Consider the following conversions:
$(i) \ O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta H_1$
$(ii) \ F_{(g)} + e^- \to F^{-}_{(g)} ; \Delta H_2$
$(iii) \ Cl_{(g)} + e^- \to Cl^{-}_{(g)} ; \Delta H_3$
$(iv) \ Na_{(g)} \to Na^{+}_{(g)} + e^- ; \Delta H_4$
Which of the following statements is incorrect?

  • A
    $\Delta H_1$ and $\Delta H_2$ are less negative than $\Delta H_3$
  • B
    $\Delta H_2$ is more negative than $\Delta H_1$
  • C
    $\Delta H_2$ and $\Delta H_3$ are negative while $\Delta H_1$ is positive
  • D
    $\Delta H_1, \Delta H_2$ and $\Delta H_3$ are negative while $\Delta H_4$ is positive

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Nitrogen has lower electron affinity than its preceding element carbon because

The correct order of electron gain enthalpies of $Cl$,$F$,$Te$,and $Po$ is .....

The correct order of electron affinity is:

Column-$I$ shows the electronic configuration of some elements and Column-$II$ shows the values of electron gain enthalpy. Match these values correctly. Match the electronic configuration with the electron gain enthalpy.
Column-$I$ Electronic Configuration Column-$II$ Electron Gain Enthalpy / $kJ \ mol^{-1}$
$(A) \ 1s^2 2s^2 2p^6$ $(1) -53$
$(B) \ 1s^2 2s^2 2p^6 3s^1$ $(2) -328$
$(C) \ 1s^2 2s^2 2p^5$ $(3) -141$
$(D) \ 1s^2 2s^2 2p^4$ $(4) +53$

Which one of the elements has the maximum electron affinity?

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