chapter 7 the quantum-mechanical model of the atom chemistry: a molecular approach, 1 st ed. nivaldo...

Chapter 7The Quantum-Mechanical

Model of the Atom

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

•Quantum Mechanics•The Nature of Light•Atomic Spectroscopy and the Bohr Model•The Wave Nature of Matter

•The de Broglie Wavelength•Uncertainty Principle

•Quantum Mechanics and the Atom•The Shapes of Atomic Orbitals

Anyone who is not shocked by quantum mechanics has not understood it

Neils Bohr

Tro 1/e Chap 7 Slide 2


The Behavior of the Very Small• electrons are incredibly small

– a single speck of dust has more electrons than the number of people who have ever lived on earth

• electron behavior determines the physical and chemical properties of atoms

• to understand this behavior we must understand how electrons exist within atoms!

• directly observing electrons in the atom is impossible, e.g. if you attempt to measure its position using light, the light itself disturbs the electron changing its position


A Theory that Explains Electron Behavior

• the quantum-mechanical model explains the manner electrons exist and behave in atoms

• helps us understand and predict the properties of atoms that are directly related to the behavior of the electrons– why some elements are metals while others are nonmetals

– why some elements gain 1 electron when forming an anion, while others gain 2

– why some elements are very reactive while others are practically inert

– and other Periodic patterns we see in the properties of the elements

• Prior to the development of QM the nature of light was viewed as being very different from that of subatomic particles such as electrons

• As QM developed, light was found to have many of the characteristics in common with electrons

• Chief among these is the wave-particle duality of light – certain properties are best described by thinking of it as a wave, while other properties are best described by thinking of it as a particle



The Nature of Lightits Wave Nature

• light is a form of electromagnetic radiation– composed of perpendicular oscillating waves, one for the

electric field and one for the magnetic field• an electric field is a region where an electrically charged particle

experiences a force

• a magnetic field is a region where an magnetized particle experiences a force

• all electromagnetic waves move through space at the same, constant speed– 3.00 x 108 m/s in a vacuum = the speed of light, c


Speed of Energy Transmission


Insert fig 5.15

The Electromagnetic Spectrum


Electromagnetic Radiation

• Electromagnetic waves consist of oscillating, perpendicular electric and magnetic fields.

• The wavelength of radiation is the distance between peaks in a wave. ()

• The frequency is the number of peaks that pass a point in a second. ( )


Electromagnetic Radiation


Characterizing Waves• the amplitude is the height of the wave

– the distance from node to crest • or node to trough

– the amplitude is a measure of how intense the light is – the larger the amplitude, the brighter the light• the wavelength, () is a measure of the distance covered by the wave

– the distance from one crest to the next • or the distance from one trough to the next, or the distance between alternate nodes


Characterizing Waves• the frequency, () is the number of waves that pass a point in a given period of time

– the number of waves = number of cycles– units are hertz, (Hz) or cycles/s = s-1

• 1 Hz = 1 s-1

• the total energy is proportional to the amplitude and frequency of the waves– the larger the wave amplitude, the more force it has– the more frequently the waves strike, the more total force there is


The Relationship Between Wavelength and Frequency

• for waves traveling at the same speed, the shorter the wavelength, the more frequently they pass

• this means that the wavelength and frequency of electromagnetic waves are inversely proportional– since the speed of light is constant, if we know

wavelength we can find the frequency, and visa versa

m

cs s

m1-


Frequency and Wavelength• Given two of the following calculate the

third: c (velocity of propagation).c

c/cis wavelength measured in length units (m, cm, nm,

etc.)

is frequency measured in Hz (s-1).

• c is the velocity• velocity of light in vacuum = 3.0 x108 ms-1


Example 7.1- Calculate the wavelength of red light with a frequency of 4.62 x 1014 s-1

the unit is correct, the wavelength is appropriate for red light

Check:

Solve:

∙ = c, 1 nm = 10-9 m

Concept Plan:

Relationships:

= 4.62 x 1014 s-1

, (nm)

Given:

Find:

s-1) (m) (nm)

c

m 10

nm 19

m 1049.6 1062.4

1000.3c 71-

s14

-1sm

8

nm 1049.6m 10

nm 1m 1049.6 2

97


Practice – Calculate the wavelength of a radio signal with a frequency of 100.7 MHz


m 98.2 10007.1

1000.3c1-

s8

-1sm

8

1-8-16

s 10007.1MHz 1

s 10MHz 00.71

Practice – Calculate the wavelength of a radio signal with a frequency of 100.7 MHz

the unit is correct, the wavelength is appropriate for radiowaves

Check:

Solve:

∙ = c, 1 MHz = 106 s-1

Concept Plan:

Relationships:

= 100.7 MHz

, (m)

Given:

Find:

Hz) (s-1) (m)

c

MHz 1

s 10 -16


Color

• the color of light is determined by its wavelength– or frequency

• white light is a mixture of all the colors of visible light– a spectrum

– RedOrangeYellowGreenBlueViolet

• when an object absorbs some of the wavelengths of white light while reflecting others, it appears colored– the observed color is predominantly the colors reflected

Amplitude & Wavelength


Electromagnetic Spectrum


The Electromagnetic Spectrum• visible light comprises only a small fraction of

all the wavelengths of light – called the electromagnetic spectrum

• short wavelength (high frequency) light has high energy– radiowave light has the lowest energy– gamma ray light has the highest energy

• high energy electromagnetic radiation can potentially damage biological molecules– ionizing radiation


Thermal Imaging using Infrared Light


Electromagnetic Spectrum• Recognize common units for wavelength frequency

• meters (m) radio Hertz Hz s-1

• micrometers m (cycles per

• (10-6 m) microwaves second)

• nanometers nm megahertz

• (10-9 m) light MHz (106 Hz)

• Å angstrom (10-10 m)


Interference• the interaction between waves is called interference

• when waves interact so that they add to make a larger wave it is called constructive interference– waves are in-phase

• when waves interact so they cancel each other it is called destructive interference– waves are out-of-phase


Diffraction• when traveling waves encounter an

obstacle or opening in a barrier that is about the same size as the wavelength, they bend around it

– this is called diffraction

– traveling particles do not diffract


Diffraction• the diffraction of light through two slits separated by a distance comparable to the wavelength results in an

interference pattern of the diffracted waves

• an interference pattern is a characteristic of all light waves


The Particle Nature of Light• Prior to the 1900s light was thought to be purely a wave phenomenon

– Principally due to the discovery of diffraction

• This classical view began to be questioned….


The Photoelectric Effect• it was observed that many metals emit electrons when a light shines on their

surface– this is called the Photoelectric Effect

• classic wave theory attributed this effect to the light energy being transferred to the electron

• according to this theory, if the wavelength of light is made shorter, or the light waves intensity made brighter, more electrons should be ejected– remember: the energy of a wave is directly proportional to its amplitude

and its frequency– if a dim light was used there would be a lag time before electrons were

emitted • to give the electrons time to absorb enough energy


The Photoelectric Effect


The Photoelectric EffectThe Problem

• in experiments with the photoelectric effect, it was observed that there was a maximum wavelength for electrons to be emitted– called the threshold frequency– regardless of the intensity

• it was also observed that high frequency light with a dim source caused electron emission without any lag time


Photoelectric Effect


Einstein’s Explanation

• Einstein proposed that the light energy was delivered to the atoms in packets, called quanta or photons

• the energy of a photon of light was directly proportional to its frequency– inversely proportional to it wavelength– the proportionality constant is called Planck’s Constant, (h)

and has the value 6.626 x 10-34 J∙s

ch

hE


Define quantum, photon.

• Quantum- A packet of energy equal to hThe smallest quantity of energy that can be emitted or absorbed.

• Photon- A quantum of electromagnetic radiation.• Thus light can be described as a particle (photon)

or as a wave with wavelength and frequency. This is called wave-particle duality.


Given one of the following, calculate the other: E,

• Ephoton = h

• where:Ephoton is the energy of each photon (packet) of light.

h is Planck’s constant = 6.626x10-34 Js

is the frequency of light in Hz (s-1).


Example 7.2- Calculate the number of photons in a laser pulse with wavelength 337 nm and total energy 3.83 mJ

Solve:E=hc/, 1 nm = 10-9 m, 1 mJ = 10-3 J, Epulse/Ephoton = # photons

Concept Plan:

Relationships:

= 337 nm, Epulse = 3.83 mJ

number of photons

Given:

Find:

hc

Ephoton nm 1

m 10 9

J 108598.5

1037.3

1000.3 10626.6hcE 19

m7

-1sm

8sJ

34

photon

m 1037.3nm 1

m 10nm 1037.3 7

92

nm) (m) Ephotonnumberphotons

photon

pulse

E

E

J 1083.3mJ 1

J 10mJ 83.3 3

3

photons 1049.6

J 108598.5

J 1083.3photons ofnumber

15

3

3


Practice – What is the frequency of radiation required to supply 1.0 x 102 J of energy from

8.5 x 1027 photons?


1-7

sJ34

J26

photon s 108.1 10626.6

10761.1

h

E

What is the frequency of radiation required to supply 1.0 x 102 J of energy from 8.5 x 1027 photons?

Solve:E=h, Etotal = Ephoton∙# photons

Concept Plan:

Relationships:

Etotal = 1.0 x 102 J, number of photons = 8.5 x 1027

Given:

Find:

h

Ephoton

(s-1)Ephotonnumberphotons

photons ofnumber

E total

J 10761.1105.8

J 100.1E 26

27

2

photon


Ejected Electrons

• 1 photon at the threshold frequency has just enough energy for an electron to escape the atom– binding energy,

• for higher frequencies, the electron absorbs more energy than is necessary to escape

• this excess energy becomes kinetic energy of the ejected electron

Kinetic Energy = Ephoton – Ebinding

KE = h -

• Although quantization of light explained the photoelectric effect, the wave explanation of light continued to have explanatory power

• What slowly emerged is what we now call the wave-particle duality of light



Exciting e-

Many ways:

• Heat, light, e- bombardment, chemical reactions


Atomic Spectra• when atoms or molecules absorb energy, that energy is

often released as light energy– fireworks, neon lights, etc.

• when that light is passed through a prism, a pattern is seen that is unique to that type of atom or molecule – the pattern is called an emission spectrum– non-continuous– can be used to identify the material

• flame tests

Hg He H


Emission Spectra


Examples of Spectra

Oxygen spectrum

Neon spectrum


Atomic Spectra

• Rydberg analyzed the spectrum of hydrogen and found that it could be described with an equation that involved an inverse square of integers

• Predicted wavelengths of emission spectrum for hydrogen

221-7

n

1

m

1m 10097.1

1

A Bar Code for AtomsFlame Tests


Na K Li Ba


Emission vs. Absorption Spectra

Spectra of Mercury

Bohr’s Model

• Bohr attempted to develop a model for the atom that explained atomic spectra…



Bohr’s Model• Neils Bohr proposed that the electrons could only have

very specific amounts of energy– fixed amounts = quantized

• the electrons traveled in orbits that were a fixed distance from the nucleus– stationary states– therefore the energy of the electron was proportional the

distance the orbital was from the nucleus• electrons emitted radiation when they “jumped” from

an orbit with higher energy down to an orbit with lower energy– the distance between the orbits determined the energy of the

photon of light produced


Bohr Model of H Atoms

e- is never observed between states, only in one state or the next


photon = E = Ehigh-Elow

E = h = hc/light = E/h = hc/E

• Since Ehigh and Elow are discreet numbers, must be a discreet number

• Therefore, and must be discreet numbers, giving rise to single

frequencies and wavelengths of light. Hence, the line spectra.


QuestionFor the yellow line in the sodium spectrum, =

589.0 nm. Calculate the energy in joules of a photon emitted by an excited sodium atom.

= 5.090 1014 /sE = 3.373 10-19 J

• Bohr’s model could not be applied to atoms with 2 or more e-

Needed a new approach…



Wave Behavior of Electrons• de Broglie proposed that particles could have wave-like

character• because it is so small, the wave character of electrons is

significant• electron beams shot at slits show an interference

pattern– A single e- produces an interference pattern– the electron interferes with its own wave (not other e-)


however, electrons actually present an interference pattern, demonstrating the behave like waves

Electron Diffraction

if electrons behave like particles, there should only be two bright spots on the target

• Explains the existence of stationary states / energy levels in the Bohr modelhttp://www.colorado.edu/physics/2000/quantumzone/debroglie.html

• Prevents e- from crashing into the nucleus as predicted by classical physics


http://www.colorado.edu/physics/2000/quantumzone/debroglie.html

http://www.colorado.edu/physics/2000/quantumzone/debroglie.html


Wave Behavior of Electrons• de Broglie predicted that the wavelength of a particle

was inversely proportional to its momentum

)s(m(kg)

ss

mkg

m1-

2

2

velocitymass

h


m 1074.2

1065.2kg 10.119

10626.6

mv

h

10

m631-

mk342

2

s

s

gs

Solve:

Example 7.3- Calculate the wavelength of an electron traveling at 2.65 x 106 m/s

=h/mv

Concept Plan:

Relationships:

v = 2.65 x 106 m/s, m = 9.11 x 10-31 kg (back leaf)

m

Given:

Find:

vm

h

m, v (m)


Practice - Determine the wavelength of a neutron traveling at 1.00 x 102 m/s

(Massneutron = 1.675 x 10-24 g)


m 1096.3

1000.1kg 10675.1

10626.6

mv

h

9

m227-

mk342

2

s

s

gs

Solve:

Practice - Determine the wavelength of a neutron traveling at 1.00 x 102 m/s

=h/mv, 1 kg = 103 g

Concept Plan:

Relationships:

v = 1.00 x 102 m/s, m = 1.675 x 10-24 g

m

Given:

Find:

vm

h

m (kg), v (m)m(g)

g 10

kg 13

kg 10675.1

g 10

kg 1g 10675.1

27

324


Practice - Determine the wavelength of a bowling ball traveling at 1 m/s

(Mass = 1 kg)

1 x 10-24 nm

1 septillionth of a nanometer. This is so ridiculously small compared to the size of the bowling ball itself that you'd never notice any wavelike stuff going on; that's why we can generally ignore the effects of quantum mechanics when we're talking about everyday objects

It's only at the molecular or atomic level that the waves begin to be large enough (compared to the size of an atom) to have a noticeable effect.

The Uncertainty Principle• How can a single entity behave as both a particle and a wave?

• How does a single e- aimed at a double-slit produce an interference pattern?

• A possible hypothesis is the e- splits into 2, travels through both slots and interferes with itself



Testing the hypothesis…

any experiment designed to observe the electron results in detection of a single electron particle and no interference pattern

Experiment designed to ‘watch’ which slit the e- travels through using a laser beam, flash is seen from one slit or the other…never both at once

• No matter how hard we try we can never see the interference pattern and simultaneously determine which hole the e- goes through

• We have encountered the absolutely small and have no way of determining what it is without disturbing it

• Wave nature and particle nature of the e- are complementary properties

• The more you know about one, the less you know about the other



Uncertainty Principle• Heisenberg stated that the product of the uncertainties in both the position

and speed of a particle was inversely proportional to its mass

– x = position, x = uncertainty in position (related to particle nature)

– v = velocity, v = uncertainty in velocity (related to wave nature)

– m = mass

• the means that the more accurately you know the position of a small particle, like an electron, the less you know about its speed

– and visa-versa

• Electron is observed EITHER as a particle OR a wave, never both at once

m

1

4

hv

x


Determinacy vs. Indeterminacy• according to classical physics, particles move in a path determined by the

particle’s velocity, position, and forces acting on it– determinacy = definite, predictable future

• because we cannot know both the position and velocity of an electron, we cannot predict the path it will follow– indeterminacy = indefinite future, can only predict probability

• the best we can do is to describe the probability an electron will be found in a particular region using statistical functions

Trajectory vs. Probability


Electron Energy• Position and velocity are complimentary

• electron energy and position are complimentary

– because kinetic energy is related to velocity, KE = ½mv2

• for an electron with a given energy, the best we can do is describe a region in the atom of high probability of finding it – called an orbital

– a probability distribution map of a region where the electron is likely to be found

– A plot of distance vs. 2 represents an orbital

• many of the properties of atoms are related to the energies of the electrons

Eψψ Η


Wave Function,

• calculations show that the size, shape and orientation in space of an orbital are determined be three integer terms in the wave function

– added to quantize the energy of the electron

• these integers are called quantum numbers

– principal quantum number, n

– angular momentum quantum number, l

– magnetic quantum number, ml


Principal Quantum Number, n• characterizes the energy of the electron in a particular orbital

– corresponds to Bohr’s energy level

• n can be any integer 1• the larger the value of n, the greater/less negative the energy• energies are defined as being negative

– an electron would have E = 0 when it just escapes the atom

• the larger the value of n, the larger the orbital• as n gets larger, the amount of energy between orbitals gets smaller

218- 1

J 10-2.18En

n for an electron in H


Principal Energy Levels in Hydrogen

Electron Transitions• in order to transition to a higher energy state, the electron must

gain the correct amount of energy corresponding to the difference in energy between the final and initial states

• electrons in high energy states are unstable and tend to lose energy and transition to lower energy states– energy released as a photon of light

• each line in the emission spectrum corresponds to the difference in energy between two energy states


Predicting the Spectrum of Hydrogen• the wavelengths of lines in the emission spectrum of

hydrogen can be predicted by calculating the difference in energy between any two states

• for an electron in energy state n, there are (n – 1) energy states it can transition to, therefore (n – 1) lines it can generate

• both the Bohr and Quantum Mechanical Models can predict these lines very accurately

2

182

18

electronhydrogen releasedphoton

1J 1018.2

1J 1018.2

hch

EEEE

initialfinal

initialfinal

nn

Hydrogen Energy Transitions

m 1046.7

104466.2

1000.3 10626.6

E

hc 6

J20-

s

m8sJ

34

Solve:

Example 7.7- Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 6 to n = 5

E=hc/En = -2.18 x 10-18 J (1/n2)

Concept Plan:

Relationships:

ni = 6, nf = 5

m

Given:

Find:

2H1

REn

ni, nf Eatom Ephoton

E

ch Eatom = -Ephoton

J 104466.26

1

5

1J 1018.2E 20

2218

atom

Ephoton = -(-2.6644 x 10-20 J) = 2.6644 x 10-20 J

the unit is correct, the wavelength is in the infrared, which is appropriate because less energy than 4→2 (in the visible)

Check:


Practice – Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 2 to n = 1

m 1021.1

1064.1

1000.3 10626.6

E

hc 7

J18-

s

m8sJ

34

Solve:

Calculate the wavelength of light emitted when the hydrogen electron transitions from n = 2 to n = 1

E=hc/En = -2.18 x 10-18 J (1/n2)

Concept Plan:

Relationships:

ni = 2, nf = 1

m

Given:

Find:

2H1

REn

ni, nf Eatom Ephoton

E

ch Eatom = -Ephoton

J 1064.12

1

1

1J 1018.2E 18

2218

atom

Ephoton = -(-1.64 x 10-18 J) = 1.64 x 10-18 J

the unit is correct, the wavelength is in the UV, which is appropriate because more energy than 3→2 (in the visible)

Check:

Shapes of Orbitals• Shapes are important because covalent bonds depend on sharing e-

• Chemical bonds consists on overlapping orbitals



The Shapes of Atomic Orbitals

• the l quantum number primarily determines the shape of the orbital

• l can have integer values from 0 to (n – 1)• each value of l is called by a particular letter

that designates the shape of the orbital– s orbitals are spherical (l = 0)– p orbitals are like two balloons tied at the knots– d orbitals are mainly like 4 balloons tied at the knot– f orbitals are mainly like 8 balloons tied at the knot


l = 0, the s orbital• each principal energy

state has 1 s orbital

• lowest energy orbital in a principal energy state

• spherical

• number of nodes = (n – 1)

Probability & Radial Distribution Functions

• 2 is the probability density (= prob. / unit vol.)– the probability of finding an electron at a particular point in space– for s orbital maximum at the nucleus?– decreases as you move away from the nucleus

• the Radial Distribution function represents the total probability at a certain distance from the nucleus– maximum at most probable radius

• nodes in the functions are where the probability drops to 0


Probability Density Function

Probability density at a point


Radial Distribution Function

Radial distribution represents total probability at radius r

2s and 3s2s

n = 2,l = 0

3sn = 3,l = 0


l = 1, p orbitals• each principal energy state above n = 1 has 3 p

orbitals– ml = -1, 0, +1

• each of the 3 orbitals point along a different axis– px, py, pz

• 2nd lowest energy orbitals in a principal energy state• two-lobed• node at the nucleus, total of n nodes


p orbitals


l = 2, d orbitals• each principal energy state above n = 2 has 5 d

orbitals– ml = -2, -1, 0, +1, +2

• 4 of the 5 orbitals are aligned in a different plane– the fifth is aligned with the z axis, dz squared

– dxy, dyz, dxz, dx squared – y squared

• 3rd lowest energy orbitals in a principal energy state• mainly 4-lobed

– one is two-lobed with a toroid• planar nodes

– higher principal levels also have spherical nodes


d orbitals


l = 3, f orbitals• each principal energy state above n = 3 has 7 d

orbitals– ml = -3, -2, -1, 0, +1, +2, +3

• 4th lowest energy orbitals in a principal energy state• mainly 8-lobed

– some 2-lobed with a toroid

• planar nodes– higher principal levels also have spherical nodes


f orbitals


Why are Atoms Spherical?


Energy Shells and Subshells

chapter 7 the quantum-mechanical model of the atom chemistry: a molecular approach, 1 st ed. nivaldo...

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