Which one is the electron configuration of "Fe"^(+)? [Ar] 3d^5 4s^1, or [Ar]3d^6?
1 Answer
Neither of them are correct! The correct electron configuration is
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,
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
That establishes the electron configuration
In diagram
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
color(blue)ul([Ar]3d^6 4s^2)
Now you may wonder,
"why is the
4s ionized instead of the3d first, even though inFe , the sixth3d electron is highest in energy?"
The radial extent of the
Graphed from H atom wave functions
The
So, the