Here are the electron affinities of the 16th and 17th groups. It is applicable to all p-block element groups. d) O-ion has comparatively larger size than oxygen atom. Use this … Electron affinity is the amount of energy required to detach one electron from a negatively charged ion of an atom or molecule. Other names: Molecular oxygen; Oxygen molecule; Pure oxygen; O2; Liquid oxygen; UN 1072; UN 1073; Dioxygen Permanent link for this species. Click here to buy a book, photographic periodic table poster, card deck, or 3D print based on the images you see here! For illustration, the overall E.A. ?H_dep (Mg) -148. kJ/mol Mg(g) first ionization energy 738. kJ/mol B. the ionization energy of O2-. When an electron is added to a neutral gaseous oxygen atom , the process is exothermic and 141 kJ energy is liberated. 1st Electron Affinity is usually exothermic as the energy released when the nucleus attracts the the additional electron is larger than the energy absorbed to overcome inter-electronic repulsion. C. the second ionization energy of O. D. twice the electron affinity of O+.E. The second electron affinity (EA 2) is always larger than the first electron affinity (EA 1) as it is hard to add an electron into a negative ion than a neutral atom. Therefore spin pair repulsion exists and energy is needed to overcome it ...however the 1st electron affinity of oxygen is still negative. The electron affinity of oxygen is equal to A. the ionization energy of O-. c) O-ion will tend to resist the addition of another electron . Calculate the second electron affinity of oxygen using the information below: O^- (g) + e^- ? for the formation of oxide or sulphide ions has been determining to be endothermic to … Solution: When an electron is added to negatively charged ion, it experiences more repulsion rather than attraction. O^2+ (g) EA_2 =? O-(g) + e- O 2-(g) EA= +780 kJ. It is indicated using the symbol Ea and is usually expressed in units of kJ/mol. a) oxygen is more electronegative . b) oxygen has high electron affinity . This can be understood with the following example of the successive electron affinity of oxygen: O(g) + e- O-(g) EA= -141 kJ. none of these. Electron affinity follows a trend on the periodic table. Use the Born-Haber cycle and data from Appendix IIB and Table 9.3 in the textbook to calculate the lattice energy of CaO. The general trend for electron affinity down the group is that it decreases because of the increase in atomic radius.The exception of $\ce{Cl}>\ce{F}$, I can understand is because fluorine has a high electron density and it is unfavourable to add more electrons as it would only increase the electron -electron repulsion. Electron affinity increases from the first element to the second element in a group. Hence, oxygen has lesser electron affinity than that of sulphur and however in a group, it decreases from top to bottom as size increases. But in the case of oxygen, the first electron is added to a 2p subshell where one electron is already present. You may need the following data: electron affinity of oxygen EA_1=-141 kJ/mol, EA_2=744 kJ/mol; ionization energy of calcium IE_1=590 kJ/mol, IE_2=1145 kJ/mol Comparatively larger size than oxygen atom to a neutral gaseous oxygen atom the addition of another electron from. 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