Download Chem 110 2014 (Chapter 6)

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1–1
Elements
Hydrogen
Nitrogen-7
Carbon-6
Oxygen-8
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1–2
Hydrogen
Carbon-6
Hydrogen
Hydrogen
Hydrogen
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1–3
Chapter 6. Electronic Structure of Atoms
•
Chemical behavior of atoms is primarily
determined by arrangement of electrons
outside atomic nucleus.
•
Introduce a model of the atom
–
–
help us understand this arrangement,
to understand why elements exhibit their
characteristic kinds of chemical behavior
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1–4
Nature of Light
• Most of our understanding of electronic structure
comes from analysis of light emitted or absorbed
by substances
• Need to look at nature of visible light & other
forms of radiant energy.
– Includes radio waves, X-rays, IR, microwaves,
visible light, etc. & are different kinds of
electromagnetic radiation
– exhibits wavelength-like behavior and travels
through space at speed of light in vacuum.
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1–5
Waves
• To understand the electronic structure of atoms, one
must understand the nature of electromagnetic
radiation.
• The distance between corresponding points on
adjacent waves is the wavelength ().
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1–6
Waves
• The number of waves
passing a given point per
unit of time is the
frequency ().
• For waves travelling at
the same velocity, the
longer the wavelength, the
smaller the frequency.
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1–7
Electromagnetic Spectrum
Differences are due to different wavelengths
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1–8
Electromagnetic Radiation
All waves are characterized by:
1. Wavelength (): distance between 2 peaks in
wave.
2. Frequency (v): number of waves per second that
pass given point in space (s-1 or hertz, Hz)
3. Speed (c): speed of light is 2.9979 x 108 m/s
(same rate for all electromagnetic radiation in
vacuum).
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1–9
FREQUENCY AND WAVELENGTH RELATED
The wavelength, frequency and speed of
electromagnetic radiation are all related by:
 = c

displacement
direction
of
This means we can…..
propagat
Calculate the wavelength
ion
From frequency and velocity.
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1–10
EXAMPLES
Calculating the wavelength of electromagnetic radiation
from its frequency and velocity.
The FOX broadcasts at 99.3 MHz.
What is the wavelength of the radiation?
99.3 x 106 Hz
The frequency is
= 99.3 x106 s-1
Radio waves are a form of electromagnetic radiation
c  
c = 2.998 x 108 m/s
1
2.998  10 ms
 
 3.019m
6 1

99.3  10 s
c
8
We can also…….
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1–11
EXAMPLES
Calculate the frequency of electromagnetic radiation
from its wavelength and velocity.
The wavelength of the yellow light from a sodium lamp
is 589 nm.
What is the frequency of the radiation?
c

9
1  10 m
7
  589nm 
 5.89  10 m
1nm
1
2.998  10 ms
14 1
 
 5.09  10 s
7

5.89  10 m
c
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8
1–12
EXCITED GROUP 1 ELEMENTS
Li
Na
K
What causes the colour?
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1–13
The Particle Nature of Light
•
The wave model of light explains many
aspects of its behavior but not all
•
It could not explain the emission of light from
electronically excited gas atoms
•
This is known as emission or atomic spectra
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1–14
Quantised Energy and Photons
• The wave nature of light does not
explain how an object can glow when
its temperature increases.
• Max Planck explained it by assuming
that energy comes in packets called
quanta.
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1–15
Quantum Theory
• Max Planck (1900): energy can be released or
absorbed by atoms only in discrete amounts
– Quantum (fixed amount): smallest quantity of
energy that can be emitted or absorbed
E = nhv
E = energy, n = + integer, v = frequency,
h (Planck’s constant) = 6.626 x 10-34 Js
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1–16
Quantised Energy and Photons
• Einstein used this assumption to explain the
photoelectric effect.
• He concluded that energy is proportional to
frequency:
E = h
where h is Planck’s constant, 6.6310−34 Js.
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1–17
Quantised Energy and Photons
• Therefore, if one knows
the wavelength of light,
one can calculate the
energy in one photon,
or packet, of that light:
c = 
E = h
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1–18
Photoelectric Effect
E = hv = hc/
Example: Calculate the energy of one photon of
yellow light whose wavelength is 589 nm.
Which has more energy: X-rays or microwaves?
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1–19
Quantised Energy and Photons
Another mystery involved
the emission spectra
observed from energy
emitted by atoms and
molecules.
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1–20
Quantised Energy and Photons
• One does not observe a
continuous spectrum as
one gets from a white
light source.
• Only a line spectrum of
discrete wavelengths is
observed.
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1–21
The line
spectra of
several
elements
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1–22
Rydberg Equation
Line spectrum of excited hydrogen atoms
Rydberg equation
1

=
RH
1
2
ni
1
nf2
R is the Rydberg constant = 1.096776 x 107 m-1
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1–23
Emission Spectra
• Atoms give off light when heated or otherwise
excited energetically
• Light given off by energetically excited atom is
not continuous distribution of ’s
• Continuous spectrum: Contains all the
wavelengths (& all energies) of light
• Line (discrete) spectrum: Contains only some
of the wavelengths of light.
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1–24
Nuclear Model of the Atom
• Rutherford showed:
– Atomic nucleus is composed of protons (+) &
neutrons (0).
– Nucleus is very small compared to size of entire
atom.
• Questions left unanswered:
– How are electrons arranged & how do they move?
– Electrons are moving charged particles
– Moving charged particles give off energy
– Atom should constantly be giving off energy
– Electrons should crash into nucleus and atom
collapse!!
This led to Neil Bohr Model of the Atom
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1–25
Bohr Model
Explained spectra of hydrogen:
•
H atom has only certain allowable energy levels
(stationary states)…fixed circular orbits.
•
Atom does not radiate energy while in one of
its stationary states
•
Atom changes stationary states by absorbing
or emitting a photon whose energy equals
difference in energy between two states (E =
hv)
Spectrum is not continuous…energy has only
certain states!
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1–26
The Bohr explanation of the series of spectral lines
When excited hydrogen atoms return to lower energy
states, they emit photons of certain energies & certain colors.
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1–27
Bohr’s Model
• Atom’s energy has only certain levels or states
• Atoms have minimum energy: ground state
• Higher energy levels: excited states
– Farther from nucleus, higher its energy
– Putting e- in excited state requires addition of
energy
– Bringing e- back to ground state releases energy
– Only specific frequencies can be
absorbed/emitted
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1–28
Quantised Energy and Photons
• Niels Bohr adopted Planck’s assumption and
explained these phenomena in this way:
1. Electrons in an atom can only occupy
certain orbits (corresponding to certain
energies).
2. Electrons in permitted orbits have
specific, “allowed” energies; these
energies will not be radiated from the
atom.
3. Energy is only absorbed or emitted in
such a way as to move an electron from
one “allowed” energy state to another;
the energy is defined by:
E = h
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1–29
Quantised Energy and Photons
The energy absorbed or emitted from the
process of electron promotion or
demotion can be calculated by the
equation:
1
1
E = −RH n 2
ni2
f
(
)
Note E = hv = hc/
where RH is the Rydberg constant,
2.18  10−18 J, and ni and nf are the initial
and final energy levels of the electron.
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1–30
Emission Spectrum of the Hydrogen Atom
Bohr showed that the energies that the electron in the H
atom can possess is given by the equation:
 RH
En 
n2
where R = Rydberg’s constant (2.18 x 10-18 J or 109 677 cm-1)
n = principal quantum number (1,2,3,…..)
For n = 1, En becomes most negative and this corresponds
to the most stable energy state called the ground state.
n = 2,3,… are called excited states and the
corresponding energies are higher in magnitude.
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1–31
Interaction of a photon and an atom results in the atom
moving to a higher energy state; atom is excited.
State of lowest energy called the ground state.
E2 (excited state)
E
E1 (ground state)
E2 – E1 =
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E12 =
hν
1–32
Absorption Spectroscopy
Atoms selectively absorb photons of light.
Spectra that results are termed line spectra as only
specific energies are absorbed due to selective absorption.
Consider the absorption spectrum of an atom that was
measured to give the following spectrum:
33250
E2 (excited state)
A
E
E1 (ground state)
wavenumber(cm-1)
Let us now calculate the difference in energy in joules.
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1–33
Use the equation:
E = hcv
Note: the unit for wavenumber is cm-1, but c is
given in m s-1
2.998 x 108 m s -1x 100 cm
c =
1s
1m
= 2.998 x 1010 cm s-1
Therefore
E12 = (6.626 x 10-34 J s)(2.998 x 1010 cm s-1)(32 250 cm-1)
= 6.406 x 10-19 J
Now, consider an atom with one electron and four allowed
energy states
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1–34
E4
E3
E2
E1
How many lines do we see in the absorption?
The answer is 3, at E12 , E13 and E14
Emission Spectroscopy
Atom emits a photon of energy.
Note: electron does not have to return directly to
the ground state; instead can return stepwise
down the energy ladder.
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1–35
E4
1
E
2
4
E3
5
E2
3
6
E1
In the above case, the emission spectrum would have 6
lines, as 6 different transitions can occur at 6 different
E34 E23 E24 E12 E13 E14
energies.
Intensity of
Energy
Energy
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1–36
The electron is initially in the excited state, ni ; during
emission it drops to a lower energy state, nf..
The change in energy is given by:
E = Ef - Ei
Therefore,
OR
OR
OR
E =
E = -RH
-RH
-RH
2
nf
ni2
1 nf2
hv =E = RH
1
ni2
1 - 1
ni2 nf2
When a photon is emitted, ni > nf , E < 0 and when energy
is absorbed nf > ni and E > 0.
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1–37
The various series in atomic H emission spectrum
Series
Lyman
Balmer
Paschen
Brackett
Pfund
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nf
1
2
3
4
5
ni
2,3,4,
3,4,5,
4,5,6,
5,6,7,
6,7,8,
….
….
….
….
….
Region
UV
Visible
IR
IR
IR
1–38
Important Points of Bohr’s Model
1. The energies of electrons (energy levels) in an
atom are quantized.
2. Quantum numbers are necessary to describe
certain properties of electrons in an atoms
(such as energy & location).
3. An electron’s energy increases with
increasing distance from the nucleus
4. The discrete energies (lines) in the spectra of
the elements result from quantized electronic
energies.
Neil Bohr won the the Nobel Prize (1922).
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1–39
Problems with the Bohr Model
Only explains hydrogen atom spectrum
– and other 1 electron systems
• Neglects interactions between electrons
– small particle circling the nucleus
• Assumes circular or elliptical orbits for electrons which is not true
Can electrons also exhibit wave-like character?
(and not just a small particle circling about the
nucleus)
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1–40
Wave Behavior of Matter
• Radiation appears to have either a wavelike or a
particle-like (photon) character.
– Louis de Broglie: Can the electron orbiting the
nucleus of a H atom be thought of not as a particle
but rather a wave with a characteristic
wavelength?
– He posited that if light can have material properties,
matter should exhibit wave properties.
– He demonstrated that the relationship between mass
and wavelength is:
h
 = mv
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1–41
The Wave Nature of Matter
– Heisenberg (uncertainty principle): impossible
to know simultaneously both the exact
momentum (mass times speed) of the electron &
its exact location in space.
x * mu ≥
h
4p
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1–42
The Uncertainty Principle
• Heisenberg showed that the more precisely the
momentum of a particle is known, the less
precisely its position is known.
• In many cases, our uncertainty of the
whereabouts of an electron is greater than the
size of the atom itself!
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1–43
Quantum Mechanics
• Erwin Schrödinger developed a
mathematical treatment into which both
the wave and particle nature of matter
could be incorporated.
• It is known as quantum mechanics.
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1–44
Quantum Mechanics
• The wave equation is designated
with a lower case Greek psi ().
• The square of the wave
equation, 2, gives a probability
density map of where an
electron has a certain statistical
likelihood of being at any given
instant in time.
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1–45
Wave Mechanics
• Treats electrons as waves and uses wave equations to
calculate probability densities of finding electron in
particular region in atom
– solutions are wave functions or atomic orbitals
– cannot know precisely location of electron at any
moment but can describe its probability
– for given energy level, we can depict probability
with electron density diagram or electron cloud.
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1–46
Quantum-Mechanical Model of an Atom
• Electron density: describes distribution of electron
in orbital
– High in those regions of orbital where probability
of finding electron is high and low in regions
where probability is low
• H atom (ground state): e- is almost always found
within sphere with certain radius, (0.529 Å)
centered about nucleus
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1–47
Electron probability in the ground-state H atom.
Electron probability density decreases with distance
from nucleus
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1–48
Orbitals and Quantum Numbers
• Each orbital describes a specific distribution of
electron density in space (We called them
“Shells”)
– as given by the orbital’s probability density.
• Each orbital has a characteristic energy &
shape.
• Electrons occupy orbitals characterized by
4 Quantum Numbers (QN)
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1–49
1. Principal Quantum Number (n)
Principal QN (n = 1, 2, 3, . . .): related to size and
distance/energy of orbital
•
•
•
Identifies how much energy electrons in orbital
have
Higher values mean higher energy & further
distance from nucleus
Less tightly bound to nucleus
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1–50
2. Angular Momentum QN (l)
•
Angular Momentum QN (l = 0 to n - 1) relates
to shape of the orbital.
•
Each principal energy level contains one or more
energy sublevels
– there are n sublevels in each principal energy
level
– each type of sublevel has a different shape and
energy
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1–51
Quantum Number, l
Value of l
0
1
2
3
Type of orbital
s
p
d
f
3. Magnetic Quantum Number, ml
• Describes the three-dimensional
orientation of the orbital.
• Values are integers ranging from -l to l:
−l ≤ ml ≤ l.
• Therefore, on any given energy level,
there can be up to 1 s orbital, 3 p orbitals,
5 d orbitals, 7 f orbitals, etc.
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1–53
Magnetic Quantum Number, ml
• Orbitals with the same value of n form a shell.
• Different orbital types within a shell are subshells.
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1–54
Observing the Effect of Electron Spin
When a beam of H atoms is shot into a powerful
magnetic field, the beam was split by the field into two
beams - (opposing directions: half are attracted, half are
repelled)
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1–55
4. Spin Quantum Number, ms
• In the 1920s, it was discovered that two
electrons in the same orbital do not have
exactly the same energy.
• The “spin” of an electron describes its
magnetic field, which affects its energy.
• This led to a fourth quantum number, the
spin quantum number, ms.
• The spin quantum number has only 2
allowed values: +½ and −½.
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1–56
Electron Spin
• Electron Spin QN (ms = +1/2, -1/2): relates to spin
states of electrons.
– Important when more than 1 electron is present
– Two e- in same orbital have opposite spins =
lower energy
– Write 4 QNs for any electron
Pauli Exclusion Principle: no two electrons in
the same atom can have the same four QNs.
Atomic orbital can hold a maximum of 2
electrons and they must have opposing spins.
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1–57
Quantum Numbers
Name
Symbol
Values
Principal
n
1,2,3,4….
Energy &
distance
Angular
Momentum
l
0,1,….,n-1
Shape
l=012 3
s p d f
Magnetic
ml
+l…….-l
Orientation
Spinmagnetic
ms
+1/2, -1/2
Spin
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Meaning
1–58
s Orbitals
• Value of l = 0
• Spherical in shape
• Radius of sphere
increases with
increasing value of n
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1–59
p Orbitals
• Value of l = 1
• Have two lobes with a node between
them- dumbbell.
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1–60
d Orbitals
• Value of l is 2
Figure
5.21
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• Four of the five
orbitals have 4
lobes; the other
resembles a p
orbital with a
doughnut around
the centre
1–61
Atomic Orbitals
Relationship between quantum numbers and atomic orbitals.
n
l
ml
1
0
0
1
1s
2
0
0
1
2s
1
-1,0,1
3
2px, 2py, 2pz
0
0
1
3s
1
-1,0,1
3
3
2
No. of orbitals AO designations
-2,-1,0,1,2
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5
3px,
3py, 3pz
3dxy, 3dyz, 3dzx,
3dx2- y2, 3dz2
1–62
What are n, l, & ml values for the 2p sublevels?
A.
B.
C.
D.
E.
n
1
2
2
2
2
l
1
0
2
1
1
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ml
+1 to -1
0
+2 to -2
+1 to -1
0
1–63
What are n, l, & ml values for the 5f sublevels?
A.
B.
C.
D.
E.
n
5
5
5
5
5
l
0
0
3
2
1
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ml
+1 to -1
0
+3 to -3
+2 to -2
0
1–64
What type of orbital is designated
n = 4, l = 2, ml = +1?
a) 4s
b) 4p
c) 4d
d) 2f
e) 4f
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1–65
What type of orbital is designated
n = 2, l = 0, ml = 0?
a) 2s
b) 2p
c) 2d
d) 2f
e) 1d
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1–66
What is maximum number of orbitals in n = 3?
a) 1
b) 3
c) 4
d) 7
e) 9
Each shell has a total of n2 orbitals.
n=3 : l= 0 : 1 x 3s
l=1 : 3 x 3p
l = 2 : 5 x 3d
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1–67
All of the following sets of quantum numbers are
allowed EXCEPT
a) n = 3, l = 1, ml = -1
b) n = 2, l = 1, ml = 0
c) n = 5, l = 0, ml = -1
d) n = 4, l = 2, ml = +2
e) n = 1, l = 0, ml = 0
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1–68
Which represents impossible combinations of n and
l?
A. 1p
B. 4s
C. 4f
D. 2d
E. 1p & 2d
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1–69
How many electrons can be described by the
following quantum numbers:
n = 3, l = 1, ml =0, ms = -1/2?
a) 0
b) 1
c) 2
d) 3
e) 6
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1–70
Which of the following sets of quantum numbers
is not allowed?
a)
b)
c)
d)
e)
n = 4, l = 0, ml = 0, ms = +1/2
n = 5, l = 3, ml = 2, ms = -1/2
n = 2, l = 1, ml = -1, ms = -1/2
n = 3, l = 2, ml = 0, ms = +1/2
n = 1, l = 0, ml = 0, ms = 0
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1–71
Energies of Orbitals
• For a one-electron hydrogen atom, orbitals
on the same energy level have the same
energy, i.e. they are degenerate.
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1–72
Energies of Orbitals
• As the number of
electrons increases, so
does the repulsion
between them.
• Therefore, in manyelectron atoms,
orbitals on the same
energy level are no
longer degenerate.
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1–73
Factors Affecting Atomic Orbital Energies
• Electrostatic effects play major role in
determining energy states of many e- atoms
– caused by nucleus-electron attractions and
electron-electron repulsions.
– reasons for differences in energy: nuclear
charge, shielding and shape
• Nuclear charge (Z): Higher nuclear charge
lowers orbital energy (stabilizes system) by
increasing nucleus-electron attractions.
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1–74
Shielding
• Effect of Electron Repulsions (shielding):
electron “feels” repulsion from other e-s
• All electrons located between given electron
& nucleus screen, or shield, that electron
from full attractive force of nucleus.
• Effective nuclear charge (Zeff): nuclear
charge electron actually experiences, thus
making it easier to remove.
• Inner electrons shield outer electrons very
effectively.
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1–75
Electron Configurations
• Distribution of all electrons in an atom.
• Consist of:
– Number denoting the energy level.
– Letter denoting the type of orbital.
– Superscript denoting the number of
electrons in those orbitals.
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1–76
Electron Configuration
• States how many e- an atom has in each of its
orbitals (or how many e- of various energies)
• A shorthand system of symbols
1s22s22p4: indicates n, l,
• For a many electron atom, build-up the energy
levels, filling each orbital in succession by
energy
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1–77
Orbital Diagrams
• Each box represents one
orbital.
• Half-arrows represent the
electrons.
• The direction of the arrow
represents the spin of the
electron.
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1–78
Orbital Diagrams
• Arrangement of electrons can also be specified in
terms of orbital occupancy
• Orbital diagram: shows e- occupancy of each
orbital about nucleus of atom
• Hund’s rule: every orbital in sublevel is singly
occupied with one electron before any orbital is
doubly occupied
– And all electrons in singly occupied orbitals
have same spin
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1–79
Hund’s Rule
“For degenerate orbitals, the lowest energy is
attained when the number of electrons with
the same spin is maximised.”
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1–80
Hund’s Rule
The electron configuration of carbon is 1s22s22p2 .
The different ways of placing 2 electrons in the three p orbitals are as
follows:
2px2
2py
2pz
2px1
2py1
2pz
2px1 2py1
2pz
None of these arrangements violate Pauli Exclusion Principle, but we
must determine which one will afford the greatest stability.
The answer is provided by Hund’s Rule which states that the most
stable arrangement of electrons in subshells is the one with
the greatest number of parallel spins.
Thus the orbital diagram for carbon is:
1s2
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2s2
2px1 2py1
2pz
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The box below represents the 1s atomic orbital:
1s1
The Pauli Exclusion Principle
Statement: no two electrons in an atom can have the same
four quantum numbers.
Only two electrons may occupy the same atomic orbital, but
they must have opposite spins.
Consider the He atom which has two electrons.
There are 3 different ways of placing the two electrons in the 1s or
1s2
1s2
1s2
(a)
(b)
(c)
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(a) and (b) do not conform to Pauli Exclusion Principle; only
(c) acceptable.
Diamagnetism and Paramagnetism
Paramagnetic substances are attracted by a magnet; an odd
number of electrons must be present.
On the other hand, if the electron spins are paired, magnetic
effects cancel out and the atom is referred to as diamagnetic.
He
Li
1s2
diamagnetic
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1s2
2s1paramagnetic
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Electronic Configurations and the Periodic Table
• Properties of the elements repeat themselves in
a regular manner
• Groups have similar chemical properties due to
similar e- configurations (outer e- configs)
• Chemical properties repeat in a regular
manner because e- configs repeat.
• Let’s look at the electronic configurations for
Group 1A elements Li, Na, K
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1–84
Electronic Configurations and the Periodic Table
• Elements in Group 1A have one outer s
electron causing similar chemical properties
No. of outer electrons = group number
• Much of the chemical reactivity of elements is
due to the No. of outer electrons (valence e-)
• Electron configuration can also be determined
from the periodic table
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Highest energy level called the valence orbital
or“shell”
– electrons in valence “shell” are valence electrons
– involved in forming compounds
– outer electrons: highest quantum number (n)
– valence and outer e- are same for main group
elements but not transition metals.
– electrons not in valence shell: core electrons
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1–86
The Effect of Orbital Shape on Orbital Energy
• Shielding and
Penetration causes an
energy level to split into
sublevels of different
energy
05_19
2
6
Subshell electron capacity
10
14
6d
5f
7s
6p
5d
Increasing energy
• Order of sublevel
energies:
s < p < d < f
4f
6s
5p
4d
5s
4p
3d
4s
3p
3s
• Penetration is greater
for 2s than 2p
– reduces electrostatic
attraction
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2p
2s
1s
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Aufbau
Aufbau principle: enormally occupy esublevels in an atom
in order of increasing
energy
Energies of orbitals
in different levels
often overlap (4s is
lower than 3d, 5s
lower than 4d)
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05_20
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
7s
7p
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What element has the following electron
configuration?
a) P
b) S
c) O
d) N
e) F
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1–89
Electronic Configurations and the Periodic Table
1. Begin with H and He and continue in order of
increasing atomic number
2. As you move across each period
a. add e- to the ns sublevel
b. add e- to the np sublevel
c. add e- to the (n-1)d sublevel
d. add e- to the (n-2)f sublevel
3. Continue until you reach your element
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1–90
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1–91
Noble Gas Configuration
One can often use symbol of previous noble
gas to represent core electrons
1s22s22p6 = [Ne]
Also called a condensed electron configuration
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Examples
•
11
Na=[Ne]3s1
•
12
Mg= [Ne]3s2
•
•
13
Al = [Ne]3s23p1
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1–93
What (–1) ion (X-) has following electron configuration?
a) Nab) Arc) Br[Ne] = 1s22s22p6
d) Ke) F• Fluorine is normally 1s22s22p5 ,but with the addition of
an electron to make F-, the config. is: 1s22s22p6
Write the electron config. for Li, Na, K when finished
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1–94
Select the correct set of quantum numbers
(n, l, ml, ms) for the highest energy electron in the
ground state of potassium, K.
Write out the electron configuration: K = [Ar] 4s1
A. 4, 1, -1, ½
B. 4, 1, 0, ½
C. 4, 0, 1, ½
D. 4, 0, 0, ½
E. 4, 1, 1, ½
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1–95
Practice
Determine the electron configuration and orbital
diagram for the following:
He, C, S, Co, Kr
What are 4 quantum numbers for the last
electron added?
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Some Anomalies
Some
irregularities
occur when
there are
enough
electrons to
half-fill s and
d orbitals on a
given row.
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Some Anomalies
• For instance, the electron configuration
**for chromium is: [Ar] 4s1 3d5 rather
than the expected: [Ar] 4s2 3d4.
• This occurs because the 4s and 3d
orbitals are very close in energy.
• These anomalies occur in f-block atoms,
as well.
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1–98
Some Anomalies
• For instance, the electron configuration
**for chromium is: [Ar] 4s1 3d5 rather
than the expected: [Ar] 4s2 3d4.
• This occurs because the 4s and 3d
orbitals are very close in energy.
• These anomalies occur in f-block atoms,
as well.
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Exceptions- Cr and Cu
•
•
•
•
One electron in the s -subshell
Cr (Z=24)
1s2 2s22p63s23p64s13d5
Cr [Ar]4s1 3d5 <=All s and d subshells are half
full
• Cu ( Z= 29)
• Cu [Ar]4s1 3d10 <=Prefers a filled d subshell,
leaving s with one electron
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1–100
Unusual Configurations
• Unusual configuration for some transition metals:
Cr, Cu, Ag, Au, Mo, Nb
• Half-filled and filled sublevels (Noble gases) are
unexpectedly stable
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• Which of the following two electronic
configuration is more stable?
a [Ar]4s1 3d5
b [Ar]4s2 3d4
• This is the electronic configuration for
chromium.
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Which of the following electron configurations
represents the ground state for an element?
A.
B.
C.
D.
E.
[Ne]3s13p1
[He]2s12p3
[Ne]3s23p23d1
[Ne]3s23p33d1
[Ne]3s23p3
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1–103
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Valence electron configuration
• Consider the electron configuration of Br:
• Br =[Ar] 4s2 3d10 4p5
• Although the 3d electrons are outer- shell
electrons, they are not involved in bonding
and are not considered valence electrons.
We do not consider completely filled d and f
subshells to be valence electrons.
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1–106
An element with the electron configuration
[Ar]ns2(n - 1)d10np4 has ____________
valence (or outer) electrons.
A. 2
B. 4
C. 6
D. 8
E. 16
X =[Ar] 4s2 3d10 4p4
X =[Ar] 3d10 4s2 4p4
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