Transcript ATOMIC ELECTRON CONFIGURATIONS AND PERIODICITY

1
CHAPTER 8
ATOMIC ELECTRON
CONFIGURATIONS
AND
PERIODICITY
2
Chapter 8 Outline
• Magnetic Spin
• Periodic Trends
–Effective nuclear charge
–Atomic Radii Size
–Ion Radii Size
–Ionization Energy and Electron
Affinities
ATOMIC ELECTRON
CONFIGURATIONS AND PERIODICITY
3
Arrangement of Electrons
in Atoms
Electrons in atoms are arranged as
SHELLS (n)
SUBSHELLS (l)
ORBITALS (ml)
4
Arrangement of Electrons
in Atoms
Each orbital can be assigned no
more than 2 electrons!
This is tied to the existence of a 4th
quantum number, the electron spin
quantum number, ms.
5
6
Electron
Spin
Quantum
Number,
ms
Can be proven experimentally that an
electron has a spin.
The two spin directions are given by ms
where ms = +1/2 and -1/2.
Electron Spin Quantum Number
Diamagnetic: substance NOT attracted to
a magnetic field.
Paramagnetic: substance is attracted to a
magnetic field. Substance has unpaired
electrons.
7
Paramagnetic -vs- Ferromagnetic
Unpaired electrons
in an electric field
alien themselves to
the applied field
Unpaired electrons in their
normal, random orientation.
ALL metals behave this way, the
exception is Cobalt, the reason
why Cobalt magnets are so
VERY expensive.
8
9
QUANTUM
NUMBERS
n ---> shell
1, 2, 3, 4, ...
l ---> subshell
0, 1, 2, ... n - 1
ml ---> orbital
-l ... 0 ... +l
ms ---> electron spin
+1/2 and -1/2
10
Pauli Exclusion Principle
No two electrons in the
same atom can have
the same set of 4
quantum numbers.
That is, each electron has a
unique address = set of 4
quantum numbers.
Electrons in Atoms
When n = 1, then l = 0
this shell has a single orbital (1s) to
which 2e- can be assigned.
When n = 2, then l = 0, 1
2s orbital
2e-
three 2p orbitals
6e-
TOTAL =
8e-
11
Electrons in Atoms
When n = 3, then l = 0, 1, 2
3s orbital
2e-
three 3p orbitals
6e-
five 3d orbitals
10e-
TOTAL =
18e-
12
Electrons in Atoms
When n = 4, then l = 0, 1, 2, 3
4s orbital
2ethree 4p orbitals 6efive 4d orbitals
10eseven 4f orbitals 14eTOTAL =
32e-
And many more!
13
14
Assigning Electrons to Atoms
• Electrons generally assigned to orbitals
of successively higher energy.
• For H atoms, E = - C(1/n2). E depends
only on n.
• For multi - electron atoms, energy
depends on both n and l.
15
Figure 8.4
16
Assigning Electrons to Subshells
• In the H atom all
subshells of same n
have same energy.
• In many-electron atom:
a) subshells increase in
energy as value of n + l
increases.
b) for subshells of same
n + l, subshell with
lower n is lower in
energy.
17
Electron
Filling Order
Figure 8.5
18
Writing Atomic Electron
Configurations
Two ways of writing
complete
configurations.
One is called the
spectroscopic
notation.
SPECTROSCOPIC NOTATION
for H, atomic number = 1
1
1s
value of n
no. of
electrons
value of l
Writing Atomic Electron
Configurations
Two ways of
writing
complete
configurations.
Other is called
the orbital
box notation.
19
ORBITAL BOX NOTATION
for He, atomic number = 2
Arrows
depict
2
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
20
21
Figure 8.7
22
Lithium
Group 1A
Atomic number = 3
1s22s1 ---> 3 total electrons
3p
3s
2p
2s
1s
23
Beryllium
3p
3s
2p
2s
1s
Group 2A
Atomic number = 4
1s22s2 ---> 4 total
electrons
24
Boron
3p
3s
2p
2s
1s
Group 3A
Atomic number = 5
1s2 2s2 2p1 ---> 5 total
electrons
25
Carbon
Group 4A
Atomic number = 6
1s2 2s2 2p2 ---> 6 total
electrons
3p
3s
2p
2s
1s
Here we see HUND’S RULE.
When placing electrons in a
set of orbitals having the
same energy, we place them
singly as long as possible.
26
Nitrogen
3p
3s
2p
2s
1s
Group 5A
Atomic number = 7
1s2 2s2 2p3 ---> 7 total
electrons
27
Oxygen
Group 6A
Atomic number = 8
1s2 2s2 2p4 ---> 8 total
electrons
3p
3s
2p
2s
1s
28
Fluorine
Group 7A
Atomic number = 9
1s2 2s2 2p5 ---> 9 total
electrons
3p
3s
2p
2s
1s
29
Neon
Group 8A
Atomic number = 10
1s2 2s2 2p6 ---> 10 total
electrons
3p
3s
2p
2s
1s
Note that we have
reached the end of
the 2nd period, and
the 2nd shell is full!
30
Electron Configurations
Noble gas configuration: filled s and p levels.
Electron configs. are shortened by using the noble
gas configuration to represent the core electrons.
The remaining electrons are valence electrons.
Notations:
Spectroscopic
Orbital Box
Examples: Boron
1s22s22p1
1s 2s
2p
Noble Gas
Valence
[He] 2s22p1
2s22p1
31
Quantum Numbers
32
If we use positive values of ml and ms
first, the set of four quantum numbers
can be determined for the “last
electron in” for a given element.
This is called the
“last-electron-in” game!!!
33
Electron Configurations
of p-Block Elements
34
Sodium
Group 1A
Atomic number = 11
1s2 2s2 2p6 3s1 or
“neon core” + 3s1
[Ne] 3s1 (uses rare gas notation)
Note that we have begun a new period.
All Group 1A elements have
[core]ns1 configurations.
35
Aluminum
Group 3A
Atomic number = 13
1s2 2s2 2p6 3s2 3p1
[Ne] 3s2 3p1
All Group 3A elements
have [core] ns2 np1
configurations where n
is the period number.
3p
3s
2p
2s
1s
36
Phosphorus
Group 5A
Atomic number = 15
1s2 2s2 2p6 3s2 3p3
[Ne] 3s2 3p3
All Group 5A
elements have
3p
3s
[core ] ns2 np3
configurations
where n is the
period number.
2p
2s
1s
37
Calcium
Group 2A
Atomic number = 20
1s2 2s2 2p6 3s2 3p6 4s2
[Ar] 4s2
All Group 2A elements have
[core]ns2 configurations
where n is the period number.
Relationship of Electron
Configuration and Region of
the Periodic Table
•
•
•
•
Gray = s block
Orange = p block
Green = d block
Violet = f block
38
Transition Metals
Table 8.4
All 4th period elements have the
configuration [argon] nsx (n - 1)dy
and so are “d-block” elements.
Chromium
Iron
Copper
39
Transition Element
Configurations
3d orbitals used for
Sc - Zn (Table 8.4)
40
41
Figure 8.9
42
43
Lanthanides and Actinides
All these elements have the configuration
[core] nsx (n - 1)dy (n - 2)fz and so are
“f-block” elements.
Cerium
[Xe] 6s2 5d1 4f1
Uranium
[Rn] 7s2 6d1 5f3
44
Lanthanide Element Configurations
4f orbitals used for
Ce - Lu and 5f for
Th - Lr (Table 8.2)
Ion Configurations
To form cations from elements remove 1
or more e- from subshell of highest n
[or highest (n + l)].
P [Ne] 3s2 3p3 ---> P3+ [Ne] 3s2 3p0 + 3e-
1s
2s
P
2p
3s
P3+
3p
45
46
Ion Configurations
For transition metals, remove ns electrons
and then (n - 1) electrons.
Fe [Ar] 4s2 3d6 ---> Fe2+ [Ar] 4s0 3d6 + 2e1s
2s
Fe2+
2p
Fe
3s
3p
4s
3d
47
Ion Configurations
How do we know the configurations of ions?
Determine the magnetic properties of ions.
Ions with UNPAIRED ELECTRONS are
PARAMAGNETIC.
Without unpaired electrons DIAMAGNETIC.
48
Sample Question
Give the electron configuration for Ni+4.
Include the valence box diagram and
state if the ion is paramagnetic or
diamagnetic.
1s
2s
Ni4+
2p
Ni
3s
3p
4s
3d
49
PERIODIC
TRENDS
Movies on these later!!
General Periodic Trends
1. Atomic and ionic size
2. Ionization energy
3. Electron affinity
Higher Z*.
Electrons held
more tightly.
Larger orbitals.
Electrons held less
tightly.
50
51
Atomic Size
• Size goes UP when going
down a group.
See Figure 8.9.
• Because electrons are
added further from the
nucleus, there is less
attraction.
• Size goes DOWN when
going across a period.
SIZE
Effective Nuclear Charge:
relating to Atom radii
The reason for the difference in energy
for 2s and 2p subshells, for example, is
the effective nuclear charge, Z*.
52
53
Screening or Shielding
effect: helps explain
periodic trends in a
period
Figure 8.6
54
Effective Nuclear Charge, Z*
• Z* is the nuclear charge experienced by
the outermost electrons.
• Explains why E(2s) < E(2p)
• Z* increases across a period owing to
incomplete shielding by inner electrons.
• 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!
The nuclear charge
increases
55
56
Atomic Size
Size decreases across a period owing to
increase in Z*. Each added electron feels a
greater and greater + charge.
57
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
Atomic Number
25
30
35
40
58
Sizes of Transition Elements
• 3d subshell is inside the 4s subshell.
• 4s electrons feel a more or less constant Z*.
• Sizes stay about the same and chemistries
are similar!
59
60
Ion Sizes
Li,152 pm
3e and 3p
Does the size go
up or+down
when
losing
an
Li + , 60 pm
electron
2e and 3 p to form
a cation?
Ion Sizes
• CATIONS are
SMALLER than
the atoms from
which they come.
• The
electron/proton
attraction has
gone UP and so
size DECREASES.
61
Ion Sizes
F,64 pm
9e and 9p
Does the size go up or
down when
gaining an
electron
to form an
F- , 136 pm
anion?
10 e and 9 p
62
63
Ion Sizes
• ANIONS are
LARGER than the
atoms from
which they come.
• The
electron/proton
attraction has
gone DOWN and
so size
INCREASES.
64
Figure 8.13
Ion Sizes
For an
isoelectronic series
• Size decreases as the atomic number
increases
• Example:
N-3 to Al+3
65
66
Redox Reactions
Why do metals lose
electrons in their
reactions?
Why does Mg form Mg2+
ions and not Mg3+?
Why do nonmetals take
on electrons?
Ionization Energy
See Figure 8.12
IE = energy required to remove an electron
from an atom in the gas phase.
Mg (g) + 738 kJ ---> Mg+ (g) + e-
A(g) ---> A+(g) + e-
I.E. = +E
67
Ionization Energy
IE = energy required to remove an electron
from an atom in the gas phase.
Mg (g) + 738 kJ ---> Mg+ (g) + eMg+ (g) + 1451 kJ ---> Mg2+ (g) + e-
Mg+ has 12 protons and only 11 electrons.
Therefore, IE for Mg+ > Mg.
68
69
Ionization Energy
Mg (g) + 735 kJ ---> Mg+ (g) + e-
Mg+ (g) + 1451 kJ ---> Mg2+ (g) + eMg2+ (g) + 7733 kJ ---> Mg3+ (g) + e-
Energy cost is very high to dip into a shell
of lower n. This is why ox. no. = Group no.
70
71
Trends in Ionization Energy
• IE increases across a
period because Z*
increases.
• Metals lose electrons
more easily than
nonmetals.
• Metals are good reducing
agents.
• Nonmetals lose electrons
with difficulty.
72
Trends in Ionization Energy
• IE decreases down a group
• Because size increases.
• Reducing ability generally
increases down the periodic
table.
• See reactions of Li, Na, K
Figure 8.11
73
74
Lithium
Periodic Trend
in the Reactivity
of Alkali Metals
with Water
Sodium
Potassium
Here are those movies promised!
Electron Affinity
75
A few elements GAIN electrons to
form anions.
Electron affinity is the energy
involved when an atom Gains an
electron.
A(g) + e- ---> A- (g)
Note: Both
EA’s and IE’s
are “-”.
E.A. = -E
Cl E.A. = -349 kJ/Mol
76
Electron Affinity of Oxygen
E is
Exothermic
because O has an
affinity for an e-.
EA = - 141 kJ
O- ion [He] 
 

+ electron
O atom [He] 
 

77
Electron Affinity of Nitrogen
EA = 0 kJ
E is zero for N
due to electronelectron
repulsions.
N- ion [He] 
 

+ electron
N atom [He] 



Trends in Electron Affinity
• See Figure 8.12
• Affinity for electron increases
across a period (EA becomes
more negative).
• Affinity decreases down a group
(EA becomes less negative).
78
Figure 8.12
79
Trends in Electron Affinity
• See Appendix
F
Atom EA
• Affinity for electron
F
-328 kJ
increases across a period
Cl -349 kJ
(EA becomes more
Br -325 kJ
negative).
-295 kJ
• Affinity decreases down a I
group (EA becomes less
negative).
These numbers are for
the ion forming the atom.
80
81
Practice Problems
1. Write the electron configurations:
a. Orbital box notation for C
b. Spectroscopic notation for Mg
c. Noble Gas notation for Sc
d. Noble Gas notation for P
e. Valence notation for Pb
82
Practice Problems
2. Answer the following about the elements in
question 1:
a. number of unpaired electron's
b. arrange according to size (small to large)
c. arrange according to ionization
energy (small to large)
d. predict ions that each elements will form
e. determine the set of 4 quantum
numbers for the last electron.
83
Practice Problems
3. Give the symbol of the element with the
lowest atomic number that has
a. a half filled p sublevel
b. 3 filled 4d orbitals
c. unpaired electrons in 2 or more sublevels
84
Practice Problems Answers
1. a.
1s
2s
2p
b. 1s22s22p63s2
c. [Ar] 4s23d1
d. [Ne] 3s23p3
e. 6s26p2 or 6s24f145d106p2
2. a. 2, 0, 1, 3, 2
b. C, P, Mg, Sc, Pb
c. Pb, Sc, Mg, P, C
d. -4, +2, +2 and +3, -3, +2 and +4
85
Practice Problems Answers
2. e. C
2, 1, 0, +1/2
Mg
3, 0, 0, -1/2
Sc
3, 2, 2, +1/2
P
3,1,-1,+1/2
Pb
6, 1, 0, +1/2
3. N, Pd, Cr
86
Electrons and Quantum Numbers
n
l
ml
1
1
0
0
0
0
ms
+1/2
-1/2
1s
87
Electrons and Quantum Numbers
n
l
ml
ms
2
2
0
0
0
0
+1/2
-1/2
2s
2
2
2
2
2
2
1
1
1
1
1
1
1
0
-1
1
0
-1
+1/2
+1/2
+1/2
-1/2
-1/2
-1/2
2p
88
Hydrogen
“Group 1A”
Atomic number = 1
1 electron
1s
1s1
89
Helium
“Group 8A”
Atomic number = 2
2 electrons
1s
1s2
90
Lithium
Group 1A
Atomic number = 3
3 electrons
1s
2s
1s22s1
91
Beryllium
Group 2A
Atomic number = 4
4 electrons
1s
2s
1s22s2
92
Boron
Group 3A
Atomic number = 5
5 electrons
1s
2s
1s22s22p1
2p
93
Carbon
Group 4A
Atomic number = 6
6 electrons
1s
2s
1s22s22p2
2p
94
Nitrogen
Group 5A
Atomic number = 7
7 electrons
1s
2s
1s22s22p3
2p
95
Oxygen
Group 6A
Atomic number = 8
8 electrons
1s
2s
1s22s22p4
2p
96
Fluorine
Group 7A
Atomic number = 9
9 electrons
1s
2s
1s22s22p5
2p
97
Neon
Group 8A
Atomic number = 10
10 electrons
1s
2s
1s22s22p6
2p
Sample Question
Give the spectroscopic notation for P.
1s22s22p63s23p3
Determine the set of 4 quantum numbers
for the “last electron in”.
(3, 1, -1, 1/2)
Give a box diagram for the valence
electrons.
3s
3p
98
99
Sodium
Group 1A
Atomic number = 11
11 electrons
1s
2s
1s22s22p63s1
2p
3s
100
Magnesium
Group 2A
Atomic number = 12
12 electrons
1s
2s
1s22s22p63s2
2p
3s
101
Aluminum
Group 3A
Atomic number = 13
13 electrons
1s
2s
2p
1s22s22p63s23p1
3s
3p
102
Silicon
Group 4A
Atomic number = 14
14 electrons
1s
2s
2p
1s22s22p63s23p2
3s
3p
103
Phosphorus
Group 5A
Atomic number = 15
15 electrons
1s
2s
2p
1s22s22p63s23p3
3s
3p
104
Sulfur
Group 4A
Atomic number = 16
16 electrons
1s
2s
2p
1s22s22p63s23p4
3s
3p
105
Chlorine
Group 7A
Atomic number = 17
17 electrons
1s
2s
2p
1s22s22p63s23p5
3s
3p
106
Argon
Group 8A
Atomic number = 18
18 electrons
1s
2s
2p
1s22s22p63s23p6
3s
3p
107
Potassium
Group 1A
Atomic number = 19
19 electrons
1s
2s
2p
1s22s22p63s23p64s1
3s
3p
4s
108
Calcium
Group 2A
Atomic number = 20
20 electrons
1s
2s
2p
1s22s22p63s23p64s2
3s
3p
4s
109
Scandium
B Group
Atomic number = 21
21 electrons
1s
2s
2p
3s
3p
4s
3d
1s22s22p63s23p64s23d1
Sample Questions
1. Write the noble gas
configuration for
[Ar] 4s23d104p2
a) Ge
b) Cs
[Xe] 6s1
c) Pb
[Xe] 6s24f145d106p2
110
Sample Questions
2. What element has (4, 2, 1, -1/2) as its last
electron in?
Use a valence box diagram and give the
valence configuration for this element.
4d
4d7
2
1
0
-1
-2
5s2 or 5s24d7
Rh
111
Sample Questions
3. Give the symbol of the element with
the lowest atomic number that
a. is a noble gas with no p electrons.
b. is in period 3 and has 2 unpaired
electrons.
c. is in period 2 and has 2 filled p
orbitals.
d. has 4 electrons in level 2.
He
Si
F
C
112