Chapter One - 國立臺灣大學sfcheng/inchem94/Chapt 1.pdf · 2005-09-21 · Slide 3 of 58...

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Slide 1 of 58 Chapter One Atomic Structure

Transcript of Chapter One - 國立臺灣大學sfcheng/inchem94/Chapt 1.pdf · 2005-09-21 · Slide 3 of 58...

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Chapter One

Atomic Structure

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The abundances of the elements in the universe

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Subatomic particle of relevance to chemistry

100Electromagnetic radiation from nucleus

γγ photo

½-10e- ejected from nucleus

ββ particle

0+2424He2+ nucleusαα particle

½+105.486 x 10-4e+Positron

½00c. 0νNeutrino

½011.0087nNeutron

½+111.0073pProton

½-105.486 x 10-4e-Electron

SpinCharge/e#

Mass number

Mass/u*SymbolParticle

*Masses are expressed in atomic mass units, u, with 1 u = 1.6605 x 10-27 kg# Elementary charge e = 1.602 x 10-19 C.

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Nuclear Binding Energyfor Helium

Ebind = (∆m)C2

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Average Binding Energies

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Rutherford’s Alpha Scattering Experiment

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Rutherford’s Model

• Ernest Rutherford discovered the positive charge of an atom is concentrated in the center of an atom,the nucleus

– An atom, can be visualized as a giant indoor football stadium

– The nucleus can be represented by a pea in the center of the stadium,

– The electrons are a few bees buzzing throughout. The roof of the stadium prevents the bees from leaving.

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Continuous Spectra

Line Spectra

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Emission Spectrum of Hydrogenin Visible Light Region

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Line Spectra of Some Elements

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Planck’s Constant

• Planck’s quantum hypothesis states that energy can be absorbed or emitted only as a quantum or as whole multiples of a quantum, thereby making variations discontinuous, changes can only occur in discrete amounts.

• The smallest amount of energy, a quantum, is given by:

E = hv

as Planck’s constant, h = 6.626 X 10-34 J s.

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The Photoelectric Effect• Albert Einstein considered electromagnetic energy

to be bundled in to little packets called photons.

Energy of photon = E = hv

– Photons of light hit surface electrons and transfer their energy hv = B.E. + K.E.

– The energized electrons overcome their attraction and escape from the surface

hv e- (K.E.)

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Bohr’s Hydrogen Atom

ve-rZ •

Orbit

Postulations•Rutherford’s nuclei model•The energy of an electron in a H atom is quantized•Planck & Einstein’s photon theory

E = hv•Electron travels in a circle•Classical electromagnetic theory is not applied

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(1) Classical physics

centripetal force = Coulombic attractionmv2/r = Ze2/r2

E = - ( 2π2 mZ2e4)/(n2 h2) when n =1, E (1) = - (2π2 mZ2e4)/( h2)

E = E (1) /n2

r = (n2 h2)/ (4π2 mZe2)

(2) Total energy

E = 1/2 mv2 - Ze2/r

(3) Quantizing the angular momentum

mvr = n (h/2π )

Quantum number

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Bohr’s Hydrogen Atom

• Niels Bohr found that the electron energy (En) was quantized, that is, that it can have only certain specified values.

• Each specified energy value is called an energy level of the atom

En = - B/n2

– n is an integer, and B is a constant which equals 2.179 x 10-18 J

– The energy is zero when the electron is located infinitely far from nucleus

– The negative sign represents the forces of attraction

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The Bohr Model

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Bohr Explains Line Spectra

• Bohr’s equation is most useful in determining the energy change (∆Elevel) that accompanies the leap of an electron from one energy level to another

• For the final and initial levels:

Ef = -B / nf2 Ei = -B / ni

2

The energy difference between nf and ni is:

∆Elevel = Ef - Ei

= ( -B / nf2 ) – (-B / ni

2 )

= B(1/ni2 – 1/nf

2)

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Energy Levels and Spectral Lines for Hydrogen

Visible

UV

IR

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Ground States and Excited States

• When an atom has its electrons in their lowest possible energy levels, it is in its ground state

• When an electron has been promoted to a higher level, it is in an excited state

– Electrons are promoted through an electric discharge, heat, or some other source of energy

– An atom in an excited state eventually emits photons as the electron drops back down to the ground state

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Problems of Bohr’s Model of Atom

•The energy levels of Bohr’s H atom cannotbe applied to other atoms.

•The orbit of electrons cannot be defined.

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The Uncertainty Principle

• Werner Heisenberg’s uncertainty principle states that we can’t simultaneously know exactly where a tiny particle like an electron is and exactly how it is moving

(∆Px) (∆x) = h/4π = Px = mvxπ2h

=h2h

•The act of measuring the particle actually interferes with the particle• In light of the uncertainty principle, Bohr’s model of the hydrogen atom fails, in part, because it tells more than we can know with certainty.

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Uncertainty Principle Illustrated

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De Broglie’s Equation• Louis de Broglie speculated that matter can behave

as both particles and waves, just like light

• He proposed that a particle move with a mass mmoving at a speed c will have a wave nature consistent with a wavelength given by the equation:

p = mc

E = mc2 = pc = hν

p = hν /c = h/ λ

• De Broglie’s prediction of matter waves led to the development of the electron microscope

λ = h/p = h/mc

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What is the speed of an electron to have a wavelength of X-ray?

Wavelength of X-ray ~ 0.1 nm = 1 x 10-10 mMass of electron = 9.11 x 10-31 kgPlanck constant h = 6.626 x 10-34 Js (kg m2/s)

<Answer>λ= h/p = h/mvv = h/mλ = (6.626 x10-34)/[(9.11 x10-31)(1 x 10-10)]

= 7 x106 m/s

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How to achieve the speed of an electron of 7 x106 m/s?

V

<Answer>E= eV= ½ mv2

V = ½ mv2 /e = ½ (9.11 x10-31)(7 x106)2 /(1.6022 x 10-19)V = 140 V(1 eV = 1.6022 x 10-19 J)

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Electron diffraction The experimental confirmation of de Broglie’s wave hypothesis was first made in 1927 by Davisson and Germer of the Bell Laboratories who investigated the scattering of electrons from various surfaces.

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Wave Functions

• Quantum mechanics, or wave mechanics, is the treatment of atomic structure through the wavelike properties of the electron

• Wave mechanics provides a probability of where an electron will be in certain regions of an atom

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Erwin Schrödinger developed a wave equation to describe the hydrogen atom

• An acceptable solution to Schrödinger’s wave equation is called a wave function

• A wave function (ψ) is characterized by an energy state of the atom

1),,( 2 =∫+∞

∞−τψ dzyx

2),,( zyxψ : the probability of finding an electron at (x, y, z) position in an atom

The probability of finding an electron in the universe is equal to 1.

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x

y Traveling wavey(x, t) = y0sin(kx−ωt)y: amplitude of the wave

If y is a function of x only,then, y(x) = y0sinkx

Standing wavey(x) = y0sinkxBoundary condition:

y = 0, when x = 0y = 0, when x = L

L

L x0

n=1

n=2

n=3kL = nπ, k = nπ/L

n = integers (quantum number)y(x) = y0sin[(nπ/L) x]

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Standing Waves

&

Quantum Number

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Wave Mechanics

• ψ : amplitude of the wave• The probability that a particle will be detected is

proportional to .

ψψ E=Η̂

( )[ ] 08

2

2

2

2

=−+ ψπψ

xUEh

mdxd

Schrödinger equation

one-dimensionalHamiltonian operator

( ) ψψψ

πExU

dxd

mh

=+− 2

2

2

2

8

( ) ψψπ

ExUdxd

mh

=

+− 2

2

2

2

8( )xU

dxd

mh

+−=Η 2

2

2

2

π

Operator for K.E. P.E.

P.E.

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( )[ ] 0,,8

2

2

2

2

2

2

2

2

=−+∂∂

+∂∂

+∂∂

ψπψψψ

zyxUEh

mzyx

three-dimensional Schrödinger equation

( )zyxUzyxm

h,,

2

2

2

2

2

2

2

2

+

∂∂

+∂∂

+∂∂

−=Ηπ

ψψ E=Η̂

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One-dimensional Schrödinger equation

Particle in a Box

( )[ ] 08

2

2

2

2

=−+ ψπψ

xUEh

mdxd

( )[ ] 08

2

2

2

2

=−+ ψπψ

xUEh

mdxd

The probability that the particle will be detected is proportional to

In the box, 0 < x < L and U = 0

ψψ

Edxd

m=− 2

22

2h

( )xUdxd

mh

+−=Η 2

2

2

2

πψψ E=Η̂

π2h

=h2

22

2

2

2

2

28ˆ

dxd

mdxd

mh h

−=−=Ηπ

[ ] [ ])sin()sin(2 2

22

kxAEkxAdxd

m=−

h

Let and A, k are constants)sin( kxA=ψ

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[ ] [ ])sin()sin(2 2

22

kxAEkxAdxd

m=−

h

[ ] [ ])sin()sin( 22

2

kxAkkxAdxd

−=Since

[ ] [ ])sin()sin(2

22

kxAEkxAkm

=h

mk

E2

22h=

Boundary conditions for the particle in the box:

1. The particle cannot be outside the box.

2. In a given state, the total probability of finding the particle in the box must be 1.

3. The wave function must be continuous.

0)( and 0)0( == Lψψ

1)(0

2 =∫ dxxLψ

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0)( and 0)0(condition Boundary == Lψψ

)sin()( kxAx =ψ

xL

nAx

nL

nknkL

kLAL

πψ

ππ

ψ

sin)(

intergers is where, or ,

0sin)(

=

==

==

1)(condition Boundary 0

2 =∫ dxxLψ

20

2

0

22

0

2

1)sin(

1)sin()(

Adxx

Ln

dxxL

nAdxx

L

LL

=

==

∫∫π

πψ

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LA

AL

dxxL

nL

2

12

)sin( 20

2

=

==∫π

xL

nL

ψ sin2

)( =

2

2

22222

822

=

==

Ln

mh

Ln

mmk

ππhh

2

22

8mLhn

E =ψ 2ψ

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• (normalization)

• ψ(x,y,z) is a single valued function w.r.t. the coordinates

• ψ(x,y,z) is a continuous function

• ψ(x,y,z) is a finite function

1),,(2 =∫

+∞

∞−τψ dzyx

Boundary condition for solving ψ in Schrödinger eq. for an electron in an atom:

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For H atom

re

rZe

U22 −

=−

=

( )( ) ( ) ( )( ) ( )φθ

φθφθψψ

,

,,),,(

Υ=ΦΘ=

=

rRrRrzyx

Z= 1

e-r

x

z

y

φ

θ

To solve the equation more easily,Cartesian coordinates x, y, z are

transformed to polar coordinates r, θ, φ.

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Wavefunctions of Hydrogen Atom

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Quantum Numbers and Atomic Orbitals

• The wave functions for the hydrogen atom contain threeparameters that must have specific integral values called quantum numbers.

• A wave function with a given set of these three quantum numbers is called an atomic orbital.

• These orbitals allow us to visualize the region in which there is a probability of find an electron.

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Quantum Numbers

The principal quantum number (n)– Can only be a positive integer

– The size of an orbital and its electron energy depend on the n number

– Orbitals with the same value of n are said to be in the same principle shell

Value of n 1 2 3 4 5Shell K L M N O

n = 1, 2, 3 ····

When values are given to quantum numbers, a specific atomic orbital is defined

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Quantum Numbers (continued)

• The orbital angular momentum quantum number (l)

– Can have positive integral values

0≤ l ≤ n-1– Determines the shape of the orbital

– All orbitals having the same value of n and the same value of l are said to be in the same subshell

– Orbitals and subshells are also designated by a letter:

Value of l 0 1 2 3

Orbital or subshell s p d f

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Quantum Numbers (continued)

• The magnetic quantum number (ml):

– Can be any integer from -l to +l

-l ≤ ml ≤ l

– Determines the orientation in space of the orbitals of any given type in a subshell

– The number of possible value for ml = 2l + 1, and this determines the number of orbitals in a subshell

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The relationship between quantum numbers0≤ l ≤ n-1For example: n= 1, l= 0

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

-l ≤ ml ≤ lFor example: l= 1, ml = -1,0,1

l= 2, ml = -2,-1,0,1,2

1s- orbital

2p, 2s- orbitals

3d, 3p, 3s- orbitals

px, py, pz- orbitals

dxy, dyz, dzx, dz2, dx2-y2

orbitals

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Quantum Numbers Summary

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The 1s Orbital

• The 1s orbital has spherical symmetry.• The electrons are more concentrated near the center

Υ0, 0(θ,φ) = 1/2π1/2

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(+)

(-)0

noder

0

0

The 2s Orbital

Υ0,0(θ,φ) = 1/2π1/2

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The 2s Orbital

• The 2s orbital has two regions of high electron probability, both being spherical

• The region near the nucleus is separated from the outer region by a spherical node- a spherical shell in which the electron probability is zero

node

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The Three p Orbitals

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The Five d Orbital Shapes

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The Seven f Orbital Shapes

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Electron Spin – the 4th Quantum Number

• The electron spin quantum number (ms) explains some of the finer features of atomic emission spectra

– The number can have two values: +1/2 and –1/2

( ms= ½ , -½ )– The spin refers to a magnetic field induced by the

moving electric charge of the electron as it spins

– The magnetic fields of two electrons with opposite spins cancel one another; there is no net magnetic field for the pair.

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The Stern-Gerlach Experiment

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Hydrogen atom and Schrödinger equation

• energy is quantized (n)• magnitude of angular momentum is

quantized (l)• the orientation of angular momentum

is quantized (ml)

Electron has spin (ms)

Page 58: Chapter One - 國立臺灣大學sfcheng/inchem94/Chapt 1.pdf · 2005-09-21 · Slide 3 of 58 Subatomic particle of relevance to chemistry Electromagnetic 0 0 1 radiation from nucleus

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hydrogen atom

22

42

2 hnmeZ

E −=