AP Chemistry 3: Chemical Bonding Name __________________________

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
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

c. measured as dipole moment

3. bond strength increases with polarity
d. 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)
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)

O–C–O [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: [: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

3. 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 e-
c. 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:::]-
valence e- 5 4 6 5 4 6 5 4 6
assigned e- 7 4 5 6 4 6 5 4 7
formal -2 0 +1 -1 0 0 0 0 -1
 [::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 180o

3 0
3 120o

2 1

4 0

4 3 1 109.5o

2 2

5 0

4 1

90o
5
120o

3 2

2 3

6 0

6 5 1 90o

4 2

2. 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 3. bond order = sigma bond + share of  bonds
both nuclei (i.e. H–H: overlap of 1s orbitals) (each N–O bond has bond order = 1 1/3)
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

c. 1 s + 2 p form 3 sp2 hybrids (3 electron domains)

ground state excited state

hybridized state
         
s p s p sp2 p

d. 1 s + 3 p form 4 sp3 hybrids (4 electron domains)

ground state excited state hybridized

state
            
s p s p sp3

e. 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)
4. 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 p
sp2 p p sp 2
2 2
sp sp sp sp
p p sp 2
sp2 p p

e.  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
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 C 2 H5
| |
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
x x
1. >C=C< : cis
x
2. >C=C<x : trans
 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.

pentane 2-methybutane 2,2-dimethylpropane

b. C4H8 Write the names for each structure drawn.
C
C C C
\ | C–C
\ / | |
C=C–C–C C=C C=C–C
C=C C–C
\
C

c. C5H10 Draw the structure and name each isomer.

2. 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 + H 2O
|| || || ||
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) = 1000 mol/(L•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. g. Calculate the mole percent salicylic acid in the aspirin.

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 aspirin acetic acid

Practice Problems a. What is the range of electronegativities?

metals metalloids nonmetals
A. Bonding
1. Consider the main group elements (1-2, 13-18).
a. Record the number of valence electrons. b. Rank the following bonds from most polar (1) to least
b. Draw the Lewis dot structure for element "X" polar (6). Place + next to the atom with the lower
c. Record the ionic charge when forming ionic bond electronegativity.
d. Record the total number of electrons surrounding the N–S O–S F–S
atom when forming covalent bond(s)
1 2 13 14 15 16 17 18
a P–S S–S Cl–S

b X X X X X X X X
5. Consider the following data for hydrogen halides.
c Bond Bond
H- Electronegativity Dipole
d Strength Length
Halide difference moment
(kJ/mol) (Å)
2. Illustrated below are four ions—A+, B+, C- and D-—showing
their relative ionic radii. HF 1.9 1.82 436 0.92
HCl 0.9 1.08 431 1.27
HBr 0.7 0.82 366 1.41
A+ B+ C– D– HI 0.4 0.44 299 1.61
Indicate whether the following correlate directly
or inversely.
What combinations of ions are impossible?
Invers
Direct
What would have the greatest lattice energy? e
What would have the least lattice energy? Electronegativity difference & dipole
moment
3. The table lists the ionic radius (x 10-10 m) of common ions.
Electronegativity difference & bond
Li+ (0.68) Be2+ (0.31) O2- (1.40) F- (1.33) strength
Na+ (0.97) Mg2+(0.66) Al3+ (0.51) S2- (1.84) Cl- (1.81)
Dipole moment & bond strength
a. Use the above information to estimate the relative
lattice energy for each ionic bond. E  Q1Q2/d Bond length & bond strength
(Q1 and Q2 = ionic charge and d  (rcation + ranion) 6. Consider the following data for C-C bonds.
Ionic Bond Relative Lattice Energy C-C bond Single Double Triple
Bond Strength (kJ/mol) 348 614 839
LiF
How does bond strength correlate with the
MgO number of shared electrons?
NaCl
7. Complete the chart with the formula or name of the binary
Al2S3 molecule.
Formula Name
b. Ionic compounds melt when the temperature
N2 O 5
is high enough to break the ionic bond. Rank
the above compounds in order of lowest to carbon tetrachloride
highest melting point. CO2

nitrogen monoxide
c. What is the relative lattice energy for the strongest
ionic bond formed from the ions listed in the table? OF2

hydrogen bromide
4. Use the electronegativity values to answer the
questions. HBr(aq)
H 2.1 B. Lewis Structures
Li 1.0 Be 1.5 B 2.0 C 2.5 N 3.0 O 3.5 F 4.0 8. Record the number of covalent bonds
typically
Na 0.9 Mg 1.2 Al 1.5 Si 1.8 P 2.1 S 2.5 Cl 3.2 formed by the main group elements.
K 0.8 Ca 1.0 Ga 1.6 Ge 1.8 As 2.0 Se 2.4 Br 2.8 1 2 13 14 15 16 17 18
Rb 0.8 Sr 1.0 In 1.7 Sn 1.8 Sb 1.9 Te 2.1 I 2.5
9. In the Lewis structure shown below, A, D, E, Q, X and Z 12. Draw Lewis structures for the following molecules where
represent non-metal elements in the first two rows of the the first atom listed is the central atom, unless
periodic table. Identify the elements.
indicated.
CH4 NH3

A D E Q X Z

10. Draw the Lewis structure for the diatomic CO2 CH2O
molecules.
H2 N2 O2 F2

POCl3 (lowest formal CNO– (lowest formal charge)

11. There are three ways to draw Lewis structures for NCO– charge)
a. Calculate the formal charge for each version
Structure [:::N–CO:]– [::N=C=O::]– [:NC–O:::]–

Formal
Charge SCN– (C is central atom) BCl3
b. Which is the preferred structure? Give two reasons.

SF6 BrF3

H2O (O is the central HF (F is the central atom)

atom)

SO3 N2O (lowest formal

charge)

IF5 AsF5

SF4 SO2Cl2 (lowest formal charge)

RnCl2 IF4–

XeF4 NO2
 C. VSEPR Model
SF4
and tooth
13. Make the following ideal molecules using clay
picks. Complete the table for each ideal SO2Cl2
molecule.
Bond
Domain Molecular RnCl2
(–) (• • ) Angl Polar?
Geometry Geometry
e
2 0 IF4-
3 0
XeF4
3 1
4 0
NO2
3 1
D. Valence-Bond Theory
4 2 15. Based on the Lewis structures from question 11, determine
5 0 the number of  bonds, the number of  bonds, number of
4 1 lone electron pairs, hybridization around the central atom,
and the bond order for the perimeter atoms.
3 2   Lone Bond
Molecule Hybridization
2 3 Bonds Bonds Pairs Order
6 0 CH4
5 1 NH3
4 2 CO2
14. Based on the Lewis structures from question 11, determine CH2O
electron domain geometry, molecular geometry, polarities and POCl3
bond angle (indicate if it is less than the ideal angle "<").
CNO-
Molecul Molecular
Domain Geo Polarity Angle SCN-
e Geo
BCl3
CH4
SF6
NH3 BrF3
H2 O
CO2 HF
SO3
CH2O
N2 O
IF5
POCl3
AsF5
CNO - SF4
SO2Cl2
SCN- RnCl2
IF4-
BCl3
XeF4
NO2
SF6
16. Label the hybridization for each carbon atom.

BrF3

H2 O

HF
17. Draw the resonance structures for the following molecules.
SO3 Molecule Resonance Structures

N2 O CO32-

IF5
SO2
AsF5
 E. Simple Organic Molecules—Hydrocarbons Cl-CH CH2
18. Draw the structure formulas, including hydrogen, and \ /
names for the first six members of the alkane H2C–CH2
series.
CH3 – CH2 – C – OH CH3 – O – C – CH2 – CH3
|| ||
O O

..
F F F
| | | CH3 – N – CH3
CH2 – CH – CH2 |
CH3

19. Draw semi-condensed structural formulas for the following

C2 H5
organic compounds. |
CH3 – CH2 – C – CH2 – CH3
|
C2 H5

ethane benzene CH3 CH3

\ /
C=C CH3 – C – CH2 – CH3
||
/ \ O
H H

1,3-pentadiene 3-ethylbutyne

HO – CH2 – CH2 – CH3 CH3 – O – C3H7

diethylamine ethanoic acid (acetic acid) OH

CH3 – CH2 – CH2 – CH |
|| CH3 – C – CH3
O |
OH

2,2-dimethylpropane methylbenzene
CH3 – CH2 – CH – CH3 H2N – CH2 – C – OH
| ||
NH2 O

ethanol (ethyl alcohol) 2-propanol (isopropyl alcohol) 21. Draw the structures and name six isomers of C4H10O,
which include 4 alcohols and 2 ethers.

propanone (acetone) diethyl ether (ether)

methanal (formaldehyde) methyl propanoate

20. Name the following organic molecules from their semi-
condensed structural formulas.
CH3 CH3
| |
CH3 – C – CH2 – CH – CH3
|
CH3 |
C2 H5

H2C–CH2 HC  C – CH2Cl
Summary
/ \ Bonding
 Bonds are classified into three broad groups: ionic bonds are Certain molecular shapes have cancelling bond dipoles,
the result of electrostatic forces between cations and anions; producing a nonpolar molecule, which is one whose dipole
covalent bonds form when electrons are shared between non- moment is zero. In other shapes the bond dipoles do not cancel
metal atoms; and metallic bonds, which bind metal cations with and the molecule is polar (a nonzero dipole moment). In general
mutually shared valence electrons. non-bonding pairs of electrons around the central atom produce
Bonds involve the interaction of valence electrons, which are polar molecules.
represented by electron-dot symbols, called Lewis symbols. The Valence-Bond Theory
tendencies of atoms to gain, lose, or share their valence Valence-bond theory is an extension of Lewis's notion of
electrons often follow the octet rule, which can be viewed as an electron-pair bonds. In valence-bond theory, covalent bonds are
attempt by atoms to achieve a noble gas electron configuration. formed when atomic orbitals on neighboring atoms overlap. The
The strength of the electrostatic attractions between ions is bonding electrons occupy the overlap region and are attracted to
measured by the lattice energy, which increases with ionic both nuclei simultaneously, which bonds the atoms together.
charge and decreases with distance between ions. To extend valence-bond theory to polyatomic molecules, s,
Electronegativity measures the ability of an atom to attract p, and sometimes d orbitals are blended to form hybrid orbitals,
electrons in a covalent bond. Electronegativity generally which overlap with orbitals on another atom to make a bond.
increases from left to right in the periodic table and decreases Hybrid orbitals also hold non-bonding pairs of electrons. A
down a column. The difference in atoms' electronegativities is particular mode of hybridization can be associated with each of
used to determine the polarity of a covalent bond; the greater the the five common electron-domain geometries (linear = sp;
difference, the more polar the bond. A polar molecule has a trigonal planar = sp2; tetrahedral = sp3; trigonal bipyramidal =
positive side (+) and a negative side (–). The separation of sp3d; and octahedral = sp3d2).
charge produces a dipole, the magnitude of which is given by the Covalent bonds formed between hybridized electrons are
dipole moment. Polar bonds are stronger and shorter than non- called sigma () bonds, where the electron density lies along the
polar bonds. line connecting the atoms. Bonds that form between non-
Bond strength and length is also affected by the number of hybridized p orbitals are called pi () bonds. A double bond
shared electrons. Sharing of one pair of electrons produces a consists of one  bond and one  bond and a triple bond consists
single bond; whereas the sharing of two or three pairs of of one  and two  bonds.
electrons produces double or triple bonds, respectively. Multiple Sometimes a  bond can be placed in more than one
bonds are stronger and shorter than single bonds. location. In such situations, we describe the molecule by using
The procedures used for naming two-element, binary, two or more resonance structures. The molecule is envisioned as
molecular compounds follow the rules below. a blend of these multiple resonance structures and the  bonds
1. The lower electronegative element is written first in the are delocalized; that is, spread among several atoms. The bond
formula and named as an element. order value represents the actual bond strength and is sum of the
2. The name of the second element is given an –ide  bond plus a share of the  bond(s).
ending. Hydrocarbons
3. Prefixes are used to indicate the number of atoms of Carbon molecules (except CO, CO2) are called organic.
each element; mono is not used with the first. Hydrocarbons are organic molecules that contain mostly carbon
Lewis Structures and hydrogen. The four groups of hydrocarbons are alkanes,
Electron distribution in molecules is shown with Lewis alkenes, alkynes, and aromatic. The naming of hydrocarbons is
structures, which indicate how many valence electrons are based on the longest continuous chain of carbon atoms in the
involved in forming bonds and how many remain as unshared structure. The locations of alkyl groups, which branch off the
electron pairs. If we know which atoms are connected to one chain, are specified by numbering along the carbon chain. Ring
another, we can draw Lewis structures for molecules and ions by structures have the prefix cyclo. The names of alkenes and
a simple procedure, where eight electrons are placed around alkynes are based on the longest continuous chain of carbon
each atom. When there are too few valence electrons, then it will atoms that contains the multiple bond, and the location of the
be necessary to add double or triple bonds. When there are too multiple bond is specified by a numerical prefix.
many valence electrons (and the central atom has at least third The chemistry of an organic compound is dominated by the
energy level electrons), then it will be necessary to place presence of the functional group.
additional electrons (up to 10 or 12) around the central atom; Isomers are substances that possess the same molecular
forming an expanded octet. When the total number of valence formula, but differ in the arrangements of atoms. In structural
electrons is an odd number, then it will be necessary to place isomers the bonding arrangements differ. Different isomers are
seven electrons around the atom with the odd number of valence given different names. Alkenes exhibit not only structural
electrons. isomerism but geometric isomerism (cis-trans) as well. In
When there are multiple valid Lewis structures for a geometric isomers the bonds are the same, but the molecules
molecule or ion, we can determine which is most likely by have different geometries. Geometric isomerism is possible in
assigning a formal charge to each atom, which is the sum of half alkenes because rotation about the C=C bond is restricted.
the bonding electrons and all the unshared electrons. Most
acceptable Lewis structures will have low formal charges with
any negative formal charge on the more electronegative atom. Practice Multiple Choice
VSEPR Model Briefly explain why the answer is correct in the space provided.
The valence-shell electron-pair repulsion (VSEPR) model 1. Types of hybridization exhibited by the C atoms in
rationalizes molecular geometries based on the repulsions propene, CH3CHCH2 include which of the following?
between electron domains, which are regions about a central I. sp II. sp2 III. sp3
atom where electrons are likely to be found. Pairs of electrons, (A) I only (B) II only (C) III only (D) II and III
bonding and non-bonding, create domains around an atom,
which are as far apart as possible. Electron domains from non-
bonding pairs exert slightly greater repulsions, which leads to 2. Which molecule contains 1 sigma () and 2 pi () bonds?
smaller bond angles than idealized values. The arrangement of (A) H2 (B) F2 (C) N2 (D) O2
electron domains around a central atom is called the electron
domain geometry; the arrangement of atoms is called the
molecular geometry. 3. Which molecule has the shortest bond length?
 (A) N2 (B) O2 (C) Cl2 (D) Br2 (A) HCN (B) NH3 (C) SO2 (D) PF 5

4. Which molecule has only one unshared pair of valence 17. Which molecule has the largest dipole moment?
electrons? (A) CO (B) CO2 (C) O2 (D) HF
(A) Cl2 (B) NH3 (C) H2O2 (D) N2

18. Which molecule has a dipole moment of zero?

5. The electron pairs in a molecule where the central atom (A) C6H6 (benzene) (B) NO
exhibits sp3d2 hybrid orbitals are directed toward the 3
(C) SO2 (D) NH
corners of
(A) a tetrahedron (B) a square pyramid
(C) a trigonal bipyramid (D) an octahedron 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
6. The SbCl5 molecule has trigonal bipyramid structure.
Therefore, the hybridization of Sb orbitals should be
(A) sp2 (B) sp3 (C) dsp2 (D) dsp
3
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-
7. For which molecule are resonance structures necessary to
describe the bonding satisfactorily?
(A) H2S (B) SO2 (C) CO2 (D) OF2
21. Which compound has the greatest lattice energy?
(A) BaO (B) MgO (C) CaS (D) MgS
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 22. Which molecule has the weakest bond?
(A) CO (B) O2 (C) Cl2 (D) N2
9. CCl4, CO2, PCl3, PCl5, SF6
Which does NOT describe any of the molecules above? 23. How are the bonding pairs arranged in the best Lewis
(A) Linear (B) Octahedral structure for ozone, O3?
(C) Square planar (D) Tetrahedral (A) O–O–O (B) O=O–O (C) OO–O (D)
O=O=O
10. The geometry of the SO3 molecule is best described as
(A) trigonal planar (B) trigonal pyramidal
(C) square pyramidal (D) bent 24. Which species has the shortest bond length?
(A) CN- (B) O2 (C) SO2 (D) SO 3
11. Pi () bonding occurs in each of the following EXCEPT
(A) CO2 (B) C2H4 (C) CN- (D) CH4 25. Which species has a valid non-octet Lewis structure?
(A) GeCl4 (B) SiF4 (C) NH4+ (D) SeCl 4
12. According to the VSEPR model, the progressive decrease
in the bond angles in the series of molecules CH4, NH3, 26. The Lewis structure for SeS2 with zero formal charge has
and H2O is best accounted for by the (A) 2 bonding pairs and 7 nonbonding pairs of electrons.
(A) increasing strength of the bonds (B) 2 bonding pairs and 6 nonbonding pairs of electrons.
(B) decreasing size of the central atom (C) 3 bonding pairs and 6 nonbonding pairs of electrons.
(C) increasing electronegativity of the central atom
(D) 4 bonding pairs and 5 nonbonding pairs of
(D) increasing number of unshared pairs of
electrons.
electrons

27. Which molecular shape cannot exhibit geometric isomerism?

Questions 13-15 refer to the following molecules. (A) tetrahedron (B) square planar
(A) PH3 (B) H2O (C) CH4 (D) C2H4 (C) trigonal bipyramid (D) octahedron
13. The molecule with only one double bond
28. Which molecule is NOT polar?
14. The molecule with the largest dipole moment (A) H2O (B) CO2 (C) NO2 (D) SO 2

15. The molecule that has trigonal pyramidal geometry 29. Which species has sp2 hybridization for the central atom?
(A) C2H2 (B) SO32- (C) O3 (D) BrI 3
16. Which molecule has a zero dipole moment?
 1. Rank the oxides in order of greatest lattice energies (1) to
least (3), without looking up any values. Explain.
30. In which species is the F-X-F bond angle the smallest? BaO CaO MgO
(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. 2. There are several oxides of nitrogen; among the more
common are N2O, NO2 and NO3-.
(D) trigonal bipyramidal and T -shaped. a. Draw the Lewis structures of these
molecules.
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+ < NH 3
N2 O NO2 NO3-
b. Which of these molecules "violate" the octet rule?
33. What is the molecular geometry and polarity of BF3? Explain.
(A) trigonal pyramidal and polar
(B) trigonal pyramidal and nonpolar
(C) trigonal planar and polar c. Draw resonance structures of N2O.
(D) trigonal planar and nonpolar

d. For each resonance structure from question 2c,

34. Which set does not contain a linear species?
calculate the formal charge and evaluate which
(A) CO2, SO2, NO2 (B) H2O, HCN, BeI2
structure is most likely. Explain.
(C) OCN-, C2H2, OF2 (D) H2S, CIO2-, NH2
-

35. The hybrid orbitals of nitrogen in N2O4 are

(A) sp (B) sp2 (C) sp3 (D) sp3d

36. How many sigma and how many pi bonds are in

e. Which side of the N–O bond is +? Explain.
CH2=CH–CH2–C–CH3?
||
O
(A) 5 sigma and 2 pi. (B) 8 sigma and 4 pi. f. Rank the strength of the N–O bond in order of
(C) 11 sigma and 2 pi. (D) 13 sigma and 2 pi. strongest (1) to weakest (3). Explain your answer.
N2 O NO2 NO3-

37. What is the best estimate of the H-O-H bond angle in H3O+?
(A) 109.5o (B) 107o (C) 90o (D) 120
o
3. Complete the chart for SeF2, SeF4, and SeF6.
SeF2 SeF4 SeF6
38. Which of the following pairs of compounds are isomers?
(A) CH3–CH2–CH2–CH3 and CH3–CH(CH3)–CH3 Lewis
(B) CH3–CH(CH3)–CH3 and CH3–CH3–C=CH2 Structure
(C) CH3–O–CH3 and CH3–CO–CH3
Se-
(D) CH3–OH and CH3–CH2–OH Hybridization
Domain
Geometry
39. CH3–CH2–CH2Br Molecular
Which of the following structural formulas represents an Geometry
isomer of the compound that has the structural formula Ideal Bond
represented above? Angle
(A) CH2Br–CH2–CH2 (B) CH3–CHBr–CH3
Polarity
(C) CH2Br–CH2–CH2Br (D) CH3–CH2–CH2–CH2Br 4. Consider the ion SF3+.
a. Draw a Lewis structure.

Practice Free Response

b. Identify the type of hybridization exhibited by sulfur. d. Predict whether the F-S-F bond angle is equal to,
greater than or less than 109.5°. Explain
c. Identify the electron-domain and molecular geometries.
Electron-domain
Molecular geometry
geometry
5. Consider the ion SF5-
a. Draw a Lewis structure.

b. Identify the type of hybridization exhibited by sulfur.

c. Identify the electron-domain and molecular geometries.

Electron domain geometry Molecular geometry

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

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