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Chapter 7
Quantum Theory and
the Electronic Structure
of Atoms
Dr. Sa’ib Khouri
AUM- JORDAN
Chemistry
By Raymond Chang
2
Properties of Waves
Wavelength (l) is the distance between identical points on
successive waves.
Amplitude is the vertical distance from the midline of a
wave to the peak or trough.
Frequency (n) is the number of waves that pass through a
particular point in 1 second. (Hz = 1 cycle/s).
The speed of the wave (u) = l x n
3
James Maxwell (1873), proposed that visible light consists of
electromagnetic waves.
Electromagnetic radiation is the
emission and transmission of energy
in the form of electromagnetic waves.
Speed of light (c) in vacuum =
3.00 x 108 m/s
All electromagnetic radiation
l x n = c
Electromagnetic Radiation
The wavelength of electromagnetic waves is usually given in
nanometers (nm). 1 nm = 1 ⨯ 10
-9
m
4
Visible light ranges from a wavelength of 400 nm
(violet) to 700 nm (red).
5
6
l x n = c
l = c/n
l = 3.00 x 108 m/s / 6.0 x 104 Hz
l = 5.0 x 103 m
What is the wavelength (in meter) of an electromagnetic wave
whose frequency is 6.0 x 104 Hz. In which region falls this
frequency?
l = 5.0 x 1012 nm
Example
7
When solids are heated, they emit electromagnetic radiation
over a wide range of wavelengths.
Radiant energy emitted by an object at a certain temperature
depends on its wavelength.
Planck’s Quantum Theory
Planck gave the name quantum to the smallest quantity of
energy that can be emitted (or absorbed) in the form of
electromagnetic radiation
The energy E of a single quantum of energy is given by
E = h x n
h = 6.63 x 10-34 J•s Planck’s constant
8
The Photoelectric Effect
a phenomenon in which electrons are
ejected from the surface of certain
metals exposed to light of at least a
certain minimum frequency, called the
threshold frequency.
Elbert Einstein (1905) suggested that
a beam of light is really a stream of
particles are now called photons.
each photon must possess energy E,
given by the equation
E = h x n
where n is the frequency of light.
9
E = h x n
E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)
E = 1.29 x 10
-15
J/photon
= 7.77 x 10
8
J/mol
E = h x c / l
When copper is bombarded with high-energy electrons, X rays
are emitted. Calculate the energy (in joules) associated with
the photons if the wavelength of the X rays is 0.154 nm.
Example
10
Bohr’s Theory of the Hydrogen Atom
Emission Spectra either continuous or line spectra
11
12
1. e- can only have specific
(quantized) energy values
2. light is emitted as e- moves from
one energy level to a lower energy
level
Bohr’s Model of the Atom (1913)
En = -RH ( )
1
n2
n (principal quantum number) = 1, 2, 3, …
RH (Rydberg constant) = 2.18 x 10-18J
13
Ephoton = DE = Ef - Ei
Ef = -RH ( )
1
n2
f
Ei = -RH ( )
1
n2
i
i f
DE = RH ( )
1
n2
1
n2
nf = 1
ni = 2
nf = 1
ni = 3
nf = 2
ni = 3
14
15
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = DE = -1.55 x 10-19 J
l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
l = 1280 nm
Calculate the wavelength (in nm) of a photon emitted
by a hydrogen atom when its electron drops from n =
5 to n = 3 states.
Ephoton = h x c / l
l = h x c / Ephoton
i f
DE = RH ( )
1
n2
1
n2
Ephoton =
Example
16
Schrödinger Wave Equation
In 1926 Schrödinger used an equation that described
both the particle and wave nature of the e-
Wave function y (psi) describes:
1. Energy of e- with a given y
2. Probability of finding e- in a volume of space
Schrödinger's equation can only be solved exactly for
the hydrogen atom.
For multi-electron systems it must
approximate its solution.
Quantum Mechanics
17
1. The principal quantum number (n)
n = 1, 2, 3, 4, ….
n=1
n=2
n=3
Quantum Numbers
Three numbers are derived from the mathematical solution of the
Schrödinger equation for the hydrogen atom: the principal quantum
number (n), the angular momentum quantum number (l), and the
magnetic quantum number (ml )
the value of n determines the energy of an orbital and also relates to
the average distance of the electron from the nucleus in a particular
orbital
18
2. The angular momentum quantum number (l)
for a given value of n, l = 0, 1, 2, 3, … n-1
n = 1, l = 0
n = 2, l = 0 or 1
n = 3, l = 0, 1, or 2
n = 4, l = 0, 1, 2 or 3
n = 5, l = 0, 1, 2, 3 or 4
etc.
Shape of the “volume” of space that the e- occupies
(orbitals)
l = 0 s orbital
l = 1 p orbital
l = 2 d orbital
l = 3 f orbital
l = 4 g orbital
l = 5 h orbital
19
l = 0 (s orbitals)
l = 1 (p orbitals)
20
l = 2 (d orbitals)
21
3. The magnetic quantum number (ml)
for a given value of l, ml = -l, …., 0, …. +l
if l = 0 (s orbital), ml = 0
if l = 1 (p orbital), ml = -1, 0, or 1
if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2
if l = 3 (f orbital), ml = -3, -2, -1, 0, 1, 2 or 3
The magnetic quantum number (m/) describes
the orientation of the orbital in space.
22
ml = -1, 0, or 1 3 orientations in space
23
ml = -2, -1, 0, 1, or 2 5 orientations in space
24
ms = +½ or -½
ms = -½ms = +½
Pauli exclusion principle - no two
electrons in an atom can have the
same four quantum numbers.
The electron spin quantum number (ms)
Experiments on the emission spectra of hydrogen and sodium
atoms indicated that lines in the emission spectra could be split
by the application of an external magnetic field.
25
For example:
when a hydrogen atom with a single electron passes through the field, it is
deflected in one direction or the other, depending on the direction of the spin
In a stream consisting of many atoms, there will be equal distributions of the
two kinds of spins, so that two spots of equal intensity are detected on the
screen.
26
Atomic Orbitals
The table below shows the relation between quantum numbers
and atomic orbitals
27
Shell: electrons with the same value of n
Subshell: electrons with the same values of n and l
Orbital: electrons with the same values of n, l, and ml
If n, l, and ml are fixed, then ms = ½ or - ½
y (n, l, ml, ½) or y (n, l, ml, -½)
An orbital can hold 2 electrons
28
How many 2p orbitals are there in an atom?
2p
n=2
l = 1
If l = 1, then ml = -1, 0, or +1 3 orbitals (total 6 e-
)
How many electrons can be placed in the 3d subshell?
3d
n=3
l = 2
If l = 2, then ml = -2, -1, 0, +1, or +2
5 orbitals (total10 e-)
29
According to equation below, the energy of e- in H-atom is
determined only by its principal quantum number n.
En = -RH ( )
1
n2
n=1
n=2
n=3
The Energies of Orbitals
Orbitals
with the
same
principal
quantum
number
(n) all
have the
same
energy.
30
Orbital energy levels in a many-electron atom
Energy level depends on both n and l
n=1 l = 0
n=2 l = 0
n=2 l = 1
n=3 l = 0
n=3 l = 1
n=3 l = 2
31
“Fill up” electrons in lowest energy orbitals
H 1 electron
H 1s1
He 2 electrons
He 1s2
Li 3 electrons
Li 1s22s1
Be 4 electrons
Be 1s22s2
B 5 electrons
B 1s22s22p1
C 6 electrons
? ?
Building-Up Principle (Aufbau principle)
32
C 6 electrons
Hund’s rule: The most stable arrangement of
electrons in subshells is the one with the greatest
number of parallel spins.
C 1s22s22p2
N 7 electrons
N 1s22s22p3
O 8 electrons
O 1s22s22p4
F 9 electrons
F 1s22s22p5
Ne 10 electrons
Ne 1s22s22p6
33
The order in which atomic subshells are filled in
a many-electron atom.
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
34
how the electrons are distributed among the various
atomic orbitals in an atom.
1s1
principal quantum
number n
angular momentum
quantum number l
number of electrons
in the orbital or subshell
Orbital diagram
H
1s1
Electron configuration
35
What is the electron configuration of Mg atom?
Mg: 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons
Abbreviated as [Ne]3s2 [Ne] 1s22s22p6
What are the possible quantum numbers for the last (outermost)
electron in Cl?
Cl: 17 electrons 1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons
Last electron added to 3p orbital
n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½
36
We conclude that there are six possible ways to designate the
electron using the (n, /, m/, ms) notation:
In these six designations we see that the values of n and / are
constant, but the values of m/ and ms can vary.
37
Paramagnetic
unpaired electrons
2p
Diamagnetic
all electrons paired
2p
Diamagnetism and Paramagnetism
38
Outermost subshell being filled with electrons
39
40
Many examples to be given in the lecture
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Halogen
Neon Inert Gas
Noble Gas

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Ch7 quantum theory and the electronic structure of atoms

  • 1. 1 Chapter 7 Quantum Theory and the Electronic Structure of Atoms Dr. Sa’ib Khouri AUM- JORDAN Chemistry By Raymond Chang
  • 2. 2 Properties of Waves Wavelength (l) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. Frequency (n) is the number of waves that pass through a particular point in 1 second. (Hz = 1 cycle/s). The speed of the wave (u) = l x n
  • 3. 3 James Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation l x n = c Electromagnetic Radiation The wavelength of electromagnetic waves is usually given in nanometers (nm). 1 nm = 1 ⨯ 10 -9 m
  • 4. 4 Visible light ranges from a wavelength of 400 nm (violet) to 700 nm (red).
  • 5. 5
  • 6. 6 l x n = c l = c/n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m What is the wavelength (in meter) of an electromagnetic wave whose frequency is 6.0 x 104 Hz. In which region falls this frequency? l = 5.0 x 1012 nm Example
  • 7. 7 When solids are heated, they emit electromagnetic radiation over a wide range of wavelengths. Radiant energy emitted by an object at a certain temperature depends on its wavelength. Planck’s Quantum Theory Planck gave the name quantum to the smallest quantity of energy that can be emitted (or absorbed) in the form of electromagnetic radiation The energy E of a single quantum of energy is given by E = h x n h = 6.63 x 10-34 J•s Planck’s constant
  • 8. 8 The Photoelectric Effect a phenomenon in which electrons are ejected from the surface of certain metals exposed to light of at least a certain minimum frequency, called the threshold frequency. Elbert Einstein (1905) suggested that a beam of light is really a stream of particles are now called photons. each photon must possess energy E, given by the equation E = h x n where n is the frequency of light.
  • 9. 9 E = h x n E = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m) E = 1.29 x 10 -15 J/photon = 7.77 x 10 8 J/mol E = h x c / l When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm. Example
  • 10. 10 Bohr’s Theory of the Hydrogen Atom Emission Spectra either continuous or line spectra
  • 11. 11
  • 12. 12 1. e- can only have specific (quantized) energy values 2. light is emitted as e- moves from one energy level to a lower energy level Bohr’s Model of the Atom (1913) En = -RH ( ) 1 n2 n (principal quantum number) = 1, 2, 3, … RH (Rydberg constant) = 2.18 x 10-18J
  • 13. 13 Ephoton = DE = Ef - Ei Ef = -RH ( ) 1 n2 f Ei = -RH ( ) 1 n2 i i f DE = RH ( ) 1 n2 1 n2 nf = 1 ni = 2 nf = 1 ni = 3 nf = 2 ni = 3
  • 14. 14
  • 15. 15 Ephoton = 2.18 x 10-18 J x (1/25 - 1/9) Ephoton = DE = -1.55 x 10-19 J l = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J l = 1280 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from n = 5 to n = 3 states. Ephoton = h x c / l l = h x c / Ephoton i f DE = RH ( ) 1 n2 1 n2 Ephoton = Example
  • 16. 16 Schrödinger Wave Equation In 1926 Schrödinger used an equation that described both the particle and wave nature of the e- Wave function y (psi) describes: 1. Energy of e- with a given y 2. Probability of finding e- in a volume of space Schrödinger's equation can only be solved exactly for the hydrogen atom. For multi-electron systems it must approximate its solution. Quantum Mechanics
  • 17. 17 1. The principal quantum number (n) n = 1, 2, 3, 4, …. n=1 n=2 n=3 Quantum Numbers Three numbers are derived from the mathematical solution of the Schrödinger equation for the hydrogen atom: the principal quantum number (n), the angular momentum quantum number (l), and the magnetic quantum number (ml ) the value of n determines the energy of an orbital and also relates to the average distance of the electron from the nucleus in a particular orbital
  • 18. 18 2. The angular momentum quantum number (l) for a given value of n, l = 0, 1, 2, 3, … n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 n = 4, l = 0, 1, 2 or 3 n = 5, l = 0, 1, 2, 3 or 4 etc. Shape of the “volume” of space that the e- occupies (orbitals) l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital l = 4 g orbital l = 5 h orbital
  • 19. 19 l = 0 (s orbitals) l = 1 (p orbitals)
  • 20. 20 l = 2 (d orbitals)
  • 21. 21 3. The magnetic quantum number (ml) for a given value of l, ml = -l, …., 0, …. +l if l = 0 (s orbital), ml = 0 if l = 1 (p orbital), ml = -1, 0, or 1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2 if l = 3 (f orbital), ml = -3, -2, -1, 0, 1, 2 or 3 The magnetic quantum number (m/) describes the orientation of the orbital in space.
  • 22. 22 ml = -1, 0, or 1 3 orientations in space
  • 23. 23 ml = -2, -1, 0, 1, or 2 5 orientations in space
  • 24. 24 ms = +½ or -½ ms = -½ms = +½ Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. The electron spin quantum number (ms) Experiments on the emission spectra of hydrogen and sodium atoms indicated that lines in the emission spectra could be split by the application of an external magnetic field.
  • 25. 25 For example: when a hydrogen atom with a single electron passes through the field, it is deflected in one direction or the other, depending on the direction of the spin In a stream consisting of many atoms, there will be equal distributions of the two kinds of spins, so that two spots of equal intensity are detected on the screen.
  • 26. 26 Atomic Orbitals The table below shows the relation between quantum numbers and atomic orbitals
  • 27. 27 Shell: electrons with the same value of n Subshell: electrons with the same values of n and l Orbital: electrons with the same values of n, l, and ml If n, l, and ml are fixed, then ms = ½ or - ½ y (n, l, ml, ½) or y (n, l, ml, -½) An orbital can hold 2 electrons
  • 28. 28 How many 2p orbitals are there in an atom? 2p n=2 l = 1 If l = 1, then ml = -1, 0, or +1 3 orbitals (total 6 e- ) How many electrons can be placed in the 3d subshell? 3d n=3 l = 2 If l = 2, then ml = -2, -1, 0, +1, or +2 5 orbitals (total10 e-)
  • 29. 29 According to equation below, the energy of e- in H-atom is determined only by its principal quantum number n. En = -RH ( ) 1 n2 n=1 n=2 n=3 The Energies of Orbitals Orbitals with the same principal quantum number (n) all have the same energy.
  • 30. 30 Orbital energy levels in a many-electron atom Energy level depends on both n and l n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2
  • 31. 31 “Fill up” electrons in lowest energy orbitals H 1 electron H 1s1 He 2 electrons He 1s2 Li 3 electrons Li 1s22s1 Be 4 electrons Be 1s22s2 B 5 electrons B 1s22s22p1 C 6 electrons ? ? Building-Up Principle (Aufbau principle)
  • 32. 32 C 6 electrons Hund’s rule: The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins. C 1s22s22p2 N 7 electrons N 1s22s22p3 O 8 electrons O 1s22s22p4 F 9 electrons F 1s22s22p5 Ne 10 electrons Ne 1s22s22p6
  • 33. 33 The order in which atomic subshells are filled in a many-electron atom. 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s
  • 34. 34 how the electrons are distributed among the various atomic orbitals in an atom. 1s1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s1 Electron configuration
  • 35. 35 What is the electron configuration of Mg atom? Mg: 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl: 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3 l = 1 ml = -1, 0, or +1 ms = ½ or -½
  • 36. 36 We conclude that there are six possible ways to designate the electron using the (n, /, m/, ms) notation: In these six designations we see that the values of n and / are constant, but the values of m/ and ms can vary.
  • 37. 37 Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p Diamagnetism and Paramagnetism
  • 38. 38 Outermost subshell being filled with electrons
  • 39. 39
  • 40. 40
  • 41. Many examples to be given in the lecture
  • 46. Boron