# Electronegativity of Ions

Is there any method to compare the electronegativity of ions such as O-, N+, S-, O+?

Sure. Think about what plays a role in the electronegativities of atoms, for example, why is F more electronegative than H?

F is more electronegative because it has smaller radius w.r.t H.
So I presume you mean size factor will play a role in electronegativity.
Consider these ions-

O (8en & 8p), O+ (7en & 8p), N+ (6en & 7p), N (7en & 7p)

Here how do you determine the sizes?
I have a method to find out the sizes of isoelectronic species. No idea about this one.

Right, it has to do with radius, but also the number of electrons shielding the nucleus and the charge of the nucleus itself. Think of it like this: F has very few electrons between the "edge" of the atom and the nucleus, and the nucleus has 9 protons (which are charged).

When dealing with ions, look at the number of electrons in the valence shell, the number of electrons in non-valence shells (i.e. how will these shield attraction effects from the nucleus on the valence electrons?) and the charge of the nucleus itself. For example, in an anion, there are more valence electrons than in its respective neutral atom, so the total nuclear charge is less than the charge of the valence electrons. This means that the electrons will not be "pulled in" towards the nucleus as much, resulting in a higher radius.

So using this information, would you guess that O is more or less electronegative than O+?

You are explaining a method to compare the radius of an atom and its ion (which I already know)
What I asked was how do you compare ( in other words arrange) ions of different atoms in the increasing order/decreasing order of electronegativies. Like-

O (8en & 8p), O+ (7en & 8p), N+ (6en & 7p), N (7en & 7p)

My fault, sorry for the misunderstanding.

When you're looking at O and N, they're very close to each other on the periodic table, so the electronegativity difference between the neutral atom and the ion is going to be about the same. You know that O+ and N+ are more electronegative than their neutral atoms, and that O is more electronegative than N. So from here, you look at the difference in electronegativity between the two neutral atoms (I think it's around 0.4-0.5). I'm not sure of the exact number, but the change in electronegativity from a neutral atom to its cation is pretty small. So for the case you gave, the order (decreasing) is O+, O, N+, N. Does that make sense?

the order (decreasing) is O+, O, N+, N. Does that make sense?
No it doesn't. The correct answer is O+ > N+ > O > N.

Alright, so it's been a while since I've done electronegativities of atoms vs. ions.

I suppose the ratio of protons to electrons comes into play here. Since you're dealing with fairly small numbers (6, 7, 8) a difference of one electron will make a bigger difference than I expected. Let me check my gen chem text and see if I can find anything, and I'll get back to you.

If anyone else has input on this feel free to help; clearly I don't know as much about this as I thought.

No it doesn't. The correct answer is O+ > N+ > O > N.
electronegativity is a measure of the ability of an atom to attract Bonding electrons in a molecule. So i believe that this term is not defined on ION. I believe you are talking about effective nuclear charge, so the arrangement is correct. Electronegativity is a hypothetical value based on Maths, so different scientists have diff ways of calculating it thus give rise to different scales.

electronegativity is a measure of the ability of an atom to attract Bonding electrons in a molecule. So i believe that this term is not defined on ION. I believe you are talking about effective nuclear charge, so the arrangement is correct. Electronegativity is a hypothetical value based on Maths, so different scientists have diff ways of calculating it thus give rise to different scales.
Yes it is true. This debate is meaningless by definition of electronegativity>