The 3^{rd} ionisation energy amongst Li,Be,B | vedclass

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Question

(C) The $3^{rd}$ ionisation energy involves the removal of an electron from a $2+$ cation.For $Li$ $(Z=3)$: The electronic configuration is $1s^2 2s^1

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$B$

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(C) The $3^{rd}$ ionisation energy involves the removal of an electron from a $2+$ cation.For $Li$ $(Z=3)$: The electronic configuration is $1s^2 2s^1$. Removing $3$ electrons is impossible as it only has $3$ electrons total.For $Be$ $(Z=4)$: The configuration of $Be^{2+}$ is $1s^2$. Removing the $3^{rd}$ electron requires breaking the stable $1s^2$ shell,which is very high.For $B$ $(Z=5)$: The configuration of $B^{2+}$ is $1s^2 2s^1$. Removing the $3^{rd}$ electron is relatively easy as it is the valence electron.For $C$ $(Z=6)$: The configuration of $C^{2+}$ is $1s^2 2s^2$. Removing the $3^{rd}$ electron requires breaking the stable $2s^2$ subshell.Comparing the energy required,$B$ has the lowest $3^{rd}$ ionisation energy because the electron is removed from the $2s^1$ orbital,which is less stable than the $1s^2$ configuration of $Be^{2+}$ or the $2s^2$ configuration of $C^{2+}$.

The $3^{rd}$ ionisation energy amongst $Li$,$Be$,$B$ and $C$ is lowest for

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