Chapter 9. Molecular Geometry and Bonding Theories · 2020. 12. 30. · Chapter 9. Molecular...

Chapter 9. Molecular Geometry and Bonding Theories

9.1 Molecular Shapes

Read p. 342-346

In Ch 8, we learned that LEWIS STRUCTURES help us understand the compositions of molecules and their

covalent bonds.

• However, Lewis structures do not show one of the most important aspects of molecules—their overall

shapes.

• The shape and size of molecules—sometimes referred to as molecular geometry––are defined by the

angles and distances between the nuclei of the component atoms.

• The SHAPE and SIZE of a molecule of a substance, together with the STRENGTH and

POLARITY of its bonds, largely determine the properties of that substance.

Our first goal is to learn the relationship between two-dimensional Lewis structures and three-dimensional

molecular shapes.

• We then can examine the nature of covalent bonds. The lines used to depict bonds in Lewis structures

provide important clues about the ORBITALS that molecules use in bonding.

• By examining these orbitals, we can gain a greater understanding of the behavior of molecules. For example, the Lewis structure of CCl4 tells us only that four Cl atoms are bonded to a central C atom: The Lewis structure is drawn with the atoms all in the same plane.

As shown in FIGURE 9.1, however, the actual arrangement is the Cl atoms at the corners of a tetrahedron, a

geometric object with four corners and four faces, each an equilateral triangle.

Ch 9: Molecular Geometry and Bonding 2

The shape of a molecule is determined by its BOND ANGLES, the angles made by the

lines joining the nuclei of the atoms in the molecule.

• The bond angles of a molecule, together with the bond lengths (Ch 8.8), define the shape and size

of the molecule.

• In Figure 9.1, you should be able to see that there are six Cl – C – Cl bond angles in CCl4 and that they

all have the same value of 109.5o, the angle size characteristic of a tetrahedron.

• In addition, all four bonds are the same length (1.78 Å).

• Thus, the shape and size of CCl4 are completely described by stating that the molecule is tetrahedral

with bonds of length 1.78 Å.

Let’s examine molecular shapes with molecules (and ions) that, like CCl4, have a single central atom bonded to

two or more atoms of the same type. • Such molecules have the general formula ABn in which the central atom A is bonded to n B atoms. • Both CO2 and H2O are AB2 molecules, for example, whereas SO3 and NH3 are AB3 molecules, and so

on. • The number of shapes possible for ABn molecules depends on the value of n. se

When A is a representative element (one from the s block or p block of the periodic table), we can answer these

questions by using the VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) model. • The model has useful predictive capabilities for determining shape names and bond angles.

9.2 The VSEPR Model

Read p. 347-356

In Ch 8.3, we examined how a single covalent bond is formed between two atoms when a pair of electrons

occupies the space between the atoms.

• A BONDING PAIR of electrons thus defines a region in which the electrons are most likely to be

found.

• We will refer to such a region as an ELECTRON DOMAIN.

• A NONBONDING PAIR (or LONE PAIR) of electrons defines an electron domain that is located

only on one atom.

• For example, the Lewis structure of NH3 has four electron domains around the central nitrogen atom

(three bonding pairs, represented as usual by short lines, and one nonbonding pair, represented by dots):

3

• Each multiple bond in a molecule also constitutes a single electron domain.

• The resonance structure for O3 has three electron domains around the central oxygen atom (a single

bond, a double bond, and a nonbonding pair of electrons):

In general, each nonbonding pair, single bond, or multiple bond produces a single electron domain around the central atom in a molecule.

• So, every electron region around the CENTRAL atom MATTERS! The VSEPR model is based on the idea that electron domains are negatively charged and therefore repel one

another.

• Electron domains try to stay out of one another’s way.

• The best arrangement of a given number of electron domains is the one that minimizes the repulsions

among them. 2 Types of Geometry’s:

1. The arrangement of electron domains about the central atom of an ABn molecule is its ELECTRON-DOMAIN GEOMETRY.

• Examines TOTAL regions of electrons around the CENTRAL atom (bonding and nonbonding). 2. MOLECULAR GEOMETRY is the arrangement of ONLY the ATOMS in the molecule or ion – any

nonbonding pairs in the molecule are not part of the description. To determine the shape of any molecule,

• First use the VSEPR model to predict the electron-domain geometry. o How many TOTAL regions of e- around the central atom?

• Then determine how many of those regions are bonding vs. nonbonding regions or e- around central atom.

• If all the electron domains in a molecule are ALL bonds, then the molecular geometry is identical to the electron-domain geometry.

• If one or more domains involve nonbonding pairs of e-, then the molecular geometry involves only electron domains due to bonds even though the nonbonding pairs contribute to the electron-domain geometry.

To predicting the shapes of molecules/ions using VSEPR:

1. Draw the Lewis structure of the molecule/ion. 2. Count the TOTAL # of e- domains around the CENTRAL ATOM. 3. Each nonbonding electron pair and each bond (single, double, triple) count as a one electron domain.

o Be careful with multiple bonds like double/triple bonds: They still only count as a region of e-. They are just sharing more e- in that ONE region.

4. Determine the electron-domain geometry by arranging the electron domains about the central atom so that the REPULSIONS among them are MINIMIZED. (See Table 9.2 and 9.3)


5. Determine the molecular geometry – bonding AND nonbonding regions of e- around CENTRAL ATOM.

TABLE 9.2 summarizes the possible molecular geometries when an ABn molecule

has four or fewer electron domains about A. These geometries are important because

they include all the shapes usually seen in molecules or ions that OBEY THE OCTET RULE.

Bond Angles

180 o

120 o

Less than 120 o

109.5 o

107 o

104.5 o

Polar or Nonpolar

Nonpolar (Some 2 or 3

atom

molecules can

be polar)

Nonpolar

Polar

Nonpolar

Polar

Polar

5

Let’s look at 2 examines and using VSEPR model to determine the electron-domain and molecular geometry names:

Ammonia (NH3) has 3 bonding regions and 1 nonbonding region in the Lewis structure, which tells us it has a

TOTAL of 4 electron domains.

• Using table 9.2, we know that the electron-domain name would be tetrahedral.

• Since one of the regions around N atom has a set of lone pairs, the molecular geometry name will be

different from the electron-domain name.

• Since there are 3 N – H domains, the molecular geometry name will be trigonal pyramidal.

• The bond angle deviates from the ideal value of 109.5o (electron-domain of tetrahedral. The lone pairs

push the bonds away and the bond angle for trigonal pyramidal would be 107o.

As one more example, let’s determine the shape of the CO2 molecule. Its Lewis structure

reveals two electron domains (each one a double bond) around the central carbon:

• Two electron domains orient in a linear electron-domain geometry.

• Because neither domain is a nonbonding pair of electrons, the molecular geometry is also linear, and the

bond angle is 180o .

Try it out:

pHET Simulation on Molecular Shapes : https://phet.colorado.edu/en/simulation/molecule-shapes

https://phet.colorado.edu/en/simulation/molecule-shapes


1. Draw all possible Lewis structures for the following molecules/ions and predict the MOLECULAR

GEOMETRY: O3

SnCl3

-1

SeCl2

Examples

7

CO3

-2

Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles

For example, consider methane (CH4), ammonia (NH3), and water (H2O). All three have a tetrahedral electron-

domain geometry, but their bond angles differ slightly:

• Notice that the bond angles decrease as the number of nonbonding electron pairs increases.

• A bonding pair of electrons is attracted by both nuclei of the bonded atoms, but a nonbonding pair is

attracted primarily by only one nucleus.

• Because a nonbonding pair experiences less nuclear attraction (WHY?? Because lone pairs have a

greater electron-electron repulsion), its electron domain is spread out more in space than is the

electron domain for a bonding pair.

• Nonbonding electron pairs therefore take up more space than bonding pairs. As a result, electron

domains for nonbonding electron pairs exert GREATER REPULSIVE FORCES and tend to DECREASE

BOND ANGLES.

Because multiple bonds contain a higher electron density region (sharing 4 or 6 e-) than single

Bonds (sharing 2e-), multiple bonds also represent enlarged electron domains. Consider the Lewis

structure of phosgene:

• Because three electron domains surround the central atom, we might expect a

trigonal planar geometry with 120o bond angles.

• The double bond, however, seems to act much like a nonbonding pair of

electrons, reducing the Cl – C – Cl bond angle to 111.4o.

• In general, electron domains for MULTIPLE bonds exert a GREATER

REPULSIVE FORCE on adjacent electron domains than do electron domains for single bonds.

Examples


Molecules with Expanded Valence Shells

Atoms from period 3 and beyond may be surrounded by more than four electron pairs. (Ch 8.7)

• Molecules with five or six electron domains around the central atom have molecular geometries based

on either a trigonal-bipyramidal (five domains) or octahedral (six domains) electron-domain

geometry (See TABLE 9.3).

The most stable electron-domain geometry for five electron domains is the trigonal

bipyramidal (two trigonal pyramids sharing a base).

• Unlike the other arrangements we have seen, the electron

domains in a trigonal bipyramid can point toward two

geometrically distinct types of positions.

• Two domains point toward axial positions and three point

toward equatorial positions (See FIGURE 9.8).

• Each axial domain makes a 90o angle with any equatorial

domain.

• Each equatorial domain makes a 120o angle with either of

the other two equatorial domains and a 90o angle with either

axial domain.

Suppose a molecule has five electron domains, and there are one or

more nonbonding pairs.

• Will the domains from the nonbonding pairs occupy axial or equatorial positions?

• To answer this question, we must determine which location minimizes repulsion between domains.

• Repulsion between two domains is much greater when they are situated 90o from each other than

when they are at 120o.

• An equatorial domain is 90o from only two other domains (the axial domains), but an axial domain is

90o from three other domains (the equatorial domains).

• Hence, an equatorial domain experiences less repulsion than an axial domain.

• Because the domains from nonbonding pairs exert larger repulsions than those from bonding pairs,

nonbonding domains always occupy the equatorial positions in a trigonal bipyramid. The most stable electron-domain geometry for six electron domains is octahedral.

• An octahedral has six vertices and eight faces, each an equilateral triangle.

• All the bond angles are 90o, and all six vertices are equivalent.

• Therefore, if an atom has five bonding electron domains and one nonbonding

domain, we can put the nonbonding domain at any of the six vertices of the

octahedron.

• The result is always a square-pyramidal molecular geometry.

• When there are two nonbonding electron domains, however, their repulsions

are minimized by pointing them toward opposite sides of the octahedral,

producing a square-planar molecular geometry, as shown in Table 9.3.

9

Table 9.3 – Expanded Octet Geometries

Bond Angles

120o and 90o

90o

Nonpolar

Nonpolar

Nonpolar

Polar

Polar

Polar

Nonpolar


2. Draw Lewis structures for the following EXPANDED OCTET molecules. Predict the ELECTRON DOMAIN & MOLECULAR GEOMETRIES:

SF4 IF5

BrF3 ICl4

-1

Examples

11

Shapes of Larger Molecules

Although the molecules and ions we have considered contain only a single central atom, the VSEPR model can

be extended to more complex molecules. What if the molecule has MORE THAN ONE CENTRAL ATOM?

For example, the acetic acid molecule (Vinegar):

we can use the VSEPR model to predict the geometry about each atom:

• The left C has four electron domains (all bonding), and so the

electron-domain and molecular geometries around that atom are both

tetrahedral.

• The central C has three electron domains (counting the double bond

as one domain), making both the electron-domain and the molecular

geometries trigonal planar.

• The O on the right has four electron domains (two bonding, two

nonbonding), so its electron-domain geometry is tetrahedral and its

molecular geometry is bent.

• The bond angles about the central C atom and the O atom are

expected to deviate slightly from the ideal values of and because of

the spatial demands of multiple bonds and nonbonding electron pairs.

• The structure of the acetic acid molecule is shown in FIGURE 9.9.


3.

4.

Examples

13

9.3 Molecular Shape and Molecular Polarity

Read p. 356 - 358

Recall that BOND POLARITY is a measure of how equally the electrons in a bond are shared between the two

atoms of the bond.

• As the difference in electronegativity between the two atoms INCREASES, so does the bond polarity.

(Ch 8.4)

For a molecule consisting of more than two atoms, the dipole moment

depends on both the polarities of the individual bonds and the

geometry of the molecule.

• For each bond in the molecule, we consider the BOND

DIPOLE, which is the dipole moment due only to the two

atoms in that bond.

• Consider the linear CO2 molecule as shown in FIGURE 9.10,

each C = O bond is polar, and because the C = O bonds are

identical, the bond dipoles are equal in magnitude (size).

• A plot of the molecule’s electron density clearly shows that the

individual bonds are polar, but what can we say about the

OVERALL dipole moment of the molecule? Bond dipoles and dipole moments are vector quantities; that is, they have both a magnitude and a direction.

• The two bond dipoles in CO2, although equal in magnitude, are

opposite in direction.

• Adding them is the same as adding two numbers that are equal

in magnitude but opposite in sign, such as 1 + (-1) = 0.

• The bond dipoles, like the numbers, “CANCEL” each other.

• Therefore, the dipole moment of CO2 is zero, even though the individual bonds are polar.

• The geometry of the molecule dictates that the OVERALL DIPOLE MOMENT be zero, making CO2

a NONPOLAR molecule.

Nonpolar molecule: all of the bond dipoles within the molecule cancel each other out.


Now consider H2O, a bent molecule with two polar bonds (FIGURE

9.11).

• Again, the two bonds are identical (both are POLAR), and the

bond dipoles are equal in magnitude.

• Because the molecule is bent, however, the bond dipoles do NOT

directly oppose each other and therefore do NOT cancel.

• Hence, the H2O molecule has an overall NONZERO dipole

moment and is therefore a POLAR molecule.

• The oxygen atom carries a partial negative charge, and the

hydrogen atoms each have a partial positive charge, as shown in

the electron-density model.

• The oxygen molecule has 2 sets on lone pairs. So not all the

dipole moments cancel out. Polar molecule: all of the bond dipoles within the molecule do NOT cancel each other out.

Examples of Nonpolar vs Polar molecules:

NON-POLAR MOLECULES - symmetry causes dipoles to cancel out

AP FRQ TIP: CANNOT JUST SAY “BECAUSE OF SYMMETRY IT IS NOT POLAR” YOU NEED

TO EXPLICITLY SAY “DIPOLES CANCEL OUT DUE TO SYMMETRY”

15

POLAR MOLECULES - dipoles DO NOT cancel out

Beware -of drawing it on paper and it looks like the dipoles cancel,

but the TRUE MOLECULAR shape does not cancel out the dipoles.

You have to take into account BOTH the molecular geometry shape

and actual atoms in the bond dipoles.

POLAR GEOMTRY - all these geometries have bond dipoles within the molecule to not cancel out


These examples give you a false sense that the dipoles cancel.

The Lewis structure does not necessarily indicate the molecular geometry

WATCH AP You tube if you need additional help: Unit 3.1-3.3

H20

NH3

CH2F2

Examples

https://www.youtube.com/watch?v=NczVUxmtAk8&t=210s

17

1. Predict whether the following are polar or nonpolar. Draw Lewis structures for the following molecules. Predict the ELECTRON DOMAIN & MOLECULAR GEOMETRIES: (p.357)

SF6 BrCl

SO2 SiCl4

BF3 HCl

Examples


9.4 Covalent Bonding and Orbital Overlap

Read p. 358-359

Lewis structures and VSEPR theory give us the shape and location of electrons in a molecule.

• They do not explain why a chemical bond forms. • How can quantum mechanics be used to account for molecular shape? • What are the orbitals that are involved in bonding?

We use VALENCE-BOND THEORY – a covalent bond forms when the orbitals on two atoms OVERLAP.

• In Lewis theory, covalent bonding occurs when atoms share electrons because the sharing concentrates electron density between the nuclei.

• In valence-bond theory, we visualize the buildup of electron density between two nuclei as occurring

when a valence atomic orbital of one atom shares space, or overlaps, with a valence atomic orbital of

another atom.

• The overlap of orbitals allows two electrons of opposite spin to share the space between the nuclei,

forming a covalent bond.

The coming together of two H atoms to form H2 is seen in FIGURE 9.13.

• Each atom has a single electron in a 1s orbital.

• As the orbitals overlap, electron density is concentrated between the

nuclei.

• Because the electrons in the overlap region are simultaneously attracted

to both nuclei, they hold the atoms together, forming a covalent bond.

The idea of orbital overlap producing a covalent bond applies equally well to

other molecules.

• In HCl, for example, chlorine has the electron configuration [Ne]3s23p5.

• All the valence orbitals of chlorine are full except one 3p orbital, which

contains a single electron.

• This 3p electron pairs with the single 1s electron of H to form a covalent

bond (Figure 9.13).

• Because the other two chlorine 3p orbitals are already filled with a pair

of electrons, they do not participate in the bonding to hydrogen.

Likewise, we can explain the covalent bond in Cl2 in terms of the overlap of the

singly occupied 3p orbital of one Cl atom with the singly occupied 3p orbital of

another.

19

Learning Objective Essential Knowledge

Unit 2.2

SAP-3.B Represent the

relationship between

potential energy and distance between atoms,

based on factors that

influence the interaction

strength.

SAP-3.B.1 A graph of potential energy versus the distance between atoms is a useful representation for describing the interactions between

atoms. Such graphs illustrate both the equilibrium bond length (the

separation between atoms at which the potential energy is lowest) and the

bond energy (the energy required to separate the atoms).

SAP-3.B.2 In a covalent bond, the bond length is influenced by both the

size of the atom’s core and the bond order (i.e., single, double, triple).

Bonds with a higher order are shorter and have larger bond energies.

SAP-3.B.3 Coulomb’s law can be used to understand the strength of interactions between cations and anions.

a. Because the interaction strength is proportional to the charge on each

ion, larger charges lead to stronger interactions. b. Because the interaction strength increases as the distance between the

centers of the ions (nuclei) decreases, smaller ions lead to stronger

interactions.

There is always an optimum distance between the two nuclei in any covalent bond.

• FIGURE 9.14 shows how the potential energy of a system consisting of two H atoms changes as the

atoms come together to form an H2 molecule.

• When the atoms are infinitely far apart, they do not “feel” each other and so the energy approaches zero.

• As the distance between the atoms decreases, the overlap between their 1s orbitals increases.

• Because of the increase in electron density between the nuclei, the potential energy of the system

decreases.

• That is, the strength of the bond increases, as shown by the decrease in the potential energy of the two-

atom system.

• However, Figure 9.14 also shows that

as the atoms come closer together than

0.74 Å, the energy increases sharply.

• This increase, which becomes

significant at short internuclear

distances, is due mainly to the

electrostatic repulsion between the

nuclei.

• The INTERNUCLEAR DISTANCE,

or BOND LENGTH, is the distance

that corresponds to the minimum of

the potential-energy curve.

• The potential energy at this minimum

corresponds to the BOND

STRENGTH.

• Thus, the observed BOND LENGTH

is the distance at which the attractive

forces between unlike charges (electrons and nuclei) are balanced by the repulsive forces between like

charges (electron–electron and nucleus–nucleus).


AP Chemistry: 2.1-2.4 Chemical Bonds, Intramolecular Force, and Structure of Solids WATCH THIS VIDEO BY MICHAEL FARABAUGH TO ASSIST IN THIS QUESTIONS(~16:30min)

❏ Internuclear distance – means the distance between the 2 nuclei atoms

The distance between the 2 atoms is large. No bond is formed. No attraction or

repulsions. PE = 0

This is the most stable arrangement.

There is a balance between the attractions and the repulsions. A stable bond is formed. PE is at a minimum (very negative PE). Remember – forming bonds RELEASE ENERGY

Atoms are closer together. No bond is formed yet, but there is

an attraction between the atoms. PE is less than 0, which means it

is becoming more stable.

How much energy needed to BREAK the bond (very positive PE) Remember – breaking

bonds NEED

ENERGY

Higher energy – less stable arrangement

of the atoms

https://www.youtube.com/watch?v=BcRVEB_6PV0&list=PLoGgviqq4845Sy3UfnNh_PljzAptMR7MQ&index=3

21

Trends in Bond Length and Bond Energy

❏ Atomic Size (Use trends)

❏ The smaller the bond, more energy required to break (larger bond

energy – more difficult to break)

❏ The longer the bond, less energy required to break (smallest bond

energy – easier to break)

❏ Bond order ( single, double or triple)

❏ Bond order means how many bonds the atoms have between them.

❏ As bond order increases, bond length decreases and bond energy

increases

❏ Single bonds are longer (sharing 2 e-) in distances between the

atoms and require LESS energy to break (compared to multiple

bonds)

❏ Triple bonds are shorter in distances between the atoms (sharing

6e-) so they require MORE energy to break (more difficult to

break)

AP Chemistry: 2.1-2.4 Chemical Bonds, Intramolecular Force, and Structure of Solids WATCH THIS VIDEO BY MICHAEL FARABAUGH TO ASSIST IN THIS QUESTIONS(~19min)

1. Use your knowledge of periodic trends in atomic radius to answer the following question.

H – F H – Cl H – Br H - I

a. Which of these bonds is the shortest?

b. Which of these bonds is the longest?

c. Which bond is the easiest to break (i.e., smallest bond energy)?

d. Which bond is the most difficult to break (i.e., largest bond energy)?

Examples



2. C – C C=C C = C

a., Which of these bonds is the shortest?

b. Which of these bonds is the longest?

Example: Diagram of Potential Energy versus Internuclear Distance

AP Chemistry: 2.1-2.4 Chemical Bonds, Intramolecular Force, and Structure of Solids WATCH THIS VIDEO BY MICHAEL FARABAUGH (~30min)

Consider the following information for the ClーCl bond:

Bond ClーCl BrーBr

Bond Length (pm) 200 ?

Bond Energy (kJ/mol) 243 ?

Make a prediction about the BrーBr bond, in terms of bond length and bond energy.

Examples


23

9.5 Hybrid Orbitals

Read p. 359 - 365

To explain molecular geometries, we can assume that the atomic orbitals on an atom (usually the central atom)

mix to form new orbitals called HYBRID ORBITALS.

• The shape of any hybrid orbital is different from the shapes of the original atomic orbitals.

• The process of mixing atomic orbitals is a mathematical operation called HYBRIDIZATION.

• The total number of atomic orbitals on an atom remains constant, so the number of hybrid orbitals on an

atom equals the number of atomic orbitals that are mixed.

As we examine the common types of hybridization, notice the connection between the type of hybridization and

certain of the molecular geometries predicted by the VSEPR model: linear, bent, trigonal planar, and

tetrahedral.

sp Hybrid Orbitals – 2 electron domains (LINEAR)

To illustrate the process of hybridization, consider the BeF2 molecule, which has the following

Lewis structure:

• The VSEPR model correctly predicts that BeF2 is linear with two identical bonds.

• How can we use valence-bond theory to describe the bonding?

• The electron configuration of F (1s22s22p5) indicates an unpaired electron in a 2p orbital.

• This electron can be paired with an unpaired Be electron to form a polar covalent bond.

• Which orbitals on the Be atom, however, overlap with those on the F atoms to form the Be – F bonds?

The orbital diagram for a ground-state Be atom is

• Because it has no unpaired electrons, the Be atom in its ground state cannot bond with the fluorine

atoms.

• The Be atom could form two bonds, however, by “promoting” one of the 2s electrons to a 2p orbital:


• The Be atom now has two unpaired electrons and can therefore form two polar covalent bonds with F

atoms.

• The two bonds would not be identical, however, because a Be 2s orbital would be used to form one of

the bonds and a 2p orbital would be used to form the other.

• Therefore, although the promotion of an electron allows two Be – F bonds to form, we still have not

explained the structure of BeF2.

• We can solve this dilemma by “mixing” the 2s orbital with one 2p orbital to generate two new orbitals,

as shown in FIGURE 9.15.

• Like p orbitals, each new orbital has two lobes. Unlike p orbitals, however, one lobe is much larger than

the other.

• The two new orbitals are identical in shape, but their large lobes point in opposite directions.

• These two new orbitals, which we color-code purple in Figure 9.15, are hybrid orbitals.

• Because we have hybridized one s and one p orbital, we call each hybrid an sp hybrid orbital.

• According to the valence-bond model, a linear arrangement of electron domains implies sp

hybridization.

sp2 Hybrid Orbitals – 3 electron domains: Trigonal planar and bent

In BF3, for example, mixing the 2s and two of the 2p atomic orbitals yields three equivalent sp2

(pronounced “s-p two”) hybrid orbitals (see FIGURE 9.17).

• The three sp2 hybrid orbitals lie in the same plane, 120o apart from one another.

• They are used to make three equivalent bonds with the three fluorine atoms, leading to the trigonal-

planar molecular geometry of BF3.

• Notice that an unfilled 2p atomic orbital remains unhybridized.

• This unhybridized orbital will be important when we discuss double bonds in Ch 9.6.

25

sp3 Hybrid Orbitals - 4 electron domains: Tetrahedral, trigonal pyramidal, and bent

An s atomic orbital can also mix with all three p atomic orbitals in the same subshell. • For example, the carbon atom in CH4 forms four equivalent bonds with the four hydrogen atoms.

• We can picture the process from the mixing of the 2s and all three 2p atomic orbitals of carbon to create

four equivalent sp3 (pronounced “s-p three”) hybrid orbitals.

• Each sp3 hybrid orbital has a large lobe that points toward one vertex of a tetrahedron (see FIGURE

9.18).


In methane, C has 4 valence e- and H has 1 valence e-.

• So, how can carbon make 4 bonds with only 3

unpaired orbitals?

• Carbon’s 4 valence e- are located in the 2s and

2p…but they don’t have the same energy. 2p e- have

slightly higher energy.

So, one of the 2s electrons moves to the 2p orbital, and this is Hybridization - A process in which these 4 orbitals (s p p p) are combined in such a way to create 4 equivalent hybrid atomic orbitals.

Now there are 4 electrons in 4 identical orbitals

• These 4 orbitals all have the same energy: s p p p ….so we call it sp3 hybridization

27

Hybridization Involving d Orbitals (NOT ON AP EXAM )

• Since there are only three p orbitals, trigonal bipyramidal and octahedral electron-pair geometries must involve d orbitals.

• Trigonal bipyramidal electron pair geometries require sp3d hybridization. • Octahedral electron pair geometries require sp3d2 hybridization. • Use of d orbitals in making hybrid orbitals corresponds well with the idea of an expanded octet.

Summary (Rodriguez’s Cliff notes version)

To assign hybridization: • Draw a Lewis structure. • Count the TOTAL electron-domain regions around the central atom. (BONDS and LONE PAIRS)

o s: 1 = 1 electron domain (only H2) o sp: 1 + 1 = 2 electron domains o sp2: 1 + 2 = 3 electron domains o sp3: 1 + 3 = 4 electron domains

Specify the hybridization required to accommodate the electron pairs based on their geometric arrangement. • The hybridization will correlate to the geometry shape name based on the positions of the atoms.

1. Go back to the 2 TABLES (9.2 and 9.3) in sections 9.2 to determine the hybridization of each molecule.

Examples


9.6 Multiple Bonds

Read p. 365 – 372 In the covalent bonds we have seen so far the electron density has been concentrated symmetrically about the internuclear axis. Sigma ( ) bonds: a covalent bond in which the electron density is concentrated in between the 2 atoms. ALL single bonds are sigma bonds. (Think of it has the initial connection between the atoms)

What about overlap in multiple bonds? Pi ( ) bonds: a covalent bond in which the electron density is concentrated above and below the internuclear axis. (Think of it as now that you have made the connection between the atoms, what keeps that connect strong?)

• A double bond consists of one sigma bond and one pi bond. • A triple bond has one sigma bond and two pi bonds.

Examples:

29

1. Ethylene, C2H4

a. How many total sigma and pi bonds does the molecule have?

b. What is the hybridization around each carbon atom?

c. What is the electron-domain geometry and molecular geometry names?

2. Propene, C3H6



c. What is the molecular geometry names around each carbon?

3. Formaldehyde, CH2O


b. What is the hybridization around the carbon atom?

c. What is the electron-domain geometry and molecular geometry names?

4. Propene, C2H3N



c. What is the molecular geometry names around each carbon?

Examples


Conclusions

On the basis of the examples we have seen, we can draw a few helpful conclusions for

using hybrid orbitals to describe molecular structures:

1. Every pair of bonded atoms shares one or more pairs of electrons.

• The lines we draw in Lewis structures represent two electrons each.

• In every bond at least one pair of electrons is localized in the space between the atoms in a σ

bond.

• The appropriate set of hybrid orbitals used to form the σ bonds between an atom and its

neighbors is determined by the observed geometry of the molecule.

2. The electrons in σ bonds are LOCALIZED in the region between two bonded atoms and do not make a

significant contribution to the bonding between any other two atoms.

3. When atoms share more than one pair of electrons, one pair is used to form a σ bond; the additional pairs

form π bonds. The centers of charge density in a π bond lie above and below the internuclear axis.

4. Molecules with two or more resonance structures can have π bonds that extend over more than two

bonded atoms. Electrons in π bonds that extend over more than two atoms are said to be

“DELOCALIZED.”

Electron Configurations and Molecular Properties

Two types of magnetic behavior are: • paramagnetism (unpaired electrons in molecule)

o strong attraction between magnetic field and molecule • diamagnetism (no unpaired electrons in molecule)

o weak repulsion between magnetic field and molecule

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AP EXAM PRACTICE FRQ PRACTICE 2.1-2.4 – HOMEWORK

Answers in BEGINNING of AP You Tube video 2.5-2.7

1. Answer the following questions about nitrogen and oxygen.

(a) Write the Lewis electron-dot structure for the diatomic molecules N2 and O2.

The potential energy as a function of internuclear distance for the diatomic molecules N2 and O2 is

shown in the graph above. Based on the data in the graph and the Lewis structures that you drew in part (a), which curve, 1 or 2, is the better representation of the N2 molecule?

Justify you answer in terms of the principles of chemical bonding and bond energy.


AP EXAM PRACTICE FRQ PRACTICE 2.5-2.7 - HOMEWORK Answers in BEGINNING of AP You Tube video 3.1-3.3

1. 2017 #1(c) S2Cl2 is a product of the reaction.

(i) In the box below, complete the Lewis electron-dot diagram for the S2Cl2 molecule by drawing in all

of the electron pairs.

(ii) What is the approximate value of the Cl一S一S bond angle in the S2Cl2 molecule that you drew in

part (c)(i) ? (If the two Cl一S一S bond angles are not equal, include both angles.)

2. 2017 #2 Answer the following questions about the isomers fulminic acid and isocyanic acid.

Two possible Lewis eletron-dot diagrams for fulminic, HCNO, are shown below.

(a) Explain why the diagram on the left is the better representation for the bonding in fulminic acid. Justify

your choice based on formal charges.

33

2. 2018 #2(d) The skeletal structure of the HNO2 molecule is shown in the box below.

(i) Complete the Lewis electron-dot diagram of the HNO2 molecule in the box below, including any

lone pairs of electrons.

(ii) Based on your completed diagram above, identify the hybridization of the nitrogen atom in the

HNO2 molecule.

AP EXAM FRQ PRACTICE 3.1-3.3 – AP EXAMFRQ 2012


Answers in BEGINNING of AP You Tube video 3.4-3.6

6. Answer the following questions in terms of principles of chemical bonding and intermolecular forces. In

each explanation where a comparison is to be made, a complete answer must include a discussion of both

substances. The following complete Lewis electron-dot diagrams may be useful in answering parts of this

question.

(a) At 1 atm and 298 K, pentane is a liquid whereas propane is a gas. Explain.

(b) At 1 atm and 298 K, methanol is a liquid whereas propane is a gas. Explain.

(c) Indicate the hybridization of the carbon atom in each of the following:

i. Methanol

ii. Methanoic (formic) acid

(e) Explain the following observations about the two carbon-oxygen bonds in the methanoate (formate) anion, HCO2

－. You may draw a Lewis electron-dot diagram (or diagrams) of the methanoate ion as part

of your explanations.

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(i) The two carbon-oxygen bonds in the methanoate (formate) anion, HCO2－

, have the same length.

(ii) The length of the carbon-oxygen bonds in the methanoate (formate) anion, HCO2－

, is intermediate between the length of the carbon-oxygen bond in methanol and the length of the carbon-oxygen bond in methanol.

Chapter 9. Molecular Geometry and Bonding Theories · 2020. 12. 30. · Chapter 9. Molecular...

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