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Unformatted text preview: Chapter 7
Chapter Atomic Structure Chapter goals
Chapter
• Describe the properties of electromagnetic
Describe
radiation.
radiation.
• Understand the origin of light from excited
Understand
atoms and its relationship to atomic
structure.
structure.
• Describe experimental evidence for waveparticle duality.
• Describe the basic ideas of quantum
Describe
mechanics.
mechanics.
• Define the three quantum numbers (n, l , and
and
ml ) and their relationship to atomic structure. Electromagnetic Radiation
Electromagnetic
• light
• dual nature: wave and particle
• transverse wave: perpendicular oscillating
transverse
electric and magnetic fields
electric
• longitudinal wave: alternating areas of
longitudinal
compression and decompression. The
direction of the wave is along the direction
of propagation
of
• sound Transverse Waves
Transverse
• light
• do not require medium for propagation Amplitude
Amplitude
height of wave at maximum
Y
Z
Amplitude
X Wavelength, λ ( lambda) distance traveled by wave in 1 complete oscillation;
distance from the top (crest) of one wave to the top
of the next wave. Y Z λ
X • λ measured in m, cm, nm, Å (angstrom)
(angstrom)
• 1 Å = 1 × 1 0 −10 m = 1 × 1 0 −8 cm
• frequency, ν (nu), measured in s−1 (hertz) (Hz):
frequency, (nu)
number of complete oscillations or cycles
complete
passing a point per unit time (s)
passing
• speed of propagation,
speed
distance traveled by ray per unit time
distance
in vacuum, all electromagnetic radiation
travels at same rate
travels
c = 2.998 x 1010 cm/s (speed of light)
2.998
= 2.998 x 108 m/s
2.998
(slower in air)
m
c ( = ν( s−1) × λ ( m)
) ν(
s What is the wavelength in nm of orange light,
which has a frequency of 4.80 x 1014 s−1?
which
c=λ× ν
c 2.998× 108 m s−1
2.998
λ = −− = −−−−−−−−−−−− = 6.25 × 10−7 m
−− −−−−−−−−−−−− −1
ν
4.80 × 1014 s
1 nm
nm
6.25 × 10−7 m × −−−−−−−−− = 625 nm
6.25
−−−−−−−−−
1 × 10−9 m Names to remember
Names
• Max Planck: quantized energy E = hν ~1900
• Albert Einstein: photoelectric effect ~1905
• Niels Bohr: 2D version of atom
• En=(RH)(1/n2) Balmer, 1885, then Bohr, 1913
• Louis de Broglie: Wavelike properties of
Louis
matter ~1915
matter
• Werner Heisenberg: Uncertainty Principle
Werner
~1923
• Erwin Schrödinger: Schrödinger Equation
Erwin
~1926
~1926 Planck’s equation
Planck’s
• Planck studied black body radiation, such
Planck
as that of a heated body, and realized that
to explain the energy spectrum he had to
assume that:
assume
1. An object can gain or lose energy by
absorbing or emitting radiant energy in
QUANTA of specific frequency (ν )
QUANTA
2. light has particle character (photons)
light
c
• Planck’s equation is E = h × ν = h × ──
Planck’s
──
E = energy of one photon
λ
h = Planck’s constant = 6.626× 10−34
Planck’s Electromagnetic Spectrum
Electromagnetic
λ 0.01nm
γ rays Electromagnetic Spectrum
Electromagnetic
λ 0.01nm
1nm
γ rays
xrays Electromagnetic Spectrum
Electromagnetic
λ 0.01nm 200nm
1nm
γ rays
xrays
vacuum UV Electromagnetic Spectrum
Electromagnetic
λ 0.01nm 200nm
1nm 400nm
γ rays UV xrays
vacuum UV Electromagnetic Spectrum
Electromagnetic
λ 0.01nm 200nm 800nm
1nm 400nm
γ rays UV Vis. xrays
vacuum UV Electromagnetic Spectrum
Electromagnetic
λ 0.01nm 200nm 800nm
1nm 400nm 25µ m
γ rays UV Vis.
near
xrays
infrared
vacuum UV Electromagnetic Spectrum
Electromagnetic
λ 0.01nm 200nm 800nm 1mm
1nm 400nm 25µ m
γ rays far IR
UV Vis.
near
xrays
infrared
vacuum UV Electromagnetic Spectrum
Electromagnetic
λ 0.01nm 200nm 800nm 1mm
1nm 400nm 25µ m
100mm
γ rays far IR
UV Vis.
µ waves
near
xrays
infrared
vacuum UV Electromagnetic Spectrum
Electromagnetic
λ 0.01nm 200nm 800nm 1mm
1nm 400nm 25µ m
100mm
γ rays far IR
UV Vis.
µ waves
near
xrays
infrared
radio waves
vacuum UV Electromagnetic Spectrum
Electro
Spectrum c =ν λ
hν E= E = h c/λ
c/ Compact disk players use lasers that emit
red light with a wavelength of 685 nm. What
is the energy of one photon of this light?
What is the energy of one mole of photons
of that red light?
of
λ , nm → λ , m → ν , s−1 → E, J/photon → E, J/mole
nm 10−9 m
10
× −−−−−−−
−−−−−−−
nm c ν = −−
−−
λ E = hν × Avogadro’s number 10−9 m
10
685 nm × −−−−−−− = 6.85 × 10−7 m
−−−−−−−
1 nm
c 2.998× 108 m s−1
2.998
ν = −− = −−−−−−−−−−−− = 4.38 × 1014 s−1
−− −−−−−−−−−−−−
λ
6.85 × 10−7 m
E = hν = (6.626× 10−34 J⋅ s/photon)× 4.38 × 1014
4.38
s−1 = 2.90× 10−19 J/photon
= (2.90× 10−19 J/photon)× 6.022× 1023
photons/mol
= 1.75 × 105 J/mol The Photoelectric Effect
The
• Light can strike the surface of some metals
Light
causing electrons to be ejected.
causing
• It demonstrates the particle nature of light. The Photoelectric Effect
The • What are some practical uses of the
What
photoelectric effect?
photoelectric
• Electronic door openers
• Light switches for street lights
• Exposure meters for cameras
• Albert Einstein explained the effect
– Explanation involved light having particlelike behavior.
The minimum energy needed to eject the e− is
E = h × ν (Planck’s equation)
It is also called ‘threshold’ energy.
It
– Einstein won the 1921 Nobel Prize in Physics
Einstein
for this work.
for Prob.: An energy of 2.0× 102 kJ/mol is required to cause a Cs
Prob.:
atom on a metal surface to loose an electron. Calculate the
longest possible λ of light that can ionize a Cs atom.
From the value of energy we calculate the frequency ( ν ) and, with this
From
and,
we calculate lambda (λ ).
we
Firstly, we need to calculate the energy in J per atom; it is given in kJ
Firstly,
per mol of atoms...
per
kJ 1000 J
1 mol
kJ
2
2.0× 10 ── x ───── x ───────────── = 3.3× 10−19 Joule per atom
── ───── ─────────────
mol
kJ
6.022 × 1023 atoms E=h× ν E
3.3 × 10−19 Joule
ν = ── = ───────────── = 5.0× 1014 s1
── ─────────────
h
6.626 × 10−34 J s Now, speed of light, c = λ ν
Now, c
2.998× 108 m s1
2.998
λ = ── = ───────────= 6.0× 10−7 m
── ───────────
ν
5.0× 1014 s1
5.0 1 nm
nm
6.0× 10 m × ──────── = 600 nm
────────
1× 10−9 m
−7 (Visible light) Prob. : A switch works by the photoelectric effect. The
metal you wish to use for your device requires 6.7× 10−19
metal
J/atom to remove an electron. Will the switch work if the
J/atom
light falling on the metal has a λ = 540 nm or greater? Why?
The energy of photon is
The
calculated with Planck’s Equation
calculated
c
E = h × ν = h × ──
If calculated E ≥
6.7× 10−19 J,
6.7
λ
the switch will work.
the
1× 10−9 m
540 nm × ────── = 5.40× 10−7 m
──────
nm
nm
2.998× 108 m s−1
2.998
E = 6.626× 10−34 J⋅ s × ────────── = 3.68× 10−19 J
6.626
──────────
5.40× 10−7 m
5.40
The switch won’t open, because E < 6.7× 10−19 J. λ has to
6.7
be less than 540 nm. Atomic Line Spectra and the Bohr Atom
Atomic
(Niels Bohr, 18851962)
(Niels
18851962
• An emission spectrum is formed by an electric
emission
current passing through a gas in a vacuum tube (at
very low pressure) which causes the gas to emit light.
very
– Sometimes called a bright line spectrum.
line Atomic Line Spectra and the Bohr
Atom
Atom
• The Rydberg
The
equation is an
empirical equation
that relates the
wavelengths of the
lines in the hydrogen
spectrum. Lines are
due to transitions
n2 ──── upper level
n1 ──── lower level
n , n = 1, 2, 3, 4, 5,… 1
1
1
= R 2 − 2 n
λ
n2 1 R is the Rydberg constant
R = 1.097 × 107 m 1
n1 < n 2
n’ s refer to the numbers
of the energy levels in the
emission spectrum of hydrogen Example 58. What is the wavelength in angstroms
of light emitted when the hydrogen atom’s energy
changes from n = 4 to n = 2?
changes n 2 = 4 and n1 = 2
1
1
1
= R 2 − 2 n
λ
n2 1 1
1
7
1 1
= 1.097 × 10 m 2 − 2 λ
4
2 1 λ =1.097 ×107 m 1 ( 0.250 − 0.0625)
1 λ =1.097 ×10 7 m 1 ( 0.1875)
1 λ = 2.057 ×106 m 1 1 1Å
λ = ─ ─ ─ ─ ─ ─ ─ ─ ─ = 4.862 × 10−7 m × ─ ─ ─ ─ ─ = 4862
Å
2.057× 106 m−1
10−10 m
2.057
That corresponds to the green line in H spectrum Atomic Line Spectra and the Bohr
Atom
Atom
• An absorption spectrum is formed by shining a
An absorption
beam of white light through a sample of gas.
beam
– Absorption spectra indicate the
Absorption
wavelengths of light that have been
wavelengths
absorbed.
absorbed Every element has a unique spectrum.
Thus we can use spectra to identify
elements.
elements.
This can be done in the lab, stars,
This
fireworks, etc.
fireworks, Bohr Model of the Atom
Bohr
• planetary model
• considers only the particle nature of the
considers
electron
electron
• p+ & n packed tightly in ‘tiny’ nucleus
• electrons traveling in circular paths,
electrons
orbits, in space surrounding nucleus
orbits,
• size, energy, and e– capacity of orbits
increase as does distance from nucleus
(orbital radius)
(orbital
• orbits quantized (only certain levels exist)
orbits quantized e– e– e– P+
n e– e– e–
e– Energy Levels E n=6
n=5
n=4
n=3
n=2 n=1
n=1 1
1
E ∝ ──
r2 1
E ∝ ──
n2 E n=6
n=5
n=4
n=3
n=2 n=1
n=1 Exciting the electron
from ground level
(n = 1) to upper
levels (n > 1)
Energy is absorbed E n=6
n=5
n=4
n=3
n=2 Decay of the electron
from upper levels to
lower levels:
Energy is emitted
Emission of Photons hν
One
photon
photon
per
per
transition
transition n=1
n=1 E n=6
n=5
n=4
n=3
n=2 n=1
n=1 Balmer Series, nf = 2,
for hydrogen.
There are other
series. hν Calculating E Difference Between two Levels
Calculating
A school teacher was the first to find this!
school
Johann Balmer
Johann 1
1
∀ ∆ E= Efinal − Einitial = RH(── − ──)
E=
n f 2 n i2
18
• RH= 2.18 x 1018 J/atom = 1312 kJ/mol
J/atom • ni and nf = principal quantum numbers of
and
the initial and final states: n f < ni
the
• 1,2,3,4…. Problem: Calculate ∆ E and λ for the violet line
and
of Balmer series of H. ninitial = 6 nfinal = 2
of
1
1
18
∀ ∆ E= RH(── − ──)
RH= 2.18 x 1018 J/atom
E= (──
J/atom
nf2 ni2
1
1
18
∆ E= 2.18 x 1018 J(── − ──) = 4.84 x 10−19 J
22
62
∆ E = hν ν = c/λ Then, ∆ E = hc/λ hc 6.626× 10−34 J⋅ s × 2.998x108 ms−1
hc
λ = ── = ─────────────────────
∆E
4.84 x 10−19 J
4.84
λ = 4.104 x 10−7 m × (1 Å/10−10m) = 4104 Å = 410.4 nm
4.104
Å/10 m)
410.4 Bohr Model of the Atom
Bohr
• Bohr’s theory correctly explains the H
Bohr’s
emission spectrum and those of hydrogenemission
like ions (He+, Li2+ … 1e− species)
• The theory fails for all other elements
The
because it is not an adequate theory: it
doesn’t take into account the fact that the
(very small) electron can be thought as
electron
having wave behavior.
having The Wave Nature of the Electron
The
• In 1925 Louis de
In
Broglie published
his Ph.D.
dissertation.
dissertation.
– A crucial element
crucial
of his dissertation
is that electrons
have wavelike
properties.
properties.
– The electron
The
wavelengths are
described by the
de Broglie
relationship.
relationship. h
λ=
mv
h = Planck’ s constant
m = mass of particle
v = velocity of particle The WaveParticle Duality of the
Electron
Electron
• Consequently, we now know that
Consequently,
electrons (in fact  all particles) have both
a particle and a wave like character.
particle
• This waveparticle duality is a
This
fundamental property of submicroscopic
particles (not for macroscopic ones.)
particles The WaveParticle Duality of the Electron
The
• Example: Determine the wavelength, in m and Å,
Example:
of an electron, with mass 9.11 x 1031 kg, having a
of
velocity of 5.65 x 107 m/s.
velocity
h = 6.626 x 10−34 Js = 6.626 x 10−34 kg m2/s h
6.626× 10−34 kg m2s−1
kg
6.626
λ = ── = ────────────────────
mv
9.11× 10−31kg × 5.65x107 ms−1
mv
−11
λ = 1.29 × 10−11 m Good:
Good: 1Å
within
within
−11
λ = 1.29 × 10−11 m ────── = 0.129 Å atomic
atomic
10−10 m
dimensions
10
dimensions The WaveParticle Duality of the Electron
The
• Example: Determine the wavelength, in m, of a
Example:
0.22 caliber bullet, with mass 3.89 x 103 kg,
0.22
having a velocity of 395 m/s, ~ 1300 ft/s.
having
h = 6.626 x 10−34 Js = 6.626 x 10−34 kg m2/s
6.626 h
6.626× 10−34 kg m2s−1
kg
6.626
λ = ── = ────────────────
mv 3.89× 10−3kg × 395 ms−1
mv
−34
−24
λ = 4.31 × 10−34 m = 4.31 × 10−24 Å too small!
It doesn’t apply to macroobjects! Quantum Mechanical Model of the Atom
Quantum
considers both particle and wave nature of electrons
• Heisenberg and Born in 1927 developed the concept
Heisenberg
of the Uncertainty Principle:
Uncertainty
It is impossible to determine simultaneously both the
position (x,y,z) and momentum (mv) of an electron (or
any other small particle).
any
• Consequently, we must speak of the electrons’
Consequently,
positions about the atom in terms of probability
functions, i.e., wave equation written for each electron.
functions
• These probability functions are represented as orbitals
These
orbitals
in quantum mechanics. They are the wave equations
squared and plotted in 3 dimensions.
squared Schrödinger’s Model of the Atom
Schrödinger’s
Basic Postulates of Quantum Theory
1. Atoms and molecules can exist only in certain
Atoms
energy states. In each energy state, the atom or
molecule has a definite energy. When an atom or
molecule changes its energy state, it must emit or
absorb just enough energy to bring it to the new
energy state (the quantum condition).
energy
2. Atoms or molecules emit or absorb radiation (light)
Atoms
as they change their energies. The frequency of
the light emitted or absorbed is related to the
energy change by a simple equation.
energy E = hν = hc λ Schrödinger’s Model of the Atom
Schrödinger’s
3. The allowed energy states of atoms and
The
molecules can be described by sets of
numbers called quantum numbers.
quantum
• Quantum numbers are the solutions of the
Quantum
Schrödinger, Heisenberg & Dirac equations.
Schrödinger,
• Four quantum numbers are necessary to
describe energy states of electrons in
atoms.
atoms.
.. Schr o dinger equation
b2 ∂2Ψ ∂2Ψ ∂2Ψ − 2 2 + 2 + 2 + VΨ = EΨ
8π m ∂ x ∂ y ∂ z Orbital
Orbital
• region of space within which one can expect to
region
find an electron
find
• no solid boundaries
• electron capacity of 2 per orbital
electron
per
• space surrounding nucleus divided up into
space
large volumes called shells
shells
• shells subdivided into smaller volumes called
shells
subshells
subshells
• orbitals located in subshells
• as shells get further from nucleus, energy, size,
as
and electron capacity increase
and
• shells, subshells, and orbitals described by
shells,
quantum numbers
quantum Quantum Numbers
Quantum
• The principal quantum number has the
The
symbol n
symbol
• n = 1, 2, 3, 4, ... indicates shell
1,
shell
•
K, L, M, N, … shells
shells
• as n increases, so does size, energy, and
as
electron capacity
electron
The electron’s energy depends principally
energy
on n .
on Quantum Numbers
Quantum
• The angular momentum (azimuthal) quantum
quantum
number has the symbol . It indicates
number
It
subshell.
subshell. = 0, 1, 2, 3, 4, 5, .......(n1)
0,
from 0 to maximum (n1) for each n
from = s, p, d, f, g, h, ....... Subshells
s,
Subshells tells us the shape of the orbitals.
tells
• These orbitals are the volume around the
These
atom that the electrons occupy 9095% of the
time.
time.
This is one of the places where Heisenberg’s
Uncertainty principle comes into play.
Uncertainty Magnetic Quantum Number, ml
Magnetic
• The symbol for the magnetic quantum number is
The
m . It specifies the orientation of the orbital. For a given l,
m = , ( + 1), ( +2), .....0, ......., ( 2), ( 1), • If = 0 (or an s orbital), then m = 0.
If
0.
– Notice that there is only 1 value of m .
Notice
This implies that there is one s orbital per n
This
value. n ≥ 1
• If = 1 (or a p orbital), then m = 1,0,+1.
– There are 3 values of m .
There
Thus there are three p orbitals per n value.
Thus
n≥ 2 Magnetic Quantum Number, m
Magnetic
• If = 2 (or a d orbital), then m = 2, 1, 0, +1, +2.
– There are 5 values of m .
There
Thus there are five d orbitals per n value. n ≥ 3
Thus
• If = 3 (or an f orbital), then
m = 3, 2, 1, 0, +1, +2, +3.
3,
– There are 7 values of m . 2x +1 orbitals
There
Thus there are seven f orbitals per n value, n ≥ 4
Thus
• Theoretically, this series continues on to g, h, i,
Theoretically,
etc. orbitals.
etc.
– Practically speaking atoms that have been
Practically
discovered or made up to this point in time
only have electrons in s, p, d, or f orbitals in
their ground state configurations.
their #orbitals
n2
2l + 1 ml
n shell l subshell
s
0
1K0
1
s
0
2L0
1
–1,0,1
1p
3
0
3 M0 s
1
–1,0,1
1p
3
2d
5
2,1,0,1,2
0
4N0 s
1
–1,0,1
1p
3
2d
5
2,1,0,1,2
3 f 3,2,1,0,1,2,3 7 Max
#e–
1
2
28
4
6
2
6 18
9
10
2
6
16 10 32
14 Maximum two electrons per orbital Electrons Indicated by Shell
and Subshell
and
Symbolism
#electrons nl principal number
4s1 3s2 4d 12 # letter: s, p, d,.. orbital
letter: s,
5p4 3p 7 4f 5 there are 4 electrons in the 5p orbitals 4f14 The Shape of Atomic Orbitals
Atomic
• s orbitals are spherically symmetric. A plot of the surface density as a function
of the distance from the nucleus for an
s orbital of a hydrogen atom
• It gives the probability of finding the electron
It
at a given distance from the nucleus
at
A node (nodal
node
point or surface)
point
is a point where
Ψ and the
and
probability
probability
density (Ψ 2) are
density
zero p orbitals
orbitals
• p orbital properties:
– The first p orbitals appear in the n = 2
The
shell.
shell.
• p orbitals are peanut or dumbbell shaped
orbitals
volumes.
volumes.
– They are directed along the axes of a
They
Cartesian coordinate system.
Cartesian
• There are 3 p orbitals per n level.
There
– The three orbitals are named px, py, pz.
– They have an = 1.
– m = 1,0,+1 3 values of m p Orbitals
Orbitals
y y
z
x y px x
z py x pz z d orbitals
orbitals • d orbital properties:
– The first d orbitals appear in the n = 3 shell.
• The five d orbitals have two different shapes:
– 4 are clover leaf shaped.
– 1 is peanut shaped with a doughnut around it.
– The orbitals lie directly on the Cartesian axes
The
or are rotated 45o from the axes.
or
• There are 5 d orbitals per n
level.
– The five orbitals are named d xy , d yz , d xz , d x 2  y 2 , d z 2
– They have an = 2.
– m = 2,1,0,+1,+2
5 values of m d Orbitals
Orbitals y y
z z z
yx x dx2–y2 dxy x
y
z x dz2 y
z x dxz dyz f orbitals
orbitals
• f orbital properties:
– The first f orbitals appear in the n = 4
The
shell.
shell.
• The f orbitals have the most complex
The
shapes.
shapes.
• There are seven f orbitals per n level.
– The f orbitals have complicated names.
– They have an = 3
– m = 3, 2, 1,0, +1,+2, +3 7 values of
3,
m
– The f orbitals have important effects in
The f orbitals
orbitals
• f orbital shapes Prob: A possible excited state for the H atom has
an electron in a 5d orbital. List all possible sets of
quantum numbers n, l , and ml for this electron
n = 5, l=2 five possible ml : 2, 1, 0, +1, +2 Then, there are five sets of (n, l , ml )
Then,
(5, 2, 2)
2)
(5, 2, 1)
(5, 1)
(5, 2, 0)
(5, 2, +1)
(5, +1)
(5, 2, +2)
(5, +2) Prob: Which of the following represent valid sets
of quantum numbers? For a set that is invalid,
explain briefly why it is not correct.
explain
a) n = 3, l = 3, ml : 0
3,
b) n = 2, l = 1, ml : 0
b) No: maximum l = n1
No:
l = 0, 1, and 2, no 3 c) n = 6, l = 5, ml : −1
d) n = 4, l = 3, ml : −4
d) No: minimum value of ml : −l,
that is ml : −3
that
−3 Prob: What is the maximum number of orbitals
that can be identified by each of the following
sets of quantum numbers? When “none” is the
correct answer, explain the reason.
correct
Answer
a) n = 4, l = 3
a) Seven b) n = 5, 25 c) n = 2, l = 2
c) None Why?
ml : 3, 2, 1, 0, 1, 2, 3
3,
n2
maximum l = n  1 d) n = 3, l = 1, ml : 1 One, just that described #s. Prob: State which of the following are incorrect
designations for orbitals according to the
quantum theory: 3p, 4s, 2f, and 1p
quantum
Answer
a) 3p
a) correct Why?
n = 3, l = 1 maximum l = 31=2 b) 4s correct
correct c) 2f
c) incorrect maximum l = 1 (l = 3 for f) d) 1p incorrect maximum l = 0 (l = 1 for p) ...
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This note was uploaded on 11/07/2011 for the course CHM 2045 taught by Professor Geiger during the Fall '08 term at University of Central Florida.
 Fall '08
 geiger
 Chemistry, Atom

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