Section Five pp. 1
Section Five: Quantum Mechanics and Chemical Periodicity
Reading: Tro Chapters 7 and 8
Recommended Problems: Mastering Chemistry Online Homework
Quantum Mechanics: An Introduction
Electromagnetic radiation exhibits wavelike properties which can be depicted in
the following representation below:
1.
2.
3.
The wavelength, λ (lower-case Greek letter lambda), is represented by number 1, or the
peak-to-peak distance, and is generally in units of meters (although usually converted to
nm). The number 2 represents the amplitude, or the height of the wave above/below the
center line. Finally, the number 3 represents the node, or the point at which an electron
occupying an orbital will NOT be found (to be discussed in more detail later). The
number of cycles per second is called the frequency, denoted with the symbol ν (the
Greek letter nu), and expressed in units of s-1 of Hz (Hertz).
Both light and waves can be characterized by wavelength and frequency, and all
electromagnetic radiation moves at the speed of light (c). Max Planck first analyzed data
from the emission of light from hot, glowing solids. He observed that the color of the
solids varied with temperature. Furthermore, Planck suggested that a relationship exists
between energy of atoms in the solid and wavelength. This led to the following equation:
νλ = c
This equation has multiple interpretations. If the wavelength is long, there will be fewer
cycles of the wave passing a point per second; thus, the frequency will be low.
Conversely, for a wave to have a high frequency, the distance between the peaks of the
wave must be small (short wavelength). In summary, an inverse relationship exists
between frequency and wavelength of electromagnetic radiation as the speed of light is
always constant.
Energy can be released (or absorbed) by atoms in discreet chunks (also known as
quantum or photons) of some minimum size; that is, energy is quantized and
emitted/absorbed in whole number units of hν. Collectively, we say that:
Section Five pp. 2
E = hν = hc/λ
where h = Planck’s constant (6.626 x 10-34 J s), ν = frequency, λ = wavelength, and c =
speed of light (3.0 x 108 m/s). Energy (E) is generally expressed in units of Joules (J),
where this actually implies J/atom.
Problem #1: Yellow light exhibits a wavelength of approximately 570 nm. Determine
the frequency of this light and the total energy of the photon being emitted in units of J
and kJ/mol.
Problem #2: When an electron beam strikes a block of copper, x-rays with a frequency
of 2.0 x 1018 Hz are emitted. How much energy is emitted at this wavelength by (A) an
excited copper atom when it generates an x-ray photon; (B) 1.00 mol of excited copper
atoms; and (C) 1.00 g of copper atoms?
Problem #3: A minimum energy of 495 kJ/mol is required to break the oxygen-oxygen
bond in O2. What is the longest wavelength of radiation that possesses the necessary
energy to break the bond?
Electromagnetic Spectrum
Electromagnetic waves are produced by a combination of electrical and magnetic
fields which are at right angles to one another (i.e. perpendicular). The various types of
waves include: radiowaves, microwaves, infrared, visible light, ultraviolet, x-rays, cosmic
Section Five pp. 3
rays, and gamma rays. All electromagnetic waves travel at the speed of light; they differ
from one another by their corresponding wavelengths and frequencies.
Increasing Frequency (ν) and Energy (E)
Radio
Micro
IR
Visible (ROYGBV)
UV
X-ray
Cosmic
Gamma
Increasing Wavelength (λ)
Problem #4: Arrange the following types of photons of radiation in order of increasing
speed: infrared radiation, visible light, x-rays, and microwaves. Which form of radiation
possesses the longest wavelength? Which form of radiation possesses the largest amount
of energy per photon?
Photoelectric Effect
Consider incident radiation which consists of a stream of photons of energy hν.
Once these particles strike the surface of a metal, the energy is absorbed by an electron.
If the energy of the photon is less than the energy required to remove an electron from the
metal, then an electron will NOT be ejected. Moreover, the energy required to remove an
electron from the surface of a metal is called the work function (Φ) of the metal.
However, if the energy of the photon is greater than the work function, then an electron is
ejected with a kinetic energy equal to the difference between the energy of the incoming
photon and the work function. That is,
KEelectron = ½ mv2 = hν - hνo
where m = mass, v = velocity, h = Planck’s constant, and ν = frequency. These results,
known as the photoelectric effect, can be summarized alongside with Einstein’s theory:
1.
2.
3.
An electron can be driven out of the metal only if it receives at least a certain
minimum energy, Φ, from the photon during the collision. Therefore, the
frequency of the radiation must have a certain minimum value if electrons are to
be ejected. This minimum frequency depends on the work function and hence on
the identity of the metal.
Provided a photon has enough energy, a collision results in the immediate ejection
of an electron.
The kinetic energy of the ejected electron from the metal increases linearly with
the frequency of the incident radiation according to the equation ½ mv2 = hν hνo.
Section Five pp. 4
Problem #5: Calculate the ionization energy of a rubidium atom, given that radiation of
wavelength 58.4 nm produces electrons with a speed of 2450 km/s.
Problem #6: The energy required for the ionization of a certain atom is 3.44 x 10-18 J.
The absorption of a photon of unknown wavelength ionizes the atom and ejects an
electron with velocity 1.03 x 108 m/s. Calculate the wavelength of the incident of
radiation.
Problem #7: The ionization energy of gold is 890.1 kJ/mol. Is light with a wavelength
of 225 nm capable of ionizing a gold atom in the gas phase?
Section Five pp. 5
de Broglie Relation and the Wave-Particle Duality of Matter
If electromagnetic radiation has a dual character, could it be that matter, which
has been regarded as consisting of particles, also have wavelike properties? On the basis
of theoretical observations, Louis de Broglie suggested that all particles should be
regarded as having wavelike properties. Therefore, it was proposed that:
λ = h/p = h/mv
where λ = wavelength, h = Planck’s constant, p = linear momentum (in units of kg m/s),
m = mass (kg), and v = velocity (m/s). BE CAREFUL: v is velocity and NOT frequency
(ν). Beginning chemistry students often confuse these symbols!
Problem #8: If an electron travels at a velocity of 1.000 x 107 m/s and has a mass of
9.109 x 10-28 g, what is its wavelength?
Heisenberg Uncertainty Principle
Heisenberg determined that there is a fundamental limitation to just how precisely
one can know both the position and the momentum of a particle at a given time. That is,
Δx • Δp > h/4π
where x = the location of particle, p = linear momentum, and h = Planck’s constant. This
relationship means that the more precisely we know a particle’s motion, the less precisely
we can know its momentum, and vice versa. The uncertainty principle implies that we
cannot know the exact path of the electron as it moves around the nucleus. It is therefore
not appropriate to assume that the electron is moving around the nucleus in a welldefined orbit.
Problem #9: An electron with a mass of 9.109 x 10-31 kg is known to have an uncertainty
of 1 pm in its position. Determine the uncertainty in its speed.
Section Five pp. 6
Bohr Model
Bohr proposed a model that included the idea that the electron in a hydrogen atom
moves around the nucleus only in certain allowed circular orbits. Furthermore, Bohr
concluded the following as applied to the bright-line spectrum of hydrogen:
1.
Hydrogen atoms exist in only specified energy states given by the Rydberg energy
equation:
E = -2.178 x 10-18 J (Z2/nfinal2 - Z2/ninitial2)
2.
3.
4.
where E = Energy (J), Z = atomic number, and n = orbital level.
Hydrogen atoms can absorb only certain amounts of energy, and no others.
When excited hydrogen atoms lose energy, they lose only certain amounts of
energy, emitted as photons.
The different photons given off by hydrogen atoms produce the color lines seen in
the bright-line spectrum of hydrogen. The greater the energy lost by the atom, the
greater the energy of the photon.
The problem with classical physics of the time was that an electron orbiting a
nucleus would lose energy and eventually collapse into the nucleus. In Bohr’s model, an
electron can travel around a nucleus without radiating energy. Furthermore, an electron
in a given orbit has a certain definite amount of energy. The only way an electron can
lose energy is by dropping from one energy level to a lower one. When this happens, the
atom emits a photon of radiation corresponding to the difference in energy levels.
However, electrons in higher levels cannot drop to a lower level if that level is filled.
Because no electrons can move to a lower level, none of them can lose energy. The atom
is energetically stable and is said to be in its ground state.
Atoms can absorb energy from an outside source (such as heat from a flame or
electrical energy from a source of voltage), causing one or more of the electrons within
the atom to move to higher energy levels. When electrons are moved to these higher
levels, the atom is said to be in an excited state. However, the atom is energetically
unstable, so it will not remain in an excited state for a long period of time. Eventually,
electrons return to lower levels. As they do so, energy is given off in the form of quanta.
Problem #10: Determine the energy of the line in the spectrum of hydrogen that
represents the movement of an electron from a Bohr orbit with n = 6 to n = 4.
Problem #11: Determine the wavelength of light that must be absorbed by a hydrogen
atom in its ground state to excite it to the n = 2 orbit.
Section Five pp. 7
At first, Bohr’s model appeared to be very promising. The energy levels
calculated by Bohr closely agreed with the values obtained from the hydrogen emission
spectrum. However, when Bohr’s model was applied to atoms other than hydrogen, it
did not work at all. It was therefore concluded that Bohr’s model is fundamentally
incorrect for atoms with more than one electron.
Schrödinger Equation
Erwin Schrödinger proposed the theory of quantum mechanics which suggested
that an electron (or any other particle) exhibiting wavelike properties should be described
by a mathematical equation called a wavefunction (denoted by the Greek letter psi, ψ).
Specifically, he developed a mathematical formalism to describe the hydrogen atom as a
wave. This infamous equation (which involves differential calculus!) gives the
probability of finding an electron at some point in a three-dimensional space at any given
instant but gives no information about the path the electron follows (recall Heisenberg).
The region in space where there is a probability of finding an electron is known as an
orbital, and in the Schrödinger equation, the wavefunction (squared mathematically) is
used to calculate the probability of finding an electron in space.
Structures of Atoms
1.
Principle Quantum Number (n): specifies the energy level of an electron and
labels the shell of an atom
2.
Angular Momentum Quantum Number (l): specifies the subshell of a given shell
in an atom and determines the shape of the orbitals in the subshell
p
s
d
f
Section Five pp. 8
3.
Magnetic Quantum Number (ml): identifies the individual orbitals of a subshell of
an atom and determines the orientation in space
4.
Spin Quantum Number (ms): distinguishes the two spin states of an electron
n
l
Orbital Designation
ml
# of orbitals
1
2
3
4
5.
Pauli Exclusion Principle: No two electrons can have the same four sets of
identical quantum numbers. Moreover, when two electrons (and NO MORE
THAN TWO!) occupy the same given orbital, their spins must be different.
Problem #12: How many subshells are there for n = 6? How many total orbitals are
there in the shell with n = 6?
Section Five pp. 9
Problem #13: Write the subshell notation and the number of electrons that can have the
following quantum numbers if all the orbitals of that subshell are filled:
A.
n = 3; l = 2
B.
n = 5; l = 0
C.
n = 7; l = 1
D.
n = 4; l = 3
Problem #14: How many electrons can have the following quantum numbers in an atom:
A.
n = 3, l = 1
B.
n = 5, l = 3, ml = -1
C.
n = 2, l = 1, ml = 0
D.
n=7
Problem #15: Which of the following sets of quantum numbers {n, l, ml, ms} are allowed
and which are not? For the sets of quantum numbers that are incorrect, state what is
wrong.
A.
{2, 2, -1, -½}
B.
{6, 0, 0, +½}
C.
{5, 4, +5, +½}
Problem #16: What is the maximum number of electrons in an atom that can have these
quantum numbers?
A.
n=3
B.
n = 2, l = 0, ml = 0
C.
n = 5, ms = +½
D.
n = 2, l = 1
Section Five pp. 10
Many-Electron Atoms
We have to examine the balance of attractions and repulsions in the atom to
explain why subshells of a given shell have different energies. As well as being attracted
by the nucleus, each electron in a many-electron atom is repelled by the other electrons
present. As a result, it is less tightly bound to the nucleus than it would be if those other
electrons were absent. We say that each electron is shielded from the full attraction of
the nucleus by the other electrons in the atom. The shielding effectively reduces the pull
of the nucleus on an electron. The effective nuclear charge, Zeff, experienced by the
electron is always less than the actual nuclear charge, Z, because the electron-electron
repulsions work against the pull of the nucleus. Note that the other electrons do not
“block” the influence of the nucleus; they simply provide additional repulsive coulombic
interactions that partly counteract the pull of the nucleus. Finally, an s-electron of any
shell can be found very close to the nucleus, so we say that it can penetrate through the
inner shells. A p-electron penetrates much less. Because a p-electron penetrates less than
an s-electron through the inner shells of the atom, it is more effectively shielded from the
nucleus and hence experiences a smaller effective nuclear charge than an s-electron does.
That is, an s-electron is bound more tightly than a p-electron and has a slightly lower
(more negative) energy. In a many-electron atom, because of the effects of penetration
and shielding, the order of energies of orbitals in a given shell is typically s < p < d < f.
6.
Aufbau Principle: The procedure for arriving at the ground-state electron
configurations of atoms and molecules in order of increasing atomic number. To
proceed from one atom to the next, we add a proton and some neutrons to the
nucleus and then describe the orbital into which the added electron goes.
7.
Hund’s Rule: Whenever orbitals of equal energy (degenerate) are available,
electrons occupy these orbitals singly before pairing begins.
8.
Core vs. Valence electrons: inner vs. outermost electrons (latter contained within
outermost shell)
9.
Diamagnetic vs. Paramagnetic species: former has all its electrons paired and is
slightly repelled by a magnetic field; latter has one or more unpaired electrons and
is attracted into a magnetic field
Problem #17: Determine the ground-state electron configuration for each of the
following elements (see the last page of this section for a sample orbital energy diagram):
A.
sulfur
B.
cesium
Section Five pp. 11
C.
polonium
D.
rhenium
E.
vanadium
Problem #18: Predict the number of valence electrons present in each of the following
atoms (include the outermost d-electrons):
A.
Bi
B.
Ba
C.
Mn
D.
Zn
Problem #19: Determine the ground-state electron configuration for each of the
following ions:
A.
Al+3
B.
Tc+4
C.
Ra+2
D.
I-
E.
Ir+
F.
Ru+4
G.
Mo+2
H.
Co+3
Section Five pp. 12
Problem #20: Predict the number of valence electrons present for each of the following
ions:
A.
In+
B.
Tc+2
C.
Ta+2
D.
Re+
Problem #21: Give the ground-state electron configuration and number of unpaired
electrons expected for each of the following ions:
A.
Ga+
B.
Cu+2
C.
Pb+2
D.
Se-2
Problem #22: For each of the following ground-state ions, predict the type of orbital that
the electrons of highest energy will occupy:
A.
Fe+2
B.
Bi+3
C.
As+3
D.
Os+
Periodic Trends
1.
Ionization Energy – energy needed to remove an electron from gaseous atom
Section Five pp. 13
2.
Atomic Radius
3.
Electron Affinity – energy released when an electron is added to gaseous atom
4.
Electronegativity – the electron pulling power of an atom when it is part of a
molecule (denoted with the Greek letter χ)
5.
Metallic Character
Section Five pp. 14
Problem #23: Arrange the following in terms of DECREASING first ionization energy:
Be, B, C, N, O, F, Ne
Problem #24: Why is the first ionization energy of aluminum slightly lower than the first
ionization energy for magnesium?
Problem #25: Why is the second ionization energy for sodium so much greater than its
first ionization energy?
Problem #26: Arrange the following in terms of DECREASING atomic (or ionic) radii:
O+, O, O-
Problem #27: Give a reason why the electronegativity for F is so much greater than the
electronegativity for Fr.
Section Five pp. 15
APPENDIX TO QUANTUM MECHANICS
Orbital Energy Levels:
__ __ __
6p
__ __ __ __ __
5d
__ __ __ __ __ __ __
4f
__
6s
E
n
e
r
g
y
__ __ __
5p
__ __ __ __ __
4d
__
5s
__ __ __
4p
__ __ __ __ __
3d
__
4s
__ __ __
3p
__
3s
__ __ __
2p
__
2s
__
1s
Example of Ionization Energies:
Al(g) → Al+(g) + eAl+(g) → Al2+(g) + eAl2+(g) → Al3+(g) + eAl3+(g) → Al4+(g) + e-
I1 = 580 kJ/mol
I2 = 1815 kJ/mol
I3 = 2740 kJ/mol
I4 = 11,600 kJ/mol