Question

The correct order of second ionization enthalpy of $C,N,O$ and $F$ :

• A. $C>N>O>F$
• B. $O>N>F>C$
• C. $O>F>N>C$
• D. $F>O>N>C$

Answer 1

The correct option is C $O>F>N>C$

Electronic configuration of;

${}_{8}O⟶1s^{2}2s^{2}2p^4$

${}_{9}F⟶1s^{2}2s^{2}2p^5$

${}_{7}N⟶1s^{2}2s^{2}2p^3$

${}_{6}C⟶1s^{2}2s^{2}2p^2$

$2ndIE⟶O^{-}⟶1s^{2}2s^{2}2p^3$

$F^{-}⟶1s^{2}2s^{2}2p^4$

$N^{-}⟶1s^{2}2s^{2}2p^2$

$C^{-}⟶1s^{2}2s^{2}2p^1$

Oxygen having half filled configuration is most stable and will require high energy to remove an $e^{-}$. Followed by fluorine being the most electronegative element, then followed by $N$ which is less electronegative than $F$ but more than $C$.

$\therefore$ Order of second ionization enthalpy is $O>F>N>C$