Step 1: General trend.
Ionisation energy generally increases across a period due to increasing nuclear charge.
Step 2: Apply known exceptions.
- \(B\) has lower IE than \(Be\) because \(B\) loses a \(2p\) electron (easier to remove).
- \(O\) has lower IE than \(N\) because \(O\) has paired electrons in \(2p\) causing extra repulsion.
Step 3: Arrange with exceptions.
\[
B<Be<O<N
\]
Final Answer:
\[
\boxed{B<Be<O<N}
\]