Chemistry 3: Chemical Bonding
A.
Bonding (8.1 to 8.4)
1. why bond?
a. attached (bonded) state is a lower energy state
b. complete valence shell = lower energy state
1. metals lose valence electrons (+ ion)
2. nonmetals with 5-7 valence electrons gain
electrons to complete valence shell (– ion)
3 nonmetals shares valence e- with nonmetals
a. 1-3 valence electrons share 1 for 1 
doubling valence number
b. 4-7 valence electrons share to fill s and p
orbitals (8 electrons = octet rule)
c. Lewis symbols
1. chemical symbol + dots for valence electrons
2. Na•, •Mg•, etc.
d. three major types of bonds
1. ionic bond: electrostatic attraction between
cations and anions
2. covalent bond: shared electrons between
non-metal atoms
3. metallic bond: metal atoms collectively share
valence electrons
2. ionic bonding
a. metal and nonmetal: Na(s) + ½ Cl2(g)  NaCl(s)
b. cations and anions: Na+(aq) + Cl-(aq)  NaCl(s)
c. cation listed first in formula and name (number of
ions is not indicated in name)
d. formula represents simplest ratio of ions to make
a neutral compound—not a molecule
e. bond strength—energy needed to break the bond:
E  Q1Q2/r, where Q1 and Q2 are ion charges and
r is distance between ions (lattice energy)
3. covalent bonding
a. bonding atoms' orbitals overlap, which maximizes
attraction between nuclei and bonding electrons
b. atoms share 2, 4 or 6 electrons
1. 2 (single), 4 (double), 6 (triple) bond
2. multiple bonds reduce bond distance
a. bond distance < sum of atomic radii
b. shorter bond distance = stronger bond
c. polar bond when electrons are not shared equally
1. electronegativity
a. measures atom's attraction for bonding
electron pair (higher # = stronger)
b. relative scale where period 2 elements
are 1.0 (Li) to 4.0 (F), with 0.5 intervals
1. noble gases are excluded
2. trend:
a. increase across period
b. decrease down groups
2. bond polarity
a. electronegativity difference between
atoms result in uneven sharing of
electrons  partially positive side, +,
and a partial negative side, –
b. notation
d.
c. measured as dipole moment
3. bond strength increases with polarity
naming binary molecules
1. two types of nonmetals covalently bonded
2. + atom is written first with element name
3. second element is given –ide ending
4. prefix used to indicate number of atoms
a. 1—mono, 2—di, 3—tri, 4—tetra, etc.
b. mono never used for first element
c. example: CO2 is carbon dioxide
5. common names: NH3 (ammonia), H2O2
(hydrogen peroxide) and H2O (water)
Name __________________________
B.
Lewis Structures (8.5 to 8.7)
1. shows the atoms in a molecule with their bonding and
non-bonding electron pairs
a. bonding electrons (– single, = double,  triple)
b. lone (non-bonding or unshared) electron pair (••)
2.


drawing Lewis structures with one central atom
count the total number of valence electrons (subtract
charge for ions)
CO2: 4 + 2(6) = 16
IF2–: 7 + 2(7) + 1 = 22
draw a skeleton structure
o first element in formula is central, except H
o single bonds to other atoms (max. 4)
[F–I–F]–
(ions are bracketed)
place electrons around each atom
o 8 total electrons
o except H, Be and B or when total number of
electrons is an odd number
O–C–O

.. .. ..
:O – C – O:
.. .. ..

3.
.. .. ..
[:F – I – F:]–
.. .. ..
count Lewis structure electrons (including bonding
electrons)
o if equal to valence electrons, stop
o if valence e- < Lewis e-, add additional bonds to
reduces # of electrons by 2's
o if valence e- > Lewis e-, add 2 or 4 electrons to
central atom (3rd period or higher); called
expanded octet
..
..
.. .... ..
O=C=O
[:F – I – F:]–
..
..
.. .. ..
added bonds
expanded octet
when more than one Lewis structure is possible use
formal charge to decide which is more likely
a. each atom is assigned its lone electrons plus half
the bonding electrons
b. formal charge = valence e- – assigned ec. preferred structure
1. atoms have formal charges closest to zero
2. negative formal charge reside on the more
electronegative atom (upper right most on the
periodic table)
d. example: NCS[::N=C=S::][:::N–CS:][:NC–S:::]5 4 6
5 4 6
5 4 6
valence e7 4 5
6 4 6
5 4 7
assigned e-2 0 +1
-1 0 0
0 0 -1
formal
 [::N=C=S::]- is preferred because formal charges
are closest to zero and negative charge is on the
nitrogen (higher electronegativity)
e. technique can produce erroneous structures
(experiments are required to determine actual
structure)
C.
VSEPR Model (9.1 to 9.3)
1. rules
a. maximum separation between electron pairs
b. atom positions define molecular geometry
c. lone electron pairs squeeze bond angle
(actual angle < ideal angle)
Electron
Domain
Molecular
Bond
–
:
Domains
Geometry
Geometry
Angle
2
2
0
3
0
180o
120o
3
4
2
1
4
0
3
1
2
2
5
0
4
1
109.5o
90o
120o
5
6
2.
3
2
2
3
6
0
5
1
4
2
90o
polar molecules
a. lone electron pairs distort symmetry except for
sp3d-linear and sp3d2-square planar
b. different perimeter atoms
c. polar interactions increase water solubility,
increase melting and boiling temperatures,
decrease evaporation (volatility)
D.
Valence-Bond Theory (9.4 to 9.5)
1. explains electron domain geometries in terms of
electron orbitals
2. atomic orbitals from bonding atoms merge, which
allows single electrons from each atomic orbital to
occupy overlapping area and simultaneously attract
both nuclei (i.e. H–H: overlap of 1s orbitals)
3. more complex molecules require a fusion of s and p
orbitals into equivalent (hybrid) orbitals
a. explains why covalent bonds around an atom are
all the same even if electrons were originally in
different shaped (s, p and d) atomic orbitals
b. 1 s + 1 p form 2 sp hybrids (2 electron domains)
ground state
excited state
hybridized state
                
s
p
s
p
sp
p
1 s + 2 p form 3 sp2 hybrids (3 electron domains)
c.
ground state
excited state
hybridized state
                
s
p
s
p
sp2
p
1 s + 3 p form 4 sp3 hybrids (4 electron domains)
d.
ground state
excited state
hybridized state
                
s
p
s
p
sp3
e.
4.
expanded octet hybridization
1. 1 s + 3 p + 1 d = 5 sp3d hybrids (5 domains)
2. 1 s + 3 p + 2 d = 6 sp3d2 hybrids (6 domains)
not all valence electrons enter hybrid orbitals
a. one electron pair per bond enters a hybrid orbital
1. sigma bond ()
2. electrons located between bonding atoms
b. lone pairs of electrons enter hybrid orbital
c. remaining bonding pairs of electrons from multiple
bonds remain in pure p orbitals
1. pi bond ()
2. electrons located above/below bonding atoms
d. example: ::O=C=O::
p
sp2
p
sp2 sp
p
sp2
sp
sp2
sp2
p
sp2
e.
p
p
p
p
 bond electrons can spread out across entire
molecule (delocalized)
1. NO3- has one  bond, which is shared evenly
and simultaneously between 3 O's
2. multiple Lewis structures show all possible
locations for  bonds = resonance forms
3.
bond order = sigma bond + share of  bonds
(each N–O bond has bond order = 1 1/3)
E.
Simple Organic Molecules—Hydrocarbons (25.1 to 25.6)
1. general properties
a. contain C and H
b. nonpolar, flammable (fuels)
2. formulas and names
a. number of carbons in parent chain
1
2
3
4
5
meth
eth
prop
but
pent
6
7
8
9
10
hex
hept
oct
non
dec
b. bond between carbons
1. alkanes (all single bonds) end in “ane”
2. alkenes (1 or more double bonds)
1 double end in “ene”, 2 double end in "diene"
3. alkynes (1 or more triple bonds) end in “yne”
4. cyclical
a. 3 to 6 carbon ring with single bonds
between carbons: prefix "cyclo"
b. 6 carbon ring with 3 shared  bonds:
benzene (called aromatic hydrocarbon)
c. branches
1. C-branches—“yl”
2. benzene branch—"phenyl"
3. location of branches
a. number of the parent carbon
b. lowest number possible
c. dash: # – word, comma: #, #
d. number of branches (2—di, 3—tri, etc.)
3. condensed structural formula
a. hydrogens are written after the carbon
b. branches are in parentheses after hydrogens
c. example: 4-ethyl-2-methyl-1-hexene
CH2C(CH3)CH2CH(C2H5)CH2CH3
d. semi-condensed (shows branches and bonds)
CH3
C2H5
|
|
CH2=C–CH2–CH–CH2–CH3
4. functional groups
a. dramatically modify properties of hydrocarbon
b. haloalkanes: halogen replaces one or more H
1. reduces reactivity (flammability)
2. named as a branch with an “o” ending
c. oxygen containing groups
1. hydroxyl group (C–OH)
a. water soluble
b. alcohols (-ol ending)
c. acids (-anoic acid ending)
2. carbonyl group (C=O)
a. aldehydes (-al ending)
b. ketones (-one ending)
c. esters (-oate ending )
3. ethers have C–O–C (-yl -yl ether ending)
4. increases polarity: C–OH > C=O > C–O–C
d. amines
1. replace H in ammonia with hydrocarbon
group = amine (CH3NH2 = methylamine)
2. when NH2 branches off hydrocarbon = amino
CH3CH(NH2)CH2CH3 (2-aminobutane)
3. weak bases (neutralize acids—absorb H+)
5. isomerism
a. structural isomers: same molecular formula,
different structure and name
1. move double/triple bond position
2. move branch
3. form cycloalkane from alkene
b. geometric isomers: same molecular formula and
structure, but different spatial arrangement
around the >C=C<, where carbons can't rotate
1. x>C=C<x : cis
2.
x>C=C< : trans
x
Experiments
1.
Isomerism Lab—Make all the isomers for the given formula using the molecular model kits and information provided.
a. C5H12 Draw the structure for each isomer named.
b.
c.
2.
pentane
C4H8 Write the names for each structure drawn.
C
\
C=C–C–C
C=C
\
C
2-methybutane
2,2-dimethylpropane
C
C
\
/
C=C
C
|
C=C–C
C–C
| |
C–C
C5H10 Draw the structure and name each isomer.
Aspirin Synthesis Lab (Wear Goggles)—Synthesize aspirin from acetic acid and salicylic acid and determine its purity.
a. In order to increase the yield of aspirin and to speed up the reaction, acetic anhydride is used, which is made by removing
water from two acetic acid molecules (dehydration synthesis). Highlight the H and OH that are removed from the two acetic
acid molecules.
CH3–C–OH
+
HO–C–CH3

CH3–C–O–C–CH3
+
H2O
||
||
||
||
O
O
O
O
acetic acid
acetic acid
acetic anhydride
water
b. Aspirin is an ester that is produced by dehydration synthesis. Highlight the H and OH from the carboxyl group (-COOH)
on acetic acid and the hydroxyl group (-OH) on salicylic acid.
CH3-C-OH + HO-C6H4-COOH
 CH3-C-O-C6H4COOH + H2O
||
||
O
O
acetic acid
salicylic acid
aspirin
water
Add 2.0 g of salicylic acid, 5.0 mL of acetic anhydride and 5 drops of 85 % H 3PO4 (catalyst) to a 125-ml Erlenmeyer flask. Place
the flask in a 600-mL beaker half filled with 75oC water and clamp in place. Heat for 15 minutes (stir the contents occasionally
with a stirring rod). Slowly add 2 ml of water to the flask to decompose any excess acetic anhydride. When the contents stop
smelling like vinegar, remove the flask from the water bath and add 20 ml of water. Put the flask in an ice bath for 5 minutes to
hasten crystallization and increase the yield. Collect the aspirin by filtering the cold mixture.
To test the purity of the aspirin, add 0.10 g aspirin to 5 mL of 95% ethanol in a 50-mL beaker and dissolve. Add 5 mL of 0.025
M Fe(NO3)3 in 0.5 M HCl and 40 ml of distilled water and stir. Fill a cuvette with the solution and measure the absorbance of the
solution at 525 nm.
c. Record the absorbance.
Salicylic acid forms a magenta complex with Fe3+ and the spectrophotometer measures its color intensity (absorbance). The
concentration of salicylic acid, an impurity, is directly proportional to the absorbance, A, according to Beer's law (A = abc),
where a (molar absorptivity) =
L/mol•cm and b (cuvette path length) = 1.0 cm.
d. Calculate c, the moles of salicylic acid per liter.
f. Calculate the moles of aspirin, C9H8O4, in 0.10 g.
e.
Calculate the moles of salicylic acid in 50 mL.
h.
Perform the following on the chemical equation for the synthesis of aspirin below.
(1) Cross out the bond on the acetic anhydride molecule that is broken during the reaction.
(2) Highlight the hydrogen atom on salicylic acid's hydroxyl group and draw a line showing where it attaches to acetic
anhydride to form acetic acid.
(3) Draw a line that bonds the remaining half of the acetic anhydride molecule to the salicylic acid molecule.
(4) Highlight the catalyst for the reaction.
acetic anhydride
salicylic acid
g.
Calculate the mole percent salicylic acid in the aspirin.
aspirin
acetic acid
5.
Practice Problems
1.
A. Bonding
Consider the main group elements (1-2, 13-18).
a. Record the number of valence electrons.
b. Draw the Lewis dot structure for element "X"
c. Record the ionic charge when forming ionic bond
d. Record the total number of electrons surrounding the
atom when forming covalent bond(s)
1
2
13
14
15
16
17
18
a
b
X
X
X
X
X
X
X
X
Electronegativity difference & bond strength
c
2.
Consider the following data for hydrogen halides.
Bond
Bond
HElectronegativity
Dipole
Strength Length
Halide
difference
moment
(kJ/mol)
(Å)
HF
1.9
1.82
436
0.92
HCl
0.9
1.08
431
1.27
HBr
0.7
0.82
366
1.41
0.4
0.44
299
1.61
HI
Indicate whether the following correlate directly or inversely.
Direct Inverse
Electronegativity difference & dipole moment
Dipole moment & bond strength
d
Illustrated below are four ions—A+, B+, C- and D-—showing
their relative ionic radii.
A+
B+
C–
D–
What combinations of ions are impossible?
Bond length & bond strength
6. Consider the following data for C-C bonds.
C-C bond
Single
Double
Triple
Bond Strength (kJ/mol)
348
614
839
How does bond strength correlate with the number of
shared electrons?
7.
What would have the greatest lattice energy?
What would have the least lattice energy?
3. The table lists the ionic radius (x 10-10 m) of common ions.
Li+ (0.68) Be2+ (0.31)
O2- (1.40) F- (1.33)
Na+ (0.97) Mg2+(0.66) Al3+ (0.51) S2- (1.84) Cl- (1.81)
a. Use the above information to estimate the relative
lattice energy for each ionic bond. E  Q1Q2/d
(Q1 and Q2 = ionic charge and d  (rcation + ranion)
Ionic Bond
Relative Lattice Energy
Complete the chart with the formula or name of the binary
molecule.
Formula
Name
N2O5
carbon tetrachloride
CO2
nitrogen monoxide
OF2
LiF
hydrogen bromide
MgO
8.
NaCl
B. Lewis Structures
Record the number of covalent bonds typically formed by the
main group elements.
1
2
13
14
15
16
17
18
Al2S3
b.
Ionic compounds melt when the temperature is high
enough to break the ionic bond. Rank the above
compounds in order of lowest to highest melting point.
c.
What is the relative lattice energy for the strongest
ionic bond formed from the ions listed in the table?
9.
In the Lewis structure shown below, A, D, E, Q, X and Z
represent non-metal elements in the first two rows of the
periodic table. Identify the elements.
A
4.
Use the electronegativity values to answer the questions.
H 2.1
Li 1.0 Be 1.5 B 2.0
C 2.5
N 3.0 O 3.5 F 4.0
Na 0.9 Mg 1.2 Al 1.5 Si 1.8 P 2.1
S 2.5 Cl 3.2
K 0.8 Ca 1.0 Ga 1.6 Ge 1.8 As 2.0 Se 2.4 Br 2.8
Rb 0.8 Sr 1.0 In 1.7 Sn 1.8 Sb 1.9 Te 2.1 I 2.5
a. What is the range of electronegativities?
metals
metalloids
nonmetals
b.
Rank the following bonds from most polar (1) to least
polar (6). Place + next to the atom with the lower
electronegativity.
N–S
O–S
F–S
P–S
S–S
Cl–S
D
E
Q
X
Z
10. Draw the Lewis structure for the diatomic molecules.
H2
N2
O2
F2
11. There are three ways to draw Lewis structures for NCO–
a. Calculate the formal charge for each version
Structure [:::N–CO:]–
[::N=C=O::]–
[:NC–O:::]–
Formal
Charge
b.
Which is the preferred structure? Give two reasons.
12. Draw Lewis structures for the following molecules where
the first atom listed is the central atom, unless indicated.
CH4
NH3
CO2
CH2O
POCl3 (lowest formal charge)
SCN–
(C is central atom)
SF6
CNO– (lowest
formal charge)
BCl3
BrF3
C. VSEPR Model
13. Make the following ideal molecules using clay and tooth
picks. Complete the table for each ideal molecule.
Domain
Bond
Molecular
(–)
(• •)
Polar?
Geometry
Angle
Geometry
2
0
3
0
2
1
4
0
3
1
2
2
5
0
4
1
3
2
2
3
6
0
5
1
4
2
14. Based on the Lewis structures from question 11, determine
electron domain geometry, molecular geometry, polarities and
bond angle (indicate if it is less than the ideal angle "<").
Molecular
Molecule Domain Geo
Polarity
Angle
Geo
CH4
H2O (O is the central atom)
HF (F is the central atom)
NH3
CO2
CH2O
SO3
N2O (lowest formal charge)
POCl3
CNO-
IF5
AsF5
SCNBCl3
SF4
SO2Cl2 (lowest formal charge)
SF6
BrF3
H2O
RnCl2
IF4–
HF
SO3
XeF4
NO2
N2O
IF5
AsF5
SF4
SO2Cl2
E. Simple Organic Molecules—Hydrocarbons
18. Draw the structure formulas, including hydrogen, and
names for the first six members of the alkane series.
RnCl2
IF4XeF4
NO2
D. Valence-Bond Theory
15. Based on the Lewis structures from question 11, determine
the number of  bonds, the number of  bonds, number of
lone electron pairs, hybridization around the central atom,
and the bond order for the perimeter atoms.
Lone
Bond


Molecule
Hybridization
Order
Bonds Bonds Pairs
19. Draw semi-condensed structural formulas for the following
organic compounds.
CH4
NH3
CO2
ethane
benzene
1,3-pentadiene
3-ethylbutyne
diethylamine
ethanoic acid (acetic acid)
2,2-dimethylpropane
methylbenzene
ethanol (ethyl alcohol)
2-propanol (isopropyl alcohol)
propanone (acetone)
diethyl ether (ether)
CH2O
POCl3
CNOSCNBCl3
SF6
BrF3
H2O
HF
SO3
N2O
IF5
AsF5
SF4
SO2Cl2
RnCl2
IF4XeF4
NO2
16. Label the hybridization for each carbon atom.
17. Draw the resonance structures for the following molecules.
Molecule
Resonance Structures
CO3
2-
methanal (formaldehyde)
methyl propanoate
20. Name the following organic molecules from their semicondensed structural formulas.
CH3
CH3
|
|
CH3 – C – CH2 – CH – CH3
|
CH3
SO2
|
C2H5
H2C–CH2
/
\
Cl-CH CH2
\
/
H2C–CH2
HC  C – CH2Cl
CH3 – CH2 – C – OH
||
O
CH3 – O – C – CH2 – CH3
||
O
F
F
F
|
|
|
CH2 – CH – CH2
..
CH3 – N – CH3
|
CH3
Practice Multiple Choice
C2H5
|
CH3 – CH2 – C – CH2 – CH3
|
C2H5
CH3 CH3
\
/
C=C
/
\
H
H
HO – CH2 – CH2 – CH3
Briefly explain why the answer is correct in the space provided.
1. Types of hybridization exhibited by the C atoms in
propene, CH3CHCH2 include which of the following?
I. sp
II. sp2
III. sp3
(A) I only
(B) II only (C) III only (D) II and III
2.
Which molecule contains 1 sigma () and 2 pi () bonds?
(A) H2
(B) F2
(C) N2
(D) O2
3.
Which molecule has the shortest bond length?
(A) N2
(B) O2
(C) Cl2
(D) Br2
4.
Which molecule has only one unshared pair of valence
electrons?
(A) Cl2
(B) NH3
(C) H2O2
(D) N2
5.
The electron pairs in a molecule where the central atom
exhibits sp3d2 hybrid orbitals are directed toward the
corners of
(A) a tetrahedron
(B) a square pyramid
(C) a trigonal bipyramid (D) an octahedron
6.
The SbCl5 molecule has trigonal bipyramid structure.
Therefore, the hybridization of Sb orbitals should be
(A) sp2
(B) sp3
(C) dsp2
(D) dsp3
7.
For which molecule are resonance structures necessary to
describe the bonding satisfactorily?
(A) H2S
(B) SO2
(C) CO2
(D) OF2
8.
Which molecules have planar configurations?
I. BCl3
II. CHCl3
III. NCl3
(A) I only
(B) II only (C) III only (D) II and III
9.
CCl4, CO2, PCl3, PCl5, SF6
Which does NOT describe any of the molecules above?
(A) Linear
(B) Octahedral
(C) Square planar
(D) Tetrahedral
CH3 – C – CH2 – CH3
||
O
CH3 – O – C3H7
CH3 – CH2 – CH2 – CH
||
O
OH
|
CH3 – C – CH3
|
OH
CH3 – CH2 – CH – CH3
|
NH2
H2N – CH2 – C – OH
||
O
21. Draw the structures and name six isomers of C4H10O,
which include 4 alcohols and 2 ethers.
10. The geometry of the SO3 molecule is best described as
(A) trigonal planar
(B) trigonal pyramidal
(C) square pyramidal
(D) bent
11. Pi () bonding occurs in each of the following EXCEPT
(A) CO2
(B) C2H4
(C) CN(D) CH4
12. According to the VSEPR model, the progressive decrease
in the bond angles in the series of molecules CH4, NH3,
and H2O is best accounted for by the
(A) increasing strength of the bonds
(B) decreasing size of the central atom
(C) increasing electronegativity of the central atom
(D) increasing number of unshared pairs of electrons
Questions 13-15 refer to the following molecules.
(A) PH3
(B) H2O
(C) CH4
(D) C2H4
13. The molecule with only one double bond
14. The molecule with the largest dipole moment
15. The molecule that has trigonal pyramidal geometry
16. Which molecule has a zero dipole moment?
(A) HCN
(B) NH3
(C) SO2
(D) PF5
17. Which molecule has the largest dipole moment?
(A) CO
(B) CO2
(C) O2
(D) HF
18. Which molecule has a dipole moment of zero?
(A) C6H6 (benzene)
(B) NO
(C) SO2
(D) NH3
19. Which pair of atoms should form the most polar bond?
(A) F and B
(B) C and O
(C) F and O
(D) N and F
20. Which pair of ions should have the highest lattice energy?
(A) Na+ and Br(B) Li+ and F(C) Cs+ and F(D) Li+ and O2-
28. Which molecule is NOT polar?
(A) H2O
(B) CO2
(C) NO2
(D) SO2
29. Which species has sp2 hybridization for the central atom?
(A) C2H2
(B) SO32(C) O3
(D) BrI3
30. In which species is the F-X-F bond angle the smallest?
(A) NF3
(B) BF3
(C) CF4
(D) BrF3
31. For ClF3, the electron domain geometry of Cl and the
molecular geometry are, respectively,
(A) trigonal planar and trigonal planar.
(B) trigonal planar and trigonal bipyramidal.
(C) trigonal bipyramidal and trigonal planar.
(D) trigonal bipyramidal and T -shaped.
32. The size of the H-N-H bond angles of the following species
increases in which order?
(A) NH3 < NH4+ < NH2(B) NH3 < NH2- < NH4+
(C) NH2- < NH3 < NH4+
(D) NH2- < NH4+ < NH3
33. What is the molecular geometry and polarity of BF3?
(A) trigonal pyramidal and polar
(B) trigonal pyramidal and nonpolar
(C) trigonal planar and polar
(D) trigonal planar and nonpolar
21. Which compound has the greatest lattice energy?
(A) BaO
(B) MgO
(C) CaS
(D) MgS
34. Which set does not contain a linear species?
(A) CO2, SO2, NO2
(B) H2O, HCN, BeI2
(C) OCN-, C2H2, OF2
(D) H2S, CIO2-, NH2-
22. Which molecule has the weakest bond?
(A) CO
(B) O2
(C) Cl2
(D) N2
35. The hybrid orbitals of nitrogen in N2O4 are
(A) sp
(B) sp2
(C) sp3
(D) sp3d
23. How are the bonding pairs arranged in the best Lewis
structure for ozone, O3?
(A) O–O–O (B) O=O–O (C) OO–O (D) O=O=O
36. How many sigma and how many pi bonds are in
CH2=CH–CH2–C–CH3?
||
O
(A) 5 sigma and 2 pi.
(B) 8 sigma and 4 pi.
(C) 11 sigma and 2 pi.
(D) 13 sigma and 2 pi.
24. Which species has the shortest bond length?
(A) CN(B) O2
(C) SO2
(D) SO3
37. What is the best estimate of the H-O-H bond angle in H3O+?
(A) 109.5o (B) 107o
(C) 90o
(D) 120o
25. Which species has a valid non-octet Lewis structure?
(A) GeCl4 (B) SiF4
(C) NH4+
(D) SeCl4
26. The Lewis structure for SeS2 with zero formal charge has
(A) 2 bonding pairs and 7 nonbonding pairs of electrons.
(B) 2 bonding pairs and 6 nonbonding pairs of electrons.
(C) 3 bonding pairs and 6 nonbonding pairs of electrons.
(D) 4 bonding pairs and 5 nonbonding pairs of electrons.
38. Which of the following pairs of compounds are isomers?
(A) CH3–CH2–CH2–CH3 and CH3–CH(CH3)–CH3
(B) CH3–CH(CH3)–CH3 and CH3–CH3–C=CH2
(C) CH3–O–CH3 and CH3–CO–CH3
(D) CH3–OH and CH3–CH2–OH
39.
27. Which molecular shape cannot exhibit geometric isomerism?
(A) tetrahedron
(B) square planar
(C) trigonal bipyramid
(D) octahedron
CH3–CH2–CH2Br
Which of the following structural formulas represents an
isomer of the compound that has the structural formula
represented above?
(A) CH2Br–CH2–CH2
(B) CH3–CHBr–CH3
(C) CH2Br–CH2–CH2Br (D) CH3–CH2–CH2–CH2Br
c. Identify the electron-domain and molecular geometries.
Electron-domain geometry
Molecular geometry
Practice Free Response
1.
2.
Rank the oxides in order of greatest lattice energies (1) to
least (3), without looking up any values. Explain.
BaO
CaO
MgO
There are several oxides of nitrogen; among the more
common are N2O, NO2 and NO3-.
a. Draw the Lewis structures of these molecules.
d.
5.
Consider the ion SF5a. Draw a Lewis structure.
b.
b.
3.
N2O
NO2
NO3Which of these molecules "violate" the octet rule?
Explain.
c.
Draw resonance structures of N2O.
d.
For each resonance structure from question 2c,
calculate the formal charge and evaluate which
structure is most likely. Explain.
e.
Which side of the N–O bond is +? Explain.
f.
Rank the strength of the N–O bond in order of
strongest (1) to weakest (3). Explain your answer.
N2O
NO2
NO3-
Complete the chart for SeF2, SeF4, and SeF6.
SeF2
SeF4
6.
Two Lewis structures can be drawn for the OPF3 molecule.
Structure 1
Structure 2
..
:O:
:O:
.. | ..
.. || ..
:F–P–F:
:F–P–F:
.. | ..
.. | ..
:F:
:F:
..
..
Which Lewis structures best represents a molecule of
OPF3? Justify your answer in terms of formal charge.
7.
a.
Draw the condensed structural formula and name four
structural isomers of C4H9Cl.
Formula
Name
b.
Draw the structural formula and name two geometric
isomers of C2H2Cl2.
Formula
Name
Lewis
Structure
b.
Identify the type of hybridization exhibited by sulfur.
Identify the type of hybridization exhibited by sulfur.
c. Identify the electron-domain and molecular geometries.
Electron domain geometry
Molecular geometry
SeF6
SeHybridization
Domain
Geometry
Molecular
Geometry
Ideal Bond
Angle
Polarity
4. Consider the ion SF3+.
a. Draw a Lewis structure.
Predict whether the F-S-F bond angle is equal to,
greater than or less than 109.5°. Explain
8.
Compounds of Xe and F form molecules where the
hybridization of Xe is sp3d and sp3d2. Write the formula and
draw the Lewis structure for the two molecules.
sp3d
sp3d2