A = 1s^2, 2s^2, 2p^4,B = 1s^2, 2s^2, 2p^5,C = | vedclass

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(A) The electronic configurations are: $A = 1s^2\, 2s^2\, 2p^4$ (Oxygen atom,$O$),$B = 1s^2\, 2s^2\, 2p^5$ (Oxygen anion,$O^-$),$C = 1s^2\, 2s^2\, 2p^

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$B_{(g)} + e^- 	o C_{(g)}$ is exothermic

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(A) The electronic configurations are: $A = 1s^2\, 2s^2\, 2p^4$ (Oxygen atom,$O$),$B = 1s^2\, 2s^2\, 2p^5$ (Oxygen anion,$O^-$),$C = 1s^2\, 2s^2\, 2p^6$ (Oxygen dianion,$O^{2-}$).$1$. The process $A + e^- 	o B$ $(O + e^- 	o O^-)$ is exothermic because energy is released when an electron is added to a neutral atom.$2$. The process $B + e^- 	o C$ $(O^- + e^- 	o O^{2-})$ is endothermic because energy is required to overcome the electrostatic repulsion between the incoming electron and the negatively charged $O^-$ ion.$3$. Therefore,the statement '$B_{(g)} + e^- 	o C_{(g)}$ is exothermic' is incorrect.$4$. Ionization energy $(A_{(g)} 	o A_{(g)}^{+} + e^-)$ is always endothermic,so option $B$ is incorrect.$5$. The process $C_{(g)} 	o A_{(g)} + 2e^-$ is the reverse of the two electron gain processes. Since the first step is exothermic and the second is endothermic,the overall process $C 	o A + 2e^-$ is endothermic.$6$. The process $B_{(g)} 	o A_{(g)} + e^-$ is the reverse of the first electron gain process $(A + e^- 	o B)$. Since $A + e^- 	o B$ is exothermic,the reverse process $B 	o A + e^-$ must be endothermic.$7$. Re-evaluating the options based on standard chemical principles,the provided image suggests $A + e^- 	o B$ is exothermic and $B + e^- 	o C$ is endothermic. None of the options provided in the prompt are technically correct based on standard definitions. However,if we assume the question implies identifying the nature of electron gain,$A + e^- 	o B$ is exothermic.

$A = 1s^2\, 2s^2\, 2p^4$,$B = 1s^2\, 2s^2\, 2p^5$,$C = 1s^2\, 2s^2\, 2p^6$. $A, B$ and $C$ are atoms/anions of the same element. Which one is correct? (One of $A/B/C$ is a neutral atom)

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