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$\ce{O}$ has the 1st electron gain enthalpy $\pu{-141 kJ mol-1}$. $-ve$ value implies that energy is released when electron is added to an isolated atom. This also means that if $\pu{141 kJ}$ energy is supplied to $\pu{1 mol}\ \ce{O-}$ it forms $\pu{1 mol}\ \ce{O}$.

The second electron gain enthalpy has value $\pu{+780 kJ mol-1}$. This implies that energy is to be supplied to form $\ce{O^2-}$. But if we supply $\pu{780 kJ}$ energy to $\ce{O-}$, would not the 1st added electron be removed; since it takes $\pu{141 kJ}$ of energy to be removed?

What am I missing?

Edit $1$- You guys are very rude. Seriously, $3$ downvotes but $0$ comments for a relatively well written question? Get over your attitude people.

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  • $\begingroup$ @Poutnik Are Ionization enthalpy and Electron Gain Enthalpy not different? $\endgroup$
    – Eisenstein
    Jun 27, 2022 at 14:20
  • $\begingroup$ @Poutnik Yes, I know they are different! My question was about electron affinities, but your comment referenced ionization energies. $\endgroup$
    – Eisenstein
    Jun 27, 2022 at 14:26
  • $\begingroup$ @Poutnik Very sorry for the misunderstanding. Please see the edit. $\endgroup$
    – Eisenstein
    Jun 27, 2022 at 14:31
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    $\begingroup$ Useful links for text and formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial. $\endgroup$
    – Poutnik
    Jun 27, 2022 at 14:39
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    $\begingroup$ E.g. write $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ to get $\ce{H2SO4}$ or $\ce{a A <=> p P}$ or $\pu{6.022E23 mol-1}$ (all eventually with double dollars in the display mode like $$\ce{H2SO4}$$. $\endgroup$
    – Poutnik
    Jun 27, 2022 at 14:40

1 Answer 1

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The correct answer is that the oxide dianion is thermodynamically unstable in vacuo (Warning: PDF autodownload) with +780kJ/mol of excess energy, i.e. it does take +780kJ/mol of energy to add an electron to an "oxygen monoanion", AND that the oxide ions we see in, say, magnesium oxide crystals, are sufficiently lowered in energy, by the magnesium cation's positive charges' attracting the oxides' electrons, that they do not have any excess energy and are therefore thermodymically stable.

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  • $\begingroup$ Thanks for the answer. I understand that $O^{2-}$ is unstable, but how can it be formed, since addition of the second electron requires an energy more than that required to remove the 1st added electron? $\endgroup$
    – Eisenstein
    Jun 28, 2022 at 4:51
  • $\begingroup$ Let one say that the neutral oxygen atom has energy 0kJ/mol. The oxygen monoanion has energy -141kJ/mol and the oxygen dianion has energy (-141)+(+780)kJ/mol. NOT (-141)-(+780) kJ/mol. $\endgroup$
    – Kanghun Kim
    Jun 28, 2022 at 4:54
  • $\begingroup$ To say it in different words; you are trying to subtract the "+780" your currenrmost result -141, when in fact you have to add it. $\endgroup$
    – Kanghun Kim
    Jun 28, 2022 at 4:56
  • $\begingroup$ But is it not true that if we add the first electron affinity to the $O^-$ it becomes $O$? $\endgroup$
    – Eisenstein
    Jun 28, 2022 at 5:02
  • $\begingroup$ O-: -141kJ/mol and O: 0kJ/mol $\endgroup$
    – Kanghun Kim
    Jun 28, 2022 at 5:48

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