{"id":350,"date":"2017-12-14T21:33:51","date_gmt":"2017-12-14T21:33:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/periodic-trends\/"},"modified":"2024-10-10T21:13:11","modified_gmt":"2024-10-10T21:13:11","slug":"periodic-trends","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-mcc-introductorychemistry\/chapter\/periodic-trends\/","title":{"raw":"4.3 Periodic Trends","rendered":"4.3 Periodic Trends"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  1. Be able to state how certain properties (effective nuclear charge, atomic radii, and ionization energy) of atoms vary based on their relative position on the periodic table.<\/li>\r\n \t
  2. Be able to explain the periodic table trends observed within a period and a group.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

    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 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

    Effective Nuclear Charge<\/h2>\r\nMany of the periodic properties of atoms depend on electron configuration; in particular, the valence electrons and their level of attraction to the nucleus.\r\n\r\nValence electrons are simultaneously attracted to the positive charge of the nucleus and screened<\/b> (repelled) by the negative charges of other electrons. This net nuclear charge felt by valence electrons is known as its Effective Nuclear Charge, Zeff<\/sub><\/b> (pronounced \u201czed-effective\u201d). The effective nuclear charge is always less than the actual nuclear charge, and can be roughly estimated using the following equation:\r\n

    [latex]Z_{eff}= Z - S[\/latex]<\/p>\r\nWhere Z is the nuclear charge (equal to the number of protons), and S is the screening constant which can be approximated to the number of non-valence, \u201ccore\u201d electrons.\r\n

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    Example 1: Effective nuclear charge<\/h3>\r\n

    Approximate the effective nuclear charge of magnesium.<\/p>\r\n[reveal-answer q=\"17728622\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"17728622\"]\r\n\r\nFirst we must determine the electron configuration of magnesium to determine the number of core electrons.\r\n\r\nMg =1s<\/i>2<\/sup> 2s<\/i>2<\/sup> 2p<\/i>6 <\/sup>3s<\/i>2<\/sup> = [Ne]3s<\/i>2<\/sup>, therefore magnesium has 10 core electrons from its 1s<\/i>2<\/sup>, 2s<\/i>2<\/sup>, 2p<\/i>6<\/sup> orbitals.\r\n\r\nMagnesium is element 12, so it has 12 protons and a nuclear charge of 12.\r\n\r\nZeff <\/sub>= 12 \u2013 10\r\n\r\nZeff <\/sub>= 2+\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nMoving left to right across a period on the periodic table, each subsequent element has an additional proton and valence electron, but the core electrons which are responsible for the majority of screening remain the same. This results in a trend that in general the effective nuclear charge increases from left to right across any period of the periodic table.\r\n\r\nMoving from top to bottom down a column of the periodic table, we might expect the elements to have a similar effective nuclear charge as they all have the same number of valence electrons. However, we actually see a slight increase in Zeff <\/sub>moving down a column of the periodic table. As the principal quantum number (n<\/i>) increases, the orbital size increases making the core electron clouds more spread out. These core electron clouds that are more diffuse do not screen as well, giving a slight increase to Zeff<\/sub> (Figure 1)\r\n\r\n[caption id=\"attachment_2465\" align=\"aligncenter\" width=\"600\"]\"Figure<\/a> Figure 1. The periodic trend for effective nuclear charge.[\/caption]\r\n

    Atomic Radii<\/h2>\r\n

    The 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\n

    As you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting a larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus. Going across a row on the periodic table, left to right, the trend is different. This is because the number of protons\u2014and hence the nuclear charge\u2014is increasing as you go across the row. The increasing positive charge leads to a larger effective nuclear charge which casts a tighter grip on the valence electrons, so as you go across the periodic table, the atomic radii decrease.<\/p>\r\n\r\n<\/div>\r\n

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    Figure 2 \"Atomic Radii Trends on the Periodic Table\"<\/a> shows spheres representing the atoms of the s<\/em> and p<\/em> blocks from the periodic table to scale, showing the two trends for the atomic radius.<\/p>\r\n\r\n<\/div>\r\n

    \r\n\r\n[caption id=\"attachment_4709\" align=\"aligncenter\" width=\"473\"]\"\"<\/a> Figure 2 Atomic Radii Trends on the Periodic Table[\/caption]\r\n

    Although 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.<\/p>\r\n\r\n

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    Example 2: Atomic Radii Trend\r\n<\/strong><\/h3>\r\n

    Referring only to a periodic table and not to figure 2, which atom is larger in each pair?<\/p>\r\na. Si or S\r\n\r\nb. S or Te\r\n\r\n[reveal-answer q=\"367562\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"367562\"]\r\n\r\na. 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.\r\n\r\nb. S is above Te on the periodic table, so Te is larger because as you go down the column, the atoms get larger.\r\n\r\n[\/hidden-answer]\r\n

    Check Your Learning<\/strong><\/h4>\r\nReferring only to a periodic table and not to figure 2, which atom is smaller, Ca or Br?\r\n[reveal-answer q=\"652575\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"652575\"]Br[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

    Ionization Energy<\/h2>\r\n

    Ionization energy (IE)<\/a><\/span>\u00a0is the amount of energy required to remove an electron from an atom in the gas phase:<\/p>\r\nA(g) \u2192 A+<\/sup>(g) + e\u2212<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u0394H \u2261 IE<\/span>\r\n

    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. However, 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:<\/p>\r\n

    Figure 3 \"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

    \r\n\r\n[caption id=\"attachment_4711\" align=\"aligncenter\" width=\"486\"]\"\"<\/a> Figure 3 Ionization Energy on the Periodic Table[\/caption]\r\n\r\n<\/div>\r\n
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    Example 2: Ionization Energy Trend\r\n<\/strong><\/h3>\r\n

    Which atom in each pair has the larger IE?<\/p>\r\na. Ca or Sr\r\n\r\nb. Na or Cl\r\n\r\n[reveal-answer q=\"3675623\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"3675623\"]\r\n\r\na. Because Sr is below Ca on the periodic table, it is easier to remove an electron from it; thus, Ca has the higher IE.\r\n\r\nBecause Cl is to the right of Na on the periodic table, it is harder to remove an electron from it, therefore it has a higher IE.\r\n\r\n[\/hidden-answer]\r\n

    Check Your Learning<\/strong><\/h4>\r\nWhich atom has the lower ionization energy, C or F?\r\n[reveal-answer q=\"6525754\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"6525754\"]C[\/hidden-answer]\r\n\r\n<\/div>\r\n
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    Key Takeaways<\/h3>\r\n