Chapter 11 Theories of Covalent Bonding If you are doing this lecture “online” then print the lecture notes available as a word document, go through this.

Chapter 11

Theories of Covalent Bonding

If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

VALENCE BOND THEORY:

DEVELOPED BY LINUS PAULING, who received the Nobel Prize in 1954 for his work

A view of chemical bonding in which bonds arise from the overlap of atomic orbitals on two atoms to give a bonding orbital of electrons localized between the bonded atoms

RULE: Realize that Valence Bond Theory and all the others don't explain everything

The Central Themes of VB Theory

Basic Principle

A covalent bond forms when the orbitals of two atoms overlap and the overlap region, which is between the nuclei, is occupied by a pair of electrons.

(The two wave functions are in phase so the amplitude increasesbetween the nuclei.)

The Central Themes of VB Theory

Themes

A set of overlapping orbitals has a maximum of two electrons that must have opposite spins.

The greater the orbital overlap, the stronger (more stable) the bond.

The valence atomic orbitals in a molecule are different from those in isolated atoms.

There is a hybridization of atomic orbitals to form molecularorbitals.

Figure 11.1 Orbital overlap and spin pairing in three diatomic molecules.

Hydrogen, H2

Hydrogen fluoride, HF

Fluorine, F2

Regular atomic orbital overlap can explain these bonds.


Ha to Hb: 1sa to 1sb overlap radius = 74 pm

As overlap increases, strength of bond increases - both electrons are mutually attracted to both atomic nuclei.

At optimum distance between nuclei with maximum overlap, a sigma bond (strong primary bond) forms. Max electron density is along the axis of the bond

Ha to Fb: 1sa to 2pb direct overlap or bond


F to F: the picture looks like a 2p orbital on one F is overlapping with a 2p orbital on the other F atom, but actually each F is sp3 hybridized & electrons are localized between two atomic nuclei


We cannot use this direct overlap picture for CH4’s bonding. The 2s and the three 2p orbitals on each C do not fit into the CH4 molecule's 109o bond angles, since the 2p orbitals are at 90° to each other

Valence Bond Theory states that HYBRID orbitals of the outermost orbitals on an atom are formed from the atoms’ atomic orbitals

Hybrid Orbitals

The number of hybrid orbitals obtained equals the number of atomic orbitals mixed.

The type of hybrid orbitals obtained varies with the types of atomic orbitals mixed.

Key Points

sp sp2 sp3 sp3d sp3d2

Types of Hybrid Orbitals

Figure 11.2 The sp hybrid orbitals in gaseous BeCl2.

atomic orbitals on Be

hybrid orbitals

orbital box diagrams

You have to know how to draw this energy hybrid formation.

Figure 11.2 The sp hybrid orbitals in gaseous BeCl2(continued).

orbital box diagrams with orbital contours

Figure 11.3 The sp2 hybrid orbitals in BF3.


Note the three sigma bonds formed between B and each F.

Figure 11.4 The sp3 hybrid orbitals in CH4.


Figure 11.5 The sp3 hybrid orbitals in NH3.


Figure 11.5 continued The sp3 hybrid orbitals in H2O.


VALENCE BOND THEORY

Expanded Valence Shells have hybrid orbitals using s, p & d atomic orbitals. Example: PCl5 P: [Ne]3s23p3

dsp3 hybridization results in 5 bonds and trigonal bipyramidal geometry

(You can write these as dsp3 or sp3d)

Figure 11.6 The sp3d hybrid orbitals in PCl5.


Figure 11.7 The sp3d2 hybrid orbitals in SF6.


Figure 11.8

The conceptual steps from molecular formula to the hybrid orbitals used in bonding.

Molecular formula

Lewis structure

Molecular shape and e-

group arrangement

Hybrid orbitals

Figure 10.1

Step 1

Figure 10.12

Step 2 Step 3

Table 11.1

SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule

SOLUTION:

PROBLEM: Use partial orbital diagrams to describe mixing of the atomic orbitals of the central atom leads to hybrid orbitals in each of the following:

(a) Methanol, CH3OH (b) Sulfur tetrafluoride, SF4

(a) (a) CH3OH

H

CH H

OH

The groups around C are arranged as a tetrahedron.

O also has a tetrahedral arrangement with 2 nonbonding e- pairs.

SFF

F

F

SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule

continued

2p

2s single C atomsingle C atom

sp3

hybridized hybridized C atomC atom

2p

2s single O atomsingle O atom

sp3

hybridized hybridized O atomO atom

(b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs.

3p

3s

3d

S atomS atomsp3d

3d

hybridized hybridized S atomS atom

VALENCE BOND THEORY

There can be more than one central atom, and each has its own hybridization and geometry

C2H6 and H2O2 and CH3COOH

C2H6: both C's are sp3 hybridized and can rotate around axis of bond.

H2O2: both O's are sp3 , etc.

Figure 11.9 The bonds in ethane(C2H6).

both C are sp3 hybridized s-sp3 overlaps to bonds

sp3-sp3 overlap to form a bondrelatively even

distribution of electron density over all bonds

VALENCE BOND THEORY: Multiple Bonds

H2CO: the Lewis structures shows a double bond between C and O, but we know it does not have twice the bond dissociation energy of a single C-O bond

Pauling proposed that there was only one sigma bond between any two atoms, and the other multiples were weaker pi bonds

If there are only 3 bonds around this carbon, it can't be sp3 hybridized - instead we have sp2 hybrid orbitals

sp2 hybridization results in only 3 bonds, and trig planar geometry, with 120° angles

bond is a sideways or parallel overlap of the p atomic orbitals rather than the direct overlap of bonds

Figure 11.10 The and bonds in ethylene (C2H4).

overlap in one position -

p overlap -

electron densityProper name is ethene.

VALENCE BOND THEORY

Look at acetylene: its geometry is linear. C is forming a triple bond to another C and a single bond to H, so that's only two bonds

Therefore sp hybridization results in only 2 bonds, and linear geometry

There are 2 bonds from the parallel overlap of the 2p orbitals remaining on both C's

Figure 11.11 The and bonds in acetylene (C2H2).

overlap in one position -

p overlap -

SAMPLE PROBLEM 11.2 Describing the Bond in Molecules

SOLUTION:

PROBLEM: Describe the types of bonds and orbitals in acetone, (CH3)2CO.

PLAN: Use the Lewis structures to ascertain the arrangement of groups and shape at each central atom. Postulate the hybrid orbitals taking note of the multiple bonds and their orbital overlaps.

H3C

C

CH3

O

spsp33 hybridized hybridized


CC

C

O

H

H

HHH

H


bondsbond

CC

C

O

sp3

sp3

sp3

sp3

sp3

sp3

sp3

sp3

sp2 sp2

sp2

sp2

sp2sp2

H

HH

HH

H

H2CO hybrid orbitals and sigma and pi bond formation

Remember the C=C double bond has sigman and pi bonds.

The C has sigma bonds from its hybrid orbitals to the two H’s and the O. The leftover p orbitals will form the pi bond.

Figure 11.13 from 4th ed.

Restricted rotation of -bonded molecules in C2H2Cl2.

CIS TRANS

This cis/trans arrangement will be important in chem 2, organic chem and biology!

VALENCE BOND THEORY: RESONANCE

Resonance Structures and Bonding:

resonance structures involve an electron pair used alternately as a bond or a LP

Ozone: O3 O==O--O or O--O==O

All are sp2, trig planar, each has 3 sp2 orbitals and a p orbital remaining.

VALENCE BOND THEORY

Benzene: C6H6 has carbons with sp2 hybrids and 120o angles, each C has 2 bonds to other C's, 1 bond to H, and 1 bond electron available

Get "ring" of delocalized e-s

SUMMARY: draw the Lewis structure; determine arrangement of electron pairs using VSEPR, specify the hybrid orbitals to accommodate the e- pairs

Benzene sigma bond formation between C’s and C-Hs

The leftover p orbitals will form alternating pi bonds as shown in sketch.

MOLECULAR ORBITAL THEORY:

- explains why H2 forms easily and He2 does not - is an alternate way of viewing e- orbitals in molecules

where pure s and pure p orbitals combine to produce orbitals that are delocalized over the molecule

- they can have different energies and are assigned electrons just like we do in an atom - Pauli exclusion principle and Hund's rule included

Pauling's Valence Bond Theory does not explain everything

MO Theory doesn't either, but it does correctly predict the electronic structure of certain molecules that do not follow Lewis's approach, including the paramagnetism of certain molecules, like O2

The Central Themes of MO Theory

A molecule is viewed on a quantum mechanical level as a collection of nuclei surrounded by delocalized molecular orbitals.

Atomic wave functions are summed to obtain molecular wave functions. The number of molecular orbitals produced is always = # of atomic orbitals brought by the combining atoms (only orbitals on different atoms are combined). If wave functions reinforce each other, a bonding MO is formed (region of high electron density exists between the nuclei).

If wave functions cancel each other, an antibonding MO is formed (a node of zero electron density occurs between the nuclei).

Atomic orbitals combine most effectively with orbitals of the same type and similar energy (s w/s, n=2 w/ n=2)

The electrons of the molecule are placed in bonding or antibonding orbitals of successively higher energy (just like Hund's rule).

Amplitudes of wave functions added

Figure 11.13

An analogy between light waves and atomic wave functions.

Amplitudes of wave functions

subtracted.

Figure 11.14 Contours and energies of the bonding and antibonding molecular orbitals (MOs) in H2.

The bonding MO is lower in energy and the antibonding MO is higher in energy than the AOs that combined to form them.

MOLECULAR ORBITAL THEORY

BOND ORDER: the number of bonding e- pairs shared by 2 atoms in a molecule

Fractional bond orders are possible in MO Theory!

Silberberg method:B.O. = ½(# of e- in bonding orbitals - # of e- in antibonding orbitals)

Figure 11.15 The MO diagram for H2.

En

erg

y

MO of H2

*1s

1s

AO of H

1s

AO of H

1s

H2 bond order = 1/2(2-0) = 1

Filling molecular orbitals with electrons follows the same concept as filling atomic orbitals.

Figure 11.16 MO diagram for He2+ and He2.

En

erg

y

MO of He+

*1s

1s

AO of He+

1s

MO of He2

AO of He

1s

AO of He

1s

*1s

1s

En

erg

y

He2+ bond order = 1/2 He2 bond order = 0

AO of He

1s

SAMPLE PROBLEM 11.3 Predicting Stability of Species Using MO Diagrams

SOLUTION:

PROBLEM: Use MO diagrams to predict whether H2+ and H2

- exist. Determine their bond orders and electron configurations.

1s1s

AO of HAO of H

1s1s

AO of HAO of H

MO of HMO of H22++

bond order = 1/2(1-0) = 1/2

HH22++ does exist does exist

MO of HMO of H22--

bond order = 1/2(2-1) = 1/2

H2- does exist

1s1s 1s1s

AO of HAO of H AO of HAO of H--

*2s

2s

2s2s

Figure 11.17

2s 2s

*2s

2s

Li2 bond order = 1 Be2 bond order = 0

Bonding in s-block homonuclear diatomic molecules.E

ner

gy

Li2Be2

Figure 11.18Contours and energies of s and p MOs through

combinations of 2p atomic orbitals.

Or the pz orbitals

Figure 11.19 Relative MO energy levels for Period 2 homonuclear diatomic molecules.

MO energy levels for O2, F2, and Ne2

MO energy levels for B2, C2, and N2

without 2s-2p mixing

with 2s-2p mixing

Memorize this!

Figure 11.20

MO occupancy and molecular properties for B2 through Ne2

Figure 11.21

The paramagnetic properties of O2

SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties

SOLUTION:

PROBLEM: As the following data show, removing an electron from N2 forms an ion with a weaker, longer bond than in the parent molecules, whereas the ion formed from O2 has a stronger, shorter bond:

Explain these facts with diagrams that show the sequence and occupancy of MOs.Explain these facts with diagrams that show the sequence and occupancy of MOs.

Bond energy (kJ/mol)Bond energy (kJ/mol)

Bond length (pm)Bond length (pm)

NN22 NN22++ OO22 OO22

++

945945

110110

498498841841 623623

112112121121112112

N2 has 10 valence electrons, so N2+ has 9.

O2 has 12 valence electrons, so O2+ has 11.

SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties

continued

2s

2s

2p

2p

2p

2p

N2 N2+ O2 O2

+

bond orders

1/2(8-2)=3 1/2(7-2)=2.5 1/2(8-4)=2 1/2(8-3)=2.5

2s

2s

2p

2p

2p

2p

bonding e- lost

antibonding e- lost

MO Theory Practice

1. Draw the bonding and antibonding molecular orbitals for H2.

2. Do Valence Bond Theory (hybridization) and MO Theory for both O2 and O2

2-. Which theory works better to explain the molecule and ion?

3. For N2, N2+ and N2

- comparea. Magnetic characterb. Net number of bondsc. Bond Orderd. Bond lengthe. Bond strength

Answers

1. See picture in text.

2. VB Theory shows O2 as sp2 hybridized with one bond and one bond. There are two lone pairs on each O. O2

2- has one bond, and each O has three lone pairs. MO Theory shows a bond order of 2 for O2 and that it is paramagnetic. MO Theory shows a bond order of 1 for O2

2- and diamagnetic. But MO Theory fits the real data that O2 is paramagnetic.

Answers con’t

N2 N2+ N2

-

a. Diamag Paramag Paramag

b. 2 1.5 1.5

c. 3 2.5 2.5

d. Short Longer Longer

e. Strong Weaker Weaker