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Classification of Elements and Periodicity in Prope rties 89 valence electron in an atom will be less than the 2s electrons of beryllium. Therefore, it is the actual charge on the nucleus because of easier to remove the 2p-electron from boron “shielding” or “screening” of the valence compared to the removal of a 2s- electron from electron from the nucleus by the intervening beryllium. Thus, boron has a smaller first core electrons. For example, the 2s electron ionization enthalpy than beryllium. Another in lithium is shielded from the nucleus by “anomaly” is the smaller first ionization the inner core of 1s electrons. As a result, the enthalpy of oxygen compared to nitrogen. This valence electron experiences a net positive arises because in the nitrogen atom, three charge which is less than the actual charge 2p-electrons reside in different atomic orbitals of +3. In general, shielding is effective when (Hund’s rule) whereas in the oxygen atom, the orbitals in the inner shells are completely two of the four 2p-electrons must occupy the filled.This situation occurs in the case of same 2p-orbital resulting in an increased alkali metals which have single outermost electron-electron repulsion. Consequently, ns-electron preceded by a noble gas electronic it is easier to remove the fourth 2p-electron configuration. from oxygen than it is, to remove one of the When we move from lithium to fluorine three 2p-electrons from nitrogen. across the second period, successive electrons are added to orbitals in the same principal Problem 3.6 quantum level and the shielding of the nuclear The first ionization enthalpy i∆) values charge by the inner core of electrons does of the third period elements, Na, Mg and not increase very much to compensate for Si are respectively 496, 737 and 786 kJ the increased attraction of the electron to the mol . Predict whether the first i H value nucleus. Thus, across a period, increasing for Al will be more close to 575 or 760 kJ nuclear charge outweighs the shielding. mol ? Justify your answer. Consequently, the outermost electrons are Solution held more and more tightly and the ionization –1 enthalpy increases across a period. As we go It will be more close to 575 kJ mol . down a group, the outermost electron being The value for Al should be lower than that of Mg because of effective shielding increasingly farther from the nucleus, there is of 3p electrons from the nucleus by an increased shielding of the nuclear charge 3s-electrons. by the electrons in the inner levels. In this case, increase in shielding outweighs the (d) Electron Gain Enthalpy increasing nuclear charge and the removal of When an electron is added to a neutral the outermost electron requires less energy gaseous atom (X) to convert it into a negative down a group. ion, the enthalpy change accompanying the From Fig. 3.6(a), you will also notice that process is defined as the Electron Gain the first ionization enthalpy of boron (Z = 5) Enthalpy (∆ H). Electron gain enthalpy is slightly less than that of beryllium (Z = 4) provides a measure of the ease with which even though the former has a greater nuclear an atom adds an electron to form anion as charge. When we consider the same principal represented by equation 3.3. quantum level, an s-electron is attracted to the X(g) + e → X (g) (3.3) nucleus more than a p-electron. In beryllium, the electron removed during the ionization is Depending on the element, the process an s-electron whereas the electron removed of adding an electron to the atom can be during ionization of boron is a p-electron. The either endothermic or exothermic. For many penetration of a 2s-electron to the nucleus is elements energy is released when an electron more than that of a 2p-electron; hence the 2p is added to the atom and the electron gain electron of boron is more shielded from the enthalpy is negative. For example, group nucleus by the inner core of electrons than 17 elements (the halogens) have very high Reprint 2026-27