2. EElleeccttrroonniicc SSttrruuccttuurree
Our goal:
•Understand why some substances behave as they do.
• For example: Why are K and Na reactive metals? Why do
H and Cl combine to make HCl? Why are some
compounds molecular rather than ionic?
Atom interact through their outer parts, their electrons.
The arrangement of electrons in atoms are referred to as their
electronic structure.
Electron structure relates to:
•Number of electrons an atom possess.
•Where they are located.
•What energies they possess.
3. TThhee WWaavvee NNaattuurree ooff LLiigghhtt
Study of light emitted or absorbed by substances has lead to the
understanding of the electronic structure of atoms.
Light made up of electromagnetic (E.M) radiation
Characteristics of light:
• All waves have a characteristic wavelength, l, and amplitude, A.
• The frequency, n, of a wave is the number of cycles which pass a
point in one second. Measured in hertz , 1 hertz = 1
cycle/second
• The speed of a wave, v, is given by its frequency multiplied by
its wavelength: λ α (1/ ν) λ ν = constant (c)
c = λ ν c : speed of light = 3 x 108 m/s
5. EElleeccttrroommaaggnneettiicc RRaaddiiaattiioonn
• Modern atomic theory arose out of studies of the
interaction of radiation with matter.
• Electromagnetic (E.M.) radiation moves through a
vacuum with a speed of 2.99792458 ´ 108 m/s.
• There are many kind of E.M. radiation with different
wavelengths and frequencies shown in the following
figure.
• Visible radiation is the only part our eye can detect. It has
wavelengths between 400 nm (violet) and 750 nm (red).
7. Example(1):
What is the wavelength of light with a frequency 5.89 x 105
Hz.
λ = c
ν = 3 x 108 m/s
5.89 x 105 s-1
= 509 m (Radio wave)
Example (2):
What is the frequency of blue light with a wavelength of
484 nm?
ν = c
λ = 3 x 108 m/s
484 x 10-9 m = 6.2 x 1014 s-1 or Hz
8. TThhee NNaattuurree ooff MMaatttteerr
In 1990 Matter and energy were seen as different from each
other in fundamental ways
Matter:
consist of particles
Particles have a mass
Its position in space can be specified.
Energy:
could come in waves, with any frequency.
Massless and delocalized.
Their position in space could not be specified.
It was assumed that there was no intermingling of matter
and light
9. At the beginning of 20 century, certain experimental results
suggested that this picture was incorrect
The first important advance came from Max Planck, he found that
the cooling of hot objects couldn’t be explained by viewing
energy as a wave.
Plank found that the results could not be explain in term of the
physics of his day (matter absorb or emit any quantity of energy) .
Plank account for these observation by postulating that:
The energy can be gained or lost only in whole-number multiple
of the quantity hhνν
ΔΔ EE == nnhhνν wwhheerree nn iiss aann iinntteeggeerr ((11,, 22,,
33……))..
hh :: iiss PPllaanncckk’’ss ccoonnssttaanntt == 66..662266 xx 1100--3344 JJ ss
IItt sseeeemmeedd cclleeaarr tthhaatt eenneerrggyy iiss qquuaannttiizzeedd aanndd ccaann ooccccuurr iinn ddiissccrreettee
uunniitt ooff ssiizzee hhνν ,, tthheessee ppaacckkeettss ooff eenneerrggyy ((hhνν)) aarree ccaalllleedd qquuaannttuumm..
A system can transfer energy only in whole quanta.
Thus energy seems to have particulate properties
10. The next development came when Einstein proposed that E.M.
radiation is itself quantized.
He suggested that E.M. radiation can be viewed as a stream of
“particles” called photons
10
Each photon has energy Ephoton = hν = hc/λ
Combine this with E = mc2 (Einstein equation)
hc/λ
c2
m = E /c2 m =
you get the apparent mass of a photon
m = h / (λc)
Does a photon really have a mass? The answer appears to be yes.
However, it is clear that photons do not have mass in the classical
sense. A photon has mass only in relativistic sense – it has no rest
mass.
11. 11
We can summarize the important conclusions from the work of
Plank and Einstien as follows:
Energy is quantized, it can occur only in discrete unit called
quanta.
E.M. radiation, which was previously thought to exhibit only
wave properties, seems to show certain characteristics of
particulate matter as well.
This phenomenon is referred to as the dual nature of light
Is the opposite is true? That is, does matter exhibit wave
properties.
de Brolie supplied the answer to this question.
m = h /(λc) for a particle with velocity v
m = h /(λv) λ = h/(mv)
This equation, called de Brolie equation, allow us to calculate the
wave lenghth of particle
12. 12
Example:
The laser light of a CD is 7.80 x 102 m. calculate
A) What is the frequency of this light?
B) What is the energy of a photon of this light?
C)What is the apparent mass of a photon of this light?
A) ν = c/λ ν = 3 x 108(m/s)/ 7.80x102 m = 3.85 x 105s-1
B) Ephoton = hν
Ephoton = 6.626 x 10-34 J s x 3.85 x 105s-1
Ephoton = 2.55 x 10-28 J
C) m = h / (λc)
m = 6.626 x 10-34 J s /7.80x102 m x 3x108(m/s)
m = 2.83x10-45 Js2/m2 = 2.83x10-45(kg m2/s2) s2/m2
m = 2.83x10-45kg
13. 13
Example:
What is the wavelength of an electron with a mass of
9.11 x 10-31 kg traveling at 1.0 x 107 m/s?
m = h / (λc)
λ = h /mc
λ = 6.626 x 10-34 J s
9.11 x 10-31 kg x 1.0 x 107 m/s = 7.2 x 10-11 m
14. 14
TThhee aattoommiicc ssppeeccttrruumm ooff hhyyddrrooggeenn
Another important experiment was the study of the emission
light by excited H-atoms. When hydrogen gas receives high-energy
spark, H2 molecules absorb energy, some of H-H
bonds are broken. The resulting H-atoms are excited; that is
they contain excess energy which they release by emitting
light of various wavelength to produce what is called the
emission spectrum of H-atoms.
To understand the significance of H-emission spectrum, we
must describe the continuous spectrum that results when
white light is passed through a prism,
15. 15
continuous spectrum
Contain all the wavelength of visible (white) light.
All the colors are possible.
Like the rainbow.
When H-emission spectrum in visible region is passed
through prism, only a few lines can be seen, each
correspond to a discrete wavelength. The H-emission
spectrum is called line spectrum.
18. 18
What is the significance of line spectrum of hydrogen?
It indicate only certain energies are allowed for the
hydrogen atom.
Energy of electron H-is quantized
Only certain energies are possible.
Use DE = hn = hc / l
19. 19
BBoohhrr MMooddeell
He developed the quantum model of the hydrogen atom.
He proposed that the atom was like a solar system, the
electron in H-atom move around the nucleus only in
certain allowed circular orbit
The electrons were attracted to the nucleus because of
opposite charges.
Didn’t fall in to the nucleus because it was moving around
20. 20
He didn’t know why but only certain energies were
allowed.
He called these allowed energies: energy levels.
Putting energy into the atom moved the electron away
from the nucleus from ground state to excited state.
When it returns to ground state it gives off light of a
certain energy
The energy levels for H-atom are shown in the following
figure.
22. 22
TThhee BBoohhrr MMooddeell
for each energy level the energy is:
E = -2.178 x 10-18 J (Z2 / n2 )
n: is the energy level
Z: is the nuclear charge, which is +1 for hydrogen.
n = 1 is called the ground state
when the electron is removed from the atom, n = ¥
When the electron moves from one energy level to
another.
ΔE = Efinal - Einitial
ΔE = -2.178 x 10-18 J Z2 (1/ nf
2 - 1/ ni
2)
23. 23
Example:
Calculate the energy need to move an electron from its
ground state to the third energy level.
ΔE = -2.178 x 10-18 J Z2 (1/ n2 - 1/ n2)
f
i
ΔE = -2.178 x 10-18 J (+1)2 (1/9 – 1/1)
ΔE = +1.936 x 10-18 J (+ mean energy absorbed)
Example:
Calculate the energy released when an electron moves from
n= 4 to n=2 in a hydrogen atom.
ΔE = -2.178 x 10-18 J (+1)2 (1/4 – 1/16)
ΔE = - 5.2125 x 10-19 J
24. 24
Example:
Calculate the energy required to remove the electron from
hydrogen atom in its ground state.
ΔE = -2.178 x 10-18 J (+1)2 (1/¥ – 1/1)
ΔE = 2.178 x 10-18 J
25. 25
Bohr model:
Only works for hydrogen atoms and other monoelectronic
species.
electrons don’t move in circles
the quantization of energy is right, but not because they
are circling like planets.
The negative sign of the energy level :
increase the energy of the electron when you make it
further to the nucleus.
the maximum energy an electron can have is zero, at an
infinite distance (n = ¥ ).
26. 26
The Quantum Mechanical MMooddeell ooff tthhee aattoomm
A totally new approach was needed.
Three physicists were at the forefront of this effort:
Heisenberg, de Broglie, and Schrödinger. The approach
they developed known as wave mechanics or quantum
mechanics
De Broglie said matter could be like a wave.
Schrödinger proposed an equation that contains both
wave and particle terms.
Much math, but what is important are the solutions.
Solving the equation leads to wave functions.
27. 27
• The wave function is a F(x, y, z) Actually F(r,θ,φ)
• Solutions to the equation are called orbitals (not Bohr orbits).
• Each solution is tied to a certain energy level.
• The wave function gives the shape of the electronic
orbital.
• The square of the wave function, gives the probability of
finding the electron, that is, gives the electron density for
the atom.
• There is a limit to what we can know from Schrödinger
equation.
• We can’t know how the electron is moving or how it gets
from one energy level to another.
29. 29
QQuuaannttuumm NNuummbbeerrss
There are many solutions to Schrödinger’s equation
Each solution can be described with quantum numbers that
describe some aspect of the solution.
Principal quantum number (n):
has an integral value: 1, 2, 3, ……, it is related to the size
and energy of an orbital.
As (n) increase:
orbital become larger, electron spends more time farther
from the nucleus
higher energy, because the electron is less tightly bound
to nucleus, energy is less negative.
30. 30
Angular momentum quantum number (ℓ):
has integer values from 0 to n-1 for each value of n
It is related to the shape of the orbital (as shown in the
following figures
the value of (ℓ) for a particular orbital is commonly
assigned a letter: ℓ = 0 is called s , ℓ = 1 is called p
ℓ =2 is called d , ℓ =3 is called ƒ , ℓ =4 is called g
Magnetic quantum number (m ℓ):
– integer values between - ℓ and + ℓ including zero.
– The value of mℓ is related to the orientation of the
orbital in space relative to the other orbitals in the
atom.
32. 32
P orbitals
There are three p-orbitals px, py, and pz.
The three p-orbitals lie along the x-, y- and z- axes of a
Cartesian system.
The letters correspond to allowed values of mℓ of -1, 0,an +1.
Electron-distribution of a 2p orbital.
35. 35
Electron spin quantum number (m s):
the electron has a magnetic moment with two possible
when the atom placed in an external magnetic field
– Can have 2 values , either +1/2 or -1/2
36. 36
For our purpose, the main significance of electron spin is
connected with the postulate of Pauli: in a given atom no
two electrons can have the same set of four quantum
numbers (n , ℓ, m ℓ, and ms ), this is called Pauli exclusion
principle. Since electrons in the same orbital have the same
value of n , ℓ, m ℓ , they must have different values of ms .
Then, since only two value of ms are allowed, an orbital can
hold only two electrons, and they must have opposite spin.
37. 37
Quantum number for the first four level of orbitals in H-atom:
n ℓ Orbital
designation
m ℓ No. of
orbitals
1 0 1s 0 1
2 0 2s 0 1
1 2p 1+ ,0 ,1- 3
3 0 3s 0 1
1 3p 1+ ,0 ,1- 3
2 3d 2+,1+ ,0 ,1- ,2- 5
4 0 4s 0 1
1 4p 1+ ,0 ,1- 3
2 4d 2+,1+ ,0 ,1- ,2- 5
3 4f ,2+ ,1+ ,0 ,1- ,2-,3-
3+
7
38. Example:
For n = 4, what are the possible values of ℓ.
38
ℓ = 0→ n -1 , so ℓ = 0 → 4-1
ℓ = 0, 1, 2, 3
s, p, d, f
For ℓ = 2. What are the possible values of mℓ
mℓ = - ℓ → +ℓ mℓ = -2 → +2
mℓ = -2, -1, 0, +1, +2
How many possible values for ℓ and mℓ are there when
n = 3
ℓ = 3-1 = 2 ℓ = 0, 1, 2
for ℓ = 0 mℓ = 0 ,, for ℓ = 1 mℓ = -1, 0, +1
for ℓ = 2 mℓ = -2, -1, 0, +1, +2
39. 39
7s
6s
5s
Increasing energy 1s
4s
3s
2s
7p 6d
6p
5p
4p
3p
2p
5d
4d
3d
5f
4f
Orbitals and Their Energies
40. 40
TThhee PPeerriiooddiicc TTaabbllee
Developed independently by German Julius Lothar
Meyer and Russian Dmitri Mendeleev (1870”s)
Didn’t know much about atom.
Put atoms in columns by similar properties.
Predicted properties of missing elements.
41. 41
Aufbau PPrriinncciippllee aanndd tthhee ppeerriiooddiicc ttaabbllee
Our main assumption is that the atoms have the same type
of orbitals as have been described from the hydrogen atom.
As protons are added one by one to the nucleus to build up
the elements, electrons are similarly added to these H-like
orbitals. This is called aufbau principle
H has one electron, occupy the 1s orbital
The configuration for H can be represent as:
H: 1s1
1s
Quantum no. for the electron is:
n=1, ℓ = 0, mℓ =0, ms =+1/2
Helium has two electron
42. 42
Helium has two electrons
He: 1s2
Lithium has three electrons
Li: 1s2 2s1 2p
Be: 1s2 2s2 2p
B: 1s2 2s2 2p1
Quantum no. for the first electron
is: n=1, ℓ = 0, mℓ =0, ms =+1/2
Quantum no. for the second electron
is: n=1, ℓ = 0, mℓ =0, ms = -1/2
1s
1s 2s 2p
1s 2s 2p
1s 2s 2p
43. 43
C: 1s2 2s2 2p2
Two electrons occupy 2p orbital, since there are three 2p
orbitals with the same energy, the mutually repulsive
electrons will occupy separate 2p orbitals
Hund’s rule: the lowest energy configuration for an atom
is the one having the maximum number of unpaired
electrons.
N: 1s2 2s2 2p3
O: 1s2 2s2 2p4
1s 2s 2p
1s 2s 2p
1s 2s 2p
44. 44
F: 1s2 2s2 2p5
Ne: 1s2 2s2 2p6
1s 2s 2p
With neon, the orbital with n =1 and n = 2 are now completely
filled.
Na: 1s2 2s2 2p63s1 can be abbreviate as Na : [Ne] 3s1
Write the symbol of the noble gas before the element
Then the rest of the electrons.
Mg: [Ne] 3s2
Al: [Ne] 3s2 3p1
1s 2s 2p
Ne
45. 45
At this point it is useful to introduce the following concepts:
Valence electrons- the electrons in the outermost principle
quantum level of an atom (not d).
Core electrons- the inner electrons
Hund’s Rule- The lowest energy configuration for an atom
is the one have the maximum number of unpaired electrons
in the orbital.
Example:
element valence electrons core electrons
O 6 2
N 5 2
Ne 8 2
Mg 2 10
46. 46
K: 1s2 2s2 2p63s13p64s1 or [Ar] 4s1 (valence electrons = 1)
Ca: 1s2 2s2 2p63s13p64s2 or [Ar] 4s2 (valence electrons = 1)
Sc: [Ar] 4s23d1 Ti: [Ar] 4s23d2 V: [Ar] 4s23d3
Valence electrons: 3 4 5
The expected configuration for chromium is:
Cr: [Ar] 4s23d4 however, the observed configuration is:
Cr: [Ar] 4s13d5 both 4s and 3d half-filled
Also the expected configuration for Cu is: Cu : [Ar] 4s23d9
The observed configuration is: Cu : [Ar] 4s13d10
4s is half-filled, 3d is filled
49. 49
Electron Configurations aanndd tthhee PPeerriiooddiicc TTaabbllee
The periodic table can be used as a guide for electron
configurations.
• the groups label (1A-8A) called the main-groups or
representative elements (no. of the group = valence electrons
• the groups label (1B-8B) called the Transion elements
• the (n+1)s orbital is always fill before nd orbitals.
• The period number is the value of n.
• Groups 1A and 2A (1 2) have the s-orbital filled.
• Groups 3A - 8A (13 - 18) have the p-orbital filled.
• Groups 3B - 2B (3 - 12) have the d-orbital filled.
• The lanthanides and actinides have the f-orbital filled.
50. 50
Elements in the same column have the same electron
configuration.
Put in columns because of similar properties.
Similar properties because of electron configuration.
Noble gases have filled energy levels.
Transition metals are filling the d orbitals
51. 51
PPeerriiooddiicc TTrreennddss iinn aattoommiicc pprrooppeerrttiieess
Ionization energy (I.E.):
Ionization energy the energy required to remove an electron
form a gaseous atom or ion in its ground state.
X(g) → X+
(g) + e
We will consider the energy required to remove several
electrons from Al in the ground state.
Al(g) → Al+
(g) + e I1 = 580 kJ/mol
Al+
(g) → Al+2
(g) + e I2 = 1815 kJ/mol
Al+2
(g) → Al+3
(g) + e I3 = 2740 kJ/mol
(g) → Al+4
(g) + e I4 = 11600 kJ/mol
Al+3
52. 52
Several points can be illustrated from these results:
Highest energy electron (the one bound least tightly) that is removed first.
The first ionization energy I1 is the energy required to remove the first
electron (highest-energy electron)
The value of I1 is considerably smaller than the value of I2 (second
ionization energy). The primary factor is simply charge, electron is removed
from +1 ion (Al+) .
The increase in positive charge bind the electron more firmly, and the
ionization energy increases. The same trend shows up in I3 and I4, where the
electron is removed from Al+2 and Al+3 ions respectively.
The increase in I.E. from I1 to I2 occur also because the first electron is
removed from 3p orbital that is higher in energy than 3s orbital from which
the second electron is removed.
The largest jump in I.E. by far occur in going from the I3 and I4 because
Al+3 has the configuration (1s2 2s2 2p6), the core electrons are bound much
more tightly than valence electrons.
53. 53
In the following table, the ionization energies for all the period 3
are given. Note the large jump in energy in each case in going
from removal of valence electrons to removal of core electrons
Symbol I1 I2 I3
H He
Li
Be
BC
NO
F
Ne
1312
2731
520
900
800
1086
1402
1314
1681
2080
5247
7297
1757
2430
2352
2857
3391
3375
3963
11810
14840
3569
4619
4577
5301
6045
54. 54
He
First Ionization energy Atomic number
H
C
Be
Li
N
B
The values of the first I.E. for the elements
are shown in the following figure:
55. 55
Note that:
As you go down a group first I.E. decreases because of electron
being removed are, on average, farther from the nucleus. As n
increases, the size of the orbital increases, and the electron is easier to
remove.
As you go across a period from left to right, first I.E. increases
because - Same shielding (same principle
quantum level) .
- Increasing nuclear charge (electrons are
strongly bound)
There are some trends in I.E. in going across period. For example,
trends occur from Be to B and from N to O. it can be explain in term
of electron repulsion. Half-filled and filled orbitals are harder to
remove electrons from
56. 56
The ionization energies for the representative elements are
summarized in the following figure
57. 57
Atomic Size or radius:
The size of the orbital cannot be specified exactly (The electron cloud
doesn’t have a definite edge), neither can the size of an atom.
We can make some arbitrary choice to obtain values for atomic radii.
These values can be obtain by measuring the distance between atoms
in chemical compounds.
For example, in Br2 molecule, the distance between the two nuclei is
228 pm. The Br atomic radius is assumed to be half this distance, or
114 pm, as shown in the following figure.
60. 60
AAss wwee ggoo ddoowwnn aa
ggrroouupp eeaacchh aattoomm
hhaass aannootthheerr
eenneerrggyy lleevveell..
SSoo tthhee aattoommss ggeett
bbiiggggeerr
H
Li
Na
K
Rb
61. 61
As you ggoo aaccrroossss aa ppeerriioodd tthhee rraaddiiuuss
ggeettss ssmmaalllleerr..
SSaammee eenneerrggyy lleevveell
MMoorree nnuucclleeaarr cchhaarrggee
OOuutteerrmmoosstt eelleeccttrroonnss aarree cclloosseerr
Na Mg Al Si P S Cl Ar
63. 63
Electron Affinity:
The energy change associated with adding an electron to a
gaseous atom.
X(g) + e X-
(g)
High electron affinity gives you energy-
exothermic
More negative
64. 64
In general, electron
affinity becomes more
exothermic as you go
from left to right across
a row.
Increase (more - ) from
left to right (greater
nuclear charge).
Decrease as we go down
a group (More
shielding)
