Ionization Enthalpy

The minimum amount of energy required to remove an electron from the valence shell of an isolated gaseous atom in its ground state resulting in the formation of positive ion is known as ionization energy or ionization potential.

M(g) →^IE M⁺(g) + e^–

In other words, the first ionization enthalpy for an element M is the enthalpy change (Δ1 H ) for the reaction depicted in the above equation.

The ionization enthalpy is expressed in units of kJ mol^-1. Smaller is the ionization enthalpy, easier for the neutral atom to change itself into a positive ion. Since energy is always required to remove the electrons from the atom, ionization enthalpy is always positive. The second ionization enthalpy will be higher than the first ionization enthalpy because it is more difficult to remove an electron from a positively charged ion than from a neutral atom.The term “ionization enthalpy” is not appropriate and is taken as the first ionization enthalpy, second ionization enthalpy, third ionization enthalpy for the enthalpy change required for the removal of first, second, third electron respectively.

For example,

_M(g) IE₁ _M⁺_{(g) + e-}

_M(g) IE₂ _M²⁺₍_{g) + e-}

_M(g) IE₃ _M³⁺₍_{g) + e-}

The ionization enthalpy goes on increasing for each successive removal of electrons. This is due to the fact that after the removal of an electron, the number of electrons decreases while the nuclear charge remains the same. Consequently the remaining electrons are held more tightly by the nucleus and removal of each successive electron becomes more difficult. Hence, successive ionization enthalpies go on increasing. So order becomes

IE₁ < IE₂ < IE₃ < …………….

The effective nuclear charge experienced by a valence electrons in an atom will be less than the actual charge on the nucleus due to a phenomenon known as ‘shielding effect’ or ‘screening effect’ where the force of attraction by the nucleus on the valence electron is reduced due to the repulsive forces exerted by the inner shell electrons.

For example, the 2s electron in lithium is shielded from the nucleus by the inner core of 1s electron. As result, the valence electron experiences a net positive charge which is less than the actual charge of +3. Shielding is the most effective whenever there is a full shell of electrons between the outermost electron and the nucleus, as in the case of noble gases. This is the reason for a sharp decrease in ionization enthalpy going from noble gases to alkali metals.

When we move across the period from lithium to fluorine in the second period, we know atomic radii go on decreasing. Here the successive electrons which are added to the orbitals in the same principal quantum level the shielding of electrons does not increase very much to compensate for the increased attraction of the electrons to the nucleus. Consequently, the outermost electrons are more tightly bounded thus increasing the ionization potential across the period.

On the other hand while moving down the group the distance of the outermost electron from the nucleus increases; there is an increase in the shielding of the nuclear charge by the electrons in the inner levels. Here in this case, the increase in the shielding outweighs the increased in nuclear charge thus facilitating the easy removal of the valence electron, reducing the ionization enthalpy down the group.

If we took at the first ionization enthalpy of the elements of the second period some anomalous trends are observed between beryllium and boron, and between nitrogen and oxygen. Let us find the reason for it.

Anomalous behavior between Beryllium and Boron:

We find that beryllium has a higher ionization enthalpy (899 kJmol^-1) in comparison to boron (801 kJmol^-1) even though Be(Z=4) has low nuclear charge than that of B(Z=5). When we consider the same principal quantum level, s-orbital electrons are attracted more towards the nucleus than p-orbital electrons. In beryllium the electron is to be removed from the s-orbital whereas in boron the electron is to be removed from the p-orbital. The penetration of a 2s-electron is more than that of the 2p-electron, hence the 2p-electron faces more shielding effect from the nucleus by the inner core electrons than the 2s-electrons of beryllium. Therefore it becomes easier to remove the 2p-electron from boron as compared to the removal of 2s-electron from beryllium.

Anomolous behavior between nitrogen and oxygen:

Nitrogen (Z=7) has anomalously high first ionization enthalpy (1402 kJ/mol) in comparison to oxygen (Z=8), (1314 kJ/mol). This arises because we know that half-filled orbitals and full-filled degenerate orbitals are more stable than the incompletely filled degenerate orbitals. Here nitrogen has half-filled p-orbitals whereas oxygen has incompletely filled p-orbitals in which four 2p electrons occupy the same 2p-orbital resulting in greater repulsion between the electrons.

Consequently it becomes easier to remove the fourth 2p-electron from oxygen than to remove one of the three 2p-electrons from nitrogen.