Ionization enthalpy generally increases across a period due to increasing effective nuclear charge.
However, nitrogen has a half-filled p-orbital configuration (2p\(^3\)), which provides extra stability, making its ionization enthalpy higher than oxygen.
Thus, the correct order is: \[ N > O > C > Be \]
If $ | \vec{a} | = 3 $, $ | \vec{b} | = 2 $, then find $ (3\vec{a} - 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) $.