Transcript CHAPTER 5

CHAPTER 4
The Structure of
Atoms
Chapter Outline
Subatomic Particles
1.
Fundamental Particles
2.
The Discovery of Electrons
3.
Canal Rays and Protons
4.
Rutherford and the Nuclear
Atom
5.
Atomic Number
6.
Neutrons
7.
Mass Number and Isotopes
8.
Mass spectrometry and Isotopic
Abundance
9.
The Atomic Weight Scale and
Atomic Weights
10. The Periodic Table: Metals,
Nonmetals, and Metalloids
The Electronic Structures of Atoms
11. Electromagnetic radiation
12. The Photoelectric Effect
13. Atomic Spectra and the Bohr
Atom
14. The Wave Nature of the
Electron
15. The Quantum Mechanical
Picture of the Atom
16. Quantum Numbers
17. Atomic Orbitals
18. Electron Configurations
19. Paramagnetism and
Diamagnetism
20. The Periodic Table and
2
Electron Configurations
Fundamental Particles
Three fundamental particles make up atoms. The
following table lists these particles together with their
masses and their charges.
Particle
Mass (amu) Charge
Electron (e-)
0.00054858
-1
Proton (p,p+)
1.0073
+1
Neutron(n,n0)
1.0087
0
3
The Discovery of Electrons
• Humphrey Davy in the early 1800’s
passed electricity through compounds and
noted and concluded that:
– the compounds decomposed into elements.
– compounds are held together by electrical forces.
• Michael Faraday in 1832-1833 realized
that the amount of reaction that occurs
during electrolysis is proportional to the
electrical current passed through the
compounds.
4
The Discovery of Electrons
• Cathode Ray Tubes experiments performed in
the late 1800’s & early 1900’s.
– Consist of two electrodes sealed in a glass tube
containing a gas at very low pressure.
– When a voltage is applied to the cathodes a glow
discharge is emitted.
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The Discovery of Electrons
• These “rays” are emitted from cathode (end) and travel to anode (+ end).
– Cathode Rays must be negatively charged!
• J.J. Thomson modified the cathode ray
tube experiments in 1897 by adding two
adjustable voltage electrodes.
– Studied the amount that the cathode ray
beam was deflected by additional electric
field.
6
The Discovery of Electrons
7
The Discovery of Electrons
8
The Discovery of Electrons
9
The Discovery of Electrons
• Thomson used his modification to measure
the charge to mass ratio of electrons.
Charge to mass ratio
e/m = -1.75882 x 108 coulomb/g
• Thomson named the cathode rays
electrons.
• Thomson is considered to be the
“discoverer of electrons”.
• TV sets and computer screens are cathode
10
ray tubes.
The Discovery of Electrons
• Robert A. Millikan
won the Nobel Prize
in 1923 for his famous
oil-drop experiment.
• In 1909 Millikan
determined the
charge and mass of
the electron.
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The Discovery of Electrons
• Millikan determined that the charge on a
single electron = -1.60218 x 10-19
coulomb.
• Using Thomson’s charge to mass ratio
we get that the mass of one electron is
9.11 x 10-28 g.
– e/m = -1.75882 x 108 coulomb
– e = -1.60218 x 10-19 coulomb
– Thus m = 9.10940 x 10-28 g
12
Canal Rays and Protons
• Eugene Goldstein noted streams of positively charged particles in
cathode rays in 1886.
– Particles move in opposite direction of cathode rays.
– Called “Canal Rays” because they passed through holes
(channels or canals) drilled through the negative electrode.
• Canal rays must be positive.
– Goldstein postulated the existence of a positive fundamental
particle called the “proton”.
13
Rutherford and the Nuclear Atom
• Ernest Rutherford
directed Hans Geiger
and Ernst Marsden’s
experiment in 1910.
– α- particle scattering
from thin Au foils
– Gave us the basic
picture of the atom’s
structure.
14
Rutherford and the Nuclear Atom
15
Rutherford and the Nuclear Atom
Rutherford’s major conclusions from the αparticle scattering experiment
1. The atom is mostly empty space.
2. It contains a very small, dense center called the
nucleus.
3. Nearly all of the atom’s mass is in the nucleus.
4. The nuclear diameter is 1/10,000 to 1/100,000
times less than atom’s radius.
16
Rutherford and the Nuclear Atom
17
Rutherford and the Nuclear Atom
• Because the atom’s mass is contained in
such a small volume:
– The nuclear density is ~1015g/mL.
– This is equivalent to ~3.72 x 109 tons/in3.
– Density inside the nucleus is almost the same
as a neutron star’s density.
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Atomic Number
• The atomic number is equal to the number of
protons in the nucleus.
– Sometimes given the symbol Z.
– On the periodic table Z is the uppermost number in
each element’s box.
• In 1913 H.G.J. Moseley realized that the atomic
number determines the element .
– The elements differ from each other by the number of
protons in the nucleus.
– The number of electrons in a neutral atom is also
equal to the atomic number.
19
Neutrons
• James Chadwick in 1932 analyzed the
results of α-particle scattering on thin Be
films.
• Chadwick recognized existence of
massive neutral particles which he called
neutrons.
– Chadwick discovered the neutron.
20
Mass Number and Isotopes
• Mass number is given the symbol A.
• A is the sum of the number of protons and
neutrons.
– Z = proton number N = neutron number
– A=Z+N
• A common symbolism used to show mass and
proton numbers is
A
12
48
Z
6
20
 Can be shortened to this symbolism.
E for example C, Ca,
14
63
N, Cu,
107
197
79
Ag, etc.
Au
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Mass Number and Isotopes
• Isotopes are atoms of the same element but with
different neutron numbers.
– Isotopes have different masses and A values but are
the same element.
• One example of an isotopic series is the
hydrogen isotopes.
1H
or protium is the most common hydrogen isotope.
• one proton and no neutrons
2H
or deuterium is the second most abundant hydrogen
isotope.
• one proton and one neutron
3H
or tritium is a radioactive hydrogen isotope.
• one proton and two neutrons
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Mass Number and Isotopes
• The stable oxygen isotopes provide another
example.
• 16O is the most abundant stable O isotope.
• How many protons and neutrons are in 16O?
8 prot onsand 8 neut rons
 17O
is the least abundant stable O isotope.
 How many protons and neutrons are in 17O?
8 prot onsand 9 neut rons
 18O
is the second most abundant stable O isotope.
How many protons and neutrons in 18O?
8 prot onsand10 neut rons
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Mass Spectrometry and
Isotopic Abundances
• Francis Aston devised the first mass
spectrometer.
– Device generates ions that pass down an evacuated
path inside a magnet.
– Ions are separated based on their mass.
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Mass Spectrometry and
Isotopic Abundances
There are four factors which determine a
particle’s path in the mass spectrometer.
1 accelerating voltage
2 magnetic field strength
3 masses of particles
4 charge on particles
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Mass Spectrometry and
Isotopic Abundances
• Mass spectrum of Ne+ ions shown below.
– How scientists determine the masses and
abundances of the isotopes of an element.
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The Atomic Weight Scale and
Atomic Weights
• If we define the mass of 12C as exactly 12 atomic
mass units (amu), then it is possible to establish
a relative weight scale for atoms.
– 1 amu = (1/12) mass of 12C by definition
– What is the mass of an amu in grams?
27
The Atomic Weight Scale and
Atomic Weights
Example 4-1: Calculate the number of atomic
mass units in one gram.
– The mass of one 31P atom has been experimentally
determined to be 30.99376 amu.
– 1 mol of 31P atoms has a mass of 30.99376 g.
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The Atomic Weight Scale and
Atomic Weights
Example 4-1: Calculate the number of atomic
mass units in one gram.
– The mass of one 31P atom has been experimentally
determined to be 30.99376 amu.
– 1 mol of 31P atoms has a mass of 30.99376 g.
 1 mole 31 P atoms 
? amu  1.000 g 

31
 30.99376 g P 
 6.022  1023 31 P atoms   30.99376 amu 

  31

31
P atom 
 1 mole P atoms

 6.022  10 amu
23
31
P
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The Atomic Weight Scale and
Atomic Weights
• The atomic weight of an element is the
weighted average of the masses of its
stable isotopes
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Atomic Weight Scale and Atomic
Weights
Example 4-2: Naturally occurring Cu consists of 2
isotopes. It is 69.1% 63Cu with a mass of 62.9
amu, and 30.9% 65Cu, which has a mass of 64.9
amu. Calculate the atomic weight of Cu to one
decimal place.
atomicweight  (0.691)(62
.9 amu)  (0.309)(64
.9 amu)


 


63
Cu isotope
65
Cu isotope
atomicweight  63.5amu for copper
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The Atomic Weight Scale and
Atomic Weights
Example 4-3: Naturally occurring chromium
consists of four isotopes. It is 4.31% 2450Cr, mass
= 49.946 amu, 83.76% 2452Cr, mass = 51.941 amu,
9.55% 2453Cr, mass = 52.941 amu, and 2.38%
54Cr, mass = 53.939 amu. Calculate the atomic
24
weight of chromium.
You do it!
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The Atomic Weight Scale and
Atomic Weights
Example 4-3: Naturally occurring chromium
consists of four isotopes. It is 4.31% 2450Cr, mass
= 49.946 amu, 83.76% 2452Cr, mass = 51.941 amu,
9.55% 2453Cr, mass = 52.941 amu, and 2.38%
54Cr, mass = 53.939 amu. Calculate the atomic
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weight of chromium.
atomicweight  (0.0431 49.946amu)  (0.8376 51.941amu)
 (0.0955 52.941amu)  (0.0238 53.939amu)
 2.153 43.506 5.056 1.284 amu
 51.998amu
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The Atomic Weight Scale and
Atomic Weights
Example 4-4: The atomic weight of boron is 10.811
amu. The masses of the two naturally occurring
isotopes 510B and 511B, are 10.013 and 11.009
amu, respectively. Calculate the fraction and
percentage of each isotope.
You do it!
• This problem requires a little algebra.
– A hint for this problem is x + (1-x) = 1
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The Atomic Weight Scale and
Atomic Weights
Example 4-4: The atomic weight of boron is 10.811
amu. The masses of the two naturally occurring
isotopes 510B and 511B, are 10.013 and 11.009
amu, respectively. Calculate the fraction and
percentage of each isotope.
10.811amu  x(10.013amu)  1  x (11.009amu)

 

10
B isotope
11
B isotope
 10.013x  11.009- 11.009x  amu
10.811-11.009 amu  10.013x - 11.009x  amu
- 0.198 -0.996x
0.199  x
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The Atomic Weight Scale and
Atomic Weights
• Note that because x is the multiplier for the
10B isotope, our solution gives us the
fraction of natural B that is 10B.
• Fraction of 10B = 0.199 and % abundance
of 10B = 19.9%.
• The multiplier for 11B is (1-x) thus the
fraction of 11B is 1-0.199 = 0.801 and the
% abundance of 11B is 80.1%.
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The Periodic Table: Metals, Nonmetals,
and Metalloids
• 1869 - Mendeleev & Meyer
– Discovered the periodic law
• The properties of the elements are periodic
functions of their atomic numbers.
37
The Periodic Table: Metals, Nonmetals,
and Metalloids
• Groups or families
– Vertical group of elements on periodic table
– Similar chemical and physical properties
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The Periodic Table: Metals, Nonmetals,
and Metalloids
• Period
– Horizontal group of elements on periodic table
– Transition from metals to nonmetals
39
The Periodic Table: Metals, Nonmetals,
and Metalloids
•
Some chemical properties of metals
1.
2.
3.
4.
Outer shells contain few electrons
Form cations by losing electrons
Form ionic compounds with nonmetals
Solid state characterized by metallic bonding
40
The Periodic Table: Metals, Nonmetals,
and Metalloids
• Group IA metals
– Li, Na, K, Rb, Cs, Fr
• One example of a
periodic trend
– The reactions with
water of Li
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The Periodic Table: Metals, Nonmetals,
and Metalloids
• Group IA metals
– Li, Na, K, Rb, Cs, Fr
• One example of a
periodic trend
– The reactions with
water of Li, Na
42
The Periodic Table: Metals, Nonmetals,
and Metalloids
• Group IA metals
– Li, Na, K, Rb, Cs, Fr
• One example of a
periodic trend
– The reactions with
water of Li, Na, & K
43
The Periodic Table: Metals, Nonmetals,
and Metalloids
• Group IIA metals
– alkaline earth metals
• Be, Mg, Ca, Sr, Ba, Ra
44
The Periodic Table: Metals, Nonmetals,
and Metalloids
•
Some chemical properties of nonmetals
1.
2.
3.
Outer shells contain four or more electrons
Form anions by gaining electrons
Form ionic compounds with metals and covalent
compounds with other nonmetals
Form covalently bonded molecules; noble gases are
monatomic
4.
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The Periodic Table: Metals, Nonmetals,
and Metalloids
• Group VIIA nonmetals
– halogens
– F, Cl, Br, I, At
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The Periodic Table: Metals, Nonmetals,
and Metalloids
• Group VIA nonmetals
– O, S, Se, Te
47
The Periodic Table: Metals, Nonmetals,
and Metalloids
• Group 0 nonmetals
– noble, inert or rare gases
– He, Ne, Ar, Kr, Xe, Rn
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The Periodic Table: Metals, Nonmetals,
and Metalloids
• Stair step function on periodic table separates
metals from nonmetals.
• Metals are to the left of
stair step.
– Approximately 80% of the
elements
• Best metals are on the far
left of the table.
49
The Periodic Table: Metals, Nonmetals,
and Metalloids
• Stair step function on periodic table separates
metals from nonmetals.
• Nonmetals are to the right
of stair step.
– Approximately 20% of the
elements
• Best nonmetals are on the
far right of the table.
50
The Periodic Table: Metals, Nonmetals,
and Metalloids
• Stair step function on periodic table separates
metals from nonmetals.
• Metalloids have one side
of the box on the stair
step.
51
The Periodic Table: Metals, Nonmetals,
and Metalloids
• Periodic trends in metallic character
52
Electromagnetic Radiation
• The wavelength of electromagnetic radiation
has the symbol λ .
• Wavelength is the distance from the top (crest)
of one wave to the top of the next wave.
– Measured in units of distance such as m,cm, Å.
– 1 Å = 1 x 10-10 m = 1 x 10-8 cm
• The frequency of electromagnetic radiation has
the symbol .
• Frequency is the number of crests or troughs
that pass a given point per second.
– Measured in units of 1/time - s-1
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Electromagnetic Radiation
• The relationship between wavelength and
frequency for any wave is velocity = λ υ.
• For electromagnetic radiation the velocity is 3.00
x 108 m/s and has the symbol c.
• Thus c = λ υ for electromagnetic radiation.
54
Electromagnetic Radiation
55
Electromagnetic Radiation
56
Electromagnetic Radiation
• Molecules interact with electromagnetic
radiation.
– Molecules can absorb and emit light.
• Once a molecule has absorbed light
(energy), the molecule can:
1. Rotate
2. Translate
3. Vibrate
4. Electronic transition
57
Electromagnetic Radiation
• For water:
– Rotations occur in the microwave portion of spectrum.
– Vibrations occur in the infrared portion of spectrum.
– Translation occurs across the spectrum.
– Electronic transitions occur in the ultraviolet portion of
spectrum.
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Electromagnetic Radiation
Example 4-5: What is the frequency of green light
of wavelength 5200 Å?
59
Electromagnetic Radiation
Example 4-5: What is the frequency of green light
of wavelength 5200 Å?
c     
c

 1 x 10-10 m 
  5.200 10-7 m
(5200Å) 
 1Å

3.00 108 m/s

5.200 10-7 m
  5.77 1014 s -1
60
Electromagnetic Radiation
• In 1900 Max Planck studied black body
radiation and realized that to explain the
energy spectrum he had to assume that:
1. energy is quantized
2. light has particle character
• Planck’s equation is
E  h  or E 
hc

h  P lanck’s constant 6.626x 10
-34
J s
61
Electromagnetic Radiation
Example 4-6: What is the energy of a photon of
green light with wavelength 5200 Å? What is the
energy of 1.00 mol of these photons?
62
Electromagnetic Radiation
Example 4-6: What is the energy of a photon of
green light with wavelength 5200 Å? What is the
energy of 1.00 mol of these photons?
From Example 4-5, we know that   5.77 x 1014 s-1
E  h
E  (6.626  10-34 J  s)(5.77  1014 s -1 )
E  3.83  10-19 J per photon
For1.00molof photons:
(6.022 1023 photons)(3
.83 10-19 J per photon) 231kJ/mol
63
The Photoelectric Effect
Light can strike the surface of some metals
causing an electron to be ejected.
64
The Photoelectric Effect
• What are some practical uses of the
photoelectric effect?
You do it!
•
•
•
•
Electronic door openers
Light switches for street lights
Exposure meters for cameras
Albert Einstein explained the photoelectric effect
– Explanation involved light having particle-like
behavior.
– Einstein won the 1921 Nobel Prize in Physics for this
65
work.
Atomic Spectra and the Bohr Atom
An emission spectrum is formed by an
electric current passing through a gas in a
vacuum tube (at very low pressure) which
causes the gas to emit light.
– Sometimes called a bright line spectrum.
66
Atomic Spectra and the Bohr Atom
67
Atomic Spectra and the Bohr Atom
• An absorption spectrum is formed by
shining a beam of white light through a
sample of gas.
– Absorption spectra indicate the wavelengths of
light that have been absorbed.
68
Atomic Spectra and the Bohr Atom
• Every element has a unique spectrum.
• Thus we can use spectra to identify
elements.
– This can be done in the lab, stars,
fireworks, etc.
69
Atomic Spectra and the Bohr Atom
• Atomic and molecular spectra are
important indicators of the underlying
structure of the species.
• In the early 20th century several eminent
scientists began to understand this
underlying structure.
– Included in this list are:
– Niels Bohr
– Erwin Schrodinger
– Werner Heisenberg
70
Atomic Spectra and the Bohr Atom
Example 4-7: An orange line of wavelength 5890
Å is observed in the emission spectrum of
sodium. What is the energy of one photon of this
orange light?
You do it!
71
Atomic Spectra and the Bohr Atom
Example 4-7: An orange line of wavelength 5890
Å is observed in the emission spectrum of
sodium. What is the energy of one photon of this
orange light?
 1  10 -10 m 
  5.890  10 7 m
  5890 Å 
Å


E  h 

6.626  10

34

hc

J  s 3.00  10 m/s
7
5.890  10 m
19
 3.375  10 J
8

72
Atomic Spectra and the Bohr Atom
• The Rydberg
equation is an
empirical
equation that
relates the
wavelengths of
the lines in the
hydrogen
spectrum.
 1
1 
 R  2  2 

 n1 n 2 
R is theRydberg constant
1
R  1.097 107 m -1
n1  n 2
n’s refer tothenumbers
of theenergy levelsin the
emission spectrumof hydrogen
73
Atomic Spectra and the Bohr Atom
Example 4-8: What is
the wavelength of light
emitted when the
hydrogen atom’s energy
changes from n = 4 to n
= 2?
74
Atomic Spectra and the Bohr Atom
Example 4-8. What is
n 2  4 and n1  2
the wavelength of light 1
 1
1 
 R 2  2 
emitted when the

 n1 n 2 
hydrogen atom’s energy 1
 1 1 
 1.097  107 m-1  2  2 
changes from n = 4 to n 
2 4 
= 2?
1
1
7
-1  1
 1.097  10 m



4
16



1

1

1

 1.097  107 m-1  0.250  0.0625 
 1.097  107 m-1  0.1875 
 2.057  106 m-1
75
Atomic Spectra and the Bohr Atom
Notice that the wavelength calculated from
the Rydberg equation matches the wavelength
of the green colored line in the H spectrum.
76
Atomic Spectra and the Bohr Atom
• In 1913 Neils Bohr incorporated Planck’s
quantum theory into the hydrogen
spectrum explanation.
• Here are the postulates of Bohr’s theory.
1.
Atom has a number of definite and discrete
energy levels (orbits) in which an electron
may exist without emitting or absorbing
electromagnetic radiation.
As the orbital radius increases so does the energy
1<2<3<4<5......
77
Atomic Spectra and the Bohr Atom
2. An electron may move from one discrete
energy level (orbit) to another, but, in so
doing, monochromatic radiation is
emitted or absorbed in accordance with
the following equation.
E 2 - E1  E  h 
hc

E 2  E1
Energy is absorbed when electrons jump to higher orbits.
n = 2 to n = 4 for example
Energy is emitted when electrons fall to lower orbits.
n = 4 to n = 1 for example
78
Atomic Spectra and the Bohr Atom
3. An electron moves in a circular orbit about
the nucleus and it motion is governed by the
ordinary laws of mechanics and
electrostatics, with the restriction that the
angular momentum of the electron is
quantized (can only have certain discrete
values).
angular momentum = mvr = nh/2π
h = Planck’s constant n = 1,2,3,4,...(energy
levels)
v = velocity of electron m = mass of electron
r = radius of orbit
79
Atomic Spectra and the Bohr Atom
• Light of a characteristic wavelength (and
frequency) is emitted when electrons move from
higher E (orbit, n = 4) to lower E (orbit, n = 1).
– This is the origin of emission spectra.
• Light of a characteristic wavelength (and
frequency) is absorbed when electrons jump
from lower E (orbit, n = 2) to higher E (orbit, n=
4)
– This is the origin of absorption spectra.
80
Atomic Spectra and the Bohr Atom
81
Atomic Spectra and the Bohr Atom
• Bohr’s theory correctly explains the H
emission spectrum.
• The theory fails for all other elements
because it is not an adequate theory.
82
The Wave Nature of the Electron
In 1925 Louis de Broglie published his Ph.D.
dissertation.
– A crucial element of his dissertation is that electrons
have wave-like properties.
– The electron wavelengths are described by the de
Broglie relationship.
h

mv
h  P lanck’s const ant
m  mass of part icle
v  velocit yof part icle
83
The Wave Nature of the Electron
84
The Wave Nature of the Electron
• De Broglie’s assertion was verified by
Davisson & Germer within two years.
• Consequently, we now know that electrons
(in fact - all particles) have both a particle
and a wave like character.
– This wave-particle duality is a fundamental
property of submicroscopic particles.
85
The Wave Nature of the Electron
Example 4-9: Determine the wavelength, in m, of an
electron, with mass 9.11 x 10-31 kg, having a velocity
of 5.65 x 107 m/s.
– Remember Planck’s constant is 6.626 x 10-34 Js which
is also equal to 6.626 x 10-34 kg m2/s.
86
The Wave Nature of the Electron
Example 4-9. Determine the wavelength, in m, of an
electron, with mass 9.11 x 10-31 kg, having a velocity
of 5.65 x 107 m/s.
– Remember Planck’s constant is 6.626 x 10-34 Js which
is also equal to 6.626 x 10-34 kg m2/s.
h

mv
6.6261034 kg m 2 / s 2 s

-31
7
9.1110 kg 5.6510 m/s




  1.2910
11
m

87
The Wave Nature of the Electron
Example 4-10: Determine the wavelength, in m, of a
0.22 caliber bullet, with mass 3.89 x 10-3 kg, having
a velocity of 395 m/s, ~ 1300 ft/s.
You do it!
88
The Wave Nature of the Electron
Example 4-10. Determine the wavelength, in m, of a
0.22 caliber bullet, with mass 3.89 x 10-3 kg, having
a velocity of 395 m/s, ~ 1300 ft/s.
You do it!
h

mv
34
2
2
6.62610 kg m / s s

-3
3.8910 kg 395 m/s




  4.3110
34
m
89
The Quantum Mechanical
Picture of the Atom
• Werner Heisenberg in 1927 developed the
concept of the Uncertainty Principle.
• It is impossible to determine
simultaneously both the position and
momentum of an electron (or any other
small particle).
– Detecting an electron requires the use of
electromagnetic radiation which displaces the
electron!
• Electron microscopes use this phenomenon
90
The Quantum Mechanical
Picture of the Atom
• Consequently, we
must speak of the
electrons’ position
about the atom in
terms of probability
functions.
• These probability
functions are
represented as
orbitals in quantum
mechanics.
91
The Quantum Mechanical
Picture of the Atom
Basic Postulates of Quantum Theory
1. Atoms and molecules can exist only in
certain 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).
92
The Quantum Mechanical
Picture of the Atom
2. Atoms or molecules emit or absorb
radiation (light) as they change their
energies. The frequency of the light
emitted or absorbed is related to the
energy change by a simple equation.
E  h 
hc

93
The Quantum Mechanical
Picture of the Atom
3. The allowed energy states of atoms and
molecules can be described by sets of
numbers called quantum numbers.
•
•
Quantum numbers are the solutions of the
Schrodinger, Heisenberg & Dirac equations.
Four quantum numbers are necessary to
describe energy states of electrons in atoms.
..
Schr o dinger equation
b2   2  2  2 
 2  2  2  2   V  E
8 m   x  y  z 
94
Quantum Numbers
• The principal quantum number has the
symbol n.
n = 1, 2, 3, 4, ...... “shells”
n = K, L, M, N, ......
The electron’s energy
depends principally on n .
95
Quantum Numbers
• The angular momentum quantum number
has the symbol .
 = 0, 1, 2, 3, 4, 5, .......(n-1)
 = s, p, d, f, g, h, .......(n-1)
•  tells us the shape of the orbitals.
• These orbitals are the volume around the
atom that the electrons occupy 90-95% of
the time.
This is one of the places where Heisenberg’s
Uncertainty principle comes into play.
96
Quantum Numbers
• The symbol for the magnetic quantum number
is m.
m = -  , (-  + 1), (-  +2), .....0, ......., ( -2), ( -1), 
• If  = 0 (or an s orbital), then m = 0.
– Notice that there is only 1 value of m.
This implies that there is one s orbital per n value. n  1
• If  = 1 (or a p orbital), then m = -1,0,+1.
– There are 3 values of m.
Thus there are three p orbitals per n value. n  2
97
Quantum Numbers
• If ℓ = 2 (or a d orbital), then mℓ = -2,-1,0,+1,+2.
– There are 5 values of m.
Thus there are five d orbitals per n value. n  3
• If ℓ = 3 (or an f orbital), then
mℓ = -3,-2,-1,0,+1,+2, +3.
– There are 7 values of mℓ.
Thus there are seven f orbitals per n value, n ≥ 4
• Theoretically, this series continues on to g,h,i,
etc. orbitals.
– Atoms that have been discovered or made up to
this point in time only have electrons in s, p, d, or f
orbitals in their ground state configurations.
98
Quantum Numbers
• The last quantum number is the spin
quantum number, ms.
• The spin quantum number only has two
possible values.
ms = +1/2 or -1/2
• This quantum number tells us the spin and
orientation of the magnetic field of the
electrons.
• Wolfgang Pauli in 1925 discovered the
Exclusion Principle.
– No two electrons in an atom can have the same
set of 4 quantum numbers.
99
Atomic Orbitals
• Atomic orbitals are regions of space
where the probability of finding an
electron about an atom is highest.
• s orbital properties:
– There is one s orbital per n level.
–=0
1 value of m
100
Atomic Orbitals
• s orbitals are spherically
symmetric.
101
Atomic Orbitals
• p orbital properties:
– The first p orbitals appear in the n = 2 shell.
• p orbitals are peanut or dumbbell shaped
volumes.
– They are directed along the axes of a Cartesian
coordinate system.
• There are 3 p orbitals per n level.
– The three orbitals are named px, py, pz.
– They have an  = 1.
– m = -1,0,+1 3 values of m
102
Atomic Orbitals
• p orbitals are peanut or dumbbell shaped.
103
Atomic 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 or
are rotated 45o from the axes.
• 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 
104
Atomic Orbitals
• d orbital shapes
105
Atomic Orbitals
• f orbital properties:
– The first f orbitals appear in the n = 4 shell.
• The f orbitals have the most complex
shapes.
• There are seven f orbitals per n level.
– The f orbitals have complicated names.
– They have an ℓ = 3
– mℓ = -3,-2,-1,0,+1,+2, +3
7 values of mℓ
– The f orbitals have important effects in the
lanthanide and actinide elements.
106
Atomic Orbitals
• f orbital shapes
107
Atomic Orbitals
• Spin quantum number effects:
– Every orbital can hold up to two electrons.
• Consequence of the Pauli Exclusion Principle.
– The two electrons are designated as having
– one spin up and one spin down
• Spin describes the direction of the
electron’s magnetic fields.
108
Atomic Orbitals
109
Paramagnetism and Diamagnetism
• Unpaired electrons have their spins
aligned or
• This increases the magnetic field of the
atom.
• Atoms with unpaired electrons are called
paramagnetic .
– Paramagnetic atoms are attracted to a
magnet.
110
Paramagnetism and Diamagnetism
• Paired electrons have their spins unaligned
.
– Paired electrons have no net magnetic field.
• Atoms with paired electrons are called
diamagnetic.
– Diamagnetic atoms are repelled by a magnet.
111
Paramagnetism and Diamagnetism
112
Paramagnetism and Diamagnetism
• Because two electrons in the same orbital
must be paired, it is possible to calculate
the number of orbitals and the number of
electrons in each n shell.
• The number of orbitals per n level is given
by n2.
• The maximum number of electrons per n
level is 2n2.
– The value is 2n2 because of the two paired
electrons.
113
Paramagnetism and Diamagnetism
Energy Level
n
# of Orbitals
n2
Max # of e2n2
1
1
2
2
4
8
3
9
18
4
16
32
114
The Periodic Table and
Electron Configurations
• The principle that describes how the
periodic chart is a function of electronic
configurations is the Aufbau Principle.
• The electron that distinguishes an
element from the previous element
enters the lowest energy atomic orbital
available.
115
The Periodic Table and
Electron Configurations
• The Aufbau Principle describes the electron
filling order in atoms.
116
The Periodic Table and
Electron Configurations
117
The Periodic Table and
Electron Configurations
• There are two ways to remember the correct
filling order for electrons in atoms.
1. You can use this mnemonic.
118
The Periodic Table and
Electron Configurations
2. Or you can use the periodic chart .
119
The Periodic Table and
Electron Configurations
• Now we will use the Aufbau Principle to
determine the electronic configurations of the
elements on the periodic chart.
• 1st row elements.
1s

1
H
2
He 
Configuration
1
1s
2
1s
120
The Periodic Table and
Electron Configurations
• 2nd row elements.
•Hund’s rule tells us that the electrons will fill the
p orbitals by placing electrons in each orbital
singly and with same spin until half-filled. Then
the electrons will pair to finish the p orbitals.
121
The Periodic Table and
Electron Configurations
• 3rd row elements
3s
3p
Configuration
Ne 
11 Na
12
Mg Ne 
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
Ne 
Ne 
Ne 
Ne 
Ne 
Ne 

 
  
  
  
  
Ne 3s1
Ne 3s 2
Ne 3s 2 3p1
Ne 3s2 3p22
Ne 3s2 3p3
Ne 3s2 3p4
Ne 3s2 3p55
Ne 3s2 3p6
122
The Periodic Table and
Electron Configurations
• 4th row elements
3d
19
K
Ar
4s

4p
Configuration
Ar 4s1
123
The Periodic Table and
Electron Configurations
3d
4s
19 K Ar 

20

Ca Ar
4p
Configuration
Ar 4s1
Ar 4s2
124
The Periodic Table and
Electron Configurations
3d
4s
19 K Ar 

20
Ca Ar

21
Sc You do it!
4p
Configuration
Ar 4s1
Ar 4s2
125
The Periodic Table and
Electron Configurations
3d
4s
19 K Ar

20
Ca Ar

Sc Ar 

21
4p
Configuration
Ar 4s1
Ar 4s2
Ar 4s2 3d1
126
The Periodic Table and
Electron Configurations
3d
19
20
21
4s
K Ar

Ca Ar

Sc Ar 

4p
Configuration
Ar 4s
Ar 4s2
Ar 4s2 3d1
1
T i You do it!
22
127
The Periodic Table and
Electron Configurations
3d
4s
19 K Ar 

20
Ca Ar

Sc Ar 

T i Ar  

21
22
4p
Configuration
Ar 4s1
Ar 4s2
2
1
Ar 4s 3d
Ar 4s2 3d2
128
The Periodic Table and
Electron Configurations
3d
4s
19 K Ar 

20
Ca Ar

Sc Ar 

22
T i Ar  

23
V Ar   

21
4p
Configuration
Ar 4s1
Ar 4s2
Ar 4s2 3d1
Ar 4s2 3d2
Ar 4s2 3d3
129
The Periodic Table and
Electron Configurations
3d
4s
19 K Ar 

20
Ca Ar

Sc Ar 

22
T i Ar  

23
V Ar   

Cr Ar     

21
24
4p
Configuration
Ar 4s1
Ar 4s2
Ar 4s2 3d1
Ar 4s2 3d2
Ar 4s2 3d3
Ar 4s1 3d5
T hereis an extrameasure of st abilityassociat ed
with half - filled and completelyfilled orbitals.
130
The Periodic Table and
Electron Configurations
3d
25 Mn Ar     
4s

4p
Configuration
Ar 4s2 3d5
131
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
4s

4p
Configuration
Ar 4s2 3d5
Fe You do it !
132
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
Fe Ar     
4s


4p
Configuration
Ar 4s2 3d5
Ar 4s2 3d6
133
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
27
4s

Fe Ar     

Co Ar     

4p
Configuration
Ar 4s2 3d5
2
6
Ar 4s 3d
Ar 4s2 3d7
134
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
27
28
4s

Fe Ar     

Co Ar     

Ni Ar     

4p
Configuration
Ar 4s2 3d5
Ar 4s2 3d6
Ar 4s2 3d7
Ar 4s2 3d8
135
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
4s

Fe Ar     

Co Ar     

28
Ni Ar     

29
Cu You do it!
26
27
4p
Configuration
Ar 4s2 3d5
Ar 4s2 3d6
Ar 4s2 3d7
Ar 4s2 3d8
136
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
27
28
29
4s
4p

Fe Ar     

Co Ar     

Ni Ar     

Cu Ar      
Configuration
Ar 4s2 3d5
Ar 4s2 3d6
Ar 4s2 3d7
Ar 4s2 3d8
Ar 4s1 3d10
Anot herexceptionlike Cr and
for essentially thesame reason.
137
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
27
28
29
30
4s

Fe Ar     

Co Ar     

Ni Ar     

Cu Ar      
Zn Ar      
4p
Configuration
Ar 4s2 3d5
Ar 4s2 3d6
Ar 4s2 3d7
Ar 4s2 3d8
Ar 4s1 3d10
Ar 4s2 3d10
138
The Periodic Table and
Electron Configurations
3d
4s
31 Ga Ar       
4p
Configuration
Ar 4s2 3d10 4p1
139
The Periodic Table and
Electron Configurations
3d
4s
31 Ga Ar        
32
4p
Configuration
Ar 4s2 3d10 4p1
Ge You do it !
140
The Periodic Table and
Electron Configurations
3d
4s
31 Ga Ar        
32
4p
Ge Ar        
Configuration
Ar 4s2 3d10 4p1
Ar 4s2 3d10 4p2
141
The Periodic Table and
Electron Configurations
3d
4s
4p
31 Ga Ar        
32
33
Ge Ar        
As Ar         
Configuration
Ar 4s2 3d10 4p1
Ar 4s2 3d10 4p2
Ar 4s2 3d10 4p3
142
The Periodic Table and
Electron Configurations
3d
4s
4p
31 Ga Ar       
32
Ge Ar        
33
As Ar         
34
Se You do it!
Configuration
Ar 4s2 3d10 4p1
Ar 4s2 3d10 4p2
Ar 4s2 3d10 4p3
143
The Periodic Table and
Electron Configurations
3d
4s
4p
31 Ga Ar        
32
33
34
Ge Ar        
As Ar         
Se Ar         
Configuration
Ar 4s2 3d10 4p1
Ar 4s2 3d10 4p2
Ar 4s2 3d10 4p3
Ar 4s2 3d10 4p4
144
The Periodic Table and
Electron Configurations
3d
4s
4p
Configuration
31 Ga Ar        
32
33
34
35
Ge Ar        
As Ar         
Se Ar         
Br Ar         
Ar 4s2 3d10 4p1
Ar 4s2 3d10 4p2
Ar 4s2 3d10 4p3
Ar 4s2 3d10 4p4
Ar 4s2 3d10 4p5
145
The Periodic Table and
Electron Configurations
3d
4s
4p
Configuration
31 Ga Ar       
Ar 4s2 3d10 4p1
2
10
2




Ge
Ar








Ar
4s
3d
4p
32
2
10
3




As
Ar









Ar
4s
3d
4p
33
2
10
4




Se
Ar









Ar
4s
3d
4p
34
2
10
5




Br
Ar









Ar
4s
3d
4p
35
2
10
6




Kr
Ar









Ar
4s
3d
4p
36
146
The Periodic Table and
Electron Configurations
• Now we can write a complete set of quantum
numbers for all of the electrons in these three
elements as examples.
– Na
– Ca
– Fe
• First for 11Na.
– When completed there must be one set of 4
quantum numbers for each of the 11 electrons in
(remember Ne has 10 electrons)
3s
11
Na Ne 
3p
Configuration
Ne 3s
1
147
The Periodic Table and
Electron Configurations
1st e-
n

m
1
0
0
ms
 1/2
148
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e- 1
0
0
1st e-
ms
 1/2
1 s elect rons
 1/2 
149
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e - 1
0
0
3rd e -
0
0
1st e -
2
ms
 1/2

1 s elect rons
 1/2 

 1/2
150
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e - 1
0
0
3rd e -
2
0
0
4 th e -
2
0
0
1st e -
ms
 1/2
1 s elect rons
 1/2 
 1/2
2 s elect rons
 1/2
151
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e - 1
0
0
3rd e -
2
0
0
4 th e -
2
0
0
5th e -
2
1
-1
1st e -
ms
 1/2
1 s elect rons
 1/2 
 1/2
2 s elect rons
 1/2
 1/2
152
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e - 1
0
0
3rd e -
2
0
0
4 th e -
2
0
0
5th e -
2
1
-1
 1/2
2 s elect rons
 1/2
 1/2
6 th e -
2
1
0
 1/2
1st e -
ms
 1/2
1 s elect rons
 1/2 
153
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e - 1
0
0
3rd e - 2
0
0
4 th e - 2
0
0
5th e - 2
1
-1
 1/2
2 s elect rons
 1/2
 1/2
6 th e - 2
1
0
 1/2
7 th e - 2
1
1
 1/2
1st e -
ms
 1/2
1 s elect rons
- 1/2 
154
The Periodic Table and
Electron Configuration
n

m
1
0
0
2 nd e - 1
0
0
3rd e -
2
0
0
4 th e -
2
0
0
5th e -
2
1
-1
 1/2
2 s elect rons
 1/2
 1/2
6 th e -
2
1
0
 1/2
7 th e -
2
1
1
 1/2
8th e -
2
1
1
 1/2
1st e -
ms
 1/2
1 s elect rons
 1/2 
155
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e- 1
0
0
3rd e- 2
0
0
4 th e- 2
0
0
5th e- 2
1
-1
 1/2
2 s elect rons
 1/2
 1/2
6 th e- 2
1
0
 1/2
7 th e- 2
1
1
 1/2
8th e- 2
1
1
 1/2
1
0
 1/2
1st e-
9 th e- 2
ms
 1/2
1 s elect rons
 1/2 
156
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e - 1
0
0
3rd e - 2
0
0
4 th e - 2
0
0
5th e- 2
1
6 th e - 2
1
7 th e - 2
1
8th e- 2
1
9 th e - 2
1
10th e - 2
1
 1/2 

0
 1/2 
 1  1/2

2 p elect rons
 1  1/2 
0
 1/2 

 1  1/2 

1st e -
ms
 1/2
1 s elect rons
 1/2 
 1/2
2 s elect rons
 1/2
-1
157
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e - 1
0
0
3rd e -
2
0
0
4 th e -
2
0
0
5th e-
2
1
6 th e -
2
1
7 th e -
2
1
8th e-
2
1
9 th e -
2
1
10th e -
2
1
11th e -
3
0
1st e -
ms
 1/2
1 s elect rons
 1/2 
 1/2
2 s elect rons
 1/2
 1/2 

0
 1/2 
 1  1/2

2 p elect rons
 1  1/2 
0
 1/2 

 1  1/2 

0
 1/23 s elect ron
-1
158
The Periodic Table and
Electron Configurations
• Next we will do the same exercise for 20Ca.
– Again, when finished we must have one set of 4
quantum numbers for each of the 20 electrons in
Ca.
• We represent the first 18 electrons in Ca with
the symbol [Ar].
3d
20 Ca [Ar]
4s

4p
Configuration
Ar 4s2
159
The Periodic Table and
Electron Configurations
n

m
ms
[Ar] 19th e - 4
0
0
 1/2
160
The Periodic Table and
Electron Configurations
n

m
[Ar]19th e -
4
0
0
20th e -
4
0
0
ms
 1/2 

4 s elect rons
 1/2

161
The Periodic Table and
Electron Configurations
• Finally, we do the same exercise for 26Fe.
– We should have one set of 4 quantum numbers for
each of the 26 electrons in Fe.
• To save time and space, we use the symbol
[Ar] to represent the first 18 electrons in Fe
3d
26 Fe Ar     
4s

4p
Configuration
Ar 4s2 3d6
162
The Periodic Table and
Electron Configurations
n

m
ms
[Ar] 19th e - 4
0
0
 1/2
163
The Periodic Table and
Electron Configurations
n

m
[Ar]19th e -
4
0
0
20th e -
4
0
0
ms
 1/2 

4 s elect rons
 1/2

164
The Periodic Table and
Electron Configurations
n

[Ar] 19th e -
4
0
0
20th e -
4
0
0
21st e -
3
2
-2
m
ms
 1/2

4 s elect rons
 1/2

 1/2
165
The Periodic Table and
Electron Configurations
n

m
[Ar] 19th e -
4
0
0
20th e -
4
0
21st e -
3
2
22nd e -
You do it !
ms
 1/2 

4 s elect rons
0
 1/2

- 2  1/2
166
The Periodic Table and
Electron Configurations
n

m
ms
[Ar] 19th e -
4
0
0
20th e -
4
0
0
21st e -
3
2
-2
 1/2

4 s elect rons
 1/2

 1/2
22nd e -
3
2
-1
 1/2
167
The Periodic Table and
Electron Configurations
n

m
ms
[Ar] 19th e -
4
0
0
20th e -
4
0
0
21st e -
3
2
-2
 1/2

4 s elect rons
 1/2

 1/2
22nd e -
3
2
-1
 1/2
23rd e -
3
2
0
 1/2
168
The Periodic Table and
Electron Configurations
n

m
[Ar] 19th e -
4
0
0
20th e -
4
0
0
21st e -
3
2
-2
 1/2

4 s elect rons
 1/2

 1/2
22nd e -
3
2
-1
 1/2
23rd e -
3
2
0
 1/2
24th e -
3
2
 1  1/2
ms
169
The Periodic Table and
Electron Configurations
n

[Ar] 19th e -
4
0
20th e -
4
0
21st e -
3
2
22nd e -
3
2
23rd e -
3
2
24th e -
3
2
25th e -
3
2
m
ms
 1/2

4 s elect rons
0
 1/2 

- 2  1/2

- 1  1/2 

0  1/2 half - filled d shell
 1  1/2 

 2  1/2

0
170
The Periodic Table and
Electron Configurations
n

m
[Ar] 19th e -
4
0
0
20th e -
4
0
0
21st e -
3
2
-2
 1/2

4 s elect rons
 1/2

 1/2
22nd e -
3
2
-1
 1/2
23rd e -
3
2
0
 1/2
24th e -
3
2
1
 1/2
25th e -
3
2
2
 1/2
26th e -
You do it !
ms
171
The Periodic Table and
Electron Configurations
n

m
[Ar] 19th e -
4
0
0
20th e -
4
0
0
21st e -
3
2
-2
 1/2

4 s elect rons
 1/2

 1/2
22nd e -
3
2
-1
 1/2
23rd e -
3
2
0
 1/2
24th e -
3
2
1
 1/2
25th e -
3
2
2
 1/2
26th e -
3
2
-2
 1/2
ms
172
Synthesis Question
• What is the atomic number of the element
that should theoretically be the noble gas
below Rn?
• The 6 d’s are completed with element 112
and the 7p’s are completed with element
118. Thus the next noble gas (or perhaps
it will be a noble liquid) should be element
118.
173
Group Question
• In a universe different from ours, the laws
of quantum mechanics are the same as
ours with one small change. Electrons in
this universe have three spin states, -1, 0,
and +1, rather than the two, +1/2 and -1/2,
that we have. What two elements in this
universe would be the first and second
noble gases? (Assume that the elements
in this different universe have the same
174
symbols as in ours.)