^"incorrect"#
rather than
#=> overbrace(color(blue)([Ar] 3d^6 4s^2))^"correct"#
and this has been proven by magnetism studies wherein iron has a large magnetic moment (i.e. has many unpaired electrons). This can be found in many textbooks, such as Chemistry: A Molecular Approach (Tro).
WHY NOT THE FIRST CONFIGURATION?
The observed electron configuration (the second one) is correct, and by observing the orbital potential energies, one can make an hypothesis as to why.
As we go across the first-row transition metals, the
However, rather than continuing to add the last two electrons in the
- The
#3d# orbitals are more compact and have no radial nodes, so for a while (for#6# post-argon-core electrons), it is favored to fill them. - The
#3d# orbitals aren't that far in energy from the#4s# orbitals (#"3.75 eV"# difference), so eventually, together with the compactness of the#3d# , it becomes favorable to fill the#4s# orbitals.
(For copper on the other hand, that's when filling the
#4s# is too unfavorable, and we get#[Ar] 3d^10 4s^1# . Zinc just says "fine, the#3d# are filled anyway.")
For perspective, the cumulative, two-electron ionization energy out of the
Why the filling of the