Chemistry C150 Spring 2018 Problem Set 2 KEY (10 points) Problems 2b, 3a, 6b, 7d, 8a were graded for accuracy. The remaining problems were checked for completeness. 1. Electronic Structure Models of Solid There are two ways to approximate electron wavefunction in a solid: Free Electron model and Tight Binding approximation. a. Describe how each model estimates the electronic structure of a solid The Free Electron Model assumes that the electrons are delocalized as an electron gas (electron-electron interactions are negligible) throughout a crystal lattice, and the electrons experience a constant potential in any point in space. The Tight Binding approximation is based on the idea that electrons are bound to specific atoms and only weakly interact with electrons on neighboring atoms. b. Suggest an example of a solid that would be well described by each type of model The Free Electron Model is suitable for materials where electrons are free to move between atoms, which is true for most metals, such as Li, Na, Mg, Al, etc. The Tight Binding model works better for materials with a more localized structure, where electrons are largely confined to the potential well around a particular atom. Many polymers and semiconductors can be described well by this model, such as silicon, boron nitride, etc. c. Why does LCAO approach gives good solution to the Tight Binding model, while not so for the Free Electron model? Linear Combination of Atomic Orbitals gives good solution to the Tight Binding model because it uses the wavefunctions of the atomic/molecular orbitals as basis set for the crystal orbitals of the solid. This means that the electrons must be bound to the potential well of atoms quite strongly, such that their wavefunctions take the form of the atomic/molecular wavefunctions (s/p/d-like wavefunctions). In Free Electron model, the electrons are loosely bound to the nuclei, and thus they do not feel the potential energy of the atomic nuclei in the same way as how atomic wavefunctions are derived. 2. 1D Band Diagram Black phosphorous, which is one of the several allotropes of elemental phosphorous, can be obtained by heating white phosphorous under high pressure. The resulting structure consists of puckered six-membered rings extended in three dimensions.
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Chemistry C150 Spring 2018
Problem Set 2 KEY (10 points)
Problems 2b, 3a, 6b, 7d, 8a were graded for accuracy. The remaining problems were
checked for completeness.
1. Electronic Structure Models of Solid
There are two ways to approximate electron wavefunction in a solid: Free Electron model and
Tight Binding approximation.
a. Describe how each model estimates the electronic structure of a solid
The Free Electron Model assumes that the electrons are delocalized as an electron gas
(electron-electron interactions are negligible) throughout a crystal lattice, and the electrons
experience a constant potential in any point in space. The Tight Binding approximation is based
on the idea that electrons are bound to specific atoms and only weakly interact with electrons
on neighboring atoms.
b. Suggest an example of a solid that would be well described by each type of model
The Free Electron Model is suitable for materials where electrons are free to move between
atoms, which is true for most metals, such as Li, Na, Mg, Al, etc.
The Tight Binding model works better for materials with a more localized structure, where
electrons are largely confined to the potential well around a particular atom. Many polymers
and semiconductors can be described well by this model, such as silicon, boron nitride, etc.
c. Why does LCAO approach gives good solution to the Tight Binding model, while not so
for the Free Electron model?
Linear Combination of Atomic Orbitals gives good solution to the Tight Binding model
because it uses the wavefunctions of the atomic/molecular orbitals as basis set for the crystal
orbitals of the solid. This means that the electrons must be bound to the potential well of atoms
quite strongly, such that their wavefunctions take the form of the atomic/molecular
wavefunctions (s/p/d-like wavefunctions). In Free Electron model, the electrons are loosely
bound to the nuclei, and thus they do not feel the potential energy of the atomic nuclei in the
same way as how atomic wavefunctions are derived.
2. 1D Band Diagram
Black phosphorous, which is one of the several allotropes of elemental phosphorous, can be
obtained by heating white phosphorous under high pressure. The resulting structure consists
of puckered six-membered rings extended in three dimensions.
Consider a simplified one-dimensional picture, consisting of a zig-zag chain of P atoms, where
each P-P-P angle is exactly 90°. The unit cell of this structure consists of two P atoms with
unit cell length a.
a. Draw the molecular orbital diagram for the diatomic P2 unit. You only need to consider
the 3p atomic orbitals of P.
The molecular orbital diagram is shown above. The energy of each orbital was derived from
Hückel approximation and is plot to scale as shown on the y axis.
b. Draw all the valence crystal orbitals corresponding to k at 0 and /a. For each crystal
orbital, draw the band dispersion diagram (E(k) vs k).
c. Based on your crystal orbitals, draw a predicted band structure diagram over the first
Brillouin zone and DOS plot for this structure.
Combine all band dispersion diagrams in (b). In doing so, we derive the energy of each
molecular orbitals from the energy diagram shown in (a):
𝛼𝜎𝑧 = 𝛼 + 𝛽𝜎
𝛼𝜎∗𝑧 = 𝛼 − 𝛽𝜎
𝛼𝜋𝑥 = 𝛼𝜋𝑦 = 𝛼 + 𝛽𝜋
𝛼𝜋∗𝑥 = 𝛼𝜋∗𝑦 = 𝛼 − 𝛽𝜋
The resulting band diagram is shown below
(Note that at k = π/2a, all crystal orbitals feature a node at every other unit cell, and the energy
of each crystal orbital is equal to the energy of the molecular orbitals.)
d. Draw the Fermi level on your band structure and DOS diagrams. Do you expect this
material to be an insulator, semiconductor, or metal?
See the diagram in (c). The Fermi energy is at EF = α as each P2 unit has six electrons to fill
six bands. We fill the whole σz band, and a combination of πx and π*x bands, and a combination
of πy and π*y bands. The DOS plot is shown on the right with total DOS plot in black. Notice
that in this case, the Fermi energy lies at where two bands intersect (πx intersects π*x and πy
intersects π*y), and there are only a few states available at EF. Therefore, we expect this
material to be semimetal.
(Note: don’t worry too much about exact ordering of the bands. Depending on how you draw
the band diagram, your answer may differ from the one shown here)
e. Draw the nearest-neighbor COOP curve for the crystal orbitals generated from the
interaction of σpz molecular orbitals of the P2 unit.
The lowest energy crystal orbitals generated from σz orbitals are π bonding in nature (k =
π/a). They correspond to in-phase overlap between nearest neighbor. Thus the COOP curve
is as depicted.
3. 2D Band Diagram
Consider a 2D square array of metal atoms in the xy plane below
a. Draw the crystal orbitals generated from d(x2-y2) atomic orbitals at the special positions
Γ, X, Y, and M in the first Brillouin zone. For each crystal orbital, identify the type of
the interactions (σ/π/δ and bonding/antibonding). Draw the band diagram with the
Γ→X→M→Γ→Y→M direction on the x axis.
b. Draw the crystal orbitals generated from d(yz) orbitals at the special positions Γ, X, Y,
and M. Identify the type of the interactions, and draw the band diagram with the
Γ→X→M→Γ→Y→M direction on the x axis.
c. Draw the band diagram of d(xz) orbitals with the Γ→X→M→Γ→Y→M direction on the
x axis. (Note: you can derive the band diagram utilizing the symmetry of d(xz) with
respect to d(yz) without redrawing the crystal orbitals)
Because d(xz) is related to d(yz) through a 90° rotation about the z axis, we can expect the Y
point of d(xz)-based crystal orbitals to have the same energy as the X point of d(yz)-based
crystal orbitals. Similarly, the X point of d(xz) should have the same energy as the Y point of
d(yz). Γ and M points for both d(xz) and d(yz) should have the same energy. The band diagram
is drawn below:
The calculated band structures of the following 2D materials are seen below:
d. The valence band is shown in blue and marked with a “V” and the conduction band is
shown in orange and marked with a “C”. For each 2D material, predict whether it is a
metal, semiconductor, or insulator. (Note: the Fermi levels for each material have been
normalized to 0 eV for each plot).
The band gap (Eg) of MoS2 is 1.8 eV, while that of ZrS2 is 1.0 eV. These are measured from
the bottom of the conduction band (C) to the top of the valence band (V). These materials
should behave as semiconductors because their band gaps are small (< 2 eV). NbS2 and PdS2
are metallic because the Fermi energy fall in the middle of the band. It takes very little energy
to excite electrons in the band.
4. Potassium and beryllium are both electrically conductive metals. Rationalize this based on the
basis of the band structure for each of these two elements.
Potassium only has one 4s electron, so it has a half-filled conduction band for solid potassium,
leading to it being a metal since electrons can be conducted through the material. Beryllium
has two 2s electrons and thus a filled band. However, electrical conduction occurs because of
the 2p band overlaps with 2s.
5. Magnetic Properties
a. On an atomic level, what causes a material to be: i) paramagnetic, ii) diamagnetic, iii)
ferromagnetic?
i) Paramagnetism occurs in all materials with unpaired electrons. The magnetic dipole in a
material due to the spin of unpaired electron(s) will orient in the direction of an applied
field.
ii) All materials are inherently diamagnetic, and this arises from the response of the electrons
in a material to an applied magnetic field. The circulation of electrons induced by the
magnetic field in turn induces a small magnetic moment in opposition to the direction of
the applied field (Lenz’s law).
iii) Above the Curie temperature, thermal energy overcomes the preference for ordered
alignment of the spins, and the spin orientations become disordered. When the temperature
is lowered below Tc, exchange interaction between spins can orient all moments in the
same direction, creating a net magnetic moment.
b. Rank i, ii, and iii by the relative magnitude of their magnetic susceptibility.