The correct order of the second ionization en | vedclass

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Question

(C) The electronic configurations of the elements are:$C (Z=6): 1s^2 2s^2 2p^2$$N (Z=7): 1s^2 2s^2 2p^3$$O (Z=8): 1s^2 2s^2 2p^4$$F (Z=9): 1s^2 2s^2 2

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$O > F > N > C$

Vedclass · Answer

(C) The electronic configurations of the elements are:$C (Z=6): 1s^2 2s^2 2p^2$$N (Z=7): 1s^2 2s^2 2p^3$$O (Z=8): 1s^2 2s^2 2p^4$$F (Z=9): 1s^2 2s^2 2p^5$For the second ionization energy,we remove one electron to form the $M^+$ ion:$C^+: 1s^2 2s^2 2p^1$$N^+: 1s^2 2s^2 2p^2$$O^+: 1s^2 2s^2 2p^3$ (Half-filled $p$-orbital,highly stable)$F^+: 1s^2 2s^2 2p^4$Due to the stable half-filled $2p^3$ configuration of $O^+$,it has the highest second ionization energy.Comparing the remaining,$F^+$ has a higher effective nuclear charge than $N^+$,and $N^+$ is higher than $C^+$.Therefore,the correct order is $O > F > N > C$.

The correct order of the second ionization energy of $C$,$N$,$O$,and $F$ is :-

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