CHE141
Chapter 9
Chapter 9
Molecular Geometry and Bonding Theories
1. For a molecule with the formula AB2 the molecular shape is __________.
(a). linear or trigonal planar
(b).linear or bent
(c). linear or T-shaped
(d). T-shaped
Explanation: When answering questions like these always try to imagine a molecule
that would fit the generic (AB2) formula. In this case you could use the examples of
CO2 or H2O. Thus in the case of CO2, the molecule would be linear since the central
atom would not have any lone pairs while with water the shape would be bent due to
the lone pairs on the central atom.
2. According to VSEPR theory, if there are five electron domains in the valence shell of
an atom, they will be arranged in a(n) __________ geometry.
(a). octahedral
(b).trigonal bipyramidal
(c). tetrahedral
(d). trigonal planar
Explanation: The trigonal bipyramidal arrangement of electron domains would be
the ideal one to minimize the repulsions involved.
3. The electron-domain geometry and molecular geometry of iodine trichloride are
__________ and __________ respectively.
(a). trigonal bipyramidal, trigonal planar
(b). tetrahedral, trigonal pyramidal
(c). trigonal bipyramidal, T-shaped
(d). octahedral, trigonal planar
Explanation: The Lewis structure of ICl3 shows that the two lone pairs occupy the
equatorial positions in this trigonal bipyramidal electron domain geometry. The
molecule itself is T-shaped. The lone pairs occupy equatorial positions and are 120°
apart, experiencing the least amount of repulsions, compared to axial positions.
Cl
Cl
I
Cl
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Chapter 9
4. According to valence bond theory, which orbitals on bromine atoms overlap in the
formation of the bond in Br2?
(a). 3s
(b). 3p
(c). 4s
(d).4p
Explanation: According to the valence bond theory, orbital overlap between two
atoms leads to a bond formation. The orbitals that overlap are usually the outermost
valence orbitals. In Br atoms, the valence electrons are held in the 4s and 4p orbitals
with the 4p being the outermost orbital.
5. The electron-domain geometry of a sulfur-centered compound is trigonal
bipyramidal. The hybridization of the central sulfur atom is __________.
(a). sp
(b). sp2
(c). sp3d
(d). sp3
Explanation: A trigonal bipyramidal electron domain geometry indicates that there
are 5 electron domains available. These 5 electron domains can be made possible
only if a sp3d set of hybrid orbitals is available on the central atom.
6. The hybridization of orbitals on the central atom in a molecule is sp. The electrondomain geometry around this central atom is __________.
(a). octahedral
(b).linear
(c). trigonal planar
(d). tetrahedral
Explanation: The presence of sp hybrid orbitals indicates that the central atom can
form only 2 bonds, making the electron domain geometry linear.
7. The hybridization of orbitals on the central atom in a molecule is sp2. The electrondomain geometry about this central atom is __________.
(a). tetrahedral
(b). linear
(c). trigonal planar
(d). trigonal bipyramidal
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Chapter 9
Explanation: The presence of the sp2 hybrid orbitals on the central atom indicate that
the electron domains can be arranged only in a trigonal planar geometry.
8. The hybridization of the carbon atom in carbon dioxide is __________.
(a). sp
(b). sp2
(c). sp3
(d). sp3d
Explanation: The Lewis structure of the molecule shows that it is a linear molecule
with no lone pairs on the central atom. As a result of this the central atom can have
only sp hybrid orbitals on it.
O
C
O
9. In order to produce sp3 hybrid orbitals, ____ s atomic orbital(s) and ____ p atomic
orbitals (s) must be mixed.
(a). one, two
(b).one, three
(c). one, one
(d). two, two
Explanation: The sp3 hybrid orbitals are formed by the combination of one s orbital
and three p orbitals. This information can be deduced by adding the number of
orbitals (one “s” + three “p”) present in the name of the hybrid orbitals.
10. There are ______ σ and _______ π bonds in the
H
C
C
H
molecule.
(a). 3 and 2
(b). 3 and 4
(c). 4 and 3
(d). 2 and 3
Explanation: The single bonds between the C and H atoms are counted as sigma (σ)
bonds while the triple bond between the 2 C atoms is made of one sigma and two pi
(π) bonds.
11. There are ______ σ and ________ π bonds in the
H2C
C
CH2
molecule.
(a). 4, 2
(b).6, 2
(c). 2, 2
(d). 2, 6
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Chapter 9
Explanation: Here the C atoms on the outside are each bonded to 2 H atoms by
single bonds and then boded to the middle carbon with a single bond, making the total
sigma bonds = 6. The number of pi bonds on the other hand is 2, bonding the middle
C atom to the outside C atoms.
12. In counting the electron domains around the central atom in VSEPR theory, a
__________ is not included.
(a). nonbonding pair of electrons
(b). single covalent bond
(c). core level electron pair
(d). double covalent bond
Explanation: According to the Valence Shell Electron Pair Repulsion (VSEPR)
theory, it is the electrons that are in the valence (outermost) shell that contribute
towards bonding. The core electrons do not participate in bonding at all.
13. The O-C-O bond angle in the CO32- ion is approximately __________.
(a). 90°
(b). 109.5°
(c). 180°
(d).120o
Explanation: The Lewis structure of the CO32- ion shows that the three O atoms are
arranged around the central C atom in a trigonal planar manner. Since the central C
atom does not have any lone pairs the bond angles would not be affected and hence
be exactly 120°.
2O
C
O
O
14. The Cl-C-Cl bond angle in the CCl2O molecule (C is the central atom) is slightly
__________.
(a). greater than 109.5°
(b). less than 109.5°
(c). less than 120°
(d). greater than 120°
Explanation: The Lewis structure of the CCl2O molecule shows that there is a
double bond between the O and the central C atom. A double bond represents a
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Chapter 9
slightly larger electron domain which disrupts the ideal 120° Cl-C-Cl bond angle,
making it smaller.
O
C
Cl
Cl
15. The molecular geometry of the left-most carbon atom in the molecule below is
__________.
O
H
C
H
C
H
O
H
(a). trigonal planar
(b).tetrahedral
(c). trigonal bipyramidal
(d). octahedral
Explanation: When working with a molecule that has more than one central atom
always concentrate on counting the number of electron domains around the particular
central atom in question. Here the left most C atom has 4 electron domains around it
and no lone pairs. As a result of this the geometry around it would be tetrahedral.
16. The molecular geometry of the right-most carbon in the molecule below is
__________.
O
H
C
H
C
H
O
H
(a). T-shaped
(b).trigonal planar
(c). tetrahedral
(d). octahedral
Explanation: When working with a molecule that has more than one central atom
always concentrate on counting the number of electron domains around the particular
central atom in question. Here the right most C atom has 3 electron domains around
it and no lone pairs. As a result of this the geometry around it would be trigonal
planar.
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17. An electron domain consists of __________
(a). a nonbonding pair of electrons
(b). a single bond
(c). a multiple bond
(a). a and b only
(b). b and c only
(c). a and c only
(d).a, b, and c
Explanation: Electron domains are regions where valence electrons are most likely
to be present. These electrons may be present as a non-bonding pair or a bonding pair
of electrons.
18. According to VSEPR theory, if there are three electron domains on a central atom,
they will be arranged such that the angles between the domains are __________.
(a). 90°
(b). 180°
(c). 109.5°
(d).120°
Explanation: According to the VSEPR theory the best arrangement of a given
number of electron domains is the one that minimizes the repulsions between the
pairs. Thus for 3 electron domains the best arrangement will be the one with angles
of 120°.
19. According to VSEPR theory, if there are four electron domains on a central atom,
they will be arranged such that the angles between the domains are __________.
(a). 120°
(b).109.5°
(c). 180°
(d). 90°
Explanation: According to the VSEPR theory the best arrangement of a given
number of electron domains is the one that minimizes the repulsions between the
pairs. Thus for 4 electron domains the best arrangement will be the one with angles
of 109.5°.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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Chapter 9
20. The electron-domain geometry and the molecular geometry of a molecule of the
general formula ABn are __________.
(a). never the same
(b). always the same
(c). sometimes the same
(d). not related to each other
Explanation: The electron domain geometry of a molecule describes the arrangement
of the electron domains in the molecule while the molecular geometry describes the
arrangement of the actual atoms in the molecule.
The presence of non-bonding electron pairs can affect the molecular geometry
making it different than the electron domain geometry.
21. The electron-domain geometry and the molecular geometry of a molecule of the
general formula ABn will always be the same if __________.
(a). there are no lone pairs on the central atom
(b). there is more than one central atom
(c). n is greater than four
(d). the octet rule is obeyed
Explanation: The presence of lone pairs on the central atom can alter the electron
domain geometry of the molecule. The two geometries can be the same only if there
are no lone pairs present on the central atom.
22. For molecules of the general formula ABn, n can be greater than four _________.
(a). for any element A
(b). only when A is boron or beryllium
(c). only when A is an element from the third period or below the third period
(d). only when A is Kr
Explanation: The value of “n” can be greater than 4 only the element “A” is from the
third or higher period of the periodic table. Elements from these periods can have
more than 8 electrons around their atoms (have expanded octets).
23. Of the molecules below, only __________ is polar.
(a). CCl4
(b). CH4
(c). SiCl4
(d).SeF4
Explanation: The polarity of a molecule is closely related to its electron domain and
molecular geometries. Both the electron domain and molecular geometries of CCl4,
Copyright © 2006 Dr. Harshavardhan D. Bapat
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Chapter 9
CH4 and SiCl4 are tetrahedral. The molecular geometry of SeF4 is see-saw while the
electron-domain geometry is trigonal bipyramidal due to a lone pair on the central Se
atom. As a result of the lone pair on the central atom, this molecule becomes polar.
F
Se
F
F
F
24. Of the molecules below, which one is nonpolar?
(a). BF3
(b). NF3
(c). IF3
(d). PBr3
Explanation: With the exception of the BF3 molecule all the other molecules will
have one or more lone pairs on the central atom. As a result of the lone pair(s) the
molecule will then be polar. In BF3 the polar B-F bonds will cancel each other,
making the molecule non-polar.
25. Three sulfur fluorides are observed: SF2, SF4, and SF6. Of these, __________ is/are
polar.
(a). SF2 only
(b).SF2 and SF4 only
(c). SF4 only
(d). SF6 only
Explanation: The polarity of a molecule is closely related to its electron domain and
molecular geometries. Here, SF6 has an octahedral geometry and does not have any
electrons on the central S atom. As a result of this the polar S-F bonds cancel each
other’s pulls. In both the other fluorides, the central S atom has lone pairs on it
making them polar molecules.
26. The molecular geometry of the BeCl2 molecule is __________, and this molecule is
__________.
(a). linear, nonpolar
(b). linear, polar
(c). bent, nonpolar
(d). bent, polar
Explanation: The polarity of a molecule is closely related to its electron domain and
molecular geometries. The Lewis structure of this molecule clearly shows that the
Copyright © 2006 Dr. Harshavardhan D. Bapat
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Chapter 9
molecule is linear and since the central Be atom does not have any lone pairs on it,
the polar Be-Cl bonds will negate each other making it a non-polar molecule.
Cl
Be
Cl
27. For molecules with only one central atom, how many lone pairs on the central atom
guarantees molecular polarity?
(a). 1
(b). 2
(c). 1 or 2
(d). Any number greater than 1.
Explanation: The polarity of a molecule is closely related to its electron domain and
molecular geometries. The presence of one lone pair of electrons on the central atom
makes the molecule polar, even if other bonds in the molecule may cancel each other.
28. According to valence bond theory, which orbitals overlap in the formation of the
bond in HBr?
(a). 1s on H and 4s on Br
(b).1s on H and 4p on Br
(c). 1s on H and 3p on Br
(d). 2s on H and 4p on Br
Explanation: According to the valence bond theory bonds are formed due to overlaps
between valence orbitals. In the case of the HBr molecule the valence electrons are
contained in the 1s orbital for H and in the 4p orbital for the Br. The 4s orbital is full
and does not have any room.
29. The combination of two atomic orbitals results in the formation of __________
molecular orbitals.
(a). 1
(b).2
(c). 3
(d). 0
Explanation: The combination of two atomic orbitals always leads to the formation
of a bonding and an antibonding molecular orbital.
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Chapter 9
30. The sp3d2 atomic hybrid orbital set accommodates __________ electron domains.
(a). 2
(b). 3
(c). 6
(d). 5
Explanation: The number of electron domains indicated by a set of hybrid orbitals
can be easily found by adding the number of atomic orbitals present in the set. Here
we have one “s” + three “p” + two “d” (= 6) orbitals present making 6 electron
domains available.
31. The sp2 atomic hybrid orbital set accommodates __________ electron domains.
(a). 2
(b).3
(c). 4
(d). 5
Explanation: The number of electron domains indicated by a set of hybrid orbitals
can be easily found by adding the number of atomic orbitals present in the set. Here
we have one “s” + two “p” (= 3) orbitals present making 3 electron domains
available.
32. The hybridization of iodine in IF3 and IF5 are __________ and __________,
respectively.
(a). sp3, sp3d
(b).sp3d, sp3d2
(c). sp3d, sp3
(d). sp3d2, sp3d
Explanation: The Lewis structures of these molecules indicate that IF3 will have a
trigonal bipyramidal electron-domain geometry while IF5 will have an octahedral
geometry. To be able to have these geometries the molecules will need the hybrid
orbitals to accommodate the electron domains.
F
F
F
F
I
F
I
F
F
F
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Chapter 9
33. The hybridizations of bromine in BrF5 and of arsenic in AsF5 are __________ and
__________, respectively.
(a). sp3, sp3d
(b). sp3d, sp3d2
(c). sp3d, sp3
(d).sp3d2, sp3d
Explanation: The Lewis structures of these molecules indicate that the BrF5 will
have an octahedral electron-domain geometry where as the AsF5 will have trigonal
bipyramidal geometry. The central Br atom will thus need 6 electron domains to
accommodate the 5 Br-F bonds and the lone pair. The central As atom on the other
hand will need only 5 electron domains to bond with the 5 F atoms.
F
F
F
F
Br
F
F
As
F
F
F
F
Consider the following species when answering questions 34 and 35:
(i) PCl3
(ii) CCl4
(iii) TeCl4
(iv) XeF4
(v) SF6
34. In which of the molecules does the central atom utilize d orbitals to form hybrid
orbitals?
(a). (i) and (ii)
(b). (iii) only
(c). (i) and (v)
(d).(iii), (iv), and (v)
Explanation: Molecules with central atoms from the third period and beyond can
have an expanded octet; meaning more than 4 pairs of electrons around them. To do
this the atoms use their empty d orbitals. Of all the species here, only Te, Xe and S fit
this requirement and hence will utilize the d orbitals to form hybrid molecular
orbitals.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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Chapter 9
35. Which of the molecules has a see-saw shape?
(a). (i) only
(b). (ii) only
(c). (iii) only
(d). (iv) only
Explanation: The Lewis structures of the species involved clearly show that only the
TeCl4 will have a see-saw shape.
Cl
Cl
P
Cl
Cl
Cl
Cl
C
Cl
Cl
Cl
Copyright © 2006 Dr. Harshavardhan D. Bapat
F
Cl
F
Cl
F
Xe
F
F
F
S
F
F
F
F
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