Bond typesBond types

Bond typeBond type electronelectron ElectronegativElectronegativity difference ity difference

IonicIonic Donate/takeDonate/take >1.7>1.7

Polar covalentPolar covalent ShareShare 0.6-1.70.6-1.7

Nonpolar Nonpolar covalentcovalent

ShareShare <0.6<0.6

Comparing ionic and covalent Comparing ionic and covalent compoundscompounds

Molecular compoundsMolecular compounds Ionic compoundsIonic compounds

smallest particlessmallest particles moleculesmolecules cations and anionscations and anions

origin of bondingorigin of bonding electron sharingelectron sharing electron transferelectron transfer

forces between forces between particlesparticles

strong bonds between atomsstrong bonds between atomsweak attractions between weak attractions between moleculesmolecules

strong attractions between anions strong attractions between anions and cationsand cationsstrong repulsions between ions of strong repulsions between ions of like chargelike charge

elements presentelements present close on the periodic tableclose on the periodic table widely separated on the periodic widely separated on the periodic tabletable

metallic elements metallic elements presentpresent

rarelyrarely usuallyusually

electrical electrical conductivityconductivity

poorpoor good, when melted or dissolvedgood, when melted or dissolved

state at room state at room temperaturetemperature

solid, liquid, or gassolid, liquid, or gas solidsolid

melting and boiling melting and boiling pointspoints

lowerlower higherhigher

other namesother names covalent compoundscovalent compounds saltssalts

Metallic bondsMetallic bonds

Metals’ valence electrons can move Metals’ valence electrons can move about between metal ions.about between metal ions.

These These nonlocalized nonlocalized electrons give electrons give metals their conductivity, metals their conductivity, malleability and strengthmalleability and strength

Intermolecular forces(van der Waals forces)

London dispersion-instantaneous dipole moment

-increases with mass -found between all molecules

Dipole-dipole- increases with polarity Hydrogen bonding

The stronger the intermolecular forces, the greater the melting and boiling points and viscosity will be.

Viscosity- resistance to flow.

stronger

weaker

Lewis Dot DiagramsLewis Dot Diagrams

HH22 H HHH bond to form H:H bond to form H:H

Each bond uses an electron from each Each bond uses an electron from each atom.atom.

To create a Lewis dot diagram. . .To create a Lewis dot diagram. . .

1.1. Count the total valence electrons Count the total valence electrons available.available.

2.2. Ensure each atom has an octet (or a Ensure each atom has an octet (or a pair for H) pair for H)

Exceptions to the Octet RuleExceptions to the Octet Rule

Less than 8: B & BeLess than 8: B & Be Expanded valence: elements in Expanded valence: elements in

period 3 or higher bonding with period 3 or higher bonding with highly electronegative elements like highly electronegative elements like F, O, and Cl.F, O, and Cl.

Odd no. of valence electronsOdd no. of valence electrons

Valence Shell Electron Pair Valence Shell Electron Pair Repulsion theory- electron pairs Repulsion theory- electron pairs repel each other, so are oriented repel each other, so are oriented

as far apart as possible.as far apart as possible.

> > > > Unshared-unshared repulsion

Unshared-shared repulsion

Shared-shared repulsion

Molecule

Total no. of

electron pairs

No. of shared pairs

No. of unshared pairs

Molecular shape

MoleculeTotal no.

of electron

pairs

No. of shared

pairs

No. of unshared

pairsMolecular shape

Bent 104.5°

Trigonal pyramidal 107.3°

PolarityPolarity

Molecule is polar if. . .Molecule is polar if. . .

there is a polar bond ANDthere is a polar bond AND

it is asymmetricalit is asymmetrical

HH++

O - O - ++H C HH C H++

H H + HH H + H++

polarnonpolar

If a central atom is symmetrically If a central atom is symmetrically surrounded by identical atoms, it will surrounded by identical atoms, it will be be nonpolarnonpolar..

Ex: linearEx: linear ABAB22

trigonal planartrigonal planar ABAB33

tetrahedral tetrahedral ABAB44

square planarsquare planar ABAB44

trigonal bipyramidal trigonal bipyramidal ABAB55

octahedraloctahedral ABAB66

HybridizationHybridization

The mixing of 2 or more orbitals of The mixing of 2 or more orbitals of similarsimilar energies on the same atom energies on the same atom to produce new orbitals of to produce new orbitals of equalequal energies.energies.

C C __ 1s1s2s 2p2s 2p 1s sp1s sp33

tetrahedraltetrahedral

3 p orbitals were used

How many bonds How many bonds andand unshared pairs unshared pairs are around the central atom? This is are around the central atom? This is the number of equal energy orbitals the number of equal energy orbitals needed.needed.

sp has 2 electron domainssp has 2 electron domains spsp22 has 3 electron domains has 3 electron domains spsp33 has 4 electron domains has 4 electron domains spsp33d has 5 electron domainsd has 5 electron domains spsp33dd22 has 6 electron domains has 6 electron domains

Sigma bonds are along the bond axis.Sigma bonds are along the bond axis.Pi bonds are sideways with parallel overlap.Pi bonds are sideways with parallel overlap.

Single bond: Single bond: 1 sigma bond1 sigma bond

Double bond: Double bond: 1 sigma, 1 pi bond1 sigma, 1 pi bond

Triple bond: Triple bond: 1 sigma, 2 pi bonds1 sigma, 2 pi bonds

Coordinate covalent bondsCoordinate covalent bonds

The shared electrons are supplied by The shared electrons are supplied by a single atoma single atom

H H H H ++

H N H + HH N H + H++ H N H H N H

HH

Heat

of

fusi

on

Heat of vaporization

Heating Curve for Water

Calculate the enthalpy change for each Calculate the enthalpy change for each stage, then total them.stage, then total them.

To get the ice to 0To get the ice to 0◦◦C:C:

∆∆HH = (10.0g)(2.09J/g= (10.0g)(2.09J/g··K)(25K)(25◦◦K)=515JK)=515J

To melt the ice:To melt the ice:

∆∆HH = (10.0g)(334 J/g)=3340J= (10.0g)(334 J/g)=3340J

To heat the water to 100To heat the water to 100◦◦C:C:

∆∆HH = (10.0g)(4.18J/g= (10.0g)(4.18J/g··K)(100K)(100◦◦K)=4180JK)=4180J

To vaporize the water:To vaporize the water:

∆∆HH = (10.0g)(2260 J/g)=22600J= (10.0g)(2260 J/g)=22600J

To heat the water to 125To heat the water to 125◦◦C:C:

∆∆HH = (10.0g)(2.02 J/g= (10.0g)(2.02 J/g··K)(25K)(25◦◦K)=505JK)=505J

Phase Diagram

Critical point- beyond this, the gas and liquid phases become indistinguishable.

Triple point- all three phases are at equilibrium

The effect of increasing temperature and Activation Energy

As the temperature increases, the peak for the most probable KE is reduced, and more significantly, moves to the right to higher values so more particles have the highest KE values. At the higher temperature T2, a greater fraction of particles has the minimum KE to react.

Enthalpy of solution- Enthalpy of solution- ∆H∆Hsolsol

The energy change during dissolving.The energy change during dissolving.

For solids in liquids, usually endothermic For solids in liquids, usually endothermic (>0)(>0)

For gases in liquids, usually exothermic For gases in liquids, usually exothermic (<0)(<0)

Henry’s LawHenry’s Law

The amount of a gas that will The amount of a gas that will dissolve in a liquid at a given dissolve in a liquid at a given temperature varies directly with the temperature varies directly with the partial pressure of that gas.partial pressure of that gas.

ConcentrationsConcentrations

Molarity (M) Molarity (M) moles solutemoles solute

L solutionL solution Molality (m) Molality (m) moles solutemoles solute

kg solventkg solvent % by mass % by mass g soluteg solute

g solutiong solution Mole fractionMole fraction moles of solutemoles of solute

moles of solutionmoles of solution

Colligative propertiesColligative properties

Properties determined by the number Properties determined by the number of particles in solution (rather than the of particles in solution (rather than the type)type) Vapor pressure decreasesVapor pressure decreases Boiling point increasesBoiling point increases Freezing point decreasesFreezing point decreases Osmotic pressureOsmotic pressure

vapor pressurevapor pressure

Decreases (with a nonvolatile solute)Decreases (with a nonvolatile solute) PPsolutionsolution = x = xsolventsolvent ∙P ∙P solvent solvent

Vapor pressure = (mole fraction) (vapor Vapor pressure = (mole fraction) (vapor pressure)pressure)

solution solution solvent solventsolvent solvent

Raoult’s Law: vapor pressure of a Raoult’s Law: vapor pressure of a solution varies directly as the mole solution varies directly as the mole fraction of the solvent.fraction of the solvent.

Determining freezing/boiling point Determining freezing/boiling point changeschanges

∆∆TTfp fp = m∙K= m∙Kfpfp

molality x freezing point constantmolality x freezing point constant

∆∆TTbp bp = m∙K= m∙Kbpbp

molality x boiling point constantmolality x boiling point constant

KKfp fp for water is 1.853 C°/1m waterfor water is 1.853 C°/1m water

KKbpbp for water is .515 C°/1m water for water is .515 C°/1m water

For other substances, see table A-8 on p.860. For other substances, see table A-8 on p.860.

Determining molecular Determining molecular massmass

If 8.02g of solute in 861g water lowers the If 8.02g of solute in 861g water lowers the freezing point to -0.430°C, calculate the freezing point to -0.430°C, calculate the molecular mass of the solute.molecular mass of the solute.

∆∆TTfp fp = m∙K= m∙Kfpfp so m = so m = ∆T∆Tfp = fp = -0.430-0.430°C °C

== .232m .232m

KKfpfp 1.853°C/m 1.853°C/m

8.02g solute8.02g solute xx 1 kg water1 kg water == 40.1 g/mol 40.1 g/mol

.861kg water .232 mol solute.861kg water .232 mol solute