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9.E: Advanced Theories of Covalent Bonding- Homework

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    Turn in your answers for the following questions - show your work

    1. Draw Lewis dot structures, determine the hybridization of each atom and sketch the orbital structure for;
      1. CH4
      2. NH3
      3. HBr
      4. PCl5
      5. CO2
      6. BF3
      7. C4H8O (3 different structures)
    2. Draw Lewis dot structures then compare and contrast the Lewis dot, hybrid, and MO models for describing the structure of these molecules.
      1. H2
      2. O2

     

    The Following Questions are for your practice - Do Not Turn In.  They include answers so you can check your work

    Valence Bond Theory

    Chapter Exercises

    1. Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.
      1. Use the bond energy found in Table 8.2.1 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)
      2. Use the enthalpy of reaction and the bond energies for \(H_2\) and \(Cl_2\) to solve for the energy of one mole of HCl bonds. \[H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \;\;\; ΔH^∘_{rxn}=−184.7\; kJ/mol\]
    2. Use valence bond theory to explain the bonding in O2. Sketch the overlap of the atomic orbitals involved in the bonds in O2.
    3. Draw the Lewis structures for CO2 and CO, and predict the number of σ and π bonds for each molecule.
      1. CO2
      2. CO

    Solutions

    1

    A graph is shown where the x-axis is labeled, “Internuclear distance ( p m ),” and the y-axis is labeled, “Energy ( J ).” The graph of the data begins at the top of the y-axis and the left side of the x-axis and dips steeply downward before rising again to almost the same height. The lowest point the graph dips to is labeled “127” on the x-axis.

    When H and Cl are separate (the x axis) the energy is at a particular value. As they approach, it decreases to a minimum at 127 pm (the bond distance), and then it increases sharply as you get closer.

    2. Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side by side to form the π bond.

    This figure shows the orbitals of two atoms being added together as they form bonds. The two atoms are shown separately on the right, each having two peanut-shaped orbitals lying perpendicularly to one another. A right-facing arrow shows that the two have moved closer together and now the upper and lower portions of the vertical peanut-shaped orbitals are shown as merging together above and below the plane of the molecule while the horizontal peanut-shaped orbitals are overlapping between the two nuclei.

    3.(a) 2 σ 2 π;

    A Lewis structure shows an oxygen atom double bonded to a carbon atom, which is double bonded to another oxygen atom. Each oxygen atom also has two lone pairs of electrons.

    (b) 1 σ 2 π;

    A Lewis structure shows a carbon atom triple bonded to an oxygen atom. Each atom also has a lone pairs of electrons.

    Hybrid Atomic Orbitals

    Chapter Exercises

    Sulfuric acid is manufactured by a series of reactions represented by the following equations:

    \(\ce{S8}(s)+\ce{8O2}(g)⟶\ce{8SO2}(g)\)

    \(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g)\)

    \(\ce{SO3}(g)+\ce{H2O}(l)⟶\ce{H2SO4}(l)\)

    Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:

    (a) circular S8 molecule

    (b) SO2 molecule

    (c) SO3 molecule

    (d) H2SO4 molecule (the hydrogen atoms are bonded to oxygen atoms)

    (a) Each S has a bent (109°) geometry, sp3

    A Lewis structure is shown in which eight sulfur atoms, each with two lone pairs of eletrons, are single bonded together into an eight-sided ring.

    (b) Bent (120°), sp2

    Two Lewis structure are shown, connected by a double-ended arrow. The left structure shows a sulfur atom with one lone pair of electrons double bonded to an oxygen atom with two lone pairs of electrons on the left and single bonded to an oxygen atom with three lone pairs of electrons on the right. The right structure shows the same molecule, except that the double bonded oxygen is on the right side of the sulfur and the single bonded oxygen is to the left of the sulfur.

    (c) Trigonal planar, sp2

    A Lewis structure of a sulfur atom singly bonded to two oxygen atoms, each with three lone pairs of electrons, and double bonded to a third oxygen atom with two lone pairs of electrons is shown.

    (d) Tetrahedral, sp3

    A Lewis structure is shown in which a sulfur atom is single bonded to four oxygen atoms. Two of the oxygen atoms have three lone pairs of electrons while the other two each have two lone pairs of electrons and are each singly bonded to a hydrogen atom.

    For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains 77.55% Xe and 22.45% F by mass.

    (a) What is the formula of the compound?

    (b) Write a Lewis structure for the compound.

    (c) Predict the shape of the molecules of the compound.

    (d) What hybridization is consistent with the shape you predicted?

    (a) XeF2

    (b)

    A Lewis structure is shown in which a xenon atom that has three lone pairs of electrons is single bonded to two fluorine atoms, each of which has three lone pairs of electrons.

    (c) linear

    (d) sp3d

     

    Write Lewis structures for NF3 and PF5. On the basis of hybrid orbitals, explain the fact that NF3, PF3, and PF5 are stable molecules, but NF5 does not exist.

    Two Lewis structures are shown. The left structure shows a nitrogen atom with one lone pair of electrons single bonded to three fluorine atoms, each of which has three lone pairs of electrons. The right structure shows a phosphorus atoms single bonded to five fluorine atoms, each of which has three lone pairs of electrons.

    Multiple Bonds

    Chapter Exercises

    A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H3CCN. It is present in paint strippers.

    (a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.

    (b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds.

    (c) Describe the atomic orbitals that form the π bonds in the molecule.

    (a)

    A Lewis structure is shown in which a carbon atom is attached by single bonds to three hydrogen atoms. It is also attached by a single bond to a carbon atom that is triple bonded to a nitrogen atom with one lone electron pair. Below the structure is a right facing arrow with its head near the nitrogen and its tail, which looks like a plus sign, located near the carbon atoms. The arrow is labeled, “dipole moment.”

    (b) The terminal carbon atom uses sp3 hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2p orbital on carbon and a nitrogen 2p orbital.

    Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:

    (a) ClNO (N is the central atom)

    (b) CS2

    (c) Cl2CO (C is the central atom)

    (a) sp2; (b) sp; (c) sp2

    Draw the orbital diagram for carbon in CO2 showing how many carbon atom electrons are in each orbital.

    A diagram is shown in two parts, connected by a right facing arrow labeled, “Hybridization.” The left diagram shows an up-facing arrow labeled, “E.” To the lower right of the arrow is a short, horizontal line labeled, “2 s,” that has two vertical half-arrows facing up and down on it. To the upper right of the arrow are a series of three short, horizontal lines labeled, “2 p.” Above both sets of lines is the phrase, “Orbitals in an isolated C atom.” There are two upward facing arrows on two of these lines. The right side of the diagram shows two short, horizontal lines placed halfway up the space and each labeled, “s p.” An upward-facing half arrow is drawn vertically on each line. Above these lines are two other short, horizontal lines, each labeled, “2 p,” and which have two upward facing arrows on them. Above both sets of lines is the phrase, “Orbitals in the s p hybridized C in C O subscript 2.”

    Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.

    Molecular Orbital Theory

    Chapter Exercises

    Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.

    (a) H2, \(\ce{H2+}\), \(\ce{H2-}\)

    (b) O2, \(\ce{O2^2+}\), \(\ce{O2^2-}\)

    (a) H2 bond order = 1, \(\ce{H2+}\) bond order = 0.5, \(\ce{H2-}\) bond order = 0.5, strongest bond is H2; (b) O2 bond order = 2, \(\ce{O2^2+}\) bond order = 3; \(\ce{O2^2-}\) bond order = 1, strongest bond is \(\ce{O2^2+}\)

    This page titled 9.E: Advanced Theories of Covalent Bonding- Homework is shared under a CC BY license and was authored, remixed, and/or curated by Scott Van Bramer.