SCE 3401
Inorg. Chem.
Key of Problem Set III
The Shape of Molecules
1) Draw the Lewis structure with lowest formal charges (if any), and determine the
charge of each atom in; (a) OCS; (b) NO. (c) IF5 (d) ClO–
2) These species do not obey the octet rule. Draw a Lewis structure for each of the
following: (a) BrO4– (b) HNO3 (c) ICl2– (d) SbF4–
3) Name the shape and give the AXmEn classification and ideal bond angle(s) for each of
the following general molecules:
4) Name the shape and give the AXmEn classification and ideal bond angle(s) for each of
the following general molecules:
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Inorg. Chem.
5) Determine the electron-group arrangement, molecular shape, and ideal bond angle(s)
for each of the following:
(a) O3
(b) H3O+
(c) NF3
2–
(d) SO4
(e) NO2–
(f) PH3
(g) SO3
(h) N2O
(i) CH2Cl2
Problem set 3 | page 2
SCE 3401
Inorg. Chem.
Problem set 3 | page 3
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Inorg. Chem.
6) Arrange the following AFn species in order of increasing F¬A¬F bond angles: BF3,
BeF2, CF4, NF3, OF2.
7) State an ideal value for each of the bond angles in each molecule.
(d)
8) Which molecule in each pair has the greater dipole moment? Give the reason for your
choice.
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SCE 3401
Inorg. Chem.
9) There are three different dichloroethylenes (molecular formula C2H2Cl2), which we can
designate X, Y, and Z. Compound X has no dipole moment, but compound Z does.
Compounds X and Z each combine with hydrogen to give the same product:
C2H2Cl2 (X or Z) + H2 → ClCH2—CH2Cl
What are the structures of X, Y, and Z? Would you expect compound Y to have a
dipole moment?
10) Chloral, Cl3C—CH=O, reacts with water to form the sedative and hypnotic agent
chloral hydrate, Cl3C—CH(OH)2. Draw Lewis structures for these substances, and
describe the change in molecular shape, if any, that occurs around each of the carbon
atoms during the reaction
Problem set 3 | page 5
SCE 3401
Inorg. Chem.
11) When XeF4 is reacted with a solution of water in CH3CN solvent, the product
F2OXeN≡CCH3 is formed. Applying a vacuum to crystals of this product resulted in
slow removal of CH3CN:
F2OXeN≡CCH3 → XeOF2 + CH3CN
Propose the structure for XeOF2 and F2OXeN≡CCH3
(Brock, D. S.; Bilir, V.; Mercier, H. P. A.; Schrobilgen, G. J. J. Am. Chem. Soc. 2007, 129, 3598.)
12) Which member of each pair of compounds forms intermolecular H bonds? Draw the
H-bonded structures in each case:
Problem set 3 | page 6
SCE 3401
Inorg. Chem.
(c)
(d)
13) What type(s) of the strongest intermolecular forces exist between each of the
following molecules?
(a) CH3OH (b) CCl4
(c) Cl2
(d) H3PO4
(e) SO2
(f) MgCl2
(g) CH3Cl
(h) CH3CH3 (i) NH3
(j) Kr
(k) BrF
(l) H2SO4
Solution
Dispersion forces are the only forces between nonpolar substances; dipole-dipole forces
exist between polar substances. Hydrogen bonds only occur in substances in which
hydrogen is directly bonded to either oxygen, nitrogen or fluorine.
(a) Hydrogen bonding will be the strongest force between methanol molecules since
they contain O–H bonds. Dipole-dipole and dispersion forces also exist.
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Inorg. Chem.
(b) Dispersion forces are the only forces between nonpolar carbon tetrachloride
molecules and, thus, are the strongest forces.
(c) Dispersion forces are the only forces between nonpolar chlorine molecules and,
thus, are the strongest forces.
(d) Hydrogen bonding
(e) Dipole-dipole interactions
(f) Ionic bonds
(g) Dipole-dipole interactions
(h) Dispersion forces
(i) Hydrogen bonding
(j) Dispersion forces
(k) Dipole-dipole interactions
(l) Hydrogen bonding
14) Which substance has the lower boiling point? Explain.
(a) LiCl or HCl
(b) NH3 or PH3
(c) Xe or I2
(d) NaBr or PBr3
(e) H2O or HBr
Solution
The weaker the intermolecular forces, the lower the boiling point.
a) HCl would have a lower boiling point than LiCl because the dipole-dipole
intermolecular forces between hydrogen chloride molecules in the liquid phase are
weaker than the significantly stronger ionic forces holding the ions in lithium chloride
together.
b) PH3, would have a lower boiling point than NH3 because the intermolecular forces
in PH3 are weaker than those in NH3. Hydrogen bonding exists between NH3 molecules
but weaker dipole-dipole forces hold PH3 molecules together.
c) Xe would have a lower boiling point than iodine. Both are nonpolar with dispersion
forces, but the forces between xenon atoms would be weaker than those between iodine
molecules are since the iodine molecules are more polarizable because of their larger
size.
d) PBr3, the dipole–dipole forces of phosphorous tribromide are weaker than the ionic
forces of sodium bromide.
e) HBr, the dipole-dipole forces of hydrogen bromide are weaker than the hydrogen
bonding forces of water.
f)
the cyclic molecule, cyclobutane, has less surface area exposed, so its
dispersion forces are less than the straight chain molecule, CH3CH2CH2CH3
15) Which substance has the higher boiling point? Explain.
(a) CH3CH2OH or CH3CH2CH3
(b) NO or N2 (c) H2S or H2Te
(d) FNO or ClNO
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Inorg. Chem.
a) CH3CH2OH, hydrogen bonding (CH3CH2OH) versus dispersion (CH3CH2CH3)
b) NO, dipole-dipole (NO) versus dispersion (N2)
c) H2Te, the larger molecule has larger dispersion forces.
d) FNO, greater polarity in FNO versus ClNO
e)
This molecule has dipole-dipole forces since the two C–F bonds do not cancel and the molecule
is polar. The other molecule has only dispersion forces since the two C–F bonds do cancel so
that the molecule is nonpolar.
Problem set 3 | page 9