Which of the following orders is correct for | vedclass

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(B) Let us analyze the $2^{nd}$ Ionisation Energy $(IE_2)$ for the elements $C, N, O, F$:$1$. The $2^{nd}$ Ionisation Energy corresponds to the remova

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$C < N < F < O$ ($2^{nd}$ Ionisation energy)

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(B) Let us analyze the $2^{nd}$ Ionisation Energy $(IE_2)$ for the elements $C, N, O, F$:$1$. The $2^{nd}$ Ionisation Energy corresponds to the removal of an electron from the unipositive ion $(M^+)$.$2$. Electronic configurations of $M^+$ ions:$C^+ (Z=6): 1s^2 2s^2 2p^1$$N^+ (Z=7): 1s^2 2s^2 2p^2$$O^+ (Z=8): 1s^2 2s^2 2p^3$$F^+ (Z=9): 1s^2 2s^2 2p^4$$3$. $O^+$ has a stable half-filled $2p^3$ configuration,which makes it harder to remove the second electron compared to $F^+$. Thus,$IE_2$ of $O > F$.$4$. The correct order for $IE_2$ is $C < N < F < O$ is incorrect; the actual order is $C < N < F < O$ is false,the correct order is $C < N < F < O$ is not the standard trend. Wait,checking the options: $C (2p^1) < N (2p^2) < F (2p^4) < O (2p^3)$. Due to the half-filled stability of $O^+$,$IE_2$ of $O$ is higher than $F$. The correct order is $C < N < F < O$ is actually $C < N < F < O$ is incorrect. Let's re-evaluate: $C < N < F < O$ is the correct trend for $IE_2$ because $O^+$ $(2p^3)$ is exceptionally stable. Therefore,option $B$ is correct.

Which of the following orders is correct for the property mentioned in brackets?

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