Comparing Succesive Ionisation Energies

Question

I am trying to understand successive ionisation energies. In particular, an explanation for why the second ionisation energy is greater than the first ionisation energy. I'm looking for a clear and concise explanation.

The common explanation is that it's harder for an electron to be removed from a cation than the atom or that the second electron feels a greater nuclear charge. However, these explanations do not satisfy me.

For starters, How can the nuclear charge be greater for the second electron when the number of protons within the nucleas remains unchanged. Assuming both the first and second electrons are on the same energy level, they experience the same amount of shielding from inner electrons. So it seems to me, both electrons feel the same nuclear charge. I can see the second electron, perhaps feeling repulsion from the first that results in less nuclear charge. Would this then be correct?

I don't particularly understand the explanation about it being harder to remove an electron from a cation. The way I see it, is that the atomic radius decreases when the first electron is removed, meaning the second electrons feels more electrostatic attraction from being closer. Is this explanation sufficient? Could someone please provide a clear explanation for this?

Answer 1

I will give a classical and a quantum mechanical consideration for a potential answer of your question. The ionization is only dependent on the potential as purely kinetic particles are considered unbound. You are right by stating that it depends on the repulsion between the particles. In the classical case of $N$ electrons as particles occupying the outermost shell, the effective potential before the ionization (1) and after the ionization (2) is :

$$ V_{eff,1} = - \sum_k^{N} \frac{Z_{eff}e^2}{r_k} + \sum_i^{N}\sum_j^{N} \frac{e^2}{|r_i-r_j|} \text{ (1)}\\ V_{eff,2} = - \sum_k^{N-1} \frac{Z_{eff}e^2}{r_k} + \sum_i^{N-1}\sum_j^{N-1} \frac{e^2}{|r_i-r_j|}\text{ (2)} $$

Assuming that the electrons in this shell are closed enough without taking into account the orthogonality and the symmetry of orbitals, the average distance between the electrons in the shell and the nucleus is approximately $r_k \approx R$,. The average distance between the electron is assumed to be $|r_i-r_j| \approx R'$ :

$$ \Delta V = V_{eff,2} - V_{eff,1} = \frac{Z_{eff}e^2}{R} - (2N-1) \frac{e^2}{R'} $$

As $R \gg |r_i-r_j|$ ($R \gg R'$), when an electron is ionized, we obtain $\Delta V \approx - 2N \frac{e^2}{R'}$ which is purely the repulsive part giving $V_{eff,2} < V_{eff,1}$ . Therefore in this simple case the ionization affects mainly the repulsive part of the potential. This means that the potential of the nucleus is more significant after ionization compared to the repulsive part, the bonds are stronger and the second ionization needs more energy.

A quantum mechanical treatment is quite similar, there is no defined position $R$ or a distance $R'$ for the electrons in this case. But the repulsive part has a significant value because $|r_i-r_j| \ll $ especially for singlet states where there is an overlap. This distance is also small for triplet states without an overlap compared to the mean distance to the nucleus (where the probability density is higher) $|r_i-r_j| \ll R \gt a_0$ giving the same result more justified in this case.

Answer 2

Assuming your knowledge level, I will try to focus to easy to understand verbose description. Let assume for the illustration the case of a helium atom.

• If we start with the helium nucleus $\ce{He^2+}$, it has the charge $+2e$.
• If we add 1 electron to have $\ce{He+}$, the electron is attracted by the charge $+2e$.
• If we add another electron to have $\ce{He}$, things get more complicated.
  • If this second electron had been hypothetically somewhere very far and weakly bonded, we would have neglected it and still assumed the first electron had been effectively attracted by $+2e$ of the helium nucleus.
  • If this second electron had been hypothetically inside the nucleus, we could have assumed the first electron had been effectively attracted by the charge $+2e - e = +1e$.
  • As it is none of these opposite extreme cases, it is clear the effective charge of nucleus will be somewhere from the interval $(+e, +2e)$.

If we qualitatively evaluate the mutual interaction of both electrons, the electrons effectively:

• Partly repulse the other electron out of the atom like if they were in the nucleus, pushing the effective nuclear charge toward $+e$ instead of $+2e$.
• Partly increase the nucleus attraction of the other electron. As they are further from the nucleus, pushing the other electron toward the nucleus and the effective nuclear charge toward $+3e$ instead of $+2e$.
• Due mutual electron repulsion is the mean distance of electrons in $\ce{He}$ atom bigger than for the single electron in $\ce{He+}$

Overall effect is expressed by semi-empirical Slater's rules for the effective nucleus charge $Z_{\mathrm{eff}}= Z - s$, where for a helium atom $Z=2$ and for $\ce{1s}$ electrons, $s = 0.30$.

Therefore, for $\ce{He}$ atom, $Z_{\mathrm{eff}} \ne 2$, $Z_{\mathrm{eff}}= 1.7$

Combination of

• the smaller effective nuclear charge (= smaller net force)
• the larger mean distance from the nucleus

leads to the lower ionization energy for the first electron and higher one for the second one.

Helium ionization energies:

$$\begin{array}{|c|c|} \hline \text{IE1} & \text{IE2} \\ \hline \pu{24.58738 eV} & \pu{54.41776 eV} \\ \hline \end{array}$$

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Fur further reading, see

• Wikipedia: Effective nuclear charge
• Wikipedia: Slater's rules

Answer 3

Whatever the structure of the atom, it is more difficult to remove a (negative) electron from a positive structure than to remove an electron from a neutral structure. And it is even more difficult to remove an electron from a doubly charged atom that has already lost two electrons. And if the atoms has already lost three electrons, it is immensely more difficult to remove a $4$th electron. Etc.