1 l 0 n = 1 → l = 0 n = 2 → l = 0, 1 n = 3 → l = 0, 1, 2

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1/15/2014
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Characterizes the energy of the electron in a
particular orbital
◦ corresponds to Bohr’s energy level
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CH2045, Gen. Chem. I, Section: 004
Spring 2014
Dr. Shengqian Ma
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n can be any integer  1
The larger the value of n, the more energy the
orbital has
Energies are defined as being negative
◦ an electron would have E = 0 when it just escapes
the atom
The larger the value of n, the larger the orbital
As n gets larger, the amount of energy between
orbitals gets smaller
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2
 Each value of l is called by a particular letter that
designates the shape of the orbital
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This quantum number defines the shape of
the orbital.
l is determined by n.
0≤l≤n–1
n = 1 → l = 0
n = 2 → l = 0, 1
n = 3 → l = 0, 1, 2
l = 0: s orbitals
l = 1: p orbitals
l = 2: d orbitals
l = 3: f orbitals
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(“sharp”)
(“principle”)
(“diffuse”)
(“fine”)
4
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This quantum number describes the threedimensional orientation of the orbital.
ml is defined by l.
Values are integers from −l to +l
-ll ≤ ml ≤ +l
l = 0: ml = 0
l = 1: ml = -1, 0, +1
l = 2: ml = -2, -1, 0, +1, +2
ml takes on 2l + 1 values for a given l.
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Given: n = 4
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Find: orbital designations, number of orbitals
Conceptual
ml
n
l
Plan:
0 → (n − 1)
−l → +l
Relationships:
Solve:

l:l 0 → (n
( − 1);
1) ml: −ll → +ll
n=4
 l : 0, 1, 2, 3

n = 4, l = 2 (d)
n = 4, l = 3 (f)
n = 4, l = 0 (s) n = 4, l = 1 (p)
ml : −1,0,+1 ml : −2,−1,0,+1,+2 ml : −3,−2,−1,0,+1,+2,+3
ml : 0
1 orbital
3 orbitals
5 orbitals
7 orbitals
4s
4p
4d
4f
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total of 16 orbitals: 1 + 3 + 5 + 7 = 42 : n2
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Each wavelength in the spectrum of an atom
corresponds to an electron transition between
orbitals
When an electron is excited, it transitions from an
orbital in a lower energy
gy level to an orbital in a
higher energy level
When an electron relaxes, it transitions from an
orbital in a higher energy level to an orbital in a
lower energy level
When an electron relaxes, a photon of light is
released whose energy equals the energy difference
between the orbitals
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To transition to a higher energy state, the
electron must gain the correct amount of
energy corresponding to the difference in
energy between the final and initial states
Electrons in high energy states are unstable
and tend to lose energy and transition to
lower energy states
Each line in the emission spectrum
corresponds to the difference in energy
between two energy states
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The wavelengths of lines in the emission
spectrum of hydrogen can be predicted
by calculating the difference in energy
between any two states
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For an electron in energy
gy state n, there
are (n – 1) energy states it can transition
to, therefore (n – 1) lines it can generate
The energy of a photon released is equal to the
difference in energy between the two levels the
electron is jumping between
It can be calculated by subtracting the energy
of the initial state from the energy of the final
state
Eelectron = Efinal state − Einitial state
Eemitted photon = −Eelectron
Both the Bohr and Quantum Mechanical
Models can predict these lines very
accurately for a 1-electron system
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Given: ni = 6, nf = 5
Find: m
Conceptual
Plan:
n i, n f
Eatom

Ephoton
Eatom = −Ephoton
p
Relationships:
E = hc/En = −2.18
2.18 x 10−18 J (1/n2)
Solve:
Ephoton = −(−2.6644 x 10−20 J) = 2.6644 x 10−20 J
Check: the unit is correct, the wavelength is in the infrared, which is
appropriate because it is less energy than 4→2 (in the visible)
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2 is the probability density
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The probability density function represents the total probability
of finding an electron at a particular point in space
◦ the probability of finding an electron at a
particular point in space
◦ for s orbital maximum at the nucleus?
◦ decreases as you move away from the nucleus
The Radial Distribution function represents
the
h totall probability
b bili at a certain
i di
distance
from the nucleus
◦ maximum at most probable radius

Nodes in the functions are where the
probability drops to 0
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2s
n = 2,
l=0
The radial distribution function represents the
total probability of finding an electron within a
thin spherical shell at a distance r from the
nucleus
3s
n = 3,
l=0
The probability at a point decreases with
i
increasing
i di
distance ffrom the
h nucleus,
l
b
but the
h
volume of the spherical shell increases
The net result is a plot that indicates the most
probable distance of the electron in a 1s orbital
of H is 52.9 pm
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
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The l quantum number primarily determines
the shape of the orbital
l can have integer values from 0 to (n – 1)
Each value of l is called by a particular letter
g
the shape
p of the orbital
that designates
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◦ s orbitals are spherical
◦ p orbitals are like two balloons tied at the knots
◦ d orbitals are mainly like four balloons tied at the
knot
◦ f orbitals are mainly like eight balloons tied at the
knot
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Each principal energy level
has one s orbital
Lowest energy orbital in a
principal energy state
Spherical
Number of nodes = (n – 1)
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Each principal energy state above n = 1
has three p orbitals
◦ ml = −1, 0, +1

Each of the three orbitals points along a
d ff
different
axis
◦ px, py, pz
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2nd lowest energy orbitals in a principal
energy state
Two-lobed
One node at the nucleus, total of n nodes
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Each principal energy state above n = 2 has
five d orbitals
◦ ml = −2, − 1, 0, +1, +2

Four of the five orbitals are aligned in a
different plane
◦ the fifth is aligned with the z axis, dz squared
◦ dxy, dyz, dxz, dx squared – y squared

3rd lowest energy orbitals in a principal energy
level
Mainly four-lobed

Planar nodes

◦ one is two-lobed with a toroid
◦ higher principal levels also have spherical nodes
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Each principal energy state above n = 3
has seven f orbitals
◦ ml = −3, −2, −1, 0, +1, +2, +3
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4th lowest energy orbitals in a principal
energy state
Mainly eight-lobed
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Planar nodes
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◦ some two-lobed with a toroid
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◦ higher principal levels also have spherical
nodes
Orbitals are determined from mathematical
wave functions
A wave function can have positive or
negative values
◦ as well as nodes where the wave function = 0
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The sign of the wave function is called its
phase
When orbitals interact, their wave functions
may be in-phase (same sign) or out-ofphase (opposite signs)
◦ this is important in bonding
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1/15/2014
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Define the term quantum mechanical model
Characterize the different regions of the
electromagnetic spectrum
Describe the evidence for the wave-particle dual
nature of light
Relate energy, frequency and wavelength
conceptually
p
y and mathematically
y
Explain and relate the concepts of threshold
frequency, binding energy and kinetic energy of an
ejected electron
Define the term emission spectrum
Relate deBroglie wavelength to mass and velocity
conceptually and mathematically
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Explain the terms complementary properties and
it’s role in Heisenberg’s uncertainty principle
Contrast deterministic with indeterminacy
Describe the purpose of each of the four quantum
numbers and use the rules that define allowable
sets
numbers
(the ffourth
t off quantum
t
b
(th
th quantum
t
number, ms is present in Chapter 8)
Solve for the energy and wavelength associated
with electron transitions in a Hydrogen atom
Describe the role of a node in a distribution
function
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CH2045, Gen. Chem. I, Section: 004
Spring 2014
Dr. Shengqian Ma
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31
Define the term periodic property
Define and apply the Pauli exclusion principle
Define the term degenerate as it applies to orbitals
Indicate the roles of Coulomb’s Law, shielding and
penetration in sublevel splitting
Define and apply the aufbau principle and Hund’s rule
Determine the expected electron configuration for any
atom on the periodic table (complete configuration and
noble gas abbreviation)
Describe an orbital filling diagram for any element on
the periodic table
Relate orbital filling diagrams, electron configurations
and quantum numbers
Determine number of valence electrons and core
electrons for any atom on the periodic table
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1/15/2014
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Compare the terms nonbonding atomic radius, bonding
atomic radius and atomic radius
Describe the trends in atomic radii on the periodic table and
relate the observed trends to the structure of the atom
Define and make predictions for diamagnetic and
paramagnetic
Relate the radius of an atom to an ion of the same element
Describe the trends in ionization energy on the periodic table
and relate the observed trends to the structure of the atom
Predict the expected trends in successive ionization energies
Define electron affinity
Describe what is meant by metallic character and relate it to
trends on the periodic table
Characterize the alkali metals, halogens and noble gases and
their trends on the periodic table
Baseball Cards:
year, team, player, card number, value ($).
Elements:
when
they weremass,
discovered,
family, reactivity,
alphabetically,
value, density,
state
of liquid
matter,ormetal
solid or
gas vs. non-metal, atomic mass,
atomic number.
Which way is CORRECT to organize the elements?
Is it possible to organize the elements correctly in more than one way?
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Dmitri Mendeleev
and Lothar Meyer
independently came
to the same
conclusion about
how elements
should be grouped.
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Ordered elements by atomic mass
Saw a repeating pattern of properties
Put elements with similar properties in the same
column
Used pattern to predict properties of undiscovered
elements (Germanium, Gallium, etc)
Where atomic mass order did not fit other properties,
he re-ordered by other properties
◦ Te & I
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1/15/2014
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This is an older style
table.
The rows are called
“periods.”
e co
u
s are
a e
The
columns
“families” or “groups.”
◦ Type A: representative
◦ Type B: transition
◦ The guys at the bottom
are inner transition
elements.
This is especially true of type A
(representative) elements.
 But, there are changes in reactivity, etc., as
one goes down a column.
 This link gives group IA as an example
example...
http://www.youtube.com/watch?v=Ft4E1eCUItI
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The rows are called periods.
How they fill gives insight into the atomic
structures.
In particular, we shall examine the idea of
electron configurations.
More about vertical trends later!
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1/15/2014
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As just stated, rows of the periodic table are
called periods.
These have the following lengths:
o2
o8
o 18
o 32
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=
=
=
=
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3 quantum numbers, n, l, and ml.
For each value of n we have n2 orbitals.
We play some number games here:
n
n
n
n
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1
2
3
4
→
→
→
→
1 orbital
4 orbitals (4 = 1 + 3)
9 orbitals (9 = 1 + 3 + 5)
16 orbitals (16 = 1 + 3 + 5 + 7)
These are all ½ the lengths of the periods!
What’s wrong?
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2=
8=
18 =
32 =
2
2
2
2
x
x
x
x
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22
32
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This gives some insight into quantum
numbers!
H has just one electron.
The Schrödinger equation seen earlier needed
another quantum number to handle multielectron atoms. (Technical detail: This
comes naturally if relativity included
included.))
This new quantum number is ms, the electron
spin. This has values of = +1/2 and -1/2.
The physical explanation is left for lecture!
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Calculations with Schrödinger’s equation show
hydrogen’s one electron occupies the lowest energy
orbital in the atom
Schrödinger’s equation calculations for multielectron
atoms cannot be exactly solved
◦ due to additional terms added for electron-electron
interactions
Approximate solutions show the orbitals to be
hydrogen-like
Two additional concepts affect multielectron atoms:
electron spin and energy splitting of sublevels
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Experiments by Stern and Gerlach
showed a beam of silver atoms is split
in two by a magnetic field
The experiment reveals that the
electrons spin on their axis
As they spin, they generate a magnetic
field
◦ spinning
charged
i i
h
d particles
ti l generate
t a
magnetic field
If there is an even number of electrons,
about half the atoms will have a net
magnetic field pointing “north” and the
other half will have a net magnetic field
pointing “south”
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