Question

What is the correct order of $2^{nd}$ ionisation energy?

• A. $C<O<N<F$
• B. $C<N<O<I$
• C. $C<F<N<O$
• D. $C<N<F<O$

Answer 1

The correct option is D $C<N<F<O$

The electronic configurations after first ionisation are given below:
$C^{+}\to 1s^{2}2s^{2}2p^1$
$N^{+}\to 1s^{2}2s^{2}2p^2$
$O^{+}\to 1s^{2}2s^{2}2p^3$
$F^{+}\to 1s^{2}2s^{2}2p^4$

The elements belong to the same period and the next electron to be withdrawn belongs to the $2p$ orbital in each case.
From $C^{+}$ to $F^{+}$ effective nuclear charge is increasing so I.E. also increases. But for $F^{+}$ less energy is required than $O^{+}$ because after losing one electron, the former will get a stable half-filled $2p^3$ configuration. On the other hand, more energy is required for removing electron from the already-stable half-filled $2p^3$ orbital for $O^{+}$.