modern atomic theory- electronic structure of atoms · 2010. 11. 22. · modern atomic...
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Modern Atomic Theory-Electronic Structure of Atoms
DR HNIMIR-CH7
Problems with the atomic model?Where should (-) electrons be found?
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First, a Little About Electromagnetic Radiation- Waves
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Another Look
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Relationship
λλλλ νννν = c
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Planck’s Relationshipblack body radiation experiment
∆E = h ν
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ProblemFor green visible light of wavelength 486 nm
a. what is the frequency of this radiation?
b. Calculate ∆E
λλλλ νννν = c
∆E = h ν
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Meaning of results?
4.086x10-19 J is the minmum amount of energy to be gotten from this green light. If you want more,
you have to take 2 x 4.086x10-19 J (or 3x, 4x, etc)
4.086x10-19 J
green light of wavelength 486.0 nm comes "packaged" in packets
(called quanta in general and photons for EM) of 4.086x10-19 J
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Two More ThingsFor red light of wavelength 725.0 nm
∆E is 2.740x10-19J
2.740x10-19J
When you calculate ∆E the units are actually J/photon
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The EM Spectrum
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Einstein’s Involvement
"particle / wave dualism"
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The Photoelectric Effect
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Frauhoffer LinesWith good optics it was noticed that
the continuous spectrum obtained from the sunhad very fine black lines in it
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Line SpectraThe Bohr Atom
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The Hydrogen Spectrum
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The Bohr ModelNiels Bohr interprets the lines as quantized energy emitted by electrons between allowed energy levels
1
2
3
4
an electron absorbs energy and is promoted to
a higher energy level
1
2
3
4
the excited electron returns to a lower energy level and emits a specific amount of energyseen as a sharp line of light
1
2
3
4
∆E
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Bohr Orbits
Bohr calculates the energy of the nth orbit by the formula
En = -B
n2
where B= 2.180x10-18J for the hydrogen atom
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ProblemCalculate the energy of the 2nd and 4th energy levels
En = -2.180x10-18J
n2
Calculate ∆∆∆∆E for the transition (∆∆∆∆E= E2-E4)
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Calculate the Wavelength (in nm) Associated with the Answer from
the Last Slide∆E= - 4.087x10-19J
∆E = h νλλλλ νννν = c
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Interpretation of the Last Slide’s Results
λλλλ= 486.0 nm
700 nm600 nm500 nm400 nm
This spectral line is due to a 4 2 electron transition
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A Useful Derived Equation∆E= Efinal - Einitial = Einner - Eouter
-2.180x10-18J
ni2
-2.180x10-18J
no2
∆E= -
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Redo the previous problem using this equation
1
ni2
1 no
2∆E= -2.180x10-18J
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Regions of Spectral Lines
1
2
3
4
UV
Lyman series
1
2
3
4
VIS
Balmer series
1
2
3
4
IR
Paschen series
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DeBroglieAssumes that if light (waves) could have par ticle proper ties
then particles could have wave properties
∆E = h ν∆E = mc2
h ν= mc2 h c = mc2
λ λ = h mc
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InterpretationThe wavelength of an electron must be
an integer multiple of the Bohr orbits (quantized)
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Evidence for Wave Nature of the Electron
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Interference Patterns Observed
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Heisenberg Uncertainty Principle
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Schroedinger’s Equation
He combines the mathematics of waves and probability
He constructs 3D probability maps for electrons, called " orbitals"
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Orbitals
"s""p"
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s,p and d orbitals
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Energy Levels, Sub-levels and Orbitals
ENERGY
1
2
3
4
energy level sub-level
1s
2s
2p
3s
3p
3d
4s4p4d4f
orbitals
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Electron Configurations-putting it all together
O8 8 electrons
start with the lowest energy level 1s sublevel is only one in 1st energy levelthe one orbital can hold 2 electrons
this fills the first energy level
next electrons goes in the 2nd energy levelthe s sub-level is lower than ptwo electrons fill the s sublevel
there is a p sublevel in the 2nd energy level
the remaining electrons can go into the p sub-level of the 2nd energy level
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Order of electron filling-first 36 elements (through krypton)
1 s 2 s 2 p 3 s 3 p 4 s 3 d 4 p
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“Out of Order”
3s
3p
3d
3rd energy level
4th energy level
4s
4p4d4f
4s
4p4d4f
3s
3p
3d
3rd energy level
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Maximum number of electrons:
1 s 2 s 2 p 3 s 3 p 4 s 3 d 4 p
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Simplifying Electron Configurations
"inert gas core" notation
19K
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Electron Spin-Stern Gerlach Experiment
furnace
Agbeam of gaseous
silver atoms
N
S
powerful magnet
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Electron Spin Pairing
representing paired electrons in an energy diagram
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Energy Diagrams- visualizingelectron configurations
1s 2s 2p
1s
2s 2p
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Example-carbon
1s
2s 2p
6C 1s2 2s2 2p2
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Q5Write electron configurations for Si and Ge
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Q5 cont.Construct energy diagrams for P and F
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Q5 cont.
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Transition Metals
Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn
4s
3d
2 2 2 2 2 2 2 2 2
0 1 2 3 5 6 7 8 10
1
5
1
10
Why the difference?
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Half-filled Sub-levels are More Stable
4s3d
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Problemdo the same for Cu
4s3d
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Magnetic Properties of Atoms
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Quantum Numbers
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Principle Q.N.n
n gives the energy level of the electron
n = 1,2,3,.... (any integer)
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Angular Momentum Q.N.l
the values of l are:
0, 1, ...., n-1
gives the number of sublevels in any energy level
(if n =2, then l= 0 and 1)
meaning:in the second energy level (n=2) there are two sub-levels (s and p)
l sub-level
0 s
1 p
2 d
3 f
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Orientation Quantum Numberml
gives the number of orbitals in a sublevel
ml = -l ....,0,.... +l
if l = 1 (p sub-level)
ml = -1,0,+1 (three p orbitals in the p sub-level)
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Spin Quantum Numberms
tells whether an electron is unpaired (the first into that orbital) or paired
ml = +1/2 for the first electron in an orbital
ml = -1/2 for the paired electron
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How it Works
8O 1s22s22p4
1s
2s 2p
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Q6Provide the four QN's for the characteristic electron of phosphorous
Which element would have the following character isticQN's?
n = 2 l = 1 ml = 0 ms = -1/2