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Chapter 7: ELECTRONS IN ATOMS AND PERIODIC PROPERTIES
Problems: 7.1-7.14, 7.19-7.26, 7.31-7.36, 7.43, 7.45-7.57, 7.59-7.62, 7.65-7.107, 7.109-7.117, 7.124-7.125
ELECTROMAGNETIC RADIATION
Electromagnetic (EM) Spectrum: a continuum of the different forms of electromagnetic
radiation or radiant energy (Fig. 7.1 on p. 298)
– The substances below are about the size of the wavelength indicated in the EM spectrum.
– e.g., an atom is about 10-10-10-9 m in size while a CD is about 10-3 m (or 1 mm) thick.
visible region: the portion of the EM spectrum that we can perceive as color
For example, a "red-hot" or "white-hot" iron bar freshly removed from a high-temperature
source has forms of energy in different parts of the EM spectrum
– red or white glow = radiation within the visible region
– warmth = radiation within the infrared region
7.1 WAVES OF LIGHT
The term electromagnetic comes from the theory
proposed by Scottish scientist James Clerk Maxwell that
radiant energy consists of waves with an oscillating
electric field and an oscillating magnetic field, which
are perpendicular to one another.
CHEM 161: Chapter 7 Notes page 1 of 25
Electromagnetic Radiation
– EM waves have the following properties: wavelength and frequency (see Fig. 7.3).
wavelength (λ=Greek “lambda”): distance between successive peaks
distance
λ= ; generally in units of m, cm, or nm
wave
frequency (ν=Greek “nu”): number of waves passing a given point in 1 s
wave cycle 1
ν= ; generally in hertz (Hz) = or
time s s
CHEM 161: Chapter 7 Notes page 2 of 25
speed of light
– The product of wavelength and frequency is equal to the speed of light in a vacuum,
c=2.99792458×108 m/s, which is usually rounded to 4 s.f. → c=2.998×108 m/s (to 4 s.f.).
c = λ × ν
distance distance wave
= ×
time wave time
Know how to convert between wavelength and frequency using the speed of light!
Ex. 1 The wavelength for the electromagnetic radiation responsible for a blue sky is about
473 nm. What is the frequency of this radiation in Hz?
Ex. 2 KUOW broadcasts at 94.9 MHz in Seattle. What is the wavelength (in nm) of this
radio wave?
THE NATURE OF MATTER
CLASSICAL Descriptions of Matter
John Dalton (1803)
– atoms are hard, indivisible, billiard-like particles
– atoms have distinct masses (what distinguishes on type of atom from another)
– all atoms of same element are the same
JJ Thomson (1890s)
– discovered the charge-to-mass ratio of electrons
→ atoms are divisible because electrons are only one part of the atom
Ernest Rutherford (1910)
– shot positive alpha particles at a thin foil of gold
→ discovery of the atomic nucleus
James Maxwell (1873)
– visible light consists of electromagnetic waves
CHEM 161: Chapter 7 Notes page 3 of 25
7.3 PARTICLES OF LIGHT AND QUANTUM THEORY
Transition between Classical and Quantum Theory
Max Planck (1900); Blackbody Radiation
– heated solids to red or white heat
– noted matter did not emit energy in continuous bursts, but in whole-number multiples of
certain well-defined quantities
→ matter absorbs/emits energy in bundles = "quanta"
(single bundle of energy= "quantum")
Albert Einstein (1905); Photoelectric Effect
– Photoelectric Effect: Light shining on a clean metal → emission of electrons only
occurs when the light has a minimum threshold frequency, ν0
– For ν < ν0 → no electrons are emitted
– For ν > ν0 → electrons are emitted, more e– emitted with greater intensity of light
– Einstein applied Planck's quantum theory to light
→ light exists as a stream of "particles" called photons.
Energy is proportional to the frequency (ν) and wavelength (λ) of radiation, and the
proportionality constant (h) is now called Planck's constant
hc
E = hν = where h = 6.626×10–34 J·s
λ
Ex. 1. Excited mercury atoms emit light strongly at a wavelength of 436 nm.
a. What is the energy (in J) for one photon of this light?
b. What is the energy (in kJ/mol) for a mole of photons of this light?
CHEM 161: Chapter 7 Notes page 4 of 25
Ex. 2. Certain elements emit light of a specific color when they are burned. When a
potassium solution is burned in a flame test, the energy of the light emitted is
4.909×10-19 J. Calculate the wavelength (in nm) for this light, and use the visible
spectrum below to determine the color of the light.
7.4 THE HYDROGEN SPECTRUM AND THE BOHR MODEL
Emission Spectra: continuous or line spectra of radiation emitted by substances
– a heated solid (e.g. the filament in an incandescent light bulb) emits light that spreads out
to give a continuous spectrum = spectrum of all wavelengths of light, like a rainbow
CHEM 161: Chapter 7 Notes page 5 of 25
Hydrogen Line Spectrum
– In contrast, when a sample of hydrogen is electrified, the resulting hydrogen emission
spectrum contains only a few discrete lines:
These discrete lines correspond to specific wavelengths → specific energies
→ The hydrogen atoms’ electrons can only emit certain energies
→ The energy of the electrons in the atom must also be quantized.
– Thus, Planck’s postulate that energy is quantized applies to the electrons within an
atom as well.
– Each element has a unique line spectrum
→ emission spectra can be used to identify unknown elements in chemical analysis
→ the element’s line spectrum is often called its "atomic fingerprint"
THE BOHR MODEL
A Danish physicist named Niels Bohr used the results from the hydrogen emission spectrum
to develop a quantum model for the hydrogen atom.
Bohr Postulates: Bohr Model of the Atom
1. Energy-level Postulate
– An electron in a hydrogen atom may only exist in discrete (quantized), circular orbits
around the nucleus
– "tennis ball and stairs" analogy for electrons and energy levels
– a ball can bounce up to or drop from one stair to another, but it can never sit
halfway between two levels
– Each orbit has a specific energy associated with it, indicated as n=1, 2, 3,...
– ground state or ground level (n = 1): lowest energy state for a one-electron atom
– when the one electron is in the lowest energy orbit
– excited state: when the electron is in a higher energy orbit (n = 2,3,4,...)
CHEM 161: Chapter 7 Notes page 6 of 25
2. Transitions Between Energy Levels
– When the atom absorbs energy, the electron can jump from a lower energy orbit to a
higher energy orbit.
– When an electron drops from a higher energy level to a lower energy level, the atom
releases energy, sometimes in the form of visible light.
– Note also that as n increases, the difference in energy between levels decreases.
→ Higher energy levels are closer to one another than the lower energy levels.
The energy absorbed or emitted by an electron when it moves from one energy level to
another can be determined using the following formula:
⎛ 1 1 ⎞ where nfinal and ninitial are the electron’s final
ΔE = −2.178×10-18 J ⎜ 2 − 2
⎜n
⎟
⎟ and initial energy levels (or orbits)
⎝ final ninitial ⎠
Ex. 1: Calculate the energy emitted by an electron when it drops from energy level 4 down
to energy level 2. (Remember to indicate the correct sign for energy lost or gained.)
CHEM 161: Chapter 7 Notes page 7 of 25
Ex. 2: Calculate the wavelength (in nm) corresponding to the energy determined in Ex. 1,
and indicate the color of the light emitted using the visible spectrum on page 4.
wavelength = __________________ color = _______________________
7.5 ELECTRONS AS WAVES
Dual Nature of the Electron
Louis de Broglie (1924)
– If light can behave like a wave and a particle
→ matter (like an electron) can behave like waves.
– If the electron behaves like a circular wave oscillating around the nucleus
→ an electron can only have specific wavelengths to form a continuous wave.
→ other wavelengths would ultimately cancel one another and not form a
continuous wave.
Topview of a standing wave Sideview of linear and circular standing waves
(Images from http://dev.physicslab.org/Document.aspx?doctype=3&filename=AtomicNuclear_deBroglieMatterWaves.xml)
If an electron can only have specific wavelengths
→ that electron can only have specific corresponding frequencies and energies.
Thus, setting the following equations equal to one another,
hc
for wave: E = mc2 for matter: E =
λ
h
and changing the c to any velocity, v, gives the de Broglie relation: λ=
mv
CHEM 161: Chapter 7 Notes page 8 of 25
The de Broglie relation is used to determine the wavelength of any matter given its velocity
and mass.
h
λ=
mv
Ex. 1 a. A baseball with mass 0.143 kg is thrown at a velocity of 45 m/s (~101 mph);
calculate the wavelength (in m) associated with the baseball’s motion.
b. How does the baseball’s λ compare in size to the baseball (diameter≈0.08 m)?
Explain.
Ex. 2 a. Calculate the wavelength (in m) of an electron also traveling at 45 m/s.
(The mass of the electron is 9.1095 ×10–31 kg.)
b. How does the electron’s λ compare in size to the electron (diameter ≈ 10–15 m)?
Explain.
Thus, although all matter can have wave properties, such properties are only significant
for submicroscopic particles.
CHEM 161: Chapter 7 Notes page 9 of 25
Quantum Mechanical Model
In 1920s, a new discipline, quantum mechanics, was developed to describe the motion of
submicroscopic particles confined to tiny regions of space.
– Quantum mechanics makes no attempt to specify the position of a submicroscopic particle
at a given instant or how the particle got there
– It only gives the probability of finding submicroscopic particles.
– Just like video footage of a location (e.g. food court at the mall) may allow you to predict
where people are likely to be but not the exact location for one person at a future time
→ Instead we “take a snapshot” of the atom at different times and “see” where the electrons
are likely to be found (See Fig. 7.20 on p. 325).
→ View “Bizarre Quantum Mechanics Explained…” animation
Werner Heisenberg (1927); Heisenberg Uncertainty Principle
– For very small particles (e.g. an electron), it is impossible to know precisely the particle’s
position and its momentum (= mass×velocity).
→ We cannot know the exact motion of an electron as it moves around the nucleus.
– Heisenberg’s Uncertainty Principle is expressed mathematically as
h where Δx=uncertainty in position,
Δx · mΔv ≥ mΔv=uncertainty in momentum, and
4π
h is Planck’s constant.
Example: a. Calculate the uncertainly in position for a baseball given the baseball’s mass
of 0.143 kg and an uncertainty in velocity of 0.45 m/s (~1 mph), then
b. Calculate the uncertainly in position for an electron (mass=9.1095 ×10–31 kg)
with the same uncertainty in velocity of 0.45 m/s.
c. How do the uncertainties in position compare to the size of the baseball
(diameter≈0.08 m)? To the size of the electron (diameter≈10-15 m)?
CHEM 161: Chapter 7 Notes page 10 of 25
Limitations of the Bohr Model → Quantum Mechanical Model
– Unfortunately, the Bohr Model failed for every other element with more than one proton or
electron. (The multiple electron-nuclear attractions, electron-electron repulsions, and
nuclear-nuclear repulsions make other atoms much more complicated than hydrogen.)
→ Most of the energy levels split into sublevels labeled s, p, d, and f.
ORBITAL ENERGY LEVELS
Orbital energy levels in the hydrogen atom
3s ___ 3p ___ ___ ___ 3d ___ ___ ___ ___ ___
Energy 2s ___ 2p ___ ___ ___
1s ___
Note that for hydrogen, all of the orbitals within the same principal
quantum number, n, have the same energy (are degenerate).
In polyelectronic atoms, the presence of more than one electron causes electron-electron
repulsions that result in a change in the energies of the various sublevels within the atom.
Orbital energy levels in polyelectronic atoms (every atom but hydrogen)
3d ___ ___ ___ ___ ___
3p ___ ___ ___
3s ___
Energy
2p ___ ___ ___
2s ___
1s ___
Note that for in polyelectronic atoms (containing more than one electron),
only the orbitals within the same sublevel are degenerate.
CHEM 161: Chapter 7 Notes page 11 of 25
7.6 QUANTUM NUMBERS AND ELECTRON SPIN
Erwin Schrödinger (1926)
– developed a differential equation to find the electron's wave function (ψ), and the square
of the wave function (ψ2) indicates the probability of finding the electron near a given point
– probability density for an electron is called the "electron cloud" or orbital
→ each atomic orbital has a distinct “shape”
– Each orbital is identified by a set of three integers called quantum numbers.
Quantum Numbers, Energy Levels, and Orbitals
– FOUR quantum numbers describe distribution and behavior of electrons in atoms
– Each wave function (ψ) corresponds to a set of 3 quantum numbers and is referred to as
an atomic orbital.
First (or Principal) Quantum Number (n): n=1,2,3,...
– relates the average distance of the electron from nucleus
– higher n means electron is further from nucleus, in a higher-energy (less stable) orbital
Second (or Angular Momentum) Quantum Number (l): l=0,…, n-1
→ Sublevels (s, p, d, f): gives "shape" of the electron clouds associated with each orbital
– The limitations on n and l
→ for n=1, l =0 → the 1s sublevel → for n=4, l =0 → the 4s sublevel
n=4, l =1 → the 4p sublevel
→ for n=2, l =0 → the 2s sublevel n=4, l =2 → the 4d sublevel
n=2, l =1 → the 2p sublevel n=4, l =3 → the 4f sublevel
→ for n=3, l =0 → the 3s sublevel
n=3, l =1 → the 3p sublevel
n=3, l =2 → the 3d sublevel
Third (or Magnetic) Quantum Number (ml): ml= -l,…,0,…,l
→ indicates the number of orbitals in each sublevel
→ l =0 (s orbital): → ml=0 → only one type of s orbital
→ l =1 (p orbitals): → ml=-1, 0, 1 → 3 types of p orbitals: px, py, pz
→ l =2 (d orbitals): → ml=-2, -1, 0, 1, 2 → 5 types of d orbitals: dxy, dyz, dxz, d z2 , d x 2 − y 2
→ l =3 (f orbitals): → ml=-3, -2, -1, 0, 1, 2, 3 → 7 types of f orbitals
Fourth (or Electron Spin) Quantum Number (ms): ms= +½ and -½
– This will be discussed in more detail later in the chapter.
CHEM 161: Chapter 7 Notes page 12 of 25
Ex. 1: What are valid values for l when n=3? ______________________________
Ex. 2: What are valid values for ml when n=4 and l=2? ______________________________
Ex. 3: Check all of the following sets of quantum numbers that are valid, and for those that are
not valid, explain why.
a. n=2, l=2, ml=1, and ms= +½.
b. n=1, l=0, ml=0, and ms= +½.
c. n=4, l=1, ml=1, and ms= −½.
d. n=3, l=3, ml=2, and ms= −½.
Note that quantum numbers allow us to determine the number of each type of orbital for each
principal energy level (n value) and sublevel (l value).
7.7 THE SIZES AND SHAPES OF ATOMIC ORBITALS
s orbitals: spherical (see Fig. 7.20 on p. 325)
– size of the orbitals increase with n
– since number of protons, neutron, and
electrons increase with n
p orbitals: dumbbell-shaped (see Fig. 7.22 on p. 327)
– 3 types: px, py, pz (where x, y, and z indicates axis on which orbital aligns)
– Figure (a) below shows the probability distribution for a pz orbital.
– Figure (b) below shows the boundary surface representations of the p orbitals.
CHEM 161: Chapter 7 Notes page 13 of 25
d orbitals: various shapes (see Fig. 7.23 on p. 327)
– 5 types: dxy, dxz, dyz,, d x 2 − y 2 , d z 2
f orbitals: You won’t be tested on f orbitals.
Be able to identify a p or d orbital given its image.
The interesting part...
– Consider the figure at the right
showing the electron distribution
for the 2px orbital.
– Note that there is a “node” (zero
probability) at the origin, so the
electron is never there.
→ How can the electron be exist in
either of the two lobes without
going through the origin?
CHEM 161: Chapter 7 Notes page 14 of 25
7.8 THE PERIODIC TABLE AND FILLING THE ORBITALS OF MULTIELECTRON ATOMS
Electron Configuration: Shorthand description of the arrangement of electrons by sublevel
according to increasing energy
Aufbau (Building-Up) Principle
– Electrons are distributed in orbitals of increasing energy, where the lowest energy orbitals
are filled first.
– Once an orbital has the maximum number of electrons it can hold, it is considered “filled.”
– Remaining electrons must then be placed into the next highest energy orbital, and so on.
– Parking garage analogy
Orbitals in order of increasing energy:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 5d < 6p
REMEMBER!
Each orbital can hold 2 electrons.
– Each s orbital can hold 2 electrons.
– A set of three p orbitals can hold 6 electrons.
– A set of five d orbitals can hold 10 electrons.
– A set of seven f orbitals can hold 14 electrons.
Ex. 1 Li → atomic number=3 → 3 e-
electron configuration for Li: ____________________________
Ex. 2 electron configuration for F: ___________________________________________
Ex. 3 electron configuration for Fe: __________________________________________
Electron configurations of atoms with many electrons can become cumbersome.
→ Abbreviated electron configurations (“noble-gas core” notation):
– Since noble gases are at the end of each row in the Periodic Table, all of their electrons
are in filled orbitals.
– Such electrons are called “core” electrons since they are more stable (less reactive)
when they belong to completely filled orbitals.
valence electrons: electrons that are in the outermost shell (unfilled orbitals)
Noble gas electron configurations are used to abbreviate the “core” electrons of all elements.
[He] = 1s2 [Kr] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
[Ne] = 1s2 2s2 2p6 [Xe] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
[Ar] = 1s2 2s2 2p6 3s2 3p6
CHEM 161: Chapter 7 Notes page 15 of 25
Writing Electron Configurations Using the Periodic Table
– The Periodic Table's shape actually corresponds to the filling of energy sublevels.
– See Fig. 7.27 (p. 306), to see how electrons for each element are distributed into the
energy sublevels.
Example: Write the electron configurations for the following using Noble Gas core notation:
[Fe] = 1s2 2s2 2p6 3s2 3p6 4s2 3d6
= __________________________________________________________
[Cd] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
= __________________________________________________________
[Ni] = __________________________________________________________
[Y] = __________________________________________________________
[I] = __________________________________________________________
CHEM 161: Chapter 7 Notes page 16 of 25
Exceptions to the Aufbau (Building-Up) Principle
Atoms gain extra stability with half-filled or completely filled d subshells.
→ If we can fill or half-fill a d subshell by promoting an electron from an s orbital to a d orbital,
we do so to gain the extra stability.
Example: Write the electron configurations for the following using Noble Gas core notation:
Transition Metal expected electron configuration actual electron configuration
chromium
copper
silver
Mo (molybdenum)
ATOMIC ORBITAL DIAGRAMS
In 1925, two graduate students in the Netherlands, Samuel Goudsmit and George Uhlenbeck
two additional electron energy states not accounted for by Schrödinger’s equations.
→ Electrons have a quantum mechanical property called spin, with two possible orientations:
spin ↑ or spin ↓.
– Two other scientists, Otto Stern and Walther Gerlach, observed experimental evidence
of electron spin when they shot a beam of Ag atoms through a non-uniform magnetic
field, and the field was split in two.
→ The Fourth (or Electron Spin) Quantum Number (ms = +½ or -½), indicates if the
electron in a specific orbital (indicated by the first 3 quantum #’s) is spin ↑ or spin ↓.
CHEM 161: Chapter 7 Notes page 17 of 25
electron configuration: shorthand description of an atom’s electrons among sublevels
atomic orbital diagram: shows the energy sublevels within an atom and the arrangement
and spin of electrons within each orbital in the sublevels
Pauli Exclusion Principle: no 2 e-s in an atom can have same four quantum #s
→ Two electrons in the same orbital must have opposite spins
– For example, with the helium atom, there are three ways to represent two electrons in
1s orbital (where spin is represented with the electron pointing up or down):
(a) (b) (c)
for He: ↑↑ ↓↓ ↑↓
1s 1s 1s
– but the Pauli exclusion principle rules out (a) and (b) since these show two electrons in
the same orbital with the same spin.
Hund's Rule: the most stable arrangement of electrons in subshells has the greatest
number of parallel spins
– i.e., distribute electrons with same spin (up or down) and do not pair
electrons until each orbital in the subshell has an electron
For example, if carbon’s electron configuration is: 1s2 2s2 2p2
→ carbon’s orbital diagram can be shown with the sublevels further from the nucleus
having higher energy and the electrons within each orbital:
↑↓
↑↓ 2p
(a) ↑↓ 2s
1s
↑ ↓
↑↓ 2p
Energy
(b) ↑↓ 2s
1s
↑ ↑
↑↓ 2p
(c) ↑↓ 2s
1s
– but using Hund's rule, we know (c) would be the most stable.
CHEM 161: Chapter 7 Notes page 18 of 25
General Rules for Assigning Electrons in Atomic Orbital Diagrams
1. First, determine the electron configuration.
2. There is only one s orbital for each level: one 1s, one 2s, one 3s, etc.
– There are 3 p orbitals for each p sublevel.
– There are 5 d orbitals for each d sublevel.
3. Each orbital orbital can only hold 2 electrons
→ Each s orbital can hold 2 e–, the 3 p orbitals can hold 6 e–, the 5 d orbitals can hold 10 e–.
4. Electrons in the same orbital must have opposite spins.
5. To fill sublevels, put one electron in each orbital (with same spin) before pairing.
Orbital energy levels in atoms with more than one electron (every atom but hydrogen)
3d ___ ___ ___ ___ ___
3p ___ ___ ___
3s ___
Energy
2p ___ ___ ___
2s ___
1s ___
Ex. 1 Use full notation to write the electron configuration then draw the atomic orbital diagram
for oxygen.
Ex. 2 Use core notation to write the electron configuration then draw the atomic orbital
diagram for the valence electrons in phosphorus (electrons in the outermost shell).
CHEM 161: Chapter 7 Notes page 19 of 25
Ex. 3 Use core notation to write the electron configuration then draw the atomic orbital
diagram for the valence electrons in cobalt (electrons in the outermost shell).
Ex. 4 Use core notation to write the electron configuration then draw the atomic orbital
diagram for the valence electrons in silver (electrons in the outermost shell).
7.9 Electron Configurations of Ions
Ions of the Main Group (Representative) Elements
– Representative elements generally form ions—ie. gain or lose electrons—to achieve a
noble gas electron configuration
→ Ions from representative metals are usually isoelectronic with—i.e. have the same
electron configuration as—one of the noble gases!
Electron Configurations of Cations and Anions
For IONS, one must account for the loss or gain of electrons:
# electrons = atomic # – (charge = change in # of valence electrons)
Or you can simply use the Periodic Table
– Find out with which element the ion is isoelectronic
– Move to the left for electrons lost or to the right for electrons gained
→ write the electron configuration for that element
Example 1: Fill in the blanks for the following ions:
Isoelectronic Isoelectronic
Electron Config. using Electron Config. using
Ion with what Ion with what
core notation core notation
element? element?
Na+ I–
P–3 Ba+2
Al+3 Ti+4
CHEM 161: Chapter 7 Notes page 20 of 25
Cations from Transition Metals, Sn, Pb
– Transition metals lose s electrons before the d electrons when forming cations
Electron Configuration Electron Configuration
Atom Ion
using core notation using core notation
Zn Zn+2
Sn Sn+4
Cu Cu+
Cd Cd+2
Ex. Write the electron configurations for the following:
Fe atom: _________________ Fe+2 ion: _______________ Fe+3 ion: _______________
Example: Given the electron configurations of Fe+2 and Fe+3, predict which ion is more stable,
and explain your choice.
7.10 PERIODIC TRENDS IN ATOMIC PROPERTIES
Atomic Radius (or Size): distance from the nucleus to the outermost electrons
CHEM 161: Chapter 7 Notes page 21 of 25
Periodic Trend for Atomic Radius
ATOMIC RADIUS
– Increases down a group: More p+, n, and e– → bigger radius
– Decreases from left to right along a period:
– Electrons that lie between the nucleus and the outermost electrons shield or screen the
outermost electrons, preventing them from experiencing the full charge of the nucleus.
→ Effective nuclear charge (Zeff) can be approximated by the following:
Zeff = # of protons – # of core electrons
– Number of p+ and e– increases, but electrons go into same subshell, and other
valence electrons cannot shield each other from the attractive force of the nucleus.
– The higher the effective nuclear charge (Zeff) → smaller radius
Compare atoms of aluminum and chlorine:
Trend from top to bottom → like a snowman
Trend from left to right → like a snowman
that fell to the right
IONIC RADII: distance from the nucleus to the
outermost electrons in an ion
– An atom loses electrons to form a cation.
→ A cation has a smaller radius than its corresponding atom.
– An atom gains electrons to form an anion.
→ An anion has a larger radius than its corresponding atom.
11 p+ loses 1 e– 11 p+ 17 p+ gains 1 e– 17 p+
11 e– 10 e– 17 e– 18 e–
Na atom Na+ ion Cl atom
Cl– ion
Example: Order the following in terms of increasing ionic radius: I−, F−, Cl−, P3−, S2−.
_______ < _______ < _______ < _______ < _______
smallest radius largest radius
CHEM 161: Chapter 7 Notes page 22 of 25
7.11 IONIZATION ENERGIES (IE)
First Ionization Energy: Energy necessary to remove the first electron from a neutral atom
in gaseous state to form the positively charged ion.
X(g) → X+(g) + e−
Consider the following ionization energies for magnesium:
Mg(g) → Mg+(g) + e− IE1 = 738 kJ/mol
Mg+(g) → Mg2+(g) + e− IE2 = 1451 kJ/mol
Thus, to completely ionize a magnesium atom requires the following:
Mg(g) → Mg2+(g) + 2 e− total IE = 738 + 1451 kJ/mol
= 2189 kJ/mol
Consider the following first ionization energies for various elements:
Periodic Trend for First Ionization Energy
– Decreases down a group:
– The bigger the atom, the farther away electrons are from the positively charged nucleus.
→ Valence electrons are less strongly attracted and are more easily removed.
– Increases from left to right along a period:
– Effective nuclear charge increases from left to right across the periodic table.
→ As the attraction between a valence electron and the nucleus increases, more
energy is required to remove a valence electron from the neutral atom.
CHEM 161: Chapter 7 Notes page 23 of 25
Variations in Successive Ionization Energies (IE)
– Recognize that it becomes more difficult to remove electrons from stable ions, so ionization
energies increase with an increasing number of electrons removed.
We can indicate first and successive ionization energies in the following way:
First ionization energy = IE1
Second ionization energy = IE2
Third ionization energy = IE3
Consider the following ionization energies for aluminum:
Al(g) → Al+(g) + e− IE1 = 580 kJ/mol
Al+(g) → Al2+(g) + e− IE2 = 1815 kJ/mol
Al2+(g) → Al3+(g) + e− IE3 = 2740 kJ/mol
Al3+(g) → Al4+(g) + e− IE4 = 11,600 kJ/mol
– Note the large jump between the 3rd and 4th IE’s for aluminum.
– Note that removing an electron from an Al3+ ion requires much more energy than
removing an electron from a neutral Al atom or the previous ions formed.
– Note that Al3+ is a stable ion (with a positive charge AND a noble gas electron
configuration), so an enormous amount of energy is required to remove an electron
from a stable ion and make it unstable.
Consider these Ionization Energies (in kJ/mol):
Note that for the elements included in the table above that the largest jump in ionization
energies occurs when an electron is being removed from a stable ion with a Noble gas
electron configuration.
CHEM 161: Chapter 7 Notes page 24 of 25
Ex. 1: Between which two ionization energies (e.g. IE1 & IE2, IE2 & IE3, etc.)
would you expect there to be the largest jump for the following?
a. Ba: Between _____ and _____ b. Ti: Between _____ and _____
Ex. 2: This 2nd period element has a large jump between IE5 and IE6: _____
EINSTEIN’S SPLIT WITH MAINSTREAM PHYSICS
“Quantum mechanics is very impressive. But an inner voice tells me that it is not yet the real thing. The theory
produces a great deal but hardly brings us closer to the secret of the Old One. I am at all events convinced that
He does not play dice.” Einstein in a letter to Max Born, quoted from R.W. Clark, Einstein: The Life and Times.
Einstein's Grand Quest for a Unified Theory (from Sept. 2004 issue of Discover magazine)
by Tim Folger
Einstein’s split with mainstream physics came at the very height of his career. In 1927, when he was 48, the world’s leading
physicists gathered at a conference in Brussels to debate an issue that remains contentious to this day: What does quantum
mechanics have to say about reality? Einstein had won the Nobel Prize in physics for research that showed that light consists
of particles of energy—research that laid the groundwork for quantum mechanics. Yet he dismissed the new theory out of
hand. At the conference, he clashed with the great Danish physicist Niels Bohr, launching a feud that would last until Einstein’s
death in 1955.
Bohr championed the strange new insights emerging from quantum mechanics. He believed that any single particle—be it an
electron, proton, or photon—never occupies a definite position unless someone measures it. Until you observe a particle, Bohr
argued, it makes no sense to ask where it is: It has no concrete position and exists only as a blur of probability.
Einstein scoffed at this. He believed, emphatically, in a universe that exists completely independent of human observation. All
the strange properties of quantum theory are proof that the theory is flawed, he said. A better, more fundamental theory would
eliminate such absurdities. “Do you really believe that the moon is not there unless we are looking at it?” he asked.
“He saw in a way more clearly than anyone else what quantum mechanics was really like,” British physicist Julian Barbour
says. “And he said, ‘I don’t like it.’” In the years after the conference in Brussels, Einstein leveled one attack after another at
Bohr and his followers. But for each attack Bohr had a ready riposte. Then in 1935 Einstein devised what he thought would be
the fatal blow. Together with two colleagues in Princeton, Nathan Rosen and Boris Podolsky, he found what appeared to be a
serious inconsistency in one of the cornerstones of quantum theory, the uncertainty principle.
Formulated in 1927 by the German physicist Werner Heisenberg, the uncertainty principle puts strict limits on how accurately
one can measure the position, velocity, energy, and other properties of a particle. The very act of observing a particle also
disturbs it, Heisenberg argued. If a physicist measures a particle’s position, for example, he will also lose information about its
velocity in the process.
Einstein, Podolsky, and Rosen disagreed, and they suggested a simple thought experiment to explain why: Imagine that a
particle decays into two smaller particles of equal mass and that these two daughter particles fly apart in opposite directions.
To conserve momentum, both particles must have identical speeds. If you measure the velocity or position of one particle, you
will know the velocity or position of the other—and you will know it without disturbing the second particle in any way. The
second particle, in other words, can be precisely measured at all times.
Einstein and his collaborators published their thought experiment in 1935, with the title “Can Quantum-Mechanical Description
of Physical Reality Be Considered Complete?” The paper was in many ways Einstein’s swan song: Nothing he wrote for the
rest of his life would match its impact. If his critique was right, quantum mechanics was inherently flawed.
Bohr argued that Einstein’s thought experiment was meaningless: If the second particle was never directly measured, it was
pointless to talk about its properties before or after the first particle was measured. But although quantum physics eventually
carried the day, it wasn’t until 1982, when the French physicist Alain Aspect constructed a working experiment based on
Einstein’s ideas, that Bohr’s argument was vindicated. In 1935 Einstein was convinced that he had refuted quantum
mechanics. And from then until his death 20 years later, he devoted nearly all his efforts to the search for a unified field theory.
CHEM 161: Chapter 7 Notes page 25 of 25
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