Transcript Chapter 8: Introduction to Quantum Mechanics

Chapter 8:
The Quantum
Mechanical Atom
Chemistry: The Molecular Nature
of Matter, 6E
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Electromagnetic Energy
Electromagnetic Radiation
 Light energy or wave
 Travels through space at speed of light in vacuum
 c = speed of light = 2.9979 x 108 m/s
 Successive series of these waves or oscillations
Waves or Oscillations
 Systematic fluctuations in intensities of electrical
and magnetic forces
 Varies rhythmically with time
 Exhibit wide range of energy
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Properties of Waves
Wavelength ()
 Distance between two successive peaks or troughs
 Unit = meter
Frequency ()
 number of waves per second that pass a given point in
space
 Unit = Hertz (Hz) = cycles/sec = 1/sec = s1
Related by
=c
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Properties of Waves
Amplitude
 Intensity of wave
 Maximum and minimum height
 Varies with time as travels through space
Nodes
 Points of zero amplitude
 Place where wave goes though axis
 Distance between nodes is always same
nodes
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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?

c


8
2.9979  10 m/s
650  10
9
m
 = 4.61 × 1014 s–1 = 4.61 × 1014 Hz
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Your Turn!
WCBS broadcasts at a frequency of 880 kHz.
What is the wavelength of their signal?
A. 341 m
c 3.00  10 8 m/s
 
B. 293 m

880  10 3 / s
C. 293 mm
D. 341 km
E. 293 mm
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Electromagnetic Spectrum
 Comprised of all frequencies of light
 Divided into regions according to wavelengths
of radiation
high energy, short waves
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low energy, long waves
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Electromagnetic Spectrum
Visible Light




Band of ’s that human eyes can see
400 to 700 nm
Make up spectrum of colors
700 nm ROYGBIV 400 nm
White light
 Equal amount of all these colors
 Can separate by passing through prism
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Important Experiments in Atomic Theory
Late 1800’s:
 Classical physics incapable of describing atoms and
molecules
 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
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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
 Pulses with frequency = 
 Energy of photon of electromagnetic radiation is
proportional to its frequency
 Energy of photon
E = h
 h = Planck’s constant
= 6.626 x 1034 J·s
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Learning Check
What is the frequency, in sec–1, of radiation which
has an energy of 3.371 x 10–19 joules per photon?
E

h

3.371  10 19 J
6.626  10  34 J  s
 = 5.087×1014 s–1
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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 x 10–30 J
B. 3.70 x 1042 J
C. 3.70 x 1036 J
D. 1.62 x 1044 J
E  h
E. 1.62 x 10–24 J
6 1
1

10
s
 34
E  6.626  10
J  s  2450 MHz 
MHz
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Photoelectric Effect
 Shine light on metal surface
 Below certain frequency ()
 Nothing happens
 Even with very intense light
 Above certain frequency ()
 number of electrons ejected increases as intensity
increases
 Kinetic energy (KE) of ejected electrons increases as
 increases
 KE = h – BE
 h = energy of light shone on surface
 BE = binding energy of electron
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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 h
E = nh
 If n = NA, then one mole of photons gained or lost
E = NAh
If light required to start reaction
 Must have light above certain frequency to start reaction
 Below minimum threshold E, brightness is NOT important
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Learning Check
How much energy is contained in one mole of
photons, each with frequency 2.00 × 1013?
E = NAh
E = (6.02×1023 mol–1)(6.626×10–34 J∙s)(2.00×1013 s–1)
E = 7.98 × 103 J/mol
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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. 1.33 × 105 Hz

E

NA h
1.60  10 3 J / mol
(6.02  10 23 mol 1 )(6.626  10  34 J  s)
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Ex. Photosynthesis
 If you irradiate plants with IR and MW radiation
 No photosynthesis
 Regardless of light intensity
 If you irradiate plants with Visible Light
 Photosynthesis occurs
 Brighter light now means more photosynthesis
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Electronic Structure of Atom
Most information comes from:
excited state
1. Study of light absorption
+h
 Electron absorbs energy
 Moves to higher energy “excited ground state
state”
2. Study of light emission
 e loses photon of light
 Drops back down to lower
energy “ground state”
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excited state
h
ground state
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Continuous Spectrum
 Continuous unbroken spectrum of all colors
 i.e., visible light through a prism
 Consider light given off when spark passes through
gas under vacuum
+
gas

 Spark (electrical discharge) excites gas molecules
(atoms)
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Line Spectrum
 Spectrum that has only a few discrete lines
 Also called atomic spectrum or emission
spectrum
 Each element has unique emission spectrum
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Atomic Spectra
 Atomic line spectra are rather complicated
 Line spectrum of hydrogen is simplest
 Single electron
 1st success in explaining quantized line spectra
 1st 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 (IR, Vis, and UV)
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Rydberg Equation
 1

1
1

 RH 

n2 n2 

2 
 1
RH = 109,678 cm1 = Rydberg
constant
 = wavelength of light emitted
n1 & n2 = whole numbers
(integers) from 1 to  where
n2 > n1
If n1 = 1, then n2 = 2, 3, 4, …
Corresponds to
allowed energy
levels for atom
 Can be used to calculate all spectral lines of
hydrogen
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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.
1
1 
1 
 1
1  1
–1
 RH 


109
,
678
cm




=24,373cm
2
2

 4 36 
6 
2

1
24,372.9cm
 = 410.3 nm
1
 4.1029  10
5
1m
1nm
cm 

100cm 1  10  9 m
Violet line in spectrum
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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  10 7 cm
  650.5nm 
 650.5  10  7 cm
1nm
1
 109,678cm1 ( 1  1 )
650.510 7 cm
22 n22
1
 (1  1 )
4 n2
7.13455
2
2
n2
1

 9.10
0.110
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1 1
1
 0.110
2
n2 4 7.13455
n2 = 3
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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
1
1 
1  1
 109,678 cm 


2
2

5 
3
C. 7.80 × 102 nm
D. 7.80 × 10–4 nm
E. 3.65 × 10–7 nm
1
 7799 cm 1

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
1
7799 cm1
1  107 nm

1cm
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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
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What Does Quantized Mean?
Potential Energy of Rabbit
 Energy is quantized if
only certain discrete
values are allowed
 Presence of
discontinuities makes
atomic emission
quantized
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Bohr Model of Atom
 1st 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
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Energy for Bohr Model of H
 Equation for energy of e in H atom
2
4
1
2

me
E 2
b
n
h2
 Ultimately b relates to RH by b = RHhc
 OR
b
RH hc
E  2 
2
n
n
 where b = RHhc = 2.1788 x 1018 J/atom
 Allowed values of n = 1, 2, 3, 4, …
 n = Quantum number
 Used to identify orbit
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Energy Level Diagram for H Atom
 Absorption of
photon
 Electron raised to
higher energy level
 Emission of photon
 Electron falls to
lower energy level
 E s are quantized
 Every time e drops from n = 3 to n = 1
 Same frequency photon is emitted
 Yields line spectra
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Bohr Model of H
 E is negative number
 Reference point E = 0 when n = 
 e not attached to nucleus
 Sign arises from Coulombic attraction between +
and – charges (oppositely charged bodies)
 Coulomb's Law
Attractive force
(charge on A)(charge on B)
E 
distance between them
 Stronger attractive force = more negative E
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Bohr Model of H
 n=1
1st Bohr orbit
 Most stable E state = ground state = Lowest E
state
 Electron remains in lowest E state unless disturbed
How to disturb the atom?
 Add
E = h
 e raised higher n orbit n = 2, 3, 4, … 
 Higher n orbits = excited states = less stable
 So e quickly drops to lower E orbit and emits
photon of E equal to E between levels
 E = Eh – El
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h = higher
l = lower
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Bohr’s Model Fails
 Theory could not explain spectra of multi- electron
atoms
 Theory doesn’t explain collapsing atom paradox
 If e– doesn’t move,
atom collapses
 Positive nucleus should
easily capture e–
 Vibrating charge should
radiate and lose energy
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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
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Light Exhibits Interference
Constructive interference
 Waves “in-phase” lead to greater amplitude
 Add
Destructive interference
 Waves “out-of-phase” lead to lower amplitude
 Cancel out
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Diffraction and Electrons
 Light
 Exhibits interference
 Has particle nature
 Electrons
 Known to be particles
 Also demonstrate interference
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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
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Bead on a Wire
 Any energy is possible, even zero
 Same chance of finding bead anywhere on wire
 Can know exact position and velocity of bead
simultaneously
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Wave on a Wire
 Integer number (n) of peaks and troughs is
required
 Wavelength is quantized:
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2L

n
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How Do We Describe an Electron?
 Has both wave and particle properties
 Confining electron makes its behavior more
wavelike
 Free electrons behave more like particles
 Energy of moving electron is E=½ mv2
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Electron on Wire—Theories
Standing wave
 Half-wavelength must occur
integer number of times
along wire’s length
2L

n
de Broglie’s equation links these
 m = mass of particle
 v = velocity of particle
h
  mv
Combining gives:
E
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n 2 h2
8 mL2
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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
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Learning Check: Calculate  for eWhat is the deBroglie wavelength associated with
an electron of mass 9.11 x 10–31 kg traveling at a
velocity of 1.0 x 107 m/s?
6.626  10
34
Js
2
1kg  m /s


1J
1.0  10 7 m/s  9.11  10  31 kg
2
 = 7.27 x 10–11 m
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Your Turn!
Calculate the deBroglie 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
6.626  10 34 J  s 1kg  m2 / s 2


35m / s  0.10kg
1J
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Wave Functions
Schrödinger’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
 Can be related to probability of finding electron at
that given point
Nodes
 Regions of wire where electrons will not be found
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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
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n = Principal Quantum Number
 Allowed values: positive integers from 1 to 
 n = 1, 2, 3, 4, 5, … 
2
Z R Hhc
E
2
n
 Determines:
 Size of orbital
 Total energy of orbital
 Number of nodes (points where * = 0)
 RHhc = 2.18 x 1018 J/atom
 For given atom,
 Lower n = Lower (more negative) E
= More stable
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ℓ = Orbital Angular Momentum QN
 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
 Specifies orbital angular momentum of e
 Determines
 Shape of orbital
 Angular variation of e path
 Kinetic energy of orbital
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mℓ = Magnetic Quantum Number
 Allowed values:
 mℓ = ℓ, ℓ+1, ℓ+2, …, 0 , …, ℓ2, ℓ1, ℓ
 Possible values of mℓ depend on ℓ
 There are 2ℓ +1 different values of mℓ for given ℓ
 z axis component of orbital angular momentum
 Determines orientation of orbital in space
 To designate specific orbital, you need three
quantum numbers
 (n, ℓ, mℓ)
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Table 8.1 Summary of Relationships
Among the Quantum Numbers n, ℓ, and m
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Energy
Energy Level Diagram for H Atom or
Other 1 e– Ion
4s
4p
4d
3s
3p
3d
2s
1s
 All orbital subshells with
same n value have same
Energy
Z 2R Hhc
E
n2
2p
E = Elo – Ehi
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 1

2

E  Z R H hc 

 n2 n2 
lo 
 hi
2
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Orbital Energies in Many Electron Atoms
1. Each orbital represented by circle or line
2. Now different subshells ( values) have different
E
3. All orbitals of same subshell = same E
4. As you go up in energy, spacing between
successive shells (n values) decreases as number
of subshells increases

Leads to overlapping of several
subshells 4s/3d 5s/4d 6s/4f/5d
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Orbitals of Many Electrons
Orbital
Designation
 Based on first 2
quantum numbers
 number for n and
letter for 
 How many e can
go in each orbital?
 Need another
quantum number
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Spin Quantum Number, ms
 Arises out of behavior of
e in magnetic field
 e acts like a top
 Spinning charge is like a
magnet
 e behave like tiny
magnets
 Leads to 2 possible
directions of e spin
 up and down
 north and south
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Possible Values:
+½

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½

54
Pauli Exclusion Principle
 No two e in same atom can have same set of all
four quantum numbers (n, , m, ms)
 Can only have 2 e per orbital
 2 e s in same orbital must have opposite spin
 e s are paired
 Odd number of es
 Not all spins paired
 Have unpaired es
 Even number of es
 Depends on number of orbitals
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Consequences of Pauli Exclusion
Principle
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Know from Magnetic Properties
 Two e s in same orbital with different spin
 Spins paired—diamagnetic
 Sample not attracted to magnetic field
 Magnetic effects tend to cancel each other
 Two e s in different orbital with same spin
 Spins unpaired—paramagnetic
 Sample pulled into magnetic field
 Magnetic effects add
 Measure extent of attraction
 Gives number of unpaired spins
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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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
58
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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
59
Ground State e Arrangements
Electron Configurations
 Distribution of es among orbitals of atom
1. List subshells that contain electrons
2. Indicate their electron population with superscript
Ex. N is 1s2 2s2 2p3
Orbital Diagrams
 Way to represent es in orbitals
1. Represent each orbital with circle (or line)
2. Use arrows to indicate spin of each electron
Ex. N is
1s
Jespersen/Brady/Hyslop
2s
2p
Chemistry: The Molecular Nature of Matter, 6E
60
Energy Level Diagram for Multi e
Atom/Ion
4f
6s
5p
4d
5s
4p
3d
4s
Energy
3p
3s
2p
2s
 How to put e– into a diagram?
 Need some rules
1s
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
61
Aufbau Principle
 Building-up principle
 Fill lowest energy subshell before going to next
highest energy subshell
 Fill lowest n first
 Within given n, fill lowest  first
 Within given , fill highest m first
 Within given m, fill highest ms (+ ½ , ) first
Pauli Exclusion Principle
 2 e per orbital
 Spins opposite
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
62
Hund’s Rule
 If you have more than 1 orbital all at the same
energy
 Put 1e into each orbital with spins parallel (all
up) until all are half filled
 Before pair up es in same orbital
Why?
 Repulsion of e in same region of space
 Empirical observation based on magnetic
properties
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
63
Orbital Diagram and e
Configurations: N
Z=7
4p
3d
4s
Energy
3p
3s
2p
Each arrow represents electron
2s
1s2 2s2 2p3
1s
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
64
Orbital Diagram and eConfigurations: V
Z = 23
4p
3d
4s
Energy
3p
3s
2p
2s
Each arrow represents an electron
1s2 2s2 2p6 3s2 3p6 4s2 3d3
1s
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
65
Learning Check
Give electron configurations and orbital diagrams
for Na and As
6s
5p
4d
5s
4p
3d
4s
Energy
3p
3s
2p
Na Z = 11
2s
1s22s22p63s1
As Z = 33
1s22s22p63s23p64s23d104p3
1s
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
66
Your Turn!
The ground state electron configuration for Ca
is:
A. [Ar] 3s1
B. 1s2 2s2 2p6 3s2 3p5 4s2
C. [Ar] 4s2
D. [Kr] 4s1
E. [Kr] 4s2
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
67
Periodic Table
 Divided into regions of 2, 6, 10, and 14 columns
 = maximum number of electrons in s, p, d, and f
sublevels
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
68
Sublevels and the Periodic Table
 Each row (period) represents different energy level
 Each region of chart represents different type of
sublevel
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
69
Now Ready to Put es into Atoms
Electron configurations must be consistent with:
Pauli Exclusion principle
 2 e per orbital, spins opposite
Aufbau principle
 Start at lowest energy orbital
 Fill, then move up
Hund’s rule
 1 e in each orbital of same, spins parallel
 Only pair up if have to
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
70
Where Are The Electrons?
 Each box represents room for electron.
 Read from left to right
n= 1 1
2
n= 2 3
4
n= 3 11
12
n= 4 19
20
21
22
23
n= 5 37
38
39
n= 6 55
56
57
n= 7 87
88
89 104 105 106 107 108 109 110 111
“ns” orbital being filled
“np” orbital being filled
“(n – 1)d” orbital being filled
“( n – 2)f” orbital being filled
H He
Li Be
Na Mg
Y
B
6
C
7
N
8
O
9
F
10
Ne
13
14
15
16
17
31
32
33
34
35
36
54
Al
Si
P
S
18
Cl Ar
V
24
25
26
27
28
29
30
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
40
41
42
43
44
45
46
47
48
49
50
51
52
53
72
73
74
75
76
77
78
79
80
81
82
83
84
85
68
69
70
71
K Ca Sc Ti
Rb Sr
5
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
Cs Ba La Hf Ta W Re Os Ir
I
Xe
86
Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg
58
59
60
61
62
63
64
65
66
67
90
91
92
93
94
95
96
97
98
99 100 101 102 103
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Th Pa
U Np Pu Am Cm Bk Cf Es Fm Md No Lr
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
71
Read Periodic Table to Determine
e- Configuration – He
 Read from left to
right
 1st e– goes into
period 1
 1st type of sublevel
to fill = “1s”
 He has 2 e–
 e– configuration for
He is: 1s2
Jespersen/Brady/Hyslop
n= 1 1
2
n= 2 3
4
n= 3 11
12
n= 4 19
20
21
22
23
n= 5 37
38
39
40
41
n= 6 55
56
57
72
73
n= 7 87
88
89 104 105 106 107 108 109 110
“ns” orbital being filled
“np” orbital being filled
“(n – 1)d” orbital being filled
“( n – 2)f” orbital being filled
H He
Li Be
Na Mg
K Ca Sc Ti
Rb Sr
Y
V
24
25
26
27
28
42
43
44
45
46
74
75
76
77
78
Cr Mn Fe Co Ni
Zr Nb Mo Tc Ru Rh Pd
Cs Ba La Hf Ta W Re Os Ir
Pt
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds
Chemistry: The Molecular Nature of Matter, 6E
72
Electron Configuration of Boron (B)
n= 1 1
2
n= 2 3
4
n= 3 11
12
n= 4 19
20
21
22
23
n= 5 37
38
39
40
41
n= 6 55
56
57
72
73
n= 7 87
88
89 104 105 106 107 108 109 110 111
H He
5
Li Be
Na Mg
B
K Ca Sc Ti
Rb Sr
Y
V
6
C
7
N
8
O
9
F
10
Ne
13
14
15
16
17
31
32
33
34
35
36
54
Al
Si
P
S
18
Cl Ar
24
25
26
27
28
29
30
42
43
44
45
46
47
48
49
50
51
52
53
74
75
76
77
78
79
80
81
82
83
84
85
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
Cs Ba La Hf Ta W Re Os Ir
I
Xe
86
Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg




B has 5 e–
Fill first shell…
Fill two subshells in 2nd shell, in order of increasing E
Electron Configuration B = 1s22s22p1
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
73
Noble Gas Core Notation for Mn
 Find last noble gas that is filled before Mn
 Next fill sublevels that follow
[Ar] 4s 2 3d 5
n= 1 1
2
n= 2 3
4
“ns” orbital being filled
“np” orbital being filled
“(n – 1)d” orbital being filled
“( n – 2)f” orbital being filled
H He
Li Be
n= 3 11 12
Na Mg
n= 4 19 20
21
22
23
n= 5 37 38
39
40
41
n= 6 55 56
57
72
73
n= 7 87 88
89 104 105 106 107 108 109 110 111
K Ca Sc Ti
Rb Sr
Y
V
5
B
6
C
7
N
8
O
9
10
17
18
F Ne
13
14
15
16
S
Cl Ar
Al Si
P
24
25
26
27
28
29
30
31
32
33
34
35
36
42
43
44
45
46
47
48
49
50
51
52
53
54
74
75
76
77
78
79
80
81
82
83
84
85
68
69
70
71
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te
I
Xe
86
Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
Fr Ra Ac Rf Db Sg Bh Hs Mt Ds Rg
58
59
60
61
62
63
64
65
66
67
90
91
92
93
94
95
96
97
98
99 100 101 102 103
Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
74
Writing Electron Configurations
 Fill pattern across row in Periodic Table is:
 ns [(n2)f ] [(n1)d ]np
 where n = row number in periodic table
 Must rearrange e configuration
 List in order of increasing n
 Within n level, list in order of increasing 
 Why?
 Once 3d and 4f are filled, they become part of core e
 Harder to remove e in core or with lower n
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
75
Learning Check
Write 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
= 1s2 2s2 2p6 3s2 3p6 4s1
2. Ni
Z = 28
= 1s2 2s2 2p6 3s2 3p6 4s2 3d8
= 1s2 2s2 2p6 3s2 3p6 3d8 4s2
3. Pb
Z = 82
= 1s22s22p63s23p64s23d104p65s24d10 5p66s24f145d106p2
= 1s22s22p63s23p63d104s24p64d104f145s25p65d106s26p2
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
76
Your Turn!
What is the correct ground state electron
configuration for Si?
A. 1s2 2s2 2p6 3s2 3p6, no unpaired spins
B. 1s2 2s2 2p6 3s2 3p4, no unpaired spins
C. 1s2 2s2 2p6 2d4, 4 unpaired spins
D. 1s2 2s2 2p6 3s2 3p2, 2 unpaired spins
E. 1s2 2s2 2p6 3s1 3p3, 4 unpaired spins
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
77
Your Turn!
An element with the electron configuration
[Xe]4f145d76s2 would belong to which class on the
periodic table?
A. transition elements
B. alkaline earth elements
C. halogens
D. rare earth elements
E. alkali metals
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
78
Chemical Reactivity
 Periodic Table arranged by chemical reactivity
 Depends on outer shell e’s (highest n)
 Arranged by n
 Each row is different n
 Core electrons
 Inner e’s = those with n < nmax
 Buried deep in atom
 Don’t normally play role in chemical bond
formation
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
79
Abbreviated Electron Configurations
= Noble Gas Notation
 [noble gas of previous row] + es filled in nth
row
 Represents core + outer shell es
 Use to emphasize that only outer shell electrons
react
Ex.
Ba = [Xe] 6s2
see above
Ru = [Kr] 4d6 5s2
S = [Ne] 3s2 3p4
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
80
Look at Group IIA
Be
Mg
Ca
Sr
Z
4
12
20
38
Electron Configuration
1s2 2s2
1s2 2s2 2p6 3s2
1s2 2s2 2p6 3s2 3p6 4s2
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2
Abbrev
[He] 2s2
[Ne] 3s2
[Ar] 4s2
[Kr] 5s2
Ba
56 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10
5s2 5p6 6s2
[Xe] 6s2
Ra
88 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10
4f14 5s2 5p6 5d10 6s2 6p6 7s2
[Rn] 7s2
 All have ns2 outer shell electrons
 Only difference is value of n
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
81
Shorthand Orbital Diagrams
 Write out lines for orbital beyond Noble gas
 Higher energy orbital to right
 Fill from left to right
Abbreviated Orbital Diagrams
Ru
[Kr]





4d
S
[Ne]

3s
Jespersen/Brady/Hyslop
 

5s

3p
Chemistry: The Molecular Nature of Matter, 6E
82
Valence Shell Electron Configurations
 One last type of electron configuration
 Use with representative elements (s and p block
elements)—longer columns
 Here only e’s in outer shell important for bonding
 Only e’s in s and p subshells
 Valence Shell = outer shell
= occupied shell with highest n
 Result - use even more abbreviated notation
for e configurations
 Sn = 5s2 5p2
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
83
Electronic Configurations:
 A few exceptions to rules
Element
Cr
Cu
Ag
Au
Expected
[Ar] 3d4 4s2
[Ar] 3d9 4s2
[Kr] 4d9 5s2
[Xe] 5d9 6s2
Experimental
[Ar] 3d5 4s1
[Ar] 3d10 4s1
[Kr] 4d10 5s1
[Xe] 5d10 6s1
 Exactly filled and exactly half-filled subshells have
extra stability
 Means that you can promote 1 electron into next
higher energy orbital to gain this extra stability
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
84
Your Turn!
The orbital diagram corresponding to the ground
state electron configuration for N is:
A.
B.
1s
2s
2p
C.
1s
2s
2p
D.
1s
2s
2p
E.
1s
2s
2p
1s
2s
2p
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
85
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]
↑↓
Jespersen/Brady/Hyslop
↑↓
↑↓
↑
Chemistry: The Molecular Nature of Matter, 6E
86
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
h
xmv 
4
x = particle position
mv = particle momentum
= mass × velocity of particle
h = Planck’s constant = 6.626 × 10–34 J∙s
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
87
Heisenberg’s Uncertainty Principle
Macroscopic scale
 Errors in measurements << value
Subatomic scale
 Errors in measurements  or > value
 If you know position exactly, know nothing about
velocity
 If you know velocity exactly, know nothing about
position
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
88
Consequence of HUP
 Can’t talk about absolute position
 Can only talk about e probabilities
 Where is e likely to be?
  = wavefunction
 Amplitude of e wave
 2 = probability of finding e at given location
 Probability of finding e in given region of space
= square of amplitude of wave at that point
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
89
Electron Cloud
e dot picture = snapshots
 Lots of dots = large amplitude of wave
 = High probability of finding e
e density
 How much of es charge packed into given volume
 High Probability
 High e charge or Large e density
 Low Probability
 Low e charge or Small e density
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
90
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 radius
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
91
e Density Distribution
 Determined by

Shape
Size
n
Orientation m
 e density
 No sharp boundary
 Gradually fades away
 “Shape”
 Imaginary surface enclosing 90% of e density of
orbital
 Probability of finding e is same everywhere on
surface
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
92
Effect of n on s Orbital
 In any given direction
probability of finding e
same
 All s orbitals are
spherically shaped
 Size  as n 
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
93
Spherical Nodes
 At higher n, now have spherical nodes
 Spherical regions of zero probability, inside orbital
 Node for e wave
 Imaginary surface where e density = 0
 2s, one spherical node, size larger
 3s, two spherical nodes, size larger yet
In general:
 Number of spherical nodes
=nℓ1
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
94
p Orbitals
 Possess one (1) nodal plane through nucleus
 e density only on 2 sides of nucleus
 2 lobes of e density
 All p orbitals have same overall shape
 Size  as n 
 For 3p get spherical node
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
95
Representations of p Orbitals
 Constant probability surface for
2p orbital
 Simplified p orbital emphasizing
directional nature of orbital
 All 3 p orbitals in p sub shell
 One points along each axis
2px
Jespersen/Brady/Hyslop
2py
2pz
Chemistry: The Molecular Nature of Matter, 6E
96
There Are Five Different d Orbitals
 Four with four lobes of e density
 One with two lobes and ring of e density
 Result of two nodal planes though nucleus
 Number of nodal planes through nucleus = ℓ
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
97
Your Turn!
Which sketch represents a pz orbital?
A.
C.
B.
D.
E.
z
y
x
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
98
Periodic Properties: Consequences of
e Configuration
 Chemical and physical properties of elements
 Vary systematically with position in periodic table
 i.e. with element's e configuration
 To explain, must first consider amount of + charge
felt by outer e s (valence e s)
 Core e s spend most of their time closer to nucleus
than valence (outer shell) e s
 Shield or cancel out (screen out, neutralize) some of +
charge of nucleus
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
99
Learning check: Li 1s2 2s1
 3 p+ in nucleus
 2 e in close (1s)
 Net + charge felt
at outer e
 ~ 1 p+
Effective Nuclear
Charge (Zeff)
 Net positive (+) charge outer e feels
 Core e s shield valence e s from full nuclear
charge
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
100
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
 Zeff determined primarily by
 Difference between charge on nucleus (Z) and
 Charge on core (number of inner e s)
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
101
Penetration
 Within same n shell
 e s in 2s orbital are closer to nucleus than e s in
2p orbital
 2s feel more of nuclear charge and partially
shield 2p from nucleus
 Likewise, 3s penetrates inside 3p which
penetrates inside 3d
 Gives rise to ordering of energy levels
 Ens < Enp < End < Enf
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
102
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
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
103
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 x 1012 m
 Range 37 – 270 pm for atoms
Jespersen/Brady/Hyslop
Chemistry: The Molecular Nature of Matter, 6E
104
Trends in Atomic Radius (r)
 Down Column (group)
 Zeff essentially constant
 n, outer e s farther away from nucleus and radius

 Across row (period)
 n constant
 Zeff, outer e s feel larger Zeff and radius 
Transition Metals and Inner Transition Metals
 Size variations less pronounced as filling core
 n same (outer e s) across row
  Zeff and r more gradually
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Atomic and Ionic Radii (in pm)
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Ionic Radii
  down column (group)
  across row (period)
Cations smaller than parent
atom
 Same Zeff, less e s,
 Radius contracts
Anions larger than parent
atom
 Same Zeff, more e s
 Radius expands
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Your Turn!
Which of the following has the smallest
radius?
A. Ar
B. K+
C. Cl–
D. Ca2+
E. S2–
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Ionization Energy
 Energy required to remove e from gas phase
atom
 Corresponds to taking e from n to n = 
M (g)  M+ (g) + e
 1st IE
 IE = E
IE 
2
R HhcZ eff
2
n
Trends: IE  down column (group) as n
IE  across row (period) as Zeff 
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Comparing 1st IE’s
 Largest 1st IEs are
in upper right
 Smallest 1st IEs
are in lower left
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Successive Ionization Energies
  slowly as
remove each
successive e
 See big  in IE
 When break
into exactly
filled or half
filled subshell
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Table 8.2: Successive Ionization Energies in
kJ/mol for H through Mg
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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
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Electron Affinity (EA)
 Potential energy change associated with
addition of 1e to gas phase atom or ion in the
ground state
 X (g) + e  X (g)
 O and F very favorable to add electrons
 1st EA usually negative (exothermic)
 Larger negative value means more favorable to
add e
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Table 8.3 Electron Affinities of Representative
Elements
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Trends in Electron Affinity (EA)
 EA becomes less exothermic down column
(group) as n
 e harder to add as orbital farther from nucleus and
feels less + charge
 EA becomes more exothermic across row
(period) as Zeff 
 Easier to attract e as + charge 
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Successive EAs
 Addition of 1st e often exothermic
 Addition of more than 1 e s requires energy
 Consider addition of electrons to oxygen:
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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