{"id":2398,"date":"2018-04-11T23:52:48","date_gmt":"2018-04-12T03:52:48","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/periodic-trends\/"},"modified":"2019-05-13T17:33:47","modified_gmt":"2019-05-13T21:33:47","slug":"periodic-trends","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/periodic-trends\/","title":{"raw":"8.5 Periodic Trends","rendered":"8.5 Periodic Trends"},"content":{"raw":"<div class=\"section\" id=\"ball-ch08_s05\" lang=\"en\">\r\n<div class=\"learning_objectives editable block\" id=\"ball-ch08_s05_n01\">\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this module, you will be able to:\r\n<ul>\r\n \t<li>Be able to state how certain properties of atoms vary based on their relative position on the periodic table.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<p id=\"ball-ch08_s05_p01\" class=\"para editable block\">One of the reasons the periodic table is so useful is because its structure allows us to qualitatively determine how some properties of the elements vary versus their position on the periodic table. The variation of properties versus position on the periodic table is called <span class=\"margin_term\"><a class=\"glossterm\">periodic trends<\/a><\/span>. There is no other tool in science that allows us to judge relative properties of a class of objects like this, which makes the periodic table a very useful tool. Many periodic trends are general. There may be a few points where an opposite trend is seen, but there is an overall trend when considered across a whole row or down a whole column of the periodic table.<\/p>\r\n\r\n<h2 class=\"para editable block\">Atomic Radii<\/h2>\r\n<p id=\"ball-ch08_s05_p02\" class=\"para editable block\">The <span class=\"margin_term\"><a class=\"glossterm\">atomic radius\u00a0<\/a><\/span>is an indication of the size of an atom. Although the concept of a definite radius of an atom is a bit fuzzy, atoms behave as if they have a certain radius. Such radii can be estimated from various experimental techniques, such as the x-ray crystallography of crystals.<\/p>\r\nThe size of atoms vary and there are two periodic trends.\u00a0 Atoms get smaller as you go from left to right across a period, and get larger as you go down a group.\u00a0 <a class=\"xref\" href=\"#ball-ch08_s05_f01\">Figure 1 \"Atomic Radii Trends on the Periodic Table\"<\/a> shows spheres representing the atoms of the <em class=\"emphasis\">s<\/em> and <em class=\"emphasis\">p<\/em> blocks from the periodic table to scale, showing the two trends for the atomic radius.\r\n<div class=\"figure large medium-height editable block\" id=\"ball-ch08_s05_f01\">\r\n\r\n[caption id=\"attachment_4709\" align=\"aligncenter\" width=\"600\"]<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Atomic-Radii-Trends.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Atomic-Radii-Trends-1.png\" alt=\"Atomic Radii Trends\" width=\"600\" height=\"652\" class=\"wp-image-4709 size-full\" \/><\/a> <strong>Figure 1.<\/strong> Atomic Radii Trends on the Periodic Table. \u00a0Although there are some reversals in the trend (e.g., see Po in the bottom row), atoms generally get smaller as you go across the periodic table and larger as you go down any one column. Numbers are the radii in pm.[\/caption]\r\n\r\n<\/div>\r\nThe atomic size is easily explained when we examine how the electron configurations change as we move on the periodic table:\r\n<ul>\r\n \t<li>As you go down a group, the valence electron configuration stays the same, but the number of shells is increasing. Each shell represents distance from the nucleus (as well as energy), thus we expect that <em>ATOMIC SIZE INCREASES as you go down a row <\/em>on the periodic table.<\/li>\r\n \t<li>As you go from left to right on the periodic table, you are adding electrons to the same shell, but, you are also adding protons (nuclear charge). These protons serve to pull the electrons closer to the nucleus. Thus, we expect that <em>as you go from left to right along each period, ATOMIC SIZE DECREASES.<\/em><\/li>\r\n<\/ul>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 1<\/h3>\r\n<p id=\"ball-ch08_s05_p06\" class=\"para\">Referring only to a periodic table and not to <a class=\"xref\" href=\"#ball-ch08_s05_f01\">Figure 1 \"Atomic Radii Trends on the Periodic Table\"<\/a>, which atom is larger in each pair?<\/p>\r\n<p class=\"para\">a) Si or S \u00a0 \u00a0 \u00a0b)\u00a0S or Te<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) Si is to the left of S on the periodic table, so it is larger because as you go across the row, the atoms get smaller.<\/p>\r\n<p class=\"simpara\">b) S is above Te on the periodic table, so Te is larger because as you go down the column, the atoms get larger.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch08_s05_p07\" class=\"para\">Referring only to a periodic table and not to <a class=\"xref\" href=\"#ball-ch08_s05_f01\">Figure 1 \"Atomic Radii Trends on the Periodic Table\"<\/a>, which atom is smaller, Ca or Br?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch08_s05_p08\" class=\"para\">Br<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2<\/h3>\r\n<p class=\"Indent\">For the following elements, write them in order of smallest to largest, using only the periodic table:<\/p>\r\n<p class=\"Indent\"><span>\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>K, As, F, N<\/p>\r\n&nbsp;\r\n<p class=\"Solution\"><strong>Solution\u00a0\u00a0 <\/strong><\/p>\r\n<p class=\"Indent\">We use the periodic table and our knowledge of the trends in atomic size; further up and to the right are the smaller atoms. The order thus becomes:<\/p>\r\n<p class=\"Indent\">Smallest<span>\u00a0\u00a0\u00a0\u00a0 <\/span>F, N, As, K<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>Largest<\/p>\r\n&nbsp;\r\n<p class=\"SelfTest\"><em><strong>Test Yourself<\/strong><\/em><\/p>\r\n<p class=\"Indent\">For the following elements, write them in order of smallest to largest, using only the periodic table:<\/p>\r\n<p class=\"Indent\">Rb, Si, Cl<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\n<span>Cl, Si, Rb<\/span>\r\n\r\n<\/div>\r\n<h2 class=\"para editable block\">Ionization Energy<\/h2>\r\n<p id=\"ball-ch08_s05_p09\" class=\"para editable block\"><span class=\"margin_term\"><a class=\"glossterm\">Ionization energy (IE)<\/a><\/span>\u00a0is the amount of energy required to remove an electron from an atom in the gas phase:<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\">A(g)\u00a0<\/span>$latex \\longrightarrow$<span class=\"informalequation block\"> A<sup>+<\/sup>(g) + e<sup>\u2212<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u0394H \u2261 IE<\/span><\/p>\r\n<p id=\"ball-ch08_s05_p10\" class=\"para editable block\">IE is usually expressed in kJ\/mol of atoms. It is always positive because the removal of an electron always requires that energy be put in (i.e., it is endothermic). IE also shows periodic trends. As you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus. \u00a0However, as you go across the periodic table and the electrons get drawn closer in, it takes more energy to remove an electron; as a result, IE increases.<span class=\"informalequation block\"><\/span><\/p>\r\n<p id=\"ball-ch08_s05_p12\" class=\"para editable block\"><a class=\"xref\" href=\"#ball-ch08_s05_f02\">Figure 2 \"Ionization Energy on the Periodic Table\"<\/a> shows values of IE versus position on the periodic table. Again, the trend isn\u2019t absolute, but the general trends going across and down the periodic table should be obvious.<\/p>\r\n\r\n<div class=\"figure large medium-height editable block\" id=\"ball-ch08_s05_f02\">\r\n\r\n[caption id=\"attachment_4711\" align=\"aligncenter\" width=\"600\"]<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Ionization-Energy.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionization-Energy-1.png\" alt=\"Ionization Energy\" width=\"600\" height=\"652\" class=\"wp-image-4711 size-full\" \/><\/a> <strong>Figure 2.<\/strong> Ionization Energy on the Periodic Table. \u00a0Values are in kJ\/mol.[\/caption]\r\n<p style=\"text-align: left\">IE also shows an interesting trend within a given atom. This is because more than one IE can be defined by removing successive electrons (if the atom has them to begin with):<\/p>\r\n\r\n<\/div>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">A(g) $latex \\longrightarrow$ A<sup class=\"superscript\">+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">1<\/sub><\/span><\/span>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">A<sup class=\"superscript\">+<\/sup>(g) $latex \\longrightarrow$ A<sup class=\"superscript\">2+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">2<\/sub><\/span><\/span>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">A<sup class=\"superscript\">2+<\/sup>(g) $latex \\longrightarrow$ A<sup class=\"superscript\">3+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\r\n<p id=\"ball-ch08_s05_p14\" class=\"para editable block\">and so forth.<\/p>\r\n<p id=\"ball-ch08_s05_p15\" class=\"para editable block\">Each successive IE is larger than the previous because an electron is being removed from an atom with a progressively larger positive charge. However, IE takes a large jump when a successive ionization goes down into a new shell. For example, the following are the first three IEs for Mg, whose electron configuration 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\">2<\/sup>:<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">Mg(g) $latex \\longrightarrow$ Mg<sup class=\"superscript\">+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">1<\/sub> = 738 kJ\/mol<\/span><\/span>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">Mg<sup class=\"superscript\">+<\/sup>(g) $latex \\longrightarrow$ Mg<sup class=\"superscript\">2+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">2<\/sub> = 1,450 kJ\/mol<\/span><\/span>\r\n<span class=\"informalequation block\"><span class=\"mathphrase\">Mg<sup class=\"superscript\">2+<\/sup>(g) $latex \\longrightarrow$ Mg<sup class=\"superscript\">3+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">3<\/sub> = 7,734 kJ\/mol<\/span><\/span><\/p>\r\n<p id=\"ball-ch08_s05_p16\" class=\"para editable block\">The second IE is twice the first, which is not a surprise: the first IE involves removing an electron from a neutral atom, while the second one involves removing an electron from a positive ion. The third IE, however, is over <em class=\"emphasis\">five times<\/em> the previous one. It suggests that there is more involved than simply overcoming a larger ionic charge. Why is it so much larger?\u00a0Because the first two electrons are removed from the 3<em class=\"emphasis\">s<\/em> subshell, but the third electron has to be removed from the <em class=\"emphasis\">n<\/em> = 2 shell, specifically, the 2<em class=\"emphasis\">p<\/em> subshell, which is lower in energy than the <em class=\"emphasis\">n<\/em> = 3 shell.\u00a0 It is evidence like this that demonstrate that electrons are organized in atoms in groups (shells and subshells).<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 3<\/h3>\r\n<p id=\"ball-ch08_s05_p17\" class=\"para\">Which atom in each pair has the larger IE?<\/p>\r\n<p class=\"para\">a) Ca or Sr \u00a0 \u00a0 \u00a0b)\u00a0K or K<sup class=\"superscript\">+<\/sup><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) Because Sr is below Ca on the periodic table, it is easier to remove an electron from it; thus, Ca has the higher IE.<\/p>\r\n<p class=\"simpara\">b) Because K<sup class=\"superscript\">+<\/sup> has a positive charge, it will be harder to remove another electron from it, so its IE is larger than that of K. Indeed, it will be significantly larger because the next electron in K<sup class=\"superscript\">+<\/sup> to be removed comes from another shell.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch08_s05_p18\" class=\"para\">Which atom has the lower ionization energy, C or F?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch08_s05_p19\" class=\"para\">C<\/p>\r\n\r\n<\/div>\r\n<h2 class=\"para editable block\">Electron Affinity<\/h2>\r\n<p id=\"ball-ch08_s05_p20\" class=\"para editable block\">The opposite of IE is described by <span class=\"margin_term\"><a class=\"glossterm\">electron affinity (EA)<\/a><\/span>, which is the energy change when a gas-phase atom accepts an electron:<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\">A(g)\u00a0+\u00a0e<sup>\u2212<\/sup>$latex \\longrightarrow$ A<sup>\u2212<\/sup>(g)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u0394H \u2261 EA<\/span><\/p>\r\n<p id=\"ball-ch08_s05_p21\" class=\"para editable block\">Electron affinity is also usually expressed in kJ\/mol.<\/p>\r\n<p class=\"para editable block\">Electron affinity also demonstrates some periodic trends, although they are less obvious than the other periodic trends discussed previously.\u00a0<a class=\"xref\" href=\"#ball-ch08_s05_f03\">Figure 3 \"Electron Affinity on the Periodic Table\"<\/a> shows EA values versus position on the periodic table for the <em class=\"emphasis\">s<\/em>- and <em class=\"emphasis\">p<\/em>-block elements.\u00a0The trend isn\u2019t absolute, especially considering the large positive EA values for the second column.<\/p>\r\n\r\n<ul>\r\n \t<li class=\"para editable block\">Electron affinity generally increases in magnitude as we move to the right across a period (row) in the periodic table (across a period).<\/li>\r\n \t<li class=\"para editable block\"><span class=\"informalequation block\"><\/span>There is not a definitive trend as you go down a group (column) on the periodic table; sometimes electron affinity increases, sometimes it decreases.<\/li>\r\n<\/ul>\r\n&nbsp;\r\n<div class=\"figure large medium-height editable block\" id=\"ball-ch08_s05_f03\">\r\n\r\n[caption id=\"attachment_4712\" align=\"aligncenter\" width=\"600\"]<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Electron-Affinity.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Electron-Affinity-1.png\" alt=\"Electron Affinity\" width=\"600\" height=\"708\" class=\"wp-image-4712 size-full\" \/><\/a> <strong>Figure 3.<\/strong> Electron Affinity on the Periodic Table. \u00a0Values are in kJ\/mol.[\/caption]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 4<\/h3>\r\n<p id=\"ball-ch08_s05_p23\" class=\"para\">Predict which atom in each pair will have the highest magnitude of EA.<\/p>\r\n<p class=\"para\">a) C or F \u00a0 \u00a0 \u00a0b)\u00a0Na or S<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) C and F are in the same row on the periodic table, but F is farther to the right. Therefore, F should have the larger magnitude of EA.<\/p>\r\n<p class=\"simpara\">b) Na and S are in the same row on the periodic table, but S is farther to the right. Therefore, S should have the larger magnitude of EA.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch08_s05_p24\" class=\"para\">Predict which atom will have the highest magnitude of EA, As or Br.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch08_s05_p25\" class=\"para\">Br<\/p>\r\n\r\n<\/div>\r\n<div class=\"key_takeaways editable block\" id=\"ball-ch08_s05_n05\">\r\n<h2>Key Concepts and Summary<\/h2>\r\nCertain properties\u2014notably effective atomic radius, IE, and EA\u2014can be qualitatively understood by the positions of the elements on the periodic table.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n<div class=\"qandaset block\" id=\"ball-ch08_s05_qs01\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch08_s05_qs01_qd01_p1\" class=\"para\">1. Write a chemical equation with an IE energy change.<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">2. State the trends in atomic radii as you go across and down the periodic table.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">3. Which atom of each pair is larger?<\/span><\/p>\r\n\r\n<\/div>\r\na) \u00a0Na or Cs \u00a0 \u00a0 \u00a0b) \u00a0N or Bi\r\n\r\n<span style=\"font-size: 1em\">4. \u00a0Which atom of each pair is larger?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0K or Cl \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Ba or Bi\r\n\r\n<\/div>\r\n5<span style=\"font-size: 1em\">. \u00a0Which atom has the higher IE?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0Na or S \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Ge or Br\r\n\r\n<\/div>\r\n6<span style=\"font-size: 1em\">. \u00a0Which atom has the higher IE?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0Li or Cs \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Se or O\r\n\r\n<\/div>\r\n7<span style=\"font-size: 1em\">. \u00a0A third-row element has the following successive IEs: 738; 1,450; 7,734; and 10,550 kJ\/mol. Identify the element.<\/span>\r\n\r\n<span style=\"font-size: 1em\">8. \u00a0For which successive IE is there a large jump in IE for Ca?<\/span>\r\n\r\n<span style=\"font-size: 1em\">9. \u00a0Which atom has the greater magnitude of EA?<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0C or F \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Al or Cl\r\n\r\n<\/div>\r\n<div class=\"question\"><\/div>\r\n<\/div>\r\n<b>Answers<\/b>\r\n\r\n1. Na(g)\u00a0$latex \\longrightarrow$ Na<sup class=\"superscript\">+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> \u0394<em class=\"emphasis\">H<\/em>\u00a0=\u00a0IE (answers will vary)\r\n\r\n2. As you go across, atomic radii decrease; as you go down, atomic radii increase.\r\n\r\n3. a) \u00a0Cs \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) Bi\r\n\r\n4. a) \u00a0K \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Ba\r\n\r\n5. a) \u00a0S \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) \u00a0Br\r\n\r\n6. a) \u00a0Li \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0O\r\n\r\n7. Mg\r\n\r\n8. The third IE shows a large jump in Ca.\r\n\r\n9. a) \u00a0F \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Cl\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"section\" id=\"ball-ch08_s05\" lang=\"en\">\n<div class=\"learning_objectives editable block\" id=\"ball-ch08_s05_n01\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this module, you will be able to:<\/p>\n<ul>\n<li>Be able to state how certain properties of atoms vary based on their relative position on the periodic table.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p id=\"ball-ch08_s05_p01\" class=\"para editable block\">One of the reasons the periodic table is so useful is because its structure allows us to qualitatively determine how some properties of the elements vary versus their position on the periodic table. The variation of properties versus position on the periodic table is called <span class=\"margin_term\"><a class=\"glossterm\">periodic trends<\/a><\/span>. There is no other tool in science that allows us to judge relative properties of a class of objects like this, which makes the periodic table a very useful tool. Many periodic trends are general. There may be a few points where an opposite trend is seen, but there is an overall trend when considered across a whole row or down a whole column of the periodic table.<\/p>\n<h2 class=\"para editable block\">Atomic Radii<\/h2>\n<p id=\"ball-ch08_s05_p02\" class=\"para editable block\">The <span class=\"margin_term\"><a class=\"glossterm\">atomic radius\u00a0<\/a><\/span>is an indication of the size of an atom. Although the concept of a definite radius of an atom is a bit fuzzy, atoms behave as if they have a certain radius. Such radii can be estimated from various experimental techniques, such as the x-ray crystallography of crystals.<\/p>\n<p>The size of atoms vary and there are two periodic trends.\u00a0 Atoms get smaller as you go from left to right across a period, and get larger as you go down a group.\u00a0 <a class=\"xref\" href=\"#ball-ch08_s05_f01\">Figure 1 &#8220;Atomic Radii Trends on the Periodic Table&#8221;<\/a> shows spheres representing the atoms of the <em class=\"emphasis\">s<\/em> and <em class=\"emphasis\">p<\/em> blocks from the periodic table to scale, showing the two trends for the atomic radius.<\/p>\n<div class=\"figure large medium-height editable block\" id=\"ball-ch08_s05_f01\">\n<figure id=\"attachment_4709\" aria-describedby=\"caption-attachment-4709\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Atomic-Radii-Trends.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Atomic-Radii-Trends-1.png\" alt=\"Atomic Radii Trends\" width=\"600\" height=\"652\" class=\"wp-image-4709 size-full\" \/><\/a><figcaption id=\"caption-attachment-4709\" class=\"wp-caption-text\"><strong>Figure 1.<\/strong> Atomic Radii Trends on the Periodic Table. \u00a0Although there are some reversals in the trend (e.g., see Po in the bottom row), atoms generally get smaller as you go across the periodic table and larger as you go down any one column. Numbers are the radii in pm.<\/figcaption><\/figure>\n<\/div>\n<p>The atomic size is easily explained when we examine how the electron configurations change as we move on the periodic table:<\/p>\n<ul>\n<li>As you go down a group, the valence electron configuration stays the same, but the number of shells is increasing. Each shell represents distance from the nucleus (as well as energy), thus we expect that <em>ATOMIC SIZE INCREASES as you go down a row <\/em>on the periodic table.<\/li>\n<li>As you go from left to right on the periodic table, you are adding electrons to the same shell, but, you are also adding protons (nuclear charge). These protons serve to pull the electrons closer to the nucleus. Thus, we expect that <em>as you go from left to right along each period, ATOMIC SIZE DECREASES.<\/em><\/li>\n<\/ul>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 1<\/h3>\n<p id=\"ball-ch08_s05_p06\" class=\"para\">Referring only to a periodic table and not to <a class=\"xref\" href=\"#ball-ch08_s05_f01\">Figure 1 &#8220;Atomic Radii Trends on the Periodic Table&#8221;<\/a>, which atom is larger in each pair?<\/p>\n<p class=\"para\">a) Si or S \u00a0 \u00a0 \u00a0b)\u00a0S or Te<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) Si is to the left of S on the periodic table, so it is larger because as you go across the row, the atoms get smaller.<\/p>\n<p class=\"simpara\">b) S is above Te on the periodic table, so Te is larger because as you go down the column, the atoms get larger.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch08_s05_p07\" class=\"para\">Referring only to a periodic table and not to <a class=\"xref\" href=\"#ball-ch08_s05_f01\">Figure 1 &#8220;Atomic Radii Trends on the Periodic Table&#8221;<\/a>, which atom is smaller, Ca or Br?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch08_s05_p08\" class=\"para\">Br<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2<\/h3>\n<p class=\"Indent\">For the following elements, write them in order of smallest to largest, using only the periodic table:<\/p>\n<p class=\"Indent\"><span>\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>K, As, F, N<\/p>\n<p>&nbsp;<\/p>\n<p class=\"Solution\"><strong>Solution\u00a0\u00a0 <\/strong><\/p>\n<p class=\"Indent\">We use the periodic table and our knowledge of the trends in atomic size; further up and to the right are the smaller atoms. The order thus becomes:<\/p>\n<p class=\"Indent\">Smallest<span>\u00a0\u00a0\u00a0\u00a0 <\/span>F, N, As, K<span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/span>Largest<\/p>\n<p>&nbsp;<\/p>\n<p class=\"SelfTest\"><em><strong>Test Yourself<\/strong><\/em><\/p>\n<p class=\"Indent\">For the following elements, write them in order of smallest to largest, using only the periodic table:<\/p>\n<p class=\"Indent\">Rb, Si, Cl<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p><span>Cl, Si, Rb<\/span><\/p>\n<\/div>\n<h2 class=\"para editable block\">Ionization Energy<\/h2>\n<p id=\"ball-ch08_s05_p09\" class=\"para editable block\"><span class=\"margin_term\"><a class=\"glossterm\">Ionization energy (IE)<\/a><\/span>\u00a0is the amount of energy required to remove an electron from an atom in the gas phase:<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\">A(g)\u00a0<\/span>[latex]\\longrightarrow[\/latex]<span class=\"informalequation block\"> A<sup>+<\/sup>(g) + e<sup>\u2212<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u0394H \u2261 IE<\/span><\/p>\n<p id=\"ball-ch08_s05_p10\" class=\"para editable block\">IE is usually expressed in kJ\/mol of atoms. It is always positive because the removal of an electron always requires that energy be put in (i.e., it is endothermic). IE also shows periodic trends. As you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus. \u00a0However, as you go across the periodic table and the electrons get drawn closer in, it takes more energy to remove an electron; as a result, IE increases.<span class=\"informalequation block\"><\/span><\/p>\n<p id=\"ball-ch08_s05_p12\" class=\"para editable block\"><a class=\"xref\" href=\"#ball-ch08_s05_f02\">Figure 2 &#8220;Ionization Energy on the Periodic Table&#8221;<\/a> shows values of IE versus position on the periodic table. Again, the trend isn\u2019t absolute, but the general trends going across and down the periodic table should be obvious.<\/p>\n<div class=\"figure large medium-height editable block\" id=\"ball-ch08_s05_f02\">\n<figure id=\"attachment_4711\" aria-describedby=\"caption-attachment-4711\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Ionization-Energy.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionization-Energy-1.png\" alt=\"Ionization Energy\" width=\"600\" height=\"652\" class=\"wp-image-4711 size-full\" \/><\/a><figcaption id=\"caption-attachment-4711\" class=\"wp-caption-text\"><strong>Figure 2.<\/strong> Ionization Energy on the Periodic Table. \u00a0Values are in kJ\/mol.<\/figcaption><\/figure>\n<p style=\"text-align: left\">IE also shows an interesting trend within a given atom. This is because more than one IE can be defined by removing successive electrons (if the atom has them to begin with):<\/p>\n<\/div>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">A(g) [latex]\\longrightarrow[\/latex] A<sup class=\"superscript\">+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">1<\/sub><\/span><\/span><br \/>\n<span class=\"informalequation block\"><span class=\"mathphrase\">A<sup class=\"superscript\">+<\/sup>(g) [latex]\\longrightarrow[\/latex] A<sup class=\"superscript\">2+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">2<\/sub><\/span><\/span><br \/>\n<span class=\"informalequation block\"><span class=\"mathphrase\">A<sup class=\"superscript\">2+<\/sup>(g) [latex]\\longrightarrow[\/latex] A<sup class=\"superscript\">3+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\n<p id=\"ball-ch08_s05_p14\" class=\"para editable block\">and so forth.<\/p>\n<p id=\"ball-ch08_s05_p15\" class=\"para editable block\">Each successive IE is larger than the previous because an electron is being removed from an atom with a progressively larger positive charge. However, IE takes a large jump when a successive ionization goes down into a new shell. For example, the following are the first three IEs for Mg, whose electron configuration 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\">2<\/sup>:<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">Mg(g) [latex]\\longrightarrow[\/latex] Mg<sup class=\"superscript\">+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">1<\/sub> = 738 kJ\/mol<\/span><\/span><br \/>\n<span class=\"informalequation block\"><span class=\"mathphrase\">Mg<sup class=\"superscript\">+<\/sup>(g) [latex]\\longrightarrow[\/latex] Mg<sup class=\"superscript\">2+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">2<\/sub> = 1,450 kJ\/mol<\/span><\/span><br \/>\n<span class=\"informalequation block\"><span class=\"mathphrase\">Mg<sup class=\"superscript\">2+<\/sup>(g) [latex]\\longrightarrow[\/latex] Mg<sup class=\"superscript\">3+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> IE<sub class=\"subscript\">3<\/sub> = 7,734 kJ\/mol<\/span><\/span><\/p>\n<p id=\"ball-ch08_s05_p16\" class=\"para editable block\">The second IE is twice the first, which is not a surprise: the first IE involves removing an electron from a neutral atom, while the second one involves removing an electron from a positive ion. The third IE, however, is over <em class=\"emphasis\">five times<\/em> the previous one. It suggests that there is more involved than simply overcoming a larger ionic charge. Why is it so much larger?\u00a0Because the first two electrons are removed from the 3<em class=\"emphasis\">s<\/em> subshell, but the third electron has to be removed from the <em class=\"emphasis\">n<\/em> = 2 shell, specifically, the 2<em class=\"emphasis\">p<\/em> subshell, which is lower in energy than the <em class=\"emphasis\">n<\/em> = 3 shell.\u00a0 It is evidence like this that demonstrate that electrons are organized in atoms in groups (shells and subshells).<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 3<\/h3>\n<p id=\"ball-ch08_s05_p17\" class=\"para\">Which atom in each pair has the larger IE?<\/p>\n<p class=\"para\">a) Ca or Sr \u00a0 \u00a0 \u00a0b)\u00a0K or K<sup class=\"superscript\">+<\/sup><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) Because Sr is below Ca on the periodic table, it is easier to remove an electron from it; thus, Ca has the higher IE.<\/p>\n<p class=\"simpara\">b) Because K<sup class=\"superscript\">+<\/sup> has a positive charge, it will be harder to remove another electron from it, so its IE is larger than that of K. Indeed, it will be significantly larger because the next electron in K<sup class=\"superscript\">+<\/sup> to be removed comes from another shell.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch08_s05_p18\" class=\"para\">Which atom has the lower ionization energy, C or F?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch08_s05_p19\" class=\"para\">C<\/p>\n<\/div>\n<h2 class=\"para editable block\">Electron Affinity<\/h2>\n<p id=\"ball-ch08_s05_p20\" class=\"para editable block\">The opposite of IE is described by <span class=\"margin_term\"><a class=\"glossterm\">electron affinity (EA)<\/a><\/span>, which is the energy change when a gas-phase atom accepts an electron:<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\">A(g)\u00a0+\u00a0e<sup>\u2212<\/sup>[latex]\\longrightarrow[\/latex] A<sup>\u2212<\/sup>(g)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u0394H \u2261 EA<\/span><\/p>\n<p id=\"ball-ch08_s05_p21\" class=\"para editable block\">Electron affinity is also usually expressed in kJ\/mol.<\/p>\n<p class=\"para editable block\">Electron affinity also demonstrates some periodic trends, although they are less obvious than the other periodic trends discussed previously.\u00a0<a class=\"xref\" href=\"#ball-ch08_s05_f03\">Figure 3 &#8220;Electron Affinity on the Periodic Table&#8221;<\/a> shows EA values versus position on the periodic table for the <em class=\"emphasis\">s<\/em>&#8211; and <em class=\"emphasis\">p<\/em>-block elements.\u00a0The trend isn\u2019t absolute, especially considering the large positive EA values for the second column.<\/p>\n<ul>\n<li class=\"para editable block\">Electron affinity generally increases in magnitude as we move to the right across a period (row) in the periodic table (across a period).<\/li>\n<li class=\"para editable block\"><span class=\"informalequation block\"><\/span>There is not a definitive trend as you go down a group (column) on the periodic table; sometimes electron affinity increases, sometimes it decreases.<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<div class=\"figure large medium-height editable block\" id=\"ball-ch08_s05_f03\">\n<figure id=\"attachment_4712\" aria-describedby=\"caption-attachment-4712\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/09\/Electron-Affinity.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Electron-Affinity-1.png\" alt=\"Electron Affinity\" width=\"600\" height=\"708\" class=\"wp-image-4712 size-full\" \/><\/a><figcaption id=\"caption-attachment-4712\" class=\"wp-caption-text\"><strong>Figure 3.<\/strong> Electron Affinity on the Periodic Table. \u00a0Values are in kJ\/mol.<\/figcaption><\/figure>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 4<\/h3>\n<p id=\"ball-ch08_s05_p23\" class=\"para\">Predict which atom in each pair will have the highest magnitude of EA.<\/p>\n<p class=\"para\">a) C or F \u00a0 \u00a0 \u00a0b)\u00a0Na or S<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) C and F are in the same row on the periodic table, but F is farther to the right. Therefore, F should have the larger magnitude of EA.<\/p>\n<p class=\"simpara\">b) Na and S are in the same row on the periodic table, but S is farther to the right. Therefore, S should have the larger magnitude of EA.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch08_s05_p24\" class=\"para\">Predict which atom will have the highest magnitude of EA, As or Br.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch08_s05_p25\" class=\"para\">Br<\/p>\n<\/div>\n<div class=\"key_takeaways editable block\" id=\"ball-ch08_s05_n05\">\n<h2>Key Concepts and Summary<\/h2>\n<p>Certain properties\u2014notably effective atomic radius, IE, and EA\u2014can be qualitatively understood by the positions of the elements on the periodic table.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch08_s05_qs01\">\n<div class=\"question\">\n<p id=\"ball-ch08_s05_qs01_qd01_p1\" class=\"para\">1. Write a chemical equation with an IE energy change.<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">2. State the trends in atomic radii as you go across and down the periodic table.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">3. Which atom of each pair is larger?<\/span><\/p>\n<\/div>\n<p>a) \u00a0Na or Cs \u00a0 \u00a0 \u00a0b) \u00a0N or Bi<\/p>\n<p><span style=\"font-size: 1em\">4. \u00a0Which atom of each pair is larger?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0K or Cl \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Ba or Bi<\/p>\n<\/div>\n<p>5<span style=\"font-size: 1em\">. \u00a0Which atom has the higher IE?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0Na or S \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Ge or Br<\/p>\n<\/div>\n<p>6<span style=\"font-size: 1em\">. \u00a0Which atom has the higher IE?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0Li or Cs \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Se or O<\/p>\n<\/div>\n<p>7<span style=\"font-size: 1em\">. \u00a0A third-row element has the following successive IEs: 738; 1,450; 7,734; and 10,550 kJ\/mol. Identify the element.<\/span><\/p>\n<p><span style=\"font-size: 1em\">8. \u00a0For which successive IE is there a large jump in IE for Ca?<\/span><\/p>\n<p><span style=\"font-size: 1em\">9. \u00a0Which atom has the greater magnitude of EA?<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0C or F \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Al or Cl<\/p>\n<\/div>\n<div class=\"question\"><\/div>\n<\/div>\n<p><b>Answers<\/b><\/p>\n<p>1. Na(g)\u00a0[latex]\\longrightarrow[\/latex] Na<sup class=\"superscript\">+<\/sup>(g) +\u00a0e<sup class=\"superscript\">\u2212<\/sup> \u0394<em class=\"emphasis\">H<\/em>\u00a0=\u00a0IE (answers will vary)<\/p>\n<p>2. As you go across, atomic radii decrease; as you go down, atomic radii increase.<\/p>\n<p>3. a) \u00a0Cs \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) Bi<\/p>\n<p>4. a) \u00a0K \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Ba<\/p>\n<p>5. a) \u00a0S \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0b) \u00a0Br<\/p>\n<p>6. a) \u00a0Li \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0O<\/p>\n<p>7. Mg<\/p>\n<p>8. The third IE shows a large jump in Ca.<\/p>\n<p>9. a) \u00a0F \u00a0 \u00a0 \u00a0 \u00a0\u00a0b) \u00a0Cl<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":330,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"8.5 Periodic Trends","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[64,63],"license":[54],"class_list":["post-2398","chapter","type-chapter","status-publish","hentry","contributor-david-w-ball","contributor-jessie-a-key","license-cc-by-nc-sa"],"part":2362,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/2398","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/users\/330"}],"version-history":[{"count":16,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/2398\/revisions"}],"predecessor-version":[{"id":4852,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/2398\/revisions\/4852"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/parts\/2362"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/2398\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/media?parent=2398"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapter-type?post=2398"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/contributor?post=2398"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/license?post=2398"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}