2.5 Periodic properties and trends  (Page 3/10)

Cations with larger charges are smaller than cations with smaller charges (e.g., V ²⁺ has an ionic radius of 79 pm, while that of V ³⁺ is 64 pm). Proceeding down the groups of the periodic table, we find that cations of successive elements with the same charge generally have larger radii, corresponding to an increase in the principal quantum number, n .

An anion (negative ion) is formed by the addition of one or more electrons to the valence shell of an atom. This results in a greater repulsion among the electrons and a decrease in Z _eff per electron. Both effects (the increased number of electrons and the decreased Z _eff ) cause the radius of an anion to be larger than that of the parent atom ( [link] ). For example, a sulfur atom ([Ne]3 s ² 3 p ⁴ ) has a covalent radius of 104 pm, whereas the ionic radius of the sulfide anion ([Ne]3 s ² 3 p ⁶ ) is 170 pm. For consecutive elements proceeding down any group, anions have larger principal quantum numbers and, thus, larger radii.

Atoms and ions that have the same electron configuration are said to be isoelectronic    . Examples of isoelectronic species are N ^3– , O ^2– , F ^– , Ne, Na ⁺ , Mg ²⁺ , and Al ³⁺ (1 s ² 2 s ² 2 p ⁶ ). Another isoelectronic series is P ^3– , S ^2– , Cl ^– , Ar, K ⁺ , Ca ²⁺ , and Sc ³⁺ ([Ne]3 s ² 3 p ⁶ ). For atoms or ions that are isoelectronic, the number of protons determines the size. The greater the nuclear charge, the smaller the radius in a series of isoelectronic ions and atoms.

Variation in ionization energies

The amount of energy required to remove the most loosely bound electron from a gaseous atom in its ground state is called its first ionization energy    (IE ₁ ). The first ionization energy for an element, X, is the energy required to form a cation with +1 charge:

The energy required to remove the second most loosely bound electron is called the second ionization energy (IE ₂ ).

The energy required to remove the third electron is the third ionization energy, and so on. Energy is always required to remove electrons from atoms or ions, so ionization processes are endothermic and IE values are always positive. For larger atoms, the most loosely bound electron is located farther from the nucleus and so is easier to remove. Thus, as size (atomic radius) increases, the ionization energy should decrease. Relating this logic to what we have just learned about radii, we would expect first ionization energies to decrease down a group and to increase across a period.

[link] graphs the relationship between the first ionization energy and the atomic number of several elements. The values of first ionization energy for the elements are given in [link] . Within a period, the IE ₁ generally increases with increasing Z . Down a group, the IE ₁ value generally decreases with increasing Z . There are some systematic deviations from this trend, however. Note that the ionization energy of boron (atomic number 5) is less than that of beryllium (atomic number 4) even though the nuclear charge of boron is greater by one proton. This can be explained because the energy of the subshells increases as l increases, due to penetration and shielding (as discussed previously in this chapter). Within any one shell, the s electrons are lower in energy than the p electrons. This means that an s electron is harder to remove from an atom than a p electron in the same shell. The electron removed during the ionization of beryllium ([He]2 s ² ) is an s electron, whereas the electron removed during the ionization of boron ([He]2 s ² 2 p ¹ ) is a p electron; this results in a lower first ionization energy for boron, even though its nuclear charge is greater by one proton. Thus, we see a small deviation from the predicted trend occurring each time a new subshell begins.