7.3: Electron Configurations of Atoms (2024)

Learning Objectives

• Using the AufbauPrinciple, the PauliExclusion Principle, and Hund's rule topredict an atom's electron configuration using the periodic table as a guide
• Differentiate between (spdf) electron configuration, orbital box diagram, and Nobel Gas Configuration
• Determine the number of valence electrons in atom based on the electron configuration and main group number
• Note the exeptionsin electron configurations when electrons shift to obtain a half filled orbital or a half filled orbital

Introduction

The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating for the loss of or gain of electrons in their subsequent orbitals (we will examine those in the next section). Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element.

Recall, we can use the periodic table to rank the energy levels of various orbitals.

Using the periodic table to determine the electron configurations of atoms is key, but also keep in mind that there are certain rules to follow when assigning electrons to different orbitals. The periodic table is an incredibly helpful tool in writing electron configurations.

Assigning Electron Configuration

We write electronic configurationsby 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 PauliExclusion Principle . That is, recognizing that each orbital can hold two electrons, one with spin up ↑, corresponding to m_s = +½, which is arbitrarily written first, and one with spin down ↓, corresponding to m_s = −½. A filled orbital is indicated by ↑↓, in which the electron spins are said to be paired. Here is a schematic orbital box 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, and the electron configuration (also known a spdfnotation) 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. 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 repulsions, 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.

Let's take a look video tutor to help you understand how to use the periodic table to write electron configuration for atoms in various elements.

Nobel Gas Configuration

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:

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 (with the greatest "n" value), 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:

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. Now we have explained why elements in the same group have similar chemical properties.

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.

Electron Configuration of Transition Metals and Inner Transition Metals

So far, we have studied the electron configuration for elements in periods 1-3 on the periodic table in which we filledsandporbitals. When we get to period 4-7 on the periodic table, we will require the use of thedandforbitals for transition metals and inner transition metals. Let's take a look at a few examples on how to write the electron configuration for such elements.

Exceptions

In several cases, the ground state electron configurations are different from those predicted using th periodic table and the Aufbau Principle. Some of these anomalies occur as the 3d orbitals are filled. For example, the observed ground state electron configuration of chromium is [Ar]4s¹3d⁵ rather than the predicted [Ar]4s²3d⁴. Similarly, the observed electron configuration of copper is [Ar]4s¹3d¹⁰ instead of [Ar]s²3d⁹. The actual electron configuration may be rationalized in terms of an added stability associated with a half-filled (ns¹, np³, nd⁵, nf⁷) or filled (ns², np⁶, nd¹⁰, nf¹⁴) subshell. Given the small differences between higher energy levels, this added stability is enough to shift an electron from one orbital to another. In heavier elements, other more complex effects can also be important, leading to some of the additional anomalies.For example, cerium has an electron configuration of [Xe]6s²4f¹5d¹, which is impossible to rationalize in simple terms. In most cases, however, these apparent anomalies do not have important chemical consequences. Table 1 summarizes some of those exceptions:

Summary

Based on the Pauli principle and a knowledge of orbital energies obtained using hydrogen-like orbitals, it is possible to construct the periodic table by filling up the available orbitals beginning with the lowest-energy orbitals (the aufbau principle), which gives rise to a particular arrangement of electrons for each element (its electron configuration). Hund’s rule says that the lowest-energy arrangement of electrons is the one that places them in degenerate orbitals with their spins parallel. For chemical purposes, the most important electrons are those in the outermost principal shell, the valence electrons.

Problems

Unless specified, use any method to solve the following problems. Answers are given in noble gas notation.

1. Find the electron configurations of the following:

1. silicon
2. tin
3. lead

2. Scenario: You are currently studying the element iodine and wish to use its electron distributions to aid you in your work.

1. Find the electron configuration of iodine
2. How many unpaired electrons does iodine have?

3. Thought Questions:

1. In your own words describe how to write an electron configuration and why it is an important skill in the study of chemistry.
2. Describe the major concepts (Hunds, Pauli...etc.) and explain why each is a key part of the "tool kit" when describing electron configurations
3. Why is it possible to abbreviate electron configurations with a noble gas in the noble gas notation?

4. Identify the following elements:

1. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
2. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁷
3. 1s² 2s² 2p⁶ 3s² 3p⁴
4. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁴

5. Without using a periodic table or any other references, fill in the correct box in the periodic table with the letter of each question. (a)The element with electron configuration: 1s² 2s² 2p⁶ 3s² 3p⁵; (b)A noble gases with f electrons; (c) a fifth-period element whose atoms have three unpaired p electrons; (d) First rowtransition metals having one 4s electron.