# 7.8: Electron Configurations

Skills to Develop

• To write the electron configuration of any element and relate its electron configuration to its position in the periodic table.
• To understand the basics of adding electrons to atomic orbitals by the Aufbau principle.

The quantum mechanical model allowed us to determine the energies of the hydrogen atomic orbitals; now we would like to extend this to describe the electronic structure of every element in the Periodic Table. The process of describing each atom’s electronic structure consists, essentially, of beginning with hydrogen and adding one proton and one electron at a time to create the next heavier element in the table; however, interactions between electrons make this process a bit more complicated than it sounds. All stable nuclei other than hydrogen also contain one or more neutrons. Because neutrons have no electrical charge, however, they can be ignored in the following discussion. Before demonstrating how to do this, however, we must introduce the concept of electron spin and the Pauli principle.

### Orbitals and their Energies

Unlike in hydrogen-like atoms with only one electron, in multielectron atoms the values of quantum numbers n and l determine the energies of an orbital. The energies of the different orbitals for a typical multielectron atom are shown in Figure $$\PageIndex{1}$$. Within a given principal shell of a multielectron atom, the orbital energies increase with increasing l. An ns orbital always lies below the corresponding np orbital, which in turn lies below the nd orbital.

Figure $$\PageIndex{1}$$: Orbital Energy Level Diagram for a Typical Multielectron Atom

These energy differences are caused by the effects of shielding (the cancelling effect of electron-electron repulsions on electron-nucleus attraction) and penetration (the extent to which a given orbital lies inside other filled orbitals). For example, an electron in the 2s orbital penetrates inside a filled 1s orbital more than an electron in a 2p orbital does. Since electrons, all being negatively charged, repel each other, an electron closer to the nucleus partially shields an electron farther from the nucleus from the attractive effect of the positively charged nucleus. Hence in an atom with a filled 1s orbital, the effective nuclear charge (Zeff) experienced by a 2s electron is greater than the Zeff experienced by a 2p electron. Consequently, the 2s electron is more tightly bound to the nucleus and has a lower energy, consistent with the order of energies shown in Figure $$\PageIndex{1}$$.

Notice in Figure $$\PageIndex{1}$$ that the difference in energies between subshells can be so large that the energies of orbitals from different principal shells can become approximately equal. For example, the energy of the 3d orbitals in most atoms is actually between the energies of the 4s and the 4p orbitals.

### Electron Spin: The Fourth Quantum Number

When scientists analyzed the emission and absorption spectra of the elements more closely, they saw that for elements having more than one electron, nearly all the lines in the spectra were actually pairs of very closely spaced lines. Because each line represents an energy level available to electrons in the atom, there are twice as many energy levels available as would be predicted solely based on the quantum numbers $$n$$, $$l$$, and $$m_l$$. Scientists also discovered that applying a magnetic field caused the lines in the pairs to split farther apart. In 1925, two graduate students in physics in the Netherlands, George Uhlenbeck (1900–1988) and Samuel Goudsmit (1902–1978), proposed that the splittings were caused by an electron spinning about its axis, much as Earth spins about its axis. When an electrically charged object spins, it produces a magnetic moment parallel to the axis of rotation, making it behave like a magnet. Although the electron cannot be viewed solely as a particle, spinning or otherwise, it is indisputable that it does have a magnetic moment. This magnetic moment is called electron spin.

Figure $$\PageIndex{2}$$: Electron Spin. In a magnetic field, an electron has two possible orientations with different energies, one with spin up, aligned with the magnetic field, and one with spin down, aligned against it. All other orientations are forbidden.

In an external magnetic field, the electron has two possible orientations (Figure $$\PageIndex{2}$$). These are described by a fourth quantum number (ms), which for any electron can have only two possible values, designated +½ (up) and −½ (down) to indicate that the two orientations are opposites; the subscript s is for spin. An electron behaves like a magnet that has one of two possible orientations, aligned either with the magnetic field or against it.

### The Pauli Exclusion Principle

The implications of electron spin for chemistry were recognized almost immediately by an Austrian physicist, Wolfgang Pauli (1900–1958; Nobel Prize in Physics, 1945), who determined that each orbital can contain no more than two electrons. He developed the Pauli exclusion principle: No two electrons in an atom can have the same values of all four quantum numbers (n, l, ml, ms).

By giving the values of n, l, and ml, we also specify a particular orbital (e.g., 1s with n = 1, l = 0, ml = 0). Because ms has only two possible values (+½ or −½), two electrons, and only two electrons, can occupy any given orbital, one with spin up and one with spin down. With this information, we can proceed to construct the entire periodic table, which was originally based on the physical and chemical properties of the known elements.

Example $$\PageIndex{1}$$

List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for electrons in a 2p orbital and predict the maximum number of electrons the 2p subshell can accommodate.

Given: orbital

Asked for: allowed quantum numbers and maximum number of electrons in orbital

Strategy:

1. List the quantum numbers (n, l, ml) that correspond to an n = 2p orbital. List all allowed combinations of (n, l, ml).
2. Build on these combinations to list all the allowed combinations of (n, l, ml, ms).
3. Add together the number of combinations to predict the maximum number of electrons the 2p subshell can accommodate.

Solution:

A For a 2p orbital, we know that n = 2, l = n − 1 = 1, and ml = −l, (−l +1),…, (l − 1), l. There are only three possible combinations of (n, l, ml): (2, 1, 1), (2, 1, 0), and (2, 1, −1).

B Because ms is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of (n, l, ml, ms): (2, 1, 1, +½), (2, 1, 1, −½), (2, 1, 0, +½), (2, 1, 0, −½), (2, 1, −1, +½), and (2, 1, −1, −½).

C Hence the 2p subshell, which consists of three 2p orbitals (2px, 2py, and 2pz), can contain a total of six electrons, two in each orbital.

Exercise $$\PageIndex{1}$$

List all the allowed combinations of the four quantum numbers (n, l, ml, ms) for a 6s orbital, and predict the total number of electrons it can contain.

(6, 0, 0, +½), (6, 0, 0, −½); two electrons

### The Aufbau Principle

The electron configuration of an element is the arrangement of its electrons in its atomic orbitals. By knowing the electron configuration of an element, we can predict and explain a great deal of its chemistry. We can predict electron configurations for the elements by following the aufbau principle (from German, meaning “building up”). First we determine the number of electrons in the atom; then we add electrons one at a time to the lowest-energy orbital available without violating the Pauli principle. We use the orbital energy diagram of Figure $$\PageIndex{1}$$, recognizing that each orbital can hold two electrons, one with spin up ↑, corresponding to ms = +½, which is arbitrarily written first, and one with spin down ↓, corresponding to ms = −½. A filled orbital is indicated by ↑↓, in which the electron spins are said to be paired. Here is a schematic orbital diagram for a hydrogen atom in its ground state:

From the orbital diagram, we can write the electron configuration in an abbreviated form in which the occupied orbitals are identified by their principal quantum number n and their value of l (s, p, d, or f), with the number of electrons in the subshell indicated by a superscript. For hydrogen, therefore, the single electron is placed in the 1s orbital, which is the orbital lowest in energy (Figure $$\PageIndex{1}$$), and the electron configuration is written as 1s1 and read as “one-s-one.”

A neutral helium atom, with an atomic number of 2 (Z = 2), has two electrons. We place one electron in the orbital that is lowest in energy, the 1s orbital. From the Pauli exclusion principle, we know that an orbital can contain two electrons with opposite spin, so we place the second electron in the same orbital as the first but pointing down, so that the electrons are paired. The orbital diagram for the helium atom is therefore

written as 1s2, where the superscript 2 implies the pairing of spins. Otherwise, our configuration would violate the Pauli principle.

The next element is lithium, with Z = 3 and three electrons in the neutral atom. We know that the 1s orbital can hold two of the electrons with their spins paired; the third electron must enter a higher energy orbital. Figure 6.29 tells us that the next lowest energy orbital is 2s, so the orbital diagram for lithium is

This electron configuration is written as 1s22s1.

The next element is beryllium, with Z = 4 and four electrons. We fill both the 1s and 2s orbitals to achieve a 1s22s2 electron configuration:

When we reach boron, with Z = 5 and five electrons, we must place the fifth electron in one of the 2p orbitals. Because all three 2p orbitals are degenerate, it doesn’t matter which one we select. The electron configuration of boron is 1s22s22p1:

At carbon, with Z = 6 and six electrons, we are faced with a choice. Should the sixth electron be placed in the same 2p orbital that already has an electron, or should it go in one of the empty 2p orbitals? If it goes in an empty 2p orbital, will the sixth electron have its spin aligned with or be opposite to the spin of the fifth? In short, which of the following three orbital diagrams is correct for carbon, remembering that the 2p orbitals are degenerate?

Because of electron-electron interactions, it is more favorable energetically for an electron to be in an unoccupied orbital than in one that is already occupied; hence we can eliminate choice a. Similarly, experiments have shown that choice b is slightly higher in energy (less stable) than choice c because electrons in degenerate orbitals prefer to line up with their spins parallel; thus, we can eliminate choice b. Choice c illustrates Hund’s rule (named after the German physicist Friedrich H. Hund, 1896–1997), which today says that the lowest-energy electron configuration for an atom is the one that has the maximum number of electrons with parallel spins in degenerate orbitals. By Hund’s rule, the electron configuration of carbon, which is 1s22s22p2, is understood to correspond to the orbital diagram shown in c. Experimentally, it is found that the ground state of a neutral carbon atom does indeed contain two unpaired electrons.

When we get to nitrogen (Z = 7, with seven electrons), Hund’s rule tells us that the lowest-energy arrangement is

with three unpaired electrons. The electron configuration of nitrogen is thus 1s22s22p3.

At oxygen, with Z = 8 and eight electrons, we have no choice. One electron must be paired with another in one of the 2p orbitals, which gives us two unpaired electrons and a 1s22s22p4 electron configuration. Because all the 2p orbitals are degenerate, it doesn’t matter which one has the pair of electrons.

Similarly, fluorine has the electron configuration 1s22s22p5:

When we reach neon, with Z = 10, we have filled the 2p subshell, giving a 1s22s22p6 electron configuration:

Notice that for neon, as for helium, all the orbitals through the 2p level are completely filled. This fact is very important in dictating both the chemical reactivity and the bonding of helium and neon, as you will see.

### Valence Electrons

As we continue through the periodic table in this way, writing the electron configurations of larger and larger atoms, it becomes tedious to keep copying the configurations of the filled inner subshells. In practice, chemists simplify the notation by using a bracketed noble gas symbol to represent the configuration of the noble gas from the preceding row because all the orbitals in a noble gas are filled. For example, [Ne] represents the 1s22s22p6 electron configuration of neon (Z = 10), so the electron configuration of sodium, with Z = 11, which is 1s22s22p63s1, is written as [Ne]3s1:

 Neon Z = 10 1s22s22p6 Sodium Z = 11 1s22s22p63s1 = [Ne]3s1

Because electrons in filled inner orbitals are closer to the nucleus and more tightly bound to it, they are rarely involved in chemical reactions. This means that the chemistry of an atom depends mostly on the electrons in its outermost shell, which are called the valence electrons. The simplified notation allows us to see the valence-electron configuration more easily. Using this notation to compare the electron configurations of sodium and lithium, we have:

 Sodium 1s22s22p63s1 = [Ne]3s1 Lithium 1s22s1 = [He]2s1

It is readily apparent that both sodium and lithium have one s electron in their valence shell. We would therefore predict that sodium and lithium have very similar chemistry, which is indeed the case.

As we continue to build the eight elements of period 3, the 3s and 3p orbitals are filled, one electron at a time. This row concludes with the noble gas argon, which has the electron configuration [Ne]3s23p6, corresponding to a filled valence shell.

Example $$\PageIndex{1}$$: Electronic Configuration of Phosphorus

Draw an orbital diagram and use it to derive the electron configuration of phosphorus, Z = 15. What is its valence electron configuration?

Given: atomic number

Asked for: orbital diagram and valence electron configuration for phosphorus

Strategy:

1. Locate the nearest noble gas preceding phosphorus in the periodic table. Then subtract its number of electrons from those in phosphorus to obtain the number of valence electrons in phosphorus.
2. Referring to Figure $$\PageIndex{1}$$, draw an orbital diagram to represent those valence orbitals. Following Hund’s rule, place the valence electrons in the available orbitals, beginning with the orbital that is lowest in energy. Write the electron configuration from your orbital diagram.
3. Ignore the inner orbitals (those that correspond to the electron configuration of the nearest noble gas) and write the valence electron configuration for phosphorus.

Solution:

A Because phosphorus is in the third row of the periodic table, we know that it has a [Ne] closed shell with 10 electrons. We begin by subtracting 10 electrons from the 15 in phosphorus.

B The additional five electrons are placed in the next available orbitals, which Figure $$\PageIndex{1}$$ tells us are the 3s and 3p orbitals:

Because the 3s orbital is lower in energy than the 3p orbitals, we fill it first:

Hund’s rule tells us that the remaining three electrons will occupy the degenerate 3p orbitals separately but with their spins aligned:

The electron configuration is [Ne]3s23p3.

C We obtain the valence electron configuration by ignoring the inner orbitals, which for phosphorus means that we ignore the [Ne] closed shell. This gives a valence-electron configuration of 3s23p3.

Exercise $$\PageIndex{1}$$

Draw an orbital diagram and use it to derive the electron configuration of chlorine, Z = 17. What is its valence electron configuration?

[Ne]3s23p5; 3s23p5

The general order in which orbitals are filled by the Aufbau process is depicted in Figure $$\PageIndex{3}$$. Subshells corresponding to each value of n are written from left to right on successive horizontal lines, where each row represents a row in the periodic table. The order in which the orbitals are filled is indicated by the diagonal lines running from the upper right to the lower left. Accordingly, the 4s orbital is filled prior to the 3d orbital because of shielding and penetration effects. Consequently, the electron configuration of potassium, which begins the fourth period, is [Ar]4s1, and the configuration of calcium is [Ar]4s2. Five 3d orbitals are filled by the next 10 elements, the transition metals, followed by three 4p orbitals. Notice that the last member of this row is the noble gas krypton (Z = 36), [Ar]4s23d104p6 = [Kr], which has filled 4s, 3d, and 4p orbitals. The fifth row of the periodic table is essentially the same as the fourth, except that the 5s, 4d, and 5p orbitals are filled sequentially.

Figure $$\PageIndex{3}$$: Predicting the Order in Which Orbitals Are Filled in Multielectron Atoms. If you write the subshells for each value of the principal quantum number on successive lines, the observed order in which they are filled is indicated by a series of diagonal lines running from the upper right to the lower left.

The sixth row of the periodic table will be different from the preceding two because the 4f orbitals, which can hold 14 electrons, are filled between the 6s and the 5d orbitals. The elements that contain 4f orbitals in their valence shell are the lanthanides. When the 6p orbitals are finally filled, we have reached the next noble gas, radon (Z = 86), [Xe]6s24f145d106p6 = [Rn]. In the last row, the 5f orbitals are filled between the 7s and the 6d orbitals, which gives the 14 actinide elements. Because the large number of protons makes their nuclei unstable, all the actinides are radioactive.

Example $$\PageIndex{2}$$: Electron Configuration of Mercury

Write the electron configuration of mercury (Z = 80), showing all the inner orbitals.

Given: atomic number

Strategy:

Using the orbital diagram in Figure $$\PageIndex{1}$$ and the periodic table as a guide, fill the orbitals until all 80 electrons have been placed.

Solution:

By placing the electrons in orbitals following the order shown in Figure $$\PageIndex{2}$$ and using the periodic table as a guide, we obtain

 1s2 row 1 2 electrons 2s22p6 row 2 8 electrons 3s23p6 row 3 8 electrons 4s23d104p6 row 4 18 electrons 5s24d105p6 row 5 18 electrons row 1–5 54 electrons

After filling the first five rows, we still have 80 − 54 = 26 more electrons to accommodate. According to Figure $$\PageIndex{2}$$, we need to fill the 6s (2 electrons), 4f (14 electrons), and 5d (10 electrons) orbitals. The result is mercury’s electron configuration:

1s22s22p63s23p64s23d104p65s24d105p66s24f145d10 = Hg = [Xe]6s24f145d10

with a filled 5d subshell, a 6s24f145d10 valence shell configuration, and a total of 80 electrons. (You should always check to be sure that the total number of electrons equals the atomic number.)

Exercise $$\PageIndex{2}$$: Electron Configuration of Flerovium

Although element 114 is not stable enough to occur in nature, atoms of element 114 were created for the first time in a nuclear reactor in 1998 by a team of Russian and American scientists. The element is named after the Flerov Laboratory of Nuclear Reactions of the Joint Institute for Nuclear Research in Dubna, Russia, where the element was discovered. The name of the laboratory, in turn, honors the Russian physicist Georgy Flyorov. Write the complete electron configuration for element 114.