Chapter 6: Electronic Structure of Atoms
Light is a form of electromagnetic radiation (EMR):
• an oscillating charge, such as an electron, gives rise to
electromagnetic radiation:
Electric Field
Magnetic Field
Chapter 6: Electronic Structure of Atoms
• Both the Electric and the Magnetic field propagate through
space
• In vacuum, both move at the speed of light (3 x 108 m/s)
Chapter 6: Electronic Structure of Atoms
Electromagnetic radiation is characterized by
• wavelength (), or frequency ()
and
• amplitude (A)

A = intensity


Chapter 6: Electronic Structure of Atoms
Frequency () measures how many wavelengths pass a
point per second:
1s
Chapter 6: Electronic Structure of Atoms
Electromagnetic radiation travels at the speed of light:
c = 3 x 108 m s-1
Relation between wavelength, frequency, and amplitude:
c=
Chapter 6: Electronic Structure of Atoms
400 nm
750 nm
Chapter 6: Electronic Structure of Atoms
Red Orange Yellow Green Blue Ultraviolet
Chapter 6: Electronic Structure of Atoms
What is the wavelength, in m, of radiowaves transmitted by
the local radio station WHQR 91.3 MHz?
c  

c

 3.29 m
Chapter 6: Electronic Structure of Atoms
A certain type of laser emits green light of 532 nm. What
frequency does this wavelength correspond to?
c  

c

 5.64  1014 s 1
 5.64 1014 Hz
Chapter 6: Electronic Structure of Atoms
Classically, electromagnetic radiation (EMR) was thought
to have only wave-like properties.
Two experimental observations challenged this view:
Blackbody radiation
Photoelectric Effect
Chapter 6: Electronic Structure of Atoms
Blackbody radiation
• Hot objects emit light
• The higher T, the higher
the emitted frequency
Chapter 6: Electronic Structure of Atoms
Blackbody radiation
prediction of classical theory
= there would be
NO DARKNESS
Brightness
“ultraviolet catastrophe”
T2
T1
wavelength ()
visible region
Chapter 6: Electronic Structure of Atoms
Blackbody radiation
• light is emitted by oscillators
• high energy oscillators require a
minimum amount of energy to be
excited:
E=h
• energy is not provided by
temperature in “black body”
Max Planck (1858 - 1947)
Chapter 6: Electronic Structure of Atoms
Blackbody radiation
frequency of oscillator
E=h
Planck’s constant = 6.63 x 10-34 J s
Energy of radiation is related to frequency, not intensity
Chapter 6: Electronic Structure of Atoms
What is the energy of a photon of electromagnetic radiation
that has a frequency of 400 kHz?
E  h
= 2.65 x 10-28 J
Chapter 6: Electronic Structure of Atoms
Photoelectric Effect
Albert Einstein (1879-1955)
e-
e- e -
Chapter 6: Electronic Structure of Atoms
Photoelectric Effect
Albert Einstein (1879-1955)
e-
e-
e- e -
• Light of a certain minimum frequency is required to
dislodge electrons from metals
Chapter 6: Electronic Structure of Atoms
Photoelectric Effect
• Ability of light to dislodge electrons from metals is
related to its frequency, not intensity
E=h
• This means that light comes in “units” of h
• Intensity is related only to the number of “units”
• The h “unit” is called a quantum of energy
• A quantum of light (EMR) energy = photon
Chapter 6: Electronic Structure of Atoms
Relationship between Energy, Wavelength, and Frequency:
E  h
c  

c


c
E

 h
E
hc

E
h
Chapter 6: Electronic Structure of Atoms
What is the energy of a photon of light of 532 nm?
E
hc

= 3.74 x 10-19 J
Chapter 6: Electronic Structure of Atoms
Electromagnetic Radiation
wave
or
stream of particles
(photons)
E=h
Whether light behaves as a wave or as a stream of photons
depends on the method used to investigate it !
Chapter 6: Electronic Structure of Atoms
Understanding light in terms of photons helped understand
atomic structure
many light sources produce a continuous spectrum
Chapter 6: Electronic Structure of Atoms
Thermally excited atoms in the gas phase emit line spectra
continuous spectrum (all wavelengths together: white light)
line spectrum (only some wavelengths: emission will have a color)
Chapter 6: Electronic Structure of Atoms
Photograph of the H2 line spectrum (Balmer series) in the visible region
(1825-1898)
Johann Balmer (1825-1898)
1 1
 RH  2  2 

 n1 n2 
1
Rydberg constant
1.097 x 107 m-1
positive integers
(e.g. 1,2,3, etc)
Chapter 6: Electronic Structure of Atoms
Niels Bohr was the first to offer an explanation for line spectra
Bohr Model of the Hydrogen Atom
• Only orbits of defined energy and radii are permitted in the
hydrogen atom
• An electron in a permitted orbit has a specific energy and will
not radiate energy and will not spiral into the nucleus
• Energy is absorbed or emitted by the electron as the
electron moves from one allowed orbit into another.
Energy is absorbed or emitted as a photon of E = h
Chapter 6: Electronic Structure of Atoms
Niels Bohr was the first to offer an explanation for line spectra
electron orbits
(1885-1962)
n=1
n=2
n=3
n=4
n=5
n=6
nucleus
Bohr’s Model of the Hydrogen Atom
Chapter 6: Electronic Structure of Atoms
Bohr’s Model of the Hydrogen Atom
n=6
n=5
n=4
n=3
Energy
n=2
absorption of a photon
n=1
Ground State
nucleus
e
Chapter 6: Electronic Structure of Atoms
Bohr’s Model of the Hydrogen Atom
n=6
n=5
n=4
n=3
n=1
Energy
n=2
Ground State
nucleus
e
Chapter 6: Electronic Structure of Atoms
Bohr’s Model of the Hydrogen Atom
n=6
n=5
n=4
e
n=3
n=1
Energy
n=2
Ground State
nucleus
“excited state”
Chapter 6: Electronic Structure of Atoms
Bohr’s Model of the Hydrogen Atom
n=6
n=5
n=4
e
n=3
n=1
Energy
n=2
Ground State
nucleus
Chapter 6: Electronic Structure of Atoms
Bohr’s Model of the Hydrogen Atom
n=6
n=5
n=4
n=3
n=1
Energy
n=2
Ground State
nucleus
e
emission of a photon
Chapter 6: Electronic Structure of Atoms
Which of these transitions represents
an absorption process?
(a) (b)
n=6
n=5
n=4
Which of these transitions involves the
largest change in energy?
n=3
n=1
Energy
n=2
(c)
Which of these transitions leads to the
emission of the longest wavelength
photon?
Ground State
nucleus
Does this wavelength correspond to a
high or low frequency?
Chapter 6: Electronic Structure of Atoms
Transitions corresponding to
the Balmer series
Chapter 6: Electronic Structure of Atoms
n = Principal Quantum Number (main energy levels)
1

Energy of electron in a given orbit: E  ( h c RH )  2 
n 
n=6
n=5
n=4
n=3
n=2
n=1
1
E  ( 2.18 10 18 J )  
9
1
E  ( 2.18 10 18 J )  
 4
E  2.18 1018 J
h=Planck’s constant, c=speed of light, RH = Rydberg constant
Chapter 6: Electronic Structure of Atoms
For an electron moving from n = 4 to n = 2:
Einitial
n=6
n=5
n=4
n=3
n=2


1
(
)


  hc R 
n

2
H
initial
E final
 1 

 ( h c RH ) 
2

 n final 
E  E final  Einitial
 1 
 1 

(
)

E  ( h c RH ) 


h
c
R
2
2 
H 


n
n
 initial 
 final 
n=1
 1
1 


E  ( h c RH ) 
2 
2

n
n
final
initial


Chapter 6: Electronic Structure of Atoms
For an electron moving from n = 4 to n = 2:
n=6
n=5
n=4
n=3
n=2
n=1
 1
1 

E  ( h c RH ) 
2 
2

 n final ninitial 
1 1
E  ( h c RH )  2  2 
2 4 
1 1 
E  ( 2.18 10 18 J )   
 4 16 
E  2.18 10 18 J  0.1875
E = - 4.09 x 10-19 J
Chapter 6: Electronic Structure of Atoms
The energy of the photon emitted is:
E = 4.09 x 10-19 J
n=6
n=5
n=4
n=3
n=2
n=1
What wavelength (in nm) does this
energy correspond to?
E
hc


hc
E
6.63 10 34 Js  3 108 ms 1

4.09 10 19 J
 = 486 x 10-9 m
= 486 nm
Chapter 6: Electronic Structure of Atoms
Balmer Series
n=6 → n=2
n=4 → n=2
n=5 → n=2
 = 486 nm
n=3 → n=2
Chapter 6: Electronic Structure of Atoms
The Wave Behavior of Matter
If light can behave like a stream of particles (photons)…
… then (small) particles should be able to behave like waves, too
For a particle of mass m, moving at a velocity v:

h
mv
De Broglie Wavelength
e.g electrons have a wavelength (electron microscope!)
Chapter 6: Electronic Structure of Atoms
The Uncertainty Principle
Werner Heisenberg (1901-1976)
and Niels Bohr
Chapter 6: Electronic Structure of Atoms
The Uncertainty Principle
It is impossible to know both the exact position and the exact
momentum of a subatomic particle
uncertainty in momentum, mv
x  mv 
uncertainty in position, x
h
4
Chapter 6: Electronic Structure of Atoms
Quantum Mechanics and Atomic Orbitals
Erwin Schrödinger (1887-1961)
Chapter 6: Electronic Structure of Atoms
Quantum Mechanics and Atomic Orbitals
• Schrödinger proposed wave mechanical model of the atom
• Electrons are described by a wave function, ψ
• The square of the wave function, ψ2, provides information on
the location of an electron (probability density or electron density)
Chapter 6: Electronic Structure of Atoms
Quantum Mechanics and Atomic Orbitals
• the denser the stippling, the
higher the probability of finding
the electron
• shape of electron density
regions depends on energy of
electron
Chapter 6: Electronic Structure of Atoms
Bohr’s model:
n=1
orbit
electron circles around nucleus
Schrödinger’s model:
orbital
z
n=1
or
electron is somewhere
within that spherical region
y
x
Chapter 6: Electronic Structure of Atoms
Bohr’s model:
• requires only a single quantum number (n) to describe an orbit
Schrödinger’s model:
• requires three quantum numbers (n, l, and m) to describe an orbital
n:
l:
ml:
principal quantum number
second or azimuthal quantum number
magnetic quantum number
Chapter 6: Electronic Structure of Atoms
Schrödinger’s model:
(1)
n = principal quantum number (analogous to Bohr model)
- the higher n, the higher the energy of the electron
1
- energy of electron in a given orbital: E  ( h c RH )  2 
n 
- is always a positive integer: 1, 2, 3, 4 ….
Chapter 6: Electronic Structure of Atoms
Schrödinger’s model:
(2)
l = azimuthal quantum number
- takes integral values from 0 to n-1
e.g.
- l is normally listed as a letter:
Value of l:
letter:
0
s
1
p
2
d
- l defines the shape of an electron orbital
3
f
n=3
Chapter 6: Electronic Structure of Atoms
Schrödinger’s model:
z
y
x
s-orbital
p-orbital
(1 of 3)
d-orbital
(1 of 5)
f-orbital
(1 of 7)
Chapter 6: Electronic Structure of Atoms
Schrödinger’s model:
(3)
ml = magnetic quantum number
- takes integral values from -l to +l, including 0
e.g.
l=2
- ml describes the orientation of an electron orbital in space
Chapter 6: Electronic Structure of Atoms
Shells:
• are sets of orbitals with the same quantum number, n
• a shell of quantum number n has n subshells
Subshells:
• are orbitals of one type within the same shell
• total number of orbitals in a shell is n2
Chapter 6: Electronic Structure of Atoms
n=3 shell
n=
1
2
3
4
l=
0
0, 1
0, 1, 2
0, 1, 2, 3
4f subshell
1s
2s, 2p
3s, 3p, 3d
4s, 4p, 4d, 4f
ml =
0
0, -1,0,1 0; -1,0,1; -2,-1,0,1,2 0; -1,0,1; -2,-1,0,1,2; -3,-2,-1,0,1,2,3
# orbitals
in subshell
1
1
Total # of
orbitals
in shell
1
3
4
1 3
5
9
1
3
5
7
16
Chapter 6: Electronic Structure of Atoms
3s-room
3p-room
3deluxe-room
3rd floor
2s-room
2nd floor
standard-room
1st floor
2promotion-room
Chapter 6: Electronic Structure of Atoms
Orbital energy levels
in the Hydrogen Atom
Chapter 6: Electronic Structure of Atoms
What is the designation for the n=3, l=2 subshell ?
How many orbitals are in this subshell ?
What are the possible values for ml for each of these orbitals ?
Chapter 6: Electronic Structure of Atoms
Which of the following combinations of quantum numbers
is possible?
n=1, l=1, ml= -1
n=3, l=0, ml= -1
n=3, l=2, ml= 1
n=2, l=1, ml= -2
Chapter 6: Electronic Structure of Atoms
Representation of Orbitals
1s
2s
3s
Chapter 6: Electronic Structure of Atoms
Representation of Orbitals
2p orbitals
Chapter 6: Electronic Structure of Atoms
Representation of Orbitals
all three p orbitals
Chapter 6: Electronic Structure of Atoms
Representation of Orbitals
3d orbitals
Chapter 6: Electronic Structure of Atoms
Which combination of quantum numbers is possible for the
orbital shown below?
(a) n=1, l=0, ml= 0
(c) n=3, l=3, ml= -2
(b) n=2, l=-1, ml= 1
(d) n=3, l=2, ml= -1
Chapter 6: Electronic Structure of Atoms
There is a fourth quantum number that characterizes electrons:
spin magnetic quantum number, ms
ms can only take two values, +1/2 or -1/2
Chapter 6: Electronic Structure of Atoms
Wolfgang Pauli (1900-1958)
A. Einstein & W. Pauli
Chapter 6: Electronic Structure of Atoms
Pauli’s Exclusion Principle:
No two electrons in an atom can have the same set of 4
quantum numbers, n, l, ml, and ms
For a given orbital, e.g. 2s, n, l, ml are fixed:
n=2, l=0, ml =0
=> an orbital can only contain two electron if they differ in ms
Chapter 6: Electronic Structure of Atoms
A maximum of 2 electron can occupy one orbital, IF these two
electrons have opposite spin:
n=2, l=0, ml =0, ms = +1/2
n=2, l=0, ml =0, ms = -1/2
2s
2p
arrows pointing up/down indicate electron spin
Chapter 6: Electronic Structure of Atoms
Energy levels in the hydrogen atom:
all subshells of a given shell
have the same energy
Chapter 6: Electronic Structure of Atoms
Energy levels in many-electron atoms:
• In many-electron
atoms, the energy of an
orbital increases with l,
for a given n
• In many-electron
atoms, the lower energy
orbitals get filled first
• orbitals with the same
energy are said to be
degenerate
Chapter 6: Electronic Structure of Atoms
Electron Configurations:
Line Notation:
1H
1s1
2He
1s2
3Li
1s22s1
4Be
1s22s2
6C
1s22s22p2
7N
1s22s22p3
10Ne
1s22s22p6
11Na
1s22s22p63s1
Chapter 6: Electronic Structure of Atoms
Electron Configurations:
Hund’s Rule:
For degenerate orbitals, the
energy is minimized when
the number of electrons with
the same spin is maximized
=> degenerate orbitals (p, d, etc)
get filled with one electron each
first (same spin).
7N
1s22s22p3
Chapter 6: Electronic Structure of Atoms
the Aufbau Principle helps you to remember the order in
which orbitals get filled:
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
6f
7s
7p
7d
7f
Chapter 6: Electronic Structure of Atoms
14Si
1s22s22p63s23p2
[Ne] 3s23p2
Line notation
Condensed line notation
orbital diagram
(no energy info)
3
d
2
p
1
“core electrons”
s
Chapter 6: Electronic Structure of Atoms
14Si
1s22s22p63s23p2
[Ne] 3s23p2
Line notation
Condensed line notation
“valence (outer shell) electrons”
orbital diagram
(no energy info)
3
d
2
p
1
s
Valence electrons take part in bonding
Chapter 6: Electronic Structure of Atoms
What is the electronic structure of Cl?
17Cl
: [Ne] 3s23p5
valence electrons (7)
3
d
2
p
1
s
core electrons
=
electron configuration
of the preceding noble gas
Chapter 6: Electronic Structure of Atoms
What is the electronic structure of Ca?
20Cl
: [Ar] 4s2
(4s orbital is filled before 3d !)
valence electrons (2)
4
f
3
d
2
p
1
s
core electrons
=
electron configuration
of the preceding noble gas
Chapter 6: Electronic Structure of Atoms
What is the electronic structure of Br?
35Br
: [Ar] 3d104s24p5
(4s orbital is filled before 3d !)
valence electrons (7)
4
f
3
d
2
p
1
s
For main group elements,
electrons in a filled d-shell
(or f-shell) are not valence
electrons
core electrons
=
electron configuration
of the preceding noble gas
Chapter 6: Electronic Structure of Atoms
Does it matter in which order the electron configuration is written ?
35Br
or:
: 1s22s22p63s23p63d104s24p5
1s22s22p63s23p64s23d104p5
ordered by orbital number
ordered by energy
4
f
3
d
2
p
1
s
NO, both are correct!
Chapter 6: Electronic Structure of Atoms
What is the electron configuration of vanadium (V)?
23V:
[Ar] 3d34s2
(4s orbital is filled before 3d !)
4
f
3
d
2
valence electrons (5)
p
1
s
core electrons
=
electron configuration
of the preceding noble gas
Chapter 6: Electronic Structure of Atoms
What is the electron configuration of chromium (Cr)?
24Cr:
[Ar] 3d54s1
4
f
3
d
2
p
1
s
[Ar] 3d44s2 is less stable than [Ar] 3d54s1
A half-filled or completely filled d-shell is a preferred configuration
Chapter 6: Electronic Structure of Atoms
1s
2s
3s
4s
3d
2p
3p
4p
4f
Chapter 6: Electronic Structure of Atoms
What is the electronic structure of the Ca ion?
2
[Ar]
4s
Ca
:
20
2+ : [Ar]
20Ca
4
f
3
d
2
p
1
s
Chapter 6: Electronic Structure of Atoms
●
Metals tend to lose electrons to form cations
●
Nonmetals tend to gain electrons to form anions
●
Atoms tend to gain or lose the number of electrons
needed to achieve the
electron configuration of the closest noble gas
Chapter 6: Electronic Structure of Atoms
What is the electronic structure of the ion formed by Se?
104s24p4
[Ar]
3d
Se
:
34
2[Ar] 3d104s24p6 = [Kr]
34Se :
4
f
3
d
2
p
1
s
Chapter 6: Electronic Structure of Atoms
What is the electronic structure of the ion formed by Br?
35Br :
35Br :
[Ar] 3d104s24p5
[Ar] 3d104s24p6 = [Kr]
4
f
3
d
2
p
1
s
Chapter 6: Electronic Structure of Atoms
What is the electronic structure of the ion formed by Rb?
1
[Kr]
5s
Rb
:
37
+
[Kr]
37Rb :
5
4
f
3
d
2
p
1
s
Chapter 6: Electronic Structure of Atoms
37Rb
+
: [Ar] 3d104s24p6
= [Kr]
-
:
[Ar] 3d104s24p6
= [Kr]
2-
: [Ar] 3d104s24p6
= [Kr]
35Br
34Se
37Rb
+
, 35Br- ,
34Se
2- ,
and 36Kr have the same electron configuration:
they are isoelectronic
Chapter 6: Electronic Structure of Atoms
Which of the four orbital diagrams written below for nitrogen
violates the Pauli Exclusion Principle?
violates Hund’s rule
(all spins must point in the same direction)
a.
b.
violates Hund’s rule
(degenerate orbitals get one electron each, first)
c.
doesn’t violate anything
d.
violates Pauli’s Exclusion Principle
there are two same spin electrons in one
orbital, i.e. all 4 quantum numbers are the
same – which is impossible
1s
2s
2p
Chapter 6: Electronic Structure of Atoms
What is the total number of orbitals in the fourth shell (n=4) ?
a. 16
b. 12
c. 4
d. 3
what is the total number of different s,p, d and f orbitals?
n=4
l=0
s
ml =
0
-1,0,1
1
p
2
d
3
f
-3,-2,-1,0,1,2,3
-2,-1,0,1,2
one s + three p
+ five d
+
=
16 orbitals (n2)
7 f orbitals
Chapter 6: Electronic Structure of Atoms
What is the number of subshells in the third shell (n=3) ?
a. 18
b. 9
c. 3
How many different types of orbitals are there?
n=3
l=0
s
1
p
2
d
d. 1
Chapter 6: Electronic Structure of Atoms
What is the electron configuration of the sodium cation, Na+ ?
a. 1s22s22p63s1
b. 1s22s22p6
c. 1s22s22p63s2
d. 1s22s22p7
11Na
+
= 11 electrons -1 = 10 electrons
1s2 2s2 2p6