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Lecture 14Professor Hicks

General Chemistry (CHE131)

Acids• H+ + anion

H+ -anion

• Ionic compounds must separate into ions to dissolve

• Acids are molecular compound because they can

dissolve without dissociating into ions

• Weak acids have a small percentage of molecules

separated into H+ and an anion, the rest stay together

as one particle

Strong acids

HCl HNO3 H2SO4

H+ Cl- H+ NO3- H+ SO4

2-

hydrochloric acid nitric acid sulfuric acid

Weak acids

HC2H3O2 HF

H+ C2H3O2- H+ F-

acetic acid hydrofluoric acid

HF ~ 95% H+ and F- 5%

Strong acids separate 100% into H+ and anion in waterHCl ~ 0 % H+ and Cl- ~ 100%

molecular

compounds

all other

molecular

compoundsdissolve

Dissolved

molecules

ionic

compounds

dissolve

cations

(+ ions)

anions

(- ions)

separated ionsH+ + F- ~5%

(separated ions)

HCl → H+ + Cl-

~ 100%

HF ~ 95%

(molecules)

acids

dissolve

Acids are

molecular compounds

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Strong Bases

+

cation OH-

any

cation

hydroxide

ion

- Ionic compounds with hydroxide ion (OH-)

Examples: NaOH, LiOH, KOH, Ca(OH)2, Ba(OH)2, Sr(OH)2,

You must know this definition

Weak Bases

• Substances that react with water to produceOH-

Base + H2O → Base-H+ + OH-

NH3 (aq) + H2O (l) → NH4+ (aq) + OH- (aq)

compounds that contain nitrogen are weak bases

Methylamine (CH3NH3), amphetamine, cocaine, or heroin

Free-base form is the form that has not accepted an H+ yet

Strong acid-strong baseneutralization reactions

acid + base → H2O (l) + salt

H+

-anion

+

cation

OH-

strong

base

acid

HCl (aq) + Ba(OH)2 (aq) → H2O (l) + BaCl2 (aq)

metathesis

reactions

ions change

partners

acid-base reactions sometimes require balancing

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+ NaClO4 (aq)HClO4 (aq) + NaOH (aq) →

How to write a complete reactionExample: a neutralization reaction

Why is water described as liquid (l) not aqueous?

Pure substances - solids that precipitate, substances

that would form separate liquid layer (oils), or are the

solvent itself are written as their physical state under

those conditions (solid, liquid, gas)

Write the molecular equation, total ionic equation, and net ionic equation

for the neutralization reaction of perchloric acid and sodium hydroxide.

strong acids and soluble ionic compounds

written as separate ions (aq) in ionic equation

Molecular compounds including weak acids

are written as molecules – not as ions

H+(aq) + ClO4-(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + ClO4

-(aq)

Complete Ionic Equation

H2O (l) Molecular Equation

strong acids

HCl, HNO3, HClO4, H2SO4

Net Ionic Equation from Complete Ionic Equation

H+(aq) + ClO4-(aq) + Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + ClO4

-(aq)

H+ (aq) + OH- (aq) → H2O (l)

Complete Ionic Equation

Net Ionic Equation

cross out spectator ions

Spectator ions are ions are dissolved

(aq) on the reactants and products side

4.29 What factors qualify a compound as a salt? Specify which of the following compounds are salts: CH4, NaF, NaOH, CaO, BaSO4, HNO3, NH3, KBr?

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4.33 Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate):

HBr(aq) + NH3(aq) →

4.33 Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate):

Ba(OH)2(aq) + H3PO4(aq) →

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4.33 Balance the following equations and write the corresponding ionic and net ionic equations (if appropriate):

HClO4(aq) + Mg(OH)2(s) →

Solubility

• Max amount of a substance that will dissolve

• Not increased with increased stirring shaking etc.

• Educated guesses of solubility can be made based on the “like dissolves like” rule for all kinds of compounds

- many exceptions to “like dissolves like”

• Detailed rules exist to predict solubility for ionic compounds based on the identity of the ions

I will give you the solubility rules for ionic compounds for the exam

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Precipitation reactions• Only some ionic compounds dissolve in water

• If solutions are mixed that contain ions that can form an insoluble compound the compound will form and then precipitate as a solid

AgClinsoluble

AgNO3

soluble

Ag+ (aq)NO3

- (aq)

NaClsoluble

Na+ (aq)Cl- (aq)

mix

Na+ (aq)NO3

- (aq)

AgCl (s)

NaNO3 issolubleAgCl solidprecipitates

initiallycloudy

Precipitation of an ionic compound is a metathesis reaction

AgNO3 + NaCl → AgCl + NaNO3

white solid formsa precipitate

metathesis reaction = ions change partners

AgNO3 (aq) + NaCl (aq) → AgCl (s)+ NaNO3 (aq)

the solubility rules tell us if a substances state is(aq) if it is soluble (s) if it will precipitate as a solid (is insoluble)

4.19 Characterize the following compounds as soluble or insoluble in water: (a) Ca3(PO4)2, (b) Mn(OH)2, (c) AgClO3, (d) K2S

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4.21 Write ionic and net ionic equations for the following reactions:

AgNO3(aq) + Na2SO4(aq) →

4.21 Write ionic and net ionic equations for the following reactions:

BaCl2(aq) + ZnSO4(aq) →

4.21 Write ionic and net ionic equations for the following reactions:

(NH4)2CO3(aq) + CaCl2(aq) →

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4.23 Which of the following processes will likely result in a precipitation reaction?

(a) Mixing a NaNO3 solution with a CuSO4 solution.

4.23 Which of the following processes will likely result in a precipitation reaction?

(b) Mixing a BaCl2 solution with a K2SO4 solution. Write a net ionic equation for the precipitation reaction.

4.24 With reference to Table 4.2, suggest one method by which you might separate (a) K+ from Ag+, (b) Ba2+ from Pb2+, (c) from Ca2+, (d) Ba2+ from Cu2+. All cations are assumed to be in aqueous solution, and the common anion is the nitrate ion.

1

Lecture 15Professor Hicks

General Chemistry (CHE131)

Molarity (M)

• Unit of concentration

molarity =moles

liter

• Conversion factor from volume of

a homogenous solution to moles

molarity

moles

literx liters

x volume

= moles

= # moles in that volume

molarity

divide

stoichiometric

number

mass

compound

moles of

formula units,

molecules,

atoms,

or ions

number

formula units,

molecules,

atoms,

or ions

multiply

molar

mass

divide

molar

mass

multiply

molar

mass

divide

molar

mass

divide

Avogadro's

number

multiply

Avogadro's

number

divide

Avogadro's

number

multiply

Avogadro's

number

equivalents

multiply by

volume litersM x V = # moles

2

Example: What is the molarity of glucose (C6H12O6) in a solution made by dissolving 100.0 grams of glucose in enough water to make 250.0 ml of solution

1) molarity has units of liters

convert mL liters 250.0 mL x10-3 liter

1.0 mL= 0.2500 L

2) Convert mass C6H12O6 to moles

100.0 grams x 1 mole

180.16 grams= 0.5551 moles

Molarity = # moles

#liters=

0.5551 mol

0.2500 L= 2.220 M glucose

Example: How many moles of NaCl are in

50.5 ml of a 0.250 M NaCl solution?

1) molarity has units of liters

convert mL liters

0.250 M NaCl x 0.0505 liters

50.5 mL x10-3 liter

1.0 mL= 0.0505 L

= 0.0126 moles NaCl

x 0.0505 liters solution = 0.0126 moles NaCl0.250 moles NaCl

liter solution

or if you write out the base units of molarity

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How many moles of CaCl2 are present in 160.0 mL of

0.36 M MgCl2 solution?

How many grams of NaOH are present in 36.0 mL of a

2.50 M solution?

4

Calculate the volume in mL of a solution required to

provide the following:

(a) 2.14 g of sodium chloride from a 0.270 M solution

(b) 4.30 g of ethanol from a 1.50 M solution

(c) 0.85 g of acetic acid (CH3COOH) from a 0.30 M

solution.

Dilution Formula MconcVconc = MdilVdil

concentrated

stock solution

Mconc = 12.0 M HCl

1) add 10 ml

HCl stock

Vconc = 0.010 lit

water

2) add water to

500 ml markMconc x Vconc = # moles HCl

12 M x 0.010 lit

= 0.12 moles HCl

added

500 ml flask

# moles HCl = # moles HCl

taken from stock after adding water

used to prepare a dilute solution

from a concentrated stock solution

Mdil =MconcVconc

Vdil

12 x 0.010

0.50= M = 0.24 M adding water did not change # moles

HCl but it did change molarity of HCl

Vdil = 0.50 lit

Mdil = 0.24 M

added

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Serial dilution

• Homogenous solutions uniform throughout

1.00 gram

of blue compound

in 1000 mL of water

1 mL

+ 999 mL

water

1 mL

+ 999 mL

water

1 mL

contains

0.000000000100 grams

(1.00 nanogram)

of blue compound

0.00100 grams

of blue compound

in 1000 mL

compared to a scale

that is limited to

0.001 g ( 1 mg)

0.00000100 grams

of blue compound

in 1000 mL

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Water is added to 125.0 mL of a 0.66 M NaNO2 solution

until the volume of the solution is exactly 250 mL. What

is the concentration of the final solution?

How would you prepare 50.0 mL of 0.100 M HCl from a

stock solution of 3.60 M HCl?

4.67 A 35.2-mL, 1.66 M KMnO4 solution is mixed with

16.7 mL of 0.892 M KMnO4 solution. Calculate the

concentration of the final solution.

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Titrations• Burette has solution of known molarity = titrant

• Flask has unknown solution = analyte

• Titrant added until the endpoint is reached

• At endpoint the titrant has been added in slight

excess

• Excess added reacts with an indicator that

changes color

• Calculations use the approximation

endpoint = equivalence point

• Equivalence point is the point where the

number of moles of added titrant exactly,

completely, reacts with all moles of analyte

titrant

analyte

In a titration

# moles = # moles

• For titrations of strong acids and bases

# moles H+ = # moles OH-

at the equivalence point

• This leads to

M1V1= M2V2

• Molarities of H+ and OH- must be used

• M2 is often the unknown

M2 =M1V1

V2

4.79 Calculate the volume in mL of a 1.420 M NaOH

solution required to titrate the following solutions:

1.25.00 mL of a 2.430 M HCl solution

2.25.00 mL of a 4.500 M H2SO4 solution

3.25.00 mL of a 1.500 M H3PO4 solution

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4.94 Acetic acid (CH3COOH) is an important ingredient

of vinegar. A sample of 50.0 mL of a commercial vinegar

is titrated against a 1.00 M NaOH solution. What is the

concentration (in M) of acetic acid present in the vinegar

if 5.75 mL of the base were required for the titration?

1

Lecture 16Professor Hicks

General Chemistry (CHE131)

What is energy?

• Ability to do work

• Work means moving something against a force

• Energy thought of as an imaginary liquid that gets moved from one container to another when processes occur

Types of energy• Kinetic energy- energy of motion.

KE = ½ mv2

• PE (opposite charges) = increases with separation (like gravity)

-+

• Potential energy- associated with position.

different types

PE gravity = mass x gravity x height

• Heat

2

Units of energy

James Joule

• SI unit Joule

2) 1 dietary calorie = 1 kilocalorie

(energy to warm 1 kg H2O 1 C)dietary and scientific calorie are not the same

dietary calorie written uppercase Calorie

1 Joule = 1 kg m2

s2

Other common units

1) 1 calorie = 4.18 Joules

First law of thermodynamics

kinetic

energypotential

energy

heat

Rudolf Clausius

applied when energy changes forms

energy like a liquid

energy cannot be created or destroyed

When things go downhill

A 10 kg bowling ball at a height of 100 m has potential energy

due to gravity

Potential energy = mgh = 10 kg x 9.8 m/s2 x10 m = 9800 J

As it falls it speeds up converting

potential energy into kinetic energy

Right before it hits ground all the potential

energy has become kinetic energy = 9800 J

After it hits the ground it is not moving so kinetic energy

is zero- all the energy has become heat = 9800 J

Heat 9800 J

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potential

energy

energy changes forms(but does not get destroyed)

1) Potential energy = 9800 J

2) Kinetic energy = 9800 J

3) Heat 9800 J

heatkinetic

energy

as ball falls

PE KE

9800 J 9800 J 9800 Jon

impact

KE heat

First Law requires

heat is like a reactant or product

in every chemical equation

• Reactions that absorb heat are said to be endothermic

• Reactions that release heat are said to be exothermic

reactants products + heat

Reactants + Heat Products

Second Law

• Second law states that

matter and energy tend to

spread out spontaneously

• A famous statement for

energy is “Heat only flows

from hotter to cooler”

• Explains why processes

that release heat tend to

not reverse themselvesSadi Carnot

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Second Law:

Energy Spreads Out

Heat 9800 J

The second law of thermodynamics says the

bowling ball is trapped on the earth because

the energy it would need to go back to the top

of building has left as heat

When the ball hit the ground and the energy

becomes heat the ball warms up a little.

Hotter CoolerSame temperature

2H2O (g)

reaction products are trapped in the lowest energy

state after heat is released, like the bowling ball

Second law:

Energy and matter spread out

Fuels

heat

CO2 (g)

Heat and matter spread out

prevents products from turning back into reactants!!!

System and Surroundings

• System – region of interest

• Surroundings – everything else

system surroundings

heat = q work = w

energy

Energy can be either…

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Internal Energy (U)

(ALL the energy that is in there)

system surroundings

q heat released

– (negative in sign)

w work done by system

- (negative in sign)

q heat absorbed

+ (positive in sign)work done on system

+ (positive in sign)

Change in internal energy (U)

U = q + w

Work• Movement performed against a force

(resistance) requires energy

• Amount of energy=Work=force × distance

Examples of work

• Lifting an object against gravity

• Stretching a spring

• Gas forming and blowing up a balloon (PV

work)

P × V =

×V = ×Area × Length

work

Force

Area

Force

Area

Enthalpy (H)

• Heat released under constant pressure = Enthalpy change (H)

• Constant pressure – PV work can be done

- Conditions for reactions of life

- Conditions of most chemical reaction unless they are run in rigid containers

• Only reflects part of the internal energy change since it is only the heat-not the work

• The alternative to constant pressure is constant volume like the aging of wine in a glass bottle

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system

surroundings

C6H12O6 (g) + 6O2 (g)

work heat

Enthalpy is heat released

under constant pressure

6CO2 (g) + 6H2O (g)

change in

enthalpy

H = q

change in

internal energy

E = q + w

Thermochemical equations

• Chemical equation with H or E included

6 CO2 + 6 H2O C6H12O6 + 6 O2H = 2830 kJ

• Endothermic = heat as a reactant = positive

• Exothermic = heat as a product = negative

• The heat is associated with the whole reaction

• Heat + moles of any substance

conversion factors

2830 kJ

6 moles CO2

2830 kJ

6 moles O2

1 mole C6H12O6

2830 kJetc.

Standard Conditions

• To be able to compare reactions Standard

Conditions are defined

• All substances at concentration of 1.0 M

gases at a partial pressure of 1.0 atm

• Heat released under constant P and

standard conditions is the Standard

Enthalpy Change (Ho)

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First law and thermochemical equations

• If a process is reversed an equal amount of

energy must flow in the opposite direction

H and E for reverse process will just have the

sign reversed

• H and E depend upon the amount of material

reacted so doubling, tripling reaction doubles,

triples the H and E

• Hess’ Law states that if a series of chemical

reactions occurs the overall H or E will just be

the sum of each step’s H and E values

First law and thermochemical equations

• Example: Determine the Ho for

2CO2 (g) 2C (s) + 2O2 (g)

given that

C (s) + O2 (g) CO2 (g) Ho = - 393 kJ

Ho = 2 x 393 kJ

= 786 kJ

Hess’ law• If a process happens in steps the Ho is the

sum of the Ho for the steps

A + B C Ho = +100 kJ

C + D E Ho = -150 kJ

__________________________________

A + B + C + D C + E Ho = +100 – 150 kJ

A + B + D E Ho = – 50 kJ

overall

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6.15 A sample of nitrogen gas expands in

volume from 1.6 L to 5.4 L at constant

temperature. Calculate the work done in joules if

the gas expands

(a) against a vacuum

(b) against a constant pressure of 0.80 atm

(c) against a constant pressure of 3.7 atm.

6.17 A gas expands and does P-V work on the

surroundings equal to 325 J. At the same time, it

absorbs 127 J of heat from the surroundings.

Calculate the change in energy of the gas.

6.19 Calculate the work done when 50.0 g of tin

dissolves in excess acid at 1.00 atm and 25°C:

Assume ideal gas behavior.

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Lecture 17Professor Hicks

General Chemistry (CHE131)

Heat Capacity and Specific Heat

• Heat capacity (C) - extensive property -units are Joule

C

• Properties that describe how much temperaturechanges when heat is absorbed

• Specific heat capacity (s) - intensive property-units are Joule

g*CSpecific heat capacity is sometimes

referred to as just specific heat

Heat capacity (C)

• Imaginary container that holds

the imaginary liquid heatcopper

water

heat capacitywater

Temperature

heat capacitycopper

same massof both

add 100 J add 100 J

100 J

T

T 100 J

water has a larger HC than copper so its temperature changes less when same amount of heat is added

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T x C = q (heat)(C) x (Joule) = (Joule)

(C)

Heat capacity (C)

• Conversion factor

change temperature heat

answers the question “if I have a certain change in temperature

for an object how much heat must have flowed to cause it”

changes in quantities are given symbol always calculated final – initialT = T final – T initial = Tf - Ti

old unit

conversion factor

new unit

Specific heat capacity (s)

Conversion factor

mass heat capacity

mass x specific heat = heat capacity

m x s = C

multiplying a mass by the specific heat capacity answers the question

“if I have a certain mass of a substance what is its heat capacity?”

specific heat capacity is heat capacity per gram

old unit

conversion factor new unit

(grams) x (Joule) = (Joule)(gramC) (C)

How much heat would be required to raise the temperature of 50.0 grams of water from 12.0 C to 22.0 C ? (s (water) = 4.18 J/g*C)

Given Information

Ti = 12.0 Tf = 22.0 calculate T = 22.0 – 12.0 = 10.0 C

mass = 50.0 grams water

s (water) = 4.18 J/g*C

s x mass = HC

4.18 J/g*C x 50.00 grams = 209 J/C (heat capacity)

HC x T = q

209 J/C x 10.0 C = 2.09 x103 J

(use this equation to calculate amount of heat to raise temperature of 50 g water 10 C )

(use this equation to calculate heat capacity of 50 g water)

these two equations can be combined:q = mass x specific heat x change temperature

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Calorimetrymeasuring heat flow

surroundingsand

system

qno heat

can escapeperfectly insulated

coffee cup calorimeter

calorimetry – measure T calculate heat

add room temperature water

10 C

add hot metal sample at about 100 C

heat

20 C

First law - energy cannot be created or destroyed

heat lost by metal was gained by the water

(and the cup which we neglect here)

qwater = - qmetal

qwater = mass water x s (water) x (Tf-Ti)

qmetal = mass metal x s (metal) x (Tf-Ti)?

Lab experiment

on specific heat

capacity of a metal

qmetal = -qwater

qmetal = mass metal × s (metal) × (Tf-Ti)

qwater = mass water × s (water) × (Tf-Ti)

mass metal × s (metal) × (Tf-Ti) = - mass water × s (water) × (Tf-Ti)

s (metal)

mass metal × (Tf-Ti)

= - mass water × s (water) × (Tf-Ti)

metal and water have same final temperature

metal and water have different initial temperatures

A sample of a pure substance with a mass 125 g has a heat capacity of 307.5 J/°C. What is the specific heat of this substance?

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A 36 kg piece of silver metal is heated from 0.0°C to 273°C. Calculate the heat absorbed (in kJ) by the silver.

Calculate the amount of heat liberated (in kJ) from 366 g of mercury when it cools from 100.0°C to 1.0°C.

A sheet of gold weighing 5.0 g and at a temperature of 99.9°C is placed flat on a sheet of iron weighing 25 g and at a temperature of 16.5°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings. (Hint: The heat gained by the gold must be equal to the heat lost by the iron. The specific heats of the metals are given in Table 6.2.)

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A 44.0-g sample of an unknown metal at 199.0°C

was placed in a constant-pressure calorimeter

containing 155 g of water at 37.0°C. The final

temperature of the system was found to be 48.4°C.

Calculate the specific heat of the metal. (The heat

capacity of the calorimeter is 12.4 J/°C.)

1

Lecture 18Professor Hicks

General Chemistry (CHE131)

Wavelength ()

wavelength distance between high points

take a snapshot of any wave

this waves wavelength

is ½ the wavelength of

the wave above

Light waves

• Many types of “light” we cannot see

• All light is called electromagnetic radiation

• Speed of all “light” is c = 2.99 x 108 m/s

aka visible spectrum

2

Why do we have to call it

electromagnetic radiation?

• water waves amplitude? height

• sound waves amplitude? pressure

• electromagnetic waves amplitude?

strength of force felt by charged object

and/or a magnet

+

--

+

e-

e-

like charges repelled

opposite charges

attracted

Diffraction

opening close to size of

wavelength of the wave

When waves collide with objects

or try to pass through openings

close to the size of their

wavelength they undergo…

Diffraction

Diffraction!

if the wave happens to be light then diffraction

spreads it out magnifying the image

for visible light a pinhole

will cause some diffraction

wave

opening wave bumps into

as it tries to pass through

d’Oh!!!

3

Class demo on pinhole diffraction of light

1. take a pin and make the smallest hole you can in a

piece of aluminum foil

2. hold the foil up to the light and try to view the tip of

the pin through the hole

3. after you see the pin move the foil out of your line of

sight and try to continue viewing the pin

4. move the pinhole back and forth

5. you will observe the pin is magnified when viewed

through the pinhole

this is the result of light undergoing diffraction at the pinhole

Electrons have wave properties

• Electrons also undergo diffraction!

Water waves Electron waves

Particle/Wave Nature of Matter

• Louis de Broglie theorized that if light can have wave and particle properties all matter should exhibit wave properties.

• He demonstrated that the relationship between mass and wavelength was

=h

mv

applies to electrons, protons, atoms, bowling balls, people, etc.

is the de Broglie wavelength

any moving object has this

wavelength

Louis de Broglie

4

Wave-particle dualityParticle view

Wave view

Homer is here

100% certain

Homer is probably here

where his wave function

is big ~99% certainHomer’s

de Broglie wave

aka “wave function”

0.1 % chance

Homer is here

or

here

or

hereor

here

or

here

or

hereor

here

the size of the bubble where the particle

probably is gets very large as the particle

gets larger. For Homer it is really 99.99999..%

certain that he is where the bubble is

For electrons this behavior is easily observed.

It is called electron tunneling.

tunneling is the science behind the

science fiction of teleportation

Standing waves

• When waves are trapped in a space they take

on wavelengths that fit into the space

1 wavelength fits in space

2 wavelengths fits in space

3 wavelengths fits in space

4 wavelengths fits in space

each is approximately a different

note of a musical chord

only these

wavelengths

can exist

others are

forbidden

movie on standing waves

5

Orbitals are the standing wave shapes

of electrons trapped in atoms

• s orbitals are spherical in shape.

• come in sets of 1

• size described by a quantum number n

• Larger n Larger orbital

• nucleus is at center

n=1

n=3

n=2

p orbitals • sets of three orbitals

• all 3 orbitals same figure 8 shape

• oriented perpendicular to each other

• each centered at the nucleusen electron has 50%

chance of being here

50% chance

of being here

d orbitals• sets of five orbitals

• 4 orbitals have double figure 8 shape

• 1 orbital is figure 8 shape with a donut belt

• all centered at the nucleus

be able to identify or sketch s, p, and d orbitals

6

Principal Quantum Number, n

• quantum number n, describes the size the orbital

• higher n values have higher energies

• The values of n are integers 1, 2, 3 …

2pn quantum

number = 2a set of 3 figure 8

shaped orbitals

orbital notation

Orbitals and Periodic Table

• Different blocks on the PT are different types of

orbitals.

memorize this!

orbital energy increases

Orbital Diagrams

• Each box or line is an orbital

• Higher quantum numbers (n) have higher energies

set of three p orbitals

set of five d orbitals

Orbitals shown

as boxesOrbitals shown

as lines

7

How electrons fill orbitals - Aufbau Principle

• “Building up” principle

• Electrons usually are in the lowest energy orbital available

• Ground State = all electrons in the lowest energy levels

possible

Lowest energy level

How electrons fill orbitals:

Pauli Exclusion Principle (PEP)

• An orbital can hold a max of 2 electrons

- set s orbitals = 1 orbital, holds 2 electrons

- set p orbitals = 3 orbitals, holds 6 electrons

- set d orbitals = 5 orbitals, holds 10 electrons

Wolfgang Pauli

• PEP and Aufbau principle together tell us that electrons are

added first to the lowest energy, but must occupy higher and higher

levels as more electrons are added

Orbital Diagrams

• Can also be drawn left-right

• Each box represents one orbital.

• Half-arrows each represent an electron

• Two electrons of opposite spin are said to be a paired

the two electrons

in the 1s orbital

are paired

Increasing energy

the one electron

in the 2s orbital

is unpaired

8

Hund’s Rule

• The most stable arrangement of electrons has

the maximum number of parallel spins

p orbitals

of a carbon

atom

Follows

Hund’s Rule

Hund’s Rule and Magnetism

• Substances with unpaired electrons will be magnetic

• Hund’s rule determines the arrangement of the spins

whether a substance is magnetic

Would not

be magnetic

Would not

be magnetic

Would

be magnetic

p orbitals

of a carbon

atom

Individual carbon atoms are observed to be magnetic

7.7 (a) What is the wavelength (in nanometers) of light

having a frequency of 8.6 × 1013 Hz? (b) What is the

frequency (in Hz) of light having a wavelength of 566

nm?

9

Scramble these 3

7.9 The average distance between Mars and Earth is

about 1.3 × 108 miles. How long would it take TV

pictures transmitted from the Viking space vehicle on

Mars' surface to reach Earth? (1 mile = 1.61 km.)

7.39 Thermal neutrons are neutrons that move at speeds

comparable to those of air molecules at room

temperature. Calculate the wavelength (in nm)

associated with a beam of neutrons moving at 7.00 ×

102 m/s. (Mass of a neutron = 1.675 × 10−27 kg.)

7.41 What is the de Broglie wavelength, in cm, of a 12.4-

g hummingbird flying at 1.20 × 102 mph? (1 mile =

1.61 km.)

10

7.82 Indicate the number of unpaired electrons present

in each of the following atoms:

B

Ne

P

Sc

Mn

Se

7.86 Which of the following species has the most

unpaired electrons? S+, S, or S−. Explain how you arrive

at your answer.

1

Lecture 19Professor Hicks

General Chemistry (CHE131)

Energy

of light

• All waves have energy

- water waves PE gravity and KE

- sound waves kinetic energy that

creates the pressure

• Light is different from other kinds of waves

• Its energy is transferred in packets called

photons

• Photons are particles

• Like a penny of energy

3

Max Planck

Planck’s Formula

• Energy of a photon of light Ephoton= hc/

• h = Planck’s constant 6.62 x 10-34 J*s

• c=2.99 x 109 m/s (speed of light)

• Photon = the smallest amount of energy

the light wave can transfer

2

Dunking tank an analogy for photons in action

if the baseball hits the target hard

enough the girl gets dunked

• Like the dunking tank

• If photon hits the DNA and has enough energy

the DNA can be broken

• Visible light does not have enough energy

• UV light ( less than 400 nm) has enough

energy 650 nm red light

Photons and DNA damage

DNA

350 nm (UV) light

What evidence do we have

for the existence of orbitals?

Flame tests and line spectra!

3

Flame tests• elements/monatomic ions produce a unique

colored light if they are placed in a flame

Symbol Name Color

As Arsenic Blue

B Boron Bright Green

Ba Barium Pale/Apple Green

Ca Calcium Orange to Red

Cs Caesium Blue

Cu(I) Copper(I) Blue

Cu(II) Copper(II) (non-halide) Green

Cu(II) Copper(II) (halide) Blue-green

Fe Iron Gold

In Indium Blue

K Potassium Lilac

Robert Bunsen

Discharge tubes

• Gases in glass tubes will conduct

electricity if very high voltage is applied

• When they do this they emit light

• Different elements give different colors

“neon” signs are

discharge tubes

helium = pink

mercury = light blueneon = reddish-orange

argon = lavender

xenon = white

Prism separates white light into Roy G. Biv

White light includes sunlight, fluorescent, and incandescent bulbs.

R ed

O range

Y ellow

G reen

B lue

I ndigo

V iolet

white

light source

white light is combination of all colors of the rainbow

4

Line Spectra

or

different line spectra for every

element like a unique fingerprint

Robert Bunsen

light

More line spectra

Line spectra can be detected from space

telling us about the elements in stars

Line Spectrum

every line is a different wavelength

E = hc486 nm

energy of electrons in atoms organized like a building

each line is like a ball thrown from a different floor

only certain potential energies

b/c floors only at certain heights

E = 6.626 x 10-34 J*s x 2.99 x108 m/s

486 x10-9 mE =

6.626 x 10-34 J*s x 2.99 x108 m/s

486 x10-9 m

6.626 x 10-34 J*s x 2.99 x108 m/s

486 x10-9 m

= 4.076 x 10–19 J

every wavelength is a different

energy that is being released

energy of a

486 nm photon

highest

energy

lowest

energy

lower

energy

lower

energy

highest

energy

lowest

energy

heat (q)

released

Wavele

ng

th (n

m)

Wavele

ng

th (n

m)

wav

elen

gth

(nm

)

656 nm

434 nm

5

Bohr’s model of the hydrogen atom

• Planetary model of the atom

• Electron(s) orbit the tiny nucleus

• Electrons jump from one fixed orbital radius to

another losing or gaining energy as a photon

Works to explain the pattern in

the line spectra of hydrogen,

but fails for all other elements

7.16 The blue color of the sky results from the scattering

of sunlight by air molecules. The blue light has a

frequency of about 7.5 × 1014 Hz. (a) Calculate the

wavelength, in nm, associated with this radiation, and (b)

calculate the energy, in joules, of a single photon

associated with this frequency.

7.100 A ruby laser produces radiation of wavelength 633

nm in pulses whose duration is 1.00 × 10−9 s.

(a) If the laser produces 0.376 J of energy per pulse, how

many photons are produced in each pulse?

(b) Calculate the power (in watts) delivered by the laser per

pulse. (1 W = 1 J/s.)

6

7.27 Is it possible for a fluorescent material to emit

radiation in the ultraviolet region after absorbing visible

light? Explain your answer.

Consider the following energy levels of a hypothetical

atom:

E4 −1.0 × 10−19 J

E3 −5.0 × 10−19 J

E2 −10 × 10−19 J

E1 −15 × 10−19 J

(a)What is the wavelength of the photon needed to excite

an electron from E1 to E4? (b) What is the energy (in

joules) a photon must have in order to excite an electron

from E2 to E3? (c) When an electron drops from the E3

level to the E1 level, the atom is said to undergo

emission. Calculate the wavelength of the photon

emitted in this process.

The first line of the Balmer series occurs at a wavelength of

656.3 nm. What is the energy difference between the two

energy levels involved in the emission that results in this

spectral line?

7

Calculate the frequency (Hz) and wavelength (nm) of the

emitted photon when an electron drops from the n = 4 to

the n = 1 level in a hydrogen atom.

When a compound containing cesium ion is heated in a

Bunsen burner flame, photons with an energy of 4.30 ×

10−19 J are emitted. What color is the cesium flame?

8/28/2011

1

2

Lecture 20Professor Hicks

General Chemistry (CHE131)

3

Electron Configurations

• Arrangement of electrons in an atoms orbitals is called its electron configuration

He = 1s2

n quantum number

orbital type

number of electronsin that orbital

4

Abbreviated Electron Configurations

core electrons abbreviatedby noble gas that came

before the element

Rb = 1s22s22p63s23p64s23d104p6 5s1

[Kr] 5s1abbreviated

electron configuration of rubidium

core electrons valenceelectrons

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5

Valence electrons

• Electrons in the highest energy levels are called valence electrons

• Exact definition is different for main group elements and the rest of the periodic table

1) For main group elements all electrons in the highest n value orbitals are valence electrons

2) For transition metals all electrons in highest n value and those in the n-1 value d orbitals are valence electrons

6

Ca = 1s22s22p63s23p64s2

Valence electrons

Br = 1s22s22p63s23p64s23d104p5

5 valenceelectrons

V = 1s22s22p63s23p64s23d3

2 valenceelectrons

7 valenceelectrons

n = 4 is highest n valueBr is in main group

n = 4 is highest n valueCa is in main group

n = 4 is highest n value V is transition metalso n-1 level 3d electrons are also included

7

Electron configurations and the Periodic Table

elements in

same group

similar electron

configurations

chemical properties are determinedby the number of valence electrons!

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8

• determine the atomic # (Z) of Mg from PT

• this is the number of electrons in the atom

Mg Z = 12, so Mg has 12 electrons

• determine the highest energy occupied orbital from its position on the PT

Example: Write the electron configuration of magnesium.

3s orbitals are highest energy orbitals for Mg

you have to remember this pattern

9

1s 2s 2p 3s

Example: Write the electron configuration of Mg.

1) 3s orbitals are highest energy orbitals for Mg so write

all orbitals up to 3s

1s22s22p63s2 or abbreviated

[Ne]3s2

2) add electrons following the Pauli Exclusion and Aufbau Principles

10

Main Group Elementselectron configuration

and ion charge

• atoms form ions that will have the same electron configuration of a noble gas

• metals lose electrons so their ions have the same electron configurations as the noble gas with a lower atomic number

• non-metals gain electrons so their ions have the same electron configurations as the noble gas with a higher atomic number

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11

Electron configurations of ions

• For positive ions valence electrons are always removedfrom the

orbital of highest n value when an ion is formed

- if more than 1 orbital has same n value e.g. 2s22p1 the electron is taken from p orbitals before s orbitals.

• for negative ions electrons are added by the Aufbau

principleExample: the magnesium atom has 2 valence electrons

Mg = 1s22s22p63s2

when it forms an ion, it loses its valence electrons from its

s orbitals since they have the highest n value

Mg2+ = 1s22s22p6

12

Electron configurations ofmain group positive ions

Na = 1s22s22p63s1

Na = [Ne]3s1

Na+ = 1s22s22p6

Na+ = [Ne]

Ne = 1s22s22p6

Ne = [Ne]

or written as the abbreviated electron configurations

Na+ has same electron configuration as the noble gas Ne

same as previous noble gas

Na forms Na+ ion

remove highest n value electron

13

Electron configurations ofmain group negative ions

S2- = 1s22s22p63s23p6

orbital diagram for S orbital diagram for S2-

same as Ar

same as nearest noble gas to the right

S = 1s22s22p63s23p4

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5

Isoelectronic Species

• Substances with the same number of valence electrons are chemically similar

• Especially if the electron configurations are the same or similar

Examples

• Elements on the PT

- Elements in a group have the same electron configurations with higher n values similar properties

- Stable monatomic ions and noble gases, N3-, O2-, F-, and Ne all have 1s22s2p6 electron configurations

- Unstable metals and ions

- Mg+1 and Na are both [Ne]3s1 and both have similar reactivity-both react by losing 1 electron Na+ or Mg2+ achieving [Ne]

7.80 The ground-state electron configurations listed here are incorrect. Explain what mistakes have been made in each and write the correct electron configurations.

Al: 1s22s22p43s23p3

B: 1s22s22p5

F: 1s22s22p6

7.83 Write the ground-state electron configurations for the following elements: B, V, Ni, As, I, Au.

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8.15 In the periodic table, the element hydrogen is

sometimes grouped with the alkali metals (as in this

book) and sometimes with the halogens. Explain

why hydrogen can resemble both the Group 1A and

the Group 7A elements.

8.17 Group these electron configurations in pairs

that would represent similar chemical properties of

their atoms:

1s22s22p63s2

1s22s22p3

1s22s22p63s23p64s23d104p6

1s22s2

1s22s22p6

1s22s22p63s23p3

Without referring to a periodic table, write the

electron configurations of elements with these

atomic numbers: (a) 7, (b) 19, (c) 28, (d) 35

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8.24 What do we mean when we say that two ions

or an atom and an ion are isoelectronic?

8.27 Write ground-state electron configurations for

(a) Li+

(b) H−

(c) N3−

(d) F−

(e) S2−

(f) Al3+

(g) Se2−

8.29 Write ground-state electron configurations for

these transition metal ions:

(a) Sc3+

(b) Ti4+

(c) V5+

(d) Cr3+

(e) Mn2+

(f) Fe2+

(g) Fe3+

m) Au+,

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8.30 Name the ions with +3 charges that have these

electron configurations: (a) [Ar]3d3, (b) [Ar], (c)

[Kr]4d6, (d) [Xe]4f145d6.

8.31 Which of these species are isoelectronic with

each other: C, Cl−, Mn2+, B−, Ar, Zn, Fe3+, Ge2+?