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<title>Building Up the Periodic Table</title>
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<div class="section" id="averill_1.0-ch06_s06" condition="start-of-chunk" version="5.0" lang="en">
<h2 class="title editable block">
<span class="title-prefix">6.6</span> Building Up the Periodic Table</h2>
<div class="learning_objectives editable block" id="averill_1.0-ch06_s06_n01">
<h3 class="title">Learning Objective</h3>
<ol class="orderedlist" id="averill_1.0-ch06_s06_l01">
<li>To write the electron configuration of any element and relate its electron configuration to its position in the periodic table.</li>
</ol>
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_p01">Now you can use the information you learned in <a class="xref" href="averill_1.0-ch06_s05#averill_1.0-ch06_s05">Section 6.5 "Atomic Orbitals and Their Energies"</a> to determine 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.<span class="footnote" id="averill_1.0-fn06_007">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.</span> Before demonstrating how to do this, however, we must introduce the concept of electron spin and the Pauli principle.</p>
<div class="section" id="averill_1.0-ch06_s06_s01">
<h2 class="title editable block">Electron Spin: The Fourth Quantum Number</h2>
<p class="para editable block" id="averill_1.0-ch06_s06_s01_p01">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 <em class="emphasis">pairs</em> 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 <em class="emphasis">n</em>, <em class="emphasis">l</em>, and <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>. 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 <span class="margin_term"><a class="glossterm">electron spin</a><span class="glossdef">The magnetic moment that results when an electron spins. Electrons have two possible orientations (spin up and spin down), which are described by a fourth quantum number (<em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>).</span></span>.</p>
<div class="figure large medium-height editable block" id="averill_1.0-ch06_s06_s01_f01">
<p class="title"><span class="title-prefix">Figure 6.31</span> Electron Spin</p>
<img src="section_10/a099e02893dbe7684308e61389fbd611.jpg">
<p class="para">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.</p>
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s01_p02">In an external magnetic field, the electron has two possible orientations (<a class="xref" href="#averill_1.0-ch06_s06_s01_f01">Figure 6.31 "Electron Spin"</a>). These are described by a fourth quantum number (<em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>), 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 <em class="emphasis">s</em> 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.</p>
</div>
<div class="section" id="averill_1.0-ch06_s06_s02">
<h2 class="title editable block">The Pauli Principle</h2>
<p class="para editable block" id="averill_1.0-ch06_s06_s02_p01">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 <span class="margin_term"><a class="glossterm">Pauli exclusion principle</a><span class="glossdef">A principle stating that no two electrons in an atom can have the same value of all four quantum numbers.</span></span>: <em class="emphasis">No two electrons in an atom can have the same values of all four quantum numbers (n, l, m</em><sub class="subscript"><em class="emphasis">l</em></sub><em class="emphasis">, m</em><sub class="subscript"><em class="emphasis">s</em></sub><em class="emphasis">).</em></p>
<p class="para editable block" id="averill_1.0-ch06_s06_s02_p02">By giving the values of <em class="emphasis">n</em>, <em class="emphasis">l</em>, and <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>, we also specify a particular orbital (e.g., 1<em class="emphasis">s</em> with <em class="emphasis">n</em> = 1, <em class="emphasis">l</em> = 0, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub> = 0). Because <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> has only two possible values (+½ or −½), two electrons, <em class="emphasis">and only two electrons</em>, 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, as you learned in <a class="xref" href="averill_1.0-ch01#averill_1.0-ch01">Chapter 1 "Introduction to Chemistry"</a>, was originally based on the physical and chemical properties of the known elements.</p>
<div class="exercises editable block" id="averill_1.0-ch06_s06_s02_n01">
<h3 class="title">Example 7</h3>
<p class="para" id="averill_1.0-ch06_s06_s02_p03">List all the allowed combinations of the four quantum numbers (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>) for electrons in a 2<em class="emphasis">p</em> orbital and predict the maximum number of electrons the 2<em class="emphasis">p</em> subshell can accommodate.</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p04"><strong class="emphasis bold">Given: </strong>orbital</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p05"><strong class="emphasis bold">Asked for: </strong>allowed quantum numbers and maximum number of electrons in orbital</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p06">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p07"><strong class="emphasis bold">A</strong> List the quantum numbers (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>) that correspond to an <em class="emphasis">n</em> = 2<em class="emphasis">p</em> orbital. List all allowed combinations of (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>).</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p08"><strong class="emphasis bold">B</strong> Build on these combinations to list all the allowed combinations of (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>).</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p09"><strong class="emphasis bold">C</strong> Add together the number of combinations to predict the maximum number of electrons the 2<em class="emphasis">p</em> subshell can accommodate.</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p10">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p11"><strong class="emphasis bold">A</strong> For a 2<em class="emphasis">p</em> orbital, we know that <em class="emphasis">n</em> = 2, <em class="emphasis">l</em> = <em class="emphasis">n</em> − 1 = 1, and <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub> = −<em class="emphasis">l</em>, (−<em class="emphasis">l</em> +1),…, (<em class="emphasis">l</em> − 1), <em class="emphasis">l</em>. There are only three possible combinations of (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>): (2, 1, 1), (2, 1, 0), and (2, 1, −1).</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p12"><strong class="emphasis bold">B</strong> Because <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> is independent of the other quantum numbers and can have values of only +½ and −½, there are six possible combinations of (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>): (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).</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p13"><strong class="emphasis bold">C</strong> Hence the 2<em class="emphasis">p</em> subshell, which consists of three 2<em class="emphasis">p</em> orbitals (2<em class="emphasis">p</em><sub class="subscript"><em class="emphasis">x</em></sub>, 2<em class="emphasis">p</em><sub class="subscript"><em class="emphasis">y</em></sub>, and 2<em class="emphasis">p</em><sub class="subscript"><em class="emphasis">z</em></sub>), can contain a total of six electrons, two in each orbital.</p>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p14">List all the allowed combinations of the four quantum numbers (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>) for a 6<em class="emphasis">s</em> orbital, and predict the total number of electrons it can contain.</p>
<p class="para" id="averill_1.0-ch06_s06_s02_p15"><strong class="emphasis bold">Answer: </strong>(6, 0, 0, +½), (6, 0, 0, −½); two electrons</p>
</div>
</div>
<div class="section" id="averill_1.0-ch06_s06_s03">
<h2 class="title editable block">Electron Configurations of the Elements</h2>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_p01">The <span class="margin_term"><a class="glossterm">electron configuration</a><span class="glossdef">The arrangement of an element’s electrons in its atomic orbitals.</span></span> 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.</p>
<div class="section" id="averill_1.0-ch06_s06_s03_s01">
<h2 class="title editable block">The Aufbau Principle</h2>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p01">We construct the periodic table by following the <span class="margin_term"><a class="glossterm">aufbau principle</a><span class="glossdef">The process used to build up the periodic table by adding protons one by one to the nucleus and adding the corresponding electrons to the lowest-energy orbital available without violating the Pauli exclusion principle.</span></span> (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 <em class="emphasis">without violating the Pauli principle</em>. We use the orbital energy diagram of <a class="xref" href="averill_1.0-ch06_s05#averill_1.0-ch06_s05_s05_f02">Figure 6.29 "Orbital Energy Level Diagram for a Typical Multielectron Atom"</a>, recognizing that each orbital can hold two electrons, one with spin up ↑, corresponding to <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> = +½, which is arbitrarily written first, and one with spin down ↓, corresponding to <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> = −½. A filled orbital is indicated by ↑↓, in which the electron spins are said to be <em class="emphasis">paired</em>. Here is a schematic orbital diagram for a hydrogen atom in its ground state:</p>
<div class="informalfigure large medium-height block">
<img src="section_10/d997d5e5ff758c530b4126fc6e573191.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p02">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 <em class="emphasis">n</em> and their value of <em class="emphasis">l</em> (<em class="emphasis">s</em>, <em class="emphasis">p</em>, <em class="emphasis">d</em>, or <em class="emphasis">f</em>), with the number of electrons in the subshell indicated by a superscript. For hydrogen, therefore, the single electron is placed in the 1<em class="emphasis">s</em> orbital, which is the orbital lowest in energy (<a class="xref" href="averill_1.0-ch06_s05#averill_1.0-ch06_s05_s05_f02">Figure 6.29 "Orbital Energy Level Diagram for a Typical Multielectron Atom"</a>), and the electron configuration is written as 1<em class="emphasis">s</em><sup class="superscript">1</sup> and read as “one-s-one.”</p>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p03">A neutral helium atom, with an atomic number of 2 (<em class="emphasis">Z</em> = 2), has two electrons. We place one electron in the orbital that is lowest in energy, the 1<em class="emphasis">s</em> 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</p>
<div class="informalfigure large medium-height block">
<img src="section_10/288de44d0f554f390dbe1541fd9a84fd.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p04">written as 1<em class="emphasis">s</em><sup class="superscript">2</sup>, where the superscript 2 implies the pairing of spins. Otherwise, our configuration would violate the Pauli principle.</p>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p05">The next element is lithium, with <em class="emphasis">Z</em> = 3 and three electrons in the neutral atom. We know that the 1<em class="emphasis">s</em> orbital can hold two of the electrons with their spins paired. <a class="xref" href="averill_1.0-ch06_s05#averill_1.0-ch06_s05_s05_f02">Figure 6.29 "Orbital Energy Level Diagram for a Typical Multielectron Atom"</a> tells us that the next lowest energy orbital is 2<em class="emphasis">s</em>, so the orbital diagram for lithium is</p>
<div class="informalfigure large medium-height block">
<img src="section_10/8fe2c07b011158a46048ec28f318ac92.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p06">This electron configuration is written as 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">1</sup>.</p>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p07">The next element is beryllium, with <em class="emphasis">Z</em> = 4 and four electrons. We fill both the 1<em class="emphasis">s</em> and 2<em class="emphasis">s</em> orbitals to achieve a 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup> electron configuration:</p>
<div class="informalfigure large medium-height block">
<img src="section_10/bccfc0b298ba9e2c9b4b28601ba5b7a0.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p08">When we reach boron, with <em class="emphasis">Z</em> = 5 and five electrons, we must place the fifth electron in one of the 2<em class="emphasis">p</em> orbitals. Because all three 2<em class="emphasis">p</em> orbitals are degenerate, it doesn’t matter which one we select. The electron configuration of boron is 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">1</sup>:</p>
<div class="informalfigure large medium-height block">
<img src="section_10/4e5657c0344f4fc4490c6812d764e5d8.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p09">At carbon, with <em class="emphasis">Z</em> = 6 and six electrons, we are faced with a choice. Should the sixth electron be placed in the same 2<em class="emphasis">p</em> orbital that already has an electron, or should it go in one of the empty 2<em class="emphasis">p</em> orbitals? If it goes in an empty 2<em class="emphasis">p</em> 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 2<em class="emphasis">p</em> orbitals are degenerate?</p>
<div class="informalfigure large medium-height block">
<img src="section_10/34fa3c4a9465e2d7c518b2475d098d4d.jpg">
</div>
<div class="informalfigure large medium-height block">
<img src="section_10/d4376862c8780c562bbfc3ac76a837c8.jpg">
</div>
<div class="informalfigure large medium-height block">
<img src="section_10/ab83394da5a5b1bf4ec147490eeef520.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p13">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 <span class="margin_term"><a class="glossterm">Hund’s rule</a><span class="glossdef">A rule stating 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.</span></span> (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 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">2</sup>, 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.</p>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p14">When we get to nitrogen (<em class="emphasis">Z</em> = 7, with seven electrons), Hund’s rule tells us that the lowest-energy arrangement is</p>
<div class="informalfigure large medium-height block">
<img src="section_10/00517047867a8e7cdda768ee84e57050.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p15">with three unpaired electrons. The electron configuration of nitrogen is thus 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">3</sup>.</p>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p16">At oxygen, with <em class="emphasis">Z</em> = 8 and eight electrons, we have no choice. One electron must be paired with another in one of the 2<em class="emphasis">p</em> orbitals, which gives us two unpaired electrons and a 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">4</sup> electron configuration. Because all the 2<em class="emphasis">p</em> orbitals are degenerate, it doesn’t matter which one has the pair of electrons.</p>
<div class="informalfigure large medium-height block">
<img src="section_10/af26b06ad87644bda0711d829ee66ce4.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p17">Similarly, fluorine has the electron configuration 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">5</sup>:</p>
<div class="informalfigure large medium-height block">
<img src="section_10/59edda5feef8276a9314152d407f4850.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p18">When we reach neon, with <em class="emphasis">Z</em> = 10, we have filled the 2<em class="emphasis">p</em> subshell, giving a 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup> electron configuration:</p>
<div class="informalfigure large medium-height block">
<img src="section_10/ac8d1e3974fc47dd378e64a86594f64d.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s01_p19">Notice that for neon, as for helium, all the orbitals through the 2<em class="emphasis">p</em> 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.</p>
</div>
<div class="section" id="averill_1.0-ch06_s06_s03_s02">
<h2 class="title editable block">Valence Electrons</h2>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s02_p01">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 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup> electron configuration of neon (<em class="emphasis">Z</em> = 10), so the electron configuration of sodium, with <em class="emphasis">Z</em> = 11, which is 1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup>3<em class="emphasis">s</em><sup class="superscript">1</sup>, is written as [Ne]3<em class="emphasis">s</em><sup class="superscript">1</sup>:</p>
<div class="informaltable block" frame="none">
<table cellpadding="0" cellspacing="0">
<tbody>
<tr>
<td>Neon</td>
<td>Z = 10</td>
<td>1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup>
</td>
</tr>
<tr>
<td>Sodium</td>
<td>Z = 11</td>
<td>1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup><strong class="emphasis bold">3</strong><em class="emphasis bolditalic">s</em><sup class="superscript">1</sup> = [Ne]<strong class="emphasis bold">3</strong><em class="emphasis bolditalic">s</em><sup class="superscript">1</sup>
</td>
</tr>
</tbody>
</table>
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s02_p02">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 <span class="margin_term"><a class="glossterm">valence electrons</a><span class="glossdef">Electrons in the outermost shell of an atom.</span></span>. 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:</p>
<div class="informaltable block" frame="none">
<table cellpadding="0" cellspacing="0">
<tbody>
<tr>
<td>Sodium</td>
<td>1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup>3<em class="emphasis">s</em><sup class="superscript">1</sup> = [Ne]3<em class="emphasis">s</em><sup class="superscript">1</sup>
</td>
</tr>
<tr>
<td>Lithium</td>
<td>1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">1</sup> = [He]2<em class="emphasis">s</em><sup class="superscript">1</sup>
</td>
</tr>
</tbody>
</table>
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s02_p03">It is readily apparent that both sodium and lithium have one <em class="emphasis">s</em> electron in their valence shell. We would therefore predict that sodium and lithium have very similar chemistry, which is indeed the case.</p>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s02_p04">As we continue to build the eight elements of period 3, the 3<em class="emphasis">s</em> and 3<em class="emphasis">p</em> orbitals are filled, one electron at a time. This row concludes with the noble gas argon, which has the electron configuration [Ne]3<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">p</em><sup class="superscript">6</sup>, corresponding to a filled valence shell.</p>
<div class="exercises editable block" id="averill_1.0-ch06_s06_s03_s02_n01">
<h3 class="title">Example 8</h3>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p05">Draw an orbital diagram and use it to derive the electron configuration of phosphorus, <em class="emphasis">Z</em> = 15. What is its valence electron configuration?</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p06"><strong class="emphasis bold">Given: </strong>atomic number</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p07"><strong class="emphasis bold">Asked for: </strong>orbital diagram and valence electron configuration for phosphorus</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p08">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p09"><strong class="emphasis bold">A</strong> 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.</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p10"><strong class="emphasis bold">B</strong> Referring to <a class="xref" href="averill_1.0-ch06_s05#averill_1.0-ch06_s05_s05_f02">Figure 6.29 "Orbital Energy Level Diagram for a Typical Multielectron Atom"</a>, 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.</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p11"><strong class="emphasis bold">C</strong> Ignore the inner orbitals (those that correspond to the electron configuration of the nearest noble gas) and write the valence electron configuration for phosphorus.</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p12">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p13"><strong class="emphasis bold">A</strong> 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.</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p14"><strong class="emphasis bold">B</strong> The additional five electrons are placed in the next available orbitals, which <a class="xref" href="averill_1.0-ch06_s05#averill_1.0-ch06_s05_s05_f02">Figure 6.29 "Orbital Energy Level Diagram for a Typical Multielectron Atom"</a> tells us are the 3<em class="emphasis">s</em> and 3<em class="emphasis">p</em> orbitals:</p>
<div class="informalfigure large medium-height">
<img src="section_10/f1911a0ce53848500c72f1a3a63fb04e.jpg">
</div>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p15">Because the 3<em class="emphasis">s</em> orbital is lower in energy than the 3<em class="emphasis">p</em> orbitals, we fill it first:</p>
<div class="informalfigure large medium-height">
<img src="section_10/e41f66a2e1e582bbf2b88f5e95dee993.jpg">
</div>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p16">Hund’s rule tells us that the remaining three electrons will occupy the degenerate 3<em class="emphasis">p</em> orbitals separately but with their spins aligned:</p>
<div class="informalfigure large medium-height">
<img src="section_10/ad37ea9d57885f324b45efd5e9cfdc0c.jpg">
</div>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p17">The electron configuration is [Ne]3<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">p</em><sup class="superscript">3</sup>.</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p18"><strong class="emphasis bold">C</strong> 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 3<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">p</em><sup class="superscript">3</sup>.</p>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p19">Draw an orbital diagram and use it to derive the electron configuration of chlorine, <em class="emphasis">Z</em> = 17. What is its valence electron configuration?</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p20"><strong class="emphasis bold">Answer: </strong>[Ne]3<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">p</em><sup class="superscript">5</sup>; 3<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">p</em><sup class="superscript">5</sup></p>
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s02_p21">The general order in which orbitals are filled is depicted in <a class="xref" href="#averill_1.0-ch06_s06_s03_s02_f01">Figure 6.32 "Predicting the Order in Which Orbitals Are Filled in Multielectron Atoms"</a>. Subshells corresponding to each value of <em class="emphasis">n</em> 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 4<em class="emphasis">s</em> orbital is filled prior to the 3<em class="emphasis">d</em> orbital because of shielding and penetration effects. Consequently, the electron configuration of potassium, which begins the fourth period, is [Ar]4<em class="emphasis">s</em><sup class="superscript">1</sup>, and the configuration of calcium is [Ar]4<em class="emphasis">s</em><sup class="superscript">2</sup>. Five 3<em class="emphasis">d</em> orbitals are filled by the next 10 elements, the transition metals, followed by three 4<em class="emphasis">p</em> orbitals. Notice that the last member of this row is the noble gas krypton (<em class="emphasis">Z</em> = 36), [Ar]4<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">d</em><sup class="superscript">10</sup>4<em class="emphasis">p</em><sup class="superscript">6</sup> = [Kr], which has filled 4<em class="emphasis">s</em>, 3<em class="emphasis">d</em>, and 4<em class="emphasis">p</em> orbitals. The fifth row of the periodic table is essentially the same as the fourth, except that the 5<em class="emphasis">s</em>, 4<em class="emphasis">d</em>, and 5<em class="emphasis">p</em> orbitals are filled sequentially.</p>
<div class="figure large medium-height editable block" id="averill_1.0-ch06_s06_s03_s02_f01">
<p class="title"><span class="title-prefix">Figure 6.32</span> Predicting the Order in Which Orbitals Are Filled in Multielectron Atoms</p>
<img src="section_10/b4a023028c90cfde8191ad2739f1cfaa.jpg">
<p class="para">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.</p>
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s02_p22">The sixth row of the periodic table will be different from the preceding two because the 4<em class="emphasis">f</em> orbitals, which can hold 14 electrons, are filled between the 6<em class="emphasis">s</em> and the 5<em class="emphasis">d</em> orbitals. The elements that contain 4<em class="emphasis">f</em> orbitals in their valence shell are the lanthanides. When the 6<em class="emphasis">p</em> orbitals are finally filled, we have reached the next (and last known) noble gas, radon (<em class="emphasis">Z</em> = 86), [Xe]6<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">f</em><sup class="superscript">14</sup>5<em class="emphasis">d</em><sup class="superscript">10</sup>6<em class="emphasis">p</em><sup class="superscript">6</sup> = [Rn]. In the last row, the 5<em class="emphasis">f</em> orbitals are filled between the 7<em class="emphasis">s</em> and the 6<em class="emphasis">d</em> orbitals, which gives the 14 actinide elements. Because the large number of protons makes their nuclei unstable, all the actinides are radioactive.</p>
<div class="exercises block" id="averill_1.0-ch06_s06_s03_s02_n02">
<h3 class="title">Example 9</h3>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p23">Write the electron configuration of mercury (<em class="emphasis">Z</em> = 80), showing all the inner orbitals.</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p24"><strong class="emphasis bold">Given: </strong>atomic number</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p25"><strong class="emphasis bold">Asked for: </strong>complete electron configuration</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p26">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p27">Using the orbital diagram in <a class="xref" href="#averill_1.0-ch06_s06_s03_s02_f01">Figure 6.32 "Predicting the Order in Which Orbitals Are Filled in Multielectron Atoms"</a> and the periodic table as a guide, fill the orbitals until all 80 electrons have been placed.</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p28">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p29">By placing the electrons in orbitals following the order shown in <a class="xref" href="#averill_1.0-ch06_s06_s03_s02_f01">Figure 6.32 "Predicting the Order in Which Orbitals Are Filled in Multielectron Atoms"</a> and using the periodic table as a guide, we obtain</p>
<div class="informaltable" frame="none">
<table cellpadding="0" cellspacing="0">
<tbody>
<tr>
<td>1<em class="emphasis">s</em><sup class="superscript">2</sup>
</td>
<td>row 1</td>
<td>2 electrons</td>
</tr>
<tr>
<td>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup>
</td>
<td>row 2</td>
<td>8 electrons</td>
</tr>
<tr>
<td>3<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">p</em><sup class="superscript">6</sup>
</td>
<td>row 3</td>
<td>8 electrons</td>
</tr>
<tr>
<td>4<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">d</em><sup class="superscript">10</sup>4<em class="emphasis">p</em><sup class="superscript">6</sup>
</td>
<td>row 4</td>
<td>18 electrons</td>
</tr>
<tr>
<td>5<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">d</em><sup class="superscript">10</sup>5<em class="emphasis">p</em><sup class="superscript">6</sup>
</td>
<td>row 5</td>
<td>18 electrons</td>
</tr>
<tr>
<td></td>
<td>row 1–5</td>
<td>54 electrons</td>
</tr>
</tbody>
</table>
</div>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p30">After filling the first five rows, we still have 80 − 54 = 26 more electrons to accommodate. According to <a class="xref" href="#averill_1.0-ch06_s06_s04_f01">Figure 6.33 "The Periodic Table, Showing How the Elements Are Grouped According to the Kind of Subshell ("</a>, we need to fill the 6<em class="emphasis">s</em> (2 electrons), 4<em class="emphasis">f</em> (14 electrons), and 5<em class="emphasis">d</em> (10 electrons) orbitals. The result is mercury’s electron configuration:</p>
<span class="informalequation">
<span class="mathphrase">1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup>3<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">p</em><sup class="superscript">6</sup>4<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">d</em><sup class="superscript">10</sup>4<em class="emphasis">p</em><sup class="superscript">6</sup>5<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">d</em><sup class="superscript">10</sup>5<em class="emphasis">p</em><sup class="superscript">6</sup>6<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">f</em><sup class="superscript">14</sup>5<em class="emphasis">d</em><sup class="superscript">10</sup> = Hg = [Xe]6<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">f</em><sup class="superscript">14</sup>5<em class="emphasis">d</em><sup class="superscript">10</sup></span>
</span>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p31">with a filled 5<em class="emphasis">d</em> subshell, a 6<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">f</em><sup class="superscript">14</sup>5<em class="emphasis">d</em><sup class="superscript">10</sup> 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.)</p>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p32">Although element 114 is not stable enough to occur in nature, two isotopes of element 114 were created for the first time in a nuclear reactor in 1999 by a team of Russian and American scientists. Write the complete electron configuration for element 114.</p>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p33"><strong class="emphasis bold">Answer: </strong>1<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">6</sup>3<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">p</em><sup class="superscript">6</sup>4<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">d</em><sup class="superscript">10</sup>4<em class="emphasis">p</em><sup class="superscript">6</sup>5<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">d</em><sup class="superscript">10</sup>5<em class="emphasis">p</em><sup class="superscript">6</sup>6<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">f</em><sup class="superscript">14</sup>5<em class="emphasis">d</em><sup class="superscript">10</sup>6<em class="emphasis">p</em><sup class="superscript">6</sup>7<em class="emphasis">s</em><sup class="superscript">2</sup>5<em class="emphasis">f</em><sup class="superscript">14</sup>6<em class="emphasis">d</em><sup class="superscript">10</sup>7<em class="emphasis">p</em><sup class="superscript">2</sup></p>
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s03_s02_p34">The electron configurations of the elements are presented in <a class="xref" href="#averill_1.0-ch06_s06_s04_f02">Figure 6.34 "Electron Configurations of the Elements"</a>, which lists the orbitals in the order in which they are filled. In several cases, the ground state electron configurations are different from those predicted by <a class="xref" href="#averill_1.0-ch06_s06_s03_s02_f01">Figure 6.32 "Predicting the Order in Which Orbitals Are Filled in Multielectron Atoms"</a>. Some of these anomalies occur as the 3<em class="emphasis">d</em> orbitals are filled. For example, the observed ground state electron configuration of chromium is [Ar]4<em class="emphasis">s</em><sup class="superscript">1</sup>3<em class="emphasis">d</em><sup class="superscript">5</sup> rather than the predicted [Ar]4<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">d</em><sup class="superscript">4</sup>. Similarly, the observed electron configuration of copper is [Ar]4<em class="emphasis">s</em><sup class="superscript">1</sup>3<em class="emphasis">d</em><sup class="superscript">10</sup> instead of [Ar]<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">d</em><sup class="superscript">9</sup>. The actual electron configuration may be rationalized in terms of an added stability associated with a half-filled (<em class="emphasis">ns</em><sup class="superscript">1</sup>, <em class="emphasis">np</em><sup class="superscript">3</sup>, <em class="emphasis">nd</em><sup class="superscript">5</sup>, <em class="emphasis">nf</em><sup class="superscript">7</sup>) or filled (<em class="emphasis">ns</em><sup class="superscript">2</sup>, <em class="emphasis">np</em><sup class="superscript">6</sup>, <em class="emphasis">nd</em><sup class="superscript">10</sup>, <em class="emphasis">nf</em><sup class="superscript">14</sup>) 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 indicated in <a class="xref" href="#averill_1.0-ch06_s06_s04_f02">Figure 6.34 "Electron Configurations of the Elements"</a>. For example, cerium has an electron configuration of [Xe]6<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">f</em><sup class="superscript">1</sup>5<em class="emphasis">d</em><sup class="superscript">1</sup>, which is impossible to rationalize in simple terms. In most cases, however, these apparent anomalies do not have important chemical consequences.</p>
<div class="callout editable block" id="averill_1.0-ch06_s06_s03_s02_n03">
<h3 class="title">Note the Pattern</h3>
<p class="para" id="averill_1.0-ch06_s06_s03_s02_p35">Additional stability is associated with half-filled or filled subshells.</p>
</div>
</div>
</div>
<div class="section" id="averill_1.0-ch06_s06_s04">
<h2 class="title editable block">Blocks in the Periodic Table</h2>
<p class="para editable block" id="averill_1.0-ch06_s06_s04_p01">As you have learned, the electron configurations of the elements explain the otherwise peculiar shape of the periodic table. Although the table was originally organized on the basis of physical and chemical similarities between the elements within groups, these similarities are ultimately attributable to orbital energy levels and the Pauli principle, which cause the individual subshells to be filled in a particular order. As a result, the periodic table can be divided into “blocks” corresponding to the type of subshell that is being filled, as illustrated in <a class="xref" href="#averill_1.0-ch06_s06_s04_f02">Figure 6.34 "Electron Configurations of the Elements"</a>. For example, the two columns on the left, known as the <span class="margin_term"><a class="glossterm"><em class="emphasis">s</em> block</a><span class="glossdef">The elements in the left two columns of the periodic table in which the <em class="emphasis">ns</em> orbital is being filled.</span></span>, consist of elements in which the <em class="emphasis">ns</em> orbitals are being filled. The six columns on the right, elements in which the <em class="emphasis">np</em> orbitals are being filled, constitute the <span class="margin_term"><a class="glossterm"><em class="emphasis">p</em> block</a><span class="glossdef">The elements in the six columns on the right of the periodic table in which the <em class="emphasis">np</em> orbitals are being filled.</span></span>. In between are the 10 columns of the <span class="margin_term"><a class="glossterm"><em class="emphasis">d</em> block</a><span class="glossdef">The elements in the periodic table in which the (<em class="emphasis">n</em> − 1)<em class="emphasis">d</em> orbitals are being filled.</span></span>, elements in which the (<em class="emphasis">n</em> − 1)<em class="emphasis">d</em> orbitals are filled. At the bottom lie the 14 columns of the <span class="margin_term"><a class="glossterm"><em class="emphasis">f</em> block</a><span class="glossdef">The elements in the periodic table in which the (<em class="emphasis">n</em> − 2)<em class="emphasis">f</em> orbitals are being filled.</span></span>, elements in which the (<em class="emphasis">n</em> − 2)<em class="emphasis">f</em> orbitals are filled. Because two electrons can be accommodated per orbital, the number of columns in each block is the same as the maximum electron capacity of the subshell: 2 for <em class="emphasis">ns</em>, 6 for <em class="emphasis">np</em>, 10 for (<em class="emphasis">n</em> − 1)<em class="emphasis">d</em>, and 14 for (<em class="emphasis">n</em> − 2)<em class="emphasis">f</em>. Within each column, each element has the same valence electron configuration—for example, <em class="emphasis">ns</em><sup class="superscript">1</sup> (group 1) or <em class="emphasis">ns</em><sup class="superscript">2</sup><em class="emphasis">np</em><sup class="superscript">1</sup> (group 13). As you will see, this is reflected in important similarities in the chemical reactivity and the bonding for the elements in each column.</p>
<div class="callout editable block" id="averill_1.0-ch06_s06_s04_n01">
<h3 class="title">Note the Pattern</h3>
<p class="para" id="averill_1.0-ch06_s06_s04_p02">Because each orbital can have a maximum of 2 electrons, there are 2 columns in the <em class="emphasis">s</em> block, 6 columns in the <em class="emphasis">p</em> block, 10 columns in the <em class="emphasis">d</em> block, and 14 columns in the <em class="emphasis">f</em> block.</p>
</div>
<div class="figure large medium-height editable block" id="averill_1.0-ch06_s06_s04_f01">
<p class="title"><span class="title-prefix">Figure 6.33</span> The Periodic Table, Showing How the Elements Are Grouped According to the Kind of Subshell (<em class="emphasis bolditalic">s</em>, <em class="emphasis bolditalic">p</em>, <em class="emphasis bolditalic">d</em>, <em class="emphasis bolditalic">f</em>) Being Filled with Electrons in the Valence Shell of Each Element</p>
<img src="section_10/cfc09e4254ec9da140ab9a0af29f322a.jpg">
<p class="para">The electron configurations of the elements are in <a class="xref" href="#averill_1.0-ch06_s06_s04_f02">Figure 6.34 "Electron Configurations of the Elements"</a>.</p>
</div>
<p class="para editable block" id="averill_1.0-ch06_s06_s04_p03">Hydrogen and helium are placed somewhat arbitrarily. Although hydrogen is not an alkali metal, its 1<em class="emphasis">s</em><sup class="superscript">1</sup> electron configuration suggests a similarity to lithium ([He]2<em class="emphasis">s</em><sup class="superscript">1</sup>) and the other elements in the first column. Although helium, with a filled <em class="emphasis">ns</em> subshell, should be similar chemically to other elements with an <em class="emphasis">ns</em><sup class="superscript">2</sup> electron configuration, the closed principal shell dominates its chemistry, justifying its placement above neon on the right. In <a class="xref" href="averill_1.0-ch07#averill_1.0-ch07">Chapter 7 "The Periodic Table and Periodic Trends"</a>, we will examine how electron configurations affect the properties and reactivity of the elements.</p>
<div class="figure full editable block" id="averill_1.0-ch06_s06_s04_f02">
<p class="title"><span class="title-prefix">Figure 6.34</span> Electron Configurations of the Elements</p>
<img src="section_10/9dc094f77bbed0a0e1c0d845af6d6168.jpg">
<p class="para">The electron configurations of elements indicated in red are exceptions due to the added stability associated with half-filled and filled subshells. The electron configurations of the elements indicated in blue are also anomalous, but the reasons for the observed configurations are more complex. For elements after No, the electron configurations are tentative.</p>
</div>
<div class="exercises editable block" id="averill_1.0-ch06_s06_s04_n02">
<h3 class="title">Example 10</h3>
<p class="para" id="averill_1.0-ch06_s06_s04_p04">Use the periodic table to predict the valence electron configuration of all the elements of group 2 (beryllium, magnesium, calcium, strontium, barium, and radium).</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p05"><strong class="emphasis bold">Given: </strong>series of elements</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p06"><strong class="emphasis bold">Asked for: </strong>valence electron configurations</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p07">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p08"><strong class="emphasis bold">A</strong> Identify the block in the periodic table to which the group 2 elements belong. Locate the nearest noble gas preceding each element and identify the principal quantum number of the valence shell of each element.</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p09"><strong class="emphasis bold">B</strong> Write the valence electron configuration of each element by first indicating the filled inner shells using the symbol for the nearest preceding noble gas and then listing the principal quantum number of its valence shell, its valence orbitals, and the number of valence electrons in each orbital as superscripts.</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p10">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p11"><strong class="emphasis bold">A</strong> The group 2 elements are in the <em class="emphasis">s</em> block of the periodic table, and as group 2 elements, they all have two valence electrons. Beginning with beryllium, we see that its nearest preceding noble gas is helium and that the principal quantum number of its valence shell is <em class="emphasis">n</em> = 2.</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p12"><strong class="emphasis bold">B</strong> Thus beryllium has an [He]<em class="emphasis">s</em><sup class="superscript">2</sup> electron configuration. The next element down, magnesium, is expected to have exactly the same arrangement of electrons in the <em class="emphasis">n</em> = 3 principal shell: [Ne]<em class="emphasis">s</em><sup class="superscript">2</sup>. By extrapolation, we expect all the group 2 elements to have an <em class="emphasis">ns</em><sup class="superscript">2</sup> electron configuration.</p>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p13">Use the periodic table to predict the characteristic valence electron configuration of the halogens in group 17.</p>
<p class="para" id="averill_1.0-ch06_s06_s04_p14"><strong class="emphasis bold">Answer: </strong>All have an <em class="emphasis">ns</em><sup class="superscript">2</sup><em class="emphasis">np</em><sup class="superscript">5</sup> electron configuration, one electron short of a noble gas electron configuration. (Note that the heavier halogens also have filled (<em class="emphasis">n</em> − 1)<em class="emphasis">d</em><sup class="superscript">10</sup> subshells, as well as an (<em class="emphasis">n</em> − 2)<em class="emphasis">f</em><sup class="superscript">14</sup> subshell for Rn; these do not, however, affect their chemistry in any significant way.</p>
</div>
<div class="callout editable block" id="averill_1.0-ch06_s06_s04_n03">
<h3 class="title">Summary</h3>
<p class="para" id="averill_1.0-ch06_s06_s04_p15">In addition to the three quantum numbers (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>) dictated by quantum mechanics, a fourth quantum number is required to explain certain properties of atoms. This is the <strong class="emphasis bold">electron spin</strong> quantum number (<em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>), which can have values of +½ or −½ for any electron, corresponding to the two possible orientations of an electron in a magnetic field. The concept of electron spin has important consequences for chemistry because the <strong class="emphasis bold">Pauli exclusion principle</strong> implies that no orbital can contain more than two electrons (with opposite spin). 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 <strong class="emphasis bold">aufbau principle</strong>), which gives rise to a particular arrangement of electrons for each element (its <strong class="emphasis bold">electron configuration</strong>). <strong class="emphasis bold">Hund’s rule</strong> 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 <strong class="emphasis bold">valence electrons</strong>. The arrangement of atoms in the periodic table results in blocks corresponding to filling of the <em class="emphasis">ns</em>, <em class="emphasis">np</em>, <em class="emphasis">nd</em>, and <em class="emphasis">nf</em> orbitals to produce the distinctive chemical properties of the elements in the <em class="emphasis bolditalic">s</em> <strong class="emphasis bold">block</strong>, <em class="emphasis bolditalic">p</em> <strong class="emphasis bold">block</strong>, <em class="emphasis bolditalic">d</em> <strong class="emphasis bold">block</strong>, and <em class="emphasis bolditalic">f</em> <strong class="emphasis bold">block</strong>, respectively.</p>
</div>
<div class="key_takeaways editable block" id="averill_1.0-ch06_s06_s04_n04">
<h3 class="title">Key Takeaway</h3>
<ul class="itemizedlist" id="averill_1.0-ch06_s06_s04_l01">
<li>The arrangement of atoms in the periodic table arises from the lowest energy arrangement of electrons in the valence shell.</li>
</ul>
</div>
<div class="qandaset block" id="averill_1.0-ch06_s06_s04_qs01" defaultlabel="number">
<h3 class="title">Conceptual Problems</h3>
<ol class="qandadiv" id="averill_1.0-ch06_s06_s04_qs01_qd01">
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs01_qd01_qa01">
<div class="question">
<p class="para">A set of four quantum numbers specifies each wave function. What information is given by each quantum number? What does the specified wave function describe?</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs01_qd01_qa02">
<div class="question">
<p class="para">List two pieces of evidence to support the statement that electrons have a spin.</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs01_qd01_qa03">
<div class="question">
<p class="para">The periodic table is divided into blocks. Identify each block and explain the principle behind the divisions. Which quantum number distinguishes the horizontal rows?</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs01_qd01_qa04">
<div class="question">
<p class="para">Identify the element with each ground state electron configuration.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>[He]2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">3</sup>
</li>
<li>[Ar]4<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">d</em><sup class="superscript">1</sup>
</li>
<li>[Kr]5<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">d</em><sup class="superscript">10</sup>5<em class="emphasis">p</em><sup class="superscript">3</sup>
</li>
<li>[Xe]6<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">f</em> <sup class="superscript">6</sup>
</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs01_qd01_qa05">
<div class="question">
<p class="para">Identify the element with each ground state electron configuration.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>[He]2<em class="emphasis">s</em><sup class="superscript">2</sup>2<em class="emphasis">p</em><sup class="superscript">1</sup>
</li>
<li>[Ar]4<em class="emphasis">s</em><sup class="superscript">2</sup>3<em class="emphasis">d</em><sup class="superscript">8</sup>
</li>
<li>[Kr]5<em class="emphasis">s</em><sup class="superscript">2</sup>4<em class="emphasis">d</em><sup class="superscript">10</sup>5<em class="emphasis">p</em><sup class="superscript">4</sup>
</li>
<li>[Xe]6<em class="emphasis">s</em><sup class="superscript">2</sup>
</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs01_qd01_qa06">
<div class="question">
<p class="para">Propose an explanation as to why the noble gases are inert.</p>
</div>
</li>
</ol>
</div>
<div class="qandaset block" id="averill_1.0-ch06_s06_s04_qs02" condition="1col" defaultlabel="number">
<h3 class="title">Numerical Problems</h3>
<ol class="qandadiv" id="averill_1.0-ch06_s06_s04_qs02_qd01">
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa01">
<div class="question">
<p class="para">How many magnetic quantum numbers are possible for a 4<em class="emphasis">p</em> subshell? A 3<em class="emphasis">d</em> subshell? How many orbitals are in these subshells?</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa02">
<div class="question">
<p class="para">How many magnetic quantum numbers are possible for a 6<em class="emphasis">s</em> subshell? A 4<em class="emphasis">f</em> subshell? How many orbitals does each subshell contain?</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa03">
<div class="question">
<p class="para">If <em class="emphasis">l</em> = 2 and <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub> = 2, give all the allowed combinations of the four quantum numbers (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>) for electrons in the corresponding 3<em class="emphasis">d</em> subshell.</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa04">
<div class="question">
<p class="para">Give all the allowed combinations of the four quantum numbers (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>) for electrons in a 4<em class="emphasis">d</em> subshell. How many electrons can the 4<em class="emphasis">d</em> orbital accommodate? How would this differ from a situation in which there were only three quantum numbers (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em>)?</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa05">
<div class="question">
<p class="para">Given the following sets of quantum numbers (<em class="emphasis">n</em>, <em class="emphasis">l</em>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub>, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub>), identify each principal shell and subshell.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>1, 0, 0, ½</li>
<li>2, 1, 0, ½</li>
<li>3, 2, 0, ½</li>
<li>4, 3, 3, ½</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa06">
<div class="question">
<p class="para">Is each set of quantum numbers allowed? Explain your answers.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>
<em class="emphasis">n</em> = 2; <em class="emphasis">l</em> = 1; <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub> = 2; <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> = +½</li>
<li>
<em class="emphasis">n</em> = 3, <em class="emphasis">l</em> = 0; <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub> = −1; <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> = −½</li>
<li>
<em class="emphasis">n</em> = 2; <em class="emphasis">l</em> = 2; <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub> = 1; <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> = +½</li>
<li>
<em class="emphasis">n</em> = 3; <em class="emphasis">l</em> = 2; <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub> = 2; <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> = +½</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa08">
<div class="question">
<p class="para">List the set of quantum numbers for each electron in the valence shell of each element.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>beryllium</li>
<li>xenon</li>
<li>lithium</li>
<li>fluorine</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa07">
<div class="question">
<p class="para">List the set of quantum numbers for each electron in the valence shell of each element.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>carbon</li>
<li>magnesium</li>
<li>bromine</li>
<li>sulfur</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa09">
<div class="question">
<p class="para">Sketch the shape of the periodic table if there were three possible values of <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> for each electron (+½, −½, and 0); assume that the Pauli principle is still valid.</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa10">
<div class="question">
<p class="para">Predict the shape of the periodic table if eight electrons could occupy the <em class="emphasis">p</em> subshell.</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa11">
<div class="question">
<p class="para">If the electron could only have spin +½, what would the periodic table look like?</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa12">
<div class="question">
<p class="para">If three electrons could occupy each <em class="emphasis">s</em> orbital, what would be the electron configuration of each species?</p>
<ol class="orderedlist" numeration="loweralpha">
<li>sodium</li>
<li>titanium</li>
<li>fluorine</li>
<li>calcium</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa13">
<div class="question">
<p class="para">If Hund’s rule were not followed and maximum pairing occurred, how many unpaired electrons would each species have? How do these numbers compare with the number found using Hund’s rule?</p>
<ol class="orderedlist" numeration="loweralpha">
<li>phosphorus</li>
<li>iodine</li>
<li>manganese</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa15">
<div class="question">
<p class="para">Write the electron configuration for each element in the ground state.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>aluminum</li>
<li>calcium</li>
<li>sulfur</li>
<li>tin</li>
<li>nickel</li>
<li>tungsten</li>
<li>neodymium</li>
<li>americium</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa14">
<div class="question">
<p class="para">Write the electron configuration for each element in the ground state.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>boron</li>
<li>rubidium</li>
<li>bromine</li>
<li>germanium</li>
<li>vanadium</li>
<li>palladium</li>
<li>bismuth</li>
<li>europium</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa17">
<div class="question">
<p class="para">Give the complete electron configuration for each element.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>magnesium</li>
<li>potassium</li>
<li>titanium</li>
<li>selenium</li>
<li>iodine</li>
<li>uranium</li>
<li>germanium</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa16">
<div class="question">
<p class="para">Give the complete electron configuration for each element.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>tin</li>
<li>copper</li>
<li>fluorine</li>
<li>hydrogen</li>
<li>thorium</li>
<li>yttrium</li>
<li>bismuth</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa18">
<div class="question">
<p class="para">Write the valence electron configuration for each element:</p>
<ol class="orderedlist" numeration="loweralpha">
<li>samarium</li>
<li>praseodymium</li>
<li>boron</li>
<li>cobalt</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa20">
<div class="question">
<p class="para">Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>barium</li>
<li>neodymium</li>
<li>iodine</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa19">
<div class="question">
<p class="para">Using the Pauli exclusion principle and Hund’s rule, draw valence orbital diagrams for each element.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>chlorine</li>
<li>silicon</li>
<li>scandium</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa22">
<div class="question">
<p class="para">How many unpaired electrons does each species contain?</p>
<ol class="orderedlist" numeration="loweralpha">
<li>lead</li>
<li>cesium</li>
<li>copper</li>
<li>silicon</li>
<li>selenium</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa21">
<div class="question">
<p class="para">How many unpaired electrons does each species contain?</p>
<ol class="orderedlist" numeration="loweralpha">
<li>helium</li>
<li>oxygen</li>
<li>bismuth</li>
<li>silver</li>
<li>boron</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa23">
<div class="question">
<p class="para">For each element, give the complete electron configuration, draw the valence electron configuration, and give the number of unpaired electrons present.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>lithium</li>
<li>magnesium</li>
<li>silicon</li>
<li>cesium</li>
<li>lead</li>
</ol>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa24">
<div class="question">
<p class="para">Use an orbital diagram to illustrate the aufbau principle, the Pauli exclusion principle, and Hund’s rule for each element.</p>
<ol class="orderedlist" numeration="loweralpha">
<li>carbon</li>
<li>sulfur</li>
</ol>
</div>
</li>
</ol>
</div>
<div class="qandaset block" id="averill_1.0-ch06_s06_s04_qs02_ans" condition="1col" defaultlabel="number">
<h3 class="title">Answers</h3>
<ol class="qandadiv">
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa01_ans">
<div class="answer">
<p class="para">For a 4<em class="emphasis">p</em> subshell, <em class="emphasis">n</em> = 4 and <em class="emphasis">l</em> = 1. The allowed values of the magnetic quantum number, <em class="emphasis">ml</em>, are therefore +1, 0, −1, corresponding to three 4<em class="emphasis">p</em> orbitals. For a 3<em class="emphasis">d</em> subshell, <em class="emphasis">n</em> = 3 and <em class="emphasis">l</em> = 2. The allowed values of the magnetic quantum number, <em class="emphasis">ml</em>, are therefore +2, +1, 0, −1, −2, corresponding to five 3<em class="emphasis">d</em> orbitals.</p>
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa02_ans" audience="instructoronly">
<div class="answer" audience="instructoronly" d="" html="http://www.w3.org/1999/xhtml" mml="http://www.w3.org/1998/Math/MathML" xlink="http://www.w3.org/1999/xlink" xml="http://www.w3.org/XML/1998/namespace">
</div>
</li>
<li class="qandaentry" id="averill_1.0-ch06_s06_s04_qs02_qd01_qa03_ans">
<div class="answer">
<p class="para"><em class="emphasis">n</em> = 3, <em class="emphasis">l</em> = 2, <em class="emphasis">ml</em> = 2, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> = <span class="inlineequation"><math xml:id="averill_1.0-ch06_m060" display="inline"><semantics><mrow><mo>+</mo><mfrac><mn>1</mn><mtext>2</mtext></mfrac></mrow></semantics></math></span>;<em class="emphasis">n</em> = 3, <em class="emphasis">l</em> = 2, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">l</em></sub> = 2, <em class="emphasis">m</em><sub class="subscript"><em class="emphasis">s</em></sub> = <span class="inlineequation"><math xml:id="averill_1.0-ch06_m061" display="inline"><semantics><mrow><mo>−</mo><mfrac><mn>1</mn><mtext>2</mtext></mfrac></mrow></semantics></math></span></p>
</div>
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