Electronic Structure of Atoms
The Nature of Light Energy
Light Energy as Waves
Electromagnetic Radiation
Atoms:
• Absorb and
• Emit energy, often in the form of electromagnetic
radiation:
• Visible light
• microwaves
• radio & TV waves
• Ultra Violet
• Infrared, etc
The Nature of Light Energy
(1) White Light is not white, it is colored: the Spectrum:
Spectroscope
VIB G.Y O R
i
o
l
e
t
n l
d u
i e
g
o
r
e
e
n
e
l
l
o
w
r
a
n
g
e
e
d
(2) Light is electrical and magnetic:
• Electromagnetic
(3) Light does not travel in truly straight lines:
• It travels in waves

short

long
 = high
 = low
Light Energy as Waves: two important characteristics
1.
wavelength ():
 The distance (m) between corresponding points on adjacent crests
2.
frequency () or (f ):
 The number of waves passing a given point per unit of time (1/s = s1-)
1


constant (c)
 

c     (m)  (1/sec)
For waves traveling at the same velocity, the longer the wavelength,
the smaller the frequency
Speed (c) = wave length (λ) x frequency ()
SPEED =
m/sec =
DISTANCE x PER UNIT TIME
m
x 1/sec
Knowing  and , you calculate the speed of light !
Electromagnetic Radiation
A form of energy characterized by waves (or pulses) of
varying frequencies () and wavelengths ().
{*Light Waves}
{3D-Wave}
{Wavelength of v. l.}
c  
c


Electromagnetic Radiation
• Speed of Light: All
electromagnetic radiation
travels at the same velocity (c),
3.00  108 m/s.
Einstein’s Theory of
Special Relativity: Energy and
mass are different forms of the
same thing
1
 α

E  mc2
c
Frequency (f )
Problem: What is the wavelength of a photon of light that has a
frequency of 3.8 x 109 s-1 ?
c
c


3.00 x 108 m s -1

3.8 x 109 s -1
= 7.89x10-2 m
The Nature of Energy:
Discrete vs. Continuous
Digital:
0110100101001
Analog
Eggs:
Water:
Quanta (Photon):
particles
Waves:
Energy as a Particle (Photon, Quanta)
When light energy shines
on a metal, an electron
current is generated.
Light
Energy
waves
particles
Light is behaving as a particle (photon)
that knocks-off valence electrons from
the metal.
Energy as a Particle (Photon, Quanta)
{Metals & EM Radiation}
• The wave nature of light does
not explain how an object can
glow when its temperature
increases.
• Max Planck explained it by
assuming that energy comes in
packets called quanta (energy
bundle, photon).
Max Planck (1848-1947)
Planck concluded that energy (E) is proportional to frequency():

E  h
For any particular frequency () there is a particular
bundle of Energy (E) that exists as a discrete quantity
(quanta) that is a multiple of Planck’s constant (h).
where h is Planck’s constant, 6.63  10−34 J-s.
Energy from electrons comes in discrete quantities (bundles) that
are whole number multiples of h. 1
2
The Nature of Energy
Since c = , then
E
c
 

c
 h  h   
 

Therefore, if one knows the wavelength
of light, one can calculate the energy in
one photon, or packet, of that light.
Problem: What is the wavelength
(in Å) of a ray whose energy is
6.16 x 10-14 erg? {1.0 erg =10-7
Joule}
The Nature of Energy
c
E = h  h  

Problem: What is the wavelength (in Å) of a ray whose energy
is 6.16 x 10-21 Joules?
{Note: Modules use erg =10-7 Joule}
8

3
.
0
x
10
m / sec 
c
- 34
  h    6.63 x 10 Joules.sec

- 21
E
 6.16 x 10 Joules 
-5
3.23
x
10
m

 1Å 
5

3
.
23
x
10
Å


? Å  3.23 x 10 m
-10
 10 m 
-5
Energy as……
(1) Waves

c = 
c
 

(2) Particle (Photon, Quanta)

ΔE =h
(3) Matter
E  mc2
E
c 
 hE  h   
 
END
The Wave-Particle Duality of Matter
•
Electromagnetic radiation can behave as
a particle or as wave phenomena
•
Louis de Broglie posited that if light can
have material properties, matter should
exhibit wave properties.
•
{ElectonWaves}
He demonstrated that the relationship
between mass (m) and wavelength ()
was:

1
∝m
h
 = mv
velocity (v)
(where h is Planck’s constant, 6.63  10−34 J-s, and v is velocity
of light)
= eq given
The Wave Nature of Matter
Problem: An electron has a mass of 9.06 x 10-25 kg and
is traveling at the speed of light. Calculate its
wavelength?
h
(6.63 x 10-34 J / s)
-18


2.44
x10
m
 = mv (9.06 x 10-25 kg ) x (3.00 x 108 m/s)
Problem: What is the wavelength of a 70.0 kg skier
traveling down a mountain at 15.0 m/s?
h
 = mv
-34
( 6.63 x 10 J / s )

 6.31 x10 -37 m
(70.0 kg ) x (15.0 m/s)
J = Joule = kg.m2
The Nature of Energy
White Light’s Continuous Spectrum:
VIB G.Y O R
The Nature of Energy
Substances both absorb and emit only certain Discrete Spectra
{Flame Tests.Li,Na,K} {Na,B}
{AtomicSpectra}
The Bohr “Planetary” Model
of the Atom (1913)
•
Niels Bohr adopted Planck’s assumption and explained atomic phenomena
in this way:
1. Electrons in an atom can only occupy certain orbits (corresponding to
certain energies, frequencies and wavelengths, because E=h=h c/λ).
2. Electrons in permitted orbits have specific, “allowed” energies; these
energies will not be radiated from the atom.
1st
EL
f=4
2nd EL
f=5
3. Energy is only absorbed or emitted in such a way as
to move an electron from one “allowed” energy state
to another; the energy is defined by
E = h
The Bohr Model of the Atom
Which series releases most energy?
{ExcitedElectrons*}
The larger the fall the
greater the energy
Atomic Spectra & Bohr Atom
The energy absorbed or emitted
from the process of electron
promotion or demotion can be
calculated by the
Rydberg formula for hydrogen (1885)
1


RH
(
1
1
- 2
nf2
ni
)
Rydberg formula for hydrogen-like
elements (He+, Li 2+, Be3+ etc., )
1


where RH is the Rydberg constant, 2.18  10−18 J,
and ni and nf are the initial and final energy levels
of the electron. Z is the atomic number
Atomic Spectra & Bohr Atom
Since energy and wavelength are mathematically
related, the Rydberg Equation can also be expressed
in terms of energy:
RH RH
1
1
1
= n2 - n2
 E = RH n 2 - n 2
f
i
f
i

(
)
 2.180 x 10 -18 J 
 E  
2
nf


(
)
 2.180 x 10 -18 J 

  E f  Ei
2
ni


The energy possessed by an electron at a particular energy level
(En) can be expressed as:

-18
2.180 x 10 Joule
En 
n2
 = eq given
where RH is the Rydberg constant, 2.18  10−18 J, and ni and nf are the initial
and final energy levels of the electron.
Atomic Spectra and the Bohr Atom
Problem: How much energy (J) is liberated when an
electron changes from n = 4 to n = 2? What is the
wavelength (m) of the light emitted?
E  Ef  Ei
 2.180 x 10-18 J 

2


nf



 2.180 x 10-18 J 
 2.180 x 10-18 J 
 
  
2
2
2f
ni




 2.180 x 10-18 J 


2
4i


E  (0.545 x 10 -18 J) - (0.1362x 10-18 J)  0.4088 x 10-18 J
To convert energy to wavelength, we must employ the equations:
c
c
  h 
E

c


c
E  h  h   
 

8
3
.
0
x
10
m/s 

-34
-7
 6.63 x 10 J.s 

4.865
x
10
m

-18
 0.4088 x 10 J 
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.
  4.865 x 10-7 m
E  0.4088 x 10-18 J
2006 (B)
Ele 1
Ele 2
Heisenberg’s Uncertainty Principle
• Heisenberg showed that the more
precisely the momentum of a particle
is known, the less precisely is its
position known:
h
(x) (mv) 
4
• In many cases, our uncertainty of the
whereabouts of an electron is greater
than the size of the atom itself!
Quantum Model of the Atom
•
•
•
•
•
•
Max Planck (energy quanta, Planck’s constant)
Albert Einstein (energy and frequency)
Niels Bohr (electrons and Spectra)
Louis de Broglie (particle-wave duality of matter)
Werner Heisenberg (electron uncertainty)
Erwin Schrödinger (probability wave function, the four
quantum numbers)
Quantum Numbers
• Describe the location of electrons within atoms.
• There are four quantum numbers:
 Principal = describes the energy level (1,2,3,etc)
Azimuthal = energy sublevel, orbital type (s2, p6,
d10, f14)
Magnetic = orbital orientation or cloud (2
electrons on each cloud) Example: three p clouds:
px, py, pz
Spin = which way the electron is spinning (↑↓)
Electron Configuration, Orbital Notation
and Quantum Numbers
Principal (n)= energy level
Azimuzal () = sublevel
orbital type
1s2 2s2 2p6 3s23p63d10 4s24p64d104f14
Magnetic (ml) = orbital
cloud orientation (2eper orbital)
Spin (ms) =
electron + or -
Electron Configuration
Two issues:
(1)Arrangement of electrons within an atom
1s2 2s2 2p6 3s23p63d10 4s24p64d104f14
(2) Order in which electrons fill the orbitals
1s22s22p63s23p64s23d104p65s24d105p66s24f14
Aufbau Principle:
States that in the ground state of an Atom, an electron enters the
orbital with lowest energy first and subsequent electrons are fed
in the order of increasing energy
Electron Configuration
Two issues:
(1)Arrangement of electrons within an atom
1s2 2s2 2p6 3s23p63d10 4s24p64d104f14
(2) Order in which electrons fill the orbitals
1s22s22p63s23p64s23d104p65s24d105p66s24f14
Aufbau Process:
Using Periodic Table Sub-blocks:
Historic Development of Atomic Theory
Bohr (1913)
Schrödinger (1926)
The Schrödinger Equation
i
• is the imaginary unit, (complex number whose square is a negative real
number)
t
• is time,
• is the partial derivative with respect to t,
• is the reduced Planck's constant (Planck's constant divided by 2π),
ψ(t) • ψ(t) is the wave function,
• is the Hamiltonian (a self-adjoint operator acting on the state
space).
Quantum Mechanics
• Developed by Erwin Schrödinger, it
is a mathematical model
incorporating both the wave &
particle nature of electrons.
• The wave function is designated
with a lower case Greek psi ().
• The square of the wave function, 2,
gives a probability density map of
where an electron has a certain
statistical likelihood of being at any
given instant in time.
{QuantumAtom}
The Schrödinger Equation
• Solving the wave equation gives a set of
wave functions ψ(t ,) or orbitals, and
their corresponding energies.
• Each orbital describes a spatial
distribution of electron density.
• An orbital is described by a set of three
quantum numbers.
Principal Quantum Number, n
1
2
3
• The principal quantum number, n, describes the
energy level on which the orbital resides.
• The values of n are integers ≥ 0.
Azimuthal Quantum Number, 
• This quantum number defines the shape of the orbital.
• Allowed values of  are integers ranging from 0 to n − 1.
• We also use letter designations:
Value of 
0
1
2
3
Type of orbital
s
p
d
f
=0
=1
=2
=3
Magnetic Quantum Number, ml
• Describes the three-dimensional orientation of the
orbital.
• Values are integers ranging from -l to l:
−l ≤ ml ≤ l.
• Therefore, on any given energy level, there can be up
to 1 s orbital, 3 p orbitals, 5 d orbitals, 7 f orbitals, etc.
0
+1
0
-1
Values of Quantum Numbers
• Principal Quantum #: values of n are integers ≥ 0.
• Azimuthal Quantum #: values of  are integers ranging from 0 to n − 1.
• Magnetic Quantum #: values are integers ranging from -  to  :
−  ≤ ml ≤  .
s Orbitals ( = 0)
• Spherical in shape.
• Radius of sphere
increases with
increasing value of n.
{1s}
{2s}
{3s}
p Orbitals ( = 1)
• Have two lobes with a node between them.
+1 {px}
0 {py}
-1 {pz}
2
2
6
Orbital Overlap: 1s 2s 2p
1s

+
2s
2p
“P” orbital electrons are
repelled by the “S” orbital
electrons and so spend more
time further from the
nucleus.
“P” orbital electrons also
repel from each others’
sublevels, so they run along
the axes.
d Orbitals ( = 2)
-1
2
-2
•Four of the five orbitals
have 4 lobes; the other
resembles a p orbital
with a doughnut around
the center.
1
0
f Orbitals ( = 3)
0
• There are seven f
orbitals per n level.
1
2
3
-1
-2
-3
 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.
Energies of Orbitals
• For a one-electron
hydrogen atom, orbitals
on the same energy
level have the same
energy.
• That is, they are
degenerate (collapsed).
Energies of Orbitals
• As the number of
electrons increases,
though, so does the
repulsion between
them.
• Therefore, in manyelectron atoms,
orbitals on the same
energy level are no
longer degenerate.
Electron Configuration & Periodic Table
Spin Quantum Number, ms
• 1920s: it was discovered that two electrons in the
same orbital do not have exactly the same energy.
{e-spin}
The “spin” of
an electron
describes its
magnetic
field, which
affects its
energy.
Electron Configurations
• Distribution of all
electrons in an atom.
• Consist of
 Number denoting the
energy level.
 Letter denoting the type
of orbital.
 Superscript denoting the
number of electrons in
those orbitals.
Orbital Diagrams
• Each box represents
one orbital.
• Half-arrows represent
the electrons.
• The direction of the
arrow represents the
spin of the electron.
Basic Principles of Electron
Configuration Notations
• Pauli Exclusion Principle
• Hund’s Rule of Maximum Multiplicity
Pauli Exclusion Principle
Only two electrons can occupy an orbital and
they must have opposite spins.
• No two electrons in the same atom can have exactly the same
energy (identical sets of quantum numbers)
Hund’s Rule of Maximum Multiplicity
One electron fills each orbital before a second of
opposite spin accompanies it.
“For degenerate orbitals, the lowest energy is
attained when the number of electrons with the same
spin is maximized.”
{Electron Configuration}
{Electron Configuration2}
Periodic Table and
Electron
Configuration
{e- filling order}
Nitrogen
1s22s22p3
Electronic configuration :
4s
3d
3p
3s
2p
Hund’s Rule
2s
1
1s
2
GROUP
3
4
5
6
7
0
He
1
H
2
Li
Be
B
C
N
O
F
Ne
3
4
Na Mg
Al
Si
P
S
Cl
Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Kr
K
Ca Sc
Ti
V
Neon
1s22s22p6
Electronic configuration:
4s
3d
3p
3s
2p
Hund’s Rule
2s
1
1s
2
GROUP
3
4
5
6
7
0
He
1
H
2
Li
Be
B
C
N
O
F
Ne
3
4
Na Mg
Al
Si
P
S
Cl
Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Kr
K
Ca Sc
Ti
V
Vanadium
1s22s22p6 3s2 3p6 4s2 3d3
Electronic configuration:
4s
3d
3p
[Ar]
2p
[Ne]
3s
2s
1
1s
2
GROUP
3
4
5
6
7
0
He
1
H
2
Li
Be
B
C
N
O
F
Ne
3
4
Na Mg
Al
Si
P
S
Cl
Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Kr
K
Ca Sc
Ti
V
Chromium
1s22s22p6 3s2 3p6 4s1 3d5
Electronic configuration:
4s
3d
3p
[Ar]
2p
[Ne]
3s
Notice that one of the 4s electrons
has been transferred to 3d so that 3d
is now a half filled shell with extra
stability. 4s and 3d contain only
unpaired electrons.
2s
1
1s
2
GROUP
3
4
5
6
7
0
He
1
H
2
Li
Be
B
C
N
O
F
Ne
3
4
Na Mg
Al
Si
P
S
Cl
Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Kr
K
Ca Sc
Ti
V
Nickel
1s22s22p6 3s2 3p6 4s2 3d8
Electronic configuration:
4s
3d
3p
[Ar]
2p
[Ne]
3s
2s
1
1s
2
GROUP
3
4
5
6
7
0
He
1
H
2
Li
Be
B
C
N
O
F
Ne
3
4
Na Mg
Al
Si
P
S
Cl
Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Kr
K
Ca Sc
Ti
V
Copper
1s22s22p6 3s2 3p6 4s13d10
Electronic configuration:
4s
3d
3p
Notice that again one of the 4s electrons
has been promoted to 3d so that 3d
is now a completely filled shell with extra
stability.
3s
2p
2s
1
1s
2
GROUP
3
4
5
6
7
0
He
1
H
2
Li
Be
B
C
N
O
F
Ne
3
4
Na Mg
Al
Si
P
S
Cl
Ar
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br
Kr
K
Ca Sc
Ti
V
Some Anomalies
Some irregularities occur when there are enough
electrons to half-fill s and d orbitals on a given row.
Some Anomalies
Electron configuration for copper is
[Ar] 4s1 3d5
rather than the expected
[Ar] 4s2 3d4.
•This occurs because the s and d orbitals are very close in energy.
Some Anomalies
• These anomalies also occur in f-block atoms, as well.
Electron Configuration
Identify elements which posses the following
electron configurations:
1s2 2s2 2p6 3s2 3p6 4s2 3d6 Fe
2
2
6
2
6
0
6
2
6
6
1s 2s 2p 3s 3p 3d 4s
[Ar] 4s 3d
[Ar] 4s 3d
Fe
2+
Fe
2
Fe
{Aufbau order of filling}
{Energy level order}
{Previous Nobel Gas Abbreviation}
{Cations formed by removal of outermost
electrons}
2-
Write Elect-Config for S
[Ne] 3s2 3p6
Periodic Table and
Electron
Configuration
Uses dots to represent
Valence Electrons = those in
outermost Energy Level
1
2
3
4
5
6
7
Transition Metals
Have additional electrons, but
they are in an energy level that is
lower than the valence electrons.