`text(Definition :)` The energy required to remove an electron from an isolated gaseous atom (`X`) in its ground state is called ionization enthalpy. In other words, the first ionization enthalpy for an element `X` is the enthalpy change (`Δ_i H`) for the reaction depicted in equation 3.1.
`X(g) → X^+ (g) + e^-` ..............(3.1)
● The ionization enthalpy is expressed in units of `kJ mol^(–1)`.
● We can define the second ionization enthalpy as the energy required to remove the second most loosely bound electron; it is the energy required to carry out the reaction shown in equation 3.2.
`X^(+) (g) → X^(2+) (g) + e^-` ..........(3.2)
`=>` Energy is always required to remove electrons from an atom and hence ionization enthalpies are 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.
● In the same way the third ionization enthalpy will be higher than the second and so on.
`=>` The first ionization enthalpies of elements having atomic numbers up to `60` are plotted in Fig. 3.5.
● You will find maxima at the noble gases which have closed electron shells and very stable electron configurations.
● On the other hand, minima occur at the alkali metals and their low ionization enthalpies can be correlated with their high reactivity.
`=>` The first ionization enthalpy generally increases as we go across a period and decreases as we descend in a group.
● These trends are illustrated in Figs. 3.6(a) and 3.6(b) respectively for the elements of the second period and the first group of the periodic table.
`=>` You will appreciate that the ionization enthalpy and atomic radius are closely related properties.
`=>` To understand these trends, we have to consider two factors : (i) the attraction of electrons towards the nucleus, and (ii) the repulsion of electrons from each other.
● The effective nuclear charge experienced by a valence electron in an atom will be less than the actual charge on the nucleus because of “shielding” or “screening” of the valence electron from the nucleus by the intervening core electrons.
`->` For example, the `2s` electron in lithium is shielded from the nucleus by the inner core of `1s` electrons. As a result, the valence electron experiences a net positive charge which is less than the actual charge of `+3`.
● In general, shielding is effective when the orbitals in the inner shells are completely filled.
`->` This situation occurs in the case of alkali metals which have a lone `ns`-outermost electron preceded by a noble gas electronic configuration.
● When we move from lithium to fluorine across the second period, successive electrons are added to orbitals in the same principal quantum level and the shielding of the nuclear charge by the inner core of electrons does not increase very much to compensate for the increased attraction of the electron to the nucleus.
`->` Thus, across a period, increasing nuclear charge outweighs the shielding. Therefore, the outermost electrons are held more and more tightly and the ionization enthalpy increases across a period.
● As we go down a group, the outermost electron being increasingly farther from the nucleus, there is an increased shielding of the nuclear charge by the electrons in the inner levels.
`->` In this case, increase in shielding outweighs the increasing nuclear charge and the removal of the outermost electron requires less energy down a group.
`=>` From Fig. 3.6(a), you will also notice that the first ionization enthalpy of boron (Z = 5) is slightly less than that of beryllium (Z = 4) even though the former has a greater nuclear charge.
● When we consider the same principal quantum level, an `s`-electron is attracted to the nucleus more than a `p`-electron.
● In beryllium, the electron removed during the ionization is an `s`-electron whereas the electron removed during ionization of boron is a `p`-electron.
● The penetration of a `2s`-electron to the nucleus is more than that of a `2p`-electron; hence the `2p` electron of boron is more shielded from the nucleus by the inner core of electrons than the `2s` electrons of beryllium.
● Therefore, it is easier to remove the `2p`-electron from boron compared to the removal of a `2s`- electron from beryllium.
● Thus, boron has a smaller first ionization enthalpy than beryllium.
`=>` Another “anomaly” is the smaller first ionization enthalpy of oxygen compared to nitrogen.
● This arises because in the nitrogen atom, three `2p`-electrons reside in different atomic orbitals (Hund’s rule) whereas in the oxygen atom, two of the four `2p`-electrons must occupy the same `2p`-orbital resulting in an increased electron-electron repulsion.
● Consequently, it is easier to remove the fourth `2p`-electron from oxygen than it is, to remove one of the three `2p`-electrons from nitrogen.