Which one is the electron configuration of #"Fe"^(+)#? #[Ar] 3d^5 4s^1#, or #[Ar]3d^6#?

1 Answer
Aug 20, 2017

Neither of them are correct! The correct electron configuration is #[Ar] 3d^6 4s^2#...


Consider the following diagram...

Inorganic Chemistry, Miessler et al., pg. 35

By the convention given in the diagram, the first electron in an orbital is taken to have spin down, #m_s = -1/2#. In #(b)#, note the relative energies of the #3d# electrons and #4s# electron of #color(red)("Fe"^(+))#:

#overbrace(E_(3d(m_s = -1//2)))^("5 electrons") < overbrace(E_(4s(m_s = -1//2)))^"1 electron" < overbrace(E_(3d(m_s = +1//2)))^"1 electron" < overbrace(E_(4s(m_s = +1//2)))^"1 electron"#

This means for #"Fe"^(+)#, so far, the #4s# orbital is such that the first electron in the #4s# orbital (with #m_s = -1//2#) is lower in energy than the sixth #3d# electron.

That establishes the electron configuration #color(red)([Ar]3d^6 4s^1)# for #color(red)("Fe"^(+))#.

In diagram #(a)#, we see #"Fe"# (not #"Fe"^(+)#), decreasing its #4s# orbital energy (the #4s# curves shifted down!), so that...

#overbrace(E_(4s(m_s = -1//2)))^"1 electron" < overbrace(E_(3d(m_s = -1//2)))^("5 electrons") < overbrace(E_(4s(m_s = +1//2)))^"1 electron" < overbrace(E_(3d(m_s = +1//2)))^"1 electron"#

This then agrees with experiment, that the electron configuration of #"Fe"# is...

#color(blue)ul([Ar]3d^6 4s^2)#

Now you may wonder,

"why is the #4s# ionized instead of the #3d# first, even though in #Fe#, the sixth #3d# electron is highest in energy?"

The radial extent of the #4s# is farther out, so its electrons are more accessible for ionization:

Graphed from H atom wave functions

The #3d# electrons are also stabilized by a greater effective nuclear charge. That rapid decrease in #Z_(eff)# across the transition metals is shown in the steeper slope of the #3d# curves in the above diagrams #(a)# and #(b)# than for the #4s# curves.

So, the #3d# is not readily ionized.