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Chapter 8 The Quantum Mechanical Atom Copyright © 2012 by John Wiley & Sons, Inc. CHEMISTRY The Molecular Nature of Matter SIXTH EDITION Jespersen • Brady • Hyslop

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Page 1: CHEMISTRY - NSCC NetID: Personal Web Spacefacweb.northseattle.edu/rreategui/CHEM161/LECTURES/Chapter 8 CHE… · Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter,

Chapter 8The Quantum Mechanical Atom

Copyright © 2012 by John Wiley & Sons, Inc.

CHEMISTRYThe Molecular Nature of Matter

SIXTH EDITION

Jespersen • Brady • Hyslop

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

The nature of Light

Electromagnetic Radiation

Light: Energy transferred between atoms/molecules

Travels through space at high speed in vacuum

c = speed of light = 2.9979 × 108 m/s

Light is radiation that carries energy through space by means of waves.

Waves or Oscillations

Systematic fluctuations in intensities of electrical and magnetic forces

Varies regularly with time

Exhibit wide range of energy 2

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Properties of Waves Wavelength ()

Distance between two successive peaks or troughs

Units are in meters, centimeters, nanometers

Frequency ()

Number of waves per second that pass a given point in space

Units are in Hertz (Hz = cycles/sec = 1/sec = s–1)

Related by = c

3

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Properties of Waves Amplitude

Maximum and minimum height

Intensity of wave, or brightness

Varies with time as travels through space

Nodes

Points of zero amplitude

Place where wave goes through axis

Distance between nodes is constant

4

nodes

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Learning Check: Converting from Wavelength to Frequency

The bright red color in fireworks is due to emission of light when Sr(NO3)2 is heated. If the wavelength is ~650 nm, what is the frequency of this light?

5

n =c

l=

3.00 ´108 m/s

650 ´10-9 m

= 4.61 × 1014 s–1 = 4.6 × 1014 Hz

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Your Turn!WCBS broadcasts at a frequency of 880 kHz. What is the wavelength of their signal?

A. 341 m

B. 293 m

C. 293 mm

D. 341 km

E. 293 mm

6

l =c

n=3.00 ´108m/s

880 ´103 / s

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Electromagnetic Spectrum

7

high energy, short waves low energy, long waves

Comprised of all frequencies of light

Divided into regions according to wavelengths of radiation

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Electromagnetic Spectrum

Visible light

Band of wavelengths that human eyes can see

400 to 700 nm

Make up spectrum of colors

8

White light

Combination of all these colors

Can separate white light into the colors with a prism

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Important Experiments in Atomic Theory

Late 1800’s:

Matter and energy believed to be distinct

Matter: made up of particles

Energy: light waves

Beginning of 1900’s:

Several experiments proved this idea incorrect

Experiments showed that electrons acted like:

Tiny charged particles in some experiments

Waves in other experiments

9

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Photosynthesis If you irradiate plants with infrared and microwave

radiation No photosynthesis

Regardless of light intensity

If you irradiate plants with visible light Photosynthesis occurs

More intense light now means more photosynthesis

10

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Line Spectrum

11

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Particle Theory of Light

Max Planck and Albert Einstein (1905)

Electromagnetic radiation is stream of small packets of energy

Quanta of energy or photons

Each photon travels with velocity = c

Waves with frequency =

Energy of photon of electromagnetic radiation is proportional to its frequency

Energy of photon E = h

h = Planck’s constant = 6.626 × 10–34 J s

12

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Atomic Spectra Atomic line spectra are rather complicated

Line spectrum of hydrogen is simplest

Single electron

First success in explaining quantized line spectra

First studied extensively

J.J. Balmer

Found empirical equation to fit lines in visible region of spectrum

J. Rydberg

More general equation explains all emission lines in H atom spectrum (infrared, visible, and UV)

13

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Rydberg Equation

Can be used to calculate all spectral lines of hydrogen

The values for n correspond to allowed energy levels for atom

RH = 109,678 cm–1 = Rydberg constant

= wavelength of light emitted

n1 and n2 = whole numbers (integers) from 1 to

where n2 > n1

If n1 = 1, then n2 = 2, 3, 4, …

14

22

21

111

nnRH

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Learning Check: Using Rydberg Equation

Consider n1 = 2 Calculate (in nm) for the transition from n2 = 6 down to n1 = 2.

15

1

l= R

H

1

22-

1

62

æ

èçç

ö

ø÷÷ = 109,678 cm-1 1

4-

1

36

æ

èçç

ö

ø÷÷

l =1

24,372.9 cm-1 = 4.1029 ´10-5 cm´

1 m

100 cm´

1 nm

1´10-9 m

= 410.3 nm Violet line in spectrum

= 24,373 cm–1

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Learning Check

A photon undergoes a transition from nhigher down to n = 2 and the emitted light has a wavelength of 650.5 nm?

1

7.13455= ( 1

4- 1

n2( )

2)

16

1650.5 ´ 10-7 cm

= 109,678 cm-1( 122

- 1

n2( )

2)

n2 = 3

1n

2( )2

= 14

- 17.13455

= 0.110

n2( )

2

=1

0.110= 9.10

l = 650.5 nm´1´10-7cm

1 nm= 650.5 ´10-7cm

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Your Turn!What is the wavelength of light (in nm) that is emitted when an excited electron in the hydrogen atom falls from n = 5 to n = 3?

A. 1.28 × 103 nm

B. 1.462 × 104 nm

C. 7.80 × 102 nm

D. 7.80 × 10–4 nm

E. 3.65 × 10–7 nm

17

22

1

5

1

3

1cm 678,109

1

l =1

7799 cm-1 ´

1´107nm

1 cm

1cm 77991

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Significance of Atomic Spectra Atomic line spectra tells us

When excited atom loses energy

Only fixed amounts of energy can be lost

Only certain energy photons are emitted

Electron restricted to certain fixed energy levels in atoms

Energy of electron is quantized

Simple extension of Planck's Theory

Any theory of atomic structure must account for

Atomic spectra

Quantization of energy levels in atom

18

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

What Does Quantized Mean?

Energy is quantized if only certain discrete values are allowed

Presence of discontinuities makes atomic emission

quantized

19

Potential Energy of Rabbit

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Bohr Model of Atom

First theoretical model of atom to successfully account for Rydberg equation

Quantization of energy in hydrogen atom

Correctly explained atomic line spectra

Proposed that electrons moved around nucleus like planets move around sun

Move in fixed paths or orbits

Each orbit has fixed energy

20

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Energy for Bohr Model of H Equation for energy of electron in H atom

Ultimately b relates to RH by b = RHhc

OR

Where b = RHhc = 2.1788 × 10–18 J/atom

Allowed values of n = 1, 2, 3, 4, …

n = quantum number

Used to identify orbit21

E µ -1

n2b =

2p 2me 4

h2

E = -b

n2= -

RHhc

n2

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Energy Level Diagram for H Atom

Absorption of photon

Electron raised to higher energy level

Emission of photon

Electron falls to lower energy level

22

Energy levels are quantized Every time an electron drops from one

energy level to a lower energy level Same frequency photon is emitted Yields line spectra

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Bohr Model of Hydrogen Atom n = 1 First Bohr orbit

Most stable energy state equals the ground state which is the lowest energy state

Electron remains in lowest energy state unless disturbed

How to change the energy of the atom? Add energy, as light (E = h) or other form.

Electron raised to higher n orbit n = 2, 3, 4, …

Higher n orbits = excited states = less stable

So electron quickly drops to lower energy orbit and emits photon of energy equal to E between levels

E = Eh – El h = higher l = lower

23

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Your Turn!In Bohr's atomic theory, when an electron moves from one energy level to another energy level more distant from the nucleus,

A. energy is emitted

B. energy is absorbed

C. no change in energy occurs

D. light is emitted

E. none of these

24

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Light Exhibits Interference

Constructive interference

Waves “in-phase” lead to greater amplitude

They add together

Destructive interference

Waves “out-of-phase” lead to lower amplitude

They cancel out

25

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Diffraction and Electrons Light

Exhibits interference

Has particle-like nature

Electrons

Known to be particles

Also demonstrate interference

26

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Standing vs. Traveling Waves

Traveling wave

Produced by wind on surfaces of lakes and oceans

Standing wave

Produced when guitar string is plucked

Center of string vibrates

Ends remain fixed

27

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Standing Wave on a Wire

Integer number (n) of peaks and troughs is required

Wavelength is quantized:

L is the length of the string

28

l =2L

n

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

How Do We Describe an Electron?

Has both wave-like and particle-like properties

Energy of moving electron on a wire is E =½mv 2

Wavelength is related to the quantum number, n, and the wire length:

29

l =2L

n

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Electron Has Quantized Energy

Electron energy quantized

Depends on integer n

Energy level spacing changes when positive charge in nucleus changes

Line spectra different for each element

Lowest energy allowed is for n =1

Energy cannot be zero, hence atom cannot collapse

30

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Wave FunctionsSchrödinger’s equation

Solutions give wave functions and energy levels of electrons

Wave function

Wave that corresponds to electron

Called orbitals for electrons in atoms

Amplitude of wave function squared

Can be related to probability of finding electron at that given point

Nodes

Regions where electrons will not be found

31

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Orbitals Characterized by Three Quantum Numbers:

Quantum Numbers:

Shorthand

Describes characteristics of electron’s position

Predicts its behavior

n = principal quantum number

All orbitals with same n are in same shell

ℓ = secondary quantum number

Divides shells into smaller groups called subshells

mℓ = magnetic quantum number

Divides subshells into individual orbitals

32

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

n = Principal Quantum Number

Allowed values: positive integers from 1 to

n = 1, 2, 3, 4, 5, …

Determines:

Size of orbital

Total energy of orbital

RHhc = 2.18 × 10–18 J/atom

For given atom,

Lower n = Lower (more negative) E

= More stable

E = -Z 2R

Hhc

n2

33

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

ℓ = Orbital Angular Momentum Quantum Number or Secondary

Quantum Number

Allowed values: 0, 1, 2, 3, 4, 5…(n – 1)

Letters: s, p, d, f, g, h

Orbital designation

number nℓ letter

Possible values of ℓ depend on n

n different values of ℓ for given n

Determines

Shape of orbital

34

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

mℓ = Magnetic Quantum Number

Allowed values: from –ℓ to 0 to +ℓ

Ex. when ℓ=2 then mℓ can be

–2, –1, 0, +1, +2

Possible values of mℓ depend on ℓ

There are 2ℓ+1 different values of mℓ for given ℓ

Determines orientation of orbital in space

To designate specific orbital, you need three quantum numbers

n, ℓ, mℓ

35

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Table 8.1 Summary of Relationships

Among the Quantum Numbers n, ℓ, and mℓ

36

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Orbitals of Many Electrons

37

Orbital Designation

Based on first two quantum numbers

Number for nand letter for ℓ

How many electrons can go in each orbital?

Two electrons

Need another quantum number

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Spin Quantum Number, ms

Arises out of behavior of electron in magnetic field

Electron acts like a top

Spinning charge is like a magnet

Electron behave like tiny magnets

Leads to two possible directions of electron spin

Up and down

North and south

38

Possible Values:

+½ ½

S

N

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Number of Orbitals and Electrons in the Orbitals

39

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Energy Level Diagram for Multi Electron Atom/Ion

4s

3s

2s

1s

Energ

y

4p

3p

2p

3d

4d5s

5p

4f6s

How to put electrons into a diagram?

Need some rules

40

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Pauli Exclusion Principle

No two electrons in same atom can have same set

of all four quantum numbers (n, ℓ, mℓ , ms)

Can only have two electrons per orbital

Two electrons in same orbital must have opposite spin

Electrons are said to be paired

41

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Hund’s Rule

If you have more than one orbital all at the same energy

Put one electron into each orbital with spins parallel (all up) until all are half filled

After orbitals are half full, pair up electrons

Why?

Repulsion of electrons in same region of space

Empirical observation based on magnetic properties

42

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Know from Magnetic Properties

Two electrons in same orbital have different spins

Spins paired—diamagnetic

Sample not attracted to magnetic field

Magnetic effects tend to cancel each other

Two electrons in different orbital with same spin

Spins unpaired—paramagnetic

Sample attracted to a magnetic field

Magnetic effects add

Measure extent of attraction

Gives number of unpaired spins

43

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Your Turn!

Which of the following is a valid set of four

quantum numbers (n, ℓ, mℓ , ms)?

A. 3, 2, 3, +½

B. 3, 2, 1, 0

C. 3, 0, 0, –½

D. 3, 3, 0, +½

E. 0, –1, 0, –½

44

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Your Turn!What is the maximum number of electrons allowed in a set of 4p orbitals?

A. 14

B. 6

C. 0

D. 2

E. 10

45

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Ground State Electron Arrangements

Electron Configurations Distribution of electrons among orbitals of atom

1. List subshells that contain electrons2. Indicate their electron population with superscripte.g. N is 1s 2 2s 2 2p 3

Orbital Diagrams Way to represent electrons in orbitals

1. Represent each orbital with circle (or line)2. Use arrows to indicate spin of each electron e.g. N is

46

1s 2s 2p

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Aufbau Principle

47

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f 5g

6s 6p 6d

7s 7p

8s

1

2 3

4 5

67

8

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Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E

Aufbau Principle and Periodic Table

Divided into regions of 2, 6, 10, and 14 columns

This equals maximum number of electrons in s, p, d, and f sublevels

48

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Each row (period) represents different energy level

Each region of chart represents different type of sublevel

49

Sublevels and the Periodic Table

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Orbital Diagram and Electron Configurations: e.g. N, Z = 7

4s

3s

2s

1s

Energ

y

4p

3p

2p

3d

Each arrow represents an electron

1s 22s 22p 3

50

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4s

3s

2s

1s

Energ

y

4p

3p

2p

3d

Orbital Diagram and Electron Configurations: e.g. V, Z = 23

Each arrow represents an electron

1s 22s 22p 63s 23p 64s 23d 3

51

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Learning Check

4s

3s

2s

1s

Energ

y

4p

3p

2p

3d

4d5s

5p6s

Give electron configurations and orbital diagrams for Na and As

Na Z = 11

As Z = 33

52

1s 22s 22p 63s 1

1s 22s 22p 63s 23p 64s 23d 104p 3

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Your Turn!What is the correct ground state electron configuration for Si?

A. 1s 22s 22p 63s 23p 6

B. 1s 22s 22p 63s 23p 4

C. 1s 22s 22p 62d 4

D. 1s 22s 22p 63s 23p 2

E. 1s 22s 22p 63s 13p 3

53

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Where Are The Electrons?

n= 1 1

H2

Hen= 2 3

Li4

Be5

B6

C7

N8

O9

F10

Nen= 3 11

Na12

Mg13

Al14

Si15

P16

S17

Cl18

Arn= 4 19

K20

Ca21

Sc22

Ti23

V24

Cr25

Mn26

Fe27

Co28

Ni29

Cu30

Zn31

Ga32

Ge33

As34

Se35

Br36

Krn= 5 37

Rb38

Sr39

Y40

Zr41

Nb42

Mo43

Tc44

Ru45

Rh46

Pd47

Ag48

Cd49

In50

Sn51

Sb52

Te53

I54

Xen= 6 55

Cs56

Ba57

La72

Hf73

Ta74

W75

Re76

Os77

Ir78

Pt79

Au80

Hg81

Tl82

Pb83

Bi84

Po85

At86

Rnn= 7 87

Fr88

Ra89

Ac104

Rf105

Db106

Sg107

Bh108

Hs109

Mt110

Ds111

Rg

58

Ce59

Pr60

Nd61

Pm62

Sm63

Eu64

Gd65

Tb66

Dy67

Ho68

Er69

Tm70

Yb71

Lu90

Th91

Pa92

U93

Np94

Pu95

Am96

Cm97

Bk98

Cf99

Es100

Fm101

Md102

No103

Lr54

Each box represents room for an electron.

Read from left to right

“ns” orbital being filled

“np” orbital being filled

“(n – 1)d” orbital being filled

“( n – 2)f” orbital being filled

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Read Periodic Table to Determine Electron Configuration – He

Read from left to right

First electron goes into period 1

First type of sublevel to fill = “1s”

He has 2 two electrons

Electron configuration for He is: 1s 2

55

n= 1 1

H2

Hen= 2 3

Li4

Ben= 3 11

Na12

Mgn= 4 19

K20

Ca21

Sc22

Ti23

V24

Cr25

Mn26

Fe27

Co28

Nin= 5 37

Rb38

Sr39

Y40

Zr41

Nb42

Mo43

Tc44

Ru45

Rh46

Pdn= 6 55

Cs56

Ba57

La72

Hf73

Ta74

W75

Re76

Os77

Ir78

Ptn= 7 87

Fr88

Ra89

Ac104

Rf105

Db106

Sg107

Bh108

Hs109

Mt110

Ds

“ns” orbital being filled

“np” orbital being filled

“(n – 1)d” orbital being filled

“( n – 2)f” orbital being filled

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Electron Configuration of Boron (B)

56

n= 1 1

H2

Hen= 2 3

Li4

Be5

B6

C7

N8

O9

F10

Nen= 3 11

Na12

Mg13

Al14

Si15

P16

S17

Cl18

Arn= 4 19

K20

Ca21

Sc22

Ti23

V24

Cr25

Mn26

Fe27

Co28

Ni29

Cu30

Zn31

Ga32

Ge33

As34

Se35

Br36

Krn= 5 37

Rb38

Sr39

Y40

Zr41

Nb42

Mo43

Tc44

Ru45

Rh46

Pd47

Ag48

Cd49

In50

Sn51

Sb52

Te53

I54

Xen= 6 55

Cs56

Ba57

La72

Hf73

Ta74

W75

Re76

Os77

Ir78

Pt79

Au80

Hg81

Tl82

Pb83

Bi84

Po85

At86

Rnn= 7 87

Fr88

Ra89

Ac104

Rf105

Db106

Sg107

Bh108

Hs109

Mt110

Ds111

Rg

B has 5 electrons Fill first shell… Fill two subshells in second shell, in order of increasing

energy Electron Configuration B = 1s 22s 22p 1

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Learning CheckWrite the correct ground state electron configuration for each of the following elements. List in order of increasing n and within each shell, increasing ℓ.

1. K Z = 19

= 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1

2. Ni Z = 28

= 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 8

= 1s 2 2s 2 2p 6 3s 2 3p 6 3d 8 4s 2

3. Pb Z = 82

= 1s 22s 22p 63s 23p 64s 23d 104p 65s 24d 10 5p 66s 24f 145d 106p 2

= 1s 22s 22p 63s 23p 63d 104s 24p 64d 104f 145s 25p 65d 106s 26p 2

57

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Abbreviated Electron Configurations - Noble Gas Notation

[noble gas of previous row] and electrons filled in next row

Represents core + outer shell electrons

Use to emphasize that only outer shell electrons participate in chemical reactions

e.g. Ba = [Xe] 6s 2

Ru = [Kr] 4d 6 5s 2

S = [Ne] 3s 2 3p 4

58

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Noble Gas Core Notation for Mn

n= 1 1

H2

Hen= 2 3

Li4

Be5

B6

C7

N8

O9

F10

Nen= 3 11

Na12

Mg13

Al14

Si15

P16

S17

Cl18

Ar

n= 4 19

K20

Ca21

Sc22

Ti23

V24

Cr25

Mn26

Fe27

Co28

Ni29

Cu30

Zn31

Ga32

Ge33

As34

Se35

Br36

Krn= 5 37

Rb38

Sr39

Y40

Zr41

Nb42

Mo43

Tc44

Ru45

Rh46

Pd47

Ag48

Cd49

In50

Sn51

Sb52

Te53

I54

Xen= 6 55

Cs56

Ba57

La72

Hf73

Ta74

W75

Re76

Os77

Ir78

Pt79

Au80

Hg81

Tl82

Pb83

Bi84

Po85

At86

Rnn= 7 87

Fr88

Ra89

Ac104

Rf105

Db106

Sg107

Bh108

Hs109

Mt110

Ds111

Rg

58

Ce59

Pr60

Nd61

Pm62

Sm63

Eu64

Gd65

Tb66

Dy67

Ho68

Er69

Tm70

Yb71

Lu90

Th91

Pa92

U93

Np94

Pu95

Am96

Cm97

Bk98

Cf99

Es100

Fm101

Md102

No103

Lr

59

“ns” orbital being filled

“np” orbital being filled

“(n – 1)d” orbital being filled

“( n – 2)f” orbital being filled

Find last noble gas that is filled before Mn

Next fill sublevels that follow [Ar] 4s 3d2 5

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Your Turn!

The ground state electron configuration for Ca is:

A. [Ar] 3s 1

B. 1s 22s 22p 63s 23p 54s 2

C. [Ar] 4s 2

D. [Kr] 4s 1

E. [Kr] 4s 2

60

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Look at Group 2A

Z Electron Configuration Abbrev

Be 4 1s 22s 2 [He] 2s 2

Mg 12 1s 22s 22p63s 2 [Ne] 3s 2

Ca 20 1s 22s 22p63s 23p64s 2 [Ar] 4s 2

Sr 38 1s 22s 22p63s 23p63d104s 24p65s 2 [Kr] 5s 2

Ba 56 1s 22s 22p63s 23p63d104s 24p64d105s 25p66s 2 [Xe] 6s 2

Ra 88 1s 22s 22p63s 23p63d104s 24p64d104f145s 25p6

5d106s 26p67s 2

[Rn] 7s 2

61

All have ns 2 outer shell electrons

Only difference is value of n

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Your Turn!An element with the electron configuration

[Xe]6s 24f 145d 7 would belong to which class on the periodic table?

A. Transition elements

B. Alkaline earth elements

C. Halogens

D. Lanthanide elements

E. Alkali metals

62

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Shorthand Orbital Diagrams

63

S [Ne]

3s 3p

Write out lines for orbital beyond Noble gas

Higher energy orbital to right

Fill from left to right

Abbreviated Orbital Diagrams

Ru [Kr]

4d 5s

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Your Turn!

Which of the following choices is the correct electron configuration for a cobalt atom?

4s 3d

A. [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑

B. [Ar] ↑ ↑↓ ↑↓ ↑↓ ↑↓

C. [Ar] ↑ ↑↓ ↑↓ ↑↓ ↑ ↑

D. [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑

E. [Ar] ↑↓ ↑↓ ↑↓ ↑ ↑ ↑

64

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Valence Shell Electron Configurations

An even more abbreviated notation for electron configurations

Use with representative elements (s and p block elements) – longer columns

Electrons in s and p subshells - important for bonding

Valence shell = outer shell

= occupied shell with highest n

Example: Sn = 5s 25p 2

65

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Electronic Configurations

A few exceptions to rules

Element Expected Experimental

Cr

Cu

Ag

Au

[Ar] 3d 44s2

[Ar] 3d 94s2

[Kr] 4d 95s2

[Xe] 5d 96s2

66

[Ar] 3d 54s 1

[Ar] 3d 104s 1

[Kr] 4d 105s 1

[Xe] 5d 106s 1

Exactly filled and exactly half-filled subshells have extra stability

Promote one electron into ns orbital to gain this extra stability

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Heisenberg’s Uncertainty Principle

Can’t know both exact position and exact speed of subatomic particle simultaneously

Such measurements always have certain minimum uncertainty associated with them

DxDmv ³h

4p

67

x = particle position

mv = particle momentum = mass × velocity of particle

h = Planck’s constant = 6.626 × 10–34 J s

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Consequence of Heisenberg’s Uncertainty Principle

Can’t talk about absolute position

Can only talk about electron probabilities

Where is e – likely to be?

ψ = wavefunction

Amplitude of electron wave

ψ2 = probability of finding electron at given location

Probability of finding an electron in given region of space equals the square of the amplitude of wave at that point

68

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1s Orbital Representations

a. Dot-density diagram

b. Probability of finding electron around given point, ψ2, with respect to distance from nucleus

c. Radial probability distribution = probability of finding electron at an “r” distance from nucleus

rmax = Bohr radius69

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Electron Density Distribution

Determined by

Electron density

No sharp boundary

Gradually fades away

“Shape”

Imaginary surface enclosing 90% of electron density of orbital

Probability of finding electrons is same everywhere on surface

Shape

Size n

Orientation m

70

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In any given direction probability of finding electron same

All s orbitals are spherically shaped

Size increases as nincreases

71

Effect of n on s Orbital

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Spherical Nodes

At higher n, now have spherical nodes

Spherical regions of zero probability, inside orbital

Node for electron wave

Imaginary surface where electron density = 0

2s, one spherical node, size larger

3s, two spherical nodes, size larger yet

In general:

Number of spherical nodes = n – 1

72

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Possess one nodal plane through nucleus

Electron density only on two sides of nucleus

Two lobes of electron density

All p orbitals have same overall shape

Size increases as n increases

For 3p have one spherical node73

p Orbitals

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Representations of p Orbitals

Constant probability surface for

2p orbital

Simplified p orbital emphasizing

directional nature of orbital

All 2p orbitals in p sub shell

One points along each axis

74

2px 2py 2pz

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There Are Five Different d Orbitals

Four with four lobes of electron density

One with two lobes and ring of electron density

Result of two nodal planes though nucleus

Number of nodal planes through nucleus =

75

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Your Turn!Which sketch represents a pz orbital?

76

y

z

x

A. B.

D. E.

C.

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Periodic Properties: Consequences of Electron Configuration

Chemical and physical properties of elements

Vary systematically with position in periodic table

i.e. with element's electron configuration

To explain, must first consider amount of positive charge felt by outer electrons (valence electrons)

Core electrons spend most of their time closer to nucleus than valence (outer shell) electrons

Shield or cancel out (screen out, neutralize) some of positive charge of nucleus

77

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Learning check: Li 1s22s1

Three protons in nucleus

Two core electrons in close (1s)

Net positive charge felt by outer electron: One proton

Effective Nuclear Charge (Zeff) Net positive charge outer electron feels Core electrons shield valence electrons from full

nuclear charge

78

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Shielding

Electrons in same subshell don't shield each other

Same average distance from nucleus

Trying to stay away from each other

Spend very little time one below another

Effective nuclear charge determined primarily by

Difference between charge on nucleus (Z ) and charge on core (number of inner electrons)

79

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Your Turn!What value is the closest estimate of Zeff for a valence electron of the calcium atom?

A. 1

B. 2

C. 6

D. 20

E. 40

80

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Atomic Size Experiment shows atoms/ions behave as if they

have definite size

C and H have ~ same distance between nuclei in large number of compounds

Atomic Radius (r)

Half of distance between two like atoms

H—H C—C etc.

Usually use units of picometer

1 pm = 1 × 10–12 m

Range 37 – 270 pm for atoms

81

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Trends in Atomic Radius (r)

Increases down Column (group) Zeff essentially constant

n increases, outer electrons farther away from nucleus and radius increase

Decreases across row (period) n constant

Zeff decreases, outer electrons feel larger Zeff and radius decreases

Transition Metals and Inner Transition Metals Size variations less pronounced as filling core

n same (outer electrons) across row

Decrease in Zeff and r more gradually

82

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Atomic and Ionic Radii (in pm)

83

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Ionic Radii

Increases down column (group)

Decreases across row (period)Anions larger than parent atom

Same Zeff, more electrons

Radius expands

Cations smaller than parent atom

Same Zeff, less electrons,

Radius contracts

84

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Your Turn!

Which of the following has the smallestradius?

A. Ar

B. K+

C. Cl–

D. Ca2+

E. S2–

85

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Ionization Energy Energy required to remove electron from gas

phase atom

Corresponds to taking electron from n to n =

First ionization energy M(g) M+(g) + e–

IE = E

Trends:

Ionization energy decreases down column (group) as n increases

Ionization energy increases across row (period) as Zeff increases

86

IE =R

HhcZ

eff

2

n2

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Comparing First Ionization Energies

87

Largest first ionization energies are in upper right

Smallest first ionization energies are in lower left

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Table 8.2: Successive Ionization Energies in kJ/mol for H through Mg

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Electron Affinity (EA)

Potential energy change associated with addition of one electron to gas phase atom or ion in the ground state

X(g) + e– X –(g)

O and F very favorable to add electrons

First electron affinities usually negative (exothermic)

Larger negative value means more favorable to add electron

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Table 8.3 Electron Affinities of Representative Elements

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Trends in Electron Affinity (EA)

Electron affinity becomes less exothermic down column (group) as n increases Electron harder to add as orbital farther from nucleus

and feels less positive charge

Electron affinity becomes more exothermic across row (period) as Zeff increases

Easier to attract electrons as positive charge increases

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Successive Electron Affinities

Addition of first electron – often exothermic Addition of more than one electron requires

energy Consider addition of electrons to oxygen:

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Change: EA(kJ/mol)

O(g) + e– O–(g) –141

O–(g) + e– O2–(g) +844

Net: O(g) + 2e– O2–(g) +703

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Your Turn!

Which of the following has the largest electron affinity?

A. O

B. F

C. As

D. Cs

E. Ba

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