Periodic Trends
Variation in Atomic Radius
The quantum mechanical picture makes it difficult to establish a definite size of an atom. However, there are several practical ways to define the radius of atoms and, thus, determine their relative sizes that give roughly similar values. We will use the atomic radii, which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond (this measurement is possible because atoms within molecules still retain much of their atomic identity). We know that as we scan down a group, the principal quantum number, n, increases by one for each element. Thus, the electrons are being added to a region of space that is increasingly distant from the nucleus. Consequently, the size of the atom (and its atomic radius) must increase as we increase the distance of the outermost electrons from the nucleus. This trend is illustrated for the atomic radii of the halogens in Table 1 and Figure 1. The trends for the entire periodic table can be seen in Figure 1.
Atom | Atomic radius (pm) | Nuclear charge |
F | 71 | +9 |
Cl | 99 | +17 |
Br | 114 | +35 |
I | 133 | +53 |
At | 166 | +85 |
As shown in Figure 1, as we move across a period from left to right, we generally find that each element has a smaller atomic radius than the element preceding it. This is expected due to the Zeff for the valence electron increasing from left to right across a period. The stronger pull (higher effective nuclear charge) experienced by valence electrons on the right side of the periodic table draws them closer to the nucleus, making the atomic radii smaller.
Example 1: Sorting Atomic Radii
Predict the order of increasing atomic radius for Ge, Fl, Br, Kr.
Solution
Radius increases as we move down a group, so Ge < Fl (Note: Fl is the symbol for flerovium, element 114, NOT fluorine). Radius decreases as we move across a period, so Kr < Br < Ge. Putting the trends together, we obtain Kr < Br < Ge < Fl.
Check Your Learning
Give an example of an atom whose size is smaller than fluorine.
Answer
Ne or He
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 (IE1). The first ionization energy for an element, X, is the energy required to form a cation with +1 charge:
IE1: X(g) → X+(g) + e–
The energy required to remove the second most loosely bound electron is called the second ionization energy (IE2).
IE2: X+(g) → X2+(g) + e–
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. Recall that Zeff for valence electrons increases from left to right on the Periodic Table but that atoms get larger down a group as the n level increases. Relating this to the IE, we would expect valence electrons that are less tightly held (i.e., experience a lower Zeff) would be easier to remove and require less energy. Therefore, as Zeff for the valence electrons increases from left to right, the first IE increases, too, because more energy will be required to remove the electrons. As the n level for the valence electrons increases down a group, the atoms get bigger and the Coulombic attraction between the nucleus and valence electrons decreases. The valence electrons become easier to remove and the first ionization energy decreases.
Figure 3 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 Figure 4.
Within a period, the IE1 generally increases with increasing Z. Down a group, the IE1 value generally decreases with atomic radius. 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. 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]2s2) is an s electron, whereas the electron removed during the ionization of boron ([He]2s22p1) 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.
Another deviation occurs as orbitals become more than one-half filled. The first ionization energy for oxygen is slightly less than that for nitrogen, despite the trend in increasing IE1 values across a period. Looking at the orbital diagram of oxygen, we can see that removing one electron will eliminate the electron–electron repulsion caused by pairing the electrons in the 2p orbital and will result in a half-filled orbital (which is energetically favorable). Analogous changes occur in succeeding periods (note the dip for sulfur after phosphorus).
Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from a cation with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table 2, there is a large increase in the ionization energies for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons due to the increase in Zeff felt by the core electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization.
Element | IE1 | IE2 | IE3 | IE4 | IE5 | IE6 | IE7 |
K | 418.8 | 3051.8 | 4419.6 | 5876.9 | 7975.5 | 9590.6 | 11343 |
Ca | 589.8 | 1145.4 | 4912.4 | 6490.6 | 8153.0 | 10495.7 | 12272.9 |
Sc | 633.1 | 1235.0 | 2388.7 | 7090.6 | 8842.9 | 10679.0 | 13315.0 |
Ga | 578.8 | 1979.4 | 2964.6 | 6180 | 8298.7 | 10873.9 | 13594.8 |
Ge | 762.2 | 1537.5 | 3302.1 | 4410.6 | 9021.4 | Not available | Not available |
As | 944.5 | 1793.6 | 2735.5 | 4836.8 | 6042.9 | 12311.5 | Not available |
Example 2: Ranking Ionization Energies
Predict the order of increasing energy for the following processes: IE1 for Al, IE1 for Tl, IE2 for Na, IE3 for Al.
Solution
Removing the 6p1 electron from Tl is easier than removing the 3p1 electron from Al because the higher n orbital is farther from the nucleus, so IE1(Tl) < IE1(Al). Ionizing the third electron from Al (Al2+ → Al3+ + e–) requires more energy because the cation Al2+ exerts a stronger pull on the electron than the neutral Al atom, so IE1(Al) < IE3(Al). The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons. Putting this all together, we obtain: IE1(Tl) < IE1(Al) < IE3(Al) < IE2(Na).
Check Your Learning
Which has the lowest value for IE1: O, Po, Pb, or Ba?
Answer
Ba
Variation in Electron Affinities
The electron affinity (EA) is the energy change for the process of adding an electron to a gaseous atom to form an anion (negative ion).
EA1: X(g) + e– → X–(g)
This process is exothermic for the majority of elements, which means that energy is released when the gaseous atom accepts an electron to form an anion. However, for certain elements the electron affinity for the addition of the first electron is >0, an indication that the atom will not bind an electron and form a stable negative ion. The EA of some of the elements is given in Figure 5. Subsequent EA values are associated with forming ions with more charge. The second EA is the energy associated with adding an electron to an anion to form a –2 ion, and so on.
As we might predict, it becomes easier to add an electron across a series of atoms as the effective nuclear charge of the atoms increases. We find, as we go from left to right across a period, EAs tend to become more negative. The exceptions found among the elements of group 2 (2A), group 15 (5A), and group 18 (8A) can be understood based on the electronic structure of these groups. The noble gases, group 18 (8A), have a completely filled shell and the incoming electron must be added to a higher n level, which is impossible to do. Group 2 (2A) has a filled ns subshell, and so the next electron added goes into the higher energy np, so, again, the observed EA value is not as the trend would predict. Finally, group 15 (5A) has a half-filled np subshell and the next electron must be paired with an existing np electron. In all of these cases, the initial relative stability of the electron configuration disrupts the trend in EA.
We also might expect the atom at the top of each group to have the largest EA; their first ionization potentials suggest that these atoms have the largest effective nuclear charges. However, as we move down a group, we see that the second element in the group most often has the greatest EA. The reduction of the EA of the first member can be attributed to the small size of the n = 2 shell and the resulting large electron–electron repulsions. For example, chlorine, with an EA value of –349 kJ/mol, has the highest value of any element in the Periodic Table. The EA of fluorine is –328 kJ/mol. When we add an electron to a fluorine atom to form a fluoride anion (F–), we add an electron to the n = 2 shell. The electron is attracted to the nucleus, but there is also significant repulsion from the other electrons already present in this small valence shell. The chlorine atom has the same electron configuration in the valence shell, but because the entering electron is going into the n = 3 shell, it occupies a considerably larger region of space and the electron–electron repulsions are reduced. The entering electron does not experience as much repulsion and the chlorine atom accepts an additional electron more readily.
The properties discussed in the previous section and this section (effective nuclear charge, size of atoms and ions, ionization energies, and electron affinities) are central to understanding chemical reactivity. For example, because fluorine has an energetically favorable EA and a large energy barrier to ionization (IE), it is much easier to form fluorine anions than cations. Metallic properties including conductivity and malleability (the ability to be formed into sheets) depend on having electrons that can be removed easily. Thus, metallic character increases as we move down a group and decreases across a period in the same trend observed for atomic size because it is easier to remove an electron that is farther away from the nucleus.
Ionic Radii
Ionic radius is the measure used to describe the size of an ion. A cation always has fewer electrons and the same number of protons as the parent atom; thus, it is smaller than the atom from which it is derived (Figure 6). For example, the radius of a neutral aluminum atom (1s22s22p63s23p1) is 118 pm, whereas the ionic radius of an Al3+ (1s22s22p6) is 68 pm. All electrons in the n=3 shell are removed, the remaining electrons occupying smaller shells. Even the removal of a single electron produces a cation that is smaller than the parent atom as more accurate methods of determining Zeff that account for the screening effect of the valence electrons, show that losing a valence electron increases Zeff for any remaining valence electrons.
Cations with larger charges are smaller than cations with smaller charges (e.g., V2+ has an ionic radius of 79 pm, while that of V3+ 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 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 valence electrons and a decrease in Zeff per valence electron, as valence electrons do contribute to screening albeit to a much lesser degree than the core electrons. Both effects (the increased number of electrons and the decreased Zeff) cause the radius of an anion to be larger than that of the neutral atom (Figure 6). For example, a sulfur atom ([Ne]3s23p4) has a covalent radius of 104 pm, whereas the ionic radius of the sulfide anion ([Ne]3s23p6) 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 N3–, O2–, F–, Ne, Na+, Mg2+, and Al3+ (all have the electron configuration 1s22s22p6). Another isoelectronic series is P3–, S2–, Cl–, Ar, K+, Ca2+, and Sc3+ ([Ne]3s23p6). 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 because the electrons on the atom with the greatest nuclear charge will experience the highest Zeff.
Electron Configurations of Ions
We have seen that ions are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a neutral atom. The outermost or valence electrons are removed since they have the highest energies, are shielded more, and are farthest from the nucleus. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the s orbital are easier to remove than the d or f electrons, and so the highest ns electrons are lost, and then the (n – 1)d or (n – 2)f electrons are removed. An anion forms when one or more electrons are added to a neutral atom. The added electrons fill in the order predicted by the Aufbau principle.
Example 3: Predicting Electron Configurations of Ions
What is the ground state electron configuration of:
- Na+
- P3–
- Al2+
- Fe2+
- Sm3+
Solution
First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable.
Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals.
(a) Na: 1s22s22p63s1. Sodium cation loses one electron, so Na+: 1s22s22p6.
(b) P: 1s22s22p63s23p3. Phosphorus anion (phosphide) gains three electrons, so P3−: 1s22s22p63s23p6.
(c) Al: 1s22s22p63s23p1. Aluminum cation loses two electrons, so Al2+: 1s22s22p63s1.
(d) Fe: 1s22s22p63s23p64s23d6. Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital. Fe2+: 1s22s22p63s23p63d6.
(e) Sm: 1s22s22p63s23p64s23d104p65s24d105p66s24f6. Samarium cation loses three electrons. The first two will be lost from the 6s orbital, and the final one is removed from the 4f orbital. Sm3+: 1s22s22p63s23p64s23d104p65s24d105p64f5.
Check Your Learning
Which ion with a +2 charge has the electron configuration 1s22s22p63s23p63d104s24p64d5? Which ion with a 3+ charge has this configuration?
Answer
Tc2+, Ru3+