Correct order of 3^{rd} Ionisation energy of | vedclass

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Question

(A) The $3^{rd}$ ionisation energy involves the removal of an electron from the $M^{2+}$ ion. The electronic configurations of the $M^{2+}$ ions for $

Vedclass · Accepted Answer

$O > N > C > B$

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(A) The $3^{rd}$ ionisation energy involves the removal of an electron from the $M^{2+}$ ion. The electronic configurations of the $M^{2+}$ ions for $B, C, N,$ and $O$ are:$B^{2+} (Z=5): 1s^2 2s^1$$C^{2+} (Z=6): 1s^2 2s^2$$N^{2+} (Z=7): 1s^2 2s^2 2p^1$$O^{2+} (Z=8): 1s^2 2s^2 2p^2$Comparing these,$C^{2+}$ has a stable fully filled $2s^2$ subshell,which makes its $3^{rd}$ ionisation energy significantly higher than expected.$N^{2+}$ and $O^{2+}$ have higher effective nuclear charge than $B^{2+}$ and $C^{2+}$.The actual order is $N > O > C > B$ is incorrect due to the stability of the $2s^2$ configuration in $C^{2+}$.Correct order is $N > O > C > B$ is wrong; the correct sequence is $O > N > C > B$ based on effective nuclear charge,but considering the stability of $C^{2+}$,the order is $N > O > C > B$ is incorrect. The correct order is $O > N > C > B$.

Correct order of $3^{rd}$ Ionisation energy of $B, C, N$ and $O$ will be

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