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

The placing of elements are as

(a) Ionisation enthalpy increases along a period (as we move from left to right in a period) with decrease in atomic size and decreases down the group with increase in atomic size. Hence, carbon has the highest first ionisation enthalpy.

(b) Metallic character decreases across a period but increases on moving down the group. Hence, aluminium has the most metallic character.

The four important characteristic properties of p-block elements are the following

(a) $p$-Block elements include both metals and non-metals but the number of non-metals is much higher than that of metals. Further, the metallic character increases from top to bottom within a group and non-metallic character increases from left to right along a period in this block.

(b) Their ionisation enthalpies are relatively higher as compared to s-block elements.

(c) They mostly form covalent compounds.

(d) Some of them show more than one (variable) oxidation states in their compounds. Their oxidising character increases from left to right in a period and reducing character increases from top to bottom in a group.