College Chemistry - Problem Drill 15: Chemical Bonding
Question No. 1 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
1. Alcohols are a class of organic compounds characterized by one or more
hydroxyl (-OH) groups connected to a carbon atoms of a hydrocarbon chain.
Similar to water structurally, the hydroxyl group is a hydrophilic (“water-loving”)
group. Because of “Like dissolves like”, most common alcohols are soluble in water.
What type of bond is O-H?
(A)
(B)
(C)
(D)
(E)
Question
Ionic
Nonpolar Covalent
Polar Covalent
Metallic
Cannot be determined from this information.
A. Incorrect.
The electronegativity difference isn’t great enough to be an ionic bond.
B. Incorrect.
The electronegativity difference is too great to be a non-polar covalent bond.
C. Correct.
Good job! The electronegativity difference is in the polar covalent zone.
Feedback
D. Incorrect.
Two non-metals don’t form a metallic bond. Think of their electronegativities.
E.
Incorrect.
Look up the electronegativities of the two atoms and determine how different they
are to determine its bond type.
Electronegativity of H = 2.1
Electronegativity of O = 3.5
Electronegativity difference = 1.4
The O – H bond is within the polar covalent bond region. All alcohols possess one or
more polar covalent bonds, making them better aqueous solvents.
The correct answer is (C).
Solution
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 2 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
2. Sigma bond is a strong covalent bond, formed by head-to-head overlapping
between two atomic orbitals. Pi bond is a weaker covalent bond, formed by side-toside overlapping. How many sigma and pi bonds are in the following structure?
Question
(A)
(B)
(C)
(D)
(E)
Sigma
Sigma
Sigma
Sigma
Sigma
=
=
=
=
=
12 and Pi = 2
8 and Pi = 2
8 and Pi = 4
10 and Pi = 2
12 and Pi = 0
A. Incorrect.
Sigma bonds are the first bonds between each pair of atoms. All other bonds are pi
bonds.
B. Incorrect.
Sigma bonds are the first bonds between each pair of atoms. All other bonds are pi
bonds.
C. Incorrect.
Sigma bonds are the first bonds between each pair of atoms. All other bonds are pi
bonds.
Feedback
D. Correct.
Good job! Sigma bonds are the first bonds between each pair of atoms. All other
bonds are pi bonds.
E. Incorrect.
Sigma bonds are the first bonds between each pair of atoms. All other bonds are pi
bonds.
A double bond has one sigma bond and one pi-bond.
Sigma bonds are the first bonds between each pair of atoms.
Pi bonds are the 2nd and 3rd bond between a pair of atoms.
There are 10 “first” bonds.
There are 2 “second” bonds.
Solution
Sigma bonds = 10; Pi bonds = 2
The correct answer is (D).
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 3 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
3. Hybridization is to mix atomic orbitals to form newly hybridized orbitals, which
in turn influences the molecular geometry and bonding properties. What type of
hybridization does the red atom have?
Question
(A) sp
(B) sp2
(C) sp3
(D) sp3d
(E) sp3d2
A. Correct.
Good job! This atom has 2 sigma bonds and 2 pi bonds with 2 atoms/lone-pairs
bonded to it in a linear geometry; hence it is indeed sp hybridization.
B. Incorrect.
The sp2 should have 3 atoms/lone-pairs bonded in a trigonal planar geometry.
However, this atom has only 2 atoms bonded in a linear geometry.
C. Incorrect.
Feedback
The sp3 should have 4 atoms/lone-pairs bonded in a tetrahedral geometry.
However, this atom has only 2 atoms bonded in a linear geometry.
D. Incorrect.
The sp3d should have 5 atoms/lone-pairs bonded to a central atom in a trigonal
bipyramidal geometry.
E. Incorrect!
The sp3d2 should have 6 atoms/lone-pairs bonded to a central atom in an
octahedral geometry.
Hybridization
Hybridization
geometry.
Hybridization
Hybridization
geometry.
Hybridization
geometry.
Solution
sp: 2 atoms/lone-pairs bonded to the C with a linear geometry.
sp2: 3 atoms/lone-pairs bonded to the C with a trigonal planar
sp3: 4 atoms/lone-pairs bonded to the C with a tetrahedral geometry.
sp3d: 5 atoms/lone-pairs bonded to the C with a trigonal bipyramidal
sp3d2: 6 atoms/lone-pairs bonded to the C with an octahedral
An atom with 4 sigma bonds = sp3.
An atom with 3 sigma bonds = sp2.
An atom with 2 sigma bonds (and 2 pi bonds) = sp
This is sp hybridization.
Answer: (A) sp
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 4 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
4. Resonance structures help to spread the charge around and stabilize the
molecule. Resonance stabilization most often occurs in cations, radicals and anions.
Compounds with resonance structures are always more stable than similar
compounds that don’t have or have less resonance structures. Given two pairs of
similar compounds (a) HNO3 and NO3- and (b) H2CO and HCO2-, which has greater
resonance stabilization in each pair?
Question
(A)
(B)
(C)
(D)
(E)
(a) HNO3 and (b) H2CO
(a) NO3- and (b) H2CO
(a) HNO3 and (b) HCO2(a) NO3- and (b) HCO2Molecules in each pair have the same stabilization.
A. Incorrect.
Draw out all possible line structures on each molecule. The one with more
resonance structures has greater resonance stabilization. The charged species are
likely more resonance than the non-charged ones.
B. Incorrect.
Draw out all possible line structures on each molecule. The one with more
resonance structures has greater resonance stabilization. The charged species are
likely more resonance than the non-charged ones.
C. Incorrect.
Feedback
Draw out all possible line structures on each molecule. The one with more
resonance structures has greater resonance stabilization. The charged species are
likely more resonance than the non-charged ones.
D. Correct.
Great job! Draw out all possible line structures on each molecule. The one with
more resonance structures has greater resonance stabilization. The charged species
are likely more resonance than the non-charged ones.
E. Incorrect!
Draw out all possible line structures on each molecule. The one with more
resonance structures has greater resonance stabilization. The charged species are
likely more resonance than the non-charged ones.
Draw out the 2d resonance structures with double bonds on each possible position.
Solution
HNO3 has two and NO3- has three equal-energy resonance structures. Therefore,
NO3- has greater resonance stability. Likewise, HCO2- has two and H2CO has none.
HCO2- has greater resonance stability in this pair.
The correct answer is (D).
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 5 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
5. The electronegativity difference predicts the extent of bond polarity. It can
determine the percentage of ionic character of a chemical bond. Which compound
has the greatest ionic character?
Question
(A)
(B)
(C)
(D)
(E)
CsF
RbF
NaF
LiF
Cannot be determined from this information.
A. Correct.
Good job! The greater the difference in electronegativity, the greater the ionic
character.
B. Incorrect.
Review the periodic trend of electronegativity. The greater the difference in
electronegativity, the greater the ionic character.
C. Incorrect.
Review the periodic trend of electronegativity. The greater the difference in
electronegativity, the greater the ionic character.
Feedback
D. Incorrect.
Review the periodic trend of electronegativity. The greater the difference in
electronegativity, the greater the ionic character.
E. Incorrect.
Review the periodic trend of electronegativity. The greater the difference in
electronegativity, the greater the ionic character.
The greater the difference in electronegativity, the greater the ionic character. In
general, the farther 2 elements are from each other on the periodic table, the
greater the difference in electronegativities. Cs & F are the farthest apart as they
can get.
The correct answer is (A).
Solution
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 6 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
6. Having a polar bond does not mean that such a molecule is polar. When there is
no polar bond in a molecule, the molecule is nonpolar, such as homonuclear
diatomic molecules (Cl2). However, a nonpolar molecule can possess polar bonds. If
the polar bonds are evenly or symmetrically distributed, the bond dipoles cancel
with no net dipole overall (such as BF3). The following molecules contain polar
bonds, but not all are polar molecules. The only nonpolar molecule is _____.
Question
(A)
(B)
(C)
(D)
(E)
H2O
HBr
CO2
H2S
NH3
A. Incorrect.
Water molecule is polar since the O-H bonds are polar and the bent geometry
makes the distribution of the two polar O-H bonds asymmetrical with a net dipole.
B. Incorrect.
For diatomic molecules, if the bond is polar, so as the molecule itself.
C. Correct.
Good job! The O=C=O is a linear molecule with C-O bonds are polar in nature, but
the dipoles created by these two polar bonds are cancelled out due to its
symmetrical arrangement. The molecule is nonpolar.
Feedback
D. Incorrect.
Similar to water molecule, this is a bent geometry with a net dipole from two polar
bonds. Therefore it is a polar molecule overall.
E. Incorrect!
This has a trigonal planar geometry with 3xH’s as a tripod and the lone pair on the
top. This creates a net dipole overall since the three polar H-N bonds cannot be
cancelled out in polarity. This is a polar molecule.
Draw out the model and consider the electronic structure. Add all the polar bonds
together while considering their spatial arrangement and symmetry.
CO2 is the only nonpolar molecule from above, since it is a linear molecule and all
dipoles cancelled out. Therefore, CO2 is a nonpolar molecule.
The correct answer is (C).
Solution
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 7 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
7. Molecular Orbital (MO) Theory provides a greater insight on chemical bonding.
For simple diatomic molecules, their MO diagrams can be easily constructed to
study their bonding properties such as bond orders or bond lengths since the
stronger the bond the shorter of its bond length. Determine which of these diatomic
molecules has the shortest bond length, CN-, CN and CN+.
(A) CN(B) CN
Question
(C) CN+
(D) Same
(E) Can’t be determined.
A. Correct.
Great job! Use the diatomic MO diagram to fill the electrons in order. Count the
bonding and antibonding electrons. Calculate the bond order using the formula. The
highest in bond order has the shortest in bond length.
B. Incorrect.
Use the diatomic MO diagram to fill the electrons in order. Count the bonding and
antibonding electrons. Calculate the bond order using the formula. The highest in
bond order has the shortest in bond length.
C. Incorrect.
Feedback
Use the diatomic MO diagram to fill the electrons in order. Count the bonding and
antibonding electrons. Calculate the bond order using the formula. The highest in
bond order has the shortest in bond length.
D. Incorrect.
Use the diatomic MO diagram to fill the electrons in order. Count the bonding and
antibonding electrons. Calculate the bond order using the formula. The highest in
bond order has the shortest in bond length.
E. Incorrect.
Use the diatomic MO diagram to fill the electrons in order. Count the bonding and
antibonding electrons. Calculate the bond order using the formula. The highest in
bond order has the shortest in bond length.
There are total of 14 electrons in CN-, 13 in CN and 12 in CN+. Fill in the electrons
(one pair per orbital) into the following diatomic MO energy diagram.
Solution
Count the total number of bonding electrons vs antibonding electrons. Bond Order
= (#bonding e- – #antibonding e-)/2. The bond orders of CN-, CN and CN+ are 3,
2.5 and 2. CN- has the strongest bond and also the shortest.
The correct answer is (B).
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 8 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
Question
8. Homenuclear molecules are the ones composed of only one type of element.
Unpaired electrons in an orbital are called paramagnetic electrons. The orbital has a
"net" spin and the entire molecule will have a "net" spin. Therefore, a molecule is
considered to be paramagnetic when it contains at least one paramagnetic electron
in its molecular orbitals. Which homonuclear diatomic molecule(s) of second period
elements, besides O2, should be paramagnetic?
(A) B2
(B) C2
(C) N2
(D) F2
(E) O2
A. Correct.
Good job! Count the total electrons and fill each orbital one by one in the MO
diagram. Split the pair for the last two electrons for two equal-energy orbitals. The
one with unpaired electrons is paramagnetic.
B. Incorrect.
Count the total electrons and fill each orbital one by one in the MO diagram. Split
the pair for the last two electrons for two equal-energy orbitals. The one with
unpaired electrons is paramagnetic.
C. Incorrect.
Feedback
Count the total electrons and fill each orbital one by one in the MO diagram. Split
the pair for the last two electrons for two equal-energy orbitals. The one with
unpaired electrons is paramagnetic.
D. Incorrect.
Count the total electrons and fill each orbital one by one in the MO diagram. Split
the pair for the last two electrons for two equal-energy orbitals. The one with
unpaired electrons is paramagnetic.
E. Incorrect!
Count the total electrons and fill each orbital one by one in the MO diagram. Split
the pair for the last two electrons for two equal-energy orbitals. The one with
unpaired electrons is paramagnetic.
Paramagnetism arises from the unpaired electrons in its molecular orbital diagram.
Homonuclear diatomic molecules always have an even number of electrons.
Solution
The only way any electrons can be unpaired is if they split the pair and occupy the
 orbitals singly. B2 and O2 are the only paramagnetic diatomic molecules there.
The correct answer is (A).
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 9 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
9. Molecular Orbital Theory can derive bond orders, which give information about
bond length and strength. Greater the bond order and shorter the bond length, due
to more bonds between the atoms. Moreover, the strength should also be greater.
If a bond order is zero, the molecule is too unstable and unlikely exists. According
to Molecular Orbital Theory, which homonuclear (of element of the same type)
diatomic molecules of the second period elements should not exist?
Question
(A)
(B)
(C)
(D)
(E)
Li2
Be2
B2
C2
N2
A. Incorrect.
First use the MO diagram to fill up the electrons in those molecules. Calculate the
bond order of each homonuclear diatomic molecules. As the result, there are only
two species in the second period with zero bond order. Therefore they do not exist
according to MO theory.
B. Correct!
Good job! First use the MO diagram to calculate the bond order of each
homonuclear diatomic molecules. As the result, there are only two species in the
second period with zero bond order – Be2 and Ne2. Therefore they do not exist
according to MO theory.
C. Incorrect.
Feedback
First use the MO diagram to fill up the electrons in those molecules. Calculate the
bond order of each homonuclear diatomic molecules. As the result, there are only
two species in the second period with zero bond order. Therefore they do not exist
according to MO theory.
D. Incorrect.
First use the MO diagram to fill up the electrons in those molecules. Calculate the
bond order of each homonuclear diatomic molecules. As the result, there are only
two species in the second period with zero bond order. Therefore they do not exist
according to MO theory.
E. Incorrect.
First use the MO diagram to fill up the electrons in those molecules. Calculate the
bond order of each homonuclear diatomic molecules. As the result, there are only
two species in the second period with zero bond order. Therefore they do not exist
according to MO theory.
Solution
Using the MO theory, bond order can determine the stability of a diatomic molecule,
or lack of it leads to unstable.
(1) Count the total electrons in a homonuclear diatomic molecule, starting from
Li, to Ne (2x of the atomic number).
(2) Fill in the MO energy-level diagram and count the bonding and antibonding
electrons.
(3) Calculate the bond order of each. There are two diatomics Be2 and Ne2 with
zero bond order among the second period of elements.
The correct answer is (B).
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved
Question No. 10 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
10. Metallic compounds are typically formed by the combination of a metal with a
metal (i.e. alloys) with the characteristics of a metal. Unlike the localized electrons
in covalent bonds, electrons in metallic bonds are delocalized and moving freely.
Because of the unique metallic bonding nature, these compounds have distinctive
physical properties. Metallic compounds tend to be ______.
Question
(A)
(B)
(C)
(D)
(E)
Soluble in water
Brittle
Malleable
Liquids
Semiconductors
A. Incorrect.
Metallic compounds are not soluble in water.
B. Incorrect.
Metallic compounds can be shaped easily - they are not brittle.
C. Correct.
Good job! Metallic compounds are often soft in certain extent, and can be
hammered into a sheet (i.e. malleable).
Feedback
D. Incorrect.
Metals tend to be solids, with an exception, i.e. Mercury (liquid metal).
E. Incorrect!
Metals are conductors, not semiconductors.
Malleable substances are capable of being shaped or formed by hammering or
pressure (i.e. soft). Metallic compounds do not dissolve in water. They are
malleable, ductile (which is the opposite of brittle), rarely liquids (at standard
temperature and pressure), and are conductors instead of semiconductors.
The correct answer is (C).
Solution
RapidLearningCenter.com
 Rapid Learning Inc. All Rights Reserved