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AP Chapter8 Chem ical Bonding HW : 1 10 11 12 15 17 20 24 31 37 39 49 53 55 61 65 67 71 88 103
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Page 1: 1 10 11 12 15 17 20 24 31 37 39 49 53 55 61 65 67 71 88 103.

AP Chapter 8Chemical Bonding

HW: 1 10 11 12 15 17 20 24 31 37 39 49 53 55 61 65 67 71 88 103

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8.1 - Chemical Bonds• Three basic types of bonds:

– Ionic• Electrostatic attraction between ions

– Covalent• Sharing of electrons

– Metallic• Metal atoms bonded to several

other atoms

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8.1 – Lewis Dot Symbols

• Atoms combine in order to form more stable electron configurations. The maximum stability is when the atom is isoelectronic with a noble gas = octet rule. Bonds involve valence electrons.

• The Symbols use one dot per valence electron. It is possible to accurately draw for s and p block elements.

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8.2 - Ionic Bonding• Ionic Bond = The electrostatic (+/-) force

holding together the cations and anions of an ionic compound. Caused by a TRANSFER of electron(s)

• Ionic compounds are composed of a crystal lattice structure

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8.2 - Ionic Bonding

• Can imagine as separate steps:– Cation loses electron(s) – Ionization Energy– Anion takes in electron(s) – Electron Affinity– Electrostatic attraction occurs(see next slides for a description of each of these for

NaCl)

Bonding Video – 4:12 –7:47: Formation of NaCl

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Energetics of Ionic Bonding

It takes 495 kJ/mol to remove an electron from sodium.

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Energetics of Ionic Bonding

We get 349 kJ/mol back by giving an electron to chlorine.

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Energetics of Ionic Bonding

• But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!

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Energetics of Ionic Bonding

• The third part is the electrostatic attraction between the newly formed sodium cationand chloride anion.

The LATTICE ENERGY of NaCl is +788 kJ/mol

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8.2 – Lattice Energy of Ionic Compounds

• Lattice Energy:The energy required to completely separate a mole of a solid

ionic compound into its gaseous ions.It is always a + valueThe higher the Lattice Energy, the more stable the compoundCannot be measured directly, so use “Born-Haber Cycle” to

calculate• The energy associated with electrostatic interactions is

governed by Coulomb’s law:

E = Q1Q2

r

The energy between two ions is directly proportional to the product of their charges and inversely proportional to the distance separating them. K = 8.99 x 109 Jm/C2

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Some Lattice Energies:

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Lattice Energy and the Formula of Ionic Compounds

-Lattice Energy explains why multi-positive ions can form

-Stability of “noble gas configuration” is not enough – Ionization energy is always + and increases with each subsequent IE-The lattice energy of the IONIC COMPOUND FORMED is high enough to compensate for the energy needed to remove the electrons from the cation

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• Metals, for instance, tend to stop losing electrons once they attain a noble gas configuration because energy to remove beyond that would be expended that cannot be overcome by lattice energies.

• Mg+2 is “worth” being formed when a compound like MgCl2 is formed

• Na+2 is not “worth” being formed because the lattice energy of NaCl2 is not high enough to compensate

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Transition Metal Ions

• Lose valence s electrons first

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Born – Haber Cycle • Born-Haber Cycle – The Born-Haber cycle

involves the formation of an ionic compound from the reaction of a metal with a non-metal (DHf). Born-Haber cycles are used primarily as a means of calculating lattice enthalpies, which cannot otherwise be measured directly.

• Relates Ionization Energy, Electron Affinity, and other properties in a Hess’ Law-type calculation

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Example of Born-Haber Cycle

Calculate the Lattice Energy of LiF:Goal: LiF(s) -> Li+1

(g) + F-(g)

Use:

Enthalpy of sublimation Lithium = +155.2 kJ/mol

Bond Dissociation Energy Fluorine Gas = +150.6 kJ/mol

Ionization Energy of Lithium = +520 kJ/mol

Electron Affinity of Fluorine = -328 kJ/mol

Enthalpy of Formation of Lithium fluoride = -594.1

Reaction:

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Example of Born-Haber CycleWork:

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The Born-Haber cycle can also be represented schematically

DHf = IE + DHsub + 1/2DHdiss + EA + - LE

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Another Example the of Born-Haber Cycle (not +1/-1)

Calculate the Lattice Energy of CaCl2:

Goal: CaCl2(s) -> Ca+2(g) + 2Cl-

(g)

Use:

Enthalpy of sublimation Calcium = +121 kJ/mol

Bond Dissociation Energy Chlorine Gas = +242.7 kJ/mol

First Ionization Energy of Calcium= +589.5 kJ/mol

Second Ionization Energy of Calcium = +1145 kJ/mol

Electron Affinity of Chlorine = -349 kJ/mol

Enthalpy of Formation of Calcium chloride= -795 kJ/mol

Reaction:

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Work:

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8.3 – Covalent Bonding• Electrons are shared• Molecular Compounds have covalent bonds• Lone Pairs = Pairs of electrons not in a covalent

bond• Lewis Structure – Representation of Covalent

Bonding• Octet Rule = Atoms bond to form a noble gas

configuration (often 8 valence e-)• Single Bond = 1 pair electrons• Double Bond = 2 pairs electrons• Triple Bond = 3 pairs electrons

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8.3Bond Length = Distance between nuclei of

two covalently bonded atoms in a molecule.

1>2>3Bond Energy = Energy needed to break a

bond3>2>1

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8.4 - Electronegativity

• Ability to pull electrons; Pauli Scale (max 4.0)• Polar Covalent = Unequal sharing of electrons –

0.3-1.7 difference• Nonpolar Covalent = Equal sharing - <0.3

difference• Ionic = Transfer of electrons - >1.7 difference(Some books give different ranges)

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Comparison of Ionic and Molecular Compounds

Ionic MolecularIonic Bonds Covalent Bonds (Polar or

Nonpolar)Very Strong Bonds Strong Bonds

High mp/bp Low mp/bp

Crystal Lattice Individual Molecule

Conducts electricity when in water or melted

Do not conduct electricity in water or when melted

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8.4 – Bond Polarity and Electronegativity

• Although atoms often form compounds by sharing electrons, the electrons are not always shared equally.

• Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does.

• Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

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Polar Covalent Bonds

• When two atoms share electrons unequally, a bond dipole results.

• The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:

= Qr• It is measured in debyes (D).

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Polar Covalent BondsThe greater the difference in electronegativity, the more polar is the bond.

H - F H - F

d+

-d

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8.5 – Drawing Lewis Structures

Lewis structures are representations of molecules showing all electrons, bonding and nonbonding.

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Writing Lewis Structures1. Find the sum of

valence electrons of all atoms in the polyatomic ion or molecule.– If it is an anion, add

one electron for each negative charge.

– If it is a cation, subtract one electron for each positive charge.

PCl35 + 3(7) = 26

Old method of connecting dots does not always work. Here is the other method:

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Writing Lewis Structures

2. The central atom is the least electronegative element that isn’t hydrogen. Connect the outer atoms to it by single bonds.

Keep track of the electrons:26 6 = 20

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Writing Lewis Structures

3. Fill the octets of the outer atoms.

Keep track of the electrons:26 6 = 20 18 = 2

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Writing Lewis Structures

4. Fill the octet of the central atom.

Keep track of the electrons:26 6 = 20 18 = 2 2 = 0

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Writing Lewis Structures

5. If you run out of electrons before the central atom has an octet…

…form multiple bonds until it does.

HCN

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Lewis Structures

• HNO3

• CO3-2

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8.5 – Formal Charges• Then assign formal charges.

– For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms.

– Subtract that from the number of valence electrons for that atom: The difference is its formal charge.

Total formal charge on a neutral molecule = 0Total formal charge on a polyatomic cation /anion= charge

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Writing Lewis Structures

• The best Lewis structure…1. Most times FOLLOW THE OCTET RULE2. When more than one structure is possible:…is the one with the fewest formal charges.…puts a negative charge on the most

electronegative atom.

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8.6 - Resonance

-Structure has more than one equal placement of electrons.

This is the Lewis structure we would draw for ozone, O3.

-

+

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Resonance• But this is at odds with

the true, observed structure of ozone, in which…– …both O—O bonds are

the same length.– …both outer oxygens

have a charge of 1/2.

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Resonance• One Lewis structure

cannot accurately depict a molecule such as ozone.

• We use multiple structures, resonance structures, to describe the molecule.

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Resonance

Just as green is a synthesis of blue and yellow…

…ozone is a synthesis of these two resonance structures.

Ozone resonance structure =

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Resonance• Formate Ion:

The electrons that form the second C—O bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon.

• They are not localized, but rather are delocalized.

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Resonance

• The organic compound benzene, C6H6, has two resonance structures.

• It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.

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8.7 - Exceptions to the Octet Rule

• There are three types of ions or molecules that do not follow the octet rule:– Ions or molecules with an odd number of

electrons.– Ions or molecules with less than an octet.– Ions or molecules with more than eight valence

electrons (an expanded octet).

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Odd Number of Electrons

Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons.

Ex – NO; NO2; ClO2

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Incomplete Octet

• Consider BF3:– Giving boron a filled octet places a negative charge on

the boron and a positive charge on fluorine.– This would not be an accurate picture of the

distribution of electrons in BF3.– B stays at less than an octet!!

OR

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Incomplete Octet

Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.

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Incomplete OctetThe lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom. (Group 13 is stable with incomplete octet)

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Expanded Octet

• The only way PCl5 can exist is if phosphorus has 10 electrons around it.

• It is allowed to expand the octet of atoms on the 3rd row or below.– Presumably d orbitals in

these atoms participate in bonding.

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Expanded Octet

Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygens.

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Expanded Octet

• This eliminates the charge on the phosphorus and the charge on one of the oxygens.

• The lesson is: When the central atom is on the 3rd row or below and expanding its octet eliminates some formal charges, do so.

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8.8 – Strengths of Covalent Bonds

• Most simply, the strength of a bond is measured by determining how much energy is required to break the bond.

• This is the bond enthalpy = Bond Dissociation Energy

• The bond enthalpy for a Cl—Cl bond,D(Cl—Cl), is measured to be 242 kJ/mol.

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Average Bond Enthalpies

• This table lists the average bond enthalpies for many different types of bonds.

• Average bond enthalpies are positive, because bond breaking is an endothermic process.

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Average Bond Enthalpies

NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than theC—H bond in chloroform, CHCl3.

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Enthalpies of Reaction

• Yet another way to estimate H for a reaction is to compare the bond enthalpies of bonds broken to the bond enthalpies of the new bonds formed.

• In other words, Hrxn = (bond enthalpies of bonds broken) +

-(bond enthalpies of bonds formed)

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Enthalpies of ReactionCH4(g) + Cl2(g)

CH3Cl(g) + HCl(g)

In this example:Broken:Four C—H bondsOne Cl—Cl bondMade:One C—Cl One H—Cl bondThree C-H bonds

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Enthalpies of Reaction

So,Hrxn = [4D(C—H) + D(Cl—Cl) + -[3D(C—H) +D(C—Cl) + D(H—Cl)]

= [4(413 kJ) + 242 kJ] + -[3(413 kJ) + 431 kJ + 328 kJ]

= (1894kJ) + (-1998 kJ)= 104 kJ

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Bond Enthalpy and Bond Length

• We can also measure an average bond length for different bond types.

• As the number of bonds between two atoms increases, the bond length decreases.


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