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Chemical Bonding II:

Molecular Geometry and

Hybridization of Atomic

Orbitals

Chapter 10

Chang & Goldsby

Modified by Dr. Juliet HahnCopyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Example 10.2

Predict whether each of the following molecules has a dipole

moment:

(a)BrCl

(b)BF3 (trigonal planar)

(c) CH2Cl2 (tetrahedral)

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9am class

Example 10.2 (1)

Strategy

Keep in mind that the dipole moment of a molecule depends on

both the difference in electronegativities of the elements

present and its geometry.

A molecule can have polar bonds (if the bonded atoms have

different electronegativities), but it may not possess a dipole

moment if it has a highly symmetrical geometry.

Example 10.2 (2)Solution

(a) Because bromine chloride is diatomic, it has a linear

geometry. Chlorine is more electronegative than bromine

(see Figure 9.5), so BrCl is polar with chlorine at the negative end Copyright © McGraw-Hill Education permission required

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Thus, the molecule does have a dipole moment. In fact, all

diatomic molecules containing different elements possess a

dipole moment.

Example 10.2 (3)

(b) Because fluorine is more electronegative than boron, each

B − F bond in BF3 (boron trifluoride) is polar and the three bond moments are equal. However, the symmetry of a

trigonal planar shape means that the three bond moments

exactly cancel one another:Copyright © McGraw-Hill Education permission required

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An analogy is an object that is pulled in the directions shown

by the three bond moments. If the forces are equal, the

object will not move. Consequently, BF3 has no dipole moment; it is a nonpolar molecule.

Example 10.2 (4)

(c) The Lewis structure of CH2Cl2 (methylene chloride) is

This molecule is similar to CH4 in that it has an overall tetrahedral shape. However, because not all the bonds are

identical, there are three different bond angles: HCH, HCCl, and ClCCl. These bond angles are close to, but not equal to, 109.5° .

Example 10.2 (5)

Because chlorine is more electronegative than carbon,

which is more electronegative than hydrogen, the bond

moments do not cancel and the molecule possesses a

dipole moment:

Thus, CH2Cl2 is a polar molecule.

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Hybridization

Hybridization – mixing of two or more atomic

orbitals to form a new set of hybrid orbitals

1. Mix at least 2 nonequivalent atomic orbitals (e.g. s and p).

Hybrid orbitals have very different shape from original

atomic orbitals.

2. Number of hybrid orbitals is equal to number of pure atomic

orbitals used in the hybridization process.

3. Covalent bonds are formed by:

a. Overlap of hybrid orbitals with atomic orbitals

b. Overlap of hybrid orbitals with other hybrid orbitals

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Formation of 𝑠𝑝3 Hybrid Orbitals

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Formation of Covalent Bonds in CH4

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𝑠𝑝3-Hybridized N Atom in NH3

Predict correct

bond angle

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Lone pair electron occupies

slightly more space than a

bond to an atom

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Formation of 𝑠𝑝 Hybrid Orbitals

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Formation of 𝑠𝑝2 Hybrid Orbitals

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How do I predict the hybridization of the central atom?

1. Draw the Lewis structure of the molecule.

2. Count the number of lone pairs AND the number of

atoms bonded to the central atom

# of Lone Pairs+

# of Bonded Atoms Hybridization Examples

2 𝑠𝑝 BeCl2

3 𝑠𝑝2 BF3

4 𝑠𝑝3 CH4, NH3, H2O

5 𝑠𝑝3𝑑 PCl5

6 𝑠𝑝3𝑑2 SF6

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Important Hybrid Orbitals

Example 10.3

Determine the hybridization state of the central (underlined)

atom in each of the following molecules:

(a)BeH2

(b)AlI3

(c) PF3

Describe the hybridization process and determine the molecular

geometry in each case.

Example 10.3 (1)

Strategy The steps for determining the hybridization of the

central atom in a molecule are:

draw Lewis structure

of the molecule ⟶

use VSEPR to determine the

electron pair arrangement

surrounding the central atom

(Table 10.1)

⟶use Table 10.4 to

determine the

hybridization state of

the central atom

Solution

(a) The ground-state electron configuration of Be is 1𝑠22𝑠2 and the Be atom has two valence electrons. The Lewis structure

of BeH2 is

H—Be—H

Example 10.3 (2)

There are two bonding pairs around Be; therefore, the electron

pair arrangement is linear. We conclude that Be uses sp hybrid

orbitals in bonding with H, because sp orbitals have a linear

arrangement (see Table 10.4). The hybridization process can

be imagined as follows. First, we draw the orbital diagram for

the ground state of Be:

By promoting a 2𝑠 electron to the 2𝑝 orbital, we get the excited state:

Example 10.3 (3)The 2𝑠 and 2𝑝 orbitals then mix to form two hybrid orbitals:

The two Be − H bonds are formed by the overlap of the Be sp

orbitals with the 1s orbitals of the H atoms. Thus, BeH2 is a linear molecule.

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Example 10.3 (4)

(b) The ground-state electron configuration of Al is Ne 3𝑠23𝑝1. Therefore, the Al atom has three valence electrons. The

Lewis structure of AlI3 is

There are three pairs of electrons around Al; therefore,

the electron pair arrangement is trigonal planar. We

conclude that Al uses 𝑠𝑝2 hybrid orbitals in bonding with I because 𝑠𝑝2 orbitals have a trigonal planar arrangement. The orbital diagram of the ground-state Al atom is

Example 10.3 (5)

By promoting a 3𝑠 electron into the 3𝑝 orbital we obtain the following excited state:

The 3𝑠 and two 3𝑝 orbitals then mix to form three 𝑠𝑝2 hybrid orbitals:

The 𝑠𝑝2 hybrid orbitals overlap with the 5𝑝 orbitals of I to form three covalent Al − I bonds. We predict that the AlI3 molecule is trigonal planar and all the lAll angles are 120° .

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Example 10.3 (6)

(c) The ground-state electron configuration of P is Ne 3𝑠23𝑝3. Therefore, P atom has five valence electrons. The Lewis

structure of PF3 is

There are four pairs of electrons around P; therefore, the

electron pair arrangement is tetrahedral. We conclude that P

uses 𝑠𝑝3 hybrid orbitals in bonding to F, because 𝑠𝑝3 orbitals have a tetrahedral arrangement. The hybridization process can

be imagined to take place as follows. The orbital diagram of the

ground-state P atom is

Example 10.3 (7)

By mixing the 3𝑠 and 3𝑝 orbitals, we obtain four 𝑠𝑝3 hybrid orbitals.

As in the case of NH3, one of the 𝑠𝑝3 hybrid orbitals is used to

accommodate the lone pair on P. The other three 𝑠𝑝3 hybrid orbitals form covalent P − F bonds with the 2p orbitals of F. We predict the geometry of the molecule to be trigonal pyramidal;

the F − F angle should be somewhat less than 109.5° .

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Example 10.4

Describe the hybridization state of phosphorus in phosphorus

pentabromide PBr5 .

PBr5

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Example 10.4 (1)

Strategy Follow the same procedure shown in Example 10.3.

Solution The ground-state electron configuration of P is

Ne 3𝑠23𝑝3.Therefore, the P atom has five valence electrons. The Lewis structure of PBr5 is

There are five pairs of electrons around P; therefore, the

electron pair arrangement is trigonal bipyramidal. We conclude

that P uses 𝑠𝑝3𝑑 hybrid orbitals in bonding to Br, because 𝑠𝑝3𝑑hybrid orbitals have a trigonal bipyramidal arrangement.

Example 10.4 (2)

The hybridization process can be imagined as follows. The

orbital diagram of the ground-state P atom is

Promoting a 3𝑠 electron into a 3𝑑 orbital results in the following excited state:

Example 10.4 (3)

Mixing the one 3𝑠, three 3𝑝, and one 3𝑑 orbitals generates five 𝑠𝑝3𝑑 hybrid orbitals:

These hybrid orbitals overlap the 4𝑝 orbitals of Br to form five covalent P − Br bonds. Because there are no lone pairs on the P atom, the geometry of PBr5 is trigonal bipyramidal.

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𝑠𝑝2 Hybridization of Carbon

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𝑠𝑝2 Hybridization of Carbon (1)

Unhybridized 2𝑝z orbital (gray), which is perpendicular to the plane of the hybrid (green) orbitals.

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Bonding in Ethylene, C2H4

Sigma bond σ – electron density between the 2 atoms

Pi bond π – electron density above and below plane of nuclei of the bonding atoms

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10 am class