Announcements

Chem 7 Final Exam Wednesday, Oct 9 1:30-3:30AMChapter 1-12All multiple choice questions

Exam III (Chapter 7-10)Thursday Sept 26, 2013Time: 6:00PM - 7:30PMChing Tan Room

Quiz 9 Tuesday Sept 24 Quiz 10 Thursday Sept 26

AnnouncementsWhat To Read from Chapter 7-10

Please note the following topics will be excluded from assessment. The page numbers refer to the 2nd Edition of the textbook.

1. Numerical problems involving the Rydberg equation (Chapter 7)2. Spectral analysis in the laboratory (Chapter 7 p. 226-227)3. Trends among the transition elements (Chapter 8 p. 261)4. Trends in electron affinity (Chapter 8 pp. 265-266)5. Pseudo-noble gas configuration (Chapter 8 p. 269)6. Lattice energy (Chapter 9 pp. 283-285)7. IR spectroscopy (Chapter 9 p. 292)8. Numerical problems involving electronegativity (Chapter 9 p. 296)9. Electronegativity and oxidation number (Chapter 9 p. 297)10. Section 11.3: MO theory and electron delocalization (Chapter 11 p.343)11. All sections in chapter 12 except 12.3 (types of intermolecular forces).

Chapter 9

Models of Chemical Bonding

MetallicBonding

CovalentBonding

IonicBonding

Atoms tend to form three types of chemical bonds: ionic, covalent and metallic.

sharingtransfer

pooling

Cadmium

Bromine

Iodine

Metals

Non-Metals

Covalent compounds result from chemical reactions between non-metals.

An ionic compound (salt) results from the chemical reaction between a metal and non-metal.

Ionic Compounds Covalent Compounds1. Solid: high melting and boiling points (>600˚C).

1. Gas, liquid or solid with low melting points (<300˚C)

2. Soluble in polar solvents (like water) 2. Insoluble in polar solvents

3. Insoluble in non-polar solvents (like organic solvents)

3. Soluble in organic solvents (not water soluble)

4. Molten compounds conduct electricity very well 4. Non-electrolytes as liquids

5. Aqueous solutions are electrolytes

5. No charged particles and (electrolytes) in water

6. Formed between elements with very different electronegativity.

6. Formed between elements of similar electronegativity.

Ionic and covalent compounds have very different physical properties.

Metals tend to loose (be oxidized) electrons while non-metals tend to gain electrons (be reduced) to acquire the noble gas configuration closest to them in the periodic table.

Chemical bonds and the chemistry of an element is related to the number of valence electrons are in the outer shell (highest value of n quantum number) of the atom.

1A 1ns1

2A 2ns2

3A 3ns2np1

4A 4ns2np2

5A 5ns2np3

6A 6ns2np4

7A 7ns2np5

Group # of valence e-e- configuration

LOOK FOR Group #

• For B group elements, the valence electrons are in the highest value ns orbital and the (n-1)d orbitals.

Lewis dot structures are used to depict valence electrons and bonding between atoms.

• A chemical symbol represents the nucleus and all core e-.

• A single dot around the symbol represents one valence e-.

+1 +2 +3 -1-2-3• Each atom attains a noble gas electronic

configuration.– metal (low IE and low EA) combines with a

non-metal (high IE and high EA)!– electron transfer gives full shell = “octet

rule” => very stable chemically – model works for Group A elements.

Gilbert Newton Lewis

Ionic bonds are formed by the transfer of electrons from a metal to non-metal (giving an oxidized metal and reduced non-metal).

.+ F: ::Li. Li+ +

.F: ::

.

Both atoms attain an octet!electron transfer

Lewis structures are used to depict how ionic bonds are formed by chemists (electron-transfer).

••

••

Mg••

Cl• ••

••

Cl•••

••

••Cl••

••••

Mg2+ -2

Formation of sodium chloride

O•••

•••

••O••

••

••Ba

2+ 2-Ba• • +

Formation of magnesium chloride

Formation of barium oxide

noble gas electronic configuration

Electron transfer

Electron transfer

1. Using spdf electron configurations

2. Orbital box diagrams

3. Lewis electron-dot symbols

Li (1s22s1) + F (1s22s22p5) Li+ 1s2 + F- (1s22s22p6)

Li1s 2s 2p

F1s 2s 2p

+

Li+

1s 2s 2p

F-

1s 2s 2p+

.+ F: ::Li. Li+ +

.F: ::

.

Chemists use 3 different ways to depict ionic bonding. Know all of them!

Use partial orbital diagrams and Lewis symbols to depict the formation of Na+ and O2- ions from the atoms, and determine the formula of the compound.

Draw orbital diagrams for the atoms and then move electrons to make filled outer levels. It can be seen that 2 sodiums are needed for each oxygen.

Use partial orbital diagrams and Lewis symbols to depict the formation of Na+ and O2- ions from the atoms, and determine the formula of the compound.

SOLUTION:

3s 3p

Na

3s 3p

2s 2p

O2s 2p

O2-

2 Na+

:Na

Na+ O

.

:

..

.

2Na+ + O 2-

::

::

Na

.

MetallicBonding

CovalentBonding

IonicBonding

Atoms tend to form three types of chemical bonds: ionic, covalent and metallic.

sharingtransfer

pooling

Non-metal elements typically share electrons between atoms forming covalent bonded compounds called molecules or molecular compounds.

share electronsforming chemical bond

1 H2 MOLECULE2 H ATOMS

H H H H

Example of Covalent Bonding

• Every covalent bond has a characteristic bond length that leads to maximum stability.

Energy Released When Bond Formed

The formation of a covalent bond releases energy and reduces the potential energy (i.e minimizes energy).

Atoms Far Apart

Atoms to Close Together

F F lone pairslone pairs

single covalent bond

F F

single covalent bond

F F+

7e- 7e-

F F

8e- 8e-

8 e- or an “octet” around each atom

Lewis Dot Structure

Lewis Skeletal Structure

We also use Lewis structures and use the “octet rule” to show covalent bonds and bonding between atoms.

Note the outer electrons are not shown!

Lewis Structure

Molecule Geometry

how chemical bonds are formed through atomic orbitals or hybrids of them.

is used to predict

which is used to predict

We use Lewis structures to predict molecular geometries of molecules (organic).

What is the Lewis dot structure and dash formula for water, H2O and for carbon dioxide?

What is the Lewis dot structure and the dash formula for carbon dioxide?

What is the Lewis dot structure and dash formula for water, H2O and for carbon dioxide?

• • C••

••

O ••• • O•

•

• •

• •CO O

•

••

•••

••

••

••

•CO O ••

•••

••

••

••CO O••

••

••

••

Bonding Pairs (double bond)

Non-bonding Pairs

H OH H O H H O H

Dash FormulaLewis Dot

H O H

Dash Formula

Electronsimplied

Lone Pair

BondingPair

• Bonded Pair of Electrons - bonding electrons

• Lone Pairs of Electrons - non-bonded electrons

• Bond Order - the number of electron pairs (i.e. bonds) shared between two atoms.– Single bond, Order = 1– Double bond, Order = 2 – Triple bond Order = 3

• Higher bond orders give rise to:– Shorter bond lengths between bonded atoms– Stronger bond energies (more energy needed to break)

CO O••••

••

••

The Language of Covalent Bonds:

H F

H O H

Bond length depends on the size of the bonded atoms and the bond order.

Internuclear distance(bond length)

Internuclear distance(bond length)

Internuclear distance(bond length)

Internuclear distance(bond length)

Covalent radius

72 pm

Covalent radius

114 pm

Covalent radius

133 pm

Covalent radius

100 pm

F2

Cl2

Br2

I2

72pm

100pm

114pm

133pm

Higher bond orders give shorter bond lengths and require more energy to break a bond.

Bond LengthsTriple bond < Double Bond < Single Bond

Comparing Bond Length and Bond Strength

The trends in atomic radius can be extended to the bond length in molecules in a family of compounds.

(a) S - F, S - Br, S - Cl

(b) C = O, C - O, C O

(c) C-Br, C-I, C-Cl

Using the periodic table, but not Tables 9.2 and 9.3, rank the bonds in each set in order of increasing bond length and bond strength:

Comparing Bond Length and Bond StrengthUsing the periodic table, but not Tables 9.2 and 9.3, rank the bonds in each set in order of decreasing bond length and bond strength:

(a) S - F, S - Br, S - Cl

(b) C = O, C - O, C O PLAN: (a) The bond order is one for all and sulfur is bonded to halogens;

bond length should increase and bond strength should decrease with increasing atomic radius. (b) The same two atoms are bonded but the bond order changes; bond length decreases as bond order increases while bond strength increases as bond order increases.

(a) Atomic size increases going down a group.

Bond length: S - Br > S - Cl > S - F

Bond strength: S - F > S - Cl > S - Br

(b) Using bond orders we get

Bond length: C - O > C = O > C O

Bond strength: C O > C = O > C - O

Bond Disassociation = +BE

Bond Association = -BE

∆H < 0

∆H > 0

The source of and the sign of !Hrxn in a chemical reaction arises from making and breaking chemical bonds!

Hess’s Law allows us to analyze the !Hrxn of bond formation in a chemical reaction.

Exothermic-energy releasedEndothermic

We can calculate !Hrxn if we know the bond energies for bond formation and bond breaking (tabulated).

H2 (g) + F2 (g) 2HF(g) !Hrxn = ?

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

H H 1 436.4 436.4F F 1 156.9 156.9

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

H F 2 568.2 1136.4

!Hrxn = [436.4 + 156.9] – [(2 x 568.2)] = - 543.1 kJ

∆Hrxn = ΣBEreactant bonds broken − ΣBEproduct bonds formed

Calculate the "Hrxn for the chlorination of methane to form chloromethane gas using bond energies.

CH4(g) + Cl2(g) " CH3Cl(g) + HCl(g) !Hrxn = ?

1. Write a balanced chemical equation and set up table below.2. Write Lewis structures break all bonds in reactants and reform all bonds in products. 3. Use the simple equation below to calculate !Hrxn

∆Hrxn = ΣBEreactant bonds broken − ΣBEproduct bonds formed

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

+ +

!Hrxn = [1656kJ/mol + 243 kJ/mol] - [1232kJ/mol + 339 kJ/mol + 431 kJ/mol]!Hrxn = -114 kJ/mol (exothermic)

∆Hrxn = ΣBEreactant bonds broken − ΣBEproduct bonds formed

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

41

11

339431 431

339H-ClC-ClC-H 3 414 1232

414243

C-HCl-Cl 243

1656

CH4(g) + Cl2(g) " CH3Cl(g) + HCl(g) !Hrxn = ?

Calculate the standard enthalpy change ∆H in kJ for the hydrogenation of ethyne (acetylene) to ethane.

+ 2H2(g) H3C CH3(g)

∆Hrxn = ΣBEreactant bonds broken − ΣBEproduct bonds formed

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

Calculate (in kJ) the standard enthalpy change ∆H for the hydrogenation of ethyne (acetylene) to ethane.

+ 2H2(g) H3C CH3(g)

C-C 1 347 347C-H 6 413 2478

C-H

H-H

212

413839432

839826

864

!Hrxn = [826kJ + 839kJ + 864 kJ] - [347 kJ/mol + 2478 kJ/mol] = -296 kJ/mol

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

Calculating Enthalpy Changes from Bond Energies

Use Table 9.2 to calculate !H0rxn for the following reaction:

CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)

1. Write the balanced chemical equation2. Write Lewis structures to see bonds broken and bonds formed. 3. Use the given Bond energies in the problem and a table account for bonds broken/formed to calculate !Hrxn

∆Hrxn = ΣBEreactant bonds broken − ΣBEproduct bonds formed

Type of bonds formed

Number of bonds formed

Bond energy (kJ/mol)

Energy change (kJ)

Type of bonds broken

Number of bonds broken

Bond energy (kJ/mol)

Energy change (kJ)

Calculating Enthalpy Changes from Bond Energies

Use Table 9.2 to calculate !H0rxn for the following reaction:

CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g)

Calculate the number of bonds broken and bond formed using the bond energies found in Table 9.2.

4C-H = 4 mol(413 kJ/mol) = 1652 kJ

3Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ

"!Hobonds broken = 2381 kJ

3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ1 C-H =1 mol(-413 kJ/mol) = -413 kJ

"!H0bonds formed = -2711 kJ

3H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ

!Horeaction = "!Ho

bonds broken + "!Hobonds formed = 2381 kJ + (-2711 kJ)

Electronegativity is an element’s inherent ability to draw electrons to itself when chemically bonded to another atom in a molecule (measured relative to Li).

Notice F, O, N, Cl, Br, C are highly electronegative!

Group I and II metals are the least electronegative.

The difference in electronegativity between bonding atoms blurs the distinction between covalent and ionic bonding types.

Covalent Bonding Polar Covalent

!+ !-

The delta’s are used toshow a partial charge on atoms.

F2 HF LiF

F F H F Li F

Ionic Bonding

No Symbol Ionic

We use arrows to symbolize “polar bonds” but not ionic.F2 is symmetric

in pull => no polar bond!

χ = |ENX - ENM|

How ionic or covalent a bond is between two atoms can be judged by looking at the the absolute value of the difference in electronegativity.

!+ !-

Polar bonds are unsymmetrical electron distribution that result from electronegativity differences between atoms in a chemical bond result.

Un-symmetric charge distribution

has a polar bond due to electronegativity

differences.

Symmetrical molecules

are not non-polar

Polar molecule

Determining Bond Polarity from EN Values

(a) Use a polar arrow to indicate the polarity of each bond: N—H, F—N, I—Cl. (b) Rank the following bonds in order of increasing polarity: H—N, H—O, H—C.

(a) Use your knowledge of EN values of F, Cl and N to determine bond polarity; the arrow should point toward the most negative end.

(b) Polarity increases across a period.

SOLUTION:

(a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0

N-H F-N I-Cl

(b) The order of increasing EN is C < N < O; all have an EN larger than that of H.

H-C < H-N < H-O

Determining Bond Polarity from EN Values

(a) Use a polar arrow to indicate the polarity of each bond: N—H, F—N, I—Cl. (b) Rank the following bonds in order of increasing polarity: H—N, H—O, H—C.

(a) Use Figure 9.20 (button at right) to find EN values; the arrow should point toward the negative end.(b) Polarity increases across a period.

For a polyatomic molecule we must consider the vector sum of polar bonds in the molecule to see if there is a net dipole moment.

DipoleMoment

No NetDipoleMoment

DipoleMoment

DipoleMoment

No Net Dipole

polar

Ionic compounds are typically solids of high melting and boiling points.

Across a given period compounds become more covalent with lower melting points than their ionic counterparts.

Melting point of compounds increase across a period as compounds become less ionic and more covalent.

Across Period 3 chloride compounds go from high melting solids to very low boiling point gases (as it gets more covalent)