Skip to main content
Chemistry LibreTexts

13.19: Solutions to Selected Problems

  • Page ID
    192892
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Exercise 13.2.1:

    When two s orbitals combine out-of-phase, destructive intereference occurs.

    There is a node between the atoms.

    The energy of the electrons increases.

    When two s orbitals combine in-phase, constructive interference occurs.

    There is no node between the atoms; the electrons are found between the atoms.

    The energy of the electrons decreases.

    Exercise 13.3.1:

    clipboard_e4a491c1f855423e0028828efc2836949.png

    Exercise 13.4.1:

    clipboard_e4d6ac7c87de5fc583f9ad3e83d9daa37.png

    Exercise 13.4.2:

    clipboard_e6e5ce1e75afa40e226b8e5a467697941.png

    Exercise 13.5.1:

    When 2 atomic orbitals are combined, 2 molecular orbitals are formed: one in-phase bonding orbital and one out-of-phase antibonding orbital.

    clipboard_e6355f4fd0427c06bc9ad67983b103011.png

    Exercise 13.5.2:

    In-phase combinations of atomic orbitals give bonding orbitals.

    Exercise 13.5.3:

    Out-of-phase combinations of atomic orbitals give antibonding orbitals.

    Exercise 13.5.4:

    The combinations of s + s OR s + p OR p + p OR s + d OR p + d atomic orbitals can lead to σ orbitals.

    clipboard_e02c0eb7e96a9e52a3de9772cdf1b976e.png

    Exercise 3.5.5:

    The combinations of side by side p + p or p + d atomic orbitals leads to π orbitals.

    clipboard_effe8e3e180973f8015200056a491808a.png

    e) σ*

    Exercise 13.5.7:

    Li+ and O2- are more similar in size than K+ and O2-, so the bond between Li+ and O2- is stronger.

    The energy difference between the 1s orbitals and 2s orbitals is too large, so they cannot interact. In order for orbitals to interact, the orbitals need to have the same symmetry, be in the same plane, and be similar in energy.

    Exercise 13.5.8:

    When two parallel p orbitals combine out-of-phase, destructive intereference occurs.

    There is a node between the atoms.

    The energy of the electrons increases.

    When two parallel p orbitals combine in-phase, constructive interference occurs.

    There is no node between the atoms; the electrons are found above and below the axis connecting the atoms.

    The energy of the electrons decreases.

    Exercise 13.6.1:

    clipboard_e7e71217e247e8922d8e92d9eab97fbe2.png

    Exercise 13.6.2:

    • Count the valence electrons on the molecule. That's the number of valence electrons on each atom, adjusted for any charge on the molecule. (eg C22- has 10 valence electrons: 4 from each carbon -- that's 8 -- and two more for the 2- charge).
    • Fill electrons into the lowest energy orbitals first.
    • Pair electrons after all orbitals at the same energy level have one electron.

    Exercise 13.6.3:

    clipboard_e8bbed67863e38fc1c1c29746bf52f758.png

    Exercise 13.6.4:

    clipboard_ec707799c232130268c34727004881db9.png

    Exercise 13.6.5:

    clipboard_e886f347f25e3096d837e87b9d1db6a65.png

    Exercise 13.6.6:

    clipboard_eded156afeba204bfd80e361ca28c9489.png

    Exercise 13.6.7:

    clipboard_e8310eb9e197120f66ae6c78a32a160b4.png

    Exercise 13.6.8:

    clipboard_e8937472f938e250da17049c1105e9cda.png

    Exercise 13.6.9:

    clipboard_e8119c8ff656b15ff1039251b9d79a77b.png

    Exercise 13.6.10

    1. Na, because Na has a lower ionization potential (and a lower electronegativity) than Al.
    2. Al
    3. 4-, because there are four Na+
    4. total e- = 2 x 3 e- (per Al) + 4 e- (for the negative charge) = 10 e-
    clipboard_e4b334829edb2dbe03741682ac221f536.png
    clipboard_eae436e3fd7df985cfbe831bec01b76b3.png

    g) \(\textrm{bond order} = \frac{\textrm{# bonding e- - # antibonding e-}}{2} = \frac{8-2}{2}=3\)

    Exercise 13.7.1:

    From MO6: 3. Diamagnetic (no unpaired electrons); 4. Paramagnetic; 5. Diamagnetic; 6. Diamagnetic; 7. Paramagnetic; 8. Diamagnetic

    Exercise 13.7.2:

    a. N2+. From the MO diagrams, N2+ has one less bonding electron. Thus, the bond order will be lower and the bond will be longer than in N2.
    b. N2-. From the MO diagrams, N2- has one more antibonding electron. Thus the bond order will be lower and the bond will be longer than N2.

    Exercise 13.7.3:

    a. O2+ . From the MO diagram, O2+ has one less antibonding electron. Thus the bond order will be higher and the bond will be stronger than in O2.
    b. O2. From the MO diagram, O2 has one less antibonding electron. Thus the bond order will be higher and the bond will be stronger than in O2-.

    Exercise 13.9.1:

    a)

    clipboard_e8cf225ef9393d3ed3eeb217138c3f3fe.png

    b)

    clipboard_e4753cd8b3044c0514378b0d72fb47855.png

    c)

    clipboard_e12e987ee0285b4dea4a66d3a8cea8745.png

    Exercise 13.11.1:

    a)

    clipboard_ecb019a9b483ffff7a00c8814c6622846.png

    b)

    clipboard_ecfaff1d5b3297dac24db035bb1faeed5.png

    c)

    clipboard_e1d9963a6fbcde30db289486d7f290c6c.png

    Exercise 13.11.2:

    a)

    clipboard_e2853228f0a16f51182aa3bb48a460e68.png

    b)

    clipboard_eac864519ba345793edc7e771052ddd8d.png

    c)

    clipboard_ebf951617447b4814ebbacb75cf9bc1c4.png

    d)

    clipboard_e2e57623740ced6ff2736aaa9b7bd2238.png

    e)

    clipboard_e1a2c85388cdbdae1ecbe570c638b4e7a.png

    Exercise 13.11.3:

    a)

    clipboard_ef64a8d62d95c5c00a2d708494f3309d6.png

    b)

    clipboard_ed401bb54b5248d37129b36b057a7096c.png

    c)

    clipboard_e94e9cf48a6655d3e157bbc6512882154.png

    d)

    clipboard_e1e61c1f82f5b9c96aeacbd625ab5dbec.png

    e)

    clipboard_e07886e2ddb77dcf2e641e89597195d12.png

    f)

    clipboard_efa7d5ab7a14398a475acbe2a1557afaa.png

    g) In each case all four H1s orbitals interact with each boron orbital.

    h) The constructive overlap of the four H1s orbitals and the B2s orbital results in the lowest energy combination as it contains no nodes. (LOWEST ENERGY)
    The constructive overlap of the four H1s orbitals with the three B2p orbitals results in bonding orbitals containing a single node. (MEDIUM-LOW ENERGY)

    The destructive overlap of the four H1s orbitals with the B2s orbital results in an anti-bonding orbital. (HIGH ENERGY)
    The destructive overlap of the four H1s orbitals with the three B2p orbitals results in anti-bonding orbitals. (HIGHEST ENERGY)

    clipboard_e09272dbf40e599fdd4bf9df8f9c42207.png

    Exercise 13.11.4:

    a)

    clipboard_e8d35917bebedab54883fd33fd72e4238.png

    b)

    clipboard_ed88cdbbb3403d0256814a68c06e2f133.png

    c)

    clipboard_edee54dec855e8573678cce284c43fa1a.png

    d)

    clipboard_e1d7233a061344b5615bee72f2189460d.png

    e)

    clipboard_eb47025ce90ac404181ceaebbf59873fb.png

    f)

    clipboard_ec8269483ddb360e315ee23a3f5f01513.png

    g) In some of the cases the one or more Hs may be in a nodal plane

    h) The constructive overlap of the H1s orbitals and the B2s orbital results in the lowest energy combination. (LOWEST ENERGY)

    The constructive overlap of the three H1s orbitals with the B2p orbital results in a bonding orbital (MEDIUM-LOW ENERGY)

    The constructive overlap of the two H1s orbitals with the B2p orbital results in a bonding orbital. (MEDIUM-LOW ENERGY)

    The B2p orbital that is perpendicular to the plane of hydrogen atoms results in a non-bonding orbital. (MEDIUM ENERGY)
    The destructive overlap of the two H1s orbitals with the B2p orbital results in an anti-bonding orbital (HIGH ENERGY)

    The destructive overlap of the three H1s orbitals with the B2p orbital results in an anti-bonding orbital (HIGH ENERGY)
    The destructive overlap of the three H1s orbitals with the B2p orbital results in an anti-bonding orbital (HIGHEST ENERGY)

    clipboard_e497caf1172c3cdee592e9b05b058a294.png

    Exercises 13.12.1 & 13.12.2:

    clipboard_e84e94eb56ad0f72fd683964ff63488da.png

    Exercise 13.12.3:

    clipboard_e0e7f476553b505092f01c7b2631e6206.png

    Exercises 13.13.1-13.13.3:

    clipboard_ef339279918ce3a1cfdaca7a6d8e65c44.png

    Exercises 13.14.1-13.14.5:

    clipboard_e29f1a9d573b5f4abba26b2f636740a1b.png
    clipboard_e710873f40f853c3f7dcf72f6ab17aa76.png

    Exercise 13.14.6:

    All of the answers depend on an understanding of the contributions of two resonance structures to the overall picture of acetaminde, or alternatively, that actetamide forms a conjugated pi system with four electrons delocalized over the O, C and N.

    clipboard_e92487b0d6a4aef180da59fbaa5a731f5.png

    a. Contribution of the second resonance structure introduces some double bond character to the C-N bond and some single bond character to the C-O bond. Thus, both of these bonds are intermediate in length between single and double bonds.
    b. Since the C, N and O atoms are sp2 hybridized, the C-N pi bond can only form if the remaining p orbitals on these atoms align. This places the atoms participating in the sp2 sigma bonds in the same plane.
    c. Because of the partial double bond character there is a larger barrier to rotation than is typically found in molecules with only single bonds.
    d. Because of the partial double bond character and the restricted rotation, the two H’s are not identical. (One is nearer the O and one is nearer the CH3 and the restricted rotation prevents their interconversion.

    Exercises 13.15.1-13.15.3:

    clipboard_e6c37839b6b5b2ecda70ea97550b0ebc6.png

    clipboard_e821ab9a0d391e0e0c1f59aa025000575.png

    Exercise 13.15.4:

    clipboard_ef4e632d7a8ce47caac6933694c206620.png

    Exercise 13.16.1:

    clipboard_efa183971c244bf7e854c51bbb95d2db0.png

    Because the two resonance structures shows double bonds in two different places, the implication is that all of the bonds in benzene have some double bond and some single bond character. You can think of them all as being about 1.5 bonds.

    Exercise 13.16.2:

    a) non-aromatic b) non-aromatic c) aromatic d) aromatic e) anti-aromatic f) anti-aromatic

    Exercise 13.16.3:

    a) aromatic b) non-aromatic c) anti-aromatic d) aromatic

    Exercise 13.16.4:

    a) aromatic b) anti-aromatic c) aromatic d) anti-aromatic

    Exercise 13.16.5:

    clipboard_e19e3ae7ae573f000aa76016e2913b8a8.png

    Exercise 13.16.6:

    clipboard_e03f8ee7d9c24fa336e81d9c09653a84a.png

    Exercise 13.16.7:

    In tropone, the resonance structure on the right shows more aromatic character, because it clearly shows a fully-conjugated ring with an odd number of electron pairs in the pi system. The carbonyl in the picture on the left makes that conjugation less obvious: does its carbonyl pi bond contribute to the conjugated system of the ring? If it does, that would mean four pairs of electrons in the pi system, and that would be an even number, so it would be anti-aromatic.

    clipboard_ed275ca400a6ba1e8766a2b6710c61359.png

    The hydroxy group in hydroxytropolone would stabilize the compound in the structure shown on the right, via an ion-dipole interaction with the anionic oxygen. That makes the right-hand, explicitly aromatic structure the dominant one.

    Exercise 13.17.1:

    All of the compounds are aromatic and have the same Hückel MO diagram.

    clipboard_e0cdc43d58b277f82c37a65da73fc43e5.png

    Exercise 13.17.2:

    All of the compounds are aromatic and have the same Hückel MO diagram.

    clipboard_e251ed561aa2baf34e9e02b7c6289b17b.png


    This page titled 13.19: Solutions to Selected Problems is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Chris Schaller via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

    • Was this article helpful?