# Which element has a higher 3rd ionization energy, Al or Mg? Why?

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:

• according to when one examines the ##Z_(eff)##
• according to when one goes against the .

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”)##

## Calculate the price of your order

550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
\$26
The price is based on these factors:
Number of pages
Urgency
Basic features
• Free title page and bibliography
• Unlimited revisions
• Plagiarism-free guarantee
• Money-back guarantee
On-demand options
• Writer’s samples
• Part-by-part delivery
• Overnight delivery
• Copies of used sources
Paper format
• 275 words per page
• 12 pt Arial/Times New Roman
• Double line spacing
• Any citation style (APA, MLA, Chicago/Turabian, Harvard)

# Our guarantees

Delivering a high-quality product at a reasonable price is not enough anymore.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.

### Money-back guarantee

You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.

### Zero-plagiarism guarantee

Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.

### Free-revision policy

Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.