Transcript Ch8lsn23Chem105

CHAPTER 8
Atomic Electron
Configurations
and Chemical Periodicity
Outline
-Collect homework
-Review
-Trends
-Ions
Kull Spring07 Lesson 23 Ch 8
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Review
Spdf notation
 Orbital box notation

Kull Spring07 Lesson 23 Ch 8
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Characteristics of Many-Electron Atoms:
The Electron-Spin Quantum Number
Name
Symbol
Permitted Values
Property
principal
n
positive integers(1,2,3,…) orbital energy (size)
angular
momentum
l
integers from 0 to n-1
orbital shape (The l values
0, 1, 2, and 3 correspond to
s, p, d, and f orbitals,
respectively.)
magnetic
ml
integers from -l to 0 to +l
orbital orientation
spin
ms
+1/2 or -1/2
direction of e- spin
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Practice Problem 23-1
 Consider
this set of quantum numbers:
n = 3, ℓ = 2, mℓ = -1, ms = +½
The maximum number of electrons in
an atom which can share the above set
of quantum numbers is
A) 1
B) 14
C) 3
D) 10
E) none of the above
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Practice Problem 23-1 Answer
 Consider
this set of quantum numbers:
n = 3, ℓ = 2, mℓ = -1, ms = +½
The maximum number of electrons in
an atom which can share the above set
of quantum numbers is
A) 1
B) 14
C) 3
D) 10
E) none of the above
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Practice Problem 23-2
 An
atom in its ground state contains 30
electrons. How many of these are in
sublevels with ℓ = 2?
A) 2
B) 4
C) 6
D) 8
E) 10
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Practice Problem 23-2 Answer
 An
atom in its ground state contains 30
electrons. How many of these are in
sublevels with ℓ = 2?
A) 2
B) 4
C) 6
D) 8
E) 10
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Practice Problem 23-3
What are the possible values for the
angular momentum quantum number ℓ ?
A) integers from -ℓ to 0 to +ℓ
B) 1, 2, 3, etc.
C) 2, 4, 6, etc.
D) +½ , -½
E) integers from 0 to n - 1
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Practice Problem 23-3 Answer
 What
are the possible values for the
angular momentum quantum number
(ℓ )?
A) integers from -ℓ to 0 to +ℓ
B) 1, 2, 3, etc.
C) 2, 4, 6, etc.
D) +½ , -½
E) integers from
0 toLesson
n -231Ch 8
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Practice Problem 23-4
 The
electron configuration of the
outermost electrons of atoms of the
halogen group is:
A) ns2np7
B) ns1
C) ns2np5
D) ns2np6(n-1)d7
E) ns2np6
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Practice Problem 23-4 Answer
 The
electron configuration of the
outermost electrons of atoms of the
halogen group is:
A) ns2np7
B) ns1
C) ns2np5
D) ns2np6(n-1)d7
E) ns2np6
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Practice Problem 23-5
 The
electronic configuration of the
element whose atomic number is 26 is:
A) 1s2 2s2 2p6 3s2 3p6 4s0 3d8
B) 1s2 2s2 2p6 3s2 3p6 3d6 4s2
C) 1s2 2s2 2p6 3s2 3p6 4s2 3d6
D) 1s2 2s2 2p6 3s2 3p6 4s2 3d4 4p2
E) none of the above
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Practice Problem 23-5 Answer
 The
electronic configuration of the
element whose atomic number is 26 is:
A) 1s2 2s2 2p6 3s2 3p6 4s0 3d8
B) 1s2 2s2 2p6 3s2 3p6 3d6 4s2
C) 1s2 2s2 2p6 3s2 3p6 4s2 3d6
D) 1s2 2s2 2p6 3s2 3p6 4s2 3d4 4p2
E) none of the above
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Practice Problem 23-6

The set of quantum numbers that correctly
describes an electron in a 3p orbital is
A) n = 3; ℓ = 0; mℓ = 0; ms = 0
B) n = 3; ℓ = 2; mℓ = -2, -1, 0, 1, or 2; ms = +½
or -½
C) n = 3; ℓ = 1; mℓ = -1, 0, or 1; ms = +½ or -½
D) n = 4; ℓ = 0; mℓ = -1 ,0, or 1; ms = +½ or -½
E) none of the above
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Practice Problem 23-6 Answer

The set of quantum numbers that correctly
describes an electron in a 3p orbital is
A) n = 3; ℓ = 0; mℓ = 0; ms = 0
B) n = 3; ℓ = 2; mℓ = -2, -1, 0, 1, or 2; ms = +½
or -½
C) n = 3; ℓ = 1; mℓ = -1, 0, or 1; ms = +½ or
-½
D) n = 4; ℓ = 0; mℓ = -1 ,0, or 1; ms = +½ or -½
E) none of the above
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Practice Problem 23- 7
 An
atom in its ground state contains 18
electrons. How many of these are in
orbitals with mℓ = 0?
A) 2
B) 4
C) 6
D) 8
E) 10
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Practice Problem 23-7 Answer
 An
atom in its ground state contains 18
electrons. How many of these are in
orbitals with mℓ = 0?
A) 2
B) 4
C) 6
D) 8
E) 10
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Practice Problem 23-8
 The
configuration for the six outer
electrons in ground state oxygen
atoms is
A) 2s3 2p3-1
B) 2p6
C) 2s2 2p-12 2p02
D) 2s2 2p-12 2p01 2p11
E) 2s4 2p2-1 Kull Spring07 Lesson 23 Ch 8
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Practice Problem 23-8 Answer
 The
configuration for the six outer
electrons in ground state oxygen
atoms is
A) 2s3 2p3-1
B) 2p6
C) 2s2 2p-12 2p02
D) 2s2 2p-12 2p01 2p11
E) 2s4 2p2-1
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Practice Problem 23-9
 Which
of the following is the electron
configuration for chromium, element 24?
A) 1s2 2s2 2p6 3s2 3p6 4s2
B) 1s2 2s2 2p6 3s2 3p6 4s2 3d4
C) 1s2 2s2 2p6 3s2 3p6 3d6
D) 1s2 2s2 2p6 3s2 3p6 4s1 3d5
E) 1s2 2s2 2p6 3s2 3p6 4s2 3d1 3d1
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Practice Problem 23-9 Answer
 Which
of the following is the electron
configuration for chromium, element 24?
A) 1s2 2s2 2p6 3s2 3p6 4s2
B) 1s2 2s2 2p6 3s2 3p6 4s2 3d4
C) 1s2 2s2 2p6 3s2 3p6 3d6
D) 1s2 2s2 2p6 3s2 3p6 4s1 3d5
E) 1s2 2s2 2p6 3s2 3p6 4s2 3d1 3d1
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Kull Spring07 Lesson 23 Ch 8
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PERIODIC
TRENDS
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Trends in Some Key Periodic Atomic
Properties: Trends in Electron Affinity
Atomic and ionic size
# electrons
# shells –
Larger orbitals,
-electrons held less
tightly
Effective nuclear charge
When higher: Electrons held
more tightly
Electron affinity: energy involved
when an atom gains an electron
to form an anion.
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Trends in Some Key Periodic Atomic
Properties: Trends in Atomic Size
Atomic radii of the maingroup and transition
elements.
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Characteristics of Many-Electron Atoms:
Electrostatic Effects and the Splitting of Energy
Levels
Factors Affecting Atomic Orbital Energies
The Effect of Nuclear Charge (Zeffective)
Higher nuclear charge lowers orbital energy (stabilizes the
system) by increasing nucleus-electron attractions.
The Effect of Electron Repulsions (Shielding)
Additional electron in the same orbital
An additional electron raises the orbital energy through
electron-electron repulsions.
Additional electrons in inner orbitals
Inner electrons shield outer electrons more effectively than
do electrons in the same sublevel.
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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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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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Trends in Some Key Periodic Atomic
Properties: Trends in Ionization Energy
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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.
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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
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Trends in Ion Sizes
Active Figure 8.15
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Practice Problem 23-10
 Which
of the following has the largest
radius?
A) F
B) N
C) C
D) O
E) Ne
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Practice Problem 23-10 Answer
 Which
of the following has the largest
radius?
A) F
B) N
C) C
D) O
E) Ne
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Practice Problem 23-11
 Which
of the following elements has the
largest ionization energy?
A) Na
B) Ne
C) F
D) K
E) Rb
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Practice Problem 23-11 Answer
 Which
of the following elements has
the largest ionization energy?
A) Na
B) Ne
C) F
D) K
E) Rb
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Practice Problem 23-12

Which of the following has the greatest electron
affinity (most negative value)?
A) Cl
B) K
C) He
D) Na
E) Rb
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Practice Problem 23-12 Answer

Which of the following has the greatest electron
affinity (most negative value)?
A) Cl
B) K
C) He
D) Na
E) Rb
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Practice Problem 23-13
 In
the periodic table, the most
nonmetallic elements will be found
A) at the top of Group 3A (13)
B) at the top of Group 1A (1)
C) at the top of Group 7A (17)
D) at the bottom of Group 1A (1)
E) at the bottom of Group 7A (17)
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Practice Problem 23-13 Answer
 In
the periodic table, the most
nonmetallic elements will be found
A) at the top of Group 3A (13)
B) at the top of Group 1A (1)
C) at the top of Group 7A (17)
D) at the bottom of Group 1A (1)
E) at the bottom of Group 7A (17)
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Practice Problem 23-14
 Which
ion has the smallest radius?
A) Li+
B) Na+
C) K+
D) Be2+
E) Mg2+
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Practice Problem 23-14 Answer
 Which
ion has the smallest radius?
A) Li+
B) Na+
C) K+
D) Be2+
E) Mg2+
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Practice Problem 23-15

Which of the following statements about periodic
properties is incorrect?
A) Both electron affinity and ionization energy
decrease down a group.
B) Atomic size increases to the right across a period.
C) Ionization energy increases to the right across a
period.
D) Atomic size increases down a group.
E) Electron affinity increases to the right across a
period.
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Practice Problem 23.15 Answer

Which of the following statements about periodic
properties is incorrect?
A) Both electron affinity and ionization energy
decrease down a group.
B) Atomic size increases to the right across a period.
C) Ionization energy increases to the right across a
period.
D) Atomic size increases down a group.
E) Electron affinity increases to the right across a
period.
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Practice Problem 23.16
 Which
of the following elements has the
greatest ionization energy?
A) Ga
B) As
C) K
D) Pb
E) Na
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Practice Problem 22.16 Answer
 Which
of the following elements has the
greatest ionization energy?
A) Ga
B) As
C) K
D) Pb
E) Na
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Practice Problem 23.17
 Which
of the following elements has
the greatest electron affinity?
A) Rb
B) Ca
C) Li
D) Na
E) I
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Practice Problem 23.17 Answer
 Which
of the following elements has
the greatest electron affinity?
A) Rb
B) Ca
C) Li
D) Na
E) I
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Practice Problem 23.18
 Which
of the following ions and atoms
has the largest radius?
A) Mg
B) Na
C) Na+
D) Mg2+
E) Al
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Practice Problem 23.18 Answer
 Which
of the following ions and atoms has
the largest radius?
A) Mg
B) Na
C) Na+
D) Mg2+
E) Al
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Practice Problem 23.19
 Which
of the following elements has
the greatest metallic character?
A) Mg
B) Al
C) Ca
D) Ba
E) Cs
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Practice Problem 23.19 Answer
 Which
of the following elements has
the greatest metallic character?
A) Mg
B) Al
C) Ca
D) Ba
E) Cs
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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
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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 23-20
 When
electrons in helium atoms fall
from the 3s orbital down to the 1s
orbital, how many different energies
could be released if all possible
pathways of decreasing energy are
considered?
A) 2
B) 4
C) 5
D) 6
E) 7
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Practice Problem 23-20 Answer
 When
electrons in helium atoms fall
from the 3s orbital down to the 1s
orbital, how many different energies
could be released if all possible
pathways of decreasing energy are
considered?
A) 2
B) 4
C) 5
D) 6
E) 7
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Practice Problem 23-21
 Of
the following four electron
configurations, which two represent
elements that would have similar chemical
properties?
1. 1s22s22p5
2. 1s22s22p6
3. [Ar]4s23d104p4 4. [Ar]4s23d104p5
A) 1 and 3
B) 1 and 4
C) 2 and 3
D) 1 and 2
E) 2 and 4
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Practice Problem 23-21 Answer
 Of
the following four electron
configurations, which two represent
elements that would have similar chemical
properties?
1. 1s22s22p5
2. 1s22s22p6
3. [Ar]4s23d104p4 4. [Ar]4s23d104p5
A) 1 and 3
B) 1 and 4
C) 2 and 3
D) 1 and 2
E) 2 and 4
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Practice Problem 23-22

The electron configuration for Fe2+ is 1s2 2s2 2p6
3s2 3p6 3d6. Therefore Fe2+ is
A) paramagnetic with two unpaired electrons.
B) paramagnetic with one unpaired electron.
C) paramagnetic with three unpaired electrons.
D) paramagnetic with four unpaired electrons.
E) diamagnetic.
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Practice Problem 23-22 Answer

The electron configuration for Fe2+ is
1s2 2s2 2p6 3s2 3p6 3d6.
Therefore Fe2+ is
A) paramagnetic with two unpaired electrons.
B) paramagnetic with one unpaired electron.
C) paramagnetic with three unpaired electrons.
D) paramagnetic with four unpaired electrons.
E) diamagnetic.
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