Why is the ionization energy of sulfur anion more than that of oxygen anion?

Question

Which of the following are in the correct order of their ionization energies? (multi-answer question)

1. $\ce{O > S > S- >O-}$
2. $\ce{F > F- > Cl- > Cl}$
3. $\ce{O > O- > S- > S}$
4. $\ce{F > Cl > Cl- > F-}$

The answers are 1 and 4.

Why is ionization energy of $\ce{S-}$ greater than $\ce{O-}$? The sulfur ion has an extra shell than an oxygen ion, so the atomic size of sulfur should be greater than oxygen. Therefore, the ionization energy of an oxygen ion should be more than a sulfur ion because the electrons in oxygen are closer to the nucleus when compared to sulfur. But why is the reverse true?

Answer 1

The ionisation energy of an anion is the same as the electron affinity of the element. $$\text{IE}(\ce{A-}) = \text{EA}(\ce{A})$$

Your reasoning is not wrong and one might expect exactly this behaviour. Another point for this reasoning is the substantial lower electronegativity of sulfur ($2.6$) versus oxygen ($3.4$). However, experimental findings contradict this reasoning. $$\begin{array}{lll} &&/\mathrm{kJ\cdot{}mol^{-1}}\\\hline \text{IE}\ce{(O)} & &=1314\\ \text{IE}\ce{(S)} & &=1000\\ \text{IE}\ce{(S− )} &=\text{EA}\ce{(S)} &=200\\ \text{IE}\ce{(O− )} &=\text{EA}\ce{(O)} &=141\\\hline \end{array}$$

Sources: Wikipedia articles on ionisation energies, electron affinities

Basically it all boils down to electron-electron repulsion. Oxygen ($152~\mathrm{pm}$) is much smaller than sulfur ($180~\mathrm{pm}$) (van der Waals radius, source). This goes hand in hand with the smaller orbitals (lower main quantum number for the valence shell) The electrons can therefore only occupy lesser space and repulse each other more. This is mainly the reason, why sulfur has a greater electron affinity than oxygen.

The same reasoning may be applied to fluorine and chlorine.

Answer 2

Write the electronic configuration for all atoms: $$\begin{array}{lll}\hline \text{atom}&\text{before ionization}&\text{after ionization}\\\hline \ce{O} &1s^22s^22p^4 &1s^22s^22p^3(^*)\\\hline \ce{O-} &1s^22s^22p^5 &1s^22s^22p^4\\\hline \ce{S} &1s^22s^22p^63s^23p^4 &1s^22s^22p^63s^23p^3(^*)\\\hline \ce{S-} &1s^22s^22p^63s^23p^5 &1s^22s^22p^63s^23p^4\\\hline \ce{F} &1s^22s^22p^5 &1s^22s^22p^4\\\hline \ce{F-} &1s^22s^22p^6(^*) &1s^22s^22p^5\\\hline \ce{Cl} &1s^22s^22p^63s^23p^5 &1s^22s^22p^63s^23p^4 \\\hline \ce{Cl-} &1s^22s^22p^63s^23p^6(^*) &1s^22s^22p^63s^23p^5 \\\hline \end{array}$$ $(^*)$:Stable Configurations

Since $\ce{O^-}$ and $\ce{S^-}$ go to stable configurations $\ce{S,O>S^- ,O^- }$.

Since size of $\ce{S}$ is larger than $\ce{O}$ $\implies$ $\ce{S < O } $. Assuming IE of $S^-$ reverse of EA of $S$ and $O^-$ reverse of EA of O, $O^-<s^-$ because S has large size to accomodate electron and less repulsion than small O.

[My incorrect order $\ce{S<O < S- < O- }$ was actually $\ce{S<O < S+ < O+ }$]

Thus, $\ce{O^- < S^- < S < O }$.

Apply same concept to others; Remember:

• a)Nuclear Charge
• b)Shielding
• c)Atomic radii
• Penetration Effect[s>p>d>f]
• d)Stability of fully and halfly filled orbitals.
• Successive Ionisations are always harder starting from neutral compound.$IE_1\ll IE_2\lll IE_3$

As a rule of Thumb:

• $IE\propto \frac{(a)(d)}{(b)(c)} $, very roughly