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Unformatted text preview: Chapter Ten CHEMICAL BONDING II: MOLECULAR
GEOMETRY AND HYBRIDIZATION OF
• Molecular Geometry and the VSEPR Model
Valence Bond Theory and Hybrid Orbitals
Molecular Orbital Theory MOLECULAR GEOMETRY AND THE VSEPR MODEL
2. Predict geometrical shapes of molecules and polyatomic ions.
Predict approximate bond angles between atoms within a molecule or polyatomic ion. Electron-Pair Repulsion. The major features of the geometry of molecules and polyatomic ions can be
predicted by applying a simple principle: "The valence shell electron pairs surrounding an atom are arranged
such that they are as far apart as possible."
According to the valence-shell electron-pair repulsion (VSEPR) theory, a particular molecular geometry
results from the orientation of electron pairs around a central atom. For example, consider the BeH2 molecule in
which Be is the central atom. Drawing its Lewis structure, we count two electron pairs in the valence shell of
the Be atom. There are no lone pairs.
H : Be : H
In order to minimize repulsion the two electron pairs are located as far apart as possible. The angle between two
bonds to the same atom is called the b ond angle . The H—Be—H bond angle is 180°. BeH2 is an example of a
molecule in which the atoms lie in a straight line. BeH2 has a linear geometry.
When the central atom of a molecule has no lone pairs in its valence shell, the possible orientations in
space around the central atom are determined by the number of bonding pairs of electrons. Table 10.1 gives the
shapes of molecules depending on the number of valence electron pairs about the central atom. Three Electron Pairs. When there are three valence electron pairs around a central atom they avoid each
other best if they are positioned at the corners of a triangle. The central atom sits in the middle of the triangle,
and all four atoms lie in the same plane. All bond angles are 120°. Such a molecule is called trigonal planar. 1 88
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Table 10.1 Bond Angles and Molecule Shapes
Number of BondingElectron
Pairs about Central Atom
Shape of Molecule
_____________________________________________________________ Four Electron Pairs. In Chapter 9 we discussed the use of the octet rule to determine the correct
formulas of covalent compounds. Lewis structures describe only the bonding within a molecule in a twodimensional manner. They are not geometrical structures. For instance, the Lewis structure for methane CH4
does not tell us whether the molecule is flat (planar) or tetrahedral. When the central atom has four bonding
electron pairs (as C does in CH4 ) these pairs try to keep as far as possible from each other, while maintaining
their distance from the nucleus. The tetrahedral arrangement has less repulsion between the four electron pairs
than the flat, two-dimensional geometry. This geometry is illustrated for methane in Figure 10.1. Four electron
pairs around a central atom will be arranged at the corners of a tetrahedron. In methane the H—C—H bond angle
is the tetrahedral angle, 109.5°.
H Figure 10.1 The tetrahedral geometry of the methane molecule. The C atom is at the
center of a four-sided three-dimensional figure called a tetrahedron. The four H atoms are
found at the four corners. The H–C–H bond angle is the "tetrahedral" angle, 109.5°. Five Electron Pairs. Central atoms with expanded octets have 5 or 6 electron pairs occupying their
valence shells. The geometrical figure defined by the preferred orientation of five electron pairs around the central
atom is called a trigonal bipyramid (Table 10.1). The electron pairs that lie around the "equator" are called
equatorial electrons. The electrons at the "north and south poles" are called axial electrons. The equatorial
electron pairs lie in the same plane. In PF5 the equatorial F—P—F bond angles are 120°. The axial atoms are at
right angles to the equatorial plane, and so any bond angle formed by an axial atom, the central atom, and an
equatorial atom is 90°.
axial F P
F Back Forward Main Menu TOC Study Guide TOC equatorial Textbook Website MHHE Website 1 90 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
Six Electron Pairs. The most stable arrangement of 6 electron pairs around the central atom is called an
octahedron (Table 10.1). The example of SF6 is shown in the textbook.
Effect of Lone Electron Pairs. When the central atom contains lone pairs (nonbonding pairs) in
addition to bonding pairs, the situation becomes slightly more complicated. The position of the lone electron
pairs helps to establish the geometry of a molecule. For instance, in ammonia (NH3 ) the nitrogen atom has one
lone electron pair. .. ← lone pair
There are 4 electron pairs in the valence shell of the central N atom. Therefore, when the electron pairs are
oriented as far from each other as possible, they will be positioned at the corners of a tetrahedron. But when
describing the shape of a molecule we consider only the positions of the atoms. Ammonia is trigonal pyramidal,
not tetrahedral. Its geometry is described as a trigonal pyramid because the pyramid has three sides. The N atom
sits at the top of the pyramid. ..
H H H H H
H trigonal pyramid Table 10.2 lists the molecular shapes that result when the central atom has the symbol A, a pair of bonding
electrons is B, and a pair of nonbonding electrons is E.
Table 10.2 Molecule Shapes As Determined by VSEPR Theory
Shape of Molecule
Distorted tetrahedron (see saw)
_____________________________________________________ Multiple Bonds. The idea of electron pair repulsion can be extended to predict geometries of molecules or
ions containing multiple bonds by assuming that as far as molecular shape is concerned, electrons in a multiple
bond affect the structure in the same way as those in a single bond. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 1 91
Consider SO3 , for example. One of its resonance structures is shown on the left below. S ..
.. S .. :
.. : ..
.. O :O: : .. There are four electron pairs in the valence shell of the central atom (S), but remember two of these are used in a
double bond. Thus, only three regions of electron density exist around the central atom. The SO3 molecule has a
geometry as if three electon pairs existed around the central S atom; that is, it is a planar triangle as shown on
the right. Bond Angles. VSEPR theory also explains why many observed bond angles are not quite equal to the
predicted angle. For instance, a true tetrahedral angle, as in CH4 , is 109.5°, but the measured H—N—H bond
angle in NH3 is 106.7°, and in water the H—O—H bond angle is 104.5°.
C H N H
106.7° H O
104.5° The deviation from 109.5° arises because there are three types of electron pair repulsion represented: (1)
repulsion between bonding pairs, (2) repulsion between lone pairs, and (3) repulsion between a bonding pair and
a lone pair. The force of repulsion is not the same among these three, but decreases as follows:
lone-pair vs. lone-pair
repulsion > lone-pair vs. bonding-pair > bonding-pair vs. bonding-pair
repulsion repulsion In other words, lone pairs require more space than bonding pairs and tend to push the bonding pairs closer
together. Bonding electron pairs are attracted by two nuclei at the same time. These pairs require less space than
lone electron pairs.
In ammonia, the lone pair on the nitrogen atom repels the bonding pairs more strongly than the bonding
pairs repel each other. The result is that the H—N—H bond angle is several degrees less than a tetrahedral angle.
In water the oxygen atom has two bonding pairs and two lone pairs. The greater lone-pair versus bondingpair repulsion causes the two O—H bonds to be pushed in toward each other. The H—O—H angle in water
should be less than the H—N—H angle in ammonia.
In general, when nonbonding electrons are present on the central atom the shape will be close to, but not
exactly, as predicted in Table 10.2.
EXAMPLE 10.1 VSEPR Theory
Predict the geometrical shapes and bond angles for the following compounds using the VSEPR theory.
a. SnCl 4
d. XeF2 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 92 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
•Method of Solution
a. First determine the number of valence shell electrons for the central Sn atom. A tin atom has four valence
electrons, and each Cl atom contributes one electron which it shares with the Sn atom, so 4 + 4(1) = 8. The
eight valence shell electrons are arranged as four electron pairs around the Sn atom. The Lewis structure is: ..
:Cl— Sn —Cl:
This structure shows that there are no lone pairs about the Sn atom. Thus, the four bonding electron pairs
and the Cl atoms, as well, are oriented at the corners of a tetrahedron. SnCl4 is an AB 4 type molecule, and
has a tetrahedral structure. The Cl—Sn—Cl bond angles are 109.5°.
b. Cl Ozone (O3 ) has as one contributing structure ..
.. .. O .. O.. : ..
.. Lewis structure nonlinear geometry When applying VSEPR theory to a molecule requiring resonance structures we can use any one of the
contributing structures. Here we will take the second O atom as the central atom. In VSEPR theory the
double bond is counted as if it were a single bond with one effective electron pair. The double bond acts as
one center of electron density to repel other electron pairs. Therefore, the central O atom has the equivalent
of 3 pairs of valence electrons. These three regions of charge are oriented at the corners of a triangle.
However, one of these pairs is a lone pair. The molecule is of the type AB2 E. The positions of the three O
atoms are described as nonlinear. The O—O—O bond angle should be close to 120°. However, the presence
of the lone pair compresses the bond angle because the lone-pair vs. bonding-pair repulsion is greater than
the bonding-pair vs. bonding-pair repulsion. The observed angle in ozone is 117°.
c. Now that we have had some practice we will abbreviate the explanation. To determine the shape of IF5
count the valence shell electrons around the central I atom.
from the I atom
from the F atoms (5 × 1)
total valence shell electrons Back Forward Main Menu =7
= 12 TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 1 93
The Lewis structure is
.. F valence shell electron pairs = 6
number of lone pairs
number of bonding pairs = 5
molecular type AB5 E
molecular shape: square pyramid
The F—I—F bond angles are 90°.
d. To predict the shape of XeF2 . First count the valence shell electrons around Xe.
from the Xe atom
from the F atoms 2 × 1
total valence shell electrons =8
= 10 Lewis structure .. . . .. ..
: .. Xe F :
valence shell electron pairs
number of lone pairs
number of bonding pairs =5
=2 molecular type AB2 E3
molecular shape: linear
1. 2. Predict the shape of the following molecules.
b . NF 3
c . SF 4
d . ClF 3
Predict the shape of the following ions.
– a. ClO 2
3. 2– b . CO 3 2– c . ZnCl 4 Which of the following molecules and ions have tetrahedral geometry?
a. CCl4 Back e. KrF4 Forward b . SCl 4 Main Menu – c . AlCl 4 TOC Study Guide TOC Textbook Website MHHE Website 1 94 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
4. Predict the following bond angles.
a. O—C—O in CO2
b . F—C—F in CF 4 c. H—O—O in H2 O2 POLAR MOLECULES
1. Use electronegativity values and molecular geometries to predict whether or not a molecule possesses a
dipole moment. Diatomic Molecules. In the preceding chapter you learned that polar bonds are those in which the centers
of positive charge and negative charge do not coincide. This charge separation is called a dipole, and it results
from electronegativity differences between two bonded atoms. A molecule with an electric dipole is said to be
polar and to possess a dipole moment.
The dipole moment µ is represented by an arrow or vector with the tail at the positive center and the head at
the negative center. The length of the arrow represents the magnitude of the dipole moment.
H—Cl Quantitatively the dipole moment is calculated from the equation
µ = Qr
where Q is the magnitude of the charge in coulombs from either end of the dipole and r is the charge separation
(the distance between the centers of positive and negative charge) in meters. In polar molecules this charge is
never as great as one unit of charge. Rather the charge is a partial charge, δ+ or δ–, which signifies a charge of
less than one unit. The SI unit for a dipole moment is the coulomb meter. Traditionally, dipole moments have
been measured in debye units, where 1 debye (D) = 3.33 × 10 –30 C m .
Molecules that possess a dipole moment are called p olar molecules . Molecules without a dipole
moment are called n onpolar molecules . All homonuclear diatomic molecules are nonpolar. In general,
heteronuclear diatomic molecules are polar. Polyatomic Molecules. The dipole moment of a molecule containing more than one bond depends both
on bond polarity and molecular geometry. In a polyatomic molecule the arrows representing polar bonds may
add together to yield a polar molecule with a resultant dipole moment µ. Conversely, the arrows may cancel
each other when added, producing a zero resultant dipole moment. Table 10.3 lists dipole moments of some
small molecules in Debye units. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 1 95
Table 10.3 Dipole Moments
of Some Molecules
An important property of the water molecule is its dipole moment of 1.85 D. In the water molecule each
bond is polar. The resultant of these bond dipoles yields a dipole moment for the molecule. The center of
negative charge lies closer to the O atom than does the center of positive charge.
resultant µ In CO2, the two bond dipoles cancel and µ = 0. ←+ +→
O C O resultant µ = 0
When the bond dipoles are equal and point in opposite directions the centers of + and – charge are both on the
central atom. Thus, the absence of a molecular dipole in CO 2 suggests that the molecule is linear. Conversely,
water has a dipole moment, and so it cannot be linear.
EXAMPLE 10.2 Polar and Nonpolar Molecules
Predict whether the following molecules are polar or nonpolar:
b. H2 CO
c. CCl 4
•Method of Solution
The polarity of a diatomic molecule depends on the electronegativity difference of the two bonded atoms.
a. Oxygen is a more electronegative element than carbon. As a result the C
O bond is polar with partial
positive and negative charges on C and O, respectively.
The dipole moment of carbon monoxide has been found to be 0.1 D. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 96 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
b. The dipolar nature of a polyatomic molecule depends on both the bond polarity and molecular geometry.
Formaldehyde is a planar molecule with a polar C O bond (∆ = 1.0) and two C—H bonds of very low
polarity (∆ = 0.4). Neglecting the low polarity C—H bond dipoles, we get:
The net effect of these bond dipoles is that the center of positive charge is located approximately on the
carbon atom and the center of negative charge lies near the oxygen atom.
\ δ+ δ–
Formaldehyde has a measured dipole moment of about 2.5 D. c. Chlorine is more electronegative than carbon. Therefore CCl4 has four bond dipoles. However, these polar
bonds are arranged in a symmetric tetrahedral fashion about the central carbon atom.
Cl Cl Cl In this situation the centers of positive and negative charge are on the carbon atom. The bond dipoles have
canceled each other and the molecule as a whole does not possess a dipole moment.
6. Which of the following molecules are polar?
a. BCl 3
b . SO 3
c . PF 3
d . SF 6 7. Back What two requirements must be met in order for a molecule to be polar? Which of the following molecules are polar?
b . CO 2
c . SO 2
d. NO2 Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 1 97 VALENCE BOND THEORY AND HYBRID ORBITALS
4. Describe covalent bond formation in terms of overlap of atomic orbitals.
Describe the formation of hybrid orbitals from atomic orbitals.
Assign a hybridization to the central atom of a given molecule.
Describe sigma and pi bond formation. Covalent Bonds. The v alence bond theory is one of two quantum mechanical descriptions of
chemical bonding currently in use. The other is the molecular orbital theory which will be discussed in the next
section. In the valence bond theory, the atomic orbitals of each bonded atom are essentially the same as in
separate isolated atoms. A covalent bond forms when the orbitals from two atoms overlap and a pair of electrons
occupies the region between the atoms. This overlap produces a region between the two nuclei where the
probability of finding an electron is greatly enhanced. The presence of greater electron density between the two
atoms tends to attract both atoms and bonding occurs. In HCl, for example, bonding results from the overlap of
the hydrogen 1s orbital with the half-filled 2px orbital of the chlorine atom. A chlorine atom has only the 2pz
orbital available for bonding because its other orbitals are already filled. The valence electron configuration of Cl
is 2p x 2 2py 2 2pz1 .
H Cl 3p z Hybridization. The concept of hybridization is used to account for bonding and geometry in molecules
with trigonal planar, tetrahedral, and more complex shapes. The process of orbital mixing is called
hybridization, and the new atomic orbitals are called h ybrid orbitals . We can illustrate the main features
of hybridization by examining the linear BeCl2 molecule. To start with an isolated Be atom has the ground state
electron configuration 1s2 2s2 . The orbital diagram for the valence electrons is
B [↑↓ ]
2p In a Be atom the 2p orbitals lie quite close in energy to the 2s orbital. Orbitals on the same atom that lie close
together in energy have an ability to combine with one another to form hybrid orbitals. The presence of two
Be—Cl bonds can be accounted for if, prior to bond formation, a 2s electron in a Be atom is promoted into an
empty 2p orbital.
B [↑ ]
2s [↑ ] ] ]
2p Now there are two unpaired electrons that could participate in two covalent bonds to two chlorine atoms. Since
one electron is in an s orbital and one is in a p orbital, we would expect two types of Be—Cl bonds. However,
both Be—Cl bonds are observed to be the same.
This equivalence is explained by hybridization or "mixing" of the 2s and the 2p orbitals to create two new
equivalent orbitals called sp hybrid orbitals. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 98 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
B [↑ ]↑ ]
2sp [ ] ] empty
2p These two hybrid orbitals point in opposite directions from the Be nucleus. The Be—Cl covalent bonds
result from overlap of beryllium sp hybrid orbitals that contain one electron each with the half-filled 3p orbital
3p z Cl Be
2sp 3p z Cl
2sp We can summarize the rules of hybridization as follows:
4. Hybridization is a process of mixing orbitals on a single atom in a molecule.
Only orbitals of similar energies can be mixed to form hybrid orbitals.
The number of hybrid orbitals obtained always equals the number of orbitals mixed together.
Hybrid orbitals are identical in shape and they project outward from the nucleus with characteristic
orientations in space. Table 10.4 shows several of the more important hybrid orbitals discussed in the textbook and their
characteristic geometrical shapes.
Table 10.4 Hybrid Orbitals and Their Geometries
linear (180° angle)
planar triangle (120° angle)
tetrahedral (109.5° angle)
CH4 ; NH4
trigonal bipyramidal (90° & 120°) PCl 5
octahedral (90° angle)
SF 6 ; SbCl6
_________________________________________________________ Multiple Bonds. Double and triple bonds can also be understood in terms of overlap of atomic orbitals.
C double bond in the ethylene molecule (see Figure 10.15 and Section 10.5 of the textbook), consist
of one sigma ( σ) bond and one pi (π ) bond. H
H σ C σ
π C H
H A s igma bond is a covalent bond in which the electron density is concentrated in a cylindrical pattern around
the line of centers between the two atoms. In the C
C bond, the σ bond results from overlap of an sp 2 hybrid
2 hybrid orbital from the other carbon atom (Figure 10.15 text).
orbital from one of the carbon atoms with a sp Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 1 99
In a p i bond the electron density is concentrated above and below the C—C internuclear axis, but not
cylindrically as in a sigma bond (Figure 10.2). The pi bond has two regions of electron density, but it should be
kept in mind that it is only one bond. The π bond in ethylene results from sideways overlap of the unhybridized
2pz atomic orbitals of the two carbon atoms.
H C π
σ C π H
H Figure 10.2 The pi bond has electron density above and below the internuclear axis.
The electron density is not symetrical apound the internuclear axis as in a σ bond. The
two parts represent one π bond.
EXAMPLE 10.3 Valence Bond Theory
Using unhybridized atomic orbitals, describe what atomic orbitals are used in the following molecules to form
a. F 2
b. H2 S
•Method of Solution
a. From the orbital diagram of fluorine we can see that a F atom has one half-filled 2p orbital that can be used
for bonding. If this orbital overlaps with the half-filled 2p orbital of another F atom, a pair of electrons (one
from each atom) spend most of the time between the two atoms and bonding occurs.
F [↑↓ ][↑↓ ] ↑ ↓ ] ↑ ]
2s b. 2p x 2p y 2p z Hydrogen atoms have a 1s1 electron configuration. Therefore, they can use the half-filled 1s orbital for
bonding. The orbital diagram of the valence electrons in a S atom is
S [↑↓ ][↑↓ ] ↑ ] ↑ ]
3s 3p x 3p y 3p z Overlap of the two half-filled 3p orbitals of a sulfur atom with the half-filled 1s orbitals of the two H atoms
results in the formation of two covalent bonds.
The directional properties of two p orbitals (say the py and pz) are such that the orbitals project out from the S
atom at right angles to each other. The H—S—H bond angle should be 90°. The measured angle is 92.2°.
_______________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 00 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
EXAMPLE 10.4 Hybrid Orbitals
Using orbital diagrams, outline the steps used to describe the formation of a set of sp2 hybridized orbitals on a
•Method of Solution
The ground state orbital diagram for the valence electrons of a carbon atom is [↑↓ ] [↑ ] ↑ ] ] 2s C 2p First promote one 2s electron to the empty 2p orbital. Mixing or hybridization of the 2s orbital with two of the
2p orbitals creates three equivalent sp2 hybrid orbitals. These hybrid orbitals point out from the carbon atom to
the corners of a triangle. One electron remains in an unhybridized 2p2 atomic orbital. [↑ ] ↑ ] ↑ ] [↑ ] 2 sp C 2p _______________________________________________________________________________
EXAMPLE 10.5 Hybrid Orbitals
Describe the hybridization of the central atom in:
b. PCl 5
•Method of Solution
First use VSEPR theory to predict a geometry, then match the geometry to the corresponding hybridization.
a. According to VSEPR theory the PH 3 molecule has trigonal pyramidal geometry and a lone pair. There are
four electron pairs around the P atom that occupy the corners of a teterahedron. ..
H H b. H Therefore, four equivalent orbitals project out from the central P atom at 109° from each other. The bonding
orbitals of the P atom must be sp3 hybrid orbitals.
According to VSEPR theory, PCl5 has the geometrical shape of a trigonal bipyramid.
Cl The type of hybridization that gives five equivalent orbitals that point to the corners of a trigonal bipyramid
is sp 3 d. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 2 01
EXAMPLE 10.6 Sigma And Pi Bonds
Account for the bonding in H2 CO, which is a triangular molecule.
a. What is the hybridization of the carbon atom?
b. What are the approximate bond angles about the carbon atom?
c. How many sigma and pi bonds are there in the molecule?
•Method of Solution
a. The central carbon atom of H2 CO must be using sp2 hybrid orbitals, since the three hybrid orbitals of this
type point to the corners of a triangle. There is also one electron in an unhybridized 2p atomic orbital. C
sp 2 hybrid
c. Overlap of the half-filled 1s orbital of a H atom with the half-filled sp2 hybrid orbitals leads to formation of
a C—H sigma bond. Two C—H bonds are formed in this way. The H—C—H bond angle should be 120°.
Oxygen atoms have the valence electron configuration [↑↓ ] [↑↓ ] ↑ ] ↑ ] 2s O 2px 2p y 2p z Overlap of one half-filled 2p orbital with the remaining sp2 hybrid yields a C—O sigma bond. The two
C—H sigma bonds and the C—O sigma bond are shown below.
H σ bonds C O 2p z H The sideways overlap of the carbon 2pz orbital with the 2pz orbital of the oxygen atom produces a C—O pi
bond. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 02 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals σ bond π overlap H σ Hσ C σ
π O 2p The C—O sigma and pi bonds constitute the C C 2p z z O double bond. O The complete molecule consists of three sigma bonds and one pi bond.
8. Using only atomic orbitals describe the bond formation in F2 and HBr.
9. What is the angle between the following two hybrid orbitals on the same atom?
a. two sp orbitals b. two sp2 orbitals c. two sp3 orbitals
10. What hybrid orbital set is used by the underlined atom in the following molecules?
a. C S 2
b . NO2
c . BF 3
d . Cl F 3 e . H 2 C O
11. What hybrid orbital set is used by the underlined atom in the following molecules?
a. S F 2
b . S F 4 c . XeF 4
12. How many sigma and pi bonds are there in a molecule of CO2 ?
13. The sulfur atom in selenium hexachloride is sp 3 d2 hybridized. What is the geometry of SeCl6 ? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 2 03 MOLECULAR ORBITAL THEORY
2. Describe the formation of bonding and antibonding molecular orbitals.
Use molecular orbital theory to predict for diatomic molecules the electron configurations, bond orders,
magnetic properties, and relative bond lengths. Molecular Orbitals. Until now we have considered covalent bonds in terms of the valence bond theory.
In that theory the electrons in a molecule occupy atomic orbitals of the individual atoms. Covalent bonds are
visualized in terms of the overlap of two half-filled atomic orbitals. This overlap increases the electron density
between both atomic centers.
The molecular orbital model has a different approach. In this model the available atomic orbitals are
combined to form orbitals that belong to the entire molecule. Like atomic orbitals, these molecular orbitals
have different shapes, sizes, and energies. Once the molecular orbitals have been derived, electrons are allotted to
them in a manner analogous to allotting electrons to the orbitals of atoms.
According to the molecular orbital (MO) theory, the result of the interaction of two 1s orbitals is the
formation of two molecular orbitals. One of these is a b onding molecular orbital , which has a lower
energy than the original 1s orbitals. The other molecular orbital is a n antibonding orbital , which is of
higher energy than the original 1s orbitals. It is the greater stability of the electrons in the bonding MO that
results in covalent bond formation. In the bonding MO, electron density is concentrated between the two atoms.
While the antibonding MO actually has a node (that is, a zero value) in the electron density midway between the
two nuclei. There is no contribution to bonding when electrons enter antibonding MOs. An energy level
diagram for the combination of two atomic orbitals of hydrogen to form the MOs in the H2 molecule is shown
in Figure 10.3.
σ1s Energy 1s 1s H
orbital σ1s H
Figure 10.3 Energy levels of bonding and antibonding molecular orbitals in the H2
molecule. The two electrons of the molecule enter the bonding molecular orbital (σ1s ).
Sigma antibonding and bonding molecular orbitals are formed from the combination of 1s
atomic orbitals. The bonding MO is lower in energy than the atomic orbitals, and the
antibonding MO is higher in energy than the atomic orbitals.
The antibonding and bonding MOs that result from combining 1s orbitals are examples of s igma
molecular orbitals . The electron density of sigma MOs is distributed symmetrically about the line of centers
between the two atoms. The sigma bonding MO and sigma antibonding MO resulting from combination of 1s
atomic orbitals are designated σ1s and σ1s *, respectively. Sigma MOs also result from the combination of Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 04 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
certain other atomic orbitals. For instance, the 2px orbitals on different atoms combine to form σ2p bonding
and σ2p * antibonding orbitals as shown in Figure 10.23 (a) of the text.
Two 2py orbitals and two 2p z orbitals on different atoms must approach each other sideways as shown in
Figure 10.23 (b) of the text. In the resulting molecular orbital, the electron density is concentrated above and
below the line joining the two bonded atoms, rather than around the line of centers. Such an MO is called a p i
molecular orbital . The symbol π 2p z stands for a bonding pi orbital formed by combination of two 2pz
atomic orbitals. The pi antibonding orbital is designated, π 2p z*.
Also, combination of two 2py atomic orbitals from different atoms forms π 2p y bonding and π 2p y *
antibonding orbitals as well. Molecular Orbital Configurations. For homonuclear diatomic molecules any two similar atomic
orbitals can merge to form two MOs, one bonding and one antibonding orbital. For H2 we can illustrate the
formation of the σ1s and σ1s * MO as the separate atoms are brought closer together in Figure 10.3. The H2
molecule has two electrons that are positioned in the lowest energy orbital, the σ1s . With both electrons in the
bonding MO the molecule is stable. We write the ground state MO configuration
H2 (σ1s )2
where the superscript "2" means two electrons occupy the sigma 1s orbital.
The molecular orbital theory predicts He2 to be unstable. A diatomic helium molecule would have four
electrons. Using the same diagram as for H2 we can see that the third and fourth electrons would enter the σ1s *
orbital, resulting in the configuration:
He2 (σ1s )2 (σ1s *)2
In this case the stability gained by having two electrons in a bonding orbital is cancelled by the presence of two
electrons in the antibonding orbitals. A stable He 2 molecule should not exist.
When comparing the stabilities of molecules, we use the concept of bond order.
bond order = 1
2 number of electrons – number of electrons in bonding MOs
in antibonding MOs The value of the bond order is a measure of the stability of the molecule. For H2
bond order = (1/2 )(2 – 0) = 1
and for He2
bond order = (1/2 )(2 – 2) = 0
A bond order of 1 represents a single covalent bond, and a bond order of 0 means the molecule is not stable.
Bond orders of 2 and 3 result for molecules with double bonds and triple bonds, respectively.
For most homonuclear diatomic molecules containing atoms of second period elements, an approximate
order of molecular energy levels is:
σ1s < σ1s * < σ2s < σ2s * < π 2p y = π 2p z < σ2p x < π 2p y * = π 2p z* < σ2p x *
The electron configurations of diatomic molecules of the second period elements and some of their known
properties are summarized in Table 10.5 in the textbook. The table shows the bond order of these molecules as
predicted by MO theory and the corresponding bond energies and bond lengths. The number of unpaired electrons
also correlates with the magnetic properties of these molecules. The molecular orbital theory satisfactorily
predicts the magnetic properties of molecules, something that the valence bond theory does not do. Example
10.7 illustrates writing an MO electron configuration and calculation of a bond order. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 2 05
A n umber of rules that apply to molecular orbitals and the stability of molecules are
s ummarized below.
1. The number of molecular orbitals formed is always equal to the number of atomic orbitals combined.
2. The more stable the bonding molecular orbital, the less stable is the corresponding antibonding molecular
3. In a stable molecule, the number of electrons in bonding molecular orbitals is always greater than that in
antibonding molecular orbitals.
4. As with atomic orbitals, each molecular orbital can accommodate up to two electrons with opposite spins,
in accordance with the Pauli exclusion principle.
5. When electrons are added to molecular orbitals having the same energy, the most stable arrangement is that
predicted by Hund's rule. That is, as far as possible, electrons occupy these orbitals singly, and with parallel
spins, rather than in pairs.
6. The number of electrons in the molecular orbitals is equal to the sum of all the electrons on the atoms. _______________________________________________________________________________
EXAMPLE 10.7 Molecular Orbital Diagram
– Represent bonding in the O2 ion by means of a molecular orbital diagram. What is the bond order?
•Method of Solution
– The O2 molecule has 16 electrons. The O 2 ion will have 17 electrons. The energy level diagram is shown
below. The 4 electrons in the σ1s and σ1s * orbitals are not shown because they are not involved in bonding.
π 2p *
π 2p σ2p π 2p π 2p
– The electron configuration of O2 ion is
– O2 ( σ1s ) 2 (σ1s *)2 (σ2s )2 (σ2s *)2 (π 2p y )2 (π 2p z)2 (σ2p x )2 (π 2p y *)2 (π 2p z*)1
bond order =
= Back Forward 1
2 number of electrons – number of electrons in bonding MOs
in antibonding MOs 1
(8 – 5) = 1.5
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EXAMPLE 10.8 Paramagnetism and Diamagnetism
Which of the following species are paramagnetic and which are diamagnetic:
+ b. N2 c. B2 •Method of Solution
a. The electron configuration of N2 is
N2 ( σ1s )2 (σ1s *)2 (σ2s )2 (σ2s *)2 (π 2p y )2 (π 2p z)2 (σ2p x )2
All orbitals contain pairs of electrons. Because of the lack of unpaired electrons, N2 is diamagnetic. b.
c. + N2 will have one less electron than N2 . Therefore it has one unpaired electron and is paramagnetic.
The electron configuration of B2 is
B2 ( σ1s )2 (σ1s *)2 (σ2s )2 (σ2s *)2 (π 2p y )1 (π 2p z)1
Because of the unpaired electrons, B2 is paramagnetic. _______________________________________________________________________________ EXERCISES
14. a. Which has the higher energy, a bonding molecular orbital or its corresponding antibonding molecular
orbital? b. Which has the higher energy, a bonding molecular orbital or the atomic orbitals from which it is
+ 15. Compare the bond order in Be2 to that in Be2 .
+ 16. Give the electron configurations for Li2 and Li2 . Which has the stronger Li—Li bond?
17. Which of the following diatomic species are paramagnetic?
a. O 2 b . O 2
c . Li 2
18. What orbitals overlap to form a delocalized π bond in ozone (O3 )?
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1. In molecules having the trigonal bipyramidal geometry why does a lone pair occupy an equitorial position
rather than an axial position? 2. Experimental evidence shows that carbon dioxide has no dipole moment. What does this suggest about its
molecular shape? 3. Both N2 and O2 can be ionized to form N2 and O2 , respectively. Explain why the bond order of N2 is
greater than that of N2 , but the bond order of of O 2 is less than that of O2 . + + PRACTICE TEST
1. Use VSEPR theory to predict the shapes of the following molecules:
b . HCN
c. BF 3
d. SO 2
e . SCl 2
2. What type of hybrid orbital is used by the central atom of each of the molecules in problem 1?
3. Use VSEPR theory to predict the shapes of the following polyatomic ions:
– a. BeF3 – b . AsF 4 – c . ClF 4 – d. NO3 2– e . SO 4 4. What type of hybridization is used by the central atom of each of the ions in problem 3?
5. Which of the following molecules would be tetrahedral?
a. SO 2
b . SiH 4
c . SF 4
6. Predict the approximate bond angles in
a. GeCl 2
b . IF 4
– 7. CCl 4 is a perfect tetrahedron, but AsCl4 is a distorted tetrahedron. Explain.
8. BeCl2 and TeCl2 are both covalent molecules, yet BeCl 2 is linear while TeCl2 is nonlinear (bent).
9. What types of hybrid orbitals can be formed by elements of the third period that cannot be formed by
elements of the second period?
10. When we describe the formation of hybrid orbitals on a central atom, must all the available atomic orbitals
enter into hybridization?
11. Which molecule should have the larger dipole moment, HBr or HI?
12. Which one of the following molecules has a dipole moment?
b. H2 S
c. CO 2
e . Cl 2
13. Which of the following molecules have no dipole moment?
d . Cl 2 O
e. H2 CO Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 08 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
14. Choose the best answer. For the water molecule:
a. The bonds are polar, and the molecule is nonpolar.
b. The bonds are nonpolar, and the molecule is polar.
c. The bonds are polar, and the molecule is polar.
d. The bonds are nonpolar, and the molecule is nonpolar.
15. Hydrogen peroxide (H2 O2 ) has a dipole moment of 2.1 D. Which of the bonds in H2 O2 are polar? Is the
16. Draw a molecular orbital energy level diagram and determine the bond order for each of the following.
a. H 2
2– 17. The compound calcium carbide, CaC2 , contains the acetylide ion C 2 .
2– a. Write molecular orbital electron configurations for C2 and C2 .
b. Compare their bond order.
+ 18. The bond length in N 2 is 109 pm and in N2 it is 112 pm. Explain why the bond lengths differ in this
14. a. linear
b. trigonal pyramidal
d. T shaped
e. square planar
b. trigonal planar c. tetrahedral
a. and c.
c. less than 109.5°
The two requirements are polar bonds and a molecular geometry in bonds dipoles cancell..
c. and d.
Each fluorine atom has the same electron configuration: 1s2 2s2 2px 2 2py 2 2pz1 . The 2p z orbitals of each F
atom are only half full. Therefore, a sigma bond can be formed by overlap of the 2pz orbital from each
atom. In HBr the 1s orbital of the H atom overlaps with the 4pz orbital of Br to form a sigma bond.
a. 180° b. 120° c. 109.5°
a. sp b. sp2 c . sp 2 d. sp3 d e. sp2
a. sp2 b. sp3 d c. sp3 d2
2 σ and 2 π
The antibonding MO.
+ 15. The bond order for Be2 is zero while for Be2 i t is 0.5.
16. Li2 ( σ1s )2 (σ1s *)2 (σ2s )2 L i 2 + ( σ1s )2 (σ1s *)2 (σ2s )1 L i 2 has the stronger bond. 17. a. and b. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals / 2 09
18. The central oxygen atom is sp 2 hybridized and forms σ bonds to the two terminal oxygen atoms. The
central O atom has an unhybridized 2pz orbital that overlaps "sideways" with the 2p z orbitals of both
terminal O atoms. Conceptual Questions
3. Recall that lone-pair electrons take up more space than electrons in a bond. A lone pair in an axial position
has three close neighbors in the bonding electron pairs in the equatorial positions. A lone pair in an
equatorial position has two close neighbors in the bonding electron pairs in the axial positions. (The
neighboring equatorial electron pairs are too far away). This means that a lone pair will have more room and
have less repulsion when it is in a equatorial position.
We know that the C—O bonds are polar. For CO2 to be nonpolar, the bond dipoles must be pointing in
opposite directions so that they cancel. The molecule must be linear.
Compare the electron configurations.
N2 ( σ1s )2 (σ1s )2 (σ2s )2 (σ2s )2 (π 2p y )2 (π 2p z)2 (σ2p x )2
N2 ( σ1s ) 2 (σ1s )2 (σ2s )2 (σ2s )2 (π 2p y )2 (π 2p z)2 (σ2p x )1
O2 ( σ1s )2 (σ1s )2 (σ2s )2 (σ2s )2 (π 2p y )2 (π 2p z)2 (σ2p x )2 (π 2p y *)1 (π 2p z*)1
O2 ( σ1s ) 2 (σ1s )2 (σ2s )2 (σ2s )2 (π 2p y )2 (π 2p z)2 (σ2p x )2 (π 2p y *)1
+ The bond order in N2 is 3. Loss of 1 electron gives N2 . The electron came out of a bonding orbital and so
the bond order is less than in N2 , b.o. = 2.5. The bond order in O2 is 2. Loss of 1 electron gives O2 .
The electron came out of an antibonding orbital and so the bond order is greater than in O2 , b.o. = 2.5. Practice Test
1. a. Square pyramid b. linear c. triangular d. bent e. bent
2. a. sp 3 d2
c. sp 2 d. sp2 e. sp 2
3. a. triangular b. distorted tetrahedral c. square planar
d. triangular e. tetrahedral
4. a. sp 2
b. sp3 d c. sp 3 d2 d. sp2 e. sp 3
5. b. SiH4
6. a. 120°
c. 90° and 120°
7. CCl 4 has four electron pairs about the central carbon atom, while AsCl 4 has 5 pairs about the As atom
with only 4 Cl atoms attached. The extra pair of unshared electrons prevents the formation of a tetrahedron
by AsCl 4 .
8. There are two electron pairs in the valence shell of Be in BeCl2 , but there are four pairs in the valence shell
of Te in TeCl 2 . These four pairs are at the corners of a tetrahedron. When two chlorine atoms bond to a Te
atom via two of the pairs, the Cl—Te—Cl bond angle is 109.5°.
9. sp3 d, sp3 d2
12. b. H 2 S
13. b. BCl 3
c. BeCl2 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 2 10 / Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
15. The O—H bonds are polar. Nonlinear.
HHe + He2 σ1s * ___ ↑ ↑ σ1s ↑ ↑↓ ↑↓ Bond
order 0.5 0.5 0.5 17. a. C 2 (σ1s )2 (σ1s *)2 (σ2s )2 (σ2s *)2 (π 2p y )2 (π 2p z)2
2– C2 ( σ1s ) 2 (σ1s *)2 (σ2s )2 (σ2s *)2 (π 2p y )2 (π 2p z)2 (σ2p x )2
2– b. For C2 the bond order = 2. For C 2 the bond order = 3.
18. The bond order in N2 is 3, while in N2 it is only 2.5. The greater the bond order, the shorter the bond
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