ELECTRONIC STRUCTURE Bohr ModelRydberg Eqn & Constant E-Levels; quantum #’s Planck’s Eqn &...
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Transcript of ELECTRONIC STRUCTURE Bohr ModelRydberg Eqn & Constant E-Levels; quantum #’s Planck’s Eqn &...
![Page 1: ELECTRONIC STRUCTURE Bohr ModelRydberg Eqn & Constant E-Levels; quantum #’s Planck’s Eqn & Constant Ionization E Electromagnetic Radiation Classic Physics.](https://reader035.fdocuments.net/reader035/viewer/2022062421/56649e5d5503460f94b55374/html5/thumbnails/1.jpg)
ELECTRONIC STRUCTURE
Bohr Model Rydberg Eqn & Constant
E-Levels; quantum #’sPlanck’s Eqn & Constant
Ionization E
Electromagnetic Radiation
Classic Physics - wave vs particle
de Broglie WavelengthHeisenberg Uncertainty Principle Quantum Mechanics
Orbital Shapes
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Classical Physics
+ nucleus === - e-
∆: smoothcontinuous spectra
Quantum Theory: aids to explain behavior of e-
describes e- arrangement in atoms
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Electromagnetic Radiation: consists of E by means of electrical & magnetic fields; inc/dec intensity as move thru space
7.5*1014 6.0*1014 4.0*1014
400 500 750
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WAVE PROPERTIES
2 Independent Variables
Frequency, : cycles /s, 1/s --- s-1 (Hertz)
Wavelength, : dist bet crest or trough of a wave dist in 1 cycle m, nm (109 nm = 1 m), pm, Å
In vacuum all electromag radiation same speed 3.00*108 m/s === speed of light (c) c = *
radiation wavelengthhigher , shorter
Amplitude: height of crest; depth of trough - higher amp, increase intensity - measure strength of fields
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amplitude
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Max Planck --- Planck’s Constant
E = nh
n: quantum #; 1, 2, 3, …h: constant 6.626*10-34 J-s: frequency
E of atom is quantized; specific amtsquantum: fixed amt of E required to move e- to next E-level quantum = h
Einstein Photon: quantized bundles of E; light - behaves as a particle
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Calculate E from
n= 1 & : m
c = == = c/E = nh = h = h(c/ )
Sr
Photoelectric Effect: light delivering E fixed amts; supports E-levels & quantum model of atom; light dual nature, behaves as waves & composed of particles
Light shining on metallic surface will emit e-if min. light frequency is met
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Calculate wavelength of yellow light that emitsa frequency of 5.10*1014 s-1.
= c/
s 10 10.5
sm 10 3.00
1-14
-18
5.88 * 10-7 m
Calculate the E, J, of a quantum of E w/ a frequency of 5.00 * 1015 s-1.
E = h *
E = (6.626 * 10-34 J-s) (5.00 * 1015 s-1) = 3.31 * 10-18 J
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ATOMIC SPECTRA
n2 > n1
n1 = 2 visible series
Ryberg Eqn -- Rydberg Constant 1.097*107 m-1
3 Postulates - Atom certain allowable E-levels (stationary states) - Atom not emit E in stationary state - Atom ∆es states when +/- photon of = difference of E bet 2 states
BOHR MODEL --- H atom
Ephoton = Estate 2 – Estate 1 = h emit specific quantum of E
22
21 n
1 -
n
1R
1
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n: quantum #, 1, 2, 3, …. n = 1 smallest atomic radius smallest E-level
lowest E-level is ground statee- + photon; photon E match diff bet n1 & n2;e- moves to 2nd E-level, excited state; (any higher E-level above ground state)e- emits same amt E as absorbed from photon;moves (falls) back to ground state E-level
e- location:
Calculate E-levels
Z: + charge on nucleusn: quantum #, ground state
E = -2.18*10-18 J *(Z2/n2)
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3 groups of lines H spectrum
1016 1015 1014
Lyman Series UV
Balmer Series VIS
Paschen Series IR
nn ---> n1 nn ---> n2 nn ---> n3
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Combine ∆E w/ Planck’s === Rydberg’s Eqn
Diff 2 E-levels
IONIZE H
H (g) ---- H+ (g) + e-ni = 1 nf = ∞
2
i2f
18-
n
1 -
n
1J 10*2.18- E
n
1 -
n
1
hc
J 10*2.18-
12i
2f
-18
2i
2f
1-7
n
1 -
n
1* m 10 *1.097-
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amt E absorbed, +, to completely remove 1 e-
J 10*2.18 1
1 -
1J 10*2.18- E 18-
2218-
J/mol 10*1.318 mol 1
atom 10*6.022J/atom 10*2.18 H of mol 1 6
2318-
What is the E of an e- in the 2nd E-level?
n
1J 10*2.18- E
218-
2
1J 10*2.18-
218- -5.45 * 10-19 J
What is the E required to excited e- from n=1 to n=2?
E = E2 - E1 = (-5.45*10-19) - (-2.18*10-18) = 1.63*10-18 J
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Calculate the wavelength (nm) of the line in the spectrum (H)corresponding to n1 = 2 to n2 = 4
Calculate the E emitted from a hydrogen atom when an e- drops from5th to 2nd E-level. Calculate frequency (Hz) & wavelength (nm).
(4.862*10-7 m)*(1 nm/1*10-9 m) = 486 nm
E = h = E/h
1-622
21
1-7 m 10*2.057 4
1 -
2
1m 10*1.097
1
m 10*4.862 m 10*2.057
1 m 10*2.057
1 7-1-6
1-6
Ryberg Eqn 1.097*107 m-1
2
i2f
18-
n
1 -
n
1J 10*2.18- E
c = / = c/
4.58*10-19 J6.91*1014 s-1
434 nm
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1905 -- Einstein -- E=mc2
matter & E are alternate forms of sameE is particle; physicists matter is wave-like
atom only has certain allowable E-levelsexplain line spectrum
Bohr Model
de Broglie
study of systems w/ only allowable motionsextended this reasoning to e- behave wave-like, restricted to fixed radii explain why e- certain E’s & freq.
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de Broglie combined: E=mc2 & E=h=hc/eqn for of any particle w/ mass (m) @ speed u
Matter moves in a wave inverse to its mass, therefore, heavy objects, their <<<<< smaller
Question was asked: if e- have properties of E, do photons have properties of matter?Can calculate the momentum (p)
um
h
h
p & p
h
cm
h
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Calculate the for an e- (9.11*10-31 kg) w/ speed of1.0*107 m/s.
um
h
6.626 * 10-34 J/s ====> 6.626*10-34 kg-m2/s
sm 100.1kg 109.11s
mmkg106.626
731-
34-
e 7.27*10-11 m
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Classical sense; moving particle has definite position know path & location of an object
Postulated that --- if e- is both particle and wave like, then impossible to know position & momentum of e- at the same timemore know position, less know of speed
Outcome for Atomic Model
Can state is the probability of e- in a given region;but still not sure??
Not assign fixed orbit; as Bohr model shows
HEISENBERG UNCERTAINTYUNCERTAINTY PRINCIPLE
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PAULI EXCLUSION PRINCIPLE
An orbital can hold only 2 e-’s & the 2 e-’s must have opposite spin
An orbital occupied by 2 e-’s w/ opposite spin is filledfilled
2 Hydrogen atoms 1s1 & 1s1 H -- H
2 e-’s w/ same spin direction cannot occupy same region of space
Will these 2 hydrogen atoms bond together ?????
Only if the 1s e-’s are of opposite spin
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e- structuree- structure e- locate: outside of nucleus e- cloud, shell, “subshell”, orbital, E-level E is quantized; a specific value
Shell ( E Level) 1st: closest to nucleus, lowest in E 7th: farthest from nucleus, highest in E
w/i shells are “subshells” s, p, d, f
w/i subshells are orbitials geometric shaped regions where the high probability to locate e- exists
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Shell
subshell
orbitial
max. e-
1 2 3 4 5 6 7
s s, p s, p, d s, p, s, p, s, p, d s, p d, f d, f
1 1-3 1-3-5 1-3-5-7 1-3-5-7 1-3-5 1-3
2 2-6 2-6-10 2-6- 2-6- 2-6-10 2-6 10-14 10-14
s: 1st 2 elements of each row; 1 pair, 2 e-; 1A - 2A
p: last 6 elements of each row; 3 pair, 6 e-; 3A - 8A
d: transition elements; 5 pair, 10 e-; B
f: rare earth elements; 7 pair, 14 e-; not labeled
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How many e- are present, group by E-levels
He
Be
N
F
Na
Al
1st ---> 2e-
1st ---> 2e- 2nd ---> 2e-
1st ---> 2e- 2nd ---> 5 e-
1st ---> 2e- 2nd ---> 7 e-
1st ---> 2e- 2nd ---> 8 e- 3rd ---> 1 e-
1st ---> 2e- 2nd ---> 8 e- 3rd ---> 3 e-
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Now, identify e- by subshellsHe
Be
N
F
Na
Al
1st :2e-
1st:2e- 2nd:2e-
1st:2e- 2nd:5 e-
1st:2e- 2nd:7 e-
1st:2e- 2nd:8 e- 3rd:1 e-
1st:2e- 2nd:8 e- 3rd:3 e-
s
s s
s s, p
s s, p
s s, p s
s s, p s, p
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1
2
3
4
5
6
7
<-------------------------1s------------------------------------------------>
<-2s->
<-7s->
<-6s->
<-5s->
<-4s->
<-3s->
<--2p------------------>
<-----3p--------------->
<--------4p------------>
<-------------5p---------><---------------6p----->
<------------------7p-->
<-------3d-------------------------->
<------------4d-------------------->
<----------------5d---------------->
<--------------------6d------------>
<-------------4f---------------------------------------->
<----------------------5f------------------------------->
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E- Notation
Form: 1s2
E-level“shell”
subshell
# e-
Be
N
F
Na
Al
1s22s2
1s22s2p3
1s22s2p5
1s22s2p63s1
1s22s2p63s2p1
1s22s22p3
1s22s22p5
1s22s22p63s1
1s22s22p63s23p1
Ca1s22s2p63s2p64s2
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After 4s2, 3d level fills till 3d10 completed, then complete 4p
4p to 5s, 4d completed, then complete 5p
5p to 6s, 4f level fills till 4f14 completed, then start 5d to 6p
Mn 1s22s2p63s2p64s23d5
Zn 1s22s2p63s2p64s23d10 Ga 1s22s2p63s2p64s23d104p1
Pd-46 1s22s2p63s2p64s23d104p65s24d8
Nd 1s22s2p63s2p64s23d104p65s24d105p66s25d14f3
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Group E-levels
Order e- Filling
Short-cut
1s1s222s2s22pp663s2p64s23d104p65s24d105p66s25d14f3
1s1s222s2s22pp663s2p6d104s2p6d10f35s2p6d1 6s2
1s1s222s2s22pp663s3s22pp664s4s223d3d10104p4p665s5s224d4d10105p5p666s25d14f3
describes noble gas @ end of row 5 (Xe)this is added on to
the noble gas notation filled inner core of e-
[Xe]4f35d1 6s2
Valance e- (2)
total # of e- in highest E-levelmethods easiest to show # val. e-
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RULESRULES & PRINCIPLES
HUND’S RULEHUND’S RULE
Describes the lowest E arrangement of e-
1. e- enter orbital of same E singularly till orbital half filled
2. Each orbital must be occupied with an e- w/ parallel spin direction before e-’s are paired w/ opposite spin
Based on results of measurements of magnetic properties
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FILLING ORBITALSS 1s1s222s2s22pp663s3s22pp44
1ss
2ss
ppXX ppYY ppZZ
3ss
ppXX ppYY ppZZ
1s1s
2s2s pp
3s3s pp
22
22
22 44
66
Valence e- level6 valence e-’s
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FILLING ORBITALSCr 1s1s222s2s22pp663s3s22pp66dd554s4s11
1ss
2ss
ppXX ppYY ppZZ
3ss
ppXX ppYY ppZZ
1s1s
2s2s pp
3s3s pp
22
22
22 66
66
4ss
4s4s 11
dd
22
44
d d d d dd d d d d
1 valenceelectron
55
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What Element??? Atom??
1ss
2ss
ppXX ppYY ppZZ
3ss
1s1s222s2s22pp663s3s22pp66
ss4 3
d d d d dd d d d d
ppXX ppYY ppZZ
ss4
3d d d d dd d d d d
NiNickel
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PRACTICE PROBLEMS
Write electron notation, long & short, for: Cobalt - Silver - I-1 ion
Draw orbital diagram for Cobalt
Identify the following elements:
1s22s2p63s2p64s23d104p5
1s22s2p63s2p6d104s2p6d75s2
Identify the number of valenceelectrons in each element
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Schrödinger Eqn: used to describe model for H atom - certain allowabel E - wave behavior - e- position not known
QUANTUM MECHANICS
E HE: E of e- : psi, wave fct H: Hamiltonian Operator
has no meaning, but, 2 is probability density
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3 QUANTUM NUMBERS
1 SIZE 2 SHAPE 3 ORIENTATION
1 Principal QN (n); + integer 1, 2, 3, …. identify E-level indicates size due from probability distr
2 Angular Momentum QN (l); 0 to (n-1) shape n limits l amt of l values = n n = 1 l = 0 n = 2 l = 0, 1 n = 5 l = 4, 3, 2, 1, 0
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3 Magnetic QN (ml);-1, 0, +1 shows orientation of orbitals around the nucleus l sets the ml values l = 0 then ml = 0 l = 1 then ml= -l, 0, +l
# of ml values = # of orbitals; 2l + 1l = 2 ml = -2, -1, 0, +1, +2
H Atom Summary
1. In quantum (wave) model, e- a standing wave. Leads to series of wave fcts (orbitals) describe possible E & spatial distributions2. As w/ Heisenberg, model cannot detail e- motions. But, the 2
shows probability distr of e- in that orbital (e- density maps)3. Orbital size defined as surface that contains 90% of total e- probability4. H atom many types of orbitals. Ground state is 1s, but can be excited w/ input of min E requirement
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l Sublevels (subshells) s, p, d, fl=0 l=1 l=2 l=3
orbitals n=1 l=0 1s 2s: n= l= 2p: n= l= 3s: n= l= 3p: n= l= 3d: n= l=
2 0 2 13 0 3 1 3 2
n Level: E-level (shells)
# of orbitals n= n2
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l = 0 then ml = 0 l = 1 then ml = -1, 0, +1l = 2 then ml = -2, -1, 0, +1, +2
SPx, Py, Pz
dxy, dyz, dz2, dxz, dx2-y2
n l m describes ? ? 0 4p 2 1 0 ? 3 2 -2 ? ? ? ? 2s
4 12p3d
2 0 0
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d orbitals dumb bell shaped; dz2: disc shape bet dumb bells
2p: n = 2 l = 1 ml = -1, 0, +1 Px Py Pz
p orbitals dumb bell shaped 2 regions either side of nucleus of high probability
l = 0l = 1l = 2l = 3
orbitalspdf