Download Problem Set 2 KEY

Due: Thursday, Feb 20
Chemistry 150
Problem Set 2 Key
1. 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.
a. Consider a simplified one-dimensional picture, consisting of a zig-zag chain of P
atoms, where each P-P-P angle is exactly 90°. For a two atom unit cell of dimension
a, draw all the valence crystal orbitals corresponding to k at 0 and π/a. When
forming the crystals orbitals, you only need to consider the 3P atomic orbitals of P.
See attached. We start with a simple MO picture of P2 to form 6 molecular orbitals
that are used as our basis set for generating 6 bands of crystal orbitals.
b. Based on your crystal orbitals, draw a predicted band structure diagram and DOS
plot for this structure over the first Brillouin zone.
c. Draw the Fermi level on your band structure and DOS diagrams. Do you expect
this material to be an insulator, semiconductor, or metal?
As drawn, the material would be a metal. Note that the relative energies of the
bands are not certain, so other answers are possible, as long as they are consistent
with the band structure diagram that is drawn.
2. 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 overlapping 2p band.
3. The calculated band structures of the following 2D materials are seen below:
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”. Explain why MoS2 and ZrS2 are
semiconductors with band gaps of 1.8 eV and 1.0 eV, respectively, while the NbS2 and
PdS2 analogs are both metals. (Note: the Fermi levels for each material have been
normalized to 0 eV for each plot).
The orbital splitting for these materials is as follows: dz2; dxy and dx2-y2; dxz and dyz.
Given that, ZrS2 (Zr4+, d0) and MoS2 (Mo4+, d2) are expected to be semiconductors
because they both have a full valence band and an empty conduction band at a higher
energy. The ZrS2 compound has no d electrons, so this is expected, while the MoS2
structure fills the dz2 orbital but not any higher energy orbitals. NbS2 is a metal because
it has a half-filled orbital since Nb4+ is a d1 metal. The Pd is Pd4+, d6, which has several
electrons available for conduction. Additionally, the band structure shows significant
overlap of the bands, which is likely due to orbitals that are of similar energy, thus
leading to its metallic properties.
4. For a 2D sheet of metal atoms in the xy plane, the special positions in the Brillouin zone
(Γ, X, Y, and M) will have σ, π, and δ interactions between orbitals. Draw the d(x2-y2) and
d(yz) orbitals at each of these positions and identify the bonding and/or antibonding
interactions.
See attached sheets
5. Both the Free Electron and Tight Binding approximations were derived in class.
a. Describe how each model estimates the electronic structure of a solid.
The Free Electron Model assumes that electron-electron interactions are negligible
and that the electrons are delocalized as an electron gas throughout a crystal lattice.
The Tight Binding Approximation (or Linear Combination of Atomic Orbitals
Model) 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 be 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 LCAO 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 boron
nitride.
6. Metallic hydrogen was recently claimed to have been observed by compressing hydrogen
gas to extremely high pressures (495 GPa) at low temperature (5.5 K) (Dias, Silvera
Science 2017; DOI 10.1126/science.aal1579). A proposed unit cell for solid metallic
hydrogen (SMH) is shown below.
a. How many hydrogen atoms are there per unit cell? What is the geometry at each
hydrogen?
There are four hydrogen atoms per unit cell. The geometry at each is tetrahedral.
b. The primary evidence supporting the observation of SMH was an increased
reflectivity of the sample upon compression to high pressure. The dielectric
function of a solid behaving as a free electron gas is given by
&' (
! " =1−
!) *" (
where n is the electron density, e is the charge of an electron, me is the mass of an
electron, and !) is the permittivity of vacuum. When the dielectric function
becomes negative, light is no longer able to penetrate the material and the material
becomes reflective. Solve for the plasma frequency ( "+ = " ) such that the
dielectric function becomes zero and evaluate it in terms of the parameters for the
above unit cell.
Assume that each hydrogen atom donates one electron to the free electron bath. The
electron density can then be taken as four electrons divided by the unit cell volume.
"+ =
"+ =
&' (
!) *
4
1.6/10145 68 (
4.5/1012) *2
9/1014( 6( 8 : ;<14 *12 9/10124 ;<
"+ = 5.3/104> ?@A8 14
B = ℏD
B = 34.88'F
Note that h/2π is used instead of h since the definition of vacuum permittivity
renders the plasmon frequency in units of radians/sec, and not in Hz. The fact that
the energy (34.88eV) is very close to that reported in the paper (32.5±2.1) indicates
that our assumption for electron density is good.
c. How does this value compare with the plasma frequencies for lithium (7.12 eV),
sodium (5.71 eV), and potassium (3.72 eV) and what does this imply about the
ability for SMH to reflect light?
The higher value for the plasmon frequency of SMH indicates that it is reflective
over a wider range of energies (that is, for frequencies less than the plasmon
frequency).
7. On an atomic level, what causes a material to be: i) paramagnetic, ii) diamagnetic, iii)
ferromagnetic? Rank i, ii, and iii by the relative magnitude of their magnetic susceptibility.
i)
ii)
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.
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 tend to orient
all moments in the same direction, creating a net magnetic moment.
Diamagnetic < Paramagnetic < Ferromagnetic
1
8. The magnetic moment of a paramagnetic can be written as µ = g S ( S +1) + L(L +1) . It
4
can also be simplified to µ = g S ( S +1) or µ = n ( n + 2 ) , where S is the total spin
quantum number, g is the gyromagnetic ratio, L is the orbital quantum number, m is the
magnetic moment, and n is the number of unpaired electrons. Explain briefly the
assumptions that are being made for both of these simplifications.
In the first simplification, we are assuming that our system does not possess orbital angular
momentum, that is, our magnetic moment is due only to the unpaired spins in the material,
and hence we can use a spin only model for the magnetic moment.
In the second expression, we are using n, the total number of unpaired electrons, rather
than the total spin S. These two expressions are really the same, then, and the only
simplification is that we are assuming the g value to be 2 (the actual value is 2.002320(4)).
This a good assumption to use when considering a spin-only system.
9. Complete the following table (assume g = 2):
Free Ion
Cr3+
Fe2+
Mn2+
Fe3+
Gd3+
# unpaired electrons
3
4
5
5
7
Total spin (S)
3/2
2
5/2
5/2
7/2
Spin-only magnetic moment (µs)
3.87
4.90
5.92
5.92
7.94
a. If placed in the octahedral holes of a solid oxide lattice, which if any of these metal
ions would be expected to have moments that deviate from the free ion value and
why?
None of the first four would be expected to have moments that deviate from the
free ion (spin only) values, as these should all be high-spin in an octahedral crystal
field of oxide donors (weak field ligands). Gd3+ is a lanthanide, which have more
ionic bonding than d-block transition metals, so you would also not expect much
deviation from the free ion value. Note that since orbital angular momentum will
be quenched, the experimental values will differ from free ion J values.
b. Gd3+ complexes are commonly used as MRI contrast agents. Explain why Gd3+ is
used, rather than other metals, especially those in the d-block.
Since there are seven f-orbitals, this allows for you to put a maximum of 14
electrons in these orbitals. Most lanthanides assume a Ln3+ oxidation state, leading
to Gd3+ having seven electrons, all of which are unpaired. This is the highest
possible spin for an element on the periodic table, which leads to its use as an MRI
contrast agent. Additionally, Gadolinium has a ferromagnetic Curie temperature of
20 °C, meaning it is paramagnetic at room temperature and thus seen within the
magnetic field of an MRI scanner. Additionally, the Gd complexes used in MRI
typically have one H2O ligand, which has fast exchange with the surrounding
aqueous environment and thus changes the relaxation time of the water molecules,
allowing them to be detected in MRI.
10. The mineral magnetite is the most magnetic naturally occurring material. It has the spinel
structure type (AB2O4) where A (tetrahedral holes) sites are occupied by Fe3+ and B
(octahedral holes) sites are equally occupied by Fe3+ and Fe2+ ions.
a. Draw crystal field splitting diagrams showing the electronic configuration of the d
electrons in: i) the Fe3+ tetrahedral holes, ii) the Fe3+ octahedral holes, and iii) the
Fe2+ holes.
b. What are the magnitudes of the (spin only) magnetic moments, µ, at each of the
individual ion locations i, ii, and iii?
µ (Td Fe3+) = 5.92
µ (Oh Fe3+) = 5.92
µ (Oh Fe2+) = 4.9
c. The total effective magnetic moment would be ~14 µb per cell if all of the spins
aligned in the same direction. What type of magnetic ordering would this be?
Ferromagnetic ordering. All spins aligned in the same direction would lead to a
total spin of S = 5/2 + 5/2 + 2 = 7. Plugging this into the spin only formula for µ
we get µeff = 14.9 µB.
d. In reality, the experimentally measured magnetic moment per cell is much lower
(~5 µb). Suggest an arrangement of the spins in the cell that would give rise to this
reduced magnetic moment. What type of magnetic ordering does this correspond
to?
µ = 5 is approximately what we would expect for a system with S = 2. If the spins
of the Fe(III) electrons are opposed to each other, they will cancel, leaving only the
Fe(II) spins contributing to the net magnetism. This corresponds well to the
experimentally observed µ = 5 value and would correspond to ferrimagnetic
ordering.
11. An organometallic, Co(II)L4, has a magnetic moment of µ = 1.73. Is it more likely to have
a square planar or tetrahedral coordination geometry?
With CoII, it is sufficient to use the spin-only expression for the magnetic moment.
Tetrahedrally coordinated CoII will have S = 3/2, yielding µ = 3.87. Square planar
coordination geometry, on the other hand, could yield S = ½ and therefore µ = 1.73, which
is in agreement with experiment.
12. The discovery that individual molecules could display properties of bulk magnets, such as
magnetic hysteresis, came about from the study of the first single molecule magnet in 1993.
This dodecanuclear manganese cluster has the formula Mn12O12(O2CMe)16(H2O)4, and the
core structure of the molecule is shown below. Four central MnIV ions (large gray spheres)
are surrounded by a eight MnIII ions (large white spheres), and the Mn ions are connected
through bridging oxide ligands. Significantly, this molecules was found to have an overall
S = 10 ground state and to display magnetic hysteresis below 4 K.
a. Rationalize the total ground state molecular spin (S = 10) based on the individual
spins of each manganese ion and relevant intermolecular magnetic interactions.
MnIV has S = 3/2 and MnIII has S = 2. If the central four MnIV ions are
ferromagnetically coupled, as are the peripheral eight MnIII ions, antiferromagnetic
coupling between these two units would lead to an overall S = 10 ground state.
S = 8*2 – 4*(3/2) = 10
b. What mechanism promotes the magnetic interaction between MnIII and MnIV in this
molecule?
The MnIII and MnIV ions are couple via superexchange through the bridging oxides..