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

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

Vedclass · Accepted Answer

$O > C > N > B$

Vedclass · Answer

(B) The $3^{rd}$ ionisation energy corresponds to the removal of an electron from the $M^{2+}$ ion.The electronic configurations of the $M^{2+}$ ions 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$Comparing these:$B^{2+}$ has a single electron in the $2s$ orbital,which is relatively easy to remove.$C^{2+}$ has a stable,fully filled $2s^2$ subshell,making it very difficult to remove an electron.$N^{2+}$ and $O^{2+}$ have electrons in the $2p$ subshell.Due to higher effective nuclear charge,$O^{2+}$ has a higher $3^{rd}$ $IE$ than $N^{2+}$.Comparing $C^{2+}$ and $N^{2+}$,the removal of an electron from the stable $2s^2$ configuration of $C^{2+}$ requires more energy than from the $2p^1$ configuration of $N^{2+}$.Thus,the order is $O > C > N > B$.

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

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