What is the correct order of electron affinity for $B, C, N$,and $O$?

  • A
    $O > C > N > B$
  • B
    $B > N > C > O$
  • C
    $O > C > B > N$
  • D
    $O > B > C > N$

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Similar Questions

When a chloride ion is formed from an isolated gaseous chlorine atom,$3.8 \ eV$ of energy is released. This value is equal to the:

The formation of the oxide ion,$O^{2-}_{(g)}$,from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta_f H^{\Theta} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^- \to O^{2-}_{(g)} ; \Delta_f H^{\Theta} = +780 \ kJ \ mol^{-1}$
Thus,the process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,

In which of the following is the electron gain enthalpy of elements correctly arranged?

Halogens have maximum negative electron gain enthalpy in the respective periods of the periodic table. Why?

The formation of the oxide ion,$O^{2-}_{(g)}$ from an oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta_f H^o = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^- \to O^{2-}_{(g)} ; \Delta_f H^o = +780 \ kJ \ mol^{-1}$
Thus,the process of formation of $O^{2-}$ in the gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. This is due to the fact that,

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