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Chapters 8,9Concepts of Chemical Bonding and Bonding

Theories

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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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Ionic Bonding Ionic compounds are formed when electrons are transferred from one atom to another to form ions with complete outer shells of electrons. In an ionic compound the positive and negative ions are attracted to each other by strong electrostatic forces, and build up into a strong lattice. Ionic compounds have high melting points as considerable energy is required to overcome these forces of attraction. Defined by formula units.

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Na + Cl

Na+ + Cl-

[Ne] 3s1

11p 11e

[Ne]

11p 10e

[Ne] 3s2 3p5

17p 17e

[Ne] 3s2 3p5

17p 18e

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Energetics of Ionic Bonding

As we saw in the last chapter, it takes 495 kJ/mol to remove electrons from sodium.

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Energetics of Ionic Bonding

We get 349 kJ/mol back by giving electrons 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!

• 495 - 349= +146 kJ

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Energetics of Ionic Bonding

• There must be a third piece to the puzzle.

• What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.

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Lattice Energy

• This third piece of the puzzle is the lattice energy:The energy required to completely separate a mole of

a solid ionic compound into its gaseous ions.• The energy associated with electrostatic

interactions is governed by Coulomb’s law:

Eel = Q1Q2

r2

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Lattice Energy

• Lattice energy, then, increases with the charge on the ions.

• It also increases with decreasing size of ions.

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Formation of Ionic Compounds

• Small ions with high ionic charges have large Coulombic forces of attraction.

• Large ions with small ionic charges have small Coulombic forces of attraction.

Al2O3 > CaO > KCl

• Use this information, plus the periodicity rules to arrange these compounds in order of increasing attractions among ions

KCl, Al2O

3, CaO

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Properties of Ionic compounds• Strong bonds due to very strong electrostatic forces.

High melting points. The higher the charge and the smaller the ions, the higher the melting point: Coulomb’s law

Very hard Low volatility

• Cleave along planes Brittle 3D structure Ions line up in a repetitive pattern that maximizes attractive

forces and minimizes repulsive forces. Not malleable or ductile

impact repulsion of like charges

+

+

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+

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+

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-

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Properties of Ionic compounds

Solubility and conductivity Most are soluble in polar solvents They conduct electricity only when molten or dissolved in a polar solvent, as the charged particles are free to move.

The higher the concentration of ions in a solution, the higher the electrical conductivity.

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Recall Dissociation• When an ionic substance dissolves

in water, the solvent pulls the individual ions from the crystal and solvates them.

• This process is called dissociation.• Water is a polar molecule. Its

positive ends (hydrogen sites) are attracted by the negative ions in the ionic compound, while the negative ends (oxygen sites) are attracted to the positive ions.

• Ionic compounds dissolve in water because the attraction of the ions to water is stronger than the attraction between the ions themselves.

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Problem 1: Which solution of each pair exhibits the stronger electrical conductivity? Explaina) 1.0 M Na2CO3 or 1.0 M NaCl

b) 1.0 M K2SO4 or 1.5 M KI

c) 2.0 M C2H5OH or 2.0 M LiF

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Melting pointWhen making predictions about melting point, Coulombic forces must be taken into account.

Problem 2: Which compound from each set has the greatest melting point? Why?a) LiF LiI

b) MgCl2 MgO

c) NaF MgI2

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Problem 3: Which of the following metal fluorides would have the highest melting point? Why?

a) b

Metal fluoride b) will have the highest melting point as the ionic radius of the metal is shorter than the ionic radius of the metal in metal fluoride a). According to Coulomb’s law, the force of attraction between opposite charged ions increases as the ionic radius of the ions decreases. Therefore a higher temperature will be needed to separate the ions in metal fluoride b)

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Covalent Bonding

• In these bonds atoms share electrons.

• There are several electrostatic interactions in these bonds:Attractions between electrons

and nucleiRepulsions between electronsRepulsions between nuclei

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This is how colleges explain covalent bond !!!

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Electronegativity:

• The ability of atoms in a molecule to attract electrons to itself.

• On the periodic chart, electronegativity increases as you go……from left to right

across a row.…from the bottom to

the top of a column.

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Polar Covalent Bonds• 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 because electrons spend more time around fluorine.

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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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• In diatomic molecules containing the same element (e.g. H2 or Cl2) the electron pair will be shared equally, as both atoms exert an identical attraction because they have the same electronegativity.

• When atoms are different, the more electronegative atom exerts a greater attraction for the electron pair. One end of the molecule will thus be more electron rich than the other end, resulting in a polar bond.

• This relatively small difference in electronegativity is represented by δ+ and δ-.

• The bigger the difference in electronegativities the more polar the bond.

Bond Polarity

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Polar Covalent Bonds

The greater the difference in electronegativity, the more polar is the bond.

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Covalent or Molecular Compounds• Defined by molecules

two or more non metals bonded together to form a compound.

Bonds between atoms are non polar covalent or polar covalent

Polyatomic ions

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Lewis Structures

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

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Writing Lewis Structures

1. 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.

PCl3

5 + 3(7) = 26

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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.

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Writing Lewis Structures

3. Fill the octets of the outer atoms.

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Writing Lewis Structures

4. Add up the electrons you have used and subtract them from the total number of valence electrons (step 1). Attach leftover electrons to the central atom as lone pairs.

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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.

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Formal Charges• In some cases, the atoms in a molecules can be

assembled in different ways when drawing a Lewis diagram.

• Formal charges are assigned to identify the most stable or likely structure.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.

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Writing Lewis Structures

• The best Lewis structure……is the one with the fewest charges.…puts a negative charge on the most

electronegative atom.

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Problem 4: Which is the most likely structure?

Cl – N – Cl or N – Cl – Cl | | Cl Cl

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Problem 5: Which is the most likely structure? [C = S = N]- or [S = C = N]-

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Problem 6: Write Lewis structures for:

CCl4

H2O

SCl2

CO32-

NH2-

SO2

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Resonance

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 oxygen

atoms have a charge of 1/2.

the effective number of bonds between each pair of oxygen atoms is 1.5

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Resonance• One Lewis structure

cannot accurately explain 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.

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Resonance• In truth, 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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Due to the delocalization of the electrons, each C – O is in between a double and a single bond. The bond order is 1.5.

Resonance Structures, Bond Length and Bond Energy

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Resonance

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

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

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Problem 7: Write all possible Lewis structures for the nitrate ion. Calculate the formal charge to determine the most likely structure. Then write all resonance structures of the ion and state the effective number of bonds between the nitrogen and each oxygen atom.

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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.

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Fewer Than Eight ElectronsJust for You

• 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.

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Fewer Than Eight Electrons

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More Than Eight Electrons

• 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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More Than Eight ElectronsJust for You

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 oxygen atoms.

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Molecular models• Space-Filling Model

Shows differences in the atomic radii of bonded atoms.Does not show the 3D positions of atoms very well.Does not show double and triple bonds.

• Ball-and Stick ModelShows the 3D position of atoms well.Shows single, double and triple bonds.Balls are not proportional to the size of the atoms.Sticks are not proportional to and greatly exaggerate the

bond length.H HC C

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Covalent Bond Strength

• Energy must be absorbed to break a chemical bond, therefore the strength of a bond is measured by determining how much energy is required to break the bond.

• The amount of energy required to break a bond is equal to the amount of energy released when that same bond forms.

• This is the bond enthalpy.• The bond enthalpy for a Cl—Cl bond,

BE(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

• Energy is required to break bonds. Endothermic process.

• Energy is released when new bonds are formed. Exothermic process.

• If the energy absorbed in breaking bonds is greater than the energy released when new bonds are formed, then the enthalpy of the reaction is endothermic. (positive)

• If the energy absorbed in breaking bonds is less than the energy released when new bonds are formed, then the enthalpy of the reaction is exothermic. (negative)

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

broken) • (bond enthalpies of bonds formed)

Enthalpies of Reaction

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Enthalpies of Reaction

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)In this example, one C—H bond and one Cl—Cl bond are broken; one C—Cl and one H—Cl bond are formed.

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Enthalpies of Reaction

So, bonds broken - bonds formed

Hrxn = [BE(C—H) + BE(Cl—Cl) [BE(C—Cl) + BE(H—Cl)

= [(413 kJ) + (242 kJ)] [(328 kJ) + (431 kJ)]

= (655 kJ) (759 kJ)

= 104 kJ

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Bond Length • Most atoms have lower potential energy when they

are bonded to other atoms than they have as they are independent particles.

• The figure below shows potential energy changes during the formation of a hydrogen-hydrogen bond.

• The distance between two bonded atoms at their minimum potential energy (the average distance between two bonded atoms) is the bond length.

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Bond Enthalpy and Bond Length

• As the number of bonds between two atoms increases, the bond length decreases, the bond energy (strength) increases and the PE decreases.

• This is because as the electron density between the positive nuclei increases, the attractive forces between the protons and the bonding electrons increase.

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Problem 8: Consider the space-filling models below.a) Which molecule has the least amount of PE associated with its bond?b) Which bond has the lowest bond energy?

I II

Lowest PE, as it has the shortest bond length

Lowest bond energy, as it has the longest bond length

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Bond Order• Bond order is the number of bonds between two

atoms.• When the bond order increases:

• the bond length decreases• the PE associated with the bond decreases• the bond energy increases

Bond type Bond order

Single 1

Double 2

triple 3

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Problem 9: Consider the ball-and-stick models below.a) The carbon-carbon bond in which molecule has the lowest PE?b) The carbon-carbon bond in which molecule has the lowest bond energy?c) What are the weaknesses of this model?

I IIH HC C

H

HH

H

C C

Lowest PE, as it has the shortest bond length

Lowest bond energy, as it has the longest bond length

Weaknesses: • Balls are not proportional to the size of the atoms.• Sticks are not proportional to the bond length and greatly

exaggerate the bond length

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Metallic Bond• A metallic solid can be represented as

positive kernels (or cores) consisting of the nucleus and inner electrons of each atom surrounded by a sea of mobile valence electrons.

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Metallic Bond• Properties of metals are linked to their

bonding.Good conductors of electricity: electrons

are delocalized and relatively free to move.Malleable and ductile: deforming the solid

does not change the environment surrounding each metal core.

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Metallic Bond. Alloys• Metals can be mixed into alloys. The properties of alloys

are linked to the size of the components atoms. Interstitial alloy: Formed between metal atoms of

different radius. The smaller atom fills the interstitial spaces between the larger atoms. The interstitial atoms make the lattice more rigid, decreasing the malleability and ductility. Example: steel, where C occupies the interstices in Fe.

Substitutional alloy: Formed between atoms of comparable radius. One of the atom substitutes the other in the lattice. The density lies between those of the component elements, but the malleability and ductility remains. Example: brass, where Cu is substituted with Zn.

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Covalent Network solids• Atoms are covalently bonded together into a two

or a three-dimensional network. (diamond, graphite, silicon dioxide, silicon carbide).

• The highest melting point of all bonding types since all atoms are covalently bonded.

• Three-dimensional covalent networks tend to be rigid and hard. The covalent angles are fixed. Example: diamond.

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Covalent Network solids• Graphite is an allotrope of C that forms sheets of

two-dimensional network.• High melting point because the atoms are

covalently bonded.• Soft because adjacent layers can slide past each

other relatively easy. London dispersion forces between the layers.

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Covalent Network solids• Silicon is a covalent network solid and a

semiconductor.• It forms a three-dimensional network similar

to diamond.• Conductivity increases as the temperature

increases.• N-type semiconductor: doping silicon with

an element with extra valence electron.• P-type semiconductor: doping silicon with

an element with one less valence electron.

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• Molecular Structure & Covalent Bonding Theories

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Stereochemistry

• Stereochemistry is the study of the three dimensional shapes of molecules.

• Some questions to examine in this chapter are:

1. Why are we interested in shapes?

2. What role does molecular shape play in life?

3. How do we determine molecular shapes?

4. How do we predict molecular shapes?

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Stereochemistry• These questions can be answered by looking at how drugs of any kind work. If the molecule of the drug does not have the right shape to plug or attach into the specific neuron or protein they are designed for, the drug will not work.• This is the basis for how most pharmaceutical drugs work. Drugs are organic compounds that bind to proteins because they have the right shape – like a lock and key.• Lewis structures do not tell us shape, but it gives some information we need to figure the shape.

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Two Simple Theories of Covalent Bonding

• Valence Shell Electron Pair Repulsion TheoryCommonly designated as VSEPRPrincipal originator

• R. J. Gillespie in the 1950’s

• Valence Bond Theory Involves the use of hybridized atomic orbitalsPrincipal originator

• L. Pauling in the 1930’s & 40’s

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VSEPR Theory

Valence Shell Electron Pair Repulsion Theory

Charged clouds (bonding or lone pairs of electrons) repel each other due to Coulombic repulsions.

Terminal atoms move as far away from one another as possible to minimize that repulsion.

This results in distinctive geometric shapes.

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Steps to identify the shape• The same basic approach will be used in every

example of molecular structure prediction:1. Draw the correct Lewis dot structure.

• Identify the central atom.

2. Count the number of charge clouds (regions of high electron density) around the central atom.

a) Each item below counts as a single cloud:• One single bond (consist of 2 electrons)• One double bond (consist of 4 electrons)• One triple bond (consist of 6 electrons)• One lone pair (consist of 2 electrons)• One single unpaired electrons (consist of 1 lone electron)

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Steps to identify the shapeb) Each item below is considered ONE BOND:

• A single bond (consist of 2 electrons)• A double bond (consist of 4 electrons)• A triple bond (consist of 6 electrons)

c) Each item below is considered a LONE PAIR:• One lone pair (consist of 2 electrons)• One single unpaired electrons (consist of 1 lone electron)

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Problem 10: Identify the number of charge clouds in the following compounds:a) NH3

b) SF4

c) NO2

d) BF3

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Overview of Chapter

3. Predict the shape- There are 15 shapes- You must know them all by name- You must know the bond angles for each.

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

2 2 0 Linear CO2, BeF2

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

3 3 0 Trigonal planar

CO32-, BeF3

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

3 2 1 Bent NO2-

. .

Bent less than 120ºThe lone pair exerts a greater repelling force on the bonding electrons as it is sitting closer to the nitrogen.

. .. .

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

4 4 0 Tetrahedral CCl4

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

4 3 1 Trigonal Pyramidal

NH3

. . . .

Trigonal pyramidal Around 107ºThe lone pair exerts a greater repelling force on the bonding electrons as it is sitting closer to the nitrogen.

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

4 2 2 Bent H2O

. . . .

Bent104.5º for water.The two lone pairs exert a greater repelling force on the bonding electrons as it is sitting closer to the oxygen.

.. ..

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

5 5 0 Trigonal Bipyramidal

PCl5

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

5 4 1 See Saw SF4

....

See sawIdeally 90º and120º.The lone pair occupies the equatorial position and it exerts a greater repelling force on the bonding electrons as it is sitting closer to the central atom.

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

5 3 2 T-shaped ICl3

T-shapedIdeally 90º and120º.The two lone pairs occupy the equatorial position and they exert a greater repelling force on the bonding electrons as they are sitting closer to the central atom.

..

..

..

..

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

5 2 3 Linear XeF2

LinearExactly 180º.The three lone pairs occupy the equatorial position and they cancel each other dipoles.

..

..

..

..

....

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

6 6 0 Octahedral SF6

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

6 5 1 Square Pyramidal

IF5

. . . .

Square pyramidalLess than 90º.The lone pair exerts a greater repelling force on the bonding electrons as it is sitting closer to the central atom.

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VSEPR TheoryCharge clouds

Bonds Lone Pairs Shape Examples

6 4 2 Square Planar

XeF4

Square planarExactly 90º and 180º.The lone pairs cancel each other out.

. . . .

. . . .

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Example of Molecules with More Than One Central Atom

• Look at each central atom on its own and everything bonded to it is considered a terminal atom.

• Count the charge clouds and predict the shape around it.

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Problem 11: Predict the shape around each carbon atom in this compound.

H H C = C = C H H

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Valence Bond Theory• It combines Lewis’ theory of filling octets by sharing

pairs of electrons with the electron configuration of atomic orbitals.

• It states that bonding occurs when atomic orbitals overlap.

• Example: Building BF3 with Valence Bond Theory.

B 1s 2s 2p 1s 2s 2p

F 1s 2s 2p 1s 2s 2p

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Valence Bond Theory• Example: Building BF3 with Valence Bond Theory.

B 1s 2s 2p 1s 2s 2p

F 1s 2s 2p 1s 2s 2p

But this does not explain all three identical B-F bonds• The bond angles are wrong since 1 bond is between an s

orbital from B and a p orbital from F.• Hybrid orbitals will explain the identical angles.

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HybridizationIs the morphologic mixture of 2 or more atomic orbitals

• 1 orbital s + 3 orbitals p = 4 hybrid sp3 orbitals• 1 orbital s + 2 orbitals p = 3 hybrid sp2 orbitals• 1 orbital s + 1 orbital p = 2 hybrid sp orbitals

• see animation

hybridization Charge cloudssp3 hybridization 4

sp2 hybridization 3

sp hybridization 2

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Hybridization• hybrid orbitals explain bonding electrons and lone

pair of electrons.• all single bonds are sigma bonds.• double bonds consist of a sigma bond and a pi bond.• triple bonds consist of a sigma bond and 2 pi bonds.• σ bond results from a direct head-to-head overlap

of orbitals.• π bond results from the side-to-side attractive

forces between unhybridized p orbitals.• σ bonds are stronger as they have shorter bond

lengths and greater bond energies.

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Problem 12: Use the Valence Bond and Hybridization theories to explain the trigonal planar geometry at each carbon atom

H H C = C H H

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102

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Problem 13: Use the Valence Bond and Hybridization theories to explain the linear geometry at each carbon atom H – C ≡ C – H

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A bond results from the head-on overlap of two sp hybrid orbitals.

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105

• The unhybridized p orbitals form two p bonds.• Note that a triple bond consists of one and

two p bonds.

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Extended π BondsExample: Benzene

• CH in every corner• Each p-orbital can overlap with two different

p-orbitals• This leads to delocalization of electrons• Can be used to explain resonance in Lewis

structures

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Polar Molecules: The Influence of Molecular Geometry

• Molecular geometry affects molecular polarity.Due to the effect of the bond dipoles and

how they either cancel or reinforce each other.

A B A

linear molecule nonpolar

A B A

angular molecule

polar

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Polar Molecules: The Influence of Molecular Geometry

• Polar Molecules must meet two requirements:

1. One polar bond or one lone pair of electrons on central atom.

2. Neither bonds nor lone pairs can be symmetrically arranged that their polarities cancel.

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Molecular Shapes and PolarityCharge clouds

Bonding / lone pairs

Shapes Polarity

2 2 / 0 linear Non polar *

3 3 / 0 trigonal planar Non polar *

3 2 / 1 bent Polar

4 4 / 0 tetrahedral Non polar *

4 3 / 1 trigonal pyramidal Polar

4 2 / 2 bent Polar

5 5 / 0 trigonal bipyramidal Non polar *

5 4 / 1 see saw Polar

5 3 / 2 T shaped Polar

5 2 / 3 linear Non polar *

6 6 / 0 octahedral Non polar *

6 5 / 1 square pyramidal Polar

6 4 / 2 square planar Non polar *

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Problem 14: For each of the molecules below identify shape, angle, hybridization, and polarity a) H2O

b) SiO2

c) SF4

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Molecular Orbital (MO) Theory In the Hybrid Orbital theory each atom in the compound retains its associated orbitals and electrons.

The MO theory views a molecule as a whole instead of a collection of individual atoms. MOs are similar to atomic orbitals. Similarities with hybridization:

They both have specific energy levels. They both have specific sizes an shapes. They can both hold a maximum of two electrons that spin

in opposite directions.

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Molecular Orbital (MO) Theory • Atomic orbitals combine to form MOs.

• When two atomic orbitals combine, two Mos are formed.• Orbitals are always conserved.

Example: Formation of H2.

σ*1s

anti-bonding

+ σ1s

bonding

+ +

+

+

+ +

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Molecular Orbital (MO) Theory Bonding Orbital

Is a MO that is lower than any atomic orbital from which it was derived.

Electrons that occupy these orbitals cause stability.Anti-Bonding Orbital

Is a MO that is higher than any atomic orbital from which it was derived.

Electrons that occupy these orbitals cause instability.Non-Bonding Orbital

Is a MO that is at the sane energy level as the one atomic orbital that it was derived from.

Electrons that occupy these orbitals do not cause stability or instability.

These are orbitals that contain lone pair of electrons.

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Molecular Orbital (MO) Diagram – H2

Energy

1s atomic orbital on 1st H atom

1s atomic orbital on 2nd H atom

σ* 1s anti-bonding

MO

σ 1s

bonding MO

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Molecular Orbital (MO) Diagram – O2

Energy

2s

σ* 2s

2p 2p

2s

σ 2s

σ 2p

π 2p

π* 2p

σ* 2p

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Problem 15: Develop the Molecular orbital diagram for a molecules of F2.

Energy

2s

2p 2p

2s

σ* 2s

σ 2s

σ 2p

π 2p

σ* 2p

π* 2p