Transcript Quantum Chemistry

Quantum
Chemistry
Chapter 6
Electromagnetic Radiation
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6|2
Electromagnetic Waves
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6|3
Electromagnetic Radiation

c

 = frequency of the wave
c = speed of light
 = wavelength
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Electromagnetic Spectrum
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Electromagnetic Spectrum
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Visible Spectrum
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Energy, Wavelength & Frequency
• The energy of a photon is given by –
E = h =
hc

h = 6.626×10-34 J.s, Plank’s constant
c = 3.00×108 m/s
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Sample Problem
What is the energy of a photon of infrared
light that has a wavelength of 850. nm?
E=
hc




6.626  10 34 Js 3.00  108 m / s

 1m 
850 nm  9

10
nm


E  2.34  10 19 J
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Hydrogen Spectra
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Emission Spectrum
• When hydrogen atoms are excited,
they emit radiation.
• The wavelengths of this radiation can
be calculated from 1

 1.0968  10
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2
 1
1
1

nm
 2
2
ni 
 n f
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Hydrogen Spectra
1

 1.0968  10
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2
 1
1
1
 2  2  nm
ni 
 n f
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Emission Spectra
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Bohr Model
• Bohr postulated that the energy an
electron has when it occupies an orbit
around the nucleus in a hydrogen atom
18
is:
2.1786  10 J
En  
n = 1, 2, 3, 4, ……..
n2
Bohr model of
the hydrogen
atom
• Ground state is the lowest energy level, n = 1.
• Excited state is a higher energy level.
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Bohr Model
• If an electron moves from a lower
energy level to a higher energy level, it
absorbs energy.
• If an electron moves from a higher
energy level to a lower energy level, it
emits energy.
• The change in energy is –
E = Ef - Ei
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Bohr Model
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Bohr Model
• For the hydrogen electron –
E  2.1786  10
18
1 1
J 2  2 
 nf ni 
and
1
1
2  1
1
 1.097  10  2  2  nm

 nf ni 
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Electronic Transitions
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Sample Problem
Calculate the wavelength of light emitted
by the transition of a hydrogen electron
from n=4 to n=1.
1
1
-2  1
 1.097  10  2  2  nm 1

ni 
 nf
1
1 1
 1.097  10-2  2  2  nm1

1 4 
  97.23 nm
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Wave - Particle Duality
• Very small, light weight particles, such
as electrons can behave like waves.
• de Broglie’s equation allows us to
calculate the wavelength of an electron.
h

mv
h = Planck’s constant
m = mass
v = velocity
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De Broglie Wavelength
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Sample Problem
What is the wavelength of an electron
traveling 5.31×106 m/s?

h
mv
6.626  10 34 J  s

9.11 10 31kg 5.31 10 6 m / s



 1.37  10 10 m
 0.137 nm
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The Wave Equation
• If an electron can behave like a wave, it
should be possible to write an equation
that describes its behavior.
• Schrödinger equation allows us to
calculate the energy available to the
electrons in an atom.
• Ψ is a wave function that describes
the position and paths of the electron in
its energy level.
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The Wave Equation
• Ψ*Ψ, the square of the wave function, is the
probability of finding the electron in some
region of space.
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Quantum Numbers
• There are four quantum numbers used to
describe the electron in the hydrogen atom
• n, principle quantum number, describes
the size and energy of the orbital
n = 1, 2, 3, 4, ………(only integers)
• l – angular momentum quantum number,
describes the shape of the orbital.
l = 0 to n-1 (only integers)
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Quantum Numbers
• ml – magnetic quantum number,
describes the spatial orientation of the
orbital.
ml = -l to 0 to +l (only integers)
• ms – spin quantum number,
describes the direction and spin of the
electron.
ms = +1/2 or -1/2 (only two values)
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Quantum Numbers
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Quantum Numbers
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Quantum Numbers
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Pauli Exclusion Principle
Pauli Exclusion Principle
• No two electrons can have the same
four quantum numbers.
• Spins of electrons in an orbital must be
opposite.
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Quantum Numbers
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Orbital Shapes: s orbital
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s Orbitals
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Orbital Shapes: s orbital
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p Orbitals
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Orbital Shapes: 2px orbitals
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Orbital Shapes: 2py orbital
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Orbital Shapes: 2pz orbital
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d Orbitals
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Orbital Shapes: 3dx -y orbital
2
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2
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Orbital Shapes: 3dz orbital
2
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Orbital Shapes: 3dxy orbital
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Orbital Shapes: 3dyz orbital
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Orbital Shapes: 3dxz orbital
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f Orbitals
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Orbital Energies
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Electron spin
Spin up
Spin down
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Electron shielding
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Orbital Energy Levels
in Multi-electron Atoms
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Electron Configurations
• Aufbau principle gives
the order of the orbitals
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Sample Problem
Write the electron configuration for Ca
using the Aufbau principle.
1s22s22p63s23p64s2
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Hund’s Rule
• Hund’s rule - maximize the number of
unpaired electrons in orbitals.
• Orbital diagram for C (z = 6) would be:
() () ( ) ( ) ( )
1s
2s
2p
not
() () ( ) ( ) ( )
1s 2s
2p
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Electron configuration
electron configurations
Three possible electron
configurations for carbon
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Periodic Table
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Electron Configurations
• Representative Elements are s orbital
and p orbital fillers.
• Transition metals fill the d orbitals.
• Lanthanides are 4f fillers.
• Actinides are 5f fillers
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Periodic Table Blocks
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Sample Problem
Write the electron configuration for Br &
Fe using the periodic table.
Br: [Ar]4s23d104p5
Fe: [Ar]4s23d6
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Homework
26, 34, 38, 46, 52, 64, 76, 82, 92, 98, 106,
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