Why is it easier to oxidise #Fe^(2+)# to #Fe^(3+)# than it is to oxidise #Mn^(2+)# to #Mn^(3+)#?

1 Answer
Truong-Son N.
Nov 13, 2017

Well, consider the NEUTRAL electron configurations:

#"Fe": [Ar] 3d^6 4s^2#
#"Mn": [Ar] 3d^5 4s^2#

The #4s# orbital is higher in energy in these atoms, so it is ionized first:

#"Fe"^(2+): [Ar] 3d^6#
#"Mn"^(2+): [Ar] 3d^5#

Drawn out:

#"Fe"^(2+): ul(uarr darr)" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))#

#"Mn"^(2+): ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))" "ul(uarr color(white)(darr))#

A single oxidation is the act of singly ionizing:

#"M"^(2+) -> "M"^(3+) + e^(-)#

The electron to be removed from #Fe^(2+)# is paired, having charge repulsions (making it easier to remove, i.e. the ionization energy is smaller).

Therefore, it is easier to ionize #"Fe"^(2+)# than #"Mn"^(2+)#.

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