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Solutions 17

  • Page ID
    47400
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    S17.1

    (a) eight electrons:

    A Lewis dot diagram shows the symbol for arsenic, A s, surrounded by eight dots and a superscripted three negative sign. ;

    (b) eight electrons:

    A Lewis dot diagram shows the symbol for iodine, I, surrounded by eight dots and a superscripted negative sign. ;

    (c) no electrons

    Be2+;

    (d) eight electrons:

    A Lewis dot diagram shows the symbol for oxygen, O, surrounded by eight dots and a superscripted two negative sign. ;

    (e) no electrons

    Ga3+;

    (f) no electrons

    Li+;

    (g) eight electrons:

    A Lewis dot diagram shows the symbol for nitrogen, N, surrounded by eight dots and a superscripted three negative sign.

    S17.2

    Write Lewis structures for the following:

    a. H2

    H2_2.jpg

    b. HBr

    HBr.png

    c. PCl3

    PCl3.png

    d. SF2

    SF2.gif

    e. H2CCH2

    C2H4.png

    f. HNNH

    N2H2.png

    g. H2CNH

    H2CNH.png

    h. NO

    NO-.png

    i. N2

    N2.png

    j. CO

    CO.png

    k. CN

    CN-.png

    S17.3

    a

    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.

    You can also find the potential energy function of H-Cl and take the derivative with respect to internuclear distance and find minimum.

    1. \(H–Cl\): $$energy of one bond=431 kJ/mol /N_A=431 \times 10^3 J/mol x \dfrac{1 mol}{6.02 \times 10^{23} bonds}=7.16 \times 10^{-19} J/bond$$

    S17.4

    The single bond present in each molecule results from overlap of the relevant orbitals: F 2p orbitals in F2, the H 1s and F 2p orbitals in HF, and the Cl 3p orbital and Br 4p orbital in ClBr.

    S17.5

    \(\ce{H–C≡N}\) has two σ (H–C and C–N) and two π (making the CN triple bond).

    A Lewis structure depicts a hydrogen atom singly bonded to a carbon atom which is triple bonded to a nitrogen atom. The nitrogen atom also has a lone pair of electrons.

    S17.6

    An ionic bond wave function takes into account the probability that the electrons of a multiatomic molecule might exist on the same atom; the covalent bond wave function assumes they exist on separate atoms.


    Solutions 17 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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