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Chapter 8Atomic Electron Configurations and
Chemical Periodicity
We know the electronic structure of the hydrogen atomstates as determined by the quantum numbers n, l and m.
How does this apply to larger atoms? i.e. multiple electron systems
How does the electron structure relate to the periodic table ?
How does the electron structure relate to the chemical properties of atoms ?
Electron Spin and Magnetism
Before we can talk about structure we need to learn a bit about the magnetic properties of particles.
Recall that electron move the nucleus in orbits corresponding to set angler momentum values
Recall, also that when electrons move they generate a magnetic field, B. v
B
This is analogous to electrical current moving through a loop
Electrons in orbit generate magnetic fields, which requires all materials to be magnetic.Is that so? Why? How is this possible?
Imagine two electrons in the same orbit moving in opposite directions.
B
v v
B
Electron Spin and Magnetism
The magnetic fields cancel !!
Do electrons occur in pairs in orbitals!!!
But not for this reason, since this is not physically correct.
Motion of electrons in their orbitals is not responsible for magnetism, even when the electron is unpaired. The net magnetic field averages to zero.
Yes!
Electron Spin and Magnetism
When a beam of atomic hydrogen is passed through a non-uniform magnetic field is splits into two beams
This Magnetism is not due to due to orbital motion
Another source of magnetism
From where?
Spin
Electron Spin and Magnetism
When an external magnetic field is applied the electron will either along or against the field.
Being aligned with the field is more stable than against, therefore the up orientation is slightly favored
Electron spin is an inherent magnetism associated with it, which has nothing to do with its translational motion.
The electron can the thought of as a little magnet
More stable Less stable
The distribution of up to down depends on strength of the applied magnetic field.
B
UP (s=1/2) DOWN (s=-1/2)
Magnetic field
Magnetic Materials
Paramagnetic Materials
Diamagnetic Materials
More electron electrons will align with the field than against the externally applied field.
Composed of atoms/molecules containing only paired electrons
They are repelled by an externally applied magnetic field.
Composed of atoms/molecules with unpaired electrons.
The result is a net bulk magnetic field parallel to the applied field, hence an attractive force
Ferromagnetic Materials – Have a permanent magnetic field
The magnetic field from each atom will add up, as long as the atoms are correctly aligned to give a one strong “bulk’ magnetic field. – i.e Magnets
When two electron on separate atoms are close, the field from one will cause the other to align with it since it is more stable
Magnetic Materials
Pauli Exclusion Principle
Fermions - particles have spin ½.
electrons protons neutrons
“Fermions cannot occupy the same space and spin coordinates”
This means that no two electrons can have the same quantum numbers, including the spin quantum numbers.
Therefore each orbital can only have 2 electrons since there are only two spin states s =1/2 and -1/2.
Ex) 1s orbital n = 1, l = 0, m = 0 and s = 1/2 or s = -1/2
1s Orbital
Atoms with more than one electronThe wavefunctions for multi electron atoms similar to those for the H atom
The ground state of such atoms requires that the lowest possible energy wavefunctions be “occupied”
box diagram - a simple tool used to add or subtract electrons from the boxes to represent the electron configuration of the element
Consider H, He, Li and Be
B 5
C 6
N 7
O 8
F 9
Ne 10
1s 2p2s
Hunds ruleElement # e’s
Electrons added to each empty orbital in parallel
When no new orbitals are available they are paired
Maximize spin
Electron Configuration
2 2Be 1 2s s1 18
1s1
1
2
13
14
15
16
17
1s2
2
2s1
3
2s2
4
2p1
5
2p2
6
2p3
7
2p4
8
2p5
9
2p6
10
3s1
11
3s2
12
3
4
5
6
7
8
9
10
11
12
3p1
13
3p2
14
3p3
15
3p4
16
3p5
17
3p6
18
4s1
19
4s2
20
3d1
21
3d2
22
3d3
23
4s13d5
24
3d5
25
3d6
26
3d7
27
3d8
28
4s13d10
29
3d10
30
4p1
31
4p2
32
4p3
33
4p4
34
4p5
35
4p6
36
5s1
37
5s2
38
4d1
39
4d2
40
5s14d4
41
5s14d5
42
4d5
43
5s14d7
44
5s14d8
45
5s04d10
46
5s14d10
47
4d10
48
5p1
49
5p2
50
5p3
51
5p4
52
5p5
53
5p6
54
6s1
55
6s2
56
La-Lu
5d2
72
5d3
73
5d4
74
5d5
75
5d6
76
5d7
77
6s15d9
78
6s25d10
79
5d10
80
6p1
81
6p2
82
6p3
83
6p4
84
6p5
85
6p6
86
7s1
87
7s2
88
Ac-Lr
6d2
104
6d3
105
6d4
106
6d5
107
6d6
108
6d7
109
110
111
4f05d1
57
4f15d1
58
4f3
59
4f4
60
4f5
61
4f6
62
4f7
63
4f75d1
64
4f9
65
4f10
66
4f11
67
4f12
68
4f13
69
4f14
70
5d1
71
5f06d1
89
5f06d2
90
5f26d1
91
5f36d1
92
5f46d1
93
5f6
94
5f7
95
5f76d1
96
5f9
97
5f10
98
5f11
99
5f12
100
5f13
101
5f14
102
6d1
103
A shorthand notation is commonly used to write out the electron configuration of the atoms based on the number of electrons within each subshell
It consists of: NUMBER LETTER SUPERSCRIPT(shell i.d.) (subshell) (occupancy)
B 5
C 6
N 7
O 8
F 9
Ne 10
1s 2p2s
Electron ConfigurationElement # e’s
1s2 2s2 2p1
1s2 2s2 2p2
1s2 2s2 2p3
1s2 2s2 2p4
1s2 2s2 2p5
1s2 2s2 2p6
Aufbau order and Energy LevelsThe sequence of subshells in the electron configurations not exactly same as the energy levels of H
The experimental sequence is known as the aufbau order
It is a consequence of electron-electron interactions have on the energies of the wavefunctions in all multi-electron atoms
Levels in subshells are still degenerate, the subshells are no longer degenerate in each shell, and differ in energy as s < p < d,
Some subshells can overlap the levels of a different shell; thus, for example, in neutral atoms 4s lies below 3d
Traditional aufbau sequence diagramInstead of filling orbitals in order of increasing n, we should really be filling them in order of increasing n + l
n is used as a ‘tiebreaker’ i.e the one with lowest n first
Ex) Fluorine 9 e’s
1s2 2s2 2p5
Ex) Scandium 21 e’s
1s2 2s2 2p63s2 3p64s23d1
Ex) Strontium 38 e’s
1s2 2s2 2p63s2 3p64s23d10
4p6 5s2
Afbau sequence from Periodic Table
1.008
H 1
2
13
14
15
16
17
4.003
He 2
6.939
Li 3
9.012
Be 4
10.811
B 5
12.011
C 6
14.007
N 7
15.999
O 8
18.998
F 9
20.183
Ne 10
22.990
Na 11
24.312
Mg 12
3
4
5
6
7
8
9
10
11
12
26.982
Al 13
28.086
Si 14
30.974
P 15
32.064
S 16
35.453
Cl 17
39.948
Ar 18
39.102
K 19
40.08
Ca 20
44.956
Sc 21
47.90
Ti 22
50.942
V 23
51.996
Cr 24
54.938
Mn 25
55.847
Fe 26
58.933
Co 27
58.71
Ni 28
63.546
Cu 29
65.37
Zn 30
69.72
Ga 31
72.59
Ge 32
74.922
As 33
78.96
Se 34
79.904
Br 35
83.80
Kr 36
85.47
Rb 37
87.62
Sr 38
88.905
Y 39
91.22
Zr 40
92.906
Nb 41
95.94
Mo 42
(98)
Tc 43
101.07
Ru 44
102.90
Rh 45
106.4
Pd 46
107.87
Ag 47
112.40
Cd 48
114.82
In 49
118.69
Sn 50
121.75
Sb 51
127.60
Te 52
126.90
I 53
131.30
Xe 54
132.91
Cs 55
137.33
Ba 56
138.91
La 57
178.49
Hf 72
180.95
Ta 73
183.85
W 74
186.21
Re 75
190.22
Os 76
192.2
Ir 77
195.09
Pt 78
196.97
Au 79
200.59
Hg 80
204.38
Tl 81
207.19
Pb 82
208.98
Bi 83
(209)
Po 84
(210)
At 85
(222)
Rn 86
(223)
Fr 87
226.025
Ra 88
227.029
Ac 89
(261)
Rf 104
(262)
Ha 105
(263)
Sg 106
(262)
Ns 107
(265)
Hs 108
(266)
Mt 109
new 110
new 111
140.12
Ce 58
140.91
Pr 59
144.24
Nd 60
(145)
Pm 61
150.36
Sm 62
151.97
Eu 63
157.25
Gd 64
158.93
Tb 65
162.50
Dy 66
164.93
Ho 67
167.26
Er 68
168.93
Tm 69
173.04
Yb 70
174.97
Lu 71
232.04
Th 90
231.04
Pa 91
238.03
U 92
237.05
Np 93
(244)
Pu 94
(243)
Am 95
(247)
Cm 96
(247)
Bk 97
(251)
Cf 98
(252)
Es 99
(257)
Fm 100
(258)
Md 101
(259)
No 102
(260)
Lr 103
s block
d block
p block
f block
We can now see that the very origin of the periodic table is the electron configurations of the elements
The periodic table can be used to determine the afbau order instead
As you increase the # electrons, the block structure indicates the sequence of subshells
A more detailed look at the block structure
1
0
2
3
2 3
4 4
4 5 55
6
6 66
7
The core electrons are represented by the noble gas followed by
configuration of the valence electrons.
Electron configurations for the larger elements are lengthy to write out.
Ex) Ne has an electron configuration of 1s22s22p6.
For Na, we can write either 1s22s22p63s1 or [Ne]3s1
Electron Configurations
Ex) Sr 38 1s2 2s2 2p63s2 3p64s23d10 4p6 5s2
1s2 2s2 2p63s2 3p64s23d104p6Kr 36
[Kr] 5s2
Core e’s
Valence e’s
noble gas notation - the symbol for a noble gas is used as an abbreviation for its electrons.
How many core and valence electrons do these atoms have?
a) ____core, ____valence c) ____core, ____valenceb) ____core, ____valence d) ____core, ____valence
Identify the elements with the following electron configurations.
a) 1s22s22p3 c) [Ne]3s23p3
b) 1s22s22p63s23p64s23d7 d) [Kr]5s24d5
Exercises
NCo
PTc
2 5 10 5
18 9 36 7
Exceptions to the aufbau order
1 18
1s1
1
2
13
14
15
16
17
1s2
2
2s1
3
2s2
4
2p1
5
2p2
6
2p3
7
2p4
8
2p5
9
2p6
10
3s1
11
3s2
12
3
4
5
6
7
8
9
10
11
12
3p1
13
3p2
14
3p3
15
3p4
16
3p5
17
3p6
18
4s1
19
4s2
20
3d1
21
3d2
22
3d3
23
4s13d5
24
3d5
25
3d6
26
3d7
27
3d8
28
4s13d10
29
3d10
30
4p1
31
4p2
32
4p3
33
4p4
34
4p5
35
4p6
36
5s1
37
5s2
38
4d1
39
4d2
40
5s14d4
41
5s14d5
42
4d5
43
5s14d7
44
5s14d8
45
5s04d10
46
5s14d10
47
4d10
48
5p1
49
5p2
50
5p3
51
5p4
52
5p5
53
5p6
54
6s1
55
6s2
56
La-Lu
5d2
72
5d3
73
5d4
74
5d5
75
5d6
76
5d7
77
6s15d9
78
6s15d10
79
5d10
80
6p1
81
6p2
82
6p3
83
6p4
84
6p5
85
6p6
86
7s1
87
7s2
88
Ac-Lr
6d2
104
6d3
105
6d4
106
6d5
107
6d6
108
6d7
109
110
111
Developed by Prof. R. T. Boeré (updated January, 1999)
4f05d1
57
4f15d1
58
4f3
59
4f4
60
4f5
61
4f6
62
4f7
63
4f75d1
64
4f9
65
4f10
66
4f11
67
4f12
68
4f13
69
4f14
70
5d1
71
5f06d1
89
5f06d2
90
5f26d1
91
5f36d1
92
5f46d1
93
5f6
94
5f7
95
5f76d1
96
5f9
97
5f10
98
5f11
99
5f12
100
5f13
101
5f14
102
6d1
103
Exception to Afbau order result of: Full shell stabilityHalf Shell stability
Stability of higher spin state
Electron configurations of ions
Electron configurations of ions can be determined from that of the neutral atom, i.e. electron configurations predict ions
Oxide forming from oxygen:
Same electron configuration as neon
This rationalizes the kinds of stable ions that are formed for certain elements
O = 1s22s22p4 O2- = 1s22s22p6 Ne = 1s22s22p6
Mg = 1s22s22p63s2 Mg2+ = 1s22s22p6
Magnesium cation from magnesium:
Ne = 1s22s22p6
Same electron configuration as neon
Electron configurations of ionsThus, cation electron configuration is obtained by removing electrons in the reverse Aufbau sequence
Anion electron configurations are obtained by adding electrons in the usual Aufbau sequence
Ions try to achieve:
(1) the closest noble gas configuration(2) a pseudo noble gas configuration (closed d or f subshell)
(3) a noble gas configuration for everything except d or f electrons
Cations always have their electron configurations in the sequence of the H.
1. Li+
2. P3-
3. Ga3+
2 11 2s s 21 [ ]s He2 2 6 2 31 2 2 3 3s s p s p 2 2 6 2 61 2 2 3 3 [ ]s s p s p Ar2 2 6 2 6 2 10 11 2 2 3 3 4 3 4s s p s p s d p 2 2 6 2 3 101 2 2 3 3 3s s p s p d
4. Sn2+
5. Sn4+
2 10 2[ ]5 4 5Kr s d p 2 10[ ]5 4Kr s d2 10 2[ ]5 4 5Kr s d p 10[ ]4Kr d
a) O2- c) Cl+ e) Pb4+
b) Mg6- d) Ca+ f) Ga3+
Which of the following ions are likely to form? For those which are not what ion would you expect to form from that element?
Exercise
O2- =1s22s22p6 O = 1s22s22p4 = Ne
Mg = 1s22s22p6 3s2
8
12 Mg6- = 1s22s22p6 3s2 3p6 = Ar
Cl = 1s22s22p4 17 Cl1+ = 1s22s22p3
20 Ca = 1s22s22p6 3s23p6 4s2 Ca1+ = 1s22s22p6 3s23p6 4s1
Pb = [Xe]6s24f145d106p2 Pb4+ = [Xe]4f145d1082
Ga = [Ar]4s23d104p131 Ga3+ = [Ar]3d10
a)
b)
c)
d)
e)
f )
10
18
16
19
78
28
Order of Energy Levels in Ions“Aufbau” energy levels: s below d
“Aufbau” energy levelsIn anion
Energy levels in cations
Energy levels are lowered due to weakened e-n interactions
Energy levels are increased due to enhanced e-n interactions
Subtleties of the shell structure of the atom
Why does the 4s level in neutral atoms lie below the 3d?
The s orbital has a small fraction of its probability density close to the nucleus.
3d orbitals do not have such inner regions, as they only have planar nodes
Hence an s electron from a higher shell will sometimes occur at lower energy than a d electron in a lower shell
Effective nuclear charge
From Li to Ne, nuclear charge increases from 3 to 10
The charge that a 2s or 2p electron feels is different due to the shielding from the electrons in the 1s orbital
2 s orbital penetrate into the 1s orbital and therefore are shielded less on average than d orbitals
Note: Shielding effect increases as the number of e’s increase.
This is the result of additional shielding from the 2 s and 2 p e’s
-1.77-1.42-2.78-3.15
-3.51-3.87
Note: As Z* increases orbital shrink towards nucleus as e’s are held more tightly dues to stronger electronic interactions.
Effect on atomic size
• Consider the size change from F to I
Decrease strongly
Increase significantly
Increase gently
Consider the change in size of the atoms from Li to F
Consider the size change from Li to Rb
Consider the size change from F to I
Size of Atoms and Ions
Atomic radius decreases along the period, and increases down the group
The radius of an anion is larger than its neutral atom.
Removing the electron decreases shielding without changing the charge of the nucleus.
Valence electrons of a cations are in a lower energy shell than in the neutral atom, decreasing the ionic radius.
Adding the extra electron increases shielding without changing the charge of the nucleus.
The radius of a cation is smaller than its neutral atom
i.e Z* is smaller.
i.e., Z* is larger.
Sizes of monatomic ions
Anions are larger than cations
This is always true across a period of the table
Ions in each group of the table get larger in size down the group
Isolectronic ions decrease in size across the period, as Z* increases dramatically. Ex) N3- to F- Na+ to Al3+
Trends in ionization energy
1 ( ) ( ) ( ) for g g gIE E E E e
Increases as Z* increases across each period
Decreases down a group since size increases
The energy that must be absorbed in order to remove a valence electron from a neutral atom in the gas phase
Z* and its effect on size and IE
3+
e-
e-
e-
9+
e-
e-
e-
e-
e-
e-e-
e-
e-
2 11 2s s 2 2 51 2 2s s pLi F
r = 152 pm
EA1 = 520 kJ/mol
r = 71 pm
EA1 = 1681 kJ/mol
Z*= 1.28 Z*= 5.13
>
<
<<
Periodic distribution of IE1 valuesList of the IE1 in kJ/mol for the elements
IE increases across the period
IE decreases down the group
Z* increases
Shielding effect increases i.e Z* decreases
Electron affinity
( ) ( )g gF e F
( ) ( )g gLi e Li
Energy released when an element attracts an extra electron into the lowest-energy unoccupied orbital to form an anion
For large Z* e’s are held closely to the nucleus therefore e-n interactions will be stronger for an additional electron coming in. Compare Li (Z* = 1.28) with F (Z* = 5.13 )
EA increases in magnitude across period
EA decreases in magnitude down the group
Negative since energy is released
Important Concepts from Chapter 8
Diamagnetic vs. paramagnetic vs. ferromagnetic substances
Electron spin
Pauli exclusion principle
Orbital box diagrams
Electron configurations
Aufbau order and its exceptions.
Predicting ions using electron configurations
Core vs. valence electrons
Effective nuclear charge
Periodic trends (atomic radius, ionization energy, electron affinity, ionic radius)
Midterm Test
Tuesday, October 10th (the Tuesday after next) at 7:30pm
Practice is test on website.
Section A Room ????
90 minute test
Chapters: 1, 2, 7, 8
Review Chapter 1
Units
Significant figures
SI System
Dimensional Analysis
Scientific notation
Prefixes and suffixes
Propagation through addition and multiplication
Precision/Acuracy
Chemical & physical properties
Dalton’s atomic theory of matter
Models of the atom – Rutherford's model
Subatomic particles - protons, neutrons, and electrons
Elemental Forms
Periodic table (groups and periods) – Common properties in each group
Elements - names and symbols
Atomic number and mass number - # of n’s, p’s, & e’s
Isotopes - calculating average atomic mass and percent abundance
Avogadro’s number and the mole
Chapter 2
Chapter 7Properties of waves - wavelength, frequency, amplitude, speed
Electromagnetic spectrum - speed of light
Planck’s equation and Planck’s constant
Wave-particle duality for light , electrons, etc.)
Atomic line spectra: Balmer and Rydberg Series
Ground vs. excited states
Heisenberg uncertainty principle
Bohr and Schrödinger models of the atom
Quantum numbers (n, l, ml)
Shells (n), subshells (s,p,d,f)
Shapes and properties of atomic orbitals –#nodes, # lobes
Chapter 8Diamagnetic vs. paramagnetic vs. ferromagnetic substances
Electron spin
Pauli exclusion principle
Orbital box diagrams
Electron configurations
Aufbau order and its exceptions.
Predicting ions using electron configurations
Core vs. valence electrons
Effective nuclear charge
Periodic trends -atomic radius, ionization energy, electron affinity, ionic radius
Sample problem - Chapter 1
90) The smallest repeating unit of a crystal of common salt is a cube with an edge of 0.563 nm.
a) What is the volume in nm3? In cm3?
V = l*w*h = (0.563 nm) (0.563 nm) (0.563 nm)
V = 0.178 nm3 [(102 cm/m)/(109 nm/m)]3
b) The density of NaCl is 2.17 g/cm3. What is the mass of a single cube?
Density =mass/Volume
Mass = Density*Volume = (2.17 g/cm3)*(1.78*10-22 cm3)
= 3.86*10-22 g
= 0.178 nm3
=1.78*10-22 cm3
Sample problem - Chapter 1
c) The cube contains four NaCl molecules. What is the mass of a single NaCl molecule?
Mass NaCl = Mass cube/(# molecules)
= (3.86*10-22 g)/(4 molecules)
= 9.66*10-23 g/molecule
Molecular Mass NaCl = (9.66*10-23 g)*(6.022*10-23g/mol)
= 58.2 g/mol
(58.4 g/mol)
= (9.66*10-23 g)/(1.661*10-24 g/u)
= 58.2 u
Sample problem - Chapter 2 64) When a sample of Phosphorous burns in air P4O10 is formed. One
experiment showed that 0.744 g of P formed 1.704 g of P4O10.
If the atomic mass of O is assumed to be 16.000 u, compute the atomic mass of P.
To make 1 molecule of P4O10 it requires 4 P’s be combined with 10 O’s
# P4O10 molecules made = # P consumed/4
= (mass P/atomic mass P)/4
# P4O10 molecules made = mass P4O10/Molecular mass P4O10
mass P/(4*AM(P)) = mass P4O10/(4*AM(P)+10*AM(O))
= mass P4O10/(4*AM(P)+10*AM(O))
mass P/ mass P4O10 = 4*AM(P)/(4*AM(P)+10*AM(O))
Let x = AM(P)
mass P/ mass P4O10 = 4x/(4x+10*AM(O))
(0.744 g) /( 1.704 g) = 4x/(4x+10*(16.000 u))
0.437 = 4x/(4x+(160.00 u))
1.75x+(70.0 u)= 4x
70.0 u = 2.25x
x = 31.1 u
Recall that AM(O) =16.000 u
mass P/ mass P4O10 = 4*AM(P)/(4*AM(P)+10*AM(O))
Sample problem - Chapter 2
= Atomic Mass of P
Sample problem - Chapter 3
a) In what group and period is Technetium found
99Tc is a transition metal at atomic number 43.
Period 5 Group 7
b) The valence shell of Tc consists of 5s and 4d. List the quantum numbers for electrons in these orbitals.
5s n =5 l = 0 m = 0
4d n =4 l = 2 m = -2,-1,0,1,2
76) The element Technetium is radioactive has to by synthesized and then prepared as NaTcO4. It is used in imaging of brain and thyroid etc.
c) Tc emits a -ray with energy of 0.141 MeV. What is the wavelength and frequency of the -ray photon?
Note: 1 MeV = 106 eV where 1 eV = 9.6485*104 J/mol
Recall that E = h = E/h
= 3.41*1019 1/s (= Hz)
Recall that C =
= (2.26*10-14 J/photon)/(6.62618*10-34 Js)
C=(2.998 *108 m/s)/(3.41*1019 1/s)
= 8.79*10-12 m = 8.79 pm
E = (0.141*106 eV)*(9.6485*104 (J/mol)/eV)/(6.022*1023 photons/mol)
= 2.26*10-14 J/photon
Sample problems - Chapter 8
41) Which of the following is not an allowed set of quantum numbers
n l m ms
a) 2 0 0 -1
e) 1 1 0 +1/2
c) 2 1 -1 -1/2
b) 3 4 +2 -1/2
f) 5 3 -1 +1/2
g) 2 -1 1 +1/2
d) 2 1 2 +1/2
48) Answer the following questions about elements with electron configurations A = [Ar]4s2 and B = [Ar]3d104s24p5
a) Is A a metal, metalloid, or nonmetal? Metal = gr. 2
b) Is B a metal, metalloid, or nonmetal? Nonmetal = gr. 17
c) Which has a larger ionization energy? B
d) Which has the larger atomic radius? A
e) Which has the largest electron affinity? B
f) What charge does A ionize to? +2
g) What charge does B ionize to? -1
