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Ch 7 Covalent Bonding

Masterson & Hurley

8.7 The Covalent Chemical Bond:

A Model (Zumdahl)

Strengths of the Bond Model

• Associates quantities of energy with the formation of bonds between elements

• Allows the drawing of structures showing the spatial relationship between atoms in a molecule

• Provides a visual tool to understanding chemical structure

Weaknesses of the Bond Model

• Bonds are not actual physical structures

• Bonds can not adequately explain some phenomena

– i.e. resonance

8.8 Covalent Bond Energies and

Chemical Reactions (Zumdahl)

Average Bond Energies

• In this section energies associated with various types of bonds will be considered and how the bonding concept is useful in energies of chemical reactions.

• The average of individual bond dissociation energies is used *Note your Ch 8 Notes By A. Allen, Average Bond Energies

Multiple Bonds

• Single bonds: 1 pair of shared e-

• Double bonds: 2 pairs of shared e-

• Triple bonds: 3 pairs of shared e-

• As the number of shared e- increase, the bond length shortens

*Note pg 372 Table 8.4 and 8.5

Bond Energy and Enthalpy…

• …using bond energy to calculate approximate energies for reactions

H2(g) + F2(g) 2 HF(g)

Consider this reaction, it involves breaking 1 H-H bond and 1 F-F bond and forming 2 H-F bonds

• For bonds to be broken, energy must be added to the system—an endothermic process (+ sign)

• The formation of a bond releases energy—an exothermic process (- sign)

H2(g) + F2(g) 2 HF(g)

H = D (bonds broken) - D (bonds formed)

energy required energy released

= represents the sum of terms

D = represents the bond energy per mole of bonds

D always has a positive sign

H = 1 DH-H + 1 DF-F –2 DH-F

H = 1 mol x 436kJ/mol + 1 mol x 159kJ/mol – 2mol x 565kJ/mol

= -535 kJ

Example: N2 + 3 H2 2 NH3

H = D (bonds broken) - D (bonds formed)

H = 1 mol DNN + 3 mol DH-H – 2 mol D 3N-H

= 1 mol 941kJ/mol + 3 mol 432kJ/mol – 2 mol (3)

391kJ/mol

= 2237 kJ – 2346 kJ

= -109 kJ

* Note: you need to know the structure of the molecule to determine the # of bonds !!

8.9 The Localized Electron Bonding Model

Localized Electron (LE) Model

• In this section a specific model is used to describe covalent bonds, called the localized electron (LE) model.

• The localized electron (LE) model assumes that a molecule is composed of atoms that are bound together by sharing pairs of e- using the atomic orbitals of the bound atoms.

• Lone e- pairs: e- localized on an atom (unshared)

• Bonding e- pairs: e- found in the space between atoms (shared pairs)

LE Model

1. Valence e- arrangement using Lewis Structures

2. Prediction of molecular geometry using VSEPR (valence shell e- pair repulsion)

3. Description of the type of atomic orbitals used to share or hold lone pairs of e-

8.10 Lewis Structures

• The second-row elements never exceed the octet rule, since their valence orbitals (2s and 2p) can accommodate only eight electrons

8.11 Exceptions to the Octet Rule

• H is satisfied with 2 e- (duet rule)

• B is satisfied with 6 e-

• Be is satisfied with 4 e-

• Third period and heavier elements often satisfy the octet rule but can exceed the octet rule by using their empty valence d orbitals

Sulfur Hexafluoride 1. Note that sulfur has 12 e- around it, exceeding the octet

rule (how can this happen?)…

2. SF6 fills the 3s and 3p orbitals with 8 of the valence e- and places the other 4 in the higher energy 3d orbital (third row elements can exceed the octet rule)

3. Sulfur hexafluoride is very stable

Rules for Writing Lewis Structures 1. Add up the total number of valence e- from all

atoms

2. Use a pair of e- to form a bond between each pair of bound atoms.

3. Arrange the remaining atoms to satisfy the duet rule for hydrogen and the octet rule for the secondary row elements

4. Add “extra e-” to third period elements

* Don’t worry about which e- come from which atoms, regard the molecule as a new entity. Valence e- belong to the molecule rather than individual atoms

Let’s consider PCl5

• P = 5

• Cl = 7 x 5 = 35

• Total = 40

Let’s consider SF4

• S = 6

• F = 7 x 4 = 28

• Total = 34

8.12 Resonance

• Sometimes more than one valid Lewis structure is possible for a given molecule called resonance.

• A structure can exist as an average of all the possible structures, called resonance structures.

Formal Charge

• Formal Charge is a form of book keeping for the overall charge of the “species”--to determine which Lewis structure is better

• (Val. e-)assigned = (# of lone pairs) + ½ (# of shared e-) • Formal Charge = (# of valence e- on the free atom) – (# of

valence e- assigned to the atom in the molecule)

• The sum of the formal charges of all atoms in a given molecule or ion must equal the overall charge on that species

• Lone pair (unshared) e- belong completely to the atom in question

• Shared e- are divided equally between the sharing atoms

Let’s consider sulfate… (Val. e-)assigned = (# of lone e-) + ½ (# of shared e-) Formal Charge = (# of valence e- on the free atom) – (# of valence e-

assigned)

Let’s consider sulfate…

• SO42- (Make sure you add 2 extra e-, due to 2-

charge)

• S = 6 e- O = 6 x 4 = 24 e- add 2 extra e-

• Total e- = 32 e-

(Val. e-)assigned = (# of lone e-) + ½ (# of shared e-)

Formal Charge =

(# of val. e- on free atom) – (# of val. e- assigned to atom in molecule)

For O atoms with single bonds (there are two O atoms):

Val. e- assigned = 6 + ½ (2) = 7

Formal charge = 6 – 7 = -1

For O atoms with double bonds (2 O atoms):

Val. e- assigned = 4 + ½ (4) = 6

Formal charge = 6 – 6 = 0

For the S, sulfur atom:

Val. e- assigned = 0 + ½ (12) = 6

Formal charge = 6 – 6 = 0

Let’s consider nitrate… (Val. e-)assigned = (# of lone e-) + ½ (# of shared e-) Formal Charge = (# of valence e- on the free atom) – (# of valence e-

assigned)

Let’s consider nitrate… (Val. e-)assigned = (# of lone e-) + ½ (# of shared e-) Formal Charge = (# of valence e- on the free atom) – (# of valence e-

assigned)

1. Any negative formal charges are expected to reside on the most electronegative atoms

Odd electron molecules are those in which there is not an even number of e-

i.e. NO (11e-) and NO2 (17 e-)

Do not add or take away e-, the overall charge is zero!

Does not fit localized e- model

• N = 5 e, O = 6 x 2 = 12 e- • Total e- = 17 e- --> odd # of e- • Note the free radical, ok to leave on molecule • Remember the best Lewis structure has formal charge as

close to zero as possible • Note, overall charge of NO2 is zero!

Lets consider ammonium…

• N = 5 e-

• H = 1 x 4 = 4 e-

• Total = 9 e- -- 1 e- = 8 e-

(due to 1+ charge)

• Check formal charge = 1+

8.13 Molecular Structure:

The VSEPR Model

• Molecular Structure is the 3-D arrangement of the atoms in a molecule

• We will use the VSEPR model in predicting molecular geometries of nonmetal molecules.

• Valence Shell Electron Pair Repulsion states that the structure around a given atom is determined principally by minimizing electron-pair repulsions…

• …basically, bonding and nonbonding pairs around a given atom will be positioned as far apart as possible

Effect of Unshared Electron Pairs The ideal tetrahedral angle is 109.5° Note methane, ammonia and water… Lone pairs require more room than bonding pairs and

compress angles b/w bonding pairs (no distortion if angles are 120 ° or greater)

109.5° 107° 104.5°

VSEPR and Multiple Bonds

• For the VSEPR model, multiple bonds count as one effective e- pair

• When a molecule exhibits resonance, ANY of the resonance structures can be used to predict the molecular structure using the VSEPR model

Molecules Containing No Single Central Atom

• Apply the principle of distancing shared and unshared e- pairs

• Look at real 3-D, rotatable models to develop predictive skills

• How well does VSEPR work? For non-ionic compounds, VSEPR works in most cases

Ch 9 Covalent Bonding: Orbitals

AP Chemistry

Zumdahl

Hybridization… • …is the mixing of two or more atomic orbitals of similar

energies on the same atom to produce new orbitals (hybrid orbitals) of equal energies (blending of orbitals)

+

= +

=

Cocker Spaniel Poodle Cockapoo

Hybridization Summary

1. A sigma bond is formed from an overlap of orbitals

2. Single bonds are sigma bonds

3. In general, the sum of the superscripts on the hybrid orbitals equal the number of e- groups (multiple bonds count as one group) around the central atom. The sum of the superscripts in sp3 = 4.

sp3

Evidence for Hybridization

Example: Structure of Methane and sp3;

1. Four bonds of equal length and strength

2. Four effective pairs of e- surround the carbon

3. “Whenever a set of equivalent tetrahedral atomic orbitals is required by an atom, this model assumes that the atom adopts a set of sp3 orbitals; the atom becomes hybridized.”

Carbon’s e-

Configuration

Hybridization of sp3

Tetrahedral Orbitals 4 sp3 orbitals of the carbon atom are used to share e- pairs

with the 4 1s orbitals of the hydrogen atom

sp2

sp2 Hybridization 1. Trigonal planar structure, 120° angle, in ethene rules out

sp3 hybridization 2. Sp2 hybridization creates 3 identical orbits of

intermediate energy and length and leaves one unhybridized p orbital

3. 3 effective pairs of electrons surround the carbon (double bond treated as one effective pair)

C

Trigonal planar requires sp2 hybridization…

sp

A linear arrangement requires sp hybridization

sp hybridization (linear arrangement)

Double Bonds

A double bond consists of a sigma bond and pi bond

Triple Bonds

A triple bond consists of a sigma bond and two pi bonds

Molecular Orbitals (MO) are classified in two ways, energy and shape:

According to energy: Bonding MO ;

1. A bonding MO is lower in energy than the original atomic orbitals —in going from free atoms to the molecule, e- occupy bonding MOs are lowered in energy (favors molecular bonds).

According to energy: Antibonding MO ;

2. An antibonding MO is higher in energy than the original atomic orbitals. In going from free atoms to the molecule, e- are raised in energy. These e- oppose bonding.

Molecular Orbitals (MO) are classified in two ways, energy and shape:

According to shape (symmetry) :

Sigma MOs ;

Have their e- probability centered about a line passing through the two nuclei.

According to shape (symmetry):

Pi MO ;

Have their e- probability concentrated above and below the line passing through the nuclei

Bond order = # of bonding electrons - # of antibonding electrons 2 The greater the difference between the # of e- lowered in energy (bonding e-) and the # of e- that are raised in energy (antibonding e-) when the molecule forms, the greater will be the stability f the molecule (the stronger the bonding).

Chapter 8 and 9 Summary

On Your Own (just read):

• Pp 432-436: Paramagnetism

• Pp 436-437: bonding in Heteronuclear Diatomic Molecules

• Pp 437-439: Combining the Localized Electron and Molecular Orbital Models