We just saw that the electron configurations of the elements correspond to their locations in the periodic table. Thus, elements in the same column of the table have related outer-shell (valence) electron configurations. As TABLE 6.4 shows, for example, all 2A elements have an ns2 outer configuration, and all 3A elements have an ns2 np outer configuration, with the value of n increasing as we move down each column.

As shown in FIGURE 6.30, the periodic table can be divided into four blocks based on the filling order of orbitals. On the left are two blue columns of elements. These elements, known as the alkali metals (group 1A) and alkaline earth metals (group 2A), are those in which the valence s orbitals are being filled. These two columns make up the s block of the periodic table.

On the right is a block of six pink columns that comprises the p block, where the valence p orbitals are being filled. The s block and the p block elements together are the representative elements, sometimes called the main-group elements.

The orange block in Figure 6.30 has ten columns containing the transition metals. These are the elements in which the valence d orbitals are being filled and make up the d block.

TABLE 6.4 • Electron Configurations of Group 2A and 3A Elements

FIGURE 6.30 Regions of the periodic table. The order in which electrons are added to orbitals is read left to right beginning in the top left corner.

The elements in the two tan rows containing 14 columns are the ones in which the valence f orbitals are being filled and make up the f block. Consequently, these elements are often referred to as the f-block metals. In most tables, the f block is positioned below the periodic table to save space:

The number of columns in each block corresponds to the maximum number of electrons that can occupy each kind of subshell. Recall that 2, 6, 10, and 14 are the numbers of electrons that can fill the s, p, d, and f subshells, respectively. Thus, the s block has 2 columns, the p block has 6, the d block has 10, and the f block has 14. Recall also that ls is the first s subshell, 2p is the first p subshell, 3d is the first d subshell, and 4f is the first f subshell, as Figure 6.30 shows. Using these facts, you can write the electron configuration of an element based merely on its position in the periodic table.

Let's use the periodic table to write the electron configuration of selenium (Se, element 34). We first locate Se in the table and then move backward from it through the table, from element 34 to 33 to 32 and so forth, until we come to the noble gas that precedes Se. In this case, the noble gas is argon, Ar, element 18. Thus, the noble-gas core for Se is [Ar]. Our next step is to write symbols for the outer electrons. We do this by moving across period 4 from K, the element following Ar, to Se:

Because K is in the fourth period and the s block, we begin with the 4s electrons, meaning our first two outer electrons are written 4s2. We then move into the d block, which begins with the 3d electrons. (The principal quantum number in the d block is always one less than that of the preceding elements in the s block, as seen in Figure 6.30.) Traversing the d block adds ten electrons, 3d10. Finally we move into the p block, whose principal quantum number is always the same as that of the s block. Counting the squares as we move across the p block to Se tells us that we need four electrons, 4p4. The electron configuration for Se is therefore [Ar]4s23d104p4. This configuration can also be written with the subshells arranged in order of increasing principal quantum number: [Ar]3d104s24p3.

As a check, we add the number of electrons in the [Ar] core, 18, to the number of electrons we added to the 4s, 3d, and 4p subshells. This sum should equal the atomic number of Se, 34: 18 + 2 + 10 + 4 = 34.

SAMPLE EXERCISE 6.8 Electron Configurations for a Group

What is the characteristic valence electron configuration of the group 7A elements, the halogens?


Analyze and Plan We first locate the halogens in the periodic table, write the electron configurations for the first two elements, and then determine the general similarity between the configurations.

Solve The first member of the halogen group is fluorine (F, element 9). Moving backward from F, we find that the noble-gas core is [He]. Moving from He to the element of next higher atomic number brings us to Li, element 3. Because Li is in the second period of the s block, we add electrons to the 2s subshell. Moving across this block gives 2s2. Continuing to move to the right, we enter the p block. Counting the squares to F gives 2p5. Thus, the condensed electron configuration for fluorine is

The electron configuration for chlorine, the second halogen, is

From these two examples, we see that the characteristic valence electron configuration of a halogen is ns2np5, where n ranges from 2 in the case of fluorine to 6 in the case of astatine.


Which family of elements is characterized by an ns2np2 electron configuration in the outermost occupied shell?

Answer: group 4A

SAMPLE EXERCISE 6.9 Electron Configurations from the Periodic Table

(a) Based on its position in the periodic table, write the condensed electron configuration for bismuth, element 83. (b) How many unpaired electrons does a bismuth atom have?


(a) Our first step is to write the noble-gas core. We do this by locating bismuth, element 83, in the periodic table. We then move backward to the nearest noble gas, which is Xe, element 54. Thus, the noble-gas core is [Xe].

Next, we trace the path in order of increasing atomic numbers from Xe to Bi. Moving from Xe to Cs, element 55, we find ourselves in period 6 of the s block. Knowing the block and the period identifies the subshell in which we begin placing outer electrons, 6s. As we move through thes block, we add two electrons: 6s2.

As we move beyond the s block, from element 56 to element 57, the curved arrow below the periodic table reminds us that we are entering the f block. The first row of the f block corresponds to the 4f subshell. As we move across this block, we add 14 electrons: 4f 14.

With element 71, we move into the third row of the d block. Because the first row of the d block is 3d, the second row is 4d and the third row is 5d. Thus, as we move through the ten elements of the d block, from element 71 to element 80, we fill the 5d subshell with ten electrons: 5d10.

Moving from element 80 to element 81 puts us into the p block in the 6p subshell. (Remember that the principal quantum number in the p block is the same as in the s block.) Moving across to Bi requires 3 electrons: 6p3. The path we have taken is Putting the parts together, we obtain the condensed electron configuration: [Xe]6s24f 145d106p3. This configuration can also be written with the subshells arranged in order of increasing principal quantum number: [Xe]4f 145d106s26p3.

Finally, we check our result to see if the number of electrons equals the atomic number of Bi, 83: Because Xe has 54 electrons (its atomic number), we have 54 + 2 + 14 + 10 + 3 = 83. (If we had 14 electrons too few, we would realize that we have missed the f block.)

(b) We see from the condensed electron configuration that the only partially occupied subshell is 6p. The orbital diagram representation for this subshell is

In accordance with Hund's rule, the three 6p electrons occupy the three 6p orbitals singly, with their spins parallel. Thus, there are three unpaired electrons in the bismuth atom.


Use the periodic table to write the condensed electron configuration for (a) Co (element 27), (b) Te (element 52).

Answers: (a) [Ar]4s23d7 or [Ar]3d74s2, (b) [Kr]5s24d105p4 or [Kr]4d105s25p4

FIGURE 6.31 gives, for all the elements, the ground-state electron configurations for the valence electrons. You can use this figure to check your answers as you practice writing electron configurations. We have written these configurations with orbitals listed in order of increasing principal quantum number. As we saw in Sample Exercise 6.9, the orbitals can also be listed in order of filling, as they would be read off the periodic table.

FIGURE 6.31 Valence electron configurations of the elements.

Figure 6.31 allow us to reexamine the concept of valence electrons. Notice, for example, that as we proceed from Cl ([Ne]3s2 3p5) to Br ([Ar]3d10 4s2 24p5) we add a complete subshell of 3d electrons to the electrons beyond the [Ar] core. Although the 3d electrons are outer-shell electrons, they are not involved in chemical bonding and are therefore not considered valence electrons. Thus, we consider only the 4s and 4p electrons of Br to be valence electrons. Similarly, if we compare the electron configurations of Ag (element 47) and Au (element 79), we see that Au has a completely full 4f14 subshell beyond its noble-gas core, but those 4f15 electrons are not involved in bonding. In general, for representative elements we do not consider the electrons in completely filled d or f subshells to be valence electrons, and for transition elements we do not consider the electrons in a completely filled f subshell to be valence electrons.

Anomalous Electron Configurations

The electron configurations of certain elements appear to violate the rules we have just discussed. For example, Figure 6.31 shows that the electron configuration of chromium (element 24) is [Ar]3d5 4s1 rather than the [Ar]3d4 4s2 configuration we might expect. Similarly, the configuration of copper (element 29) is [Ar]3d10 4s1 instead of [Ar]3d94s2.

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 form precisely half-filled sets of degenerate orbitals (as in chromium) or 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.


The elements Ni, Pd, and Pt are all in the same group. By examining the electron configurations for these elements in Figure 6.31, what can you conclude about the relative energies of the nd and (n + 1)s orbitals for this group?


Boron, atomic number 5, occurs naturally as two isotopes, 10B and 11B, with natural abundances of 19.9% and 80.1%, respectively. (a) In what ways do the two isotopes differ from each other? Does the electronic configuration of B differ from that of B? (b) Draw the orbital diagram for an atom of B. Which electrons are the valence electrons? (c) Indicate three major ways in which the 1s electrons in boron differ from its 2s electrons. (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) is –1135.6 kJ/mol 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, the two react to form hydrochloric acid and boric acid, H3BO3, a very weak acid in water. Write a balanced net ionic equation for this reaction.


(a) The two isotopes of boron differ in the number of neutrons in the nucleus. (Sections 2.3 and 2.4) Each of the isotopes contains five protons, but 10 contains five neutrons, whereas B contains six neutrons. The two isotopes of boron have identical electron configurations, 1s 22s 22p1, because each has five electrons.

(b) The complete orbital diagram is

The valence electrons are the ones in the outermost occupied shell, the 2s2 and 2p1 electrons. The 1s2 electrons constitute the core electrons, which we represent as [He] when we write the condensed electron configuration, [He]2s22p1.

(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 orbital. Second, the average distance of the 2s electrons from the nucleus is greater than that of the 1s electrons, so the 1s orbital is smaller than the 2s. Third, the 2s orbital has one node, whereas the 1s orbital has no nodes (Figure 6.18).

(d) The balanced chemical equation is

2 B(s) + 3 F2(g) → 2BF3(g)

(e) ΔH° = 2(–1135.6) – [0 + 0] = –2271.2 kJ. The reaction is strongly exothermic.

(f) BCl3(g) + 3 H2O(l) → H3BO3(aq) + 3 H+(aq) + 3 Cl(aq). Note that because H3BO3 is a very weak acid, its chemical formula is written in molecular form, as discussed in Section 4.3.