Generally, ionization enthalpy increases across a period from left to right. However, there are two important anomalies in the second period involving the elements given:

1. Boron (B) vs. Beryllium (Be): Beryllium ($Z=4$) has a stable, fully filled $2s$ subshell ($1s^2 2s^2$). Boron ($Z=5$) has the configuration $1s^2 2s^2 2p^1$. The $2s$ electron penetrates closer to the nucleus than the $2p$ electron and is more strongly attracted. Therefore, it is easier to remove the $2p$ electron from Boron than the $2s$ electron from Beryllium. Thus, $IE_1(B) < IE_1(Be)$.
2. Oxygen (O) vs. Nitrogen (N): Nitrogen ($Z=7$) has a stable, half-filled $2p$ subshell ($1s^2 2s^2 2p^3$). Oxygen ($Z=8$) has the configuration $1s^2 2s^2 2p^4$. The repulsion between the two electrons occupying the same $2p$ orbital in oxygen makes it easier to remove one electron compared to the stable half-filled configuration of nitrogen. Thus, $IE_1(O) < IE_1(N)$.

Combining these trends with the general increase across the period, the correct order is: B < Be < C < O < N.