ELECTRONIC STRUCTURE Bohr ModelRydberg Eqn & Constant E-Levels; quantum #’s Planck’s Eqn &...

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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

Classical Physics

+ nucleus === - e-

∆: smoothcontinuous spectra

Quantum Theory: aids to explain behavior of e-

describes e- arrangement in atoms

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

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

amplitude

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

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

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

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

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)

3 groups of lines H spectrum

1016 1015 1014

Lyman Series UV

Balmer Series VIS

Paschen Series IR

nn ---> n1 nn ---> n2 nn ---> n3

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-

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

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

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.

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

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

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

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

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

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

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-

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

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------------------------------->

E- Notation

Form: 1s2

E-level“shell”

subshell

# e-

Be

N

F

Na

Al

1s22s2

1s22s2p3

1s22s2p5

1s22s2p63s1

1s22s2p63s2p1

1s22s22p3

1s22s22p5

1s22s22p63s1

1s22s22p63s23p1

Ca1s22s2p63s2p64s2

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

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-

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

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

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

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

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

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

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

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

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

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

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