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4.E: Periodic Properties (Exercises)

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    44379
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    These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

    Q4.50

    Elements can be identified by their electron configuration and referencing it to the periodic table. Identify each element below based on their electron configurations.

    1. [Ne]3s23p2
    2. [Ne]3s23p6
    3. [Kr]5s1
    4. [Kr]5s24d105p3

    S4.50

    Strategy:

    1. Identify the noble gas and the row in which it lies within the periodic table. This represents the inner electron configuration which is represented by [noble gas symbol].
    2. Move down one row below the aforementioned period to then use the outer electron configuration, namely the configuration after [noble gas symbol], to identify the element.
    3. Move across the row, moving from left to right, assigning the electrons within the outer electron configuration to the appropriate orbitals, s, p, d, and f orbitals, to identify the element. For example, an element with [He]2s22p4 will have 2 electrons in the 2s orbital and 4 electrons in the 2p orbitals and as you move across period 2, it will correspond to oxygen (O).

     

    PeriodicTable2.jpg

    Therefore:

    1. Si
    2. Ar
    3. Rb
    4. Sb

    Q4.62

    Which set of elements would you expect to be different? Explain?

    1. Lithium and Potassium
    2. Carbon and Neon
    3. Sulfur and Oxygen
    4. Aluminum and Thallium

    Important things to consider: The number of valence electrons each element contains and whether each element is a Nonmetal, Metal, Halogen, or noble a gas

    S4.62

    B. Carbon and Neon, because they belong to different groups. Neon is a noble gas with 8 valence electrons. Carbon is a nonmetal with only 4 valence electrons.

    Q4.72

    Determine which atom is larger in each pair of elements.

    1. Na or Mg
    2. Ga or Al
    3. As or Cs
    4. Br or Fe

    Things you need to complete this problem:

    To complete this type of problem you will need a periodic table of elements handy to easily access each element to be able to tell where it is located in comparison to other elements.

    S4.72

    Strategy:

    To complete this problem there are a few rules and guidelines you will need to know about before starting.

    Firstly, we need to be able to locate each element in comparison to the other on our periodic table of elements. We know when we look at the periodic table that there are rows (horizontal) and there are columns (vertical).

    For this problem, we are locating elements and determining their atomic radius to see which is larger of the elements chosen. The atomic radius for each individual element is defined as a measure of the size of each individual atom, normally the mean (average) or the distance from the middle of the nucleus to the boundary of the neighboring cloud of electrons.

    The rules for deciding which element has a larger atomic radius are the following:

    It decreases as you go left to right on the periodic table because of the greater amount of protons. The higher the atomic number, the greater number of protons. The protons will pull the electrons closer to its nucleus thus resulting in a smaller atomic radius. It increases as you go down the group because as you go down the group, you add another principal energy level. The more energy levels, the further the electrons will be from the nucleus resulting in a a smaller atomic radius. Also, the elements in the same column as they move down closer to the bottom of the table, the radius gets larger. For example, if you look at the two elements: Iodine and Bromine, out of the two elements Iodine would have the larger radius because it is located below Bromine on the periodic table.

    So to recap: On the periodic table the elements starting from the left side have higher atomic radius, it decreases as you go across to the right and as you go down the table the radius increases. So usually, the closer to the top of the periodic table in comparison to the ones below it is smaller in radius!

    Answers:

    A) Na or Mg:

    If we look at the periodic table to the far left you will find both Sodium and Magnesium. If you look in the first row you will find sodium and to the right of it you will find magnesium. When looking back at our rules, left to right the atomic radius decreases, thus sodium has a higher atomic radius!

    Na is larger than Mg.

    B) Ga or Al

    When locating the two elements on the periodic table you will notice these two are in the same column. From our rules we know that the further down the element is on the periodic table the larger the radius! Thus, out of the two elements Gallium is the larger of the two when it comes to the atomic radius!

    Ga is larger than Al.

    C) As or Cs

    With Cesium on the first row, and almost all the way down the table of elements, we assume from our rules listed above that Cesium carries a high atomic radius. We locate the As which is two rows below and all the way to the right side of the table. The rules state left to right atoms decrease, and down the gradient of the table they increase. So for both cases of these rules makes Cesium have a larger radius thus making it the larger atom.

    Cs is larger than As.

    D) Br or Fe

    Located on the same row, we find that Iron is a located closer to the left of the periodic table than our other element Bromine. The rules state once again that as elements move left to right the atomic radius decreases, making bromine the smaller atom because of its location on the periodic table compared to Iron. Iron in this case is the bigger atom.

    Fe is larger than Br.

    Q4.75

    Write the electron configuration for each ion.

    1. Li+
    2. I-
    3. S2-
    4. Al3+
    5. Co2+

    S4.75

    Strategy: Help can be found in the following link. 6.8: Electron Configurations

    Solution:

    1. [He]
    2. [Xe]
    3. [Ar]
    4. [Ne]
    5. [Ar] 3d7

    Q4.79

    Which has the largest atomic radius in each pair?

    1. Cl or Fe
    2. Mg or B
    3. K+ or K
    4. Rb or Xe

    S4.79

    • Atomic radius becomes larger going from top to bottom in each column.
    • Atomic radius becomes smaller going left to right across the periodic table.
    • If an atom in the first family (group 1) loses an electron, the atomic radius will become smaller.

    a) Cl or Fe

    Fe has the larger atomic radius

    b) Mg or B

    Mg has the larger atomic radius

    c) K+ or K

    K has the larger atomic radius

    d) Rb or Xe

    Rb has the larger atomic radius

    Q4.86

    List these elements in order of decreasing first ionization energy: In, F, Se, Br.

    S4.86

    Strategy:

    1. Determine the location of the elements on the periodic table
    2. Remember the increasing trend of ionization energy in KJ/mol (exception in case of Boron) from left to right on the periodic table
    3. Remember the decreasing trend of ionization energies (Kj/mol) from top to bottom (Cs is the exception in the first group)

    periodic table.jpg

    Therefore in order of decreasing first ionization energy:

    F > Br > Se > In

    For more information: http://chemwiki.ucdavis.edu/Physical...ization_Energy

    Q4.88a

    To which second period element do these ionization values belong?

    • \(IE_{1}=899\;kJ/mol\)
    • \(IE_{2}=1,757\;kJ/mol\)
    • \(IE_{3}=14,849\;kJ/mol\)
    • \(IE_{4}=21,006\;kJ/mol\)

    S4.88a

    To understand and solve this problem, you must first understand the definition of ionization energy. Ionization energy is the minimum amount of energy required in order to successively remove an electron from an element. You also need to know which elements belong to which period. A period is a horizontal row on a periodic table. The second period elements include the following: lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon. It is also helpful to know the ionization trends for the periodic table. From left to right, ionization energy increases, which makes sense because metals(left side) lose electrons easily while nonmetals(right side) do not. With that being said each ionization energy an element has means we have lost one electron.The question gives you 4 values meaning this particular element can lose 4 electrons. This eliminates lithium since it only has 3 all together. Also, when there is a large jump from one value to another, this signifies that you have transitioned from valence electrons to core electrons. This is because core electrons are harder to remove. Notice that between the second and third values there is a significant jump in energy. This means that after the removal of two valence electrons, we have entered core electron territory. The only element that has 2 valence electrons in the second period is Beryllium.

    Q4.88b

    Based on the given Ionization energies, how many valence electrons does the following atom have?

    • IE1=1,402.3 kJ/mol
    • IE2=2,856.0 kJ/mol
    • IE3=4,578.1 kJ/mol
    • IE4=7,475.0 kJ/mol
    • IE5=9,444.9 kJ/mol
    • IE6=53,266.6 kJ/mol
    • IE7=64,360.0 kJ/mol

    S4.88b

    We know that ionization energy is much larger for inner orbitals than outer orbitals. So by looking at the ionization energies given, we can see that the 6th Ionization energy is much larger than the 5th. This means that the 5th electron is the last electron in the outer shell, which means this atom has 5 valence electrons.

    Q4.95

    Arrange the following elements from highest to lowest metallic character: S, Al, Cs, F, Sr, Cl

    S4.95

    1. In order to solve this problem, one must first know that metallic character increases as you go down a group in the periodic table, and as you go left within a period on the periodic table.
    2. Then, in order to determine the metallic character on each of the elements, you have to look at their location on the periodic table.
    3. Since the question asks for the elements to be listed in order of decreasing metallic character, it would make more sense to find the element with the highest metallic character first, then the second highest, etc.
    4. Double check to make sure they are in the correct order and that you have not skipped any elements.

    A. When moving from left to right, metallic character decreases because the elements tend to gain electrons to fill valence shells rather than lose them. Elements on the right side, like chlorine, will gain electrons to fill their valence shells, while elements on the left side, like cesium, will lose electrons in order to have a full valence shell. When moving down the periodic table, metallic character increases because elements can lose electrons more easily as the atomic radius increases. This is because the attraction between the nucleus and valence electrons decreases as there is more distance between them.

    B. Pull up a periodic table on a computer, or one that is in your Chemistry book. This is so that you can visually compare where each element is at on the periodic table.

    C. Look on the bottom left side of the table and try to locate some of the elements listed. You will see that cesium (Cs) and strontium (Sr) are placed on the bottom left side, so they will have the highest metallic characters. Cesium is farther down and more left than strontium, so cesium will have the highest metallic character and strontium will have the second highest metallic character. Next, you will see that aluminum (Al) is located farther left than sulfur, chlorine, and fluorine. This means that aluminum will be the element with the third highest metallic character. The only elements left are sulfur (S), chlorine (Cl), and fluorine (F). They are all located within a very close distance of one another. Sulfur is the farthest left and farthest down of the three, so it will have the fourth highest metallic character. Chlorine is right below fluorine, making it the element with the fifth highest metallic character. This leaves fluorine as the last element, meaning it has the lowest metallic character.

    D. We found cesium, strontium, aluminum, sulfur, chlorine, and fluorine on the periodic table. Cesium is the farthest left and the lowest, while fluorine is the farthest right and the highest, so we know they have the highest metallic character and the lowest metallic character, respectively.

    Answer: Cs, Sr, Al, S, Cl, F


    4.E: Periodic Properties (Exercises) is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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