1. Identify the elements

Given:

• $\mathrm{A} = \mathrm{O}$
• $\mathrm{B} = \mathrm{N}$
• $\mathrm{C} = \mathrm{Be}$
• $\mathrm{D} = \mathrm{F}$
• $\mathrm{E} = \mathrm{B}$

We need the correct order of first ionization enthalpy.

2. General trend across Period 2

Across a period, first ionization enthalpy generally increases from left to right due to increasing effective nuclear charge.

So a rough trend in Period 2 is:

But we must include the known exceptions carefully.

3. Important exceptions

(i) $\mathrm{B}$ vs $\mathrm{Be}$

• $\mathrm{Be}: 1s^2 2s^2$
• $\mathrm{B}: 1s^2 2s^2 2p^1$

The electron removed from boron is a $2p$ electron, which is higher in energy and easier to remove than the $2s$ electron of beryllium.

Therefore,

(ii) $\mathrm{O}$ vs $\mathrm{N}$

• $\mathrm{N}: 1s^2 2s^2 2p^3$ (half-filled $2p$ subshell, extra stable)
• $\mathrm{O}: 1s^2 2s^2 2p^4$

In oxygen, one $2p$ orbital has paired electrons, causing extra electron-electron repulsion. So removing one electron from oxygen is easier than from nitrogen.

Therefore,

4. Now arrange all elements in increasing order

Using the trend and exceptions:

Now convert into the given labels:

• $\mathrm{Boron} = \mathrm{E}$
• $\mathrm{Beryllium} = \mathrm{C}$
• $\mathrm{Oxygen} = \mathrm{A}$
• $\mathrm{Nitrogen} = \mathrm{B}$
• $\mathrm{Fluorine} = \mathrm{D}$

So,

5. Match with the options

Option C is:

This matches exactly.

6. Final answer

The correct option is: