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I know that metals have low ionization energies and non-metals have high ionization energies, but once a metal/non-metal becomes a cation, why does the cation itself have a ionization energy higher than the original neutral atom?

What is the scientific reason behind this increase in ionization energy?

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A cation always have high ionisation energy whether it comes from a metal or a non metal.We can understand it in this way:

As the atom become a cation,its number of electrons decreases(but notice that number of protons are same,and hence positive charge is more as compared to negetive charge).Since the number of electrons is less now, the effective nuclear charge has increased due to decrease in electron-electron repulsion.

As the number of electron decreases (which happen in a cation),the repulsion between the remaining electrons decreases. Since electron-electron repulsion make it easier to pull out a electron hence a decrease in electron-electron repulsion will contribute towards an increase in resistance of being pulled outside.

Conclusion: As you decrease number of electrons in atom by taking them out,effective nuclear charge increases, due to which electrons are bounded to nucleus strongly, making them hard to remove.

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  • $\begingroup$ This is very loose terminology and is potentially misleading. The protons don't attract the electrons harder, because the nuclear charge is constant. The increased ionisation energy results from the decrease in electron-electron repulsion, not an increase in electron-nucleus attraction. $\endgroup$ – orthocresol Nov 14 '16 at 0:32
  • $\begingroup$ But the number of electron has decreased,shouldn't it affect the attraction?? $\endgroup$ – I am Back Nov 14 '16 at 0:34
  • $\begingroup$ As a pre-emptive statement I will say that there is no doubt that Zeff increases. If you want to say that that represents an increase in the net attraction of electrons to nucleus, fine by me. However the increase in Zeff is due to a decrease in σ (an e-e repulsion term) and not an increase in Z (an e-p attraction term). Which is why I said that it was potentially misleading (not outright misleading). $\endgroup$ – orthocresol Nov 14 '16 at 1:26
  • $\begingroup$ Right, I like this much better. I'm not the person who downvoted though, so I can't help you with that. $\endgroup$ – orthocresol Nov 14 '16 at 1:44
  • $\begingroup$ That's okay.I just want to give a correct answer,a loss of 2 reputations does not matter. $\endgroup$ – I am Back Nov 14 '16 at 1:46
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Consider a simple thought experiment. Your ionisation process is essentially removing a (negatively charged) electron from a remainderatom.

$$\ce{Atom^n+ -> Remainderatom^{(n+1)+} + e-}$$

Throughout this process, you are separating charges, thus there is an electrostatic force driving the electron back to its remainderatom.

If the remainderatom has a charge of $2\ (\Rightarrow n=1)$, the attractive force is twice as strong in simplistic theories than if remainderatom’s charge is $1\ (\Rightarrow n=0)$.

Thus, it is twice as hard to remove a second electron.


This thought experiment makes use of a number of simplifications. For example, it does not consider electron energy levels (orbitals), it does not consider orbital contraction upon ionisation and much more. But it does a good job at explaining the general trend.

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Yes, it has a higher ionization energy. Cations are also smaller than the neutral atom, and the reason for both is the same. The positive charge in the core stays the same, if you remove electrons the interaction between the core and your electrons increases, due to lower shielding, lower Pauli repulsion,and so on, which lowers the energy of those electrons. This directly converts to smaller orbitals and higher ionization energy.

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