Transcript Lecture Ch#7 Electrons - Seattle Central College

Chapter 7
Electrons in Atoms
Properties of Electrons
• Electrons display both particle properties and
wave properties.
• Electrons were discovered by JJ Thompson
• Thompson also measured the charge/mass
ratio
• Milikan was able to determine the charge on
an electron, thus allowing the mass to be
determined.
• Davisson and Germer discovered the wave
nature of an electron at Bell Labs in 1927,
thus showing the particle/wave nature of an
electron
The Atom Model
Different models of atoms
• Thompson developed the plumb pudding model
of an atom 1898
• Rutherford suggested the planetary model (i.e.
electrons orbit the nucleus)
• Bohr applied concepts of quantization to
Rutherford’s model to develop the Bohr model
• Bohr model lasted 10 years, and was replaced by
a wave model, called the quantum mechanical
model, based on the wave nature of electrons.
Properties of Waves
Energy has properties of waves, for example
electromagnetic energy
Wavelength (m)
Amplitude (m)
Speed 3.0X108 m/s
Energy (j)
Frequency (1/s, Hz)
Electromagnetic Radiation
10-12
10-10
Gamma rays X- rays
10-8
Uv-rays
10-7
Visible
rays
10-4
Infrared
rays
10-2
Micro
Waves
(Radar)
100
102
Radio and Television waves
Increasing wave length in meters
Increasing energy
Visible Radiation
Electromagnetic Radiation
Our major source of
EM comes from our
sun.
EM travels at the
speed of light
3.0x108m/s
Has both wave
properties and
particle properties
Photons are the
particles possessed
by EM
Behavior of Waves
• Waves refract or bend when they pass from one
medium to another with different densities.
• Diffraction is the bending of electromagnetic
radiation as it passes around the edge of an object
or through narrow openings.
• Interference is the interaction of waves that results
in either reinforcing their amplitudes or canceling
them out.
Diffraction and Interference
Refraction
White Light
The shortest wave
lengths bend longer
ones, thus violet is the
shortest
Davisson and Germer discovered the wave nature of an
electron at Bell Labs in 1927 by observing electron
diffraction.
R
O
Y
G
B
I
V
Evidence of Quantization
The red-orange light from hydrogen gas passes
through a prism to form a line spectra. Each
different colored light has its own unique energy.
Atomic Spectrum of Sodium
Absorption Spectra
Types of Spectra
• Atomic emission spectra consist of bright
lines on a dark background.
• Atomic absorption spectra consist of
characteristic series of dark lines produced
when free gaseous atoms are illuminated by
external sources of radiation.
Hydrogen Line Spectrum
Quantum Theory
• Max Planck proposed that light can have
both wavelike and particle-like properties.
• A quantum is the smallest discrete quantity
of a particular form of energy.
• Particles of radiant energy are known as
quanta.
• Quantum theory is based on the idea that
energy is absorbed and emitted in discrete
quanta.
Quantum Theory
• Something that is quantized has values that
are restricted to whole-number multiples of a
specific base value.
• The energy of a quantum of radiation is:
 E = h where h is Planck’s constant
 h = 6.6260755 x 10-34 J•s
 Or E = hc/
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
sec
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
sec
10-10 m
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
sec
10-10 m 512Å
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m
sec
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
10-10 m
sec
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
10-10 m 512Å
sec
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
6.63 X 10-34 kg-m2-s
s2
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
6.63 X 10-34 kg-m2-s
s2
0.145 kg
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
6.63 X 10-34 kg-m2-s
cm
s2
0.145 kg 10-2 m
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
in
6.63 X 10-34 kg-m2-s
cm
s2
0.145 kg 10-2 m 2.54 cm
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
ft
in
6.63 X 10-34 kg-m2-s
cm
s2
0.145 kg 10-2 m 2.54 cm 12 in
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
mi
ft
in
6.63 X 10-34 kg-m2-s
cm
s2
0.145 kg 10-2 m 2.54 cm 12 in 5280 ft
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
mi
ft
in
6.63 X 10-34 kg-m2-s
cm
s2
0.145 kg 10-2 m 2.54 cm 12 in 5280 ft
hr
101 mi
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
mi
ft
in
6.63 X 10-34 kg-m2-s
cm
s2
0.145 kg 10-2 m 2.54 cm 12 in 5280 ft
60 min
hr
101 mi hr
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
mi
ft
in
6.63 X 10-34 kg-m2-s
cm
s2
0.145 kg 10-2 m 2.54 cm 12 in 5280 ft
60 min 60 sec
hr
min
101 mi hr
SAMPLE PROBLEMS
1. Find the frequency of green light with a wavelength of 512Å.
3.00 X 108 m
Å
= 5.86 X 1015 s-1
sec
10-10 m 512Å
2. Find the energy of green light with a wavelength of 512Å
6.63 X 10-34 j-s 3.00 X 108 m Å
-18 j
=
3.88
X
10
10-10 m 512Å
sec
3. Find the wavelength of a 145 g baseball traveling at 101 mi/hr
mi
ft
in
6.63 X 10-34 kg-m2-s
cm
s2
0.145 kg 10-2 m 2.54 cm 12 in 5280 ft
60 min 60 sec
hr
min
101 mi hr
= 1.01 X 10-34 m
Particle Nature
• Each packet of
electromagnetic
radiation energy is
called a quantum.
• Einstein called the
packets photons.
Photoelectric Effect
• The photoelectric effect is
the release of electrons
from a metal as a result of
electromagnetic radiation.
• The photoelectric effect
can be explained if
electromagnetic radiation
consists of tiny particles
called photons.
The Hydrogen Spectrum
• Johannes Rydberg revised Balmer’s equation
to describe the complete hydrogen spectrum.
 1
1
= (1.097 x10-2 (nm)-1) 2 - 2 

 n1 n2 
1
N1 is a whole number that remains fixed for a
series of calculations in which n2 is also a whole
number with values of n1+1, n1+2,… for
successive line in the spectrum.
Example
What is the wavelength of the line in the
visible spectrum corresponding to n1 = 2
and n2 = 4?
1
1
1
)
(
-2
= 1.097 X 10 pm 4
9
λ
Example
What is the wavelength of the line in the
visible spectrum corresponding to n1 = 2
and n2 = 4?
1
1
1
)
(
-2
= 1.097 X 10 pm 4
9
λ
λ = 486.1 nm
Example
What is the wavelength of the line in the
visible spectrum corresponding to n1 = 2
and n2 = 4?
1
1
-2
= 1.097 X 10 pm(
4
λ
λ = 486.1 nm
-
1
9)
The Bohr Model
• The electron in a hydrogen atom occupies a
discrete energy level and may exist only in
the available energy levels.
• The electron may move between energy
levels by either absorbing or emitting
specific amounts of energy.
• Each energy level is designated by a specific
value for n, called the principal quantum
number.
Energy of Electronic Transitions
• Neils Bohr derived the following formula for the
possible energy differences (E) be any pair of
energy levels with values n1 and n2.
2 2me 4  1
1 
E =
 2 - 2 
2
h
n2 
n1
m and e is the mass and charge of the electron.

Hydrogen Spectrum
• An energy level is
an allowed state that
an electron can
occupy in an atom.
• Movements of
electrons between
energy levels are
called electron
transitions.
Electronic States
• The lowest energy level available to an electron in
an atom is its ground state.
• An excited state of an electron in an atom is any
energy state above the ground state.
Excited and Relaxed Electrons
In terms of the Bohr model absorption and emission looks
like this.
Excited and Relaxed Electrons
Electrons move between energy levels by absorbing and
emitting energy in the form of light.
We call the lowest energy level the ground state. The higher
energy level is called the excited state.
Problems with the Bohr Model
• The Bohr model applies only to one electron atoms.
• The Bohr model doesn’t account for the observed
spectra of multielectron elements or ions.
• The movement of electrons in atoms is much less
clearly defined than Bohr allowed.
Particle or Waves?
• If electromagnetic radiation behaves as a
particle, de Broglie reasoned, why couldn’t a
particle in motion, such as an electron,
behave as a wave?
• de Broglie’s Equation
  = h/mu (m in kg and u in m/s)
Electrons as Waves
• De Broglie reasoned that
an electron in a hydrogen
atom could behave as a
circular wave oscillating
around the nucleus.
• If electrons are moving
around the nucleus in a
continuous manor, the
state of the electron must
be described by a
quantum number, n.
Tacoma Narrows Bridge
http://www.youtube.com/watch?v=P0Fi1Vcb
pAI
The Uncertainty Principle
• Quantum mechanics allows us to predict the
probabilities of where we can find an electron.
• We cannot map out on the path an electron
travels.
 The Heisenberg’s uncertainty principle
says that you cannot determine the
position and momentum of an electron at
the same time.
Electron Wave Equations
• The description of the behavior of particles as
waves is called wave mechanics or quantum
mechanics.
• The mathematical description of an electron
wave is called the wave equation.
• Wave functions, , are mathematical
descriptions of the motion of electron waves
as they vary with location and with time.
Quantum Numbers
• The principle quantum number, n, is a
positive integer that indicates the shell and
relative size of orbital(s).
• The angular momentum quantum number, l,
is an integer from zero to n-1. It defines the
shape of the orbital and subshell.
Value of l
0
1
2
3
4
Letter identifier
s
p
d
f
g
Quantum Numbers
• The magnetic quantum number, ml, is an
integer with a value from -l to +l. It defines
the orientation of an orbital in the space
around the nucleus of an atom.
• The spin magnetic quantum number, ms, is to
account for the two possible spin orientations.
The values for ms are +1/2 and -1/2.
Quantum Number Relationships
Electron Identifier
It takes a total of 4 quantum numbers to identify an
electron in a particular atom. Like it’s student ID no.
4py+1/2
spin QN; ms=1/2 (clockwise or counterclockwise
magnetic QN; ml=0 (shape orientation)
angular momentum QN; l=1 (volume shape)
principal QN; n=4 (size and energy)
Quantum Numbers
n
l
ml
# orbitals
1
0
0
1
2
0
0
1
1
-1,0,+1
3
0
0
1
1
-1,0,+1
3
2
-2.-1,0,+1,+2
5
0
0
1
1
-1,0,+1
3
2
-2.-1,0,+1,+2
5
3
-3,-2.-1,0,+1,+2,+3
7
3
4
Practice
What are the letter designations of all the subshells
in the n = 5 energy level or shell? What is total
number of orbitals in the n = 5 shell?
Shape and Sizes of Orbitals
• Psi squared, 2, defines the space, called an
orbital, in atom where the probability of
finding an electron is high.
• A radial distribution plot is a graphical
representation of the probability of finding an
electron in a thin spherical layer near the
nucleus of an atom.
Probability Electron Density for
1s Orbital
Comparison of s Orbitals
The Three 2p Orbitals
The Five 3d Orbitals
Assigning Quantum Numbers
• Pauli’s exclusion principle - no two
electrons in an atom may have the same set
of four quantum numbers.
 An orbital can only hold two electrons and
they must have opposite spins.
Practice
Write the set of quantum numbers which
describe each electron in the three 2p orbitals.
Practice
Which of the following combinations of
quantum number are allowed?
1. n = 1, l = 1, ml = 0
2. n = 3, l = 0, ml = 0
3. n = 1, l = 0, ml = -1
4. n = 2, l = 1, ml = 2
Orbital Energy Notation
Hydrogen Atom
E
3s
3p
2s
2p
1s
3d
Many Electron Atoms
• They do not follow the diagram for the
hydrogen atom.
• As l changes the energy of the orbital
changes
 The lower the value of l the lower in
energy the subshell
Sublevel Relative Energies
Beyond the 3p subshell the orbitals don’t fill in an obvious way.
For example the 4s level lies lower in energy than the 3d .
Multi-electron Orbital
Notation
Terms
• Orbitals that have the exact same
energy level are degenerate.
• Core electrons are those in the filled,
inner shells in an atom and are not
involved in chemical reactions.
• Valence electrons are those in the
outermost shell of an atom and have the
most influence on the atom’s chemical
behavior.
Electron Configuration
The way in which electrons are organized into shells, subshells
and orbitals in an atom is called the electronic configuration.
The electronic configuration of an atom can be determined
using the “Aufbau rule” also known as the “building up
principle”.
Aufbau comes from the German
meaning construction although it was
the Danish physicist Neils Bohr who
came up with the idea !!
Aufbau Principle
The Aufbau Principle states that:
“The orbitals of lower energy are filled in first with the electrons
and only then the orbitals of high energy are filled.”
What is the lowest energy orbital of an atom?
What is the third lowest energy orbital of an atom?
Aufbau Principle
The Aufbau Principle states that:
“The orbitals of lower energy are filled in first with the electrons
and only then the orbitals of high energy are filled.”
What is the lowest energy orbital of an atom?
1s orbital
What is the third lowest energy orbital of an atom?
Aufbau Principle
The Aufbau Principle states that:
“The orbitals of lower energy are filled in first with the electrons
and only then the orbitals of high energy are filled.”
What is the lowest energy orbital of an atom?
1s orbital
What is the third lowest energy orbital of an atom?
2p orbital
Aufbau Principle
How would we use our rules to “build up” the electron
configuration of a Li atom?
Li has Z = 3 so has 3 e-.
Aufbau Principle
How would we use our rules to “build up” the electron
configuration of a Li atom?
Li has Z = 3 so has 3 e-.
1s subshell
Aufbau Principle
How would we use our rules to “build up” the electron
configuration of a Li atom?
Li has Z = 3 so has 3 e-.
1s subshell
Aufbau Principle
How would we use our rules to “build up” the electron
configuration of a Li atom?
Li has Z = 3 so has 3 e-.
1s subshell
Aufbau Principle
How would we use our rules to “build up” the electron
configuration of a Li atom?
Li has Z = 3 so has 3 e-.
2s subshell
1s subshell
Aufbau Principle
How would we use our rules to “build up” the electron
configuration of a Li atom?
Li has Z = 3 so has 3 e-.
2s subshell
1s subshell
Aufbau Principle
How would we use our rules to “build up” the electron
configuration of a Li atom?
Li has Z = 3 so has 3 e-.
2s subshell
1s subshell
We can write this in shorthand as 1s22s1
Hund’s Rule
If there are multiple orbitals with the same energy how do we
decide which orbital to put an electron?
We use Hund’s rule which states:
“Electrons will not join other electrons in an
orbital if an unoccupied orbital of the same
energy is available”
Degenerate Orbitals
As we have seen previously for p, d and f subshells there are
multiple orbitals with the same energy.
In particular:
• p subshells have three orbitals with the same energy
• d subshells have five orbitals with the same energy
• f subshells have seven orbitals with the same energy
Each of these orbitals may accommodate a maximum of two
electrons.
Hund’s Rule
Using Hund’s rule how would we put three electrons in a p
subshell ?
px
px
py
py
p subshell

p subshell

pz
pz
Pauli Exclusion Principle
When we do put two electrons in one orbital then they obey the
Pauli exclusion principle.
“only electrons with opposite spin
can occupy the same orbital”
px
px
py
py
p subshell

p subshell

pz
pz
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
1s subshell
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
1s subshell
We can write this in shorthand as 1s22s22p3
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
1s subshell
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
2s subshell
1s subshell
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
2s subshell
1s subshell
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
2p subshell
2s subshell
1s subshell
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
2p subshell
2s subshell
1s subshell
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
2p subshell
2s subshell
1s subshell
Orbital Notation
How would we use our rules to “build up” the electron
configuration of a N atom?
N has Z = 7 so has 7 e-.
2p subshell
2s subshell
1s subshell
We can write this in shorthand as 1s22s22p3
Orbital Diagram Configuration
1s
H: 1s1
He: 1s2
Li: 1s22s1
Be: 1s22s2
B: 1s22s22p1
2s
2p
Orbital Diagram Configuration
1s
2s
2p

C: 1s22s22p2
or
C: 1s22s22p2

Hund’s Rule tells us which configuration is correct.
Orbital Diagram Configuration
Electron Configuration 1s
C: 1s22s22p2
N: 1s22s22p3
O: 1s22s22p4
F: 1s22s22p5
Ne: 1s22s22p6
2s
2p
Electron Configurations
Fourth Period Elements
K
1s22s22p63s23p64s1
or [Ar]4s1
Ca
1s22s22p63s23p64s2
or [Ar]4s2
Sc
1s22s22p63s23p64s23d1
or [Ar]4s23d1
Ti
1s22s22p63s23p64s23d2
or [Ar]4s23d2
V
1s22s22p63s23p64s23d3
or [Ar]4s23d3
Cr
1s22s22p63s23p64s13d5
or [Ar]4s13d5
Mn
1s22s22p63s23p64s23d5
•
•
•
1s22s22p63s23p64s13d10
or [Ar]4s23d5
Cu
or [Ar]4s13d10
Anomalies in Configurations
• Chromium and Copper do not follow the
pattern of the other elements.
 You should remember these two families,
because other elements in these families
exhibit the same types of configurations
• You can use the Periodic Table to guide you in
writing electron configurations.
Diagonal Rule
There is an easy way to
remember the sequence of the
energies of the subshells.
Using the Periodic Table
Practice
Write the electron configuration for Pt.
Write the electron configuration for I-.
Write the electron configuration for Rh2+.
Electron Configurations of Ions
• Start with the configuration for the neutral
atom, then add or remove electrons from the
valence shells to make the desired ion.
• Atoms or ions that are isoelectronic with
each other have identical numbers and
configurations of electrons.
Sizes of Atoms and Ions
Orbital Penetration and
Orbital penetration occurs when an electron in
an outer orbital has some probability of being
close to the nucleus
•Penetration ability follows this order:
s > p > d > f.
Penetration Ability of s Orbitals
Effective Nuclear Charge
Effective nuclear charge (Zeff) is the attractive force
toward the nucleus experienced by an electron in an
atom, usually the outer (valence electrons).
• Electrons in between the nucleus and the
designated electron shield the designated electron
from the nucleus.
• Electrons between the nucleus and the designated
electron also repelled the designated electron
• Shielded electrons are further away from the
nucleus than they would be if not shielded
Radii of Atoms and Ions
Ionization Energy
The quantity of energy required to
remove 1 mole of electrons from 1 mole
of the gaseous atom or ion.
X(g) ---> X+(g) + e-(g)
Ionization Energy Trends
Ionization Energies
Periodic Trends
•
First ionization energy:
increases from left to right across a period;
decreases going down a group.
Successive Ionization Energies (kJ/mol)
Elements
IE1
IE2
IE 3
IE4
IE5
IE6
H
1312
He
2372 5249
Li
520
7296 12040
Be
897
1758 15050 21070
B
801
2426
3660
24682 32508
C
1087 2348
4617
6201
37926 46956
N
1402 2860
4581
7465
9391
O
1314 3383
5298
7465
10956 13304
52976
IE1is the first
ionization energy,
which is the
energy to remove
a valence electron
from an atom to
produce a cation.
IE2 is the second
ionization energy
and is the energy
required to
remove the
second electron
from a cation.
Note: IE2>IE1, since more energy is required to remove an electron from a positive ion
compared to a neutral atom.
Successive Ionization Energies (kJ/mol)
Elements
IE1
IE2
IE 3
IE4
IE5
IE6
H
1312
He
2372 5249
Li
520
7296 12040
Be
897
1758 15050 21070
B
801
2426
3660
24682 32508
C
1087 2348
4617
6201
37926 46956
N
1402 2860
4581
7465
9391
O
1314 3383
5298
7465
10956 13304
52976
Note: The large jump
in ionization energy
to the right of the red
line is due to
destroying the noble
gas configuration,
which we know to be
very stable. Also
notice that ionization
energy increases as
we remove an
electron from a more
positive cations,
going left to right
THE END
ChemTour: Electromagnetic
Radiation
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PC | Mac
This ChemTour explores the relationship of frequency,
wavelength, and energy using animations, interactive
graphs, and equations. The quantitative exercises include
graph reading and calculations using Planck’s constant and
the speed of light.
ChemTour: Light Diffraction
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PC | Mac
This animation recreates Thomas Young’s double-slit
experiment and demonstrates how constructive and
destructive interference occur.
ChemTour: Doppler Effect
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A boat moving with or against the direction of wave
movement demonstrates the motion-induced shifts in
wavelengths and frequency that are examples of the
Doppler effect.
ChemTour: Light Emission
and Absorption
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This ChemTour examines the emission and absorption
spectra for sodium and hydrogen and relates them to
energy level transitions.
ChemTour: Bohr Model of the
Atom
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This ChemTour explores the idea that energies of electrons
surrounding atomic nuclei are quantized. In Practice
Exercises, students learn to calculate the energies of
specific states of hydrogen, and the energies involved in
electronic transitions.
ChemTour: de Broglie
Wavelength
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In this ChemTour, students learn to apply the de Broglie
equation to calculate the wavelength of moving objects
ranging from baseballs to electrons. Includes Practice
Exercises.
ChemTour: Quantum Numbers
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In this ChemTour, students explore the rules for designating
quantum numbers. Includes Practice Exercises.
ChemTour: Electron
Configuration
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This ChemTour explains how electrons are distributed
within atomic orbitals. Students learn how to determine an
element’s electron configuration and learn how to complete
an orbital box diagram. Includes practice exercises.
Suppose two photons combine in a crystal to form a single
photon of green light or "green photon."
Which of the following could be the colors of the two
combining photons?
A) Green & green
B) Blue & yellow
Combining Two Photons
C) Infrared & infrared
Please consider the following arguments for each
answer and vote again:
A. A green photon can only be produced by the
combination of two other green photons of the same
wavelength.
B. The color green is the result of combining the colors
blue and yellow, just as a green photon will result
from the combination of blue and yellow photons.
C. Only two infrared photons have the proper total
energy needed to form a green photon.
Combining Two Photons
An electron in the ground state absorbs a single photon of
light and then relaxes back to the ground state by emitting
an infrared photon (1200 nm) followed by an orange
photon (600 nm).
What is the wavelength of the absorbed photon?
A) 400 nm
B) 600 nm
Absorption and Fluorescence of Light
C) 1800 nm
Please consider the following arguments for each
answer and vote again:
A. The wavelength is inversely proportional to the energy,
so for energy to be conserved the absorbed photon
must have a wavelength of 400 nm.
B. The wavelength of the absorbed photon is the
difference of the wavelength of the two emitted
photons, which is 600 nm.
C. For the energy to be conserved, the sum of the
wavelengths must be conserved. So the wavelength of
the absorbed photon is 1800 nm.
Absorption and Fluorescence of Light
The diagram to the left
depicts the interference
pattern that results from
the constructive and
destructive interference of light waves that are diffracted
as they pass through two slits. If the pattern is the result
of green light passing through two slits, which of the
following patterns would be the result of blue light
passing through the same two slits?
A)
B)
Two-Slit Diffraction and Interferometry
C)
Please consider the following arguments for each
answer and vote again:
A. The wavelength of blue light is shorter than that of
green light, so constructive and destructive interference
occurs at smaller intervals.
B. The interference pattern is dependent only on the width
of and distance between the two slits. Therefore, the
interference pattern should not change.
C. Blue light is higher in energy than green light and
therefore would be less affected by the two slits.
Two-Slit Diffraction and Interferometry
When a photon of red light hits metal X, an electron is
ejected. Will an electron be ejected if a photon of yellow
light hits metal X?
A) Yes
B) No
Photoelectric Effect: Red and Yellow Light
C) Can't tell
Please consider the following arguments for each
answer and vote again:
A. Photons of yellow light possess more energy than
photons of red light, so a yellow photon also must eject
an electron.
B. Each metal has a specific wavelength of light that will
cause electrons to be ejected. If red light has the correct
wavelength, yellow cannot.
C. Whether a yellow photon will eject an electron from the
metal will depend on how tightly the electron is bound
to the metal.
Photoelectric effect: Red and Yellow Light
When a photon of blue light hits metal X, an electron is
ejected. Will an electron be ejected if a photon of green
light hits metal X?
A) Yes
B) No
Photoelectric Effect: Blue and Green Light
C) Can't tell
Please consider the following arguments for each
answer and vote again:
A. So long as enough photons of light hit the metal, an
electron will always be ejected, regardless of the
wavelength of the light.
B. The energy of a blue photon is higher than the energy of
a green photon so an electron removed with blue light
will not be removed with green light.
C. Whether a green photon will eject an electron from the
metal will depend on how tightly the electron is bound
to the metal.
Photoelectric effect: Blue and Green Light
A 300-nm photon can eject an electron from
a metal surface with a certain kinetic energy.
What photon wavelength would be required
to eject an electron from the same metal
surface with twice the kinetic energy?
A) 150 nm B) 200 nm C) 600 nm
Photoelectric Effect: Kinetic Energies of Electron
Please consider the following arguments for each
answer and vote again:
A. To eject an electron with twice the kinetic energy,
twice the energy must be provided by the photon, so
the photon wavelength must be halved.
B. A photon with a wavelength of 200 nm will overcome
the work function and provide twice the kinetic energy.
C. To double the kinetic energy of the ejected electron,
the wavelength of the impacting photon also must be
doubled.
Photoelectric Effect: Kinetic Energies of Electrons
Suppose a hydrogen molecule (1H2) is traveling at 800 m/s
and a deuterium molecule (2H2) is traveling at 400 m/s.
What can be said of the de Broglie wavelengths of the two
molecules?
A) λH > lD B) λH < lD C) λH = lD
De Broglie Wavelengths of H 0 Molecules
Please consider the following arguments for each
answer and vote again:
A. The kinetic energy of the deuterium molecule is twice
that of the hydrogen molecule.
Therefore, the
deuterium molecule will have a shorter de Broglie
wavelength.
B. Because the speed of the hydrogen molecule is greater
than the speed of the deuterium molecule, the de
Broglie wavelength of the hydrogen molecule will be
shorter.
C. The hydrogen molecule and the deuterium molecule
have the same momentum and therefore will have the
same de Broglie wavelength.
De Broglie Wavelengths of H O Molecules
One method for decreasing the temperature of atoms,
known as laser cooling, involves bombarding an atom
with photons of light, decreasing its overall momentum
and thus its kinetic energy (just like one could slow a fastmoving car by colliding it with another car).
A sodium atom at a temperature of 60 K has a de Broglie
wavelength of 66 pm (6.6x10-11 m). Approximately how
many photons of red light (at λ = 660 nm) would it take to
stop a sodium atom at 60 K?
A) ~1
Laser Cooling of Sodium Atoms
B) ~102
C) ~104
Please consider the following arguments for each
answer and vote again:
A. A photon travels ~105 times faster than a sodium atom.
Therefore, only one photon is required.
B. The kinetic energy of a sodium atom is ~100 times less
than the kinetic energy of a red photon.
C. The de Broglie wavelength of a sodium atom at 60 K is
~104 times shorter than the wavelength of a red photon,
so it will take 104 photons to stop a single sodium atom.
Laser Cooling of Sodium Atoms
What color will a yellow object
appear when it is seen through a
filter with the absorption
spectrum shown to the left?
A) Yellow
B) Blue
Transmission of Light through a Color Filter
C) Black
Please consider the following arguments for each
answer and vote again:
A. The filter absorbs no yellow light, so the object will
appear yellow.
B. Blue light is absorbed by the filter, so an object seen
through the filter will appear blue.
C. No yellow light is absorbed by the filter, so the object
will appear black.
Transmission of Light through a Color Filter
Photon emission from a system
possessing the energy level diagram
to the left would produce which of
the following spectra?
A)
Emission Spectra
B)
C)
Consider the following arguments for each answer and
vote again:
A. The photon wavelength depends only on the energy of
the lowest state, so only 1 wavelength is possible.
B. There are 2 possible transitions—one from each of the
2 upper levels. Thus, 2 wavelengths of light are
emitted.
C. The 3 energy levels lead to 2 high-energy transitions
and 1 low-energy transition. Therefore, 3 different
photon wavelengths are possible.
Emission Spectra
Emission from which of the following
energy level diagrams would produce
the spectrum shown to the left?
A)
Energy Levels
B)
C)
Consider the following arguments for each answer and
vote again:
A. The arrangement of the energy levels reflects the
arrangement of the lines in the emission spectrum.
B. This energy level diagram allows only 1 low-energy
transition, consistent with the emission spectrum.
C. Only this energy level diagram allows 3 high-energy
transitions and 1 low-energy transition.
Energy Levels
The diagram to the left shows the
spacing of the first five energy levels
for a hydrogen atom. Which of the
following transitions in He+ has the
same wavelength as the 4→2 transition
in H?
A) 4→2
Transition in H and He+
B) 8→4
C) 16→8
Consider the following arguments for each answer and vote
again:
A. He+ has the same electron configuration as H; therefore,
the energy level diagram will be the same.
B. The atomic number of He+ is twice that of H.
Therefore, to produce the same energy splitting, the
energy levels must be twice that of H.
C. The energy of the electron is proportional to Z2, which
is 4 for He+. Therefore, the two levels, 4 and 2, must be
increased by a factor of 4 to 16 and 8, respectively.
Transition in H and He+
Periodic Table
Which atom or ion can have the
electron configuration 1s22s22p1?
A) Li
Electron Configurations
B) Be-
C) B+
Consider the following arguments for each answer
and vote again:
A. The answer must be lithium because it is the first
element in row 2 to possess only one unpaired
electron.
B. Beryllium in its ground state has the electron
configuration 1s22s2, so Be- in its ground state will
have
the
configuration
1s22s22p1.
C. In its ground state, boron has the electron
configuration 1s22s22p1, so B+ must also have this
configuration.
Electron Configurations
Which of the following has the
lowest ionization energy?
A) H(1s1)
Ionization Energies
B) He(1s13p1)
C) He+(4p1)
Consider the following arguments for each answer
and vote again:
A. Hydrogen has a lower nuclear charge than helium, so
it always has a lower ionization energy than any
helium atom or ion.
B. He(1s13p1) has almost the same ionization energy as
H(3p1), which has a lower ionization energy than
either H(1s1) or He+(4p1).
C. Because the electron in He+(4p1) is in the fourth
shell, the ionization energy of He+(4p1) is the
lowest.
Ionization Energies
How does the ionization energy of
He(1s2) compare to the ionization
energies of H(1s1) and He+(1s1)?
A) Higher
B) Lower
Ionization Energies of He(1s2)
C) In-between
Consider the following arguments for each answer
and vote again:
A. It is harder to remove an electron from a doubly
occupied orbital than from a singly occupied orbital.
B. Each electron offsets the charge of one of the
protons, giving an effective nuclear charge of zero.
C. Each electron partially shields the other, leading to
an effective nuclear charge that is between 1 and 2.
Ionization Energies of He(1s2)
Which of the following atoms
(or ions) has the smallest
radius?
A) K+
Atomic and Ionic Radii
B) Ar
C) Cl-
Consider the following arguments for each answer
and vote again:
A. K+ has the highest nuclear charge and so has the
smallest atomic radius.
B. Because it is a noble gas, Ar has the smallest atomic
radius.
C. Cl- has the nucleus with the lowest mass, so it has
the smallest atomic radius.
Atomic and Ionic Radii
Suppose an electron is transferred
from a potassium atom to an
unknown halogen atom.
For
which of the following halogen
atoms would this process require
the least amount of energy?
A) Cl
Electron Affinity of Halogen Atoms
B) Br
C) I
Consider the following arguments for each answer
and vote again:
A. Chlorine has the greatest affinity for electrons and so
would release the most energy when an electron is
added.
B. Electron donation is most favorable energetically
when it occurs between atoms on the same row of
the periodic table.
C. Because of its massive nuclear charge and large
electron cloud, an iodine atom can most easily accept
an additional electron.
Electron Affinity of Halogen Atoms