View
168
Download
18
Category
Tags:
Preview:
DESCRIPTION
Chapter 8: The Quantum Mechanical Atom. Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop. Electromagnetic Energy. Electromagnetic Radiation Light energy or wave Travels through space at speed of light in vacuum c = speed of light = 2.9979 x 10 8 m/s - PowerPoint PPT Presentation
Citation preview
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 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 energy2
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 = s1
Related by = c
3
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
4
nodes
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
m10650
m/s109979.29
8
c
= 4.61 × 1014 s–1 = 4.61 × 1014 Hz
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
s/10880
m/s 1000.33
8
c
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Electromagnetic SpectrumVisible Light
Band of ’s that human eyes can see 400 to 700 nm Make up spectrum of colors 700 nm ROYGBIV 400 nm
8
White light Equal amount of all these colors Can separate by passing through prism
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Important Experiments in Atomic TheoryLate 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
9
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 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 1034 J·s
10
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning CheckWhat is the frequency, in sec–1, of radiation which has an energy of 3.371 x 10–19 joules per photon?
11
hE
s J10626.6
J10371.334
19
= 5.087×1014 s–1
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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.1.62 x 10–24 J
12
hE
MHzs101
MHz 2450sJ10626.6E16
34
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
13
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 = NAhIf light required to start reaction
Must have light above certain frequency to start reaction
Below minimum threshold E, brightness is NOT important
14
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning CheckHow much energy is contained in one mole of photons, each with frequency 2.00 × 1013?
E = NAh
15
E = (6.02×1023 mol–1)(6.626×10–34 J∙s)(2.00×1013 s–1)
E = 7.98 × 103 J/mol
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
16
hNE
A
)sJ10626.6)(mol 1002.6(
mol/J1060.134123
3
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
17
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Electronic Structure of Atom Most information comes from:
1. Study of light absorption Electron absorbs energy
Moves to higher energy “excited state”
2. Study of light emission e loses photon of light
Drops back down to lower energy “ground state”
18
ground state
excited state
+h
h
excited state
ground state
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Spark (electrical discharge) excites gas molecules (atoms)
19
+ gas
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Line Spectrum Spectrum that has only a few discrete lines Also called atomic spectrum or emission
spectrum Each element has unique emission spectrum
20
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 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)21
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Rydberg Equation
Can be used to calculate all spectral lines of hydrogen
RH = 109,678 cm1 = Rydberg
constant = wavelength of light emittedn1 & n2 = whole numbers
(integers) from 1 to where n2 > n1
If n1 = 1, then n2 = 2, 3, 4, …
22
22
21
111
nnRH
Corresponds to allowed energy levels for atom
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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.
23
361
41
cm678,1096
1
2
11 122HR
m101
1nm100cm
1mcm101029.4
cm9.372,24
19
51
= 410.3 nm Violet line in spectrum
=24,373cm–1
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning CheckA photon undergoes a transition from nhigher down to n = 2 and the emitted light has a wavelength of 650.5 nm?
)n1
41(
13455.71
22
24
)n1
21(cm678,091
cm105.6501
22
21
7
n2 = 3
110.0.134557
141
n122
10911001
n22 .
.
cm105.650nm1
cm101nm5.650 7
7
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
25
221
5
1
3
1cm 678,109
1
1cmnm101
cm 7997
1 7
1
1cm 77991
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
26
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
27
Potential Energy of Rabbit
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
28
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Energy for Bohr Model of H Equation for energy of e in H atom
Ultimately b relates to RH by b = RHhc
OR
where b = RHhc = 2.1788 x 1018 J/atom
Allowed values of n = 1, 2, 3, 4, … n = Quantum number Used to identify orbit
29
2
1
nE
2
422
h
meb
22 n
hcR
n
bE H
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
30
E s are quantized Every time e drops from n = 3 to n
= 1 Same frequency photon is emitted Yields line spectra
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Stronger attractive force = more negative E
31
them between distance
B) on A)(chargeon (chargeE
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 h = higher l = lower
32
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
33
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
34
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
35
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Diffraction and Electrons Light
Exhibits interference Has particle nature
Electrons Known to be particles Also demonstrate interference
36
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
38
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Wave on a Wire Integer number (n) of peaks and troughs
is required Wavelength is quantized:
39
n2L
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
40
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Electron on Wire—Theories Standing wave Half-wavelength must occur
integer number of times along wire’s length
de Broglie’s equation links these m = mass of particle v = velocity of particle
Combining gives:
2
22
8
mLhnE
41
mvh
n2L
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Learning Check: Calculate for e-What 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?
43
1J/sm1kg
kg109.11m/s101.0
sJ106.626 22
317
34
= 7.27 x 10–11 m
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
44
J1s/mkg1
kg10.0s/m35sJ10626.6 2234
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 Can be related to probability of finding
electron at that given point
Nodes Regions of wire where electrons will not be
found45
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 46
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
Number of nodes (points where * = 0)
RHhc = 2.18 x 1018 J/atom
For given atom, Lower n = Lower (more negative) E
= More stable
2H
2
n
hcRZE
47
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
48
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
mℓ = Magnetic Quantum Number Allowed values:
mℓ = ℓ, ℓ+1, ℓ+2, …, 0 , …, ℓ2, ℓ1, 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)49
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Table 8.1 Summary of Relationships Among the Quantum Numbers n, ℓ,
and m
50
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Energy Level Diagram for H Atom or Other 1 e– Ion
4s
3s
2s
1s
Ene
rgy
4p
3p
2p
3d
4d
All orbital subshells with same n value have same Energy
2lo
2hi
2
n
2
n
1ZE hcRHE = Elo – Ehi
2H
2
n
hcRZE
51
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
52
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Orbitals of Many Electrons
53
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
54
Possible Values:
+½ ½
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 es Not all spins paired Have unpaired es
Even number of es Depends on number of orbitals
55
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Consequences of Pauli Exclusion Principle
56
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
57
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
58
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
59
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ground State e ArrangementsElectron Configurations
Distribution of es among orbitals of atom 1. List subshells that contain electrons2. Indicate their electron population with
superscriptEx. N is 1s2 2s2 2p3
Orbital Diagrams Way to represent es in orbitals
1. Represent each orbital with circle (or line)2. Use arrows to indicate spin of each electron Ex. N is
60
1s 2s 2p
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Energy Level Diagram for Multi e Atom/Ion
4s
3s
2s
1s
Ene
rgy
4p
3p
2p
3d
4d5s
5p
4f6s
How to put e– into a diagram?
Need some rules
61
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
62
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 es in same orbital
Why? Repulsion of e in same region of space Empirical observation based on magnetic
properties
63
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Orbital Diagram and e Configurations: N Z = 7
4s
3s
2s
1s
Ene
rgy
4p
3p
2p
3d
Each arrow represents electron
1s2 2s2 2p3
64
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
4s
3s
2s
1s
Ene
rgy
4p
3p
2p
3d
Orbital Diagram and e- Configurations: V Z = 23
Each arrow represents an electron1s2 2s2 2p6 3s2 3p6 4s2 3d3
65
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
66
1s22s22p63s1
1s22s22p63s23p64s23d104p3
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
67
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Periodic Table Divided into regions of 2, 6, 10, and 14
columns = maximum number of electrons in s, p, d,
and f sublevels
68
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Each row (period) represents different energy level
Each region of chart represents different type of sublevel
69
Sublevels and the Periodic Table
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Now Ready to Put es 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 to70
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
71
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
72
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
“ns” orbital being filled “np” orbital being filled “(n – 1)d” orbital being filled “( n – 2)f” orbital being filled
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Electron Configuration of Boron (B)
73
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 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
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
74
“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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Writing Electron Configurations Fill pattern across row in Periodic Table is:
ns [(n2)f ] [(n1)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 n75
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
= 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
76
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
77
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
78
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
79
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Abbreviated Electron Configurations = Noble Gas
Notation [noble gas of previous row] + es filled in nth row
Represents core + outer shell es Use to emphasize that only outer shell
electrons react
Ex. Ba = [Xe] 6s2 see above
Ru = [Kr] 4d6 5s2
S = [Ne] 3s2 3p4
80
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Look at Group IIA
Z Electron Configuration Abbrev
Be 4 1s2 2s2 [He] 2s2
Mg 12 1s2 2s2 2p6 3s2 [Ne] 3s2
Ca 20 1s2 2s2 2p6 3s2 3p6 4s2 [Ar] 4s2
Sr 38 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 5s2 [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
81
All have ns2 outer shell electrons
Only difference is value of n
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Shorthand Orbital Diagrams
82
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
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 83
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Electronic Configurations: A few exceptions to rules
Element Expected Experimental
Cr
Cu
Ag
Au
[Ar] 3d4 4s2
[Ar] 3d9 4s2
[Kr] 4d9 5s2
[Xe] 5d9 6s2
84
[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
Your Turn!The orbital diagram corresponding to the ground state electron configuration for N is:
A.
B.
C.
D.
E.
85
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
1s 2s 2p
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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] ↑↓ ↑↓ ↑↓ ↑ ↑ ↑
86
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
4h
mvx
87
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
Heisenberg’s Uncertainty PrincipleMacroscopic 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
88
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
89
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Electron Cloude dot picture = snapshots
Lots of dots = large amplitude of wave
= High probability of finding e
e density How much of es charge packed into given
volume
High Probability High e charge or Large e density
Low Probability Low e charge or Small e density
90
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 radius91
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
e Density Distribution Determined by
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
Shape Size nOrientatio
nm
92
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Effect of n on s Orbital
In any given direction probability of finding e same
All s orbitals are spherically shaped
Size as n
93
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 194
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
95
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
96
x
y
z2px
x
y
z2py 2pz
x
y
z
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
97
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Which sketch represents a pz orbital?
98
x
y
x
z
y
z
x
xy
z
y
z
x
A. B.
D. E.
C.
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
99
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
100
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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)
101
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
102
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
103
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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 1012 m Range 37 – 270 pm for atoms 104
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
105
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Atomic and Ionic Radii (in pm)
106
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
107
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Which of the following has the smallest radius?
A.Ar
B.K+
C.Cl–
D.Ca2+
E.S2–
108
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Ionization Energy Energy required to remove e from gas
phase atom Corresponds to taking e from n to n = 1st IE M (g) M+ (g) + e
IE = E
Trends: IE down column (group) as n IE across row (period) as Zeff
109
2
2effH
n
hcZRIE
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Comparing 1st IE’s
110
Largest 1st IEs are in upper right
Smallest 1st IEs are in lower left
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Successive Ionization Energies
slowly as remove each successive e
See big in IE When break
into exactly filled or half filled subshell
111
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Table 8.2: Successive Ionization Energies in kJ/mol for H through Mg
112
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
113
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
114
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Table 8.3 Electron Affinities of Representative Elements
115
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
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
116
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Successive EAs Addition of 1st e often exothermic Addition of more than 1 e s requires
energy Consider addition of electrons to oxygen:
117
Change: EA(kJ/mol)
O(g) + e– O–(g) –141
O–(g) + e– O2–(g) +844
Net:
O(g) + 2e– O2–(g) +703
Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E
Your Turn!Which of the following has the largest electron affinity?
A.O
B.F
C.As
D.Cs
E.Ba
118
Recommended