Why is the second ionization energy of Magnesium higher than the first

[September24]

I was looking at some numbers and it seems like the second ionization energy of magnesium is higher than the first. Something like:

1st:700
2nd:1500.

It doesn't seem to make sense to me. For sodium or something it makes sense, since after loosing one electron, it becomes stable as a "noble gas electron configuration". However, after loosing one electron, magnesium still has one left in the valence shell to loose and maintain a "full octet".

For lack of a better term, wouldn't loosing the second electron make magnesium even more stable? If so, why is the second IA more than the first?

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[Beargryllz]

If it loses an electron, wouldn't Mg have a positive charge? Why on earth would it be easier to lose another electron if it's all positive? What kind of electron would leave such paradise? It would take more energy, I think

 

[DrknoSDN]

All second ionization energies are higher than the first. Your making an already charged atom more charged.

http://www.ptable.com/#Property/Ionization

{edit} this trend continues, all 3rd ionization is greater than the 2nd for that atom. etc/

 

[Nabeel06]

I was looking at some numbers and it seems like the second ionization energy of magnesium is higher than the first. Something like:

1st:700
2nd:1500.

It doesn't seem to make sense to me. For sodium or something it makes sense, since after loosing one electron, it becomes stable as a "noble gas electron configuration". However, after loosing one electron, magnesium still has one left in the valence shell to loose and maintain a "full octet".

For lack of a better term, wouldn't loosing the second electron make magnesium even more stable? If so, why is the second IA more than the first?

Think about it like this, if Magnesium has a total of 2 electrons in its most outer sub-shell, they "feel" an attraction to the nucleus where the positive charges are. That attractive force is "pulling" on the 2 (3s) electrons. When Magnesium ionizes to Mg+ (from 3s2 to 3s1) not the attractive force is stronger on that 1 electron. So for every electron that is removed, the next electron requires more energy in order for that electron to be removed.

Does that help? And to answer your question about the nobel gas configuration being more stable, while this is partially true, it isn't 100% correct because the FIRST-Ionization energies increase as you move to the right, but Ionization energies DECREASE as you move down a group (because of radius size)

 

[funtertaining15]

I was looking at some numbers and it seems like the second ionization energy of magnesium is higher than the first. Something like:

1st:700
2nd:1500.



For lack of a better term, wouldn't loosing the second electron make magnesium even more stable? If so, why is the second IA more than the first?

If you use Coulomb's law, the closer the electron is to the nucleus, r, the greater the force of attraction, so like the other SDNer said, there is a greater attraction b/w the 3s1 e- and the nucleus...plus, it's unpaired, which is a stable state. You don't want to mess with a stable e- config....