Arrange the following elements in increasing order of first ionization enthalpy:

Li, Be, B, C, N

Choose the correct answer from the options given below:

The correct answer is Li < B < Be < C < N.

To understand the order of the first ionization enthalpy for the elements lithium (Li), beryllium (Be), boron (B), carbon (C), and nitrogen (N), we need to look at periodic trends and their electronic configurations.

1. General Periodic Trend:
As we move from left to right across a period in the periodic table, the nuclear charge increases while the electrons are added to the same valence shell. This increases the electrostatic attraction between the nucleus and the outermost electrons, causing the atomic radius to decrease and making it harder to remove an electron. Consequently, the first ionization enthalpy generally increases across a period.

Based on this general trend, the expected order would be:
Li < Be < B < C < N

2. Exception Between Beryllium (Be) and Boron (B):
However, there is an exception when comparing beryllium (atomic number 4) and boron (atomic number 5). Let's examine their electronic configurations:

For Beryllium (Be):
$\mathrm{Be}=1s^{2}2s^2$
For Boron (B):
$B=1s^{2}2s^{2}2p^1$

In beryllium, the electron to be removed during first ionization comes from the completely filled, stable $2s$ orbital. Electrons in $s$ orbitals are closer to the nucleus and experience a stronger nuclear pull (greater penetration effect) compared to electrons in $p$ orbitals.
In boron, the electron is removed from the $2p$ orbital, which is higher in energy and more shielded from the nuclear charge by the inner $2s$ electrons. Therefore, it requires less energy to remove the outermost electron from boron than from beryllium.

Thus, the first ionization enthalpy of boron is less than that of beryllium:
${\mathrm{IE}}_1\left(B\right)<{\mathrm{IE}}_1\left(\mathrm{Be}\right)$

3. Conclusion:
Combining the general trend with this exception, the correct increasing order of first ionization enthalpy is:

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