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6.E: Lewis Acids and Molecular Shapes (Exercises)

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    44381
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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.

    Q6.23

    Determine if a bond between each pair of atoms would be pure covalent, polar covalent or ionic.

    1. Ca and Ca.
    2. He and Ar.
    3. Al and S.
    4. K and Cl.

    Strategy

    1. Identify the elements are metal, nonmetal, or gas.
    2. Know how to determine polar covalent, pure covalent, and ionic.
    3. Combine step 1 and step 2's solution, then identify what kind of bond these elements are.

    Hints

    1. Ca is nonmetal.
    2. He and Ar are gasses.
    3. Al is metal, S is nonmetal.
    4. K is metal, and Cl is nonmetal.

    Solution

    1. Ca and Ca bond together is Pure Covalent. Because Ca is nonmetal, when Ca bond with it self would has no difference in electronegativity.
    2. He and Ar bond together is Pure Covalent. Because He and Ar are nonmetal, the electronegativity is very less.
    3. Al and S bond together is Ionic. Al and S has one metal and one nonmetal, so their bond is ionic.
    4. K and Cl bond together is Ionic.K is metal, and Cl is nonmetal, so their bond is ionic

    Q6.29

    Draw Lewis Structures for the following molecular compounds: NH3, CO2, BF3

    Strategy:

    What do we know?

    A We know to draw a Lewis structure, we will be using valence electrons. We know how to find these from looking at the periodic table. The group number-10 gives us valence electrons, or we can use the atomic number-core electrons.

    Element Valence Electrons
    N 5
    H 1
    C 4
    O 6
    B 3
    F 7

    B The next step involves the steps to draw a Lewis structure.

    1. Draw the correct skeleton structure (arrange the elements without any bonds yet). Hydrogens should always be on the outside, and the least electronegative atom should be in the center.
    2. Calculate the total number of valence electrons. If you are presented with a + ion, subtract one from the total for each positive charge. If you are presented with a - ion, add one to the total for each negative charge. Remember to multiply the valence electrons by subscripts given to the compounds. For example, when calculating CO2, we would add 4 for carbon + (6)(2) for oxygen because there are two of them.
    • NH3 has 8 valence electrons total
    • CO2 has 16 valence electrons total
    • BF3 has 24 valence electrons total
    1. Place a pair of bonding electrons (2) between each element. If you have extra electrons, begin to fill the octet of the other atoms with LONE PAIRS, starting with the terminal atoms.
    2. If you have used all of your valence electrons and some structures are not "happy", do not have 8 valence electrons, use lone pairs to form double or triple bonds.
    3. Check your work! Make sure the number of electrons in the structure match the amount in step 2.

    Draw the structures.

    Answer:

    ammonia1.gif CO2c.gif valel_bf3.png

    Q6.30

    Write the Lewis structure for each molecule

    1. \(Al_2O_3\)
    2. \(CH_3Cl\)

    Solution

    \(Al_2O_3\)

    Add up valence electrons

    • Al= 3e- x 2=6e-
    • O= 6e- x 3=18e-
    • Total e- =24e-

    Aluminum will now transfer its electrons to oxygen, making oxygen happy
    with a full octet
    -keeping this in mind draw Lewis Dot Structure :Ö =Al-O-Al=Ö:

    \(CH_3Cl\)

    Add up valence electrons

    • C= 4e-
    • H= 1e- x 3=3e-
    • Cl=7e-
    • Total e- =14e-

    place carbon as the central atom and donate electrons accordingly to make carbon a full octet, hydrogen will be happy with 2 electrons, and chlorine will have 6 electrons left over
    ..
    :Cl:
    |
    H-C-H
    |
    H

    Q6.35a

    For finding the skeleton structure for BF_{3} we must first figure out the number of electrons for Boron and Florine.\left [ \left [ Number of valence electrons for Boron \right ] \right ]+\left [Number of electrons for Florine \right ]. Which is 3+7(3)=24 valence electrons(AX3).

    After determining the total number of electrons place then around the central atom to complete its octet, which is Boron in this case. Because Boron is the least electronegative atom. But Boron is an exception and only needs 6 valence electrons to make its octet.

    Then, you check the formal charges for both Boron and Florine; (B)=3-3-0=0( F)=7-1-6=0. So with this strategy we know that this is a good structure because the formal charges equal 0. Skeleton structure B--F--F--F

    Q6.35b

    Draw all of the possible Lewis structures for the following ions or molecules: include resonance structures and label formal charges on each ion.

    1. nitrate ion (NO3-)
    2. formate ion (CHO2-)
    3. cyclobutadiene (C4H4)
    4. ozone (O3)

    The octet rule states that each atom needs to have 8 electrons. Each lone pair counts as 2. A bond also counts as 2 for each atom in the bond. Then you have to look at how many bonds each element likes to have. That is based on their column or period on the periodic table. Carbon likes to have 4 bonds, Oxygen prefers 2, Nitrogen prefers 3, and Chlorine prefers 1. If they have more or less then that it gives them a formal charge. The formula for formal charge is the [number of valance electrons] -[electrons in lone bonding pairs+1/2 the number of bonding electrons]. The Lewis structures will be as follows:

    1.) Screen shot 2015-04-30 at 4.52.44 PM.png<--> Screen shot 2015-04-30 at 4.49.18 PM.png<--> Screen shot 2015-04-30 at 4.52.56 PM.png

    2.) Screen shot 2015-04-30 at 4.59.39 PM.png<--> Screen shot 2015-04-30 at 5.01.27 PM.png

    3.) Screen shot 2015-04-30 at 5.04.11 PM.png<--> Screen shot 2015-04-30 at 5.04.39 PM.png

    4.) Screen shot 2015-04-30 at 5.06.20 PM.png<--> Screen shot 2015-04-30 at 5.08.09 PM.png

    Q6.45

    Cooler X had more ice after 2 hours because most of the ice in cooler Y was melted in order to cool the drinks that started at room temperature. In cooler X the drinks were already cold and so the ice only needed to maintain its cool temperature.

    Q6.49

    Write The Lewis Structure for each atom or molecule

    1. \(H_2O\)
    2. \(SF_6\)
    3. \(SO_2\)
    4. \(S^{2-}\)

    Strategy

    1. Add up the electrons in each molecule so you can pair it correctly
    2. Find the central atom (least electronegative)
    3. Fill up the bonded atoms with electrons
    4. Fill up peripheral with left over atoms
    5. H atoms can only have electron and can only be shared with 1 other atom per Hydrogen
    6. For each (-) anion you add an electron and for each (+) cation you take away an electron

    For 1st and 2nd row elements, the octet rules typically apply (8 electrons to be happy), however, after 2 row the octet rule is often violated.

    Solutions

    \(H_2O\)

    Count the electrons in each atom.

    2 Hydrogen atoms = 2 electrons + 1 Oxygen atom 6 electrons Total: 8 electrons

    If you count all the electrons around each atoms, youll see there all filled up to be happy

    \(SF_6\)

    Count the electrons in each atoms

    1 Sulfur Atom= 6 electrons + 6 Fluoride Atoms= 42 electrons Total= 48 electrons

    -Find the central atom and fill up each bonding atom first then peripheral sites.

    - Central atom is sulfur (Least electronegative)

    \(SO_2\)

    Count electrons

    1 Sulfur atom : 6 electrons + 2 Oxygen atoms : 12 electrons Total:

    central atom : Sulfur

    -Because the central atom would not be happy while filling up the atoms,in order to do so you have to make double bonds to fulfill the octet. Once you do that you should come up with these 3 options called resonance structures.

    Image result for so2 lewis structure

    \(S^{2-}\)

    Sulfur originally has 6 electrons so now that there is a minus two sign above it, you then add another 2 electrons.. 6electrons + (2-) = 8 electrons in the sulfer atom.

    Now fill the orbitals with electrons untill empty.

    Image result for lewis dot structure s-2

    Q6.53

    Methane (\(CH_4\)) has a tetrahedral geometry. How many electron groups are on the central atom?

    Strategy

    • Valence Shell Electron Pair Repulsion (VSEPR) theory is based on the idea that electrons groups--- defined as lone pairs, single bonds, multiple bond, and single bonds--- repel one another.
    • The repulsion between the electron groups determines the geometry of the molecule.
    • The molecules that have one central atom molecular geometry depend on:
      • The number of electron groups around the central atom.
      • However many of those electron groups are bonding groups and how many are lone pairs.

    Solution

    \(CH_4\) has a tetrahedral geometry with four bonded pairs with no lone pairs, so there are four electron groups.

    Q6.55

    Identify the following properties for each figure below:

    1. total number of electron groups
    2. number of bonding pairs
    3. number of lone pairs
    4. electron geometry
    5. molecular geometry

    Trigonal Pyramidal.png Linear.png Seesaw.png

    Figure 1:Figure 2: Figure 3:

    Strategy

    1. Begin with Figure 1, find the total number of electron groups, each line, wedge, or dash represents an electron group. Use this information to determine the electron geometry. Then determine how many bonding pairs it has (X represents bonding pairs). Then, determine how many lone pairs it has (E represents lone pairs). Use this information to determine the molecular geometry.
    2. Move on to Figure 2, and follow the same procedure as you did for Figure 1.
    3. Move on to Figure 3, and follow the same procedure as you did for Figures 1 and 2.

    Solution

    1. As you can see, Figure 1 appears to have 4 electron groups in total. This gives you the information that Figure 1 has a tetrahedral electron geometry. Tetra is the prefix for 4, so you can use that to assist you in coming to an answer. Since X represents a bonding pair, you can determine that Figure 1 has 3 bonding pairs. There is only one E, so it has one lone pair. This makes the AXE notation AX3E. This information allows you to determine the molecular geometry of the figure. An AXE notation of AX3E means that this figure is trigonal pyramidal. The 3 bonding pairs make the triangular base of the pyramid because there are 3 sides to a triangle, and the one lone pair forms the tip, making it a pyramid.
      1. 4 electron groups
      2. 3 bonding pairs
      3. 1 lone pair
      4. tetrahedral
      5. trigonal bipyramidal
    2. Figure 2 appears to have 5 electron groups in total. This means it has a trigonal bipyramidal electron geometry. There are 2 Xs, meaning it has 2 bonding pairs. There are 3 Es, meaning it has 3 lone pairs. This makes the AXE notation out to be AX2E3. This notation leads us to the conclusion that it has a linear molecular geometry. The two bonding pairs make a line, which is the simpler way of determining they form a linear molecular geometry.
      1. 5 electron groups
      2. 2 bonding pairs
      3. 3 lone pairs
      4. trigonal bipyramidal
      5. linear
    3. Figure 3 has 5 electron groups in total, so we can determine it has a trigonal bipyramidal electron geometry, like Figure 2. This figure has 4 Xs, meaning it has 4 bonding pairs. There is only 1 E, so it only has one lone pair. This makes the AXE notation out to be AX4E. There are many different molecular geometries within the category of trigonal bipyramidal, but this one in particular has 4 bonding pairs and 1 lone pair. This means it has a seesaw molecular geometry. One of the bottom two Xs comes towards the front and one goes towards the back. If you picture the figure in 3D without the lone pair, it would look kind of like a seesaw.
      1. 5 electron groups
      2. 4 bonding pairs
      3. 1 lone pair
      4. trigonal bipyramidal
      5. seesaw

    6.E: Lewis Acids and Molecular Shapes (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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