The correct order of first ionization potential among following elements, Be, B, C, N and O is

The correct option is B < Be < C < O < N.

To understand the order of the first ionization potentials (or first ionization energies) for the given elements (Beryllium, Boron, Carbon, Nitrogen, and Oxygen), we must examine their positions in the periodic table and their electronic configurations.
These elements all belong to the second period of the periodic table, ordered by atomic number as follows: Be (4), B (5), C (6), N (7), and O (8).

Generally, as we move from left to right across a period, the first ionization potential increases because the nuclear charge increases while the outermost electrons are added to the same main energy level. This causes the atomic radius to decrease and the outer electrons to be held more tightly by the nucleus.

However, there are two key exceptions in this period due to the extra stability of fully filled and half-filled electronic subshells:

1. Beryllium (Be) versus Boron (B):
The electronic configurations are:
Be (Z = 4): 1s² 2s² (fully filled 2s subshell)
B (Z = 5): 1s² 2s² 2p¹
It is easier to remove an electron from the higher-energy 2p subshell of boron than from the stable, fully filled 2s subshell of beryllium. Thus, the first ionization potential of Boron is lower than that of Beryllium:

$$B<\mathrm{Be}$$

2. Nitrogen (N) versus Oxygen (O):
The electronic configurations are:
N (Z = 7): 1s² 2s² 2p³ (half-filled 2p subshell, which is highly stable)
O (Z = 8): 1s² 2s² 2p⁴
Removing an electron from the half-filled 2p subshell of nitrogen requires more energy compared to removing one from the 2p subshell of oxygen (where electron-electron repulsion also facilitates removal). Therefore, the first ionization potential of Oxygen is lower than that of Nitrogen:

$$O<N$$

Combining these trends with the general left-to-right increase (where Carbon lies between Be and O), we obtain the overall correct order:

$$B<\mathrm{Be}<C<O<N$$