Question

Among the second-period elements, the actual ionization enthalpies are in increasing order: lithium Boron Beryllium Carbon Oxygen Nitrogen Fluorine and neon. Then explain why beryllium has higher ionization enthalpy than Boron and oxygen has lower ionization enthalpy than nitrogen and fluorine?

Answer 1

To answer this question, consider the concept of the symmetry of the atom. Electrons are difficult to remove from an atom when they are present closer to the nucleus and also it is difficult to remove the electrons from an atom having a small size and high nuclear charge.

Complete step by step solution:
-So, for the first question i.e. why beryllium has higher ionization enthalpy than boron, symmetry factor can be used for explaining this. The electronic configuration of beryllium ($1{{s}^{2}}2{{s}^{2}}$) is more symmetrical than that of boron ($1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$) because both the occupied orbitals are filled in the case beryllium while boron has one half-filled orbital in the last subshell.
So, during the ionization process, the electron to be removed from the beryllium atom is $2s$electron, whereas the electron to be removed from the boron atom is a $2p$ electron. Now, $2s$ electrons are more strongly attached to the nucleus than $2p$ electrons and $p$-orbitals are also slightly higher energy than the $s$-orbital. Therefore, more energy is required to remove a $2s$ electron of beryllium than that is required to remove a $2p$ electron of boron. Hence, beryllium has a higher ionization enthalpy than boron.
-Coming to the second explanation, according to the symmetry factor, the nitrogen atom ($1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$) has all the three $2p$ orbitals half-filled, while oxygen ($1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$) has one filled and two half-filled orbitals in $2p$ subshell. So, in nitrogen, the three $2p$ electrons occupy different atomic orbitals. However, in oxygen, two of the four $2p$ electrons occupy the same atomic orbital and this results in increased electron-electron repulsion in the oxygen atom. Thus, the energy required to remove the fourth $2p$ electron from oxygen is less as compared to the energy required to remove electrons from the nitrogen atom. Hence, oxygen has a lower ionization enthalpy than nitrogen.
-While, Fluorine ($1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$) contains one electron and one proton more than oxygen. As the electron is being added to the same shell, the increase in nuclear attraction is more than the increase in electronic repulsion. Therefore, the valence electrons in fluorine atoms experience a more effective nuclear charge than that by the electron of oxygen. This results in increased energy to remove an electron from fluorine than that required from an oxygen atom. Hence, oxygen will have lower ionization enthalpy than the fluorine atom.

Note: Remember that ionization enthalpy is affected by the symmetry, atomic size and the nuclear charge of an atom. Closer the electrons present to the nucleus, difficult is to remove the electrons and higher the nuclear charge and smaller the atom, higher will be the ionization energy.