Advanced Chem Worksheet on bonding Lewis Dot Diagrams by aam35572

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									Advanced Chem: Worksheet on bonding: Lewis Dot Diagrams                                    Page 1
1. Write the electron configuration for these ions. Which one(s) do not have a noble gas configuration?
Ba+2                  I–                   Fe+2                     Zr+4                  Au+3                        Se–2
2. Explain the following trends in lattice energy: a. CaS>KCl
b. LiF>CsBr                                                    c. MgO>MgS


3. Check your handout and !H chart for values needed for this question.
a. Use the Born Haber cycle to calculate !Hf° for KI(s)                                                            (–328 kJ/mol) .




b. An approximate value for the lattice energy for a hypothetical compound KI2 is 1900 kJ/mol. Show
calculations to explain why this compound doesn’t form.




4. Write the Lewis (dot) diagrams for the following molecular compounds or polyatomic ions.
H2S               PH3                       OF2                   CCl4              CHClF2 (freon)




CO                  H3CCH3                          H2CCH2                       HCCH                         CH2O (C is central)
                    ethane                          ethene                       ethyne                       (formaldehyde)


 OH–
 hydroxide                                              NH2–                                     BrO2–
 ion                                                    amide ion                                bromite ion

1. Fe+2 , Au+3 ; 2. a +2 vs. +1, b& c. smaller size. on back 5. a, g = odd; c = less, b, d, e, f, h = more; 6. O
4 (cont). Write Lewis diagrams for the following polyatomic ions.                            Page 2
                                          SO3–2                               CN-
   NH4+
                                          sulfite ion                         cyanide ion
   ammonium
   ion

 5. Write the Lewis structures for the following species and explain how they violate the octet rule.
 (odd number of electrons, less than an octet, more than an octet).

 a. NO2                      b. TeF4                      c. BCl3                d. XeF4




 e. I3–                      f. AsF6–                       g. O2–               h. ICl3




6. Which of the following can not have more than an octet? Why not?
    Ar             O                 Cl                 Br
7. Draw resonance structures for each of the following:
SO3                                                   ClO3 (hint: move the odd electron)




CO3–2                                                   NO2–



Bond energies (covalent)
8. Which of the following bonds is the most stable? (highest bond energy) Least stable?
        C–H          C– O     C–F       C–I

9. Which of the following bonds would be the shortest?
        C–C          C=C      C! O

10. Using Bond Energies, estimate !H for this gas phase reaction. (Bonds broken – bonds made)

C! O + H–O–H –––> H–H + O=C=O                                                                   (–36 kJ)


11. Photons of EM radiation can break chemical bonds. What is the longest wavelength of light that
can break a Cl–Cl bond (242 kj/mol)? Can you see it? Hint: First change kJ/mol of bonds to J/bond.
                                                                                              (494 nm)
Adv Chem Bonding worksheet: Part 2                                                  Page 3
1. Identify these bonds as non-polar (eneg. diff = 0 to 0.2), polar (0.21 to 1.8) or ionic (above 1.8)?
    In the polar bonds, identify the ! + and ! – element of each. Which bond is the most polar
    (but not ionic)? Electronegativities are on the back of your periodic table.
a. B–Cl                 b. Cl–Cl                 c. As–O                      d. Na–Cl                    e. C–Br


2. The molecules NF3, BF3, and ClF3 all have molecular formulas of the type XF3 but have
   different molecular geometries. Write the dot diagrams and predict the shape for each. Why
   are their shapes different?




3. You did the Lewis diagrams for these molecules and ions on the previous worksheet. Use
   them to fill in the following table.
                  # e– Electron pair            bond.      non–    Molecular              Hybrid         Polar/NP
                  pairs geometry                e–         bond e– geometry               orbitals       molecule?
  ex. SF4            5 Trigonal                    4          1    Seesaw                 sp3d           polar
                        bipyramidal
  a. CClHF2

  b. PH3

  c. H2S

  d. CH2O

  e. TeF4

  f. BCl3

  g. XeF4

  h. I3–

  i. AsF6–

  j. ICl3

  k. SO3


 1. a, c, e = P (c most); b= NP; d= I; 2. Trig Pyramidal, Trig Planar, T-shaped; 3. a. tetra./tetra, sp3, P; b. tetra/ trig.
pyramid, sp3, NP; c. tetra/bent, sp3, P; d. trig planar/trig. planar, sp2, P; e. trig. bipyramid./seesaw, sp3d, P; f. trig.
planar/trig planar, sp2, NP; g. octa./sq. planar, sp3d2, NP; h. trig. bipyramid./ linear, sp3d, NP; i. octa/octa,
sp3d2, NP, j. trig, bipyramid, T-shaped, sp3d, P; k. trig.planar/trig planar, sp2, NP
 4. Describe the hybridization of the Carbon (and Oxygen) atoms in these molecules. Page 4
 How many sigma and pi bonds are there in the molecules?
      H H      HH                  C1 ____       C2 ____    C3 ____      C4 ____
      | |      | |                 C5 ____       C6 ____                         sp3, sp3,sp, sp,
 H–C–C–C"C–C=C–H                    sigma bonds = _____
                                                                                sp2, sp2, 13, 3;
      | |                                pi bonds = _____
      H H
       HH         H               O1 ____       C1 ____    C2 ____      C3 ____
    .. | |        |     ..        C4 ____       C5 ____    C6 ____
H–O–C=C–C "C–C=C=O                sigma bonds = _____                         sp3, sp2, sp2, sp, sp,
   ˙˙                   ˙˙             pi bonds = _____                       sp2, sp, 11, 5
5. Fill in Molecular Orbital energy level diagram for these species and calculate their bond order.
   Bond order = 1 (bonding e- – antibonding e-) (Yes, they can have “1”)
Then predict the magnetic properties for each: paramagnetic = unpaired electrons (attracted to
magnetic field), diamagnetic = all electrons are paired (weakly repelled by magnetic field).
            CO                   CN–                NO                O2                  O2–
#*2p        ____                 ____               ____              ____                ____
$ *2p    ____ ____            ____ ____          ____ ____         ____ ____          ____ ____
# 2p        ____                 ____               ____               ____              ____
$ 2p     ____ ____            ____ ____          ____ ____         ____ ____          ____ ____
#*2s        ____                 ____               ____               ____              ____
# 2s        ____                 ____               ____               ____              ____

Bond order____                   ____               ____               ____              ____
Mag. prop. ____                  ____               ____               ____              ____
Experimentally, NO molecules fairly easily make NO+ ions. Explain this result using your
Molecular Orbital diagram.


            O2–2                 Ne2                Ne2+               Li2               Li2 –2
#*2p        ____                 ____               ____               ____              ____
$ *2p    ____ ____            ____ ____          ____ ____         ____ ____          ____ ____
# 2p        ____                 ____               ____               ____              ____
$ 2p     ____ ____            ____ ____          ____ ____         ____ ____          ____ ____
#*2s        ____                 ____               ____               ____              ____
# 2s        ____                 ____              ____              ____              ____
Bond order____                   ____              ____              ____              ____
Mag prop ____                    ____              ____              ____              ____
Stable? (Yes/no) ___             ____              ____              ____              ____
Experimentally, superoxide ion, O2–, has a shorter bond length than the peroxide ion, O2–2.
Explain this experimental result using your MO diagrams. Which bond would be a bit “stronger?”
AP Chem Bonding Worksheet: Organic molecules                                    Page 5
. 1. Here are 3 structural isomers of pentane, C5H12 .                     H HH
                                                                            \|/
  H HH H H                  H HH                                         H C     H
                                      iso -pentane
  | | | | |                  \|/                                          \ | /
                                      <–––
H–C–C–C–C–C–H               H CH H                                      H–C– C–C–H
  | | | | |                 | | | |                                       / | \
  H HH H H                H–C–C–C–C–H        neo- pentane–––>            H C     H
                            | | | |                                         /|\
     n-pentane              H HH H                                         H HH

 a. Are these molecules Polar or Non-polar? Why do you think so?



 b. Which is a “straight chain”? __________ Which are “branched chains?” ______________
 c. Which would you predict to have the greatest intermolecular attraction (and hence the
    highest Boiling Point?) ______________ The lowest? ___________________ Why?



2. Here are 4 structural isomers of C4H8O.
A. Circle any Carbons with sp2 hybridization.
B. Which of these molecules has cis and trans isomers?
C. Which one(s) are Polar?
D. Which of these molecules would have hydrogen bonding? Put a box around the
   applicable parts of the molecule(s).
E. Which would probably dissolve in water? Why?



F. Which would probably have the highest Boiling point? Why?




 a H HH H             b H H   H                 c H   H H                 d   H   HH
   | | | |              | |   |                   |   | |                     |   | |
 H–C–C–C–C=O          H–C–C=C–C–O–H             H–C–C–C–C–H               H–C=C–O–C–C–H
   | | |                |   | |                   | | | |                   |     | |
   H HH                 H   H H                   H OH H                    H     HH




1. a. NP, all C’s Hs’; b. n=straight; iso&neo= branched; c. n- highest, neo lowest; straight= more
attraction; 2. A. a=C4 , b=C2 &C3, c=C2 , d =C1 &C2 ; B. b only; C. a, b, c; D. b only (H bonded to
O); E. a, b, c, Polar; F. b, hydrogen bonding is “strong”.
                                                                                     Page 6
3. A. Which of the following molecules would have Hydrogen bonding? Circle applicable
   parts of the molecule(s). _______________
Label Carbons with sp hybridization, sp2 hybridization and sp hybridization.

Label tetrahedral Carbons, Trigonal planar C’s and Linear C’s
                                                                                    e
a  H HH               b                c Br   H              d HH O                 H       H
   | | |                                  \  /                  | | ||                \     /
Cl–C–C–C–O–H          Br–C ! C–H          C=C                Cl–C–C–C–O–H              C=C=C
   | | |                                  /  \                  | |                   /     \
   H HH                                  H    Br                H H                 Cl      Cl



    4. What’s wrong with the following molecules? (more than 1 thing in some)

    a H HH      H                      b Br     H                  c H HH
      | | |     |                         \    /                     | | |
    H–C=C–C–O–H–C–H                       C = C– H                 H=C–C–C!O=C–H
      | | |     |                         /    \                       | |
      H ClH     H                        H      H                     Cl H



 5. Draw 3 isotopes of C8H12O2 that has:
 a. Has a “branched chain” and would not have Hydrogen bonding.




 b. Has a Carbon with sp hybridization, and would have Hydrogen bonding




 c. Has a “branched chain” and at least one C to O double bond .




3. A. a, d; B. sp=b, e, sp2 = c, d, e ; sp3 = a, d ; 4. a. 5 bonds around 1st&2nd C,2 on H ; b. 5 on 2nd
C; c. 2 on 1st H, 5 on 3rd C, 4 on O, 3 on 4th C .

								
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