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    • Abstract: Molecular Orbital Theory. Molecular Orbital Theory. Lecture 1 The Bohr Model. Prof G. W. ... Classical electrodynamical theory rejected (charged particles. undergoing acceleration ...

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Objectives of the course
• Wave mechanics / Atomic orbitals (AOs)
– The basis for rejecting classical mechanics (the Bohr Model) in treating
j g ( ) g
An introduction to electrons
– Wave mechanics and the Schrödinger equation
Molecular Orbital Theory – Representation of atomic orbitals as wave functions
i f i bi l f i
– Electron densities and radial distribution functions
– Understanding the effects of shielding and penetration on AO energies
6 Lecture Course • Bonding
– Review VSEPR and Hybridisation
Prof G. W. Watson – Linear combination of molecular orbitals (LCAO), bonding / antibonding
Lloyd Institute 2 36
Ll d I tit t 2.36 – L b lli of molecular orbitals (MOs) (σ, π and g, u)
Labelling f l l bi l (MO ) ( d )
[email protected] – Homonuclear diatomic MO diagrams – mixing of different AO’s
– More complex molecules (CO, H2O ….)
(CO )
– MO diagrams for Inorganic complexes
2
Lecture schedule Literature
Lecture 1 Revision of Bohr model of atoms • Book Sources: all titles listed here are available in the Hamilton Library
Lecture 2 Schrödinger equation, atomic wavefunctions and radial – 1. Chemical Bonding, M. J. Winter (Oxford Chemistry primer 15)
distribution functions of s orbitals Oxford Science Publications ISBN 0 198556942 – condensed text,
excellent diagrams
Lecture 3 More complex wavefunctions and radial distribution
functions and electron shielding
g – 2. Basic Inorganic Chemistry (
g y (Wiley) F.A.Cotton, G. Wilkinson, P. L.
y) , ,
Gaus – comprehensive text, very detailed on aufbau principle
Lecture 4 Lewis bonding, Hybridisation, and molecular orbitals
– 3 Inorganic Chemistry (Prentice Hall) C. Housecroft, A. G. Sharpe –
3. I i Ch i (P i H ll) C H f A G Sh
Lecture 5 Labelling MO’s. 1st row homonuclear diatomics comprehensive text with very accessible language. CD contains
interactive energy diagrams
Lecture 6 MO approach to more complex molecules and CO bonding
in transition metals complexes – Additional sources:
http://winter.group.shef.ac.uk/orbitron/ - gallery of AO and MO
h // i h f k/ bi / ll f AOs d MOs
3 4
Tutorials
• Expectation
– Tutorials are to go through p
g g problems that students are having with the
g
course An introduction to
– Tutorials are NOT for the lecturer to give you the answers to the
i l f h l i h h
Molecular Orbital Theory
questions – or to give you another lecture.
– All student must BEFORE the tutorial
• Look at the notes for the course and try to understand them Lecture 1 The Bohr Model
• Attempt the questions set – and hence find out what you can not do!
• Bring a list of questions relating to aspects of the course which you could
not understand (either from looking at the notes or attempting the Prof G. W. Watson
questions) Lloyd Institute 2.05
[email protected]
• It is a waste of both the lecturers and students time if the tutorial to ends up
being a lecture covering questions.
5
Adsorption / Emission spectra for Hydrogen Bohr model of the atom (1913)
http://www.youtube.com/watch?v=R7OKPaKr5QM
Johann Balmer (1885) measured line spectra for hydrogen Assumptions
364.6 nm (uv), 410.2 nm (uv), 434.1 nm (violet), 486.1 nm (blue), and 1) Rutherford (1912) model of the atom (Planetary model with central
656.3 nm (red). nucleus + electrons in orbit)
2) Planck (1901), Einstein (1905) – the energy electromagnetic waves is
(1901)
quantised into packets called photons (particle like property).
E = hν
Balmer discovered these lines occur in a series - both absorption and emission -
where ℜ is the Rydberg constant (3.29 ×1015 Hz) Velocity, c Wavelength, λ
ting electrric
netic field
⎛ 1 1⎞
v = ℜ⎜ 2 − 2 ⎟
⎜n n ⎟
⎝ 1 2⎠
/magn
Fluctuat
Balmer series n1=2 and n2=n1+1, n1+2, n1+3 …..
Other series for n1=1 (Lyman – UV), n1=3 (Paschen – IR) etc.
1 3
An stationary observer counts
} i.e. frequency = v Hz , cycles/sec,
ν waves passing per second
Electrons must have specific energies – no model of the atom could explain this sec-1
7 8
Bohr model of the atom Bohr model of the atom
Speed of electromagnetic waves (c) is constant (ν and λ vary) 3) Electron assumed to travel in circular orbits.
c = ν λ, ν = c / λ, E = h ν, E = h c / λ 4) Apply quantisation to orbits - only orbits allowed have quantised
angular momentum (comes from observation of spectra)
As frequency increases, wavelength decreases. Given λ ν ⎛ h ⎞
mvr = n⎜ ⎟
⎝ 2π ⎠
e.g.
e g radiowaves: λ=01m
0.1 X rays: λ = 1 x 10-12 m
X-rays:
ν = 3 x 109 Hz ν = 3 x 1020 Hz 5) Classical electrodynamical theory rejected (charged particles
E = 2 x 10-24 J E = 2 x 10-13 J undergoing acceleration must emit radiation)
6) Radiation adsorbed or emitted only when electrons jump from one
orbit to another
E – energy (J), h – Plancks constant (J s), ν – frequency (Hz),
ΔE = Ea − Eb
c – speed of light (m-1), λ – wavelength (m)
where a and b represent the energy of the start and finish orbits
9 10
Bohr model – calculating the energy and radius
Bohr model
NOT EXAMANABLE
Electron travelling around nucleus in circular − Ze 2 1
• Energy = − mv 2
orbits – must be a balance between attraction 8πε 0 r 2
to nucleus and flying off (like a planets orbit)
⎛ h ⎞
mvr = n⎜ ⎟
Electron feels two forces – must be balanced • Quantised angular momentum ⎝ 2π ⎠
− Ze 2 − Ze 2
1) Centrapedal (electrostatic) F= PE = − Ze 2 1 − (mvr )2 − n 2 h 2
4πε 0 r 2 4πε 0 r = − mv 2 = = 2 2
• Combining the two 8πε 0 r 2 2mr 2 8π mr
2) Centrafugal mv 2
1
F= KE = mv 2
r 2 r 2 − n 2 h 2 8πε 0 n 2 h 2ε 0
= r=
Equalize magnitude of forces Resulting energy
• Rearranging to give r
r (
8π 2 m − Ze 2 ) πmZe 2
mv 2 Ze 2 Ze 2 1 − Ze 2 − Ze 2 1 − Ze 2 − mZ 2e 4
= ⇒ mv 2 = E = mv 2 + = = − mv 2 • Substitute r into energy gives =
r 4πε 0 r 2 4πε 0 r 2 4πε 0 r 8πε 0 r 2 8πε 0 r 8n 2 h 2ε 0
2
-Ze – nuclear charge, e – electron charge, ε0- permittivity of free space, • Energy is dependent on n2 and Z2 (2s and 2p the same – only true for 1
r - radius of the orbit, m – mass of electron, v – velocity of the electron electron systems
11 12
Energy levels of Hydrogen Energy levels of Hydrogen
n=∞
Substitute quantised momentum into energy expression and rearrange n=5
in terms of r (radius) (see previous slide) n=4
For hydrogen (Z 1)
F h d (Z=1) n 3
n=3
n 2 h 2ε 0 n 2 a0
r= = energy n=2
π mZe 2 Z
− 13.6056 n=1
1
a0 (Bohr) radius of the 1s electron on Hydrogen 52.9 pm (n =1, Z =1) r = n 2 a0 En =
n2
1
Radius (r) depends on n2 and
() p
Z
n energy (eV) nucleus
Substitute r back into energy expression gives 1 -13.6056
2 4
− mZ e 2 -3.4014
En = 13.6056 × Z 2
2 2 2 = (in eV) 3 -1.5117
8n h ε 0 n2
4 -0.8504
0 8504
Energy of 1s electron in H is 13.6056 eV = 0.5 Hartree (1eV = 1.602 × 10-19 J) 5 -0.3779
1 ∞ 0.0000
Energy (E) depends on and Z2 Note. The spacing reflects the energy
n2 not the radius of the orbit.
Ionization energy = -13.6056 eV
13 14
n=4
Radius of orbits Emission spectra
( p
(http://www.youtube.com/watch?v=5z2ZfYVzefs)
y )
Energy of emission is Einitial - Efinal= hv
For hydrogen (Z=1)
(Z 1) n=3 ⎛ 1 1 ⎞
distance ΔE = 13.6056 ⎜ 2 − 2 ⎟
⎜n ⎟
⎝ inital n final ⎠
− 13.6056
r = n 2 a0 En = n=2
n2 Same form as fitted to emission specta
n=1 Balmer series ( nfinal=2) 5 4 3 2 1
n energy (eV) r (pm) nucleus
1 -13.6056 52.9
2 -3.4014
3 4014 211
3 -1.5117 476
n=3 n=2 λ = 656 nm
4 -0.8504
0.8504 847
n=4
4 n=2
2 λ = 486 nm
5 -0.3779 1322
n=5 n=2 λ = 434 nm
∞ 0.0000 ∞
Note.
Note The spacing reflects the radius of the
Note. The spacing reflects the energy
Orbit – not the energy. ℜ = 13.6056 eV / c = 3.29 ×1015 Hz
not the radius of the orbit.
15 16
Problems with the Bohr Model Wave / particle duality
http://www.youtube.com/watch?v=IsA_oIXdF_8
p y
• Only works for 1 electron systems de Broglie (1923)
– E.g. H, He+, Li2+
g , By this time it was accepted that EM radiation can have wave and particle
y p p
properties (photons)
• Can not explain splitting of lines in a magnetic field
– Modified Bohr-Sommerfield (elliptical orbits - not satisfactory) de
d Broglie proposed that particles could have wave properties (wave /
li d h i l ld h i (
particle duality). Particles could have an associated wavelength (l)
• Can not apply the model to interpret the emission spectra of complex atoms hc h
E = mc 2 , E = ⇒ λ=
λ mc
• Electrons were found to exhibit wave-like properties
p p
– e.g. can be diffracted as they pass through a crystal (like x-rays) No experimental at time.
– considered as classical particles in Bohr model
1925 Davisson and Germer showed electrons could be diffracted according
to Braggs Law (used for X-ray diffraction)
Numerically confirm de Broglie’s equation
17 18
Wave Mechanics Wave mechanics and atoms
• For waves: it is impossible to determine the position and momentum of the • What does this mean for atoms
electron simultaneously – Heisenberg ‘Uncertainty principle’
• Electrons in “orbits” must have an integer number of
• Use probability of finding an electron from ψ2 (actually ψ*ψ – but functions wavelengths
we will deal with are real)
• E.g. n=4 and n=5 are allowed
Where ψ is is a wave function and a solution of the Schrödinger equation
g q – These create continuous or standing waves (like on a
(1927). The time-independent form of the Schrödinger equation for the guitar string)
hydrogen atom is:
• E.g. n=4.33 is not allowed
−h 2
−e 2
∂ 2
∂ ∂ 2 2
– The wavefunction is not continuous
∇ Ψ 2
Ψ = EΨ ∇ =
2
+ +
8π m
2
4πε r
0
∂x ∂y ∂z
2 2 2
• The wave nature of electrons brings in the quantized
Kinetic Potential Total nature of the orbital energies.
energy energy energy
19 20
Atomic solutions of the Schrödinger equation for H Solution of the Schrödinger equation for H
• Schrödinger equation can be solved exactly for 1 electron systems l has values 0 to (n-1) m has values from +l through 0 to –l
– Solved by trial and error manipulations for more electrons
• 3 quantum numbers describing a three dimensional space called an atomic n 1 2 2 2 2
orbital: n, l, m (and spin quantum number describing the electron s)
l 0 0 1 1 1
n = principal quantum number, defines the orbital size with values 1 to ∞ ml 0 0 -1 0 1
Orbital 1s 2s 2p 2p 2p
l= azimuthal or angular momentum quantum number, defines shape.
For a given value of n, l has values 0 to (n-1).
g , ( )
n 3 3 3 3 3 3 3 3 3
ml = magnetic quantum number, defines the orbital orientation. l 0 1 1 1 2 2 2 2 2
For i
F a given value of l ml h values from +l th
l f l, has l f through 0 to –l.
h t l
ml 0 -1 0 1 -2 -1 0 1 2
Oribtal 3s 3p 3p 3p 3d 3d 3d 3d 3d
21 22
Last Lecture
• Recap of the Bohr model
– Electrons
An introduction to – Assumptions
– Energies / emission spectra
Molecular Orbital Theory – Radii
• Problems with Bohr model
– Only works for 1 electron atoms
– Can not explain splitting by a magnetic field
Lecture 2 – Representing atomic orbitals - The
Schrödinger equation and wavefunctions. • Wave-particle duality
Prof G. W. Watson
P fG W W t • Wave mechanics
Lloyd Institute 2.05 – Schrödinger
[email protected] – Solutions give quantum number n l ml
n, l, atomic orbitals
24
Representations of Orbitals: Polar Coordinates
For an atomic system containing one electron (e.g. H, He+ etc.) • To describe the wavefunction of atomic orbitals we must describe it in
The wavefunction, Ψ, is a solution of the Schrödinger equation three dimensional space
It describes the behaviour of an electron in region of space called an atomic
orbital (φ - phi) • For an atom it is more appropriate to use spherical polar coordinates:
Each orbital wavefunction (φ ) is most easily described in two parts
radial term – which changes as a function of distance from the nucleus
angular terms – which changes as a function of angles
l hi h h f i f l Location f i t
L ti of point P
φ xyz= φradial(r) φangular(φ,θ)
() (φ, ) = Rnl(r) Ylm(φ,θ)
() (φ, ) Cartesian = x, y, z
r, φ, θ
Orbitals have
• SIZE determined by Rnl(r) - radial part
• SHAPE determined by Ylm(φ,θ) - angular part (spherical harmonics)
y (φ ) g p (p )
• ENERGY determined by the Schrödinger equation
java applet on polar coordinates at http://qsad.bu.edu/applets/SPCExp/SPCExp.html
25 26
Wavefunctions for the AO’s of H Radial Wavefunction
General hydrogen like orbitals • R(r) of the 1s orbital of H
it decays exponentially with r
Rnl(r) Ylm(φ,θ) R(r) = 2e (−r ) it has a maximum at r = 0
3 1
⎛Z⎞ 2 ⎛ 1 ⎞ 2 ⎛ 2Z ⎞
1s 2⎜ ⎟ e ( − ρ r
⎜a ⎟
2)
⎜ ⎟ ρ =⎜
⎜ ⎟
⎟ • R(r) has no physical meaning
⎝ 4π ⎠
4
⎝ 0⎠ ⎝ n a0 ⎠
• Probability depends on R(r)2
3 (1s)2
For hydrogen this simplifies as Z=1 and ao=1 (in atomic units) and thus ρ = 2.
R(r) in a.u.
Hence – Misleading – does not take
Normalisation into account the volume 2
Constants are such that
C h h
Rnl(r) Ylm(φ,θ)
– R(r)2 increases toward r = 0
2
∫ ϕ ∂τ = 1
1
1 1s
1s (−r )
2e 2 π – Volume very small so
that is the probability of the electron probability of


Use: 0.4163