Transcript Matter and Measurement

CHAPTER 8
Atomic Electron
Configurations
and Chemical Periodicity
Road Map - Lesson 22
Recap of Ch 7 problems
Area of Focus: Chapter 8
The Exclusion Principle – introduction of the 4th q.n.
Electron Spin
Subshell filling order
Atomic electron configuration
Kull Spring07 Lesson 22 Ch 7/8
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Equations
speed of light = wavelength x frequency
c = λ X  = 3.00 x 108 m/s
 E = nh = nh(c/) n= positive integer

Planck’s constant(h) = 6.626 x 10–34 J s
 Eatom = Eemitted (or absorbed) radiation = nh

Rydberg equation




=R
n2 > n1
R = 1.096776 x 107 m-1
ΔE = Efinal – Einitial = –2.18 x 10–18 J
Ephoton = Estate A – Estate B = hν
Kull Spring07 Lesson 22 Ch 7/8
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Quantum mechanics
Quantum
number
n–
shell/level
ℓ - subshell/
sublevel
mℓ - orbital
ms - spin
Values
Total number
1,2,3,
…∞
0, 1,…
n-1
- ℓ, 0, + ℓ
n=#
subshells
n-1
+½,
n2 = # orbitals
in a shell
2 ℓ +1 (orbitals
in a subshell)
-½ 2 possible
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Practice Problem 22-1

According to the quantum mechanical
treatment of the hydrogen atom, which
set of quantum numbers is not allowed?
A) n = 3, ℓ = 2, mℓ = 0
B) n = 3, ℓ = 0, mℓ = 0
C) n = 3, ℓ = 1, mℓ = 1
D) n = 3, ℓ = 1, mℓ = -1
E) n = 3, ℓ = 1, mℓ = 2
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Practice Problem 22-1 Answer

According to the quantum mechanical
treatment of the hydrogen atom, which
set of quantum numbers is not allowed?
A) n = 3, ℓ = 2, mℓ = 0
B) n = 3, ℓ = 0, mℓ = 0
C) n = 3, ℓ = 1, mℓ = 1
D) n = 3, ℓ = 1, mℓ = -1
E) n = 3, ℓ = 1, mℓ = 2
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Practice Problem 22-2

For an excited hydrogen atom with the quantum
number n equal to 9, which of the following
statements is true?
A) The energy of the atom is less than the energy for the
state in which n is equal to 8.
B) If ℓ = 0, there are nine possible values for the magnetic
quantum number mℓ.
C) The smallest value of the magnetic quantum number
mℓ is -9.
D) The electron must be in one of the p orbitals.
E) The angular momentum quantum number ℓ can have
any of the values 0, 1, 2, 3, 4, 5, 6, 7, 8.
Kull Spring07 Lesson 22 Ch 7/8
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Practice Problem 22-2 Answer

For an excited hydrogen atom with the quantum number n
equal to 9, which of the following statements is true?
A) The energy of the atom is less than the energy for the
state in which n is equal to 8.
B) If ℓ = 0, there are nine possible values for the magnetic
quantum number mℓ.
C) The smallest value of the magnetic quantum number
mℓ is -9.
D) The electron must be in one of the p orbitals.
E) The angular momentum quantum number ℓ can have
any of the values 0, 1, 2, 3, 4, 5, 6, 7, 8.
Kull Spring07 Lesson 22 Ch 7/8
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Practice Problem 22-3
 According
to the quantum-mechanical
model, how many orbitals in a given
atom have n = 3?
A) 4
B) 7
C) 9
D) 10
E) 18
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Practice Problem 22-3 Answer
 According
to the quantum-mechanical
model, how many orbitals in a given
atom have n = 3?
A) 4
B) 7
C) 9
D) 10
E) 18
Kull Spring07 Lesson 22 Ch 7/8
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Chapter 8
 The Exclusion
4th q.n.
Principle – introduction of the
 Hund’s
rule
 Electron Spin
 Subshell filling order
 Atomic electron configuration
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Chapter 8
 Pauli
Exclusion Principle - No two electrons in
the same atom can have the same set of 4
quantum numbers.
 each electron has a unique address.
 HUND’S RULE - When placing electrons in a
set of orbitals having the same energy, we
place them singly as long as possible.
 aufbau principle is the building up process that
results in ground-state electron configurations
Kull Spring07 Lesson 22 Ch 7/8
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Electron Configurations
and the Periodic Table
Active Figure 8.7
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Electron
Filling
Order
Figure 8.5
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Quantum mechanics
Quantum
number
n–
shell/level
ℓ - subshell/
sublevel
mℓ - orbital
ms - spin
Values
Total number
1,2,3,
…∞
0, 1,…
n-1
- ℓ, 0, + ℓ
n=#
subshells
n-1
+½,
n2 = # orbitals
in a shell
2 ℓ +1 (orbitals
in a subshell)
-½ 2 possible
Kull Spring07 Lesson 22 Ch 7/8
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Writing Atomic Electron
Configurations
Two ways of writing
configs.
1)
the spdf notation.
2)
the orbital box
notation.
spdf notation
for H, atomic number = 1
1
1s
value of n
no. of
electrons
value of l
ORBITAL BOX NOTATION
for He, atomic number = 2
Arrows
2
depict
electron
spin
1s
1s
One electron has n =
1, l = 0, ml = 0,
ms = + 1/2
Other electron has n =
1, l = 0, ml = 0,
ms = - 1/2
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Phosphorus
Yellow P
Group 5A
Atomic number = 15
1s2 2s2 2p6 3s2 3p3
[Ne] 3s2 3p3
All Group 5A elements
have [core ] ns2 np3
configurations where n
is the period number.
Red P
3p
3s
2p
2s
1s
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Kull Spring07 Lesson 22 Ch 7/8
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PERIODIC
TRENDS
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General Periodic Trends

Atomic and ionic size
 # electrons
 # shells –
Larger orbitals,
-electrons held less tightly
Ionization energy


Electron affinity
 energy involved when an
atom gains an electron to form an anion.
 Effective nuclear charge
 When higher: Electrons held
Kull Spring07 Lesson 22 Ch 7/8
more tightly
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Trends in Atomic Size
Radius (pm)
250
K
1st transition
series
3rd period
200
Na
2nd period
Li
150
Kr
100
Ar
Ne
50
He
0
0
5
10
15
20
25
30
35
40
Atomic Number
Kull Spring07 Lesson 22 Ch 7/8
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Effective Nuclear Charge, Z*








Z* is the nuclear charge experienced by the outermost
electrons.
Z* increases across a period owing to incomplete shielding by
inner electrons.
The 2s electron PENETRATES the region occupied by the 1s
electron.
2s electron experiences a higher positive charge than
expected.
Estimate Z* by --> [ Z - (# inner electrons) ]
Charge felt by 2s e- in Li
Z* = 3 - 2 = 1
Be
Z* = 4 - 2 = 2
B
Z* = 5 - 2 = 3
and so on!
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Ion Configurations
To form cations, always remove electrons
of highest n value first!
P [Ne] 3s2 3p3 - 3e- ---> P3+ [Ne] 3s2 3p0
3p
3p
3s
3s
2p
2p
2s
2s
1s
1s
Kull Spring07 Lesson 22 Ch 7/8
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Ion Sizes
+
Li,152 pm
3e and 3p
Li + , 78 pm
2e and 3 p
Forming
a cation.
 CATIONS
are SMALLER than the
atoms from which they come.
 The electron/proton attraction has
gone UP and so size DECREASES.
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Ion Sizes
F, 71 pm
9e and 9p
F- , 133 pm
10 e and 9 p
Forming
an anion.
 ANIONS
are LARGER than the atoms
from which they come.
 The electron/proton attraction has gone
DOWN and so size INCREASES.
 Trends in ion sizes are the same as atom
sizes.
Kull Spring07 Lesson 22 Ch 7/8
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Trends in Ion Sizes
Active Figure 8.15
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Ionization Energy
IE = energy required to remove an electron
from an atom in the gas phase.
Mg (g) + 738 kJ ---> Mg+ (g) + e-
Mg+ (g) + 1451 kJ ---> Mg2+ (g) + e-
Mg+ has 12 protons and only 11 electrons. Therefore, IE for Mg+ > Mg
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