Which of the following orders is correct acco | vedclass

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(C) Let us analyze each option:$A$: The first ionization potential $(IP_1)$ decreases down the group. Thus,the order should be $N > P > As > Sb$. This

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$O > F > N > C$ $(IP_2)$

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(C) Let us analyze each option:$A$: The first ionization potential $(IP_1)$ decreases down the group. Thus,the order should be $N > P > As > Sb$. This is incorrect.$B$: These are isoelectronic species with $10$ electrons. For isoelectronic species,size decreases as the nuclear charge $(Z)$ increases. The order of $Z$ is $N(7)  O^{2-} > F^{-} > Na^{+}$. This is incorrect.$C$: The second ionization potential $(IP_2)$ involves removing an electron from a cation. The electronic configurations are: $C^{+} (2s^2 2p^1)$,$N^{+} (2s^2 2p^2)$,$O^{+} (2s^2 2p^3)$,$F^{+} (2s^2 2p^4)$. Due to the stable half-filled $p$-orbital in $O^{+}$,its $IP_2$ is very high. The correct order is $F > O > N > C$. This is incorrect.$D$: $SBCR$ refers to Single Bond Covalent Radius. Across a period,size decreases. The order of atomic size is $Mg > Al > Si > P$. Therefore,the reverse order $P > Si > Mg > Al$ is incorrect. However,re-evaluating the options,there might be a typo in the question's intended correct answer. Given the standard trends,none of these are strictly correct as written. Assuming the question asks for the correct trend,$IP_2$ for $O > F > N > C$ is the closest to a standard periodic trend exception.

Which of the following orders is correct according to the given property?

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