High School Chemistry - Problem Drill 17: Molecular Geometry
Question No. 1 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
1. The Lewis Structure approach provides the insight into the bonding of a
molecule. VSEPR theory gives a simple model on how to predict the geometry of a
molecule. The shape of a molecule is determined mainly by the electrons
surrounding the central atom. VSEPR is based upon what principle?
Question
(A) The size of the electrons in the valence shell of a molecule determines the
shape of the molecule.
(B) The core electrons of a molecule determine the shape of the molecule.
(C) The attraction of electrons in the valence shell of a molecule determines the
shape of the molecule.
(D) The mass of the electrons in the valence shell of a molecule determines the
shape of the molecule.
(E) The repulsion of electrons in the valence shell of a molecule determines the
shape of the molecule.
A. Incorrect.
The size of the electrons does not determine the geometric shape of a molecule.
Think about the acronym of VSEPR. The 3D structure is determined by minimizing
the repulsion of electron pairs.
B. Incorrect.
The core electrons do not determine the geometric shape and electronic structure
of a molecule. Think about the acronym of VSEPR. The 3D structure is determined
by minimizing the repulsion of electron pairs.
C. Incorrect.
Feedback
Electrons repel, not attract, each other. The 3D structure is determined by
minimizing the repulsion of electron pairs.
D. Incorrect.
The mass does not determine the geometric shape and electronic structure of a
molecule. Think about the acronym of VSEPR. The 3D structure is determined by
minimizing the repulsion of electron pairs.
E. Correct.
Good job! Valence Shell Electron Pair Repulsion (VSEPR) states that the 3D
structure is determined by minimizing repulsion of electron pairs.
VSEPR theory states that the geometric arrangement of terminal atoms, or groups
of atoms about a central atom in a covalent compound, or charged ion, is
determined solely by the repulsions between electron pairs present in the valence
shell of the central atom.
Solution
The number of electron pairs around the central atom can be determined by writing
the Lewis structure for the molecule. The geometry of the molecule depends on the
number of bonding groups (pairs of electrons) and the number of nonbonding
electrons (lone pairs) on the central atom.
The correct answer is (E).
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Question No. 2 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
2. For which of the molecules is the molecular geometry (shape) the same as
electronic geometry?
(i) PCl3 (ii) CCl4 (iii) TeCl4 (iv) XeF4 (v) SCl6
Question
(A)
(B)
(C)
(D)
(E)
(i) and (iii)
(i) and (iv)
(ii) and (v)
(ii) only
(v) only
A. Incorrect.
Draw out the Lewis structures for five molecules and determine the electronic
structures by counting the electron regions. If there is no lone pair left in the
central atom, their molecular and electronic geometry are identical according to
VSEPR.
B. Incorrect.
Draw out the Lewis structures for five molecules and determine the electronic
structures by counting the electron regions. If there is no lone pair left in the
central atom, their molecular and electronic geometry are identical according to
VSEPR.
C. Correct.
Feedback
Good job! Draw out the Lewis structures for five molecules and determine the
electronic structures by counting the electron regions. If there is no lone pair left in
the central atom, their molecular and electronic geometry are identical according to
VSEPR.
D. Incorrect.
Draw out the Lewis structures for five molecules and determine the electronic
structures by counting the electron regions. If there is no lone pair left in the
central atom, their molecular and electronic geometry are identical according to
VSEPR.
E. Incorrect.
Draw out the Lewis structures for five molecules and determine the electronic
structures by counting the electron regions. If there is no lone pair left in the
central atom, their molecular and electronic geometry are identical according to
VSEPR.
Solution
Follow the four-step process in determining the molecular geometry: (1) Write out
the Lewis structure; (2) Count the total electron regions; (3) Determine the
electronic geometry based on the electron region total; (4) Determine the
molecular geometry by ignoring any lone pair(s) at the central atom.
(i) PCl3: Electronic Geometry = Tetrahedral; Molecular Geometry = Trigonal
Pyramidal.
(ii) CCl4: Electronic Geometry = Tetrahedral; Molecular Geometry = Tetrahedral
(iii) TeCl4: Electronic Geometry = Trigonal Bipyramidal; Molecular Geometry =
See-Saw.
(iv)XeF4: Electronic Geometry = Octahedral; Molecular Geometry = Square
Planar
(v) SCl6: Electronic Geometry = Octahedral; Molecular Geometry = Octahedral
The correct answer is (C).
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Question No. 3 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
3. Molecular geometry can be determined using the VSEPR model. The bond angles
can then be estimated based on geometric arrangement of the terminal atoms
around the central atom. Phosgene is a colorless gas used as a chemical weapon
during World War I. Estimate the Cl-C-Cl bond angle in the phosgene (CCl2O)
molecule.
Question
(A) 120
(B) 90
(C) 180
(D) 109.5
(E) 60
A. Correct.
Good job! There are 3 electron regions. The molecular geometry is trigonal planar
with 120 angle. In fact, due to the two lone pairs in the oxygen atom, the bond
angle Cl-C-Cl is slightly smaller than 120.
B. Incorrect.
Follow the four steps of geometry determination based on VSEPR. The molecular
geometry is trigonal planar. The bond angle is not 90.
C. Incorrect.
Feedback
Follow the four steps of geometry determination based on VSEPR. The molecular
geometry is trigonal planar. The bond angle is not 180.
D. Incorrect.
Follow the four steps of geometry determination based on VSEPR. The molecular
geometry is trigonal planar. The bond angle is not 109.5.
E. Incorrect.
Follow the four steps of geometry determination based on VSEPR. The molecular
geometry is trigonal planar. The bond angle is not 60.
To estimate the bond angle around the central atom (C), follow the four-step
geometry determination process:
(1) Count the total valence electrons (=26) and draw out the Lewis structure.
Solution
(2) Count the total electron regions (=3). The double bond is counted as 1
region.
(3) Determine the electronic geometry. 3 Electron Regions = Trigonal Planar.
(4) Determine the molecular geometry. There is no lone pair in the central
atom. Therefore the molecular geometry is the same as electronic geometry,
i.e. trigonal planar 120.
Note: There are two lone pairs on oxygen. Due to the lone pair repulsion to
two C-Cl bonds, the Cl-C-Cl bond angle should be slightly less than 120.
The correct answer is (A).
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Question No. 4 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
4. Carbonate ion is a polyatomic anion in limestone as calcium carbonate. Sodium
bicarbonate is the key ingredient of baking soda. What is the VSEPR molecular
geometry for CO32-?
Question
(A)
(B)
(C)
(D)
(E)
Trigonal Planar
Trigonal Pyramidal
Tetrahedral
Bent
Linear
A. Correct.
Good job! There are three bonding regions and no lone pairs in the Lewis structure.
It is trigonal planar geometry.
B. Incorrect.
This molecule has 3 bonding regions and no lone pairs. It is not trigonal pyramidal.
C. Incorrect.
This molecule has 3 bonding regions and no lone pairs. It is not tetrahedral.
Feedback
D. Incorrect.
This molecule has 3 bonding regions and no lone pairs. It is not bent.
E. Incorrect.
This molecule has 3 bonding regions and no lone pairs. It is not linear.
CO32- has three bonding regions and no lone pair regions. This is trigonal planar
shape.
Some molecules involve two or more structures that are equivalent. Resonance
represents the multiple forms that a molecule can exist.
Solution
Because of three identical oxygen atoms, CO32- has three alternative forms
(resonance structures) to represent the overall electronic structure.
The correct answer is (A).
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Question No. 5 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
5. The VSEPR model does a good job of predicting molecule geometry, based on
electron regions and all-important idea of electron pair repulsion. Lone pairs tend to
take up more space than bonding pairs. Apply this idea, which has a larger Cl-P-Cl
bond angle, PCl3 or PCl4+?
Question
(A)
(B)
(C)
(D)
(E)
PCl3
PCl4+
They are both the same.
Cannot be determined from the given information.
None of the above
A. Incorrect.
PCl3 has a lone pair while the ion PCl4+ does not. The lone pair on PCl3 takes up
more space and pushes the P-Cl bonds closer. Its Cl-P-Cl (100) is therefore less
than the tetrahedron’s 109.5.
B. Correct!
PCl3 has a lone pair while the ion PCl4+ does not. PCl4+ has the tetrahedral
geometry with no lone pair on the central atom. The Cl-P-Cl is 109.5.
C. Incorrect.
Feedback
PCl3 has a lone pair while the ion does not. Their bond angles are not the same.
D. Incorrect.
PCl3 has a lone pair while the ion does not. There is sufficient information given.
E. Incorrect.
There is one correct answer above.
They both have 4 electron regions. PCl 3 has 1 lone pair while PCl4+ has none. Lone
pairs distort bond angles and push toward each other. Therefore PCl4+, with no
lone pair on P, has a larger bond angle.
PCl3 has the trigonal pyramidal geometry. Due to the lone pair on P, the P-Cl bonds
are closer to give more space for the lone pair. The bond angle is about 100.
Solution
PCl4+ has the tetrahedral geometry with no lone pair on the central atom. The bond
angle is expected to be a perfect 109.5.
The correct answer is (B).
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Question No. 6 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
6. Sulfur trioxide (SO3) is the primary agent in acid rain. According to VSEPR, the
molecular geometry of SO3 molecule is ____.
Question
(A) Linear
(B) Trigonal Planar
(C) Bent
(D) Trigonal Pyramidal
(E) Tetrahedral
A. Incorrect.
Draw the Lewis structure with one double bond. There are three bonding electron
regions with no lone pair on the central atom. This is not linear.
B. Correct.
Good job! Draw the Lewis structure with one double bond. There are three bonding
electron regions with no lone pair on the central atom. The molecular geometry is
trigonal planar.
C. Incorrect.
Feedback
Draw the Lewis structure with one double bond. There are three bonding electron
regions with no lone pair on the central atom. This is not bent.
D. Incorrect.
Draw the Lewis structure with one double bond. There are three bonding electron
regions with no lone pair on the central atom. This is not trigonal pyramidal.
E. Incorrect.
Draw the Lewis structure with one double bond. There are three bonding electron
regions with no lone pair on the central atom. This is not tetrahedral.
There are 24 valence electrons. Draw out the Lewis structure and double up one
bond to meet the Octet rule.
There are three electron regions with no lone pair on the central atom. Its molecule
geometry is trigonal planar.
Solution
The correct answer is (B).
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Question No. 7 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
7. Hydrogen sulfide occurs in volcanic gases and natural gas. It is a colorless gas
with the characteristic foul odor of rotten eggs. What is the molecular geometry for
H2S?
Question
(A)
(B)
(C)
(D)
(E)
Linear
Bent
Trigonal Planar
Tetrahedral
Octahedral
A. Incorrect.
Write out the Lewis structure and count the electron regions (4). There are two
bonding regions and two lone pair regions.
B. Correct.
Good job! There are two bonding regions and two lone pair regions. The electronic
geometry is tetrahedral and the molecular geometry is bent after ignoring the two
lone pairs.
C. Incorrect.
Feedback
Write out the Lewis structure and count the electron regions (4). There are two
bonding regions and two lone pair regions.
D. Incorrect.
Write out the Lewis structure and count the electron regions (4). There are two
bonding regions and two lone pair regions.
E. Incorrect.
Write out the Lewis structure and count the electron regions (4). There are two
bonding regions and two lone pair regions.
Write out the Lewis structure.
There are 2 lone pairs and 2 ligands – total of four electron regions.
Electronic geometry = tetrahedral
Molecular geometry = bent
Solution
The correct answer is (B).
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Question No. 8 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
8. A simple model of VSEPR can predict the molecular geometry and in turn the
polarity of a molecule. Symmetry of a molecule determines its polarity. Which
molecular geometry always results in polar molecules?
Question
(A)
(B)
(C)
(D)
(E)
Tetrahedral
Trigonal Bipyramidal
Trigonal Pyramidal
Octahedral
Linear
A. Incorrect.
Although all of them can produce polar molecules, one of them has a lone pair of
electrons that always results in a net dipole. Tetrahedron has the symmetry to
cancel all bond polarities out, resulting in net dipole of zero.
B. Incorrect.
Although all of them can produce polar molecules, one of them has a lone pair of
electrons that always results in a net dipole. Trigonal bipyramidal has the symmetry
to cancel all bond polarities out, resulting in net dipole of zero.
C. Correct.
Good job! Although all of them can produce polar molecules, trigonal pyramidal has
a lone pair of electrons that always results in a net dipole for the molecule.
Feedback
D. Incorrect.
Although all of them can produce polar molecules, one of them has a lone pair of
electrons that always results in a net dipole. Octahedron has the symmetry to
cancel all bond polarities out, resulting in net dipole of zero.
E. Incorrect.
Although all of them can produce polar molecules, one of them has a lone pair of
electrons that always results in a net dipole. Some linear molecules have the
symmetry to cancel all bond polarities out, resulting in net dipole of zero.
To determine the molecular polarity:
(1) Draw out the Lewis structure;
(2) Assign the polarity of each bond;
(3) Apply the symmetry and see if the bond dipoles have any net molecular
dipole. Highly symmetrical molecules tend to have zero molecular polarity
since their bond polarities can all cancel out.
Although all of them can produce polar molecules, trigonal pyramidal has a lone
pair of electrons that always results in a net dipole for the molecule.
Solution
The correct answer is (C).
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Question No. 9 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
9. To determine the Lewis structure, count the valence electrons. To determine the
geometry, count the total electron regions. Which one of the following molecules
has tetrahedral molecular geometry?
Question
(A)
(B)
(C)
(D)
(E)
CF4
XeF4
SCl4
IF4C2H2
A. Correct.
Good job! Draw out the Lewis structure and count the electron regions. Determine
the electronic geometry based on the electron region total. Ignore any lone pair on
the central atom and determine the molecular geometry. Adjust the bond angle to
give away more space for lone pairs. ClF4 has the same molecular geometry as its
electronic geometry since there is no lone pair left on the central atom.
B. Incorrect.
Draw out the Lewis structure and count the electron regions. Determine the
electronic geometry based on the electron region total. Ignore any lone pair on the
central atom and determine the molecular geometry. Adjust the bond angle to give
away more space for lone pairs. XeF4 has trigonal planar molecular geometry with
two lone pairs on Xe.
C. Incorrect.
Feedback
Draw out the Lewis structure and count the electron regions. Determine the
electronic geometry based on the electron region total. Ignore any lone pair on the
central atom and determine the molecular geometry. Adjust the bond angle to give
away more space for lone pairs. SCl4 has see-saw shape with one lone pair at the
equatorial position.
D. Incorrect.
Draw out the Lewis structure and count the electron regions. Determine the
electronic geometry based on the electron region total. Ignore any lone pair on the
central atom and determine the molecular geometry. Adjust the bond angle to give
away more space for lone pairs. IF4- has the square planar geometry with two lone
pairs.
E. Incorrect.
Draw out the Lewis structure and count the electron regions. Determine the
electronic geometry based on the electron region total. Ignore any lone pair on the
central atom and determine the molecular geometry. Adjust the bond angle to give
away more space for lone pairs. C2H2 has the linear geometry with carbon-carbon
triple bond.
CF4 has the tetrahedral molecular geometry like methane.
XeF4 has trigonal planar molecular geometry with two lone pairs on Xe.
SCl4 has see-saw shape with one lone pair at the equatorial position.
IF4- has the square planar geometry with two lone pairs.
C2H2 has the linear geometry with carbon-carbon triple bond.
Solution
The correct answer is (A).
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Question No. 10 of 10
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10. Bond angles can be determined from the molecular geometry by VSEPR with
adjustment from the repulsion of lone pairs. What are the approximate bond angles
of Cl-I-Cl in ICl3?
Question
(A)
(B)
(C)
(D)
(E)
90 only
90 and 180
109.5 only
120 only
90 and 120
A. Incorrect.
The central atom I has 7 valence electrons and an I-Cl bond is always single. There
are three I-Cl single bonds with two lone pairs left for the central atom. Ignore the
electron pairs and determine the molecular geometry as T-shape.
B. Correct.
Good job! The central atom I has 7 valence electrons and an I-Cl bond is always
single. There are three I-Cl single bonds with two lone pairs left for the central
atom. Ignore the electron pairs and determine the molecular geometry as T-shape.
C. Incorrect.
Feedback
The central atom I has 7 valence electrons and an I-Cl bond is always single. There
are three I-Cl single bonds with two lone pairs left for the central atom. Ignore the
electron pairs and determine the molecular geometry as T-shape.
D. Incorrect.
The central atom I has 7 valence electrons and an I-Cl bond is always single. There
are three I-Cl single bonds with two lone pairs left for the central atom. Ignore the
electron pairs and determine the molecular geometry as T-shape.
E. Incorrect.
The central atom I has 7 valence electrons and an I-Cl bond is always single. There
are three I-Cl single bonds with two lone pairs left for the central atom. Ignore the
electron pairs and determine the molecular geometry as T-shape.
Follow the four-step process to determine a molecular geometry by VSEPR.
(1) Draw the Lewis structure. There are total 28 valence electrons.
(2) Count the total electron regions. There are 5 electron regions.
(3) Determine the electronic geometry. 5 electron regions give the trigonal
bipyramidal electronic geometry.
(4) Determine the molecular geometry. There are two lone pairs at the
equatorial position. Ignore them and you will get the T-shape.
T-shape will give two different Cl-I-Cl angles 90 and 180, approximately.
Solution
The correct answer is (B).
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