The correct order of IE_2 is | vedclass

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Question

(C) The second ionization energy $(IE_2)$ is the energy required to remove an $e^-$ from a $1+$ cation in the gaseous state: $A^{+}_{(g)} \rightarrow

Vedclass · Accepted Answer

$Ne > O > F > N$

Vedclass · Answer

(C) The second ionization energy $(IE_2)$ is the energy required to remove an $e^-$ from a $1+$ cation in the gaseous state: $A^{+}_{(g)} \rightarrow A^{2+} + e^-$. $IE_2$ is influenced by atomic size,effective nuclear charge,and electronic configuration. For the elements $N, O, F, Ne$,the electronic configurations of their $1+$ ions are: $N^{+} (2s^2 2p^2)$,$O^{+} (2s^2 2p^3)$,$F^{+} (2s^2 2p^4)$,$Ne^{+} (2s^2 2p^5)$. As we move from $N$ to $Ne$,the effective nuclear charge increases and the size of the ion decreases,which generally increases the $IE_2$. However,$O^{+}$ has a stable half-filled $2p^3$ configuration,making it harder to remove an electron compared to $F^{+}$ $(2p^4)$. Therefore,the correct order of $IE_2$ is $Ne > O > F > N$.

The correct order of $IE_2$ is

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