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