The first ionization enthalpy, also known as ionization energy, is the energy required to remove the most loosely bound electron from a neutral atom in the gaseous phase to form a cation. The trend of first ionization energies generally increases across a period from left to right in the periodic table. This is due to the increasing nuclear charge and the decreasing atomic radius, which cause the valence electrons to be attracted more strongly to the nucleus.

However, there are notable exceptions based on the electron configuration stability and electron pairing in orbitals. Let's analyze the given elements:

• Lithium (Li): Being the first element in the period, it has the smallest nuclear charge and only one electron in its outer shell, which makes it easy to remove an electron.
• Beryllium (Be): This element has two electrons in the 2s orbital. The removal of one electron slightly disturbs the fully filled 2s sub-shell, creating more stability than having an unpaired electron. Therefore, Be has a higher ionization energy than Li.
• Boron (B): This element has a half-filled 2p orbital configuration (one electron in the 2p orbital), which is relatively less stable compared to a full or empty p orbital, leading to a slightly lower ionization energy than Be.
• Carbon (C): With two electrons in separate 2p orbitals (following Hund's rule), C experiences more effective nuclear shielding and electron-electron repulsion compared to a single electron in Boron's 2p orbital. This makes it relatively easier to remove an electron from B than from C, but harder than removing one from Be.
• Nitrogen (N): It has exactly half-filled 2p orbitals, which provides extra stability and hence has a higher ionization energy than Carbon. Contrarily, the configuration of three p electrons is stable owing to the exchange energy and symmetric distribution in space.

Given these points, we can order the elements by increasing first ionization enthalpy as follows:

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

This matches with Option B. Thus, the correct answer is:

Option B

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

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