Question
Ionisation enthalpies of elements of second period are given below:
Ionisation enthalpy/ kcal $mol^{-1}, 520, 899, 801, 1086, 1402, 1314, 1681, 2080$.
Match the correct enthalpy with the elements and complete the graph given in also write symbols of elements with their atomic number.

Answer

To match the correct enthalpy with the elements and to complete the graph, the following points are taken into consideration. As we move from left to right across a period, the ionization enthalpy keeps on increasing due to increased nuclear charge and simultaneous decrease in atomic radius. However, there are some exceptions given below:
  1. In spite of increased nuclear charge, the first ionisation enthalpy of B is lower than that of Be. This is due to the presence of fully filled 2s-orbital of Be $[1s^22s^2]$ which is a stable electronic arrangement. Thus, higher energy is required to knock out the electron from fully filled 2.v-orbital. While $B [1s^2 2s^2 2p^1]$ contains valence electrons in 2s and 2p-orbitals. It can easily lose its one e– from 2p-orbital in order to achieve noble gas configuration. Thus, first ionisation enthalpy of B is lower than that of Be.
Since the electrons in 2s-orbital are more tightly held by the nucleus than those present in 2p-orbital, therefore, ionisation enthalpy of B is lower than that of Be.
  1. The first ionisation enthalpy of N is higher than that of O though the nuclear charge of O is higher than that of N. This is due to the reason that in case of N, the electron is to be removed from a more stable, exactly half-filled electronic configuration $\big(1\text{s}^2\ 2\text{s}^2\ 2\text{p}^1_\text{x}\ 2\text{p}^1_\text{y}\ 2\text{p}^1_\text{z}\big)$ which is not present in $\text{O}\big(1\text{s}^2\ 2\text{s}^2\ 2\text{p}^2_\text{x}\ 2\text{p}^1_\text{y}\ 2\text{p}^1_\text{z}\big).$
Therefore, the first ionisation enthalpy of N is higher than that of O. The symbols of elements along with their atomic numbers are given in the following graph.

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