Written Assignment 9: Chemical Bonding
Answer all assigned questions and problems, and show all work.
1. Use the second member of each group from Group IA to Group 7A to show that the number of valence electrons on an atom of the element is the same as its group number.
(8 points)
Li has 1 valence electron
Cl has 7 valence electrons
(Reference: Chang 9.2)
2. Explain the following: (3 points)
a. ionic bond=electrostatic force that holds ions together in an ionic compound.
b. nonpolar covalent bond=electrons are shared equally between the two atoms.
c. polar covalent bond= the electrons spend more time in the vicinity of one atom than the other.
3. Specify which compound in the following pairs of ionic compounds has the higher lattice energy: Explain your choice. (6 points)
a. KCl or MgO=MgO
b. LiF or LiBr=LiBr
c. Mg3N2 or NaCl=NaCl
(Reference: Chang 9.23)
4. Compare the properties of ionic compounds and covalent compounds. (6 points)
Electronegativity is a property of atoms, not bonds. An ionic bond is formed between elements with very different electronegativities, a covalent bond between atoms with similar electronegativities.
(Reference: Chang 9.32)
5. Four atoms are arbitrarily labeled D, E, F, and G. Their electronegativities are as follows: D = 3.8, E = 3.3, F = 2.8, and G = 1.3. If the atoms of these elements form the molecules DE, DG, EG, and DF, how would you arrange these molecules in order of increasing covalent bond character? (4 points)
DG,EG,DF,DE
(Reference: Chang 9.37)
6. Classify the following bonds as ionic, polar covalent, or nonpolar covalent, and
give your reasons: (4 points)
a. the CC bond in H3CCH3- Covalent
b. in KI the KI bond- Ionic
c. the NB bond in H3NBCl3– Polar Covalent
d. the CF bond in CF4– Polar Covalent
(Reference: Chang 9.39)
7. Write the correct Lewis structures for the following compounds (relative positions of atoms are shown correctly).(21 points)
a. H C N
b. H C C H
c. O Sn O
d. F F
B
F
e. H O F
f. H
C F
O
g. F F
N
F
(Reference: Chang 9.47)
8. Classify the following substances as ionic compounds or covalent compounds containing discrete molecules: CH4, KF, CO, SiCl4, BaCl2.(5 points)
CH4=Covalent KF=Ionic CO=Covalent SiCl4=Covalent BaCl2=Ionic
(Reference: Chang 9.73)
9. Predict the geometries of the following species using the VSEPR method: (a) PCl3,
(b) CHCl3, (c) SiH4, (d)TeCl4. (12 points)
a. 1 lp AX3E trigonal pyramidal
b. 0 lp AX4 tetrahedral
c. 0 lp AX4 regular tetrahedral
d. 1 lp AX4E a see-saw structure
(Reference: Chang 10.7)
10. Predict the geometry of the following molecules using the VSEPR method: (a) HgBr2,
(b) N2O, (c) SCN– (arrangement of atoms is SCN). (9 points)
(Reference: Chang 10.11)
a.linear
b.linear
c.linear
11. Describe the geometry around each of the three central atoms in the CH3COOH
molecule.
(5 points)
1. Tetrahedral, four bonds, 109.5 degrees apart, sp3 hybridization
2. Trigonal planar, three bonds (one double, to O), sp2 hybridization
3. Bent (at about 109.4 with two lone pairs, sp3 hybridization
(Reference: Chang 10.13)
12. What is the relationship between the dipole momentand bond moment? How is it
possible to have bond moments and yet be nonpolar? (3 points)
1.The dipole moment can be thought of as the sum of the bond dipoles in a molecule. If you have a molecule that consists of only one bond, then the dipole moment will be the same as the bond dipole. Most molecules have a dipole moment owing to the uneven distribution of charges about the molecule.
2.Because those molecules are symmetrical. CO2 and CCl4 are examples. the dipole moments of each bond cancel.
(Reference: Chang 10.16)
13. Specify which hybrid orbitals are used by carbon atoms in the following species:
(a) CO,
(b) CO2, (c) CN–. (6 points)
a. sp + p + p.
b. sp
c. sp + p + p.
(Reference: Chang 10.37)
14. How many sigma bonds and pi bonds are there in each of the following
Molecules? (6 points)
|
a.4 sigma bonds |
|
|
b.5 sigma bonds and 1 pi bond |
|
|
c.7 sigma bonds and 3 pi bonds |
|
(Reference: Chang 10.41)
15. Which of the following species is not likely to have a tetrahedral shape? (a) SiBr4 (b) NF4+, (c) SF4, (d) BeCl4–2, (e) BF4–, (f) AlCl4–(6 points)
c.SF4
(Reference: Chang 10.71)
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