CHAPTER 5

1
The Structure of Atoms
Chapter Outline
1.
2.
3.
4.
5.
6.
7.
8.
2
Subatomic Particles
Fundamental Particles
The Discovery of Electrons
Canal Rays and Protons
Rutherford and the Nuclear Atom
Atomic Number
Neutrons
Mass Number and Isotopes
Mass spectrometry and Isotopic Abundance
Chapter Goals
9.
10.
11.
12.
13.
14.
3
The Atomic Weight Scale and Atomic Weights
The Electronic Structures of Atoms
Electromagnetic radiation
The Photoelectric Effect
Atomic Spectra and the Bohr Atom
The Wave Nature of the Electron
The Quantum Mechanical Picture of the Atom
Chapter Goals
15.
16.
17.
18.
19.
4
Quantum Numbers
Atomic Orbitals
Electron Configurations
Paramagnetism and Diamagnetism
The Periodic Table and Electron Configurations
Fundamental Particles

Three fundamental particles make up atoms. The following
table lists these particles together with their masses and their
charges.
Particle
5
Mass (amu) Charge
Electron (e-)
0.00054858
-1
Proton (p,p+)
1.0073
+1
Neutron(n,n0)
1.0087
0
The Discovery of Electrons

Humphrey Davy in the early 1800’s passed
electricity through compounds and noted:
–
–

6
that the compounds decomposed into elements.
Concluded that 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.
The Discovery of Electrons

Cathode Ray Tubes experiments performed in the late
1800’s & early 1900’s.
–
–
7
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.
The Discovery of Electrons

These “rays” are emitted from cathode (- end)
and travel to anode (+ end).
–

J.J. Thomson modified the cathode ray tube
experiments in 1897 by adding two adjustable
voltage electrodes.
–
8
Cathode Rays must be negatively charged!
Studied the amount that the cathode ray beam was
deflected by additional electric field.
The Discovery of Electrons

9
Modifications to the basic cathode ray tube
experiment.
The Discovery of Electrons

Thomson used his modification to measure the
charge to mass ratio of electrons.
Charge to mass ratio
e/m = -1.75881 x 108 coulomb/g of e-



10
Thomson named the cathode rays electrons.
Thomson is considered to be the “discoverer of
electrons”.
TV sets and computer screens are cathode ray
tubes.
The Discovery of Electrons


11
Robert A. Millikan won the 1st American Nobel
Prize in 1923 for his famous oil-drop experiment.
In 1909 Millikan determined the charge and
mass of the electron.
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.
–
–
–
12
e/m = -1.75881 x 108 coulomb
e = -1.60218 x 10-19 coulomb
Thus m = 9.10940 x 10-28 g
Canal Rays and Protons


13
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”.
Rutherford and the Nuclear Atom

Ernest Rutherford directed Hans Geiger and
Ernst Marsden’s experiment in 1910.
–
–
14
- particle scattering from thin Au foils
Gave us the basic picture of the atom’s structure.
Rutherford and the Nuclear Atom

In 1912 Rutherford decoded the -particle
scattering information.
–
15
Explanation involved a nuclear atom with electrons
surrounding the nucleus .
Rutherford and the Nuclear Atom

Rutherford’s major conclusions from the -particle
scattering experiment
1.
2.
3.
4.
16
The atom is mostly empty space.
It contains a very small, dense center called the nucleus.
Nearly all of the atom’s mass is in the nucleus.
The nuclear diameter is 1/10,000 to 1/100,000 times less
than atom’s radius.
Rutherford and the Nuclear Atom

Because the atom’s mass is contained in such
a small volume:
–
–
–
17
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.
Atomic Number

The atomic number is equal to the number of protons
in the nucleus.
–
–

In 1913 H.G.J. Moseley realized that the atomic
number determines the element .
–
–
18
Sometimes given the symbol Z.
On the periodic chart Z is the uppermost number in each
element’s box.
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.
Neutrons


James Chadwick in 1932 analyzed the results
of -particle scattering on thin Be films.
Chadwick recognized existence of massive
neutral particles which he called neutrons.
–
19
Chadwick discovered the neutron.
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
197
Z
6
20
79
 Can be shortened to this symbolism.

E for example
14
20
63
N, Cu,
C, Ca,
107
Ag, etc.
Au
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.

2H
or deuterium is the second most abundant hydrogen isotope.

3H
one proton and one neutron
or tritium is a radioactive hydrogen isotope.

21
one proton and no neutrons
one proton and two neutrons
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 protons and 8 neutrons
 17O is the least abundant stable O isotope.
 How many protons and neutrons are in 17O?
8 protons and 9 neutrons
 18O is the second most abundant stable O isotope.
How many protons and neutrons in 18O?
8 protons and 10 neutrons
•
22
Mass Spectrometry and
Isotopic Abundances

Francis Aston devised the first mass spectrometer.
–
–
23
Device generates ions that pass down an evacuated path inside
a magnet.
Ions are separated based on their mass.
Mass Spectrometry and
Isotopic Abundances

There are four factors which determine a
particle’s path in the mass spectrometer.
1
2
3
4
24
accelerating voltage
magnetic field strength
masses of particles
charge on particles
Mass Spectrometry and
Isotopic Abundances

Mass spectrum of Ne+ ions shown below.
–
25
How scientists determine the masses and
abundances of the isotopes of an element.
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.
–
–

Example 5-1: Calculate the number of atomic mass
units in one gram.
–
–
26
1 amu = (1/12) mass of 12C by definition
What is the mass of an amu in grams?
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.
The Atomic Weight Scale and
Atomic Weights
 6.022  10 23 31P atoms
(1.000 g) 
31
30.99376
g
P

27



The Atomic Weight Scale and
Atomic Weights
 6.022  10 23 31P atoms 

(1.000 g) 
31
30.99376 g P


 30.99376 amu 
23
31
 31
  6.022  10 amu P
P atom 

–
–
28
Thus 1.00 g = 6.022 x 1023 amu.
This is always true and provides the conversion factor
between grams and amu.
The Atomic Weight Scale and
Atomic Weights


29
The atomic weight of an element is the
weighted average of the masses of its stable
isotopes
Example 5-2: 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.
The Atomic Weight Scale and
Atomic Weights
atomic weight  (0.691)(62.9 amu)



63
30
Cu isotope
The Atomic Weight Scale and
Atomic Weights
atomic weight  (0.691)(62 .9 amu)  (0.309)(64 .9 amu)


 


63
31
Cu isotope
65
Cu isotope
The Atomic Weight Scale and
Atomic Weights
atomic weight  (0.691)(62 .9 amu)  (0.309)(64 .9 amu)


 


63
Cu isotope
atomic weight  63.5 amu for copper
32
65
Cu isotope
The Atomic Weight Scale and
Atomic Weights

33
Example 5-3: 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% 2454Cr, mass = 53.939 amu.
Calculate the atomic weight of chromium.
You do it!
The Atomic Weight Scale and
Atomic Weights
atomic weight  (0.0431 49.946 amu)  (0.8376  51.941 amu)
 (0.0955  52.941 amu)  (0.0238  53.939 amu)
 2.153  43.506  5.056  1.284 amu
 51.998 amu
34
The Atomic Weight Scale and
Atomic Weights


Example 5-4: 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.
–
35
A hint for this problem is x + (1-x) = 1
The Atomic Weight Scale and
Atomic Weights
10.811 amu  x(10.013 amu)  1  x (11.009 amu)

 

10
B isotope
11
B isotope
 10.013 x  11.009 - 11.009 x  amu
10.811 - 11.009 amu  10.013 x - 11.009 x  amu
- 0.198  -0.996 x
0.199  x
36
The Atomic Weight Scale and
Atomic Weights



37
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%.
The Electronic Structures of Atoms
Electromagnetic Radiation


The wavelength of electromagnetic radiation has the
symbol .
Wavelength is the distance from the top (crest) of one
wave to the top of the next wave.
–
–


The frequency of electromagnetic radiation has the
symbol .
Frequency is the number of crests or troughs that pass
a given point per second.
–
38
Measured in units of distance such as m,cm, Å.
1 Å = 1 x 10-10 m = 1 x 10-8 cm
Measured in units of 1/time - s-1
Electromagnetic Radiation



39
The relationship between wavelength and frequency for
any wave is velocity = .
For electromagnetic radiation the velocity is 3.00 x 108
m/s and has the symbol c.
Thus c =  for electromagnetic radiation.
Electromagnetic Radiation

Molecules interact with electromagnetic
radiation.
–

Once a molecule has absorbed light (energy),
the molecule can:
1.
2.
3.
4.
40
Molecules can absorb and emit light.
Rotate
Translate
Vibrate
Electronic transition
Electromagnetic Radiation

For water:
–
–
–
–
41
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.
Electromagnetic Radiation

Example 5-5: What is the frequency of green
light of wavelength 5200 Å?
c     
c

 1 x 10-10 m 
  5.200  10-7 m
(5200 Å) 
 1Å

42
3.00  108 m/s

5.200  10-7 m
  5.77  1014 s -1
Electromagnetic Radiation

In 1900 Max Planck studied black body
radiation and realized that to explain the
energy spectrum he had to assume that:
1.
2.

energy is quantized
light has particle character
Planck’s equation is
E  h  or E 
43
hc

h  Planck’ s constant  6.626 x 10-34 J  s
Electromagnetic Radiation

Example 5-6: 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 5 - 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
For 1.00 mol of photons :
44
(6.022  1023 photons)(3 .83  10-19 J per photon)  231 kJ/mol
The Photoelectric Effect

45
Light can strike the surface of some metals
causing an electron to be ejected.
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
–
–
46
Explanation involved light having particle-like behavior.
Einstein won the 1921 Nobel Prize in Physics for this 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.
–
47
Sometimes called a bright line spectrum.
Atomic Spectra and the Bohr Atom

An absorption spectrum is formed by shining
a beam of white light through a sample of gas.
–
48
Absorption spectra indicate the wavelengths of light that
have been absorbed.
Atomic Spectra and the Bohr Atom


Every element has a unique spectrum.
Thus we can use spectra to identify elements.
–
49
This can be done in the lab, stars, fireworks, etc.
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.
–
–
–
–
50
Included in this list are:
Niels Bohr
Erwin Schrodinger
Werner Heisenberg
Atomic Spectra and the Bohr Atom
-10
Example 5-7: 1An
10orange
m  line of wavelength
  5.890 10 7 m

 5890 Å 
5890 Å is -10
observed
emission
spectrum of
-10
Å in1the

710 m 


7
1

10
m


sodium.
What
is
the
energy
of
one
photon
of


5890
Å

5
.
890

10
m
  5.890 10
  5890 Å 

hc m
Å
Å light?


 E  h 
this orange

hc
You do it!
hc
E  h 
E  h 


8
34
8


6.626 10 34 J  s 3.00

10
m/s
6.626 10 J  s 3.00  10 m/s

 7
7
5.890 10 m
5.890 10 m
19
 3.375 10 J


51


Atomic Spectra and the Bohr Atom

52
The Rydberg
 1
1
1 
 R  2  2 
equation is an

n1 n 2 

empirical equation
R is the Rydberg constant
that relates the
wavelengths of the
R  1.097  107 m -1
lines in the
n

n
1
2
hydrogen spectrum.
n’ s refer to the numbers
of the energy levels in the
emission spectrum of hydrogen
Atomic Spectra and the Bohr Atom

nthe
2wavelength
4 and n1 of2light

4
and
n

2
Examplenn5-8.
What
is
2

4
and
n

2
1
2
1
emitted when the hydrogen atom’s energy




1
1
1
1
1
1


changes from
1  Rn= 14 to n1=2?
R  

2 
2  2

R

 nn12 nn22   n1
2 
 1
53

n 
1 7 -1  1 71 -1  1 1 
1
1.097
 1.097  10 m
 2m  2  2 
 2 10


2 4  2 4 
1
1
7
-1  1
 1.097  10 m   

 4 16 
2
2
Atomic Spectra and the Bohr Atom


Notice
calculated from
1 that the wavelength
7
1
7
-1

1.097

10
m
0.250
0.0625
the Rydberg equation matches
thewavelength
 green colored line in the H spectrum.
of the
1

 1.097  10 m 0.1875
77
1
54
--11
 2.057  10 m
66
--11

-7
  4.862 10 m
Atomic Spectra and the Bohr Atom


1.
In 1913 Neils Bohr incorporated Planck’s
quantum theory into the hydrogen spectrum
explanation.
Here are the postulates of Bohr’s theory.
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
55
1<2<3<4<5......
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
56
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
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/2
h = Planck’s constant n = 1,2,3,4,...(energy levels)
v = velocity of electron m = mass of electron
r = radius of orbit
57
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).
–

58
This is the origin of emission spectra.
Light of a characteristic wavelength (and
frequency) is absorbed when electron jumps from
lower E (orbit, n = 2) to higher E (orbit, n= 4)
– This is the origin of absorption spectra.
Atomic Spectra and the Bohr Atom


59
Bohr’s theory correctly explains the H
emission spectrum.
The theory fails for all other elements
because it is not an adequate theory.
The Wave Nature of the Electron

In 1925 Louis de Broglie published his Ph.D.
dissertation. h

A crucialelement of his dissertation is that electrons
have wave-like
mv properties.
– The electron wavelengths are described by the de
 Broglie
Planck’
s constant
relationship.
–
h
m  mass of particle
v  velocity of particle
60
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.
–
61
This wave-particle duality is a fundamental property
of submicroscopic particles.
The Wave Nature of the Electron

Example 5-9. Determine the wavelength, in m, of an
electron, with mass 9.11hx 10-31 kg, having a velocity of
5.65 x 107 m/s.  
mv
6.626 10 kg m  s

9.1110-31 kg 5.65 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/s2.
34
2
2


  1.29 10
62
11
m

The Wave Nature of the Electron

Example 5-10. 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
  so small compared to the
• Why is the bullet’s wavelength
mv
electron’s wavelength?
34
2
2
6.626  10 kg m  s

-3
33.89
.89 10 kg 395 m/s 

63

  4.3110 34 m
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!

64
Electron microscopes use this phenomenon
The Quantum Mechanical
Picture of the Atom


65
Consequently, we must must speak of the
electrons’ position about the atom in terms of
probability functions.
These probability functions are represented as
orbitals in quantum mechanics.
The Quantum Mechanical
Picture of the Atom
1.
66
Basic Postulates of Quantum Theory
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).
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 
67
hc

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
68
b2   2  2  2 
 2  2  2  2   V  E
8 m   x  y  z 
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 .
69
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.
70
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
71
Quantum Numbers

If  = 2 (or a d orbital), then m = -2,-1,0,+1,+2.
–
There are 5 values of m.
Thus there are five d orbitals per n value. n  3

If  = 3 (or an f orbital), then m = -3,-2,-1,0,+1,+2, +3.
–
There are 7 values of m.
Thus there are seven f orbitals per n value, n  4

Theoretically, this series continues on to g,h,i, etc.
orbitals.
–
72
Practically speaking 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.
Quantum Numbers


The last quantum number is the spin quantum
number which has the symbol ms.
The spin quantum number only has two possible
values.
–
–


This quantum number tells us the spin and
orientation of the magnetic field of the electrons.
Wolfgang Pauli in 1925 discovered the Exclusion
Principle.
–
73
ms = +1/2 or -1/2
ms = ± 1/2
No two electrons in an atom can have the same set of 4
quantum numbers.
Atomic Orbitals


Atomic orbitals are regions of space where
the probability of finding an electron about an
atom is highest.
s orbital properties:
–
–
74
There is one s orbital per n level.
=0
1 value of m
Atomic Orbitals

75
s orbitals are spherically symmetric.
Atomic Orbitals

p orbital properties:
–

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.
–
–
–
76
The first p orbitals appear in the n = 2 shell.
The three orbitals are named px, py, pz.
They have an  = 1.
m = -1,0,+1 3 values of m
Atomic Orbitals

77
p orbitals are peanut or dumbbell shaped.
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 , d , d , d 2 2 , d 2
xy
yz
xz
x -y
z
have an  = 2.
–m = -2,-1,0,+1,+2 5 values of m 
–They
78
Atomic Orbitals

79
d orbital shapes
Atomic Orbitals

f orbital properties:
–


The f orbitals have the most complex shapes.
There are seven f orbitals per n level.
–
–
–
–
80
The first f orbitals appear in the n = 4 shell.
The f orbitals have complicated names.
They have an  = 3
m = -3,-2,-1,0,+1,+2, +3
7 values of m
The f orbitals have important effects in the
lanthanide and actinide elements.
Atomic Orbitals

81
f orbital shapes
Atomic Orbitals

Spin quantum number effects:
–
Every orbital can hold up to two electrons.

–
–

82
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.
Paramagnetism and Diamagnetism

Unpaired electrons have their spins aligned
 or 
–

Atoms with unpaired electrons are called
paramagnetic .
–
83
This increases the magnetic field of the atom.
Paramagnetic atoms are attracted to a magnet.
Paramagnetism and Diamagnetism

Paired electrons have their spins unaligned
.
–

Atoms with paired electrons are called
diamagnetic.
–
84
Paired electrons have no net magnetic field.
Diamagnetic atoms are repelled by a magnet.
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.
–
85
The value is 2n2 because of the two paired electrons.
Paramagnetism and Diamagnetism
Energy Level
n
1
2
86
# of Orbitals
n2
1
4
You do it!
Max. # of e2n2
2
8
3
9
18
4
16
32
The Periodic Table and
Electron Configurations


87
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.
The Periodic Table and
Electron Configurations

88
The Aufbau Principle describes the electron filling
order in atoms.
The Periodic Table and
Electron Configurations

1.
89
There are two ways to remember the correct filling
order for electrons in atoms.
You can use this mnemonic.
The Periodic Table and
Electron Configurations
2.
90
Or you can use the periodic chart .
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
91

1
H
2
He 
Configurat ion
1
1s
1s
2
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.
92
The Periodic Table and
Electron Configurations

3s
3p
3s
3p
3rd row elements
3s
3p
Na Ne
Ne 

Na


Na Ne 
Mg Ne
Ne 
12 Mg
12
Mg Ne

11
11
11
12
Al
13 Al
13
13 Al
14 Si
14
14 Si
15 P
15
S
17 Cl
17
16
16
93
18
Ar
Ne
Ne  

Ne
Ne  

Ne 
Ne 
Ne 
Ne 



Configurat ion
ion
Configurat
Configurat ion
1
Ne
Ne 3ss1 1
Ne 3s2
Ne
Ne33ss2 2
Ne 3s

 
  
  
  
Ne
Ne 3ss22 23p
3p11 1
Ne 3s 3p
22 3p 222
Ne

3
s
Ne 3s 2 3p 2
Ne 3s22 3p33
Ne 3s22 3p44
Ne 3s22 3p55
Ne 3s2 3p6
The Periodic Table and
Electron Configurations

3d
4th row elements
19 K Ar 
94
4s

4p
Configurat ion
Ar  4s1
The Periodic Table and
Electron Configurations
3d
95
4s
19 K Ar 

20

Ca Ar 
4p
Configurat ion
Ar  4s1
Ar  4s2
The Periodic Table and
Electron Configurations
3d
96
4s
19 K Ar 

20
Ca Ar 

21
Sc You do it!
4p
Configurat ion
Ar  4s1
Ar  4s2
The Periodic Table and
Electron Configurations
3d
19 K Ar 

20
Ca Ar 

Sc Ar  

21
97
4s
4p
Configurat ion
Ar  4s1
Ar  4s2
2
1
Ar  4s 3d
The Periodic Table and
Electron Configurations
3d
19
20
21
K Ar 

Ca Ar 

Sc Ar  

Ti You do it!
22
98
4s
4p
Configurat ion
Ar  4s
Ar  4s2
Ar  4s2 3d1
1
The Periodic Table and
Electron Configurations
3d
K Ar 

Ca Ar 

Sc Ar  

Ti Ar   

19
20
21
22
99
4s
4p
Configurat ion
Ar  4s
Ar  4s2
Ar  4s2 3d1
Ar  4s2 3d 2
1
The Periodic Table and
Electron Configurations
3d
19 K Ar 

20
Ca Ar 

Sc Ar  

22
Ti Ar   

23
V Ar    

21
10
0
4s
4p
Configurat ion
Ar  4s1
Ar  4s2
2
1
Ar  4s 3d
Ar  4s2 3d 2
Ar  4s2 3d 3
The Periodic Table and
Electron Configurations
3d
19 K Ar 

20
Ca Ar 

Sc Ar  

22
Ti Ar   

23
V Ar    

Cr Ar      

21
24
10
1
4s
4p
Configurat ion
Ar  4s1
Ar  4s2
Ar  4s2 3d1
Ar  4s2 3d 2
Ar  4s2 3d 3
Ar  4s1 3d5
There is an extra measure of stability associated
with half - filled and completely filled orbitals.
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
10
2
4s

4p
Configurat ion
Ar  4s2 3d5
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
10
3
Fe You do it!
4s

4p
Configurat ion
Ar  4s2 3d5
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
10
4
Fe Ar      
4s


4p
Configurat ion
Ar  4s2 3d5
Ar  4s2 3d 6
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
27
10
5
4s

Fe Ar      

Co Ar      

4p
Configurat ion
Ar  4s2 3d5
Ar  4s2 3d 6
Ar  4s2 3d 7
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
27
28
10
6
4s

Fe Ar      

Co Ar      

Ni Ar      

4p
Configurat ion
Ar  4s2 3d5
Ar  4s2 3d 6
Ar  4s2 3d 7
Ar  4s2 3d8
The Periodic Table and
Electron Configurations
3d
Mn Ar      

Fe Ar      

Co Ar      

28
Ni Ar      

29
Cu You do it!
25
26
27
10
7
4s
4p
Configurat ion
Ar  4s 3d
Ar  4s2 3d 6
Ar  4s2 3d 7
Ar  4s2 3d8
2
5
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
27
28
29
10
8
4s
4p

Fe Ar      

Co Ar      

Ni Ar      

Cu Ar       
Another exception like Cr and
for essentiall y the same reason.
Configurat ion
Ar  4s2 3d5
Ar  4s2 3d 6
Ar  4s2 3d 7
Ar  4s2 3d8
Ar  4s1 3d10
The Periodic Table and
Electron Configurations
3d
25 Mn Ar      
26
27
28
29
10
9
30
4s

Fe Ar      

Co Ar      

Ni Ar      

Cu Ar       
Zn Ar       
4p
Configurat ion
Ar  4s2 3d5
Ar  4s2 3d 6
Ar  4s2 3d 7
Ar  4s2 3d8
Ar  4s1 3d10
Ar  4s2 3d10
The Periodic Table and
Electron Configurations
3d
4s
31 Ga Ar        
11
0
4p
Configurat ion
Ar  4s2 3d10 4p1
The Periodic Table and
Electron Configurations
3d
4s
31 Ga Ar        
32
11
1
Ge You do it!
4p
Configurat ion
Ar  4s2 3d10 4p1
The Periodic Table and
Electron Configurations
3d
4s
31 Ga Ar        
32
11
2
4p
Ge Ar         
Configurat ion
Ar  4s2 3d10 4p1
Ar  4s2 3d10 4p2
The Periodic Table and
Electron Configurations
3d
4s
4p
31 Ga Ar        
32
33
11
3
Ge Ar         
As Ar          
Configurat ion
Ar  4s2 3d10 4p1
Ar  4s2 3d10 4p2
Ar  4s2 3d10 4p3
The Periodic Table and
Electron Configurations
3d
4s
4p
31 Ga Ar        
32
11
4
Ge Ar         
33
As Ar          
34
Se You do it!
Configurat ion
Ar  4s2 3d10 4p1
Ar  4s2 3d10 4p2
Ar  4s2 3d10 4p3
The Periodic Table and
Electron Configurations
3d
4s
4p
31 Ga Ar        
32
33
34
11
5
Ge Ar         
As Ar          
Se Ar          
Configurat ion
Ar  4s2 3d10 4p1
Ar  4s2 3d10 4p2
Ar  4s2 3d10 4p3
Ar  4s2 3d10 4p4
The Periodic Table and
Electron Configurations
3d
4s
4p
Configurat ion
31 Ga Ar        
32
33
34
35
11
6
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
The Periodic Table and
Electron Configurations
3d
4s
31 Ga Ar        
4p
Configurat ion
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
11
7
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 Na
(remember Ne has 10 electrons)
3s
11
8
11 Na Ne 
3p
Configurat ion
Ne 3s1
The Periodic Table and
Electron Configurations
1st e-
11
9
n

m
1
0
0
ms
 1/2
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e- 1
0
0
1st e-
12
0
ms
 1/2 
1 s electrons
 1/2 
The Periodic Table and
Electron Configurations
n

m
1
0
0
2 nd e - 1
0
0
3rd e -
0
0
1st e -
12
1
2
ms
 1/2 
1 s electrons
 1/2 
 1/2
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-
12
2
ms
 1/2 
1 s electrons
 1/2 
 1/2 
2 s electrons
 1/2 
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 -
12
3
ms
 1/2 
1 s electrons
 1/2 
 1/2 
2 s electrons
 1/2 
 1/2
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 electrons
 1/2 
 1/2
6 th e-
2
1
0
 1/2
1st e-
12
4
ms
 1/2 
1 s electrons
 1/2 
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 electrons
 1/2 
 1/2
6 th e - 2
1
0
 1/2
7 th e - 2
1
1
 1/2
1st e -
12
5
ms
 1/2 
1 s electrons
- 1/2 
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 electrons
 1/2 
 1/2
6 th e -
2
1
0
 1/2
7 th e -
2
1
1
 1/2
12 8th e- 2
6
1
1
 1/2
1st e -
ms
 1/2 
1 s electrons
 1/2 
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 electrons
 1/2 
 1/2
6 th e- 2
1
0
 1/2
7 th e - 2
1
1
 1/2
1
1
 1/2
1
0
 1/2
1st e-
12 8th e- 2
7
9 th e- 2
ms
 1/2 
1 s electrons
 1/2 
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
10 th e - 2
1
 1/2 

0
 1/2 
 1  1/2 

2 p electrons
 1  1/2 
0
 1/2 

 1  1/2 

1st e -
12
8
-1
ms
 1/2 
1 s electrons
 1/2 
 1/2 
2 s electrons
 1/2 
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
10 th e -
2
1
11th e -
3
0
1st e -
12
9
ms
 1/2 
1 s electrons
 1/2 
 1/2 
2 s electrons
 1/2 
 1/2 

0
 1/2 
 1  1/2 

2 p electrons
 1  1/2 
0
 1/2 

 1  1/2 

0
 1/23 s electron
-1
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]
13
0
4s

4p
Configurat ion
Ar  4s2
The Periodic Table and
Electron Configurations
13
1
n

m
ms
[Ar ] 19 th e - 4
0
0
 1/2
The Periodic Table and
Electron Configurations
13
2
n

m
[Ar ]19 th e -
4
0
0
20 th e -
4
0
0
ms
 1/2 

4 s electrons
 1/2 

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      
13
3
4s

4p
Configurat ion
Ar  4s2 3d6
The Periodic Table and
Electron Configurations
13
4
n

m
ms
[Ar ] 19 th e - 4
0
0
 1/2
The Periodic Table and
Electron Configurations
13
5
n

m
[Ar ]19 th e -
4
0
0
20 th e -
4
0
0
ms
 1/2 

4 s electrons
 1/2 

The Periodic Table and
Electron Configurations
13
6
n

[Ar ] 19 th e -
4
0
0
20 th e -
4
0
0
21st e -
3
2
-2
m
ms
 1/2 

4 s electrons
 1/2 

 1/2
The Periodic Table and
Electron Configurations
13
7
n

m
[Ar ] 19 th e -
4
0
0
20 th e -
4
0
21st e -
3
2
22 nd e -
You do it!
ms
 1/2 

4 s electrons
0
 1/2 

- 2  1/2
The Periodic Table and
Electron Configurations
n
13
8

m
ms
[Ar] 19 th e -
4
0
0
20 th e -
4
0
0
21st e -
3
2
-2
 1/2 

4 s electrons
 1/2 

 1/2
22 nd e -
3
2
-1
 1/2
The Periodic Table and
Electron Configurations
n
13
9

m
ms
[Ar] 19 th e -
4
0
0
20 th e -
4
0
0
21st e -
3
2
-2
 1/2 

4 s electrons
 1/2 

 1/2
22 nd e -
3
2
-1
 1/2
23rd e -
3
2
0
 1/2
The Periodic Table and
Electron Configurations
14
0
n

m
[Ar] 19 th e -
4
0
0
20 th e -
4
0
0
21st e -
3
2
-2
 1/2 

4 s electrons
 1/2 

 1/2
22 nd e -
3
2
-1
 1/2
23rd e -
3
2
0
 1/2
24 th e -
3
2
 1  1/2
ms
The Periodic Table and
Electron Configurations
14
1
n

[Ar] 19 th e -
4
0
20 th e -
4
0
21st e -
3
2
22 nd e -
3
2
23rd e -
3
2
24 th e -
3
2
25 th e -
3
2
m
ms
 1/2 

4 s electrons
0
 1/2 

- 2  1/2 

- 1  1/2 

0  1/2 half - filled d shell
 1  1/2 

 2  1/2 

0
The Periodic Table and
Electron Configurations
14
2
n

m
[Ar] 19 th e -
4
0
0
20 th e -
4
0
0
21st e -
3
2
-2
 1/2 

4 s electrons
 1/2 

 1/2
22 nd e -
3
2
-1
 1/2
23rd e -
3
2
0
 1/2
24 th e -
3
2
1
 1/2
25 th e -
3
2
2
 1/2
26 th e -
You do it!
ms
The Periodic Table and
Electron Configurations
14
3
n

m
[Ar] 19 th e -
4
0
0
20 th e -
4
0
0
21st e -
3
2
-2
 1/2 

4 s electrons
 1/2 

 1/2
22 nd e -
3
2
-1
 1/2
23rd e -
3
2
0
 1/2
24 th e -
3
2
1
 1/2
25 th e -
3
2
2
 1/2
26 th e -
3
2
-2
 1/2
ms