Electron gain enthalply of F is less negative than that of Cl because when an electron is added to $F$, the added electron goes to the smaller $n=2$ quantum level and suffers repulsion from other electrons present in this level.

In case of Cl , the added electron goes to the larger $n=3$ quantum level and suffers much less repulsion from other electrons.

Elements in which the last electron enters in the $d$-orbitals, are called $d$-block elements or transition elements. These elements have the general outer electronic configuration $(n-1) d^{1-10} n s^{0-2} . \mathrm{Zn}, \mathrm{Cd}$ and Hg having the electronic configuration $(n-1) d^{10} n s^2$ do not show most of the properties of transition elements.

The $d$-orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. Thus, on the basis of properties, all transition elements are $d$-block elements but on the basis of electronic configuration, all $d$-block elements are not transition elements.

The present set up of the Long form of the periodic table can accommodate maximum 118 elements. Thus, in accordance with aufbau principle, the filling of $8 s$-orbital will occur. In other words 119th electron will enter 8s-orbital. As such its outmost electronic configuration will be $8 s^1$.

Since, it has only one electron in the valence shell, i.e., 8 s , therefore, its valency will be 1 and it will lie in the group IA along with alkali metals and the formula of its oxide will be $\mathrm{M}_2 \mathrm{O}$ where $M$ represents the element.

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 ionisation enthalpy keeps on increasing due to increased nuclear charge and simultaneous decrease in atomic radius.

However, there are some exceptions given below

(a) 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 $2 s$ orbital of $\mathrm{Be}\left[1 s^2 2 s^2\right]$ which is a stable electronic arrangement. Thus, higher energy is required to knock out the electron from fully filled $2 s$ orbitals. While B $\left[1 s^2 2 s^2 2 P^1\right]$ contains valence electrons in $2 s$ and $2 p$ orbitals. It can easily lose its one $e^{-}$from $2 p$ orbital in order to achieve noble gas configuration. Thus, first ionisation enthalpy of B is lower than that of Be .

Since, the electrons in $2 s$-orbital are more tightly held by the nucleus than these present in $2 p$-orbital, therefore, ionisation enthalpy of B is lower than that of Be .

(b) 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 $\left(1 s^2 2 s^2 2 p_x^1 2 p_y^1 2 p_z^1\right)$ which is not present in $\mathrm{O}\left(1 s^2 2 s^2 2 p_x^2 2 p_y^1 2 p_z^1\right)$.

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