The increasing order of electron affinity for the given electronic configurations of elements is:
$I$. $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^5$ $(Cl)$
$II$. $1s^2 \ 2s^2 \ 2p^3$ $(N)$
$III$. $1s^2 \ 2s^2 \ 2p^5$ $(F)$
$IV$. $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1$ $(Al)$

  • A
    $II < IV < III < I$
  • B
    $I < II < III < IV$
  • C
    $I < III < II < IV$
  • D
    $IV < III < II < I$

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Similar Questions

Which of the following electron affinity orders is incorrect?

The formation of the $O^{2-}$ ion is first exothermic and then endothermic,as shown by the following reaction steps:
$O_{(g)} + e^- \to O^-_{(g)}; \Delta H^o = -142 \ kJ \ mol^{-1}$
$O^-_{(g)} + e^- \to O^{2-}_{(g)}; \Delta H^o = 844 \ kJ \ mol^{-1}$
This is due to:

When the first electron gain enthalpy $(\Delta _{eg}H)$ of oxygen is $-141 \ kJ/mol,$ its second electron gain enthalpy is

Which group elements easily accept $1$ and $2$ electrons to achieve noble gas electronic configurations? Why?

Which of the following processes is endothermic?

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