Chapter 7

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Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 7
The QuantumMechanical Model
of the Atom
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2007, Prentice Hall
Electron Energy
Ηψ  E ψ
 electron energy and position are complimentary
 because KE = ½mv2
 for an electron with a given energy, the best we can do is
describe a region in the atom of high probability of
finding it – called an orbital
 a probability distribution map of a region where the electron
is likely to be found
 distance vs. y2
 many of the properties of atoms are related to the
energies of the electrons
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2
Wave Function, y
 calculations show that the size, shape and orientation in space
of an orbital are determined be three integer terms in the
wave function
 added to quantize the energy of the electron
 these integers are called quantum numbers
 principal quantum number, n
 angular momentum quantum number, l
 magnetic quantum number, ml
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3
Principal Quantum Number, n
 characterizes the energy of the electron in a particular
orbital
 corresponds to Bohr’s energy level
 n can be any integer 1
 the larger the value of n, the more energy the orbital has
 energies are defined as being negative
 an electron would have E = 0 when it just escapes the atom
 the larger the value of n, the larger the orbital
 as n gets larger, the amount of energy between orbitals
gets smaller
E n  -2.1810
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-18
 1 
J 2 
n 
for an electron in H
4
Principal Energy Levels in Hydrogen
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5
Electron Transitions
 in order to transition to a higher energy state, the electron
must gain the correct amount of energy corresponding to
the difference in energy between the final and initial states
 electrons in high energy states are unstable and tend to
lose energy and transition to lower energy states
 energy released as a photon of light
 each line in the emission spectrum corresponds to the
difference in energy between two energy states
6
Predicting the Spectrum of Hydrogen
 the wavelengths of lines in the emission spectrum of
hydrogen can be predicted by calculating the difference in
energy between any two states
 for an electron in energy state n, there are (n – 1) energy
states it can transition to, therefore (n – 1) lines it can
generate
 both the Bohr and Quantum Mechanical Models can
predict these lines very accurately
E phot on
released
   E hydrogen

 18
h 
    2 . 18  10



hc
Tro, Chemistry: A Molecular Approach
elect ron

 E
 1
J
2
n
 final
final
 E initial

 
    2 . 18  10 18
 
 

1
J
n
 initial
2
 
 
 
 
7
Hydrogen Energy Transitions
8
Calculate the wavelength of light emitted when the
hydrogen electron transitions from n = 2 to n = 1
Given: ni = 2, nf = 1
Find: ,m
Concept Plan:
Relationships:
ni, nf
 1 
E  R H  2 
n 
Ephoton
Eatom
hc
Eatom = -Ephoton
 

E
E=hc/,En = -2.18 x 10-18 J (1/n2)
Solve:
 E atom   2 . 18  10
Ephoton = -(-1.64 x 10-18 J) =
1.64 x 10-18 J
 
hc
E
18
1 
 1
18
J  2  2    1 . 64  10
J
2 
1
6 .626  10
3 .00  10 

 1 . 21  10
1 .64  10 
 34
J

8 m
s
s
-18
7
J
Check: the unit is correct, the wavelength is in the UV, which is
appropriate because more energy than 3→2 (in the visible)
m
Probability & Radial Distribution
Functions
2
 y is the probability density
 the probability of finding an electron at a particular point in space
 for s orbital maximum at the nucleus?
 decreases as you move away from the nucleus
 the Radial Distribution function represents the total
probability at a certain distance from the nucleus
 maximum at most probable radius
 nodes in the functions are where the probability drops to 0
10
Probability Density Function
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Radial Distribution Function
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12
The Shapes of Atomic Orbitals
 the l quantum number primarily determines the shape of the
orbital
 l can have integer values from 0 to (n – 1)
 each value of l is called by a particular letter that designates
the shape of the orbital
 s orbitals are spherical
 p orbitals are like two balloons tied at the knots
 d orbitals are mainly like 4 balloons tied at the knot
 f orbitals are mainly like 8 balloons tied at the knot
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13
l = 0, the s orbital
 each principal energy state has 1 s
orbital
 lowest energy orbital in a principal
energy state
 spherical
 number of nodes = (n – 1)
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14
2s and 3s
2s
n = 2,
l=0
3s
n = 3,
l=0
15
l = 1, p orbitals
 each principal energy state above n = 1 has 3 p orbitals
 ml = -1, 0, +1
 each of the 3 orbitals point along a different axis
 px, py, pz
 2nd lowest energy orbitals in a principal energy state
 two-lobed
 node at the nucleus, total of n nodes
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p orbitals
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17
l = 2, d orbitals
 each principal energy state above n = 2 has 5 d orbitals
 ml = -2, -1, 0, +1, +2
 4 of the 5 orbitals are aligned in a different plane
 the fifth is aligned with the z axis, dz squared
 dxy, dyz, dxz, dx squared – y squared
 3rd lowest energy orbitals in a principal energy state
 mainly 4-lobed
 one is two-lobed with a toroid
 planar nodes
 higher principal levels also have spherical nodes
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d orbitals
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l = 3, f orbitals
 each principal energy state above n = 3 has 7 d orbitals
 ml = -3, -2, -1, 0, +1, +2, +3
 4th lowest energy orbitals in a principal energy state
 mainly 8-lobed
 some 2-lobed with a toroid
 planar nodes
 higher principal levels also have spherical nodes
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20
f orbitals
Tro, Chemistry: A Molecular Approach
21
Why are Atoms Spherical?
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22
Energy Shells and Subshells
Tro, Chemistry: A Molecular Approach
23
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