How would you explain this exception to the general trend in terms of electron arrangements and attraction/repulsion?

The trend for ionization energy is a general increase from left to right across a periodowever, magnesium (Mg) is found to have a higher first ionization energy value than aluminum (Al).

1 Answer
Mar 12, 2018

Answer:

Well, write out the individual electronic configurations and see what you gets...

Explanation:

#Mg, Z=12; 1s^(2)2s^(2)2p^(6)3s^2#...#"1st ionization energy, 737"*kJ*mol^-1...#

#Al, Z=13; 1s^(2)2s^(2)2p^(6)3s^(2)3p^1#...#"1st ionization energy, 577.5"*kJ*mol^-1...#

I include the data, because as chemists, as physical scientists, we should always interrogate the data before we shoot our mouths off. And the premise of your question is correct. A priori, we would expect that aluminum, #Z=13#, to have a higher ionization enthalpy...the which represents this reaction...

#M(g) + Delta rarr M(g)^(+)+e^(-)#

...than that of magnesium, #Z=12#, for the simple reason that the aluminum atom has 13 nuclear charges, versus 12 nucular charges. However, as the electronic configurations clearly show, the first ionization of aluminum requires removal of a #"p orbital electron"#, the which has ZERO probability of occurring at the core, and thus a p electron should be LESS strongly bound and therefore easier to remove. Do the data support this interpretation? (And of course these data underlie the electronic configurations.)

The successive ionization energies (and these are available) should reflect the orthodoxy...I leave it to you to find these data.