Download Chapter 7: ELECTRONS IN ATOMS AND PERIODIC PROPERTIES

Famous Opinions of QM
“A scientific truth does not triumph by convincing its opponents
and making them see the light, but rather because its opponents
eventually die and a new generation grows up that is familiar
with it.”
(Max Planck, 1920)
“All these fifty years of conscious brooding have brought me no
nearer to the answer to the question, 'What are light quanta?‘”
(Albert Einstein, 1954)
“Those who are not shocked when they first come across
quantum physics cannot possibly have understood it.”
(Niels Bohr, 1958)
Famous Opinions of QM
The one great dilemma that nails us…day and
night is the wave-particle dilemma.
(Erwin Schrodinger, 1959)
I think I can safely say that nobody understands
quantum mechanics.
(Richard Feynman, 1965)
Particles and Waves
• EM radiation can behave as either a wave or a particle
depending on the situation
• “Light has properties that have no analogy at the
macroscopic level, and thus, we have to combine two
different ideas to describe its behavior.”
Wavefunctions
• A wavefunction is a probability
amplitude. The “square” of a
wavefunction gives the probability
density…the likelihood of finding the
particle in region of space.
• The wavefunctions and kinetic
energies available to a quantum
particle are quantized if the particle
is subject to a constraining potential.
• We can determine the wavefunctions
and KEs available to our system by
considering the field of force (the PE)
our system is subject to.
The Hamiltonian
• Erwin Schrodinger developed a mathematical formalism
that incorporates the wave nature of matter.
• H, the “Hamiltonian,” is a special kind of function that
gives the energy of a quantum state, which is described
by the wavefunction, Y.
• H contains a KE part and a PE part:
2
2
2
2


h
d
d
d
ˆ
H   2  2  2  2   V ( x, y, z )
8 m  dx dy dz 
• By solving the Schrodinger equation (below) with a
known Hamiltonian, we can determine the
wavefunctions and energies for quantum states.
Ĥ  E
H-atom wavefunctions
• In the H atom, we are interested in describing the regions in
space where it is likely we will find the electron, relative to the
nucleus…we want the wavefunction for the electron.
• We can model the attraction of the H atom’s single electron
to its single proton with a “Coulombic” potential curve:
er
r
0
P+
e 2
V (r) 
r
• The V(r) potential becomes part of the Hamiltonian for
the electron.

H-atom wavefunctions (cont.)
• The radial dependence of the potential suggests
that we should from Cartesian coordinates to spherical
polar coordinates.
r = interparticle distance
(0 ≤ r ≤ )
e-
p+
 = angle from “xy plane”
(/2 ≤  ≤ - /2)
 = rotation in “xy plane”
(0 ≤  ≤ 2)
H-atom wavefunctions (cont.)
• Then the Schrodinger equation Ĥ  E for
the hydrogen atom becomes:
3-dimensional KE operator in
spherical polar coordinates
   2  
1  
 
1  2 
 2 2  r

 sin 
 2
8  r  r  r  sin   
  sin   2 
Ze2

  E
r
h2
Radial Coulombic
PE operator
H-atom wavefunctions (cont.)
• If we solve the Schrodinger equation using this
potential, we find that the energy levels are
quantized:
2


Z 2  me 4 
Z
18
En   2  2 2   2.178 x10 J  2 
n  8 0 h 
n 
n is the principle quantum number, and can
have integer numbers ranging from 1 to
infinity.
The higher n, the greater the distance between
the nucleus (+) and the electron (-).
H-atom wavefunctions (cont.)
• In solving the Schrodinger Equation, two other
quantum numbers become evident:
l…the orbital angular momentum quantum number.
Ranges in value from 0 to (n - 1 ).
ml … the “z component” of orbital angular momentum.
Ranges in value from - l to 0 to +l .
• We can characterize the hydrogen-atom orbitals
using the quantum numbers: n, l , ml
Orbitals and Quantum Numbers
• Naming the electron orbitals is done as follows
– n is simply referred to by the quantum number
– l (0…n… - 1) is given a letter value as follows:
•
•
•
•
0
1
2
3
=
=
=
=
s
p
d
f
- ml (- l …0…l ) is usually “dropped”
For example: for n = 3, l = 2  “3d orbital”
Quantum Mechanical Model
The Bohr model is deterministic…uses fixed “orbits” around a central
nucleus to describe electron structure of atoms.
The QM model is probabilistic…uses probabilities to describe electron
structure.
A probabilistic electron structure is much more difficult to visualize.
HOWEVER, the electronic energy levels are still quantized.
Deterministic vs. Probabilistic
• In the Bohr model, you can
always find the electron in an
atom, just like you can always
find the moon as it orbits the
earth.
• You can always determine the
relative location of the nucleus
and electron in Bohr’s model.
• This is because the electron
follows a particular orbit
around the nucleus.
• In the QM model, the electron
does not travel along a
particular path around the
nucleus.
• You can never determine the
electron’s exact location…you
can only find where it is likely
to be.
• The Bohr orbit is replaced by
orbital which describes a
volume of space in which the
electron is likely to be found.
Quantum Numbers and Orbitals
(cont.)
• Table 7.1: Quantum Numbers and Orbitals
1 0
2 0
1
3 0
1
2
Orbital
1s
2s
2p
3s
3p
3d
ml
0
0
-1, 0, 1
0
-1, 0, 1
-2, -1, 0, 1, 2
# of Orb.
1
1
3
1
3
5
Increasing
Energy
n
Orbital Shapes (cont.)
• Example: Write down the orbitals associated with n = 4.
l = 0 to (n - 1)
= 0, 1, 2, and 3
= 4s, 4p, 4d, and 4f
Ans: n = 4
4s
4p
4d
4f
(1 ml sublevel)
(3 ml sublevels)
(5 ml sublevels
(7 ml sublevels)
Which of the following sets of quantum numbers (n, l, m)
is not allowed?
A. (3, 2, 2).
B. (0, 0, 0).
C. (1, 0, 0).
D. (2, 1, 0).
Electron Orbitals
s orbitals
Orbitals represent a
probability space where
an electron is likely to be
found.
All atoms have all
orbitals, but many of
them are not occupied.
p orbitals
d orbitals
The shapes of the
orbitals are determined
mathematically...they are
not intuitive.
Electron Orbital Shapes
• The “1s” wavefunction has no angular dependence
(i.e., it only depends on the distance from the
nucleus).
3
1  Z 
1s 
  e
 ao 

2 Z r
a0
Probability =
3
1  Z  2 

  e
 ao 

*
Probability is spherical

Electron Orbital Shapes (cont)
s (l = 0) orbitals
as n increases, orbitals
demonstrate n - 1
nodes.
Node: an area of
space where the
electron CAN’T be,
ever, no matter how
much it wants to.
Aside: What’s a node??
• Remember the guitar-string standing wave analogy?
• A standing wave is a motion in
which translation of the wave
does not occur.
• In the guitar string analogy
(illustrated), note that standing
waves involve nodes in which no
motion of the string occurs.
Electron Orbital Shapes (cont.)
2p (l = 1) orbitals
2px
not spherical, but lobed.  2 p z
2pz
2py
1

4 2
3
Z  2 
  e 2 cos
ao 
labeled with respect to orientation along x, y, and z.

Electron Orbital Shapes (cont.)
3p (l = 1) orbitals
3
3 p
z

2  Z  2
2

  6   e 3 cos
81  ao 

• more nodes as compared to 2p (expected.).
• still can be represented by a “dumbbell” contour.
Orbital Shapes (cont.)
3d (l = 2) orbitals labeled as dxz, dyz, dxy, dx2-y2 and dz2.
Orbital Shapes (cont.)
4f (l = 3) orbitals
We will not
show the
exceedingly
complex
probability
distributions
associated
with f orbitals.
Electron Orbital Energies in
the H-atom
• energy increases as 1/n2
• orbitals of same n, but
different l are considered
to be of equal energy
(“degenerate”).
• the “ground” or lowest
energy orbital is the 1s.
Orbital Summary
• Orbital E increases with n.
– At higher n, the electron is farther away from the nucleus...this is a
higher energy configuration.
• Orbital size increases with n.
– There is a larger area of space where you are likely to find an
electron at higher E’s.
• Orbital shape is the same no matter the value of n.
– 3s looks like 1s, except it’s bigger and has more nodes. Same for p, d,
f, etc.
• Number of nodes in an orbital goes as n - 1.
– 1s has zero nodes, 2s has one node, 3s has two nodes...
– 2px, 2py, 2pz each have one node, 3px, 3py, 3pz each have two nodes
– the 3d orbitals each have two nodes, 4d have three, etc.
– Note that number of nodes indicates relative energy!
• All atoms have all orbitals…but in an unexcited atom, only those closest
to the nucleus will be occupied by electrons
Orbital Quiz
F•
The shape of a given type of orbital changes as
n increases.
T•
The number of types of orbitals in a given
energy level is the same as the value of n.
T•
The hydrogen atom has a 3s orbital.
F•
The number of lobes on a p-orbital increases as
n increases. That is, a 3p orbital has more lobes
than a 2p orbital.
F•
The electron path is indicated by the surface of
the orbital.
Electron Spin
• Experiments demonstrated the
need for one more quantum
number.
• Specifically, some particles
(electrons in particular)
demonstrated inherent angular
momentum…
“spin up”
• Basically, this means that
electrons have two ways of
interacting with an applied
magnetic field.
Interpretation: the electron is a
bundle of “spinning” charge
“spin down”
Electron Spin (cont.)
ms = 1/2
ms = -1/2
• The new quantum
number is ms
(analogous to ml).
• For the electron, ms has
two values:
+1/2 and -1/2
Pauli Exclusion Principle
Defn: No two electrons may occupy
the same quantum state
simultaneously.
In other words: electrons are very
territorial. They don’t like other
electrons horning in.
In practice, this means that only two
electrons may occupy a given
orbital, and they must have
opposite spin.
Quantum Number Summary
• n: principal quantum number
– index of size and energy of electron orbital
– can have any integral value: 1, 2, 3, 4, …
• l: angular momentum quantum number
– related to the shape of the orbitals
– can have integral values 0 … n - 1
• ml: magnetic quantum number
– related to orbital orientation (relative to the other l-level
orbitals)
– can have integral values –l … 0 … +l
• ms: electron spin quantum number
– related to the “magnetic moment” of the electron
– can have half-integral values –1/2 or +1/2
Polyelectronic Atoms
• For polyelectronic atoms, a direct solution of the
Schrodinger Eq. is not possible.
   2  
1  
 
1  2 
 2 2  r

 sin 
 2
8  r  r  r  sin   
  sin   2 
2
Ze
No solution for polyelectronic atoms!!

  E
r
h2
• When we construct polyelectronic atoms, we use the
hydrogen-atom orbital nomenclature to discuss in
which orbitals the electrons reside.
• This is an approximation (and it is surprising how well
it actually works).
The Aufbau Principle
• When placing electrons into orbitals in the
construction of polyelectronic atoms, we use the
Aufbau Principle.
• This principle states that in addition to adding
protons and neutrons to the nucleus, one simply adds
electrons to the hydrogen-like atomic orbitals
• Pauli exclusion principle: No two electrons may have
the same quantum numbers. Therefore, only two
electrons can reside in an orbital (differentiated by ms).
Orbital Energies
4s
Energy
3s
2s
1s
3p
2p
3d
H has only one electron, so all
of the sublevels in a given
principal level have the same
energy...they are degenerate.
In many-electron atoms, a given electron is
simultaneously attracted to the nucleus and
repelled by other electrons, causing the energies
of the sublevels to change relative to H.
When we put electrons in orbitals, we fill
them in order of increasing energy, not n.
Let’s fill some orbitals
RULES
• Orbitals are filled starting from
the lowest energy.
• The two electrons in an orbital
must have opposite spin.
• Example: Hydrogen (Z = 1)
1s1
1s
2s
• Example: Helium (Z = 2)
2p
1s2
1s
2s
2p
Let’s fill some more orbitals
• Lithium (Z = 3)
1s22s1
1s
2s
• Berillium (Z = 4)
2p
1s22s2
1s
2s
2p
• Boron (Z = 5)
1s22s22p1
1s
2s
2p
Filling Orbitals (cont.)
• Carbon (Z = 6)
1s22s22p2
1s
2s
2p
Hund’s Rule: Lowest energy configuration is the one in
which the maximum number of unpaired electrons are
distributed amongst a set of degenerate orbitals.
REVISED RULES
• Orbitals are filled starting from the lowest energy.
• The two electrons in an orbital must have opposite spin.
• Hund’s Rule: the orbitals in degenerate series (such as
2p in the example above) must each have an electron
before any of them can have two.
Filling Orbitals (cont.)
• Carbon (Z = 6)
1s22s22p2
1s
2s
2p
• Nitrogen (Z = 7)
1s22s22p3
1s
2s
2p
Filling Orbitals (cont.)
• Oxygen (Z = 8)
1s22s22p4
1s
• Fluorine (Z = 9)
2s
2p
1s22s22p5
1s
2s
2p
• Neon (Z = 10)
1s22s22p6
1s
2s
2p
full
Filling Orbitals (cont.)
• Sodium (Z = 11)
1s
2s
2p
Ne
3s
1s22s22p63s1
3p
[Ne]3s1
3s
• Compare to Neon (Ne) (Z = 10)
1s22s22p6
1s
2s
2p
full
Filling Orbitals (cont.)
• Sodium (Z = 11)
Ne
1s22s22p63s1
[Ne]3s1
3s
• Phosphorus (P) (Z = 17)
[Ne] 3s23p3
Ne
3s
3p
• Argon (Z = 18)
[Ne] 3s23p6
Ne
3s
3p
Filling Orbitals (cont.)
We now have the orbital configurations for the first 18 elements.
Elements in same column have the same # of valence electrons!
Valence Electrons: The total number of s and p
electrons in the highest occupied energy level.
The Aufbau Principal (cont.)
• Similar to Sodium, we begin the next row of the periodic
table by adding electrons to the 4s orbital.
• Why not 3d before 4s?
• 3d is closer to the nucleus
• 4s allows for closer
approach; therefore, is
energetically preferred.
Back to Filling Orbitals
• Elements Z=19 and Z= 20:
Z= 19, Potassium:
1s22s22p63s23p64s1 = [Ar]4s1
Ar
4s
4p
1s22s22p63s23p64s2 = [Ar]4s2
Z= 20, Calcium:
Ar
4s
4p
Filling Orbitals (cont.)
• Elements Z=21 to Z=30 have occupied d orbitals:
Z= 21, Scandium: 1s22s22p63s23p64s23d1 = [Ar] 4s23d1
Ar
4s
Z= 30, Zinc:
4p
3d
1s22s22p63s23p64s23d10 = [Ar] 4s23d10
Ar
4s
4p
3d
The Aufbau Principal (cont.)
• Elements Z=19 and Z= 20:
Z= 19, Potassium: 1s22s22p63s23p64s1 = [Ar]4s1
Z= 20, Calcium: 1s22s22p63s23p64s2 = [Ar]4s2
• Elements Z = 21 to Z = 30 have occupied d orbitals:
Z= 21, Scandium: 1s22s22p63s23p64s23d1 = [Ar] 4s23d1
Z = 24, Chromium: [Ar] 4s13d5 exception
Z= 30, Zinc: 1s22s22p63s23p64s23d10 = [Ar] 4s23d10
What if you forget the orbital-filling order?
1. Write down the
orbitals for each n on
separate lines.
2. Arrows drawn as
shown will give you
the order in which the
orbitals should be
filled.
Note that this scheme fills
4s before 3d, as
expected.
Polyelectronic Atoms
+
e-
“Screening”: The presence of other
electrons means a given electron does
not feel the attraction of the nucleus as
strongly as it would in hydrogen.
“Penetration”: Orbitals that have some
probability density close to the nucleus
will be energetically favored over
orbitals that do not.
Periodic Table
This orbital filling scheme gives rise to
the modern periodic table.
Periodic Table
After Lanthanum ([Xe]6s25d1), we
start filling 4f.
Periodic Table
After Actinium ([Rn]7s26d1), we
start filling 5f.
Periodic Table
Row headings correspond to the highest
occupied energy level for any element in
that period.
Periodic Table
Column headings give total number of
valence electrons for any element in that
group.
“Valence” only refers
to s and p electrons in
the highest occupied
energy level.
What is the electron configuration for the indicated element?
A. 1s22s22p63s23p64s23d3
B. 1s22s22p63s23p64s24d3
C. 1s22s22p63s23p64s23d2
D. 1s22s22p73s23p64s23d2
Valence Electrons
• The total number of s and p electrons in the highest
occupied energy level.
• As we’ll see, all the “action” happens at the valence
electrons.
• Elements in the same group (column) in the periodic
table have the same number of valence electrons.
• This means elements in the same group tend to have
similar chemical properties.
Valence Electrons (cont.)
Chemists use Lewis dot symbols to
indicate the number of valence
electrons in an atom.
The valence electrons are drawn as
dots around the atomic symbol, with
orbital occupancy indicated...that is,
electrons that occupy the same
orbital appear as paired dots.
HOWEVER, we will encounter
situations where it is more
convenient to spread the dots out
around the element symbol.
C
C
Which of the following electron
configurations represents N?
A. 1s21p32s2
B. 1s22s22p23s1
C. 1s22s32p2
D. 1s22s22p2
But we wrote the
electron
configuration of N as
1s22s22p3 the other
day!
The electron
configuration
1s22s22p33s1 represents
an excited state of N.
Excited States
• By putting the “right” amount (a quantized amount)
of energy into an atom, we can move electrons from
a low energy orbital to a higher-energy orbital…
• Such an electron is said to be in an excited state.
Ground State of Nitrogen (Z = 7)
hν
1s22s22p3
1s
2s
2p
An Excited State of Nitrogen (Z = 7)
1s22s22p23s1
1s
2s
2p
3s
Periodic Trends
• The valence electron structure of atoms can be
used to explain various properties of atoms.
• In general, properties correlate down a group
of elements.
• A warning: such discussions are by nature
very generalized…exceptions do occur.
Ionization Energy
• The energy required to remove an electron from
a gaseous atom or ion.
ZZ+
(Z-1) -
Energy
Z+
e-
• The electron is completely “removed” from
the atom.
Ionization Energy (IE)
• Ionization is an endothermic process...we must
put energy in to separate the negatively-charged
electron from the positively-charged nucleus.
IE + X
X+ + e-
• Generally done using photons, with energy
measured in eV (1 eV = 1.6 x 10-19 J).
• The greater the propensity for an atom to “hold
on” to its electrons, the higher the ionization
energy will be.
Ionization Examples
• Removal of valence versus core electrons
Na(g)
Na+(g) + e-
IE1 = 5.14 eV
(removing “valence” electron)
[Ne]3s1
[Ne]
Na+(g)
Na2+(g) + e- IE2 = 47.3 eV
[Ne]
WOW!
1s22s22p5 (removing “core” electron)
• Takes significantly more energy to remove a
core electron…. core electron configurations
tend to be energetically stable.
Ionization Examples
• We can perform multiple ionizations of valence
electrons:
Al(g)
Al+(g) + e-
I1 = 580 kJ/mol
first
Al+(g)
Al2+(g) + e-
I2 = 1815 kJ/mol
second
Al2+(g)
Al3+(g) + e-
I3 = 2740 kJ/mol
third
Al3+(g)
Al4+(g) + e-
I4 = 11,600 kJ/mol fourth
WOW!
Ionization
• First Ionization Energies:
Column 8
Column 1
First Ionization E Trends
Increases from left
to right across a
period.
 Reason: increasing Z+ (the
number of protons in the nucleus)
which attracts the valence electron
Decreases down a
group.
 Reason: increasing distance
between electron and nucleus
Which reaction represents the ionization of F?
A. 1s22s22p5
1s22s22p6
B. 1s22s22p5
1s22s22p43s1
C. 1s22s22p5
1s22s22p4
D. 1s22s22p5
1s22s12p6
Electron Affinity
Electron Affinity: the energy change associated with the
addition of an electron to a gaseous atom.
eZ-
(Z+1)-
+Z
+Z
• We will stick with our
thermodynamic definition,
with energy released being a
negative quantity.
Energy
Electron Affinity
• Some elements have very high electron affinity:
Wow!
• Group 7 (the halogens) and Group 6 (O and S specifically).
Electron Affinity
• Some elements have essentially no electron
affinity:
N?
• Orbital configurations can explain both observations.
Electron Affinity
• Why is EA so great for the halogens?
F(g) + e1s22s22p5
F-(g)
EA = -327.8 kJ/mol
1s22s22p6
[Ne]
• Why is EA so poor for nitrogen?
N(g) + e-
1s22s22p3
N-(g)
EA > 0 (unstable)
1s22s22p4
(e- must go into occupied orbital)
Electron Affinity
• How do these arguments do for O?
O(g) + e1s22s22p4
O-(g)
1s22s22p5
EA = -140 kJ/mol
Bigger Z+ overcomes
e- repulsion.
• What about the second EA for O?
O-(g) + e1s22s22p5
O2-(g)
EA > 0 (unstable)
1s22s22p6
[Ne] configuration, but electron
repulsion is just too great.
Atomic Radii
• Atomic Radii are defined as the covalent radii,
and are obtained by taking 1/2 the distance of a bond:
r = atomic radius
Atomic Radii
• Decrease across a row
due to increase in Z+
• Increase down column
due to population of
orbitals of greater n.
Ionization and Atomic Radii
Which atom would you expect to have the lowest ionization
energy?
A. 1s22s22p3
B. 1s22s22p63s23p5
C. 1s22s22p63s23p64s2
D. 1s22s22p63s23p64s23d104p65s1
We can partition the periodic table
into general types of elements.
1A
8A
3A 4A 5A 6A 7A
2A
Non-metals
Metals
• Metals … Tend to lose electrons.
– Good conductors of heat, electricity; malleable solids
• Non-metals … Tend to gain electrons.
– Poor conductors; not malleable
• Metalloids … Can lose or gain electrons.
– Both metallic and nonmetallic properties
What about Hydrogen?
• Hydrogen (H). Valence: 1s1
H
• H is not really a metal or a halogen, although it has
some properties of both
– forms compounds in the same ratio as the alkali
metals, but the bonding mechanism is different
– can gain an electron to form H- like a halogen
• Elemental H is found as a diatomic gas: H2
• Under 3 million atm of pressure, H exhibits metallic
properties…
Metallic Hydrogen??
Under the millions of atm of pressure exerted
in a diamond-anvil cell, elemental H changes
from a gas to an opaque solid that conducts
electricity.
The covalently-bonded diatomics turn into a
network of protons in a “sea” of electrons.
This is just a photograph of a
diamond anvil cell…that’s not
metallic H in there.
Chemical Bonds
• In broad terms, a chemical bond is a term used to
characterize an interaction between two atoms that
results in a reduction in the energy of the system
relative to the isolated atoms.
• The degree of energy reduction or “stabilization” is
given by the energy required to break the bond (known
as the “bond energy”)
Chemical Bonds (cont.)
Stabilization