chapter 8: the quantum mechanical atom chemistry: the molecular nature of matter, 6e...

Chapter 8: The Quantum

Mechanical Atom

Chemistry: The Molecular Nature of Matter, 6E

Jespersen/Brady/Hyslop

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

Electromagnetic Energy Electromagnetic Radiation

Light energy or wave

Travels through space at speed of light in vacuum

c = speed of light = 2.9979 × 108 m/s

Successive series of these waves or oscillations

Waves or Oscillations Systematic fluctuations in intensities of

electrical and magnetic forces

Varies regularly with time

Exhibit wide range of energy2


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


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 though axis Distance between nodes is constant

4

nodes


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

= 4.61 × 1014 s–1 = 4.6 × 1014 Hz


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


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


Electromagnetic SpectrumVisible 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


Important Experiments in Atomic TheoryLate 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


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

10


Learning CheckWhat is the frequency, in sec–1, of radiation which has an energy of 3.371 × 10–19 joules per photon?

11

= 5.087 × 1014 s–1


Your Turn!A microwave oven uses radiation with a frequency of 2450 MHz (megahertz, 106 s–1) to warm up food. What is the energy of such photons in joules?

A. 1.62 × 10–30 J

B. 3.70 × 1042 J

C. 3.70 × 1036 J

D. 1.62 × 1044 J

E. 1.62 × 10–24 J

12


Photoelectric Effect Shine light on metal surface Below certain frequency ()

Nothing happens Even with very intense light (high amplitude)

Above certain frequency () Number of electrons ejected increases as

intensity increases Kinetic energy (KE) of ejected electrons

increases as frequency increases

KE = h – BE h = energy of light shining on surface BE = binding energy of electron

13


Means that Energy is Quantized Can occur only in discrete units of size h

1 photon = 1 quantum of energy Energy gained or lost in whole number multiples

of hE = nh

If n = NA, then one mole of photons gained or lost

E = 6.02 × 1023 hIf light is required to start reaction

Must have light above certain frequency to start reaction

Below minimum threshold energy, intensity is NOT important

14


Learning CheckHow much energy is contained in one mole of photons, each with frequency 2.00 × 1013?

E = 6.02 × 1023 h

15

E = (6.02×1023 mol–1)(6.626×10–34 J∙s)(2.00×1013 s–1)

E = 7.98 × 103 J/mol


Your Turn!If a mole of photons has an energy of 1.60 × 10–3 J/mol, what is the frequency of each photon? Assume all photons have the same frequency.

A. 8.03 × 1028 Hz

B. 2.12 × 10–14 Hz

C. 3.20 × 1019 Hz

D. 5.85 × 10–62 Hz

E. 4.01 × 106 Hz

16


For Example: 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

17


Electronic Structure of Atom Clues come from:

1. Study of light absorption Electron absorbs energy

Moves to higher energy “excited state”

2. Study of light emission Electron loses photon of light

Drops back down to lower energy “ground state”

18

ground state

excited state

+h

h

excited state

ground state


Continuous Spectrum Continuous unbroken spectrum of all

colors i.e., visible light through a prism Sunlight Incandescent light bulb Very hot metal rod

19


Discontinuous or Line Spectrum Consider light given off when spark passes

through gas under vacuum

Spark (electrical discharge) excites gas molecules or atoms

Spectrum that has only a few discrete lines Also called atomic spectrum or emission

spectrum Each element has unique emission spectrum

20

+ gas


Line Spectrum

21


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)

22


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 emittedn1 and n2 = whole numbers (integers) from 1 to

where n2 > n1

If n1 = 1, then n2 = 2, 3, 4, …

23

22

21

111

nnRH


Learning Check: Using Rydberg Equation

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

24

= 410.3 nm Violet line in spectrum

= 24,373 cm–1


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

25

n2 = 3


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

26

221

5

1

3

1cm 678,109

1

1cm 77991


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

27


What Does Quantized Mean?

Energy is quantized if only certain discrete values are allowed

Presence of discontinuities makes atomic emission quantized

28

Potential Energy of Rabbit


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

29


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 orbit

30


Energy Level Diagram for H Atom Absorption of

photon Electron raised to

higher energy level

Emission of photon Electron falls to

lower energy level

31

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


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 in the form of light: E = h 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 = lower32


Bohr’s Model Fails Theory could not explain spectra of multi-electron atoms Theory doesn’t explain collapsing atom paradox If electron doesn’t move,

atom collapses

Positive nucleus should easily capture electron

Vibrating charge should radiate and lose energy

33


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

34


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

35


Diffraction and Electrons Light

Exhibits interference Has particle-like nature

Electrons Known to be particles Also demonstrate interference

36


Standing vs. Traveling WavesTraveling 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

37


Standing Wave on a Wire Integer number (n) of peaks and troughs

is required Wavelength is quantized: L is the length of the string

38


How Do We Describe an Electron? Has both wave-like and particle-like properties

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

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

39


Electron on Wire—Theories Standing wave Half-wavelength must occur integer number

of times along wire’s length

de Broglie’s equation relates the mass and speed of the particle to its wavelength

m = mass of particle v = velocity of particle

40


Electron on Wire—Theories Starting with the equation of the standing wave

and the de Broglie equation

Combining with E = ½mv 2, substituting for v and then λ, we get

Combining gives:

41


de Broglie Explains 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

42


Learning Check: Calculate Wavelength for an Electron

What is the de Broglie wavelength associated with an electron of mass 9.11 × 10

–31 kg traveling at a velocity of 1.0 × 107 m/s?

43

J1/sm kg 1

kg) 109.11m/s) 10(1.0s J106.626 22

317

34

(

= 7.27 × 10–11 m


Your Turn!Calculate the de Broglie wavelength of a baseball with a mass of 0.10 kg and traveling at a velocity of 35 m/s.

A. 1.9 × 10–35 m

B. 6.6 × 10–33 m

C. 1.9 × 10–34 m

D. 2.3 × 10–33 m

E. 2.3 × 10–31 m

44


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

45


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 46


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 stable47


ℓ = Orbital Angular Momentum 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

48


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, mℓ49


Table 8.1 Summary of Relationships Among the Quantum Numbers n, ℓ,

and mℓ

50


Orbitals of Many Electrons

51

Orbital Designation

Based on first two quantum numbers

Number for n and letter for ℓ

How many electrons can go in each orbital? Two electrons Need another

quantum number


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

52

Possible Values:

+½ ½


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

53


Number of Orbitals and Electrons in the Orbitals

54


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

55


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

56


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

57


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

58

1s 2s 2p


Energy Level Diagram for Multi Electron Atom/Ion

4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

4d5s

5p

4f6s

How to put electrons into a diagram?

Need some rules

59


Aufbau Principle Building-up principle

Pauli Exclusion Principle Two electrons per orbital Fill following the order suggested by the

periodic table Spins must be paired

60


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

61


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

4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

Each arrow represents electron

1s 2 2s

2 2p 3

62


4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

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

Each arrow represents an electron1s

2 2s 2 2p

2 3s 2 3p

2 4s 2 3d 3

63


Learning Check

4s

3s

2s

1s

Ene

rgy

4p

3p

2p

3d

4d5s

5p6s

Give electron configurations and orbital diagrams for Na and As

Na Z = 11

As Z = 33

64

1s 2 2s

2 2p 2 3s

1

1s 22s

22p 63s

23p 64s

23d 104p

3


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

65


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

66


Each row (period) represents different energy level

Each region of chart represents different type of sublevel

67

Sublevels and the Periodic Table


Now Ready to Put Electrons into Atoms

Electron configurations must be consistent with:

Pauli Exclusion principle Two electrons per orbital, spins opposite

Aufbau principle Start at lowest energy orbital Fill, then move up

Hund’s rule One electron in each orbital of same energy,

spins parallel Only pair up if have to

68


Where Are The Electrons?

n= 1 1H

2He

n= 2 3Li

4Be

5B

6C

7N

8O

9F

10Ne

n= 3 11Na

12Mg

13Al

14Si

15P

16S

17Cl

18Ar

n= 4 19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

29Cu

30Zn

31Ga

32Ge

33As

34Se

35Br

36Kr

n= 5 37Rb

38Sr

39Y

40Zr

41Nb

42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

49In

50Sn

51Sb

52Te

53I

54Xe

n= 6 55Cs

56Ba

57La

72Hf

73Ta

74W

75Re

76Os

77Ir

78Pt

79Au

80Hg

81Tl

82Pb

83Bi

84Po

85At

86Rn

n= 7 87Fr

88Ra

89Ac

104Rf

105Db

106Sg

107Bh

108Hs

109Mt

110Ds

111Rg

58Ce

59Pr

60Nd

61Pm

62Sm

63Eu

64Gd

65Tb

66Dy

67Ho

68Er

69Tm

70Yb

71Lu

90Th

91Pa

92U

93Np

94Pu

95Am

96Cm

97Bk

98Cf

99Es

100Fm

101Md

102No

103Lr

69

Each box represents room for 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


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 70

n= 1 1H

2He

n= 2 3Li

4Be

n= 3 11Na

12Mg

n= 4 19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

n= 5 37Rb

38Sr

39Y

40Zr

41Nb

42Mo

43Tc

44Ru

45Rh

46Pd

n= 6 55Cs

56Ba

57La

72Hf

73Ta

74W

75Re

76Os

77Ir

78Pt

n= 7 87Fr

88Ra

89Ac

104Rf

105Db

106Sg

107Bh

108Hs

109Mt

110Ds



Electron Configuration of Boron (B)

71

n= 1 1H

2He

n= 2 3Li

4Be

5B

6C

7N

8O

9F

10Ne

n= 3 11Na

12Mg

13Al

14Si

15P

16S

17Cl

18Ar

n= 4 19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

29Cu

30Zn

31Ga

32Ge

33As

34Se

35Br

36Kr

n= 5 37Rb

38Sr

39Y

40Zr

41Nb

42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

49In

50Sn

51Sb

52Te

53I

54Xe

n= 6 55Cs

56Ba

57La

72Hf

73Ta

74W

75Re

76Os

77Ir

78Pt

79Au

80Hg

81Tl

82Pb

83Bi

84Po

85At

86Rn

n= 7 87Fr

88Ra

89Ac

104Rf

105Db

106Sg

107Bh

108Hs

109Mt

110Ds

111Rg

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


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

2 2s 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

72


Chemical Reactivity Periodic table arranged by chemical

reactivity Depends on outer shell electrons (highest n)

Each row is different n

Core electrons Inner electrons are those with n < nmax

Buried deep in atom

73


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 react

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

Ru = [Kr] 4d 6

5s 2

S = [Ne] 3s 2

3p 4

74


Noble Gas Core Notation for Mn

n= 1 1H

2He

n= 2 3Li

4Be

5B

6C

7N

8O

9F

10Ne

n= 3 11Na

12Mg

13Al

14Si

15P

16S

17Cl

18Ar

n= 4 19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

29Cu

30Zn

31Ga

32Ge

33As

34Se

35Br

36Kr

n= 5 37Rb

38Sr

39Y

40Zr

41Nb

42Mo

43Tc

44Ru

45Rh

46Pd

47Ag

48Cd

49In

50Sn

51Sb

52Te

53I

54Xe

n= 6 55Cs

56Ba

57La

72Hf

73Ta

74W

75Re

76Os

77Ir

78Pt

79Au

80Hg

81Tl

82Pb

83Bi

84Po

85At

86Rn

n= 7 87Fr

88Ra

89Ac

104Rf

105Db

106Sg

107Bh

108Hs

109Mt

110Ds

111Rg

58Ce

59Pr

60Nd

61Pm

62Sm

63Eu

64Gd

65Tb

66Dy

67Ho

68Er

69Tm

70Yb

71Lu

90Th

91Pa

92U

93Np

94Pu

95Am

96Cm

97Bk

98Cf

99Es

100Fm

101Md

102No

103Lr

75


Find last noble gas that is filled before Mn Next fill sublevels that follow [Ar] 4s 3d2 5


Your Turn!The ground state electron configuration for Ca is:

A. [Ar] 3s 1

B. 1s 2

2s 2

2p 6

3s 2

3p 5

4s 2

C. [Ar] 4s 2

D. [Kr] 4s 1

E. [Kr] 4s 2

76


Look at Group 2A

Z Electron Configuration Abbrev

Be 4 1s 22s

2 [He] 2s 2

Mg 12 1s 22s

22p 63s

2 [Ne] 3s 2

Ca 20 1s 22s

22p 63s

23p 64s

2 [Ar] 4s 2

Sr 38 1s 22s

22p 63s

23p 63d

104s 24p

65s 2 [Kr] 5s 2

Ba 56 1s 22s

22p 63s

23p 63d

104s 24p

64d 105s

25p 66s

2 [Xe] 6s 2

Ra 88 1s 22s

22p 63s

23p 63d

104s 24p

64d 104f

145s 25p

6

5d 106s

26p 67s

2

[Rn] 7s 2

77

All have ns 2 outer shell electrons

Only difference is value of n


Your Turn!An element with the electron configuration[Xe]4f

145d 76s

2 would belong to which class on the periodic table?

A. Transition elements

B. Alkaline earth elements

C. Halogens

D. Lanthanide elements

E. Alkali metals

78


Shorthand Orbital Diagrams

79

S [Ne]

3s 3p

Write out lines for orbital beyond Noble gas

Higher energy orbital to right Fill from left to rightAbbreviated Orbital Diagrams

Ru [Kr]

4d 5s


Valence Shell Electron Configurations One last type of electron configuration

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

Here only electrons in outer shell important for bonding

Only electrons in s and p subshells Valence shell = outer shell

= occupied shell with highest n

Result – use even more abbreviated notation for electron configurations

Sn = 5s 25p

2 80


Electronic Configurations A few exceptions to rules

Element Expected Experimental

Cr

Cu

Ag

Au

[Ar] 3d 44s

2

[Ar] 3d 94s

2

[Kr] 4d 95s

2

[Xe] 5d 96s

2

81

[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


Your Turn!The orbital diagram corresponding to the ground state electron configuration for nitrogen is:

A.

B.

C.

D.

E. 82

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p

1s 2s 2p


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] ↑↓ ↑↓ ↑↓ ↑ ↑ ↑

83


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

84

x = particle position

mv = particle momentum = mass × velocity of particle

h = Planck’s constant = 6.626 × 10–

34 J s


Heisenberg’s Uncertainty PrincipleMacroscopic scale

Errors in measurements much smaller than measured value

Subatomic scale Errors in measurements equal to or greater

than measured value If you know position exactly, know nothing

about velocity

If you know velocity exactly, know nothing about position

85


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

86


Electron CloudElectron dot picture = snapshots

Lots of dots shown by large amplitude of wave function High probability of finding electrons

Electron density How much of electrons charge packed into

given volume High Probability

High electron density or Large electron density

Low Probability Low electron density or Small electron density

87


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 between r and r + x from nucleus

rmax = Bohr radius88


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 nOrientatio

nm

89


In any given direction probability of finding electron same

All s orbitals are spherically shaped

Size increases as n increases

90

Effect of n on s Orbital


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

91


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 node

92

p Orbitals


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

93

x

y

z2px

x

y

z2py 2pz

x

y

z


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 =

94


Your Turn!Which sketch represents a pz orbital?

95

x

y

x

z

y

z

x

xy

z

y

z

x

A. B.

D. E.

C.


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

96


Learning check: Li 1s 22s

1 Three protons in

nucleus Two core electrons

in close (1s) Net positive charge

felt by outer electron Approximately oneproton

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

full nuclear charge 97


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)

98


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

99


Atomic Size Theory suggests sizes of atoms and ions

indistinct 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 100


Trends in Atomic Radius (r) Increases down Column (group)

Zeff essentially constant n increases, outer electrons farther away from

nucleus and radius increaseDecreases across row (period)

n constant Zeff decreases, outer electrons feel larger Zeff and

radius decreasesTransition 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

101


Atomic and Ionic Radii (in pm)

102


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

103


Your Turn!Which of the following has the smallest radius?

A. Ar

B. K+

C. Cl–

D. Ca2+

E. S2–

104


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 105


Comparing First Ionization Energies

106

Largest first ionization energies are in upper right

Smallest first ionization energies are in lower left


Successive Ionization Energies

Increases slowly as remove each successive electron

See big increase in ionization energy When break

into exactly filled or half filled subshell

107


108

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


Your Turn!Place the elements C, N, and O in order of increasing ionization energy.

A. C, N, O

B. O, N, C

C. C, O, N

D. N, O, C

E. N, C, O

109


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 Comparing first electron affinities usually

negative (exothermic) Larger negative value means more

favorable to add electron

110


Table 8.3 Electron Affinities of Representative Elements

111


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

112


Successive Electron Affinities Addition of first electron – often exothermic Addition of more than one electron requires

energy Consider addition of electrons to oxygen:

113

Change: EA(kJ/mol)

O(g) + e – O–(g) –141

O–(g) + e – O2–(g) +844

Net:

O(g) + 2e – O2–(g) +703


Your Turn!Which of the following has the largest electron affinity?

A. O

B. F

C. As

D. Cs

E. Ba

114

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