Ionization Energy and Stable Ions

[STAii]

Greetings !
I have a little question.
My teacher was talking about stable ions some days ago, he wanted to explain why do we have certain stable ions in ionic compounds and we don't have other ions.
For example, we notice that the stable ion of Na (in its compunds) is Na⁺¹ and not Na⁺² or Na⁺³.
The teacher explained this depending on the ionization energy, he says that there is a big difference between the first and second ionization energy of Na (since we are taking the second electron from a completed shell nearer to the necleus), So it is easier (during a reaction) to take off one electron than 2 of them.
When he tried to use the same way with Mg, he said there is a big difference between the second and third ionization energy, therefore it is easier to take off 2 electrons than 3. So the stable ion is Mg⁺².

But .. thinking about this his way, it doesn't make a lot of sense.
Since it is even easier (in Mg) to remove only one electron, and make Mg⁺¹, right ?
Before he said that, i used to think that the stable ion charge depends on how many electron the atom needs to loose/gain to actually become stable (you know, normally have a last shell with 8 electrons).

Any comment/help in making me understand where i went wrong, or where i am missing sth would be appreciated.

 

[Monique]

Hi Staii, didn't see you in a while :) could you tell me how the orbital shells are occupied for Mg and Na?

 

[STAii]

Umm ..
₁₁Na : 1s² 2s² 2p⁶ 3s¹
₁₂Mg : 1s² 2s² 2p⁶ 3s²

But what's the point ?

EDIT : btw, cool new Avatar

 

[Monique]

Well, there is your answer right?

Na has 1 electron in 3s orbital that can be taken, all other orbitals will be filled = stable.

Mg has 2 electrons in 3s, if you take one, the orbital is only partially filled = not stable, thus two are taken.

This is an out-of-my-head explanation, so anyone feel free to correct me.

 

[STAii]

Well, this is what i used to think !
see :

Before he said that, i used to think that the stable ion charge depends on how many electron the atom needs to loose/gain to actually become stable (you know, normally have a last shell with 8 electrons).

But my teacher's explanation has nothing to do with having a stable last shell ...

 

[Monique]

Oh, I see what you mean:

When he tried to use the same way with Mg, he said there is a big difference between the second and third ionization energy, therefore it is easier to take off 2 electrons than 3. So the stable ion is Mg+2.

I think he is talking about the same thing, the above example would be the same as for taking 1 or 2 electrons off of Na. You are saying that in this reasoning it would take less ionization energy to just take one..

Maybe the ionization energies of the 3s1 and 3s2 are very similar? That would explain it I guess... they are in the same shell. Where as the third ionization energy would require getting an electron from another shell..

 

[STAii]

Umm ... doesn't seems so.
First ionization energy of Mg : 737.6 (kilojoule/mole)
Second ionization energy of Mg : 1450 (kilojoule/mole)
(i am not sure of the unit though .. but i guess i am right).
And Monique, even if the first and second ionization energies are close, remember that you have to take off the first electron anyway to be able to take the second one away, so you have to calculate the first ionization energy plus the second one, so the sum up will be bigger than the first ionization energy of Na anyway !

Any other thing ?

 

[Monique]

I think this whole ionization story is strange.. in my opinion charge depends on the number of valence electrons.

 

[Chemicalsuperfreak]

Originally posted by STAii
Umm ... doesn't seems so.
First ionization energy of Mg : 737.6 (kilojoule/mole)
Second ionization energy of Mg : 1450 (kilojoule/mole)
(i am not sure of the unit though .. but i guess i am right).
And Monique, even if the first and second ionization energies are close, remember that you have to take off the first electron anyway to be able to take the second one away, so you have to calculate the first ionization energy plus the second one, so the sum up will be bigger than the first ionization energy of Na anyway !

Any other thing ?

So what's the third ionization energy of magnesium? How does the second ionization energy of magnesium compare to the second i.e. of sodium?

 

[Zargawee]

So what's the third ionization energy of magnesium? How does the second ionization energy of magnesium compare to the second i.e. of sodium?

The 3rd ionization energy for Mg is 7732 KJ/mole
and the 2nd ionization energy for Mg is 1450 KJ/mole while Na 2nd ionization energy is 4565 KJ/mole , which means it's much higher than Mg ionization energy .


ionization energy depends on effective nuclear charge, which increases each time you take an electron from the atom.

When Mg reacts with a metalloid , the metalloid will become a negative ion , and if we assumed that Mg would loose only one electron , then the compound would be Mg⁺X^- , the electron remaining in Mg last orbit would be under a very repulsion force from the negative ion, therefore the compound might not reach the least energy situation.
I guess this is a reason , I'm not sure , this is my own explanation.

 

[STAii]

Good explanation Zargawee (So, you don't support that the stable ion charge depends on the ionization energy).
So, can anyone explain how the stable ion charge can depend on the ionization energy (if not, then i will just go back to my old beleif; it is all about valence electrons).

 

[Chemicalsuperfreak]

Originally posted by Zargawee
The 3rd ionization energy for Mg is 7732 KJ/mole
and the 2nd ionization energy for Mg is 1450 KJ/mole while Na 2nd ionization energy is 4565 KJ/mole , which means it's much higher than Mg ionization energy .

There you go Staii, it's much easier to remove the second electron from Mg than from Na. So why doesn't Mg stay at Mg(I)? My guess is that 1450 kJ/mole isn't too much energy to ask for and/or there are more factors going on than just ionization energy.

 

[STAii]

Originally posted by Chemicalsuperfreak
There you go Staii, it's much easier to remove the second electron from Mg than from Na. So why doesn't Mg stay at Mg(I)? My guess is that 1450 kJ/mole isn't too much energy to ask for ...

I was not suggesting having Na⁺² (so it is meaningless to compare the second ionization energy of Na with the second ionization energy of Mg), as i said, there is no problem with this explanation when u look at Na, the problem is when u look at Mg.
Well, 1450 KJ/mole actually is a lot to ask for ! And even if it wasn't, it would be even better not to ask for it from the first place (and form Mg⁺¹).
I think that using this method, all positive ions should be +1, which doesn't make sense (And therefore, i conclude, this way should be wrong).

... there are more factors going on than just ionization energy

I agree.

 

[qatadah]

Perfect Answer 2 STaii

The electron distribution of Na is as follow:
11Na: 1s2 2s2 2p6 3s1 or
11Na: [Ne] 3s1

as you know Noble gases are the most stability an atom can have. So the atom tends to get rid of the 1 electron in 3s, having low IE and be like the noble gas Ne.
when it has the electronic distribution of the noble gas, it is stable, and holds on 2 the electrons tight, having a very high IE for three 3 reasons:
1- when u take the 3s electron, (n) decreases to be 2 and it is closer to the atom so electrons get more attracted, needing a very high IE.
2- when u take out the valence electrons (1e in this case), the efficient nuclear charge increases decreasing the size of the atom, requiring a very high IE.
3- taking out valence electrons, the atom has an electronic distribution similar to the noble gas, which is stable, requiring a high IE.
So Na is happy with Na+1

But in the case of Mg:
12mg: 1s2 2s2 2p6 3s2 or
12Mg: (Ne) 3s2
if it gets rid of 1e it would still be unstable, so it gets rid of the other one, getting rid of the 2e in 3s and for the same previous reasons it is happy with noble gas electronic distribution. forming Mg+2 and not other charges.