Recall the trend of first ionization enthalpy across a period

First ionization enthalpy generally increases from left to right across a period because effective nuclear charge increases.

Here the elements are in the second period: $\text{Be},\ \text{B},\ \text{N},\ \text{O}$

Write their electronic configurations

• Be: $1s^2 2s^2$
• B: $1s^2 2s^2 2p^1$
• N: $1s^2 2s^2 2p^3$
• O: $1s^2 2s^2 2p^4$

Compare Be and B

Although ionization enthalpy usually increases across the period, Be has a filled $2s$ subshell which is relatively stable.

B has its outermost electron in the $2p$ subshell, which is higher in energy and easier to remove.

Therefore, $\text{IE}_1(\text{B}) < \text{IE}_1(\text{Be})$

Compare N and O

N has configuration $2p^3$, which is a half-filled $p$ subshell, giving extra stability.

O has configuration $2p^4$, so one of the $p$ orbitals contains a paired electron. Electron-electron repulsion makes removal easier than in N.

Therefore, $\text{IE}_1(\text{O}) < \text{IE}_1(\text{N})$

Combine the comparisons

Using the above exceptions and overall trend: $\text{B} < \text{Be} < \text{O} < \text{N}$

Match with the given options

This corresponds to: $\boxed{\text{D: } B < Be < O < N}$

Comparison with stored answer

Stored correct answer: D

Derived answer: D

So the derived answer agrees with the stored answer.