Chapter 7: ELECTRONS IN ATOMS AND PERIODIC PROPERTIES
Problems: 7.1-7.62, 7.66-7.74, 7.77-7.129, 7.131, 7.133, 7.135-7.138, 7.140-7.142, 7.144, 7.148
ELECTROMAGNETIC RADIATION
Electromagnetic (EM) Spectrum: a continuum of the different forms of electromagnetic
radiation or radiant energy (Fig. 7.1 on p. 318)
– 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 LIGHT WAVES
English chemist and physicist Michael Faraday
established the basis for electromagnetism, and
Scottish scientist James Clerk Maxwell expanded our
understanding of electromagnetism, explaining 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 v1214 page 1 of 26 Electromagnetic Radiation
– EM waves have the following properties: wavelength and frequency (see Fig. 7.3).
wavelength (λ=Greek “lambda”): distance between successive peaks
λ=
distance
; generally in units of m, cm, or nm
cycle
frequency (ν=Greek “nu”): number of waves passing a given point in 1 s
ν=
1
cycle
cycle
; generally in hertz (Hz) =
or
s
time
s
CHEM 161: Chapter 7 Notes v1214 page 2 of 26 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
×
length length
=
time
cycle
ν
×
cycle
time
Know how to convert between wavelength and frequency using the speed of light!
Ex. 1 The frequency of solar rays ranges from 6.0×1013 Hz to 2.0×1015 Hz,
a. What is the corresponding range of wavelengths (in m) for this solar radiation?
b. In what part(s) of the EM spectrum does this radiation fall? ____________________
CHEM 161: Chapter 7 Notes v1214 page 3 of 26 Ex. 2 The least expensive type of red laser pointer has a wavelength of about 635 nm.
What is the frequency of this light in Hz?
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), Scottish scientist
– all forms of radiant energy consist of electromagnetic waves
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 (ν )
– When ν < ν → no electrons are emitted
– When ν > ν → 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.
0
0
0
CHEM 161: Chapter 7 Notes v1214 page 4 of 26 Energy is proportional to the frequency (ν) and wavelength (λ) of radiation, and the
proportionality constant (h) is now called Planck's constant
where
h = 6.626×10–34 J·s
Ex. 1. Green laser pointers emit light strongly at a wavelength of 532 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?
The Photoelectric Effect
When ν < ν → no electrons are emitted
0
CHEM 161: Chapter 7 Notes v1214 When ν > ν → electrons are emitted and
0
page 5 of 26 The Photoelectric Effect and Work Function (Φ ) of a Metal
– The work function of a metal (symbolized using the Greek letter Phi, Φ) is the minimum
amount of energy required to emit an electron from the surface of a metal.
Φ = hν
Work function:
where
0
h = 6.626×10–34 J·s
Ex. 2 a. The work function for lithium is 4.65×10-19 J. Calculate the longest wavelength
(in nm) of light that can cause an electron to be ejected from a lithium atom.
b. Use the visible spectrum below to determine the color of the light.
c. If excess energy absorbed by an emitted electron is converted into kinetic energy, calculate
the velocity of an electron ejected from a lithium atom after it absorbs violet light with a
wavelength of 402 nm. (K.E. = ½ mv2 and melectron=9.109×10-31 kg)
CHEM 161: Chapter 7 Notes v1214 page 6 of 26 7.2 ATOMIC SPECTRA
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
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"
CHEM 161: Chapter 7 Notes v1214 page 7 of 26 THE BOHR MODEL (Fabric of the Cosmos video: t=6:45)
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,...)
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.
CHEM 161: Chapter 7 Notes v1214 page 8 of 26 The energy absorbed or emitted by an electron when it moves from one energy level to
another can be determined using the following formula:
where nfinal and ninitial are the electron’s final
and initial energy levels (or orbits)
⎛ 1
1 ⎞
ΔE = −2.178×10-18 J ⎜⎜ 2 − 2 ⎟⎟
⎝ n final ninitial ⎠
Sign conventions: E > 0 when an atom gains energy; E < 0 when an atom loses energy.
Ex. 1: Calculate the energy emitted by an electron when it drops from energy level 5 down
to energy level 2. (Remember to indicate the correct sign for energy lost or gained.)
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 6.
wavelength = __________________ color = _______________________
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 to explain 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 determine where its
electrons are likely to be found (See Fig. 7.24 on p. 343).
→ View Fabric of the Cosmos (Double-Slit Experiment): t=12:05
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
4π
Δx · mΔv ≥
CHEM 161: Chapter 7 Notes v1214 where Δx=uncertainty in position,
mΔv=uncertainty in momentum,
page 9 of 26 and h is P
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
compare to the size of the baseball (diameter≈0.08 m).
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, then compare to the size of
the electron (diameter≈10-15 m).
Thus, the Uncertainty Principle is only relevant for very small particles, like an electron.
7.5 ELECTRON 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 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)
CHEM 161: Chapter 7 Notes v1214 page 10 of 26 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,
for matter: E = mc2
for wave: E =
hc
λ
and changing the c to any velocity, v, gives the de Broglie relation:
λ=
h
mv
This is used to determine the wavelength of any matter given its velocity and mass.
Ex. 1 a. A baseball with mass 0.143 kg is thrown at a velocity of 42 m/s (~95 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)?
Ex. 2 a. Calculate the wavelength (in m) of an electron also traveling at 42 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)?
Thus, although all matter can have wave properties, such properties are only significant
for submicroscopic particles. The wavelength in these very small particles is about the
same size or much larger than the particle, so the wave properties affect the behavior of the
particle.
CHEM 161: Chapter 7 Notes v1214 page 11 of 26 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 v1214 page 12 of 26 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 (ψ ) 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.
2
Quantum Numbers, Energy Levels, and Orbitals
– FOUR quantum numbers describe the distribution and behavior of electrons in atoms.
– Each wave function (ψ) corresponds to a set of 3 quantum numbers and refers to a specific
atomic orbital.
First (or Principal) Quantum Number (n): n=1,2,3,...
– relates the average distance of the electron from the nucleus
– The higher n is, the farther the electron is from the nucleus and the higher in energy
(less stable) the 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 , dx2 − y2
→ 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.
Ex. 1: What are valid values for ℓ when n=4? ______________________________
CHEM 161: Chapter 7 Notes v1214 page 13 of 26 Ex. 2: What are valid values for mℓ when n=5 and ℓ =2? ______________________________
Ex. 3: Fill in the missing set(s) of quantum numbers given the quantum numbers provided:
a. If n=2 and ℓ= 1, mℓ can be____________ given that ms= +½.
b. If n=3, ℓ can be ____________ given that mℓ=+2 and ms= +½.
c. If n=4, ℓ can be ____________ given that mℓ=-2 and ms can be ____________.
Note that quantum numbers allow us to determine the number of each type of orbital in each
principal energy level (n value) and sublevel (ℓ value).
7.7 THE SIZES AND SHAPES OF ATOMIC ORBITALS
The images below are boundary surface representations within which there’s a 90% probability
of finding the electron in a given orbital.
– These representations show the relative size, shape, and orientation of the various orbitals.
s orbitals: spherical (see Fig. 7.24 on p. 343)
– size of the orbitals increase with n
– since number of protons, neutron, and electrons increase with n
Boundary surface representations of the 1s, 2s, and 3s orbitals
showing the increase in size with n value.
A cross section of the hydrogen 1s
orbital probability distribution
divided into thin spherical shells.
p orbitals:
dumbbell-shaped (see Fig. 7.26 on p. 345)
– 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 v1214 page 14 of 26 d orbitals:
boundary surface representations of the d orbitals (see Fig. 7.27 on p. 345)
– 5 types: dxy, dxz, dyz,, d x2 −y2 , dz2
f orbitals: You won’t be tested on f orbitals.
Be able to identify a p or d orbital given its image.
CHEM 161: Chapter 7 Notes v1214 page 15 of 26 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 and near the
origin, so the electron is never
there.
→ How can the electron be present
in either of the two lobes without
going through the origin?
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 eelectron configuration for Li: ____________________________
Ex. 2
electron configuration for S: ___________________________________________
Ex. 3
electron configuration for Co: __________________________________________
CHEM 161: Chapter 7 Notes v1214 page 16 of 26 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
[Ne] = 1s2 2s2 2p6
[Ar] = 1s2 2s2 2p6 3s2 3p6
[Kr]
[Xe]
= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
Writing Electron Configurations Using the Periodic Table
– The Periodic Table's shape actually corresponds to the filling of energy sublevels.
– See Fig. 7.32 (p. 350), to see how electrons for each element are distributed into the
energy sublevels.
CHEM 161: Chapter 7 Notes v1214 page 17 of 26 Example: Write the electron configurations for the following using Noble Gas core notation:
[Co] = 1s2 2s2 2p6 3s2 3p6 4s2 3d7
= __________________________________________________________
[Cd] = 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10
= __________________________________________________________
[Ni]
= __________________________________________________________
[I]
= __________________________________________________________
Exceptions to the Aufbau (Building-Up) Principle (for Cr, Mo, W, Cu, and Ag)
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
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 ↓.
In 1925, two graduate students in the
Netherlands, Goudsmit and Uhlenbeck,
found two additional electron energy
states not accounted for by
Schrödinger’s equations.
→ Electrons have a quantum
mechanical property called spin,
with two orientations: spin ↑ or ↓.
– Two other scientists, Otto Stern
and Walther Gerlach, observed
experimental evidence of spin
when they shot a beam of Ag
atoms through a non-uniform
magnetic field, and the field split
in two.
CHEM 161: Chapter 7 Notes v1214 page 18 of 26 ATOMIC ORBITAL DIAGRAMS
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 (↑ or ↓) 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):
for He:
(a)
(b)
(c)
↑↑
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
↑
Energy
(b)
2p
↑↓
2s
↑↓
↓
1s
↑
2p
↑↓
(c)
↑
2s
↑↓
1s
– but using Hund's rule, we know (c) would be the most stable.
CHEM 161: Chapter 7 Notes v1214 page 19 of 26 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 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.
6. Finally, write the quantum numbers for the outermost electrons in the atom.
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 fluorine, then indicate the set of quantum numbers for each electron.
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)
and indicate the quantum numbers for each valence electron.
CHEM 161: Chapter 7 Notes v1214 page 20 of 26 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) then
indicate the quantum numbers for the valence electrons in the highest energy orbitals.
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:
Ion
Isoelectronic
with what
element?
Electron Config. using
core notation
Ion
Na+
I–
P–3
Ba+2
Al+3
Ti+4
CHEM 161: Chapter 7 Notes v1214 Isoelectronic
with what
element?
Electron Config. using
core notation
page 21 of 26 Cations from Transition Metals, Sn, Pb
– Transition metals lose s electrons before the d electrons when forming cations
Atom
Electron Configuration
using core notation
Electron Configuration
using core notation
Ion
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 THE SIZES OF ATOMS AND IONS
Atomic Radius (or Size): distance from the nucleus to the outermost electrons
CHEM 161: Chapter 7 Notes v1214 page 22 of 26 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 RADIUS: 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+
11 e–
loses 1 e–
Na atom
11 p+
10 e–
17 p+
17 e–
Na+ ion
Cl atom
17 p+
18 e–
gains 1 e–
Cl– ion
Example: Order the following in terms of increasing ionic radius: I , F , Cl , P , S .
−
−
−
3−
2−
_______ < _______ < _______ < _______ < _______
smallest radius
CHEM 161: Chapter 7 Notes v1214 largest radius
page 23 of 26 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) + e
X(g)
−
Consider the following ionization energies for magnesium:
→ Mg+(g) + e
Mg+(g) → Mg2+(g) + e
Mg(g)
IE1 = 738 kJ/mol
−
IE2 = 1451 kJ/mol
−
Thus, to completely ionize a magnesium atom requires the following:
Mg(g)
kJ/mol
→ Mg2+(g) + 2 e
total IE = 738 + 1451
−
= 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 v1214 page 24 of 26 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
Al+(g) → Al2+(g) + e
Al2+(g) → Al3+(g) + e
Al3+(g) → Al4+(g) + e
−
−
−
−
IE1 = 580 kJ/mol
IE2 = 1815 kJ/mol
IE3 = 2740 kJ/mol
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 v1214 page 25 of 26 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: _____
7.12 ELECTRON AFFINITIES
electron affinity (EA): energy associated with 1 mole of gaseous atoms to gain 1 mole of
electrons
X(g) + e– → X–(g)
– A quantitative measure of an atom’s ability to accept an electron
– The more negative an atom’s electron affinity
→ the more likely an atom can gain an electron
– Consider the electron affinities shown below.
– Fluorine has the most negative electron affinity, but notice there is no clear trend for
electron affinity, like there is with electronegativity or atomic radius.
Example: Explain the difference between electronegativity and electron affinity.
CHEM 161: Chapter 7 Notes v1214 page 26 of 26