Download e -1 - Milwaukie High

Notes One
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Electronic Structure
Valence Electrons vs kernel Electrons
Structure of the atom
Electromagnetic Radiation
Calculating Frequency for Electromagnetic Radiation
Calculating Energy of Radiation
Matter and Energy
Calculate Smallest Increment of Energy
Electronic Structure
• Valence electrons
• Kernel electrons
• Nucleus
•
Protons
•
Neutrons
Valence Electrons?
kernel Electrons
Kernel e-1
Valence e-1
Gold [Xe]6s26p45d94f14
Iodine [Kr]5s25p64d10
Structure of the atom
Rutherford
Nucleus
-dense
-heavy
-small.
Bohr
-e-1 orbit nucleus
-e-1 are Particles
Planck
h=6.626 x 10-34 Js
Planck 1900
Quantum loss or gain
 E=h n
Electrons
De Broglie – Wave Nature.
Wavelength()
Lower
Higher Energy ????
Higher!
Electromagnetic Radiation
amplitude
Electromagnetic Spectrum
gamma
ray
X-ray
Short
wavelength
High
Energy
Visible
UV Light IR
radar
FM TV
Short
wave AM
Long
wavelength
Low
Energy
Electromagnetic Radiation
Waves have a frequency
n Is frequency.
•n = c
C= 3.00 x 108 m/sec
Long   low n
Short   high n
Calculating Frequency for Electromagnetic Radiation
Example: Red light has  = 700 nm.
Calculate the frequency, n.
n • = C
C
n=

8
3.00x10
14 Hz
=
4.29x10
n=
7.00x10-7
Calculating Energy of Radiation
• Calculate the energy of 1.00 mol of photons of red
light, when  = 700 nm and n = 4.29 x 1014 sec-1.
 E=h n
 E = (6.63x10-34 J•s)(4.29x1014sec-1)
 E= 2.85 x 10-19 J per photon
 E =(2.85 x10-19J/ph)x(6.02x1023ph/mol)
 E per mol = 171.6 kJ/mol
Matter and Energy
Until 1900's
matter particles
energy waves.
In 1901, Max Planck
multiples of hn.
h = 6.626 x 10-34Js
E = nhn n = 1,2,3...
It is quantized.
quanta
Calculate Smallest Increment of Energy
Calculate the smallest increment of energy (the quantum of
energy) that an object can absorb from yellow light whose
wavelength is 589 nm.
n • = C
C
n=

8m/s
2.998x10
14s-1
n=
=
5.09x10
5.98x10-7m
 E=h n
 E = (6.626x 10-34J-s)(5.09x1014s-1)
 E = 3.37 x 10-19J
The Bohr Model
Energy quantized
light quantized.
Bohr showed that


1
E  - 2.18 10 J  2 
n 
-18
n = 1, 2, 3…. (principal quantum numbers)
Limited to hydrogen
The Bohr Model
Hydrogen only.
Electrons described as
particles.
E=h•n
•n=c
Ionization Energy
• Calculate the ionization energy of the ground
state hydrogen atom (n=1) per mole.
E=
-2.18x10-18J
__
1
-18J
=-2.18x10
12
E=(-2.18x10-18J)x 6.022x1023 atoms/mol
E=-1312KJ/mol
Notes Two
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Photoelectric Effect
Emission Spectra Structure of the atom
Line Spectra (Energy)
Limitations of the Bohr Model
Energy Levels For Hydrogen
Ingredients of a laser
The Wave Behavior of Matter
Quantum Mechanics and Atomic Orbitals
Photoelectric Effect
THE PHOTOELECTRIC EFFECT
Different metals hold on to their electrons more strongly
or weakly due to atomic structure.
hn = Ek + W
h is Planck’s constant.
n is the frequency.
Ek is the kinetic energy.
W is the work function. Minimum ionization energy
1 electron volt = 1.60217646 × 10-19 joules
Ek = 1/2mv2
1joule = 1Coul x 1 volt
Ek = q x volts
Momentum(photon) p = h /
Momentum(electron) p = h / = 2m(Ek)
THE PHOTOELECTRIC EFFECT
hn = Ek
Ek
W
hn = Ek + W
Ek = 1/2mv2
n
slope= h
e-1
↑nth
W
n
eV
3.60x10-7
8.33x1014
4.00x10-7
7.50x1014
4.40x10-7
6.82x1014
4.90x10-7
6.12x1014
5.50x10-7
4.45x1014
5.80x10-7
5.17x1014
1.45
1.12
0.95
0.65
0.40
0.25
n= C

hn = Ek + W
Kinteic Energy eV
m n
+7
+6
+5
+4
+3
+2
+1
0
-1
-2 W
↑nth
-3
-4
0 1 2 3 4 5 6 7 8 9 10
Frequency Hz(1014)
If the threshold frequency required for emission of
photoelectron is 4.00x1014 Hz, what is the work
function of the surface in Joules and eV?
W = h•n
W = (6.63x10-34J-s) (4.00x1014 Hz)
W = (2.65x10-19J) ÷(1.60x10-19J)
W = 1.66eV
If the work function of a metal surface is 3.00x10-19J,
and it is illuminated by light with a wavelength of 500
nm, what is the speed of the escaping photoelectrons?
hn = Ek + W
hn - W = Ek
h• C - W = Ek

n = C
8m/s)
-34
(3.00x10
(6.63x10 J-s)•
(500x10-9m)
-(3.00x10-19J) = Ek
1
2
m
v
Ek
=2
1 (9.11x10-31Kg) v2
-20
=
9.78x10 J 2
V= 4.63x105m/s
= 9.78x10-20J
A metal surface has a work function of 2.06eV. If the
fastest photoelectron emitted from is 6.00x105m/s, what
is the frequency of the light source striking the surface?
1
Ek = m v 2
2
hn = Ek + W
Ek + W
______
n=
h 2
÷h
1/2 (9.11x10-31Kg)(6.00x105m/s) + ( (2.06eV)x1.60×10-19j)
(6.63x10-34J-s)
n=
7.44x1014s-1
Light of wavelength 400 nm is incident on a metal
surface, as shown above. Electrons are ejected from
the metal surface with a maximum kinetic energy of 1.1
x 10-19 J. (a) Calculate the frequency of the incoming
light. (b) Calculate the work function of the metal
surface. (c) Calculate the stopping potential for the
emitted electrons. (d) Calculate the momentum of
emitted electrons.
(a) Calculate the frequency of the incoming light.
n • = C
C
n=

8m/s
2.998x10
14s-1
n=
=
7.50x10
4.00x10-7m
(b) Calculate the work function of the metal surface.
hn = Ek + W
hn - Ek = W
(6.63x10-34J-s x7.50x1014s-1) - 1.1x10-19J = 3.87x10-19J
(c) Calculate the stopping potential for the emitted
electrons. Ek =q x V
1.1x10-19J = 1.60×10-19C x volts
V = 0.687 volts
(d) Calculate the momentum of emitted electrons.
Momentum P= 2m(Ek)
p = 2x 9.11x10-31kg (1.1x10-19J) =4.1x10-6kg-m/s
Emission Spectra
Emission
Spectrum
Slit
Prism
Hydrogen Lamp
Photographic
Plate
Exciting Gas Atoms to Emit Light
with Electrical Energy
Hg
He
H
Examples of Spectra
Helium
Barium
Oxygen
Neon
Identifying Elements with
Flame Tests
Na
K
Li
Ba
Ingredients of a laser
The Wave Behavior of Matter
Light  particle nature,
wave property
Einstein’s / Planck’s equations Broglie 
  h/mv
particle property
effects on objects are small.
Heisenberg’s Uncertainty Principle
• The position, direction of motion, and
speed simultaneously for an electron.
Quantum Mechanics and Orbitals
Schrödinger equation wave and particle terms.
Solving  wave functions.
(wave function )2  electron density
Notes Three
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Quantum Numbers (QN)
Pauli Exclusion Principle
Orbital Shapes and Energies
The History of the Periodic Table
Mendeleev’s Periodic Table
The Aufbau Principle and the Periodic Table
Hund’s Rule
Valence electrons
Filling out the Periodic Table
Orbital Filling Diagram
Write the OFD for Cu.
How many e-1 for Cu?
29 e-1

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 

[Ar] 4s 3d
9
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
 
2

 
 
Short Hand Element?
Ar
How far up does Ar
fill?
Final Answer?
2
2
6
2
6 2
9
1s 2s 2p 3s 3p 4s 3d
Short Hand Answer

Electron Dot Diagram for Selenium


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





●
Se
●


●
●

●●

[Ar]4s 3d 4p
4
 
 
10


 
2

 
 





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
Quantum Number for Last e-1 in Cd

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


ms = +1/2 or -1/2=
l = 0(s), 1(p), 2(d), 3(f)
=
How many e-1 does Cd have?
48 e-1
Short hand element?
Kr
Many e-1 does Kr have?
36 e-1
How many e-1 to account for?
12 e-1
Short-hand fills to where?
Fill-in Valence12 e-1
Quantum #’s for Last e-1
n=4 l= 2 ml=-2 ms=-1/2
Quantum Numbers

n= 4 l= 3 ml = +2 ms = -1/2
Principal Quantum #(n)
n= 1, 2, 3, . .
Distance from
3d__ __ __ __ __
the nucleus
Weaker Attraction
-1
3p__
__ __
e more NRG
3s__
e-1 far
Larger n  higher energy
Larger n less strongly
bound to nucleus
stronger
Attraction
2p__
__
__
2s__
e-1-1less NRG
1s__
 e near
Angular Momentum Quantum #(l )
relates to orbital shape. l = 0 to n - 1
n=1  l = 0
n=2  l = 0, 1
n=3  l = 0, 1, 2
n=4  l = 0, 1, 2, 3
n=5  l = 0, 1, 2, 3, 4
n=6  l = 0, 1, 2, 3, 4, 5
n=7  l = 0, 1, 2, 3, 4, 5, 6
l = 0 for s
l = 1 for p
l = 2 for d
l = 3 for f
10-8 m
1s
Probability density Ψ2
Probability density Ψ2
Probability density Ψ2
The s Orbital
10-8 m
2s
l=0
?
?
???
No Nodal Plane
10-8 m
3s
The p Orbital
Angular Momentum QN
(Shape)
?? ?
l=1 ??
One Nodal Plane
The d Orbital
Angular Momentum QN
(Shape)
l=2 ?
?
TwoNodal Planes ??
The f Orbital
Angular Momentum QN
(Shape)
l=3
?
?? ?
?
n
1
2
3
4
5
l
Magnetic Quantum #(ml )
0 1s
0 2s
1 2p
0 3s
1 3p
2 3d
0 4s
1 4p
2 4d
3 4f
0 5s
1 5p
2 5d
3 5f
4 5g
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
0
+1 0 -1
+2 +1 0 -1 -2
+3+2 +1 0 -1 -2 -3
+4+3+2 +1 0 -1 -2 -3 -4
Spin Quantum #(ms)
Relates to electron spin states.
spinning charges.
magnetic field
Two spins
north
+1/2
south
+1/2
Pauli Exclusion Principle
No two electrons can have the same set of four
quantum numbers (n, l, ml, ms).
An orbital can hold only two electrons,
Electrons can have the same n, l, and ml values
If n = 3,
l = 2 (d-orbital),
ml = -2 (a single d-orbital)
ms = +1/2, and one with ms =-1/2
Energy Order of Orbitals
What determines the order of filling?
Penetration
2p
2s
3s
3p
3d
Distance from nucleus
Distance from nucleus
To find energy order between orbitals (Penetration).
Determine the lowest n+l
Determine the lowest n+l
Which orbital has the greatest penetration?
3d or 4s?
n+l
3d 3 + 2= 5
3d 3 + 2= 5
4f 4 + 3= 7
4s 4 + 0= 4
4 + 2= 6
2p
2 + 1= 3
6s 6 + 0= 6
Lowest n, if a tie.
3s
3 + 0= 3
7p
7 + 1= 8
4f
4 + 3= 7
5f
5 + 3= 8
5s
5 + 0= 5
4d
end
The History of the Periodic Table
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Patterns in element properties were recognized
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Dobereiner (1780-1849) found “triads” of similar elements: Cl, Br, I
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Newlands suggested in 1864 that elements should be arranged in
“octaves” because similarities occurred every 8th element
The Modern Periodic Table
•
The German Meyer (1830-1895) and Russian Mendeleev (18341907) independently developed the current arrangement of
elements
•
Mendeleev predicted the properties of “missing” elements
Aufbau Principle
Lowest energy (penetration) orbital gets electron first.
Hunds Rule
Distribute electrons in multiple orbital energy levels
C
F
Elect. Configuration
1s
2s
1s2 2s2 2p 2
1s2 2s2 2p 5
↑↓
↑↓
2px
2py
↑↓ ↑
↑↓ ↑↓
2pz
↑
↑↓
↑
Elect. Configuration 4s 3d 3d 3d 3d 3d 2p 2p 2p
2 2
x -y
2
10
2
6
2
z
xy
xz
1 ↑↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓
Ga
[Ar]4s 3d 4p
Fe
[Ar]4s 3d
yz
↑↓ ↑
↑↓ ↑ ↓ ↑ ↑ ↑ ↑
x
y
z
•Mendeleev’s Periodic Table
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Left blank spaces for the elements that he predicted would be
discovered
Arranged elements by mass, instead of by atomic number
Four Quantum Numbers
Ionization Energy (I.E.)
The quantity of energy required to remove an electron from the gaseous
atom or ion.
1. X(g) X+ + e2. It is easiest to take away the first e-, more energy needed for others
3. Greater charge makes it harder to extract e4. First e- comes from the outermost (weakest bound) orbital
5. I. E. increases from left to right on periodic table because larger
elements have larger +/- attraction for electrons
6. I. E. decreases down a group because outer e- becomes more
weakly bound
Ionization Energy (I.E.)
Electron Affinity (EA)
Energy change when e- is added to a gas atom
X(g) + e- -------> X-(g)
1.
2.
3.
EA generally become more negative (exothermic) from left to right
a. Interactions between the +nucleus and the added electrons is
favorable
b. The larger the nucleus, the better the interaction
Exceptions to the trend is a function of electron—electron interactions
a. N- doesn’t form because N has 1s22s22p3 configuration
b. The extra electron would have to go into an occupied 2p orbital
c. C- does form because it has an empty orbital: 1s22s22p2
d. This phenomenon doesn’t always hold true. O- can form 1s22s22p4
EA diminishes down a group
a. Added e- is farther from the nucleus, so less interaction is observed
b. The difference in EA is small and has exceptions (small F 2p orbital)
Atomic Radius:
• C. half the distance between the
nuclei
• molecule consisting of identical
atoms.
1. Easy to understand for diatomic
X2 molecules
2. Can be calculated from known
radii of XY molecules
3. Decreases from left to right on
periodic table because +/attraction becomes greater
4. Increases down a group
because more e- and larger
shells filled
Names for groups of the Periodic Table
Metals: tend to give up electrons to form cations
found on the lower left
Nonmetals: tend to gain electrons to become anions
found on the upper right
Metalloids: have properties of both metals and nonmetals
depending on the conditions