[Needaask]

We relate the ionization energy with the effective nuclear charge. So shouldn't the second ionization require the same amount of energy to remove the electron?

The explanation given in my notes are that it is harder to remove an electron from a positive charged atom than a neutral atom. However I don't quite understand this because if we purely use effective nuclear charges, it would be the same.

I would think that a better explanation would to be that with less electrons in the valence shell, there is less repulsion from any one electron in the valence shell. So for an electron to escape, more energy is required to remove it.

Would these two given explanations (the one in my notes) and the one I gave mean the same thing? Or are they different explanations?

Thanks

[Corribus]

Electrons shield each other from the core nuclear charge.  Remove 1 electron = less shielding.  Also, it takes more work to remove electrons that are closer to the nucleus.  Therefore ionization energy increases as the n quantum number decreases.

[Needaask]

Electrons shield each other from the core nuclear charge.  Remove 1 electron = less shielding.  Also, it takes more work to remove electrons that are closer to the nucleus.  Therefore ionization energy increases as the n quantum number decreases.

Hi Corribus

I agree but I was thinking, since effective nuclear charge=number of protons-electrons in all but valence shell, so shouldn't they be the same? So if we were to change the formula to effective nuclear charge=number of protons-all but 1 electron in the valence shell, then all of them would have the same effective nuclear charge so actually why would a group 2 atom be harder to ionize than a group 1 atom?

[Corribus]

Remember, electrons are  probability distributions.  Let's say you have helium, which has two electrons in a 1s orbital.  At any given time, the two electrons are unlikely to be the same distance from the nucleus.  Most times, one will be closer than the other.  Thus electrons in the same orbital can shield each other - and they affect each other through mutual repulsion.

When you remove one electron from helium, the electron configuration changes quite a bit.  You should not find it too surprising that the energy required to remove them sequentially changes.

[Needaask]

Remember, electrons are  probability distributions.  Let's say you have helium, which has two electrons in a 1s orbital.  At any given time, the two electrons are unlikely to be the same distance from the nucleus.  Most times, one will be closer than the other.  Thus electrons in the same orbital can shield each other - and they affect each other through mutual repulsion.

When you remove one electron from helium, the electron configuration changes quite a bit.  You should not find it too surprising that the energy required to remove them sequentially changes.

Hi Corribus

I agree with the notion that if i remove one electron there's less repulsion experienced per electron after that. But actually now I don't quite understand the effective nuclear charge.

The formula for effective nuclear charge is=total proton(nuclear charge)-all electrons except those in the valence electron. So essentially for group 1 it would roughly give me 1 and for group 2, 2. But now if we were to include electrons in the valence shell making it effective nuclear charge for 1 valence electron =total protons-all electrons but that 1 valence electron. Then the formula would just give us 1 for everything. So now it doesn't really make sense.

So actually what's wrong with using the effective nuclear charge formula this way?

Thanks Corribus

[Sossy]

Have in mind that electrons in different atomic oritals have different shielding effect and hence the resulting effective nuclear charge will depend on the distribution of electrons in the orbitals. A simple way to calculate the effective nuclear charge in an atom or ion is by using Slater's rules.

[Needaask]

Have in mind that electrons in different atomic oritals have different shielding effect and hence the resulting effective nuclear charge will depend on the distribution of electrons in the orbitals. A simple way to calculate the effective nuclear charge in an atom or ion is by using Slater's rules.

Ohh i read through the wikipedia page. So is the rule like a more complicated variant of the simple formula taught to me of Proton no-electrons except in valence shell?

So in this case if i were to use the Slater's Rule to find out the effective nuclear charge on any one of those valence electrons I wouldn't get just 1 like when i used Proton number-All electrons except 1 in the valence shell=1 right?

[Corribus]

All electrons are going to shield all other electrons by some degree.  The amount is not something that can be easily determined because of uncertainty in electron-electron interaction energies.  We can only estimate.

[Needaask]

All electrons are going to shield all other electrons by some degree.  The amount is not something that can be easily determined because of uncertainty in electron-electron interaction energies.  We can only estimate.

Oh but using the slater's rules would give me a number that's not going to be 1 for any one valence electron right?

[Corribus]

Using Slater's rules will give you a better approximation of shielding than simply taking 1 or 0.  The nuclear shielding for an electron is never precisely 1 and never precisely 0.  It can approach these values (the nuclear shielding of the 6s by the 1s electron is virtually unity, for example, just as the shielding of a 1s electron by a 6s electron is virtually nothing), but the wavefunction tails off to infinity, so the probability value is never precisely 0, no matter how far away from the nucleus you are.

[Needaask]

Using Slater's rules will give you a better approximation of shielding than simply taking 1 or 0.  The nuclear shielding for an electron is never precisely 1 and never precisely 0.  It can approach these values (the nuclear shielding of the 6s by the 1s electron is virtually unity, for example, just as the shielding of a 1s electron by a 6s electron is virtually nothing), but the wavefunction tails off to infinity, so the probability value is never precisely 0, no matter how far away from the nucleus you are.

Ohh! I think I'll appreciate these rules a lot better because of this post but going back to my simpler level can i just explain it just like this: As there are less electrons there is less shielding effect. Thus the pull of electrons by the nucleus is stronger making it harder to remove the 2nd electron?

[Corribus]

As there are less electrons there is less shielding effect. Thus the pull of electrons by the nucleus is stronger making it harder to remove the 2nd electron?

Yes this is a perfectly reasonable explanation.