Question

Ionisation enthalpies of elements of second period are given below Ioniisation enthalpy / k cal $mo{l}^{-1}$ : 520, 899, 801, 1086, 1402, 1314, 1681,2080. Match the correct enthalpy with the elements and complete the graph given in figure. Also write symols of elements with their atomic number.

Answer 1

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 decreases in atomic radius.
How ever there are some exceptions given below
(a) In spite increasesd nuclear charge, the firest Ionisation enthalpy of B is lower than that of Be. This is due to the presence of fully filled 2s orbital of Be $\left[1s^{2}2s^2\right]$ which is a stable elelctronic arrangement . Thus, higher energy is required to knock out the electron from fully filled 2s orbitals while B $\left[1s^{2}2s^{2}2p^1\right]$ contins 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 ae more tightly held by the nucleus than these present in 2p - 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 $\left(1s^{2}2s^{2}2p_x^{1}2p_y^{1}2p_z^1\right)$ Which is not present in $O\left(1s^{2}2s^{2}2p_x^{2}2p_y^{1}2p_z^1\right)$
Therefore, the first ionisation enthalpy of N is higher than that o fO the symbols of elements along with their atomic numbers are given in the folloowing graph