Oxygen ions, electron affinity

By introducing reasonable values (about 2 for nitrogen, 4 for oxygen) for the electron affinity parameter relative to carbon, 8, and for the induced electron affinity for adjacent atoms (32/8i = Vio), we have shown that the calculated permanent charge distributions for pyridine, toluene, phenyltrimethylammonium ion, nitrobenzene, benzoic acid, benzaldehyde, acetophenone, benzo-nitrile, furan, thiophene, pyrrole, aniline, and phenol can be satisfactorily correlated qualitatively with the observed positions and rates of substitution. For naphthalene and the halogen benzenes this calculation does not lead to results... 

Use the following data and data given in Tables 6.2 and 6.3 to calculate the second electron affinity, Eea2, of oxygen. Is the O2- ion stable in the gas phase Why is it stable in solid MgO ... 

In the case of the reactions with the aromatic compounds, the nature of the reaction was established by observing the optical absorption spectrum of the transient product. Thus, in the reaction with diphenyl and p-terphenyl, the observation of the spectra of the mononegative ions (3) of these compounds establishes the occurrence of an attachment reaction. The rate constants for the reactions with diphenyl, naphthalene, p-terphenyl, and naphthacene increase in the order of increasing electron affinities of these compounds. In the case of the attachment reaction to oxygen, no attempt has been made to study the product ion in the alcohols. We have, however, obtained (7) the absorption spectrum of the O2 ion in aqueous solution. 

The electron affinity E of the halogens is indeed positive, but for oxygen and sulphur energy is necessary to form a two-fold negative ion, namely about 150 and 90 kcal/gr.atom. (Actually 71 and 48 keal/gr. atom respectively, are set free in the formation of O- and S ). The electron affinity of hydrogen is weakly positive 16.5 kcal/gr.atom are set free in the formation of H. ... 

Which has the more negative electron affinity-, the oxygen atom or the O- ion Explain your answer. 

---