Our rather brief survey of electron configurations of the elements has taken us through the periodic table. We have seen that the electron configurations of elements are related to their location in the periodic table. The periodic table is structured so that elements with the same pattern of outer-shell (valence) electron configuration are arranged in columns. For example, the electron configurations for the elements in groups 2A and 3A are given in Table 6.4. We see that the 2A elements all have ns2 outer configurations, while the 3A elements have ns2np1 configurations.
Earlier, in Table 6.2, we saw that the total number of orbitals in each shell is equal to n2: 1, 4, 9, or 16. Because each orbital can hold two electrons, each shell can accommodate up to 2n2 electrons: 2, 8, 18, or 32. We see that the beautiful structure of the periodic table reflects this orbital structure. The first row has two elements, the second and third rows have eight elements, the fourth and fifth rows have 18 elements, and the sixth row has 32 elements (including the lanthanide metals). Some of the numbers repeat because we reach the end of a row of the periodic table before a shell completely fills. For example, the third row has eight elements, which corresponds to filling the 3s and 3p orbitals. As noted earlier, the remaining orbitals of the third shell, the 3d orbitals, do not begin to fill until the fourth row of the periodic table (and after the 4s orbital is filled). Likewise, the 4d orbitals don't begin to fill until the fifth row of the table, and the 4f orbitals don't begin filling until the sixth row.
All of these observations are evident in the structure of the periodic table. For this reason, we will emphasize that the periodic table is your best guide to the order in which orbitals are filled. You can easily write the electron configuration of an element based on its location in the periodic table. The pattern is summarized in Figure 6.29. Notice that the elements can be grouped in terms of the type of orbital into which the electrons are placed. On the left are two columns of elements. These elements, known as the alkali metals (group 1A) and alkaline earth metals (group 2A), are those in which the outer-shell s orbitals are being filled. On the right is a block of six columns. These are the elements in which the outermost p orbitals are being filled. The s block and the p block of the periodic table contain the representative (or main-group) elements. In the middle of the table is a block of ten columns that contain the transition metals. These are the elements in which the d orbitals are being filled. Below the main portion of the table are two rows that contain fourteen columns. These elements are often referred to as the f-block metals because they are the ones in which the f orbitals are being filled. Recall that the numbers 2, 6, 10, and 14 are precisely the number of electrons that can fill the s, p, d, and f subshells, respectively. Recall also that the 1s subshell is the first s subshell, the 2p is the first p subshell, the 3d is the first d subshell, and the 4f is the first f subshell.
Figure 6.29 Block diagram of the periodic table showing the groupings of the elements according to the type of orbital being filled with electrons.
What is the characteristic outer-shell electron configuration of the group 7A elements, the halogens?
SOLUTION The first member of the halogen group is fluorine, atomic number 9. The abbreviated form of the electron configuration for fluorine is
Similarly, the abbreviated form of the electron configuration for chlorine, the second halogen, is
From these two examples we see that the characteristic outer-shell electron configuration of a halogen is ns2np5, where n ranges from 2 in the case of fluorine to 6 in the case of astatine.
What family of elements is characterized by having an ns2np2 outer-electron configuration? Answer: Group 4A
(a) Write the complete electron configuration for bismuth, element number 83. (b) Write the abbreviated electron configuration for this element, showing the appropriate noble-gas core. (c) How many unpaired electrons does each atom of bismuth possess?
SOLUTION (a) We write the complete electron configuration by simply moving across the periodic table one row at a time and writing the occupancies of the orbital corresponding to each row (refer to Figure 6.29).
Note that 3 is the lowest possible value that n may have for a d orbital and that 4 is the lowest possible value of n for an f orbital.
The total of the superscripted numbers should equal the atomic number of bismuth, 83. The electrons may be listed, as shown here, in the order of increasing major quantum number. However, it is equally correct to list the orbitals in an electron configuration in the order in which they are read from the periodic table: 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p3.
(b) It is a simple matter to write the abbreviated electron configuration of an element using the periodic table. First locate the element of interest (in this case element 83) and then move backward until the first noble gas is encountered (in this case Xe, element 54). Thus the inner core is [Xe]. The outer electrons are then read from the periodic table as before. Moving from Xe to Cs, element 55, we find ourselves in the sixth row. Moving across this row to Bi gives us the outer electrons. Thus the abbreviated electron configuration is as follows: [Xe]6s24f145d106p3 or [Xe]4f145d106s26p3.
(c) We can see from the abbreviated electron configuration that the only partially occupied subshell is the 6p. The orbital diagram representation for this subshell is as follows:
In accordance with the Hund's rule, the three 6p electrons occupy three 6p orbitals singly, with their spins in parallel. Thus, there are three unpaired electrons in each atom of bismuth.
Use the periodic table to write the electron configurations for the following atoms by giving the appropriate noble-gas inner core plus the electrons beyond it: (a) Co (atomic number 27); (b) Te (atomic number 52). Answers: (a) [Ar]4s23d7 or [Ar]3d74s2; (b) [Kr]5s24d105p4 or [Kr]4d105s25p4
Table 6.5 gives a complete list of the ground electron configurations of the elements. You should use this table to check your answers as you practice writing electron configurations. We have written these configurations as they would be read off the periodic table. As we have seen in Sample Exercise 6.9, they are sometimes written with orbitals of a given principal quantum number grouped together.
If you inspect Table 6.5 closely, you will see that the electron configurations of certain elements appear to violate the rules we have just discussed. For example, the electron configuration of chromium is [Ar]4s13d5 rather than [Ar]4s23d4, as we might have expected. Similarly, the configuration of copper is [Ar]4s13d10 instead of [Ar]4s23d9. This anomalous behavior is largely a consequence of the closeness of the 3d and 4s orbital energies. It frequently occurs when there are enough electrons to lead to precisely half-filled sets of degenerate orbitals (as in chromium) or to a completely filled d subshell (as in copper). There are a few similar cases among the heavier transition metals (those with partially filled 4d or 5d orbitals) and among the f-block metals. Although these minor departures from the expected are interesting, they are not of great chemical significance.
Boron, atomic number 5, occurs naturally as two nuclides, 10B and 11B, with natural abundances of 19.9 percent and 80.1 percent, respectively. (a) In what ways do the two nuclides differ? How do the electronic configurations of 10B and 11B differ? (b) Draw the complete orbital diagram representation for an atom of 11B. (c) Indicate the major ways in which the 1s and 2s electrons in boron differ. (d) Elemental boron reacts with fluorine to form BF3, a gas. Write a balanced chemical equation for the reaction of solid boron with fluorine gas. (e) for BF3(g) is -1135.6 kJ mol–1. Calculate the standard enthalpy change in the reaction of boron with fluorine. (f) When BCl3, also a gas at room temperature, comes into contact with water, it reacts to form hydrochloric acid, and boric acid, H3BO3, a very weak acid in water. Write a balanced net ionic equation for this reaction.
SOLUTION (a) The two nuclides of boron differ in the number of neutrons in the nucleus (Sections 2.3 and 3.3). Both nuclides contain five protons. 10B contains five neutrons, 11B contains six neutrons. The two isotopes of boron have identical electron configurations, 1s22s22p1 because both have five electrons in the space outside the nucleus.
(b) The complete orbital diagram is:
(c) The 1s and 2s orbitals are both spherical, but they differ in three important respects: First, the 1s orbital is lower in energy than the 2s; second, the 2s orbital has one node, whereas the 1s orbital has no nodes (Figure 6.20). Third, the average distance of the 2s electrons from the nucleus is greater than that of the 1s electrons, so the latter shield the 2s electrons from the full nuclear charge. Indeed, the larger average distance of the 2s from the nucleus is the reason that the 2s electrons are higher in energy than the 1s.
(d) The balanced chemical equation is as follows:
(e) = 2(-1135.6) - [0 + 0] = -2271.2 kJ mol–1(g). The reaction is strongly exothermic.
(f) BCl3(g) + 3H2O(l) H3BO3(aq) + 3H+(aq)+ 3Cl–(aq). Note that because H3BO3 is a very weak acid, it is shown in molecular form, as discussed in Section 4.4.