Why is the first ionization energy of O lower than for N, but the second ionization energy for O higher?

Hint: We know that the chemical symbol of oxygen is O and that for nitrogen is N. Ionization energy is defined as the energy required to remove a loosely bonded electron from the valence orbital of an isolated gaseous atom. The half-filled and fully filled electronic configurations have the largest ionization energy. As the size increases, the electron gain enthalpy becomes more negative. The size of an element increases as the number of electrons in its valence shell increases.

Complete Step By Step Answer:
We know that ionization energy is defined as the energy required to remove a loosely bonded electron from the valence orbital of an isolated gaseous atom. We can answer this question by looking at the electronic configurations of both oxygen and nitrogen. The electronic configurations of nitrogen and oxygen are given as follows:
 $ N = 1{s^2}2{s^2}2{p^3} $
 $ O = 1{s^2}2{s^2}2{p^4} $
We can see that the nitrogen atom has a half-filled atomic orbital and hence, it will be more stable than the oxygen atom. Therefore, the first ionization energy of O is lower than that of N.
But when we consider second ionization energy, the oxygen ion $ ({O^ + }) $ attains half-filled electronic configuration and becomes more stable. Therefore, the second ionization energy of O is greater than that of nitrogen.

Note:
As per the trends of the modern periodic table, the ionization energy increases as we move from left to right in a period (horizontal row). This happens because along the period, the size of the element decreases and its electronegativity increases. While when we move down a period, the ionization energy decreases because size of the element increases.