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Report (any type)/Brief report
22/09/2021
Magnesium!
There is also an alternative way to approach this that is a bit different but builds on core principles you’ve been taught in General Chemistry, and distinguishes between the magnitude of increase in ionization energy:
As stefan mentions, knowing the electron configurations is a great idea for this question; it helps accompany how you visualize the electron distributions.
THIRD IONIZATION ENERGY
Just as the first ionization energy involves removing the first electron from the highest energy orbital, the third ionization energy involves removing a third electron from the highest energy orbital.
How might one express this? Well, if we let more “##Delta##” mean the application of heat and ##M## be a general metal:
##M(g) + Delta -> cancel(M^(+)(g)) + e^(-)##
##cancel(M^(+)(g)) + Delta -> cancel(M^(2+)(g)) + e^(-)##
##cancel(M^(2+)(g)) + Delta -> M^(3+)(g) + e^(-)##
##”———————————“##
##M(g) + Delta -> M^(3+)(g) + 3e^(-)##
The question is, which element is less likely to be dandy with losing an electron?
ELECTRON CONFIGURATIONS & OCTETS
If you write out the electron configurations, note that ##Mg## and ##Al## are adjacent on the third period/row, and so the answer would have something to do with either the ##Z_(eff)## or a more substantial effect. Let’s see…
##Mg##: ##1s^2 2s^2 2p^6 3s^2##
##Al##: ##1s^2 2s^2 2p^6 3s^2 3p^1##
or…
##Mg##: ##[Ne] 3s^2##
##Al##: ##[Ne] 3s^2 3p^1##
Well, if we remove THREE electrons from each of these metals…
…what we end up doing after removing the third electron is turning ##Al## into a ##Ne##-like cation. That’s pretty favorable, because it gives the atom an octet, which you’ve been taught is in line with stability and fairly inert qualities. That’s why the common oxidation state for ##Al## is ##Al^(3+)##.
However, with ##Mg##, what you end up doing is trying to remove the third electron from ##Mg^(2+)##, which is already ##mathbf(Ne)## -like. That disrupts the stable octet it already has, and is unfavorable.
Hence, we would expect a drastic increase in the third ionization energy of ##mathbf(Mg)## compared to that of ##mathbf(Al)##, and there is!
##”IE”_(Mg)^((2)) = “1450.6 kJ/mol”##
##”IE”_(Mg)^((3)) = color(blue)(“7732.6 kJ/mol”)##
##”IE”_(Al)^((2)) = “1816.6 kJ/mol”##
##”IE”_(Al)^((3)) = color(blue)(“2744.7 kJ/mol”)##
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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