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. 35Inorganic 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 functionsGraphed 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.