quantum chemistry chapter 6. copyright © houghton mifflin company. all rights reserved.6 | 2...

Quantum Chemistry

Chapter 6

Copyright © Houghton Mifflin Company. All rights reserved. 6 | 2

Electromagnetic Radiation


Electromagnetic Waves


Electromagnetic Radiation

c

= frequency of the wave

c = speed of light

= wavelength


Electromagnetic Spectrum


Visible Spectrum


Energy, Wavelength & Frequency

• The energy of a photon is given by –

hcE = h =

h = 6.626×10-34 J.s, Plank’s constant

c = 3.00×108 m/s


Sample Problem

What is the energy of a photon of infraredlight that has a wavelength of 850. nm?

34 8

9

19

6.626 10 Js 3.00 10 m/ shcE =

1 m850 nm

10 nm

E 2.34 10 J


Hydrogen Spectra


Emission Spectrum

• When hydrogen atoms are excited, they emit radiation.

• The wavelengths of this radiation can be calculated from -

.f in n

2 12 2

1 1 11 0968 10 nm


Hydrogen Spectra

.f in n

2 12 2

1 1 11 0968 10 nm


Emission Spectra


Bohr Model

• Bohr postulated that the energy an electron has when it occupies an orbit around the nucleus in a hydrogen atom

is:

18

n 2

2.1786 10 JE

n

n = 1, 2, 3, 4, ……..

• Ground state is the lowest energy level, n = 1.

• Excited state is a higher energy level.

Bohr model of the hydrogen

atom


• 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

Bohr Model


Bohr Model


Bohr Model

• For the hydrogen electron –

182 2f i

2 12 2f i

1 1E 2.1786 10 J

n n

and

1 1 11.097 10 nm

n n


Electronic Transitions


Sample Problem

Calculate the wavelength of light emitted by the transition of a hydrogen electron from n=4 to n=1.

-2 12 2

f i

-2 12 2

1 1 11.097 10 nm

n n

1 1 11.097 10 nm

1 4

97.23 nm


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 constantm = massv = velocity


De Broglie Wavelength


Sample Problem

What is the wavelength of an electron traveling 5.31×106 m/s?

34

31 6

10

h

mv

6.626 10 J s

9.11 10 kg 5.31 10 m/ s

1.37 10 m

0.137 nm


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.


The Wave Equation

• Ψ*Ψ, the square of the wave function, is the probability of finding the electron in some region of space.


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)


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)


Quantum Numbers


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.


Quantum Numbers


Orbital Shapes: s orbital


s Orbitals


Orbital Shapes: s orbital


p Orbitals


Orbital Shapes: 2px orbitals


Orbital Shapes: 2py orbital


Orbital Shapes: 2pz orbital


d Orbitals


Orbital Shapes: 3dx2-y2 orbital


Orbital Shapes: 3dz2 orbital


Orbital Shapes: 3dxy orbital


Orbital Shapes: 3dyz orbital


Orbital Shapes: 3dxz orbital


f Orbitals


Orbital Energies


Electron spin

Spin up Spin down


Electron shielding


Orbital Energy Levels in Multi-electron Atoms


Electron Configurations

• Aufbau principle gives the order of the orbitals


Sample Problem

Write the electron configuration for Ca using the Aufbau principle.

1s22s22p63s23p64s2


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


Electron configuration

Three possible electron configurations for carbon

electron configurations


Periodic Table


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


Periodic Table Blocks


Sample Problem

Write the electron configuration for Br & Fe using the periodic table.

Br: [Ar]4s23d104p5

Fe: [Ar]4s23d6


Homework

26, 34, 38, 46, 52, 64, 76, 82, 92, 98, 106,

quantum chemistry chapter 6. copyright © houghton mifflin company. all rights reserved.6 | 2...

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