Chemistry Questionnaire

CHEM3x19 – Materials Chemistry, Week 1 Feedback (1)References (e-Text available via our Canvas site): Chapter 7, pages 243-261 in Solid State Materials
Chemistry, Patrick Woodward, Pavel Karen, John Evans, Thomas Vogt, 2021.
Also, Chapter 20 in Inorganic Chemistry by Atkins, Weller, Overton, Rourke and Armstrong, 6th Edition
(any earlier addition will also include similar content).
1. Identify the ground term for each set of terms:
i) 3F, 3P, 1P, 1G
ii) 5D, 3H, 3P, 1G, 1I
iii) 6S, 4G, 4P, 2I
2. See below the ball and stick representations of the two phases of the cubic [MII(C4O4)(H2O)2] sodalite
cage showing the different relative orientations of the squarate units: (a) ‘staggered,’ and (b) ‘eclipsed’. The
UV-Vis-NIR diffuse reflectance spectra of [MII(C4O4)(H2O)2] (MII = MnII, FeII, CoII, NiII, ZnII, and CdII) are
also shown. All frameworks are in the eclipsed phase except for the Cd analogue.
How many bands would you expect to observe for the Fe analogue based on our discussion of Orgel
diagrams? For Fe2+, d6, we would expect 1 transition based on the Orgel diagram.
In reality, two overlapping bands at 8590 and 10700 cm-1 are observed. Why? For the octahedral Fe2+
centre with a high spin d6 configuration, there is degeneracy in the ground state electronic configuration. A
Jahn-Teller distortion will result which removes the degeneracy.
/1
3. For the complex [Cr(NH3)6]3+:
Why are there multiple d-d bands?
How can we explain and describe them?
i)
What is the ground state term symbol?
4
F
ii) Draw the relevant Orgel diagram.
(iii)
Transition is from 4A2g to 4T2g (21550 cm-1)
Transition is from 4A2g to 4T1g (28500 cm-1)
Note: There are additional possible d-d transitions from the ground 4A2g state to the excited states (the first
excited 4P term for example is shown above), however, these will be of higher energy (i.e., >28500 cm-1)
and will be overlapped by more intense CT transitions. Therefore, in the visible region of the absorption
spectrum for [Cr(NH3)6]3+, we observe just two of the possible d-d transitions arising from the ground 4F
term. For the purposes of our discussion, you only need to be able to determine transitions from the ground
term, but you should be aware that higher energy d-d transitions will exist.
iv) Considering the Spin and Laporte selection rules, are the transitions allowed or forbidden? The
transitions are spin allowed (S = 0), but they are Laporte forbidden (g  g and L = 0). The Laporte
rule is relaxed by (i) deviation from perfect octahedral symmetry due to non-spherical ligands, (ii)
asymmetric vibrations and (iii) spin-orbit coupling.
v) What is the most likely origin of the charge transfer (CT) transition above 50000 cm-1? Ligand-to-Metal
Charge Transfer (LMCT) since the NH3 ligands are good  donors while the d3 metal centre has
available orbitals to accept electron density.
/2
4. [FeF6]3- is colourless whereas [CoF6]3- is coloured and exhibits one band in the visible region of its
electronic absorption spectrum.
i)
Classify F- as a -donor/acceptor and/or -donor/acceptor ligand. Are these complexes high or low
spin? F- is a weak field ligand and the complexes are high spin. F- is a good -donor ligand and a poorer
-donor.
ii) What are the ground state term symbols for the two complexes? 6S for [FeF6]3- and 5D for [CoF6]3iii) Construct an Orgel diagram for the cobalt complex and assign the electronic transition.
The transition is from 5T2g to 5Eg
/3
5. In the electronic absorption spectrum of [Co(OH2)6]2+, three bands are observed at 8,100, 16,000 and
19,400 cm-1. Oxidation of the complex by one electron leads to a significant change in the electronic
absorption spectrum.
What is the ground state term symbol for [Co(OH2)6]2+? Co2+ is d7, so we have a 4F ground state term
symbol
ii) Assign the three electronic transitions in the electronic absorption spectrum of [Co(OH2)6]2+ with the aid
of an Orgel diagram.
i)
Co2+ is d7 (an additional two electrons past a fully half-filed set of d orbitals), so the Orgel diagram is like
that for d2 and we can assign the three electronic transitions as shown below:
iii) How many electronic transitions would you expect to observe in the electronic absorption spectrum of
the oxidised complex? Determine the ground state term symbol for the oxidised complex and use an
Orgel diagram and assign the transition.
The oxidised complex is [Co(OH2)6]3+ where Co3+ is d6 and the ground state term symbol is 5D. The Orgel
diagram for d6 (an additional one electron past a fully half-filed set of d orbitals) can be considered to be like
that for d1, so we have:
One transition is therefore predicted from 5T2g to 5Eg.
/4
CHEM3x19 – Materials Chemistry, Week 1 Worksheet (1)
References (e-Text available via our Canvas site): Chapter 7, pages 243-261 in Solid State Materials
Chemistry, Patrick Woodward, Pavel Karen, John Evans, Thomas Vogt, 2021.
Also, Chapter 20 in Inorganic Chemistry by Atkins, Weller, Overton, Rourke and Armstrong, 6th Edition
(any earlier addition will also include similar content).
1. Identify the ground term for each set of terms:
i) 3F, 3P, 1P, 1G
ii) 5D, 3H, 3P, 1G, 1I
iii) 6S, 4G, 4P, 2I
2. See below the ball and stick representations of the two phases of the cubic [MII(C4O4)(H2O)2] sodalite
cage showing the different relative orientations of the squarate units: (a) ‘staggered,’ and (b) ‘eclipsed’. The
UV-Vis-NIR diffuse reflectance spectra of [MII(C4O4)(H2O)2] (MII = MnII, FeII, CoII, NiII, ZnII, and CdII) are
also shown. All frameworks are in the eclipsed phase except for the Cd analogue.
How many bands would you expect to observe for the Fe analogue based on our discussion of Orgel
diagrams?
In reality, two overlapping bands at 8590 and 10700 cm-1 are observed. Why?
/1
3. For the complex [Cr(NH3)6]3+:
Why are there multiple d-d bands?
How can we explain and describe them?
i)
What is the ground state term symbol?
ii) Draw the relevant Orgel diagram.
iii) Assign the transitions in the spectrum above.
iv) Considering the Spin and Laporte selection rules, are the transitions allowed or forbidden?
v) What is the most likely origin of the charge transfer (CT) transition above 50000 cm-1?
/2
4. [FeF6]3- is colourless whereas [CoF6]3- is coloured and exhibits one band in the visible region of its
electronic absorption spectrum.
i)
Classify F- as a -donor/acceptor and/or -donor/acceptor ligand. Are these complexes high or low
spin?
ii) What are the ground state term symbols for the two complexes?
iii) Construct an Orgel diagram for the cobalt complex and assign the electronic transition.
/3
5. In the electronic absorption spectrum of [Co(OH2)6]2+, three bands are observed at 8,100, 16,000 and
19,400 cm-1. Oxidation of the complex by one electron leads to a significant change in the electronic
absorption spectrum.
i)
What is the ground state term symbol for [Co(OH2)6]2+?
ii) Assign the three electronic transitions in the electronic absorption spectrum of [Co(OH2)6]2+ with the aid
of an Orgel diagram.
iii) How many electronic transitions would you expect to observe in the electronic absorption spectrum of
the oxidised complex? Determine the ground state term symbol for the oxidised complex and use an
Orgel diagram and assign the transition.
/4
CHEM3x19 – Materials Chemistry, Week 1 Feedback (2)
References (e-Text available via our Canvas site): Chapter 7, pages 274-276 in Solid State Materials
Chemistry, Patrick Woodward, Pavel Karen, John Evans, Thomas Vogt, 2021. Also, Chapter 23 in Inorganic
Chemistry by Atkins, Weller, Overton, Rourke and Armstrong, 6th Edition (any earlier addition will also
include similar content).
SPLITTING OF ATOMIC ORBITALS IN A CRYSTAL FIELD
1.
By considering the shapes of the orbitals, sketch qualitative splitting diagrams in an octahedral crystal
field for a central element with f-orbitals (see over for pictures of these orbitals). Note that these splittings
are very small for the lanthanoids but become significant for the actinoids.
The fx3, fy3 and fz3 orbitals have their major lobes directed along x, y and z respectively. Electrons in these
orbitals are most affected by an octahedral crystal field. They are equally affected and form a set of three
degenerate orbitals.
The eight lobes of the fz(x2-y2), fy(x2-z2) and fx(y2-z2) orbitals are directed between pairs of axes (x & y, x & z and
y & z respectively). They are less affected than the first set and are also affected equally and form a second
set of three.
The eight lobes of the particularly beautiful fxyz orbital are directed between all three axes. The lobes are
furthest from the charges along the axes in an octahedral crystal field and this orbital is the lowest in energy.
Thus, the overall splitting for the seven f-orbitals in an octahedral crystal field is:
2.
With this crystal field splitting diagram in mind, comment on the observations that the electronic spectra
of lanthanoid complexes contain many absorptions some of which are weak and sharp and similar to those
of the gas-phase metal ions, and some of which are broad and are affected by the ligands present.
The interpretation of the electronic spectra of 4fn metal ions is similar to what we have already discussed for
the d-block, but there are some important differences. As we have discussed, spin-orbit coupling is very
pronounced for the f-block ions and is more important than crystal field splitting. There are many possible
transitions giving rise to absorption bands. Since the 4f electrons are well shielded and not influenced by the
environment of the ion, bands from f-f transitions are sharp (rather than broad like d-d transitions) and their
positions in the spectrum are not affected very much by complex formation. Intensities of the absorptions are
typically low, indicating that the probabilities of f-f transitions are low (i.e. little d-f mixing). Absorptions due
to 4f-5d transitions are broad and are affected by ligand environment.
3. The lanthanoid elements have Z = 58 – 71 (Ce – Lu). Unlike the d-block elements, the effects of ligand-fields
on the energy levels of lanthanoid ions are very small. As a result, the magnetic properties of lanthanoid
complexes and materials are very similar to those of the free ions and not sensitive to the complex geometry or
ligand type.
Cs
Ba
La
Ce
Pr
Nd
Pm Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
Their chemistry is almost completely dominated by the trivalent (3+) oxidation state in which all the lanthanoid
ions have 4fn configurations. To work out n:
• count across from Cs to the lanthanoid to obtain the number of valence electrons
• take away 3 for the charge
Work out the ground level for each possible fn ion. To determine J, use Hund’s 3rd rule.
mℓ
3
2
1
0
-1
-2
-3
4fn
L
S
J
ground level
Ce3+
1
3
1/2
5/2
2F
5/2
Pr3+
2
5
1
4
3H
4
Nd3+
3
6
3/2
9/2
4I
Pm3+
4
6
2
4
5I
Sm3+
5
5
5/2
5/2
6H
5/2
Eu3+
6
3
3
0
7F
0
Gd3+
7
0
7/2
7/2
8S
Tb3+
8
3
3
6
7F
6
Dy3+
9
5
5/2
15/2
6H
15/2
Ho3+
10
6
2
8
5I
Er3+
11
6
3/2
15/2
4I
9/2
4
7/2
8
15/2
Tm3+
12
5
1
6
3H
6
Yb3+
13
3
1/2
7/2
2F
7/2
Lu3+
14
0
0
0
1S
0
CHEM3x19 – Materials Chemistry, Week 1 Worksheet (2)
References (e-Text available via our Canvas site): Chapter 7, pages 274-276 in Solid State Materials
Chemistry, Patrick Woodward, Pavel Karen, John Evans, Thomas Vogt, 2021. Also, Chapter 23 in Inorganic
Chemistry by Atkins, Weller, Overton, Rourke and Armstrong, 6th Edition (any earlier addition will also
include similar content).
1.
By considering the shapes of the orbitals, sketch qualitative splitting diagrams in an octahedral crystal
field for a central element with f-orbitals (see over for pictures of these orbitals). Note that these splittings
are very small for the lanthanoids but become significant for the actinoids.
2.
With this crystal field splitting diagram in mind, comment on the observations that the electronic spectra
of lanthanoid complexes contain many absorptions some of which are weak and sharp and similar to those
of the gas-phase metal ions, and some of which are broad and are affected by the ligands present.
The lanthanoid elements have Z = 58 – 71 (Ce – Lu). Unlike the d-block elements, the effects of ligandfields on the energy levels of lanthanoid ions are very small. As a result, the magnetic properties of
lanthanoid complexes and materials are very similar to those of the free ions and not sensitive to the
complex geometry or ligand type.
Cs Ba
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
Their chemistry is almost completely dominated by the trivalent (3+) oxidation state in which all the lanthanoid
ions have 4fn configurations. To work out n:
• count across from Cs to the lanthanoid to obtain the number of valence electrons
• take away 3 for the charge
3.
Work out the ground level for each possible fn ion. To determine J, use Hund’s 3rd rule.
mℓ
3
2
1
0
-1
-2
-3
4fn
L
S
J
Ce3+
1
3
1/2
5/2
2F
Pr3+
2
5
1
Nd3+
3
6
3/2
Pm3+
4
6
2
Sm3+
5
5
5/2
Eu3+
6
3
3
Gd3+
7
0
7/2
Tb3+
8
3
3
6
7F
Dy3+
9
5
5/2
Ho3+
10
6
2
Er3+
11
6
3/2
Tm3+
12
5
1
Yb3+
13
3
1/2
Lu3+
14
0
0
ground level
5/2
6
f-Orbitals
fx 3
fx 3
fz3
fz(x2-y2)
fy(x2-z2)
fx(y2-z2)
fxyz
CHEM3x19 – Materials Chemistry, Week 2 Feedback
Note that the first question on this worksheet was for the purpose of an example in the lecture and is beyond
what would be asked in an exam.
1. SrTiO3 has what is known as the ideal perovskite structure. Strontium atoms are the corners of cubes
(gray – note that this should be green), titanium atoms are at cube body centres (green – note that this
should be grey), and oxygen atoms are at cube face centres (red). Take the cube edge length to be a.
a) What is the Bravais lattice type?
As discussed in the lecture, we can draw out the repeating atoms in 3-dimensional space for the Sr, Ti
and O. If we do this, we find that each of the atoms is in a cubic arrangement, i.e., a simple cubic lattice
type.
b) Verify that there are 3 oxygen atoms, 1 titanium atom and 1 strontium atom for each primitive unit
cell in the crystal.
Consider each of the atoms in turn with respect to the composition of the cell shown above:
O (red): there are 6 O atoms each on a face of the cube (and therefore shared by 2 cubes) so 6 x ½ = 3
Sr (green): there are 8 Sr atoms each at a corner of the cube (and therefore shared by 8 cubes) so 8 x 1/8
=1
Ti (gray): there is 1 in the middle (not shared with any other cubes) so 1 x 1 = 1
We have therefore verified that the composition is SrTiO3.
c) Write down a set of primitive lattice vectors for each of the atoms in this structure.
For Sr: (0,0,0) (it’s at the origin – corner of the cube if we consider a simple x, y, z coordinate system,
remembering that in crystallography we call these axes a, b, c. Because it’s a cubic structure, a = b = c.)
For Ti: (a/2, a/2, a/2)
For O: (a/2, a/2, 0), (a/2, 0, a/2), (0, a/2, a/2)
2. A piezoelectric ceramic material is suspected to have oxygen anions (O2-) at locations:
0 ½ ½ , ½ 0 ½ , ½ ½ 0 , barium cations (Ba2+) at 0 0 0 and titanium cations (Ti4+) at ½ ½ ½ .
a) Draw and label the contents of a unit cell of this material.
Ba
O
Ti
The structure is the same as that discussed in Question 1 with the oxygen atoms (red) at the faces of the
cube, the Ba atoms at each of the corners of the cube (green) and the Ti atom at the very centre of the
cube (gray).
b) What is the chemical formula?
We use an identical method to that described in Question 1b to achieve the formula BaTiO3.
c) Can you suggest what the Bravais lattice is that defines
this structure?
Simple cubic.
3. Identify each of the lattice directions and planes in the following figures relative to the coordinate axes
shown. All intersections between the vectors and the unit cells occur at corners or midpoints of cell
edges or face centres.
(i)
(ii)
(iii)
As shown in lectures, the lattice directions can be determined by the difference between the starting and
ending points of the vectors. The origin is (0,0,0) and is located at the back left-hand side of each cube.
(i)
(ii)
(iii)
Starting point = (0,0,1) and Ending point = (1/2, 1, 0). The difference between these (i.e., Ending
point – starting point) = (1/2,1,-1)
Starting point = (0,1/2,0) and Ending point = (1,1,1). The difference is (1,1/2,1)
Starting point = (1,1/2,1) and Ending point = (1/2,0,1). The difference is (1/2,1/2,0)
(i)
(i)
(ii)
(iii)
(ii)
(iii)
A single (0 0 1/2) plane – we consider in turn where the plane intercepts the x, y and z axes.
A single (0 1 1) plane
This one’s a trickier one (and beyond what we discussed in lectures (i.e., not for examination), but
added here for a challenge!). A (1/2 1/2 1) plane, but by convention, we use the reciprocals (once
again, for your interest only and not examinable), so it’s a (2 2 1) plane.
4. Suggest a mechanism by which nonstoichiometry occurs in Fe1-xO.
Initially Fe2+ ions occupy octahedral sites. A number of vacancies are created on these normally occupied
octahedral sites, but over the whole structure, this number is greater than required by the compound
stoichiometry. The additional iron required occupies tetrahedral sites adjacent to the vacancies. In order to
maintain charge balance, the interstitial iron is oxidised to +3 and a number of iron atoms surrounding the
defect, but on normal lattice sites, must also be oxidised to Fe3+. Note that Fe3+, as a high spin d5 ion, has
no preference for octahedral or tetrahedral sites. The relative ease of oxidising Fe2+ to Fe3+ accounts for
the fairly broad range of compositions of Fe1-xO.
5. Solid solutions occur most frequently for materials incorporating d-metal ions. As x increases in
La1-xSrxFeO3 and La(III) is replaced by Sr(II), what needs to occur to maintain overall charge balance?
The oxidation state of iron must change from Fe(III) to Fe(IV). This change can occur through a gradual
replacement of one exact oxidation state, here Fe(III) by another, Fe(IV), on a proportion of the cation
sites within the structure.
6. Which of the following form solid solutions?
(a) Na and K are chemically similar and have bcc structures; their atomic radii are 191 and 235 pm,
respectively.
No
(b) Cu and Ni have similar electropositive character, similar crystal structures (ccp) and similar atomic
radii (128 and 125 pm, respectively).
Yes
Extra Practice Questions to test your skills
The electronic absorption spectrum of [FeBr4]2- is characterised by one broad band centred at 3000 cm-1, while
the spectrum of [Fe(OH2)6]2+ is characterised by a band at 10400 cm-1 which has a shoulder peak at 8300 cm-1.
(i) What is ground term for these complexes? 5D
(ii) Construct an Orgel diagram for these two complexes and assign the transition that gives rise to the
single absorption band in [FeBr4]2-.
The transition in [FeBr4]2- is from 5E to 5T2
(iii) Explain why the spectrum of [FeBr4]2- consists of a single absorption band whilst that of [Fe(OH2)6]2+
consists of two overlapping bands.
Consider the crystal field splitting diagrams for the two complexes (remember that both Br- and H2O are weak
field ligands and give high spin complexes):
There is only one possible absorption band for the tetrahedral complex (as shown in part (ii), however for the
octahedral complex [Fe(OH2)6]2+, there is degeneracy in the ground state that will lead to a Jahn-Teller
distortion. The geometry of the complex will thus distort to lift this degeneracy (as per the definition of the
Jahn-Teller effect), and we observe 2 d-d bands that are closely spaced in energy.
CHEM3x19 – Materials Chemistry, Week 2 Questions
1. SrTiO3 has what is known as the ideal perovskite structure. Strontium atoms are the corners of cubes
(gray), titanium atoms are at cube body centers (green), and oxygen atoms are at cube face centers (red).
Take the cube edge length to be a.
a) What is the Bravais lattice type?
b) Verify that there are 3 oxygen atoms, 1 titanium atom and 1 strontium atom for each primitive unit
cell in the crystal.
c) Write down a set of primitive lattice vectors for each of the atoms in this structure.
2. A piezoelectric ceramic material is suspected to have oxygen anions (O2-) at locations:
0 ½ ½ , ½ 0 ½ , ½ ½ 0 , barium cations (Ba2+) at 0 0 0 and titanium cations (Ti4+) at ½ ½ ½ .
a) Draw and label the contents of a unit cell of this material.
b) What is the chemical formula?
c) Can you suggest what the Bravais lattice is that defines
this structure?
3. Identify each of the lattice directions and planes in the following figures relative to the coordinate axes
shown. All intersections between the vectors and the unit cells occur at corners or midpoints of cell
edges or face centres.
4. Suggest a mechanism by which nonstoichiometry occurs in Fe1-xO.
5. Solid solutions occur most frequently for materials incorporating d-metal ions. As x increases in
La1-xSrxFeO3 and La(III) is replaced by Sr(II), what needs to occur to maintain overall charge balance?
6. Which of the following form solid solutions?
(a) Na and K are chemically similar and have bcc structures; their atomic radii are 191 and 235 pm,
respectively.
(b) Cu and Ni have similar electropositive character, similar crystal structures (ccp) and similar atomic
radii (128 and 125 pm, respectively).
Extra Practice Questions to test your skills
The electronic absorption spectrum of [FeBr4]2- is characterised by one broad band centred at 3000 cm-1, while
the spectrum of [Fe(OH2)6]2+ is characterised by a band at 10400 cm-1 which has a shoulder peak at 8300 cm-1.
(i) What is ground term for these complexes?
(ii) Construct an Orgel diagram for these two complexes and assign the transition that gives rise to the
single absorption band in [FeBr4]2-.
(iii) Explain why the spectrum of [FeBr4]2+ consists of a single absorption band whilst that of [Fe(OH2)6]2+
consists of two overlapping bands.
CHEM3x19 – Materials Chemistry, Week 3 Questions (1)
Consider this example first: For ionic structures, the empirical formula is determined by counting the atoms and partial
atoms within the boundary of the unit cell.
Na+:
(8 x 1/8) + (6 x ½) = 4
Cl-:
(12 x ¼) + 1 x 1 = 4
Na Cl = NaCl
4
4
The number of formula units present in the unit cell is denoted Z, so Z = 4
Coordination = 6:6 (the first number is for the cation and the second the anion).
Note: recall our discussion and models in Lecture 4 where we saw that an atom at the corner of the cube is shared by 8
other cubes; an atom at the face of a cube is shared between 2 cubes; an atom on the side of a cube is shared by 4 other
cubes.
1. An alloy of copper and gold has the structure shown below (note that once copper atom occupies each face of the
unit cell). Calculate the composition of this unit cell. What is the lattice type of the structure?
Remember that the empirical formula is determined by counting the atoms and partial atoms within the boundary
of the unit cell. There are 6 x Cu atoms are on each of the faces of the cube (since each ion on a face is shared by
one other cube, just like a wall between two rooms, the occupancy is ½). In addition, there are 8 Au atoms at the
corners of the unit cell, each of these being shared between 8 cubes (therefore, 1/8 occupancy).
Cu: (6 x ½) = 3
Au: (8 x 1/8) = 1
composition is therefore Cu3Au and the lattice is face-centred cubic (fcc). The gold atoms are in a primitive
array, and the copper atoms form an octahedron inside the unit cell.
2. Predict structures for the ionic compounds RbI using the radius-ratio rules (below) and the ionic radii below for
Rb+ (0.148 nm) and I- (0.220 nm) (for six-fold coordination).
We calculate the following ratio: rcation/ranion = 0.148/0.220 = 0.673, from which we predict that RbI adopts a rock salt
(sodium chloride) structure type with coordination numbers of 6:6.
3. Using Zachariasen’s rules, explain why SiO2 forms a glass whereas MgO does not.
We will address Zachariasen’s rules in Lecture 7 (Week 4). Pre-empting our discussion, Zachariasen’s rules are empirical
rules that help us to rationalise why some inorganic oxides will form glasses (and why others cannot).
Zachariasen’s Rules for Glass Formation: (empirical observations for oxides)
1. No oxygen atom may be linked to more than two cations
2. The cation coordination number is small: 3 or 4.
3. Oxygen polyhedra share corners, not edges or faces.
4. For 3D networks, at least three corners must be shared
In general, all four rules should be satisfied for glass formation to occur.
Low coordination numbers, corner-sharing rules imply that glass formation is more likely with open, low density
polyhedral structures.
Here are the structures of SiO2 (cations are 4 coordinate and anions are 2 coordinate) and MgO (cations and anions are all
6 coordinate):
SiO2
MgO
Qualitatively, we have already seen the structure of SiO2 (i.e., we know that Si4+ is tetrahedral and 4 coordinate, so it
obeys Rule #2). We also know that each O2- is shared between two Si4+ ions which satisfies Rule #1. In addition, the
polyhedra share corners (Rule #3). In summary, the rules are satisfied and a glass is formed.
For MgO, we need to consider each of the rules in turn. In considering Rule #2, we can already see that the coordination
number of 6 violates the small coordination numbers that are characteristic of glasses. Therefore, it does not form a glass.
4. Again considering Zachariasen’s rules, why doesn’t Al2O3 form a glass, but aluminosilicates with compositions
11-16 mole percent (mol%) Al2O3, 52-60 mol% SiO2 and 9-11 mol% K2O will?
In Al2O3, Al3+ is in an octahedral coordination (not 3- or 4-coordinated), so it violates Rule #1, and O2− is 4-coordinated
(not 2-coordinated). In aluminosilicates, Al3+ and Si4+ are both 4-coordinated and Al requires charge balance by alkalimetal or alkaline-earth ions.
Adding Al2O3 increases the connectivity of an alkali-modified SiO2 glass by replacing non-bridging oxygen (bonded to
the alkali-metal cation) with cross-linked Al–O–Si bonds.
CHEM3x19 – Materials Chemistry, Week 3 Questions (1)
Consider this example first: For ionic structures, the empirical formula is determined by counting the atoms and partial
atoms within the boundary of the unit cell.
Na+:
(8 x 1/8) + (6 x ½) = 4
Cl-:
(12 x ¼) + 1 x 1 = 4
Na Cl = NaCl
4
4
The number of formula units present in the unit cell is denoted Z, so Z = 4
Coordination = 6:6 (the first number is for the cation and the second the anion).
Note: recall our discussion and models in Lecture 4 where we saw that an atom at the corner of the cube is shared by 8
other cubes; an atom at the face of a cube is shared between 2 cubes; an atom on the side of a cube is shared by 4 other
cubes.
1. An alloy of copper and gold has the structure shown below (note that once copper atom occupies each face of the
unit cell). Calculate the composition of this unit cell. What is the lattice type of the structure?
2. Predict structures for the ionic compounds RbI using the radius-ratio rules (below) and the ionic radii below for
Rb+ (0.148 nm) and I- (0.220 nm) (for six-fold coordination).
3. Using Zachariasen’s rules, explain why SiO2 forms a glass whereas MgO does not.
4. Again considering Zachariasen’s rules, why doesn’t Al2O3 form a glass, but aluminosilicates with compositions
11-16 mole percent (mol%) Al2O3, 52-60 mol% SiO2 and 9-11 mol% K2O will?
CHEM3x19 – Materials Chemistry, Week 3 Feedback (2)
1. The relative cracking rates of various hexanes using the zeolite HZSM-5 at 340 oC are given below. Explain the
trends in the data.
Considering the structures of these hexanes, n-hexane is linear, and the others are branched (with 2,2-dimethyl-butane)
being the most highly branched of the series. Thus, the least sterically-hindered isomer would be expected to diffuse into
the zeolite pores relatively rapidly compared with the others, and undergo cracking readily.
2. Zeolites are known to exhibit reactive-selective catalysis. With this in mind, rationalise the data presented in the
following graph for the dehydration of butanols.
dehydration of butanols
a, b = iso-butanol, n-butanol in
Ca-X (channel ~ 7 Å, cavity ~ 10 Å)
c, d = n-butanol, iso-butanol in
Ca-A (channel ~ 4 Å , cavity ~ 10 Å)
As noted in the lecture, the data represented in curve d is for iso-butanol in Ca-A. Iso-butanol is relatively more
branched than n-butanol, so it is too large to fit within the Ca-A channel to undergo catalysis. Thus, the
percentage conversion is very low. Note that when Ca-X is instead used, the channel size is larger (7 vs. 4 Å), so
the iso-butanol is catalysed.
3. Look at the following data for the diffusion coefficients of simple hydrocarbons in ZSM-5. How would you
explain the observation that 1,4-dimethylbenzene is selectively extracted from the product mixture?
1,4-Dimethylbenzene has a relatively more linear structure than the other hydrocarbons and thus would be
expected to diffuse relatively more quickly through the ZSM-5 pores.
CHEM3x19 – Materials Chemistry, Week 3 Questions (2)
1. The relative cracking rates of various hexanes using the zeolite HZSM-5 at 340 oC are given below.
Explain the trends in the data.
2. Zeolites are known to exhibit reactive-selective catalysis. With this in mind, rationalise the data presented in the
following graph for the dehydration of butanols.
dehydration of butanols
a, b = iso-butanol, n-butanol in
Ca-X (channel ~ 7 Å, cavity ~ 10 Å)
c, d = n-butanol, iso-butanol in
Ca-A (channel ~ 4 Å , cavity ~ 10 Å)
3. Look at the following data for the diffusion coefficients of simple hydrocarbons in ZSM-5. How would you
explain the observation that 1,4-dimethylbenzene is selectively extracted from the product mixture?
CHEM3x19 – Materials Chemistry, Week 4 Feedback
1. The cuboctahedra can be interconnected in two different ways, as shown below. The first involves interconnection
of copper-based paddlewheels by a neutral diamine ligand; the second involves interconnection of two carbazole
moieties. How would you describe the topologies of the two nets?
At right, we would describe the net as a primitive cubic lattice. At left, while slightly difficult to see in this 2D
figure, we have a face centred cubic structure.
2. Comment on the observation that the MOF at right could not be activated for porosity tests, while the MOF at left
is a highly porous, gas-accessible framework with the record gravimetric methane uptake.
The MOF at right is less stable to desolvation of pore solvent (i.e. ‘activation’), while the MOF at left is stable to
desolvation. This implies that the carbazole linker for the face centred cubic structure helps to ‘bolster’ the
structure and maintain the structural integrity upon desolvation. By comparison, the primitive cubic lattice that
does not have this additional structural component collapses when the pore solvent is removed.
3. We have seen that MOFs are excellent immobilisers for molecular catalysts. In this regard, consider a heterogeneous
enantioselective catalyst based on a catalytically active metal centre attached to a MOF linker.
a) How would you expect the reaction rate to compare with the equivalent catalyst in solution? How might
reaction rate vary as pore size is varied?
We might anticipate that the reaction may be relatively slower for the MOF as the analyte needs to diffuse
into the pores and the product needs to diffuse out of the pores. In homogeneous solution, this potential
hindrance would not exist. If the pore size is smaller, we might expect the reaction rate to decrease due to
impeded diffusion of the analyte into the pores and diffusion of product out of the pores.
b) How would you prove that catalysis is truly heterogeneous and not caused by catalyst leaching from
the MOF into solution?
As an example of one possible experiment to test the efficacy of the MOF catalyst, you could recover the
solid MOF catalyst by filtering it off from the solution, and then perform another catalytic reaction with
fresh solution. If the catalyst had leached into solution in the first experiment, a decrease in the product
yield would be expected. Note that there are a number of other answers to this question that would be
equally correct.
4. Solubilised alkanolamines in aqueous solution are excellent molecules for absorbing CO2 by virtue of acid-base
interactions. This process was proposed in the 1930’s to remove CO2 from flue gases. More recently, attention has
turned to high surface area solid MOFs that overcome some of the limiations of traditional absorbents.
Unfortunately, it is virtually impossible to synthesise MOFs containing pendant amine groups in a one-pot
synthesis from the starting amine-based ligand and metal ion precursors. Suggest why this may be so and explain
how you would design a synthesis to overcome the problem.
Pendant amine groups are very good coordinating groups, and can themselves coordinate metal ions, leading to
polymeric and non-crystalline (amorphous) products, rather than a desired MOF. To overcome this issue, you
could protect the amine, perform the MOF synthesis and then deprotect it. Alternatively, you could install the
amine post-synthetically.
5. The heat of adsorption provides an indication of the strength of an interaction between an adsorbate and a solid
adsorbent. The following data (Long et al., Chem. Sci., 2014, 5, 4569) relates to CO2 adsorption on the open metal
sites of [M2(dobdc)] at 298 K (where H4dobdc = 2,5-dihydroxy-1,4-benzenedicarboxylic acid). This 3-D MOF
exhibits 1-D hexagonal channels aligned along the c axis. Why do we observe a variation in amount of CO2
adsorbed (mol/mol M2+) and a variation in the heat of adsorption?
The capacity of the MOF for CO2 and the heat of adsorption are both a function of the relative strength of
association between CO2 and the open metal sites. The variability of the data with the identity of the metal ion
reflects the different Lewis acidities of the metal ions. You would be familiar with the concept of the IrvingWilliams series, and it is interesting to note here that the trends do somewhat follow this series. For example, Cu2+
which is d9 (Jahn-Teller distorted), has a relatively weaker association with CO2.
CHEM3x19 – Materials Chemistry, Week 4 Questions
1. The cuboctahedra can be interconnected in two different ways, as shown below. The first involves interconnection
of copper-based paddlewheels by a neutral diamine ligand; the second involves interconnection of two carbazole
moieties. How would you describe the topologies of the two nets?
2. Comment on the observation that the MOF at right could not be activated for porosity tests, while the MOF at left
is a highly porous, gas-accessible framework with the record gravimetric methane uptake.
3. We have seen that MOFs are excellent immobilisers for molecular catalysts. In this regard, consider a heterogeneous
enantioselective catalyst based on a catalytically active metal centre attached to a MOF linker.
a) How would you expect the reaction rate to compare with the equivalent catalyst in solution? How might
reaction rate vary as pore size is varied?
b) How would you prove that catalysis is truly heterogeneous and not caused by catalyst leaching from
the MOF into solution?
4. Solubilised alkanolamines in aqueous solution are excellent molecules for absorbing CO2 by virtue of acid-base
interactions. This process was proposed in the 1930’s to remove CO2 from flue gases. More recently, attention has
turned to high surface area solid MOFs that overcome some of the limiations of traditional absorbents.
Unfortunately, it is virtually impossible to synthesise MOFs containing pendant amine groups in a one-pot
synthesis from the starting amine-based ligand and metal ion precursors. Suggest why this may be so and explain
how you would design a synthesis to overcome the problem.
5. The heat of adsorption provides an indication of the strength of an interaction between an adsorbate and a solid
adsorbent. The following data (Long et al., Chem. Sci., 2014, 5, 4569) relates to CO2 adsorption on the open metal
sites of [M2(dobdc)] at 298 K (where H4dobdc = 2,5-dihydroxy-1,4-benzenedicarboxylic acid). This 3-D MOF
exhibits 1-D hexagonal channels aligned along the c axis. Why do we observe a variation in amount of CO2
adsorbed (mol/mol M2+) and a variation in the heat of adsorption?
CHEM3x19 – Materials Chemistry, Week 4 Feedback (2)
1. If you were considering the design of a MOF for a gas storage application, what structural and physicochemical
properties would you desire? Consider this ‘design’ from the perspective of both thermodynamics and kinetics.
For a gas storage application, we would need a MOF with a high thermodynamic adsorption capacity for the gas as
well as an appropriate pore aperture (i.e., the gas being stored needs to be able to enter the pores). In terms of
kinetics, we would ideally like to be able to adsorb and desorb the stored gas quickly, so we would desire high
kinetics for adsorption/desorption, and reversibility for the adsorption/desorption process. In terms of structural
properties of the MOF, it would need to have appropriate mechanical and chemical stability.
2. Inerpenetration is common in MOF structures. Considering the schematic below, what could you change in the
synthesis of a MOF to prevent interpenetration? Hint: remember that it’s not only the metal centres and organic
linkers that are involved in the supramolecular assembly process that generates MOFs.
A couple of options would be to incorporate a bulky substituent on the linker to prevent interpenetration. It is
possible that with the correct selection of the substituent, it can be postsynthetically removed (e.g. a thermolabile
substituent). Another option is to change the synthesis solvent to a relatively bulky solvent. Since the solvent
templates the pores, we might expect that a bulkier solvent, when occluded within the pores, would prevent another
inteterpenetrating network forming.
3. Here is a pillared layered MOF, [Ni(L){Ni(CN)4]. When L = dpac (red diamonds) or bpene (black circules), the
following sorption isotherms are observed (soild = adsorption, open = desorption):
Why is the adsorption pressure different for the MOFs with the two different pillaring ligands? Why is there
hysteresis when L = bpene, but not when L = dpac?
Looking at the ligands dpac and bpene, the former is relatively more rigid, and we might expect that the MOF is
relatively more structurally robust. By comparison, the bpene ligand is relatively more flexible. Thus, in the dpac
MOF, there is available micropore space into which the CO2 molecules adsorb and desorb. Remember that initially,
there will be solvent molecules located within the pores that must be removed before we can commence the gas
adsorption measurements.
When the ligand is changed to bpene however, the MOF is flexible. Initially, the solvents are removed and the
structural flexibility of the ligand causes the channels to become hindered, as shown in the schematic picture below.
When CO2 molecules begin to enter the hindered pores, it is not until we reach a threshold pressure of > 5 atm that
the MOF suddenly ‘opens’ its pores. The channels are now accessible. When the pressure is reduced for desorption,
the structure remains in its ‘open’ state until ~ 5 atm, at which point the structure again distorts and the channels
are hindered.
CHEM3x19 – Materials Chemistry, Week 4 Questions (2)
1. If you were considering the design of a MOF for a gas storage application, what structural and physicochemical
properties would you desire? Consider this ‘design’ from the perspective of both thermodynamics and kinetics.
2. Inerpenetration is common in MOF structures. Considering the schematic below, what could you change in the
synthesis of a MOF to prevent interpenetration? Hint: remember that it’s not only the metal centres and organic
linkers that are involved in the supramolecular assembly process that generates MOFs.
3. Here is a pillared layered MOF, [Ni(L){Ni(CN)4]. When L = dpac (red diamonds) or bpene (black circules), the
following sorption isotherms are observed (soild = adsorption, open = desorption):
Why is the adsorption pressure different for the MOFs with the two different pillaring ligands? Why is there
hysteresis when L = bpene, but not when L = dpac?
Q 1The diagrams below illustrate rotation axis. Do any of these also contain a mirror plane?
Q 2 Go to the website https://www.doitpoms.ac.uk/tlplib/crystallography3/symmetry.php
and find a lattice
Q 3 Below shows the unit cell of the semiconductor tin arsenide. What is the empirical formula for
this compound?
Q 1. Why do crystals diffract X-rays?
Q 2. Would X-ray or neutron diffraction be more appropriate to distinguish between the
following pairs of atoms:
(a) C and U
(b) W and Re
(c) N and Cl
Q 3. What additional information would you need to verify your answers in Q 2
Q 4. What is the “phase problem” and how do we solve it?
Q 5. How would a X-ray diffraction pattern change if the wavelength of the X-rays was
doubled?
Chemistry 3i19 Worksheet 11
Q 1 Consider the phase diagram on Slide 8:
(a) What phases are present at point B
(b) What is the weight % of Ni in each of the phases present at point B
(c) What is the ratio of the phases present at point B
Q 2. Consider the data on Slide 11
(a) Would you predict Zn to dissolve more Al or Ag?
(b) Would you predict more Zn or Al to dissolve in Cu?
Q3 Consider the data on Slide 14
(a) For a 40wt%Sn-60wt%Pb alloy at 150 oC, what are the compositions of the phases present.
Q 4. Would X-ray diffraction be sensitive to the lattice distortion when a small amount of Ni is
dissolved in Cu?
Chemistry 3i19 Worksheet 12
Q1. The phase diagram for SiO2, shown below, includes representations of the structures:
(a) What is the coordination number of Si in each of the four illustrated structures ( Hint the Si is
at the centre of the polyhedral).
(b) What is the most likely order (first or second) of the Stishovite to α-Quartz phase transition?
(c) Why it may be possible to recover crystalline samples of Stishovite at room temperature?
Q2 The structures of three polymorphs of ZrO2 are illustrated below:
(a) What is the likely nature of the fluorite to baddeleyite phase transition?
(b) Considering the bond distances listed in the table on slide 16 suggest why the addition of a
small cation would increase the stability of the cubic ZrO2 structure.
Q3. Based on the tolerance factor for perovskites (𝑡𝑡 =
𝑟𝑟𝐴𝐴 +𝑟𝑟𝑋𝑋
√2(𝑟𝑟𝐵𝐵 +𝑟𝑟𝑋𝑋 )
= 1 why does the inclusion of a
small cation on the A-site result in cooperative tilting of the octahedra?
Chemistry 3i19 Worksheet 13
Q 1. The following table lists the radius ratios for the alkali halides, using ionic radii for CN
=6
Ion
Radii (Å)
F-Cl-Br-I–
Radii
(Å
1.19
1.67
1.82
2.06
Li+
Na+
K+
Rb+
Cs+
0.79
0.66
0.47
0.43
0.38
1.07
0.90
0.64
0.59
0.52
1.38
1.16
0.82
0.76
0.67
1.66
1.39
0.99
0.91
0.81
1.81
1.52
1.08
0.99
0.88
(a) Whet structure do you expect these to exhibit based on the optimal packing efficiency
as described by the Radius Ratio Rules
(b) Only the seven salts indicated in bold have the expected structure. However, all the
other compounds also adopt this structure. Suggest a reason why the Radius Ratio
Rule may break-down for these salts.
Q2. In slide 14 the effective ionic radii for high and low spin Co(II) in an octahedral
environment are different – with the high spin state having a larger effective ionic radius
(0.745 vs 0.65 Å). Why does the spin state impact the effective ionic radius?
Q3. Using the information given on slide 15 what is the ideal Mg-O distance for a regular
tetrahedra?
Q3. CuScO2 adopts the delafossite structure with linear Cu (two equidistant oxygen
neighbours @ 1.80 Å) and the Sc is bonded to six equidistant oxygen atoms @2.13 Å). The
corresponding R0 values are 1.61 and 1.85 respectively with B = 0.37.
(a) What is the likely formal oxidation state of the Cu cation.
(b) What are the Bond Valence Sums for the Cu and Sc cations? Does this agree with
your expectation above?
Chem 3i19 Lecture Worksheet 15
1. Why are there no perfect crystals?
2. What is the difference between intrinsic and extrinsic defects? Which are likely to lead to
higher ionic conductivity?
3. What are the differences between cubic stabilised zirconia (CSZ) and partially stabilised
zirconia (PSZ)? Compare and contrast.
4. Which are the general features required for a material to be a good ionic conductor?
5. What type of materials are suitable as electrolytes in solid oxide fuel cells?
6. Are anion conductors more common than cation conductors?
7. What determines the cell potential of a battery?
8. Why are ‘anode’ and ‘cathode’ not suitable terms for a rechargeable Li-ion battery (e.g.
compared to standard non-rechargeable battery)?
Lecture 17
Assign an inner or outer sphere mechanism to each of the following:
(a) When [VO(edta)]2- reacts with [V(edta)]2-, a transient red colour is observed.
The appearance of a red colour during the reaction suggests that a coloured intermediate is
formed. Since there is no intermediate in the outer sphere mechanism, it suggests electron
transfer occurs via the formation of an oxo VII-O-VIV bridged complex.
(b) The rate of reduction of [Co(NCS)(NH3)5]2+ by Ti3+(aq) is 36000 times smaller than the rate of
[Co(N3)(NH3)5]2+ reduction.
It is hard to imagine that the change in ligand would have such an effect on the rate of an outer
sphere mechanism, since these are not involved. Presumably, NCS- is a poorer bridging ligand than
N3- and so the [Co(NCS)(NH3)5]2+ reduction proceeds much slower than the reduction of
[Co(N3)(NH3)5]2+. Note that it is also plausible here that [Co(NCS)(NH3)5]2+ reduction occurs via an
outer sphere mechanism as these are always significantly slower than inner sphere mechanisms.
(c) The reduction of [(NH3)5Co(µ-O)C(O)R]2+ (R = Me, Et) by V2+ and Cr2+ decreases as the pH
decreases.
Protonation of the carbonyl oxygen atom on the carboxylate ligand at low pH would prevent it
from forming a bridged complex – the drop in rate is thus consistent with an inner sphere
mechanism. There is no clear reason why this would affect the rate of an outer sphere
mechanism.
Lecture 18
3p
Energy
1. The plot at right shows the relationship between the energies
of atomic orbitals and the bands they form as they are
brought together in a close-packed array. Classify the
electrical properties (insulating, semiconducting or metallic) of
magnesium at the three interatomic distances ra, rb and rc
indicated on the plot by dashed lines.
3s
2p
2s
The electronic configuration of Mg is [Ne] 3s2. At ra, the 3s orbitals
1s
do not overlap sufficiently to form a band, so it will be an
rc
rb
ra
insulator. At rb, the 3s orbitals do overlap sufficiently to form a
Interatomic distance r
band, but that band will be full, so it will be an insulator (or
possibly a semiconductor, depending on the gap to the 3p band). At rc, the 3s orbitals overlap with
the 3p orbitals to form a larger band that is not full, so it will be a metal.
2. CdS, PbSe and ZnO are all semiconductors. Predict the order of the size of the band gaps in
these three compounds. Explain your reasoning.
Band gap should be proportional to the difference in electronegativity between the two elements.
The order should therefore be ZnO > CdS > PbSe.
3. Silicon, which has a diamond-type crystal structure, is by far the most commercially important
semiconducting material. It has a band gap of 1.1 eV. How do the band gaps of diamond-type
carbon and germanium compare to that of silicon? What is the most important factor in the
comparison?
Carbon will have a larger band gap, while germanium will have a smaller band gap. This is because
the principal factor at play is the size of the orbitals. Bigger orbitals  better overlap  wider
bands  smaller band gap.
Lecture 19
1. The rutile structure consists of edge-sharing MO6
octahedra as shown below. A simplified picture
of its band structure is shown at right. Which dorbitals on the transition metal M could form
bands through direct overlap with the d-orbitals
on neighbouring M atoms?
The t2g orbitals, which point between the M–O bonds
in octahedral coordination environments, and
therefore towards neighbouring M cations.
2. ReO3 and WO3 have the same structure type, shown at right.
Re/W atoms are grey, O atoms are white, and the unit cell is
shown by dashed lines. The electrical properties of ReO3 are
metallic, while those of WO3 are insulating. Explain the
difference in terms of the electronic structures of these
materials.
W6+ has no d-electrons, so its conduction band is empty and it is an
insulator. Re6+ has one d-electron that can delocalize in the
conduction band, making it a metal. The conduction band is formed
by the π-type overlap of t2g orbitals on the transition metal cation
and p orbitals on the O anion.
3. The figure below shows the structure of LiCoO2, commonly used as the positive electrode in
lithium-ion batteries. Light grey polyhedral represent CoO6 octahedra, dark grey circles
represent Li atoms, and dashed lines represent the unit cell boundaries. This composition and
structure corresponds to the resting (discharged) state of the battery.
a) What is the oxidation state of Co in LiCoO2?
Co3+
b) When this battery is charged, Li is removed from LiCoO2 to give Li1–xCoO2. The positive
electrode in its maximum charged state corresponds to x ≈ 0.35. What is the oxidation state of
Co in Li1–xCoO2, x = 0.35?
Co3.35+
c) As Li is removed from Li1–xCoO2 (i.e., as x increases), the unit cell parameters show almost no
change. Considering effective ionic radii (http://abulafia.mt.ic.ac.uk/shannon/), what does this
tell us about the d-electron configuration of Co in LiCoO2?
Octahedral Co4+ is always HS and has IR = 0.53 Å. Octahedral Co3+ can be HS with IR = 0.61 Å or LS
with IR = 0.545 Å. The lack of unit cell change implies that Co3+ was LS in LiCoO2.
Lecture 20
1. Propose a type of lattice defect that could turn rock-salt type CdS into a p-type semiconductor.
Justify your choice and explain how the doping works.
Introducing a vacancy onto the Cd site (i.e., Cd1–xS) would increase the average oxidation state of
Cd, because S will always effectively be S2–. This will create localised energy levels that can accept
electrons from the valence band, creating mobile positive holes in the valence band, which
conduct.
2. Silicon has a semiconducting band gap of 1.1 eV. Its conductivity can be improved by doping
with small amounts of gallium. Explain, with the aid of simple band structure diagrams, how
this works.
The valence electrons of Ga have slightly higher energy than those of Si (higher principal quantum
number n). Ga also has one fewer valence electron than Si (group III vs group IV). Ga will therefore
have some empty energy levels just above the valence band of Si. These can accept electrons
excited from the Si valence band, introducing holes into the valence band – i.e., forming a p-type
semiconductor. The energy required to excite these electrons is less than that required to excite
them to the conduction band in pure Si, therefore the doped version is a better semiconductor.
3. The band gap of rock-salt type PbSe can be tuned by doping small proportions of In for Pb,
converting it to an extrinsic semiconductor. Explain how this works and what the active charge
carrier(s) are. Assume that no other defects, such as vacancies or interstitials, are introduced
at the same time.
If no other defects are introduced at the same time then we can think of In as being incorporated
as In2+ rather than its preferred oxidation state In3+. In2+ would have one more electron than Pb2+,
which will go into a narrow dopant band somewhere between the valence band and the
conduction band.
The charge carriers are thus electrons in the valence band, i.e., it is an n-type semiconductor.
Undoped
Doped
Empty conduction
band
Empty conduction
band
Dopant band
Filled valence
band
ρ(E)
Filled valence
band
ρ(E)
Lecture 22
1. A polycrystalline sample of a solid-state compound gives the magnetic susceptibility (dashed
line) and inverse susceptibility (solid line) curves shown below. Graphically estimate the Weiss
constant θ and the Néel temperature TN from these curves, to the nearest 10 K.
6.0 x 10-5
3.5 x 10
-5
3.0 x 104
5.5 x 10
4
2.5 x 104
-5
c (emu/g)
1.5 x 104
-5
4.0 x 10
-1
2.0 x 104
4.5 x 10-5
c (g/emu)
5.0 x 10
1.0 x 104
-5
3.5 x 10
5.0 x 10
3.0 x 10-5
3
0.0
-100
0
100
200
300
T (K)
From the downturn in susceptibility (χ), TN ≈ 110 K.
From a linear fit to the inverse susceptibility (χ–1) in the paramagnetic region, θ ≈ –130 K
CHEM3X19 Tutorial
1. The electronic absorption spectrum of the [Cr(OH2)6]2+ ion is shown below and displays a single band at
around 14000 cm-1. The magnetic moment of the [Cr(OH2)6]2+ ion is 4.09 B.M.
[Cr(OH2)6]2+
(i) What is the ground term for this complex?
(ii) Construct an Orgel diagram.
(iii) Would you predict any distortion from octahedral coordination geometry for this complex?
(iv) Explain the origin of the magnetic moment for the complex, including the presence or absence of an
orbital contribution.
2. Following one-electron oxidation of an aqueous solution containing [V(OH2)6]2+, the electronic
absorption spectrum shows absorptions centred at 17200 and 25600 cm-1. Determine the ground-state
term symbol and state the origins of the two transitions in the oxidised complex.
3. The trivalent ions Nd3+ and Dy3+ have the valence electron configurations 4f3 and 4f9 respectively.
(a) Using Hund’s rules, calculate S, L and J, and give the electronic ground term for these ions.
(b) Describe how the UV-visible spectra for complexes of these ions are expected to be similar or
different. Include in your answer consideration of both the breadth and intensity of the spectral
bands.
(c) What other oxidation state(s) are readily accessible within coordination complexes of these two
elements? Briefly explain any similarities or differences between the two.
4. Considering the Lanthanoids:
(a) Account for the similar electronic spectra of Eu3+ complexes with various ligands.
(b) Why is a discussion of the trend of ionic radii for the first row d-block metal ions less simple than a
discussion of that of the Ln3+ ions?
5. On the left hand side of the following table are various quantitative features of 3d (first row transition metal)
and 4f (lanthanoid) complexes, each of which may be in general greater (denoted 4f > 3d) or lower in
magnitude (4f < 3d) for lanthanoid complexes. Which of the options gives the correct relative magnitude of
each of these features?
Ionic radius of metal ion
Ligand field splitting energy
Coordination bond covalency
Number of accessible oxidation states
Intensity of optical spectral bands (d-d
vs. f-f)
Width of optical spectral bands (d-d
vs. f-f)
Maximum possible magnetic moment
1
4f > 3d
4f < 3d 4f < 3d 4f < 3d 4f < 3d 2 4f > 3d
4f < 3d 4f < 3d 4f < 3d 4f > 3d
3
4f > 3d
4f > 3d
4f > 3d
4f > 3d
4f > 3d
4
4f < 3d 4f > 3d
4f > 3d
4f < 3d 4f < 3d 4f < 3d 4f > 3d
4f > 3d
4f < 3d 4f > 3d
4f > 3d
4f > 3d
4f < 3d 6. NiO adopts the cubic rock salt NaCl-type structure while PtO adopts the cooperite structure shown. What factor do you think is responsible for the differing crystal chemistry preferences of the two compounds? 7. One of the simplest types of solid solution involves the substitution of ions of the same charge as those already present in the lattice. Why might the substitution of octahedral iron(II) with copper(II) have structural implications? 8. What is the composition of wurtzite, the high temperature form of zinc sulfide? In the picture, yellow spheres represent S2- and grey spheres represent Zn2+. 9. By considering the effect on the lattice energies of incorporating large numbers of defects and the resultant changes in oxidation numbers of the ions making up the structure, predict whether Zn1+xO and/or Fe1-xO would show nonstoichiometry over a large range of x. 10. Comment on the following statement, which may be correct, incorrect, or partly correct: water substitution rates, k(H2O), for Ln3+(aq) ions are lower than those of M3+(aq) ions of the 3d elements. Consider the following data in formulating your answer. 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test CHEM3X19 In-Semester Test Started: Mar 31 at 22:49 Quiz Instructions CHEM3119/3919 In-Semester Test For information about this test (also referred to as quiz), please check the recent announcements. The quiz consists of 5 multiple choice questions and 7 short answer questions and is worth 10% of your final unit grade. You are allowed 25min for the quiz which can be taken once only any time between 9am Monday March 28 and 5pm Friday April 1. The quiz is open book and you can refer to your lecture notes during the quiz. Students are advised, however, not to rely entirely on being able to search through their notes for every question. Time management is important to ensure you’re able to answer all questions. This is an individual task. You may not communicate with anyone else during the quiz. You may not transmit information about the questions to anyone else or seek assistance from anyone else. Please direct any inquiries to the Unit Coordinator, Prof Brendan Kennedy (brendan.kennedy@sydney.edu.au (mailto:brendan.kennedy@sydney.edu.au) ). https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take 1/10 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test Question 1 1 pts Which definition best describes the Lanthanoid Contraction for Ln3+? For the metals, there is a uniform decrease in the ionic radius caused by the decreasing nuclear charge because the 5d and 6s electrons electrons poorly shield the 4f electrons. For the metals, there is a uniform decrease in the ionic radius caused by the decreasing nuclear charge because the 4f electrons poorly shield the 5d and 6s electrons. For the metals, there is a uniform increase in the ionic radius caused by the increasing nuclear charge because the 4f electrons poorly shield the 5d and 6s electrons. For the metals, there is a uniform decrease in the ionic radius caused by the increasing nuclear charge because the 4f electrons poorly shield the 5d and 6s electrons. https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take 2/10 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test Question 2 1 pts What is the ground state term symbol for the complex [Co(NH3)6]2+? Recall that we use the notation 2S+1L, where S is the total spin angular momentum, L is the total orbital angular momentum. 4F 3F 6S 2D Question 3 1 pts What is the ground-state term symbol for V2+? 4F 2F 6F https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take 3/10 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test 3F Question 4 1 pts For the Er3+ ion, work out the number of 4f electrons, n. Using Hund’s rules, then calculate L, S, and J, and work out the electronic ground term. n=8, L=3, S=3, J=6, ground term = 7F6 n=11, L=15/2, S=4, J=9/2, ground term = 4I15/2 n=11, L=6, S=3/2, J=9/2, ground term = 4I9/2 n=11, L=6, S=3/2, J=15/2, ground term = 4I15/2 Question 5 1 pts Which definition best describes the Lanthanoid Contraction for Ln3+? For the metals, there is a uniform increase in the ionic radius caused by the increasing nuclear charge because the 4f electrons poorly shield the 5d and 6s electrons. For the metals, there is a uniform decrease in the ionic radius caused by the increasing nuclear charge because the 4f electrons poorly shield https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take 4/10 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test the 5d and 6s electrons. For the metals, there is a uniform decrease in the ionic radius caused by the decreasing nuclear charge because the 4f electrons poorly shield the 5d and 6s electrons. For the metals, there is a uniform decrease in the ionic radius caused by the decreasing nuclear charge because the 5d and 6s electrons electrons poorly shield the 4f electrons. Question 6 1 pts Ternary oxides based on 3d metals and magnesium such as FexMg1-xO have proven very useful in catalysis. Suggest a mechanism by which non-stoichiometry occurs in this solid solution. Edit 12pt View Insert Format Tools Table Paragraph Initially Fe2+ ions occupy octahedral sites. A number of vacancies are created on these normally occupied octahedral sites, but over the whole structure, this number is greater than required by the compound stoichiometry. The additional iron required occupies tetrahedral sites adjacent to the vacancies. In order to maintain charge balance, the interstitial iron is oxidised to +3 and a number of iron atoms surrounding the defect, but on normal lattice sites, must also be oxidised to Fe3+. https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take 5/10 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test p 80 words Question 7 1 pts By considering the effect on the lattice energies of incorporating large numbers of defects and the resultant changes in oxidation numbers of the ions making up the structure, predict whether Zn1+xO and/or Fe1-xO would show nonstoichiometry over a large range of x. Edit 12pt View Insert Format Tools Table Paragraph it will show nonstoichiometry over a large range of x, because the defects increase will cause the charge balance was broken, so the numbers of Fe 3+ increase for balance the charge. https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take 6/10 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test p 33 words Question 8 1 pts Comment on the following statement, which may be correct, incorrect, or partly correct: "water substitution rates for Ln3+(aq) ions are lower than those of M3+(aq) ions of the 3d elements." Edit 12pt View Insert Format Tools Table Paragraph it is partly correct. https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take according to this, q8, Ln3+ is faster. These complexes make strong metal-oxygen bonds and have water exchange rates in the range. 7/10 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test p span 4 words Question 9 1 pts The linking of TO4 tetrahedra (e.g. AlO4 and SiO4) may occur in numerous ways to produce a variety of zeolite structures, with a number of structural features being common to these materials. Briefly describe the molecular sieving properties of zeolites and how these can be exploited for separations. Edit 12pt View Insert Format Tools Table Paragraph Mechanisms of adsorptive separation: Molecular sieving – size/shape exclusion Molecular sieving with such size selectivity demands zeolites with highly tunable pore diameters and adsorption properties. Upon heating, the zeolites lose their water content with little or no change in their crystal structure. The dehydrated zeolite can reversibly absorb water or other molecules that are small enough to pass through the channels or pores. Thermodynamic effects – preferential interactions (trade-off for energy of desorption) Kinetic effects – different diffusion rates of gases Quantum sieving effects – different diffusion rates of very light molecules in narrow pores Zeolites are molecular sieves, it can use to separate the substance based on the size and shape of the molecules being adsorbed. The critical diameter of the adsorbent molecule must be smaller than the size of the entrance window. The size of entrance window is to decide the sieves ability. https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take 8/10 3/31/22, 8:09 PM Quiz: CHEM3X19 In-Semester Test p 88 words Question 10 1 pts The linking of TO4 tetrahedra may occur in numerous ways to produce a variety of zeolite structures, with a number of structural features being common to these materials. Briefly describe the shape selective catalysis properties of zeolites and how this property can be related to their structures. Edit View 12pt Insert Format Tools Table Paragraph Excellent catalytic activity at all temperatures - highly robust, so high surface area is maintained. Shape-selectivity very important Only molecules with dimensions less than a critical size can enter the pores and reach the catalytic sites, and so react there. Only products less than a certain dimension can leave the active sites and diffuse out through the channels Zeolites can be shape-selective catalysts either by transition state selectivity or by exclusion of competing reactants on the basis of molecular diameter. They have also been used as oxidation catalysts. The reactions can take place within the pores of the zeolite, which allows a greater degree of product control. https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take 9/10 3/31/22, 8:09 PM p Quiz: CHEM3X19 In-Semester Test 57 words Saved at 20:09 https://canvas.sydney.edu.au/courses/38943/quizzes/170713/take Submit Quiz 10/10 Shriver & Atkins: Inorganic Chemistry 5e Answers to self-tests and exercises CHAPTER 1 1.7 Take the summation of the rest masses of all the nuclei of the products minus the masses of the nuclei of the reactants. If you get a negative number, energy will be released. But what you have calculated is the mass difference, which in the case of a nuclear reaction is converted to energy. 1.8 0.25 1.9 –13.2 eV 1.10 1524nm, 1.524 X 104 cm-1 Self-tests 1 0 n + γ S1.1 80 35 Br S1.2 d orbitals, 5 orbitals S1.3 4 S1.4 3p S1.5 The added p electron is in a different (p) orbital, so it is less shielded. S1.6 Ni :[Ar]3d 8 4s 2 , Ni 2+ :[Ar]3d 8 S1.7 Period 4, Group 2, s block S1.8 Going down a group the atomic radius increases and the first ionization energy generally decreases. λ S1.9 Group 16. The first four electrons are removed with gradually increasing values. Removing the fifth electron requires a large increase in energy, indicating breaking into a complete subshell. S1.10 S1.11 + 81 35 Br 1.11 ⎛1 1 = R⎜ − 1 = R⎜ − 1 = R⎜ − 1 = R⎜ − λ λ λ ⎝ 12 ⎛1 ⎝ 12 ⎛1 ⎝ 12 ⎛1 ⎝ 12 Adding another electron to C would result in the stable half filled p subshell. 1.12 0 up to n-X Cs+ 1.13 n2 1 ⎞ 7 −1 ⎟ = 1.0974 X10 m 2 ⎠ ∞ 1 ⎞ 7 −1 ⎟ = 1.0288X10 m 2 ⎠ 4 1⎞ 6 −1 ⎟ = 9.7547X10 m 2 ⎠ 3 1⎞ 6 −1 ⎟ = 8.2305X10 m 2 ⎠ 2 1.14 Exercises 1.1 (a) 147 N+ 42 He→178 O+11p + γ (b) 12 6 (c) 14 7 C+11p→137 N + γ 1 0 3 1 12 6 N+ n→ H+ C 1 Cm+126 C→257 112 Uub+ 0 n 1.2 246 96 1.3 The higher value of I2 for Cr relative to Mn is a consequence of the special stability of halffilled subshell configurations and the higher Zeff of a 3d electron verses a 4s electron. 1.15 1.16 Ne + He → Mg + n 4 2 4 2 1 0 ml Orbital designation 2 1 2p 3 2 3d 5 4 4 0 3 +1, 0, −1 +2, +1, …, −2 0 +3, +2, …, −3 Number of orbitals 3 4s 4f 1 7 n=5, l = 3, and ml = -3,-2,-1,0,1,2,3 Li: σ = Z – Zeff ; σ = 3-1.28 = 1.72 Be: σ = Z – Zeff ; σ = 4-1.19 = 2.09 B: σ = Z – Zeff ; σ = 5-2.42 = 2.58 1.4 1.5 9 4 Be+ Be→ C+ He +2 n C: σ = Z – Zeff ; σ = 6-3.14 = 2.86 1.6 Since helium-4 is the basic building block, most additional fusion processes will produce nuclei with even atomic numbers. O: σ = Z – Zeff ; σ = 8-4.45 = 3.55 12 6 1 0 l 22 10 9 4 25 12 N N: σ = Z – Zeff ; σ = 7-3.83 = 3.17 F: σ = Z – Zeff ; σ = 9-5.10 = 3.90 1.17 1 The 1s electrons shield the positive charge form the 2s electrons. Shriver & Atkins: Inorganic Chemistry 5e ANSWERS TO SELF-TESTS AND EXERCISES 2 1.18 See Figs 1.11 through 1.16 1.19 See Table 1.6 and discussion. 1.20 Table 1.6 shows SR > Ba < Ra. Ra is anomalous because of higher Zeff due to lanthanide contraction. 1.21 Anomalously high value for Cr is associated with the stability of a half filled d shell. 1.22 (a) [He]2s22p2 1.30 (b) [He]2s22p5 (c) [Ar]4s2 (d) [Ar]3d10 (e) [Xe]4f145d106s26p3 (f) [Xe]4f145d106s2 1.23 (a) [Ar]3d14s2 CHAPTER 2 Self-tests S2.1 (b) [Ar]3d2 (c) [Ar]3d5 (d) [Ar]3d4 (e) [Ar]3d S2.2 6 (b) square planar (f) [Ar] 10 (g) [Ar]3d 4s 1 S2.3 Linear S2.4 S22– : 1σg22σu23σg21πu42πg4 ; (h) [Xe]4f 7 1.24 (a) Angular Cl2– : 1σg22σu23σg21πu42πg44σu1. (a) Xe]4f145d46s2 S2.5 1σg22σu23σg21πu42πg4 (b) [Kr]4d6 S2.6 ½[2-2+4+2] = 3 S2.7 Bond order: C≡N, C=N, and C–N; Bond strength: C≡N > C=N > C–N.
S2.8
If it contains 4 or fewer electrons.
S2.9
–21 kJ mol–1
S2.10
(a) +1/2
6
(c) [Xe]4f
(d) [Xe]4f7
(e) [Ar]
(f) [Kr]4d
1.25
2
(a) S
(b) Sr
(b) +5
(c) V
(d) Tc
Exercises
(e) In
(f) Sm
2.1
(a) angular
1.26
See Figure 1.4.
(b) tetrahedral
1.27
(a) I1 increases across the row except for a dip
at S; (b) Ae tends to increase except for Mg
(filled subshell), P (half filled subshell), and
at AR (filled shell).
(c) tetrahedral
1.28
Radii of Period 4 and 5 d-metals are similar
because of lanthanide contraction.
1.29
2s2 and 2p0
2.2
(a) trigonal planar
(b) trigonal pyramidal
(c) square pyramidal
2.3
(a) T-shaped
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
2.4
(b) square planar
(b) one
(c) linear
(c) none
(a)
(d) two
2.15
3
(a) 1σg22σu2
(b) 1σg22σu21πu2
Cl
Cl
(c) 1σg22σu21πu43σg1
I
Cl
Cl
(d) 1σg22σu23σg21πu42πg3
2.16
The configuration for the neutral C2 would be
1σg21σu2 1πu4. The bond order would be ½[22+4] = 2.
2.17
(a) 1σg22σu23σg21πu42πg4
(b)
(b) 1
F
(c) There is no bond between the two atoms.
F
S
1800
2.18
120 0
2.5
2.6
(b) 1
(c) 2
F
F
(a) 2
2.19
(a) +0.5
(a) tetrahedral
(b) –0.5
(b) octahedral
(c) +0.5
2.20
(a) 176 pm
(b) 217 pm
(c) 221 pm
2.7
2(Si–O) = 932kJ > Si=O = 640kJ; therefore
two Si–O are preferred and SiO2 should (and
does) have four single Si–O bonds.
2.8
Multiple bonds are much stronger for period 2
elements than heavier elements
2.9
–483 kJ difference is smaller than expected
because bond energies are not accurate.
2.10
2.21
(a) 0
(a-c)
-1
(b) 205 kJ mol
2.11
Difference in electronegativities are AB 0.5,
AD 2.5, BD 2.0, and AC 1.0. The increasing
covalent character AD < BD < AC < AB. 2.12 (a) covalent (b) ionic (d) Possibly stable in isolation (only bonding and nonbonding orbitals are filled; not stable in solution because solvents would have higher proton affinity than He. (c) ionic 2.13 (a) sp2 (b) sp3 (c) sp3d or spd3 2 2 2 (d) p d or sp d. 2.14 (a) one 2.22 1 2.23 HOMO exclusively F; LUMO mainly S. 2.24 (a) electron deficient Shriver & Atkins: Inorganic Chemistry 5e ANSWERS TO SELF-TESTS AND EXERCISES 4 (b) electron precise CHAPTER 3 3.4 3.5 XA2 K3C60 3.6 281 pm 3.7 429 pm 3.8 CuAu. Primitive 12 carat. 3.9 Zintl phase region 3.10 (a) 6:6 and 8:8 Self-tests S3.1 S3.2 See Fig. 3.7 and Fig. 3.32. See Figure 3.35. S3.3 52% S3.4 rh = ((3/2)1/2 – 1) r = 0.225 r (b) CsCI S3.5 409 pm 3.11 6 S3.6 401 pm 3.12 S3.7 FeCr3 2B type and 4A type – in a distorted octahedral arrangement. S3.8 X2A3. 3.13 S3.9 Ti CN = 6 (thought these are as two slightly different distances it is often described as 4 + 2) and O CN = 3. (a) ρ = 0.78, so fluorite (b) FrI? ρ = 0.94, so CsCl (c) BeO? ρ = 0.19, so ZnS (d) InN? ρ = 0.46, so NaCl 3.14 CsCl. S3.10 LaInO3 3.15 S3.11 2421 kJ mol–1 Lattice enthalpies for the di- and the trivalent ions. Also, the bond energy and third electron gain enthalpy for nitrogen will be large. S3.12 Unlikely 3.16 4 four times the NaCl value or 3144 kJmol–1. S3.13 MgSO4 < CaSO4 < SrSO4 < BaSO4. 3.17 (a) 10906 kJmol–1 S3.14 NaClO4 (b) 1888 kJ mol–1 S3.15 Schottky defects. (c) 664 kJ mol–1 S3.16 Phosphorus and aluminium. S3.17 The dx2-y2 and dz2 have lobes pointing along the cell edges to the nearest neighbor metals. S3.18 3.18 (b) NaBF4 (a) n-type (b) p-type Exercises 3.1 a ≠ b ≠ c and α = 90°, 3.2 Points on the cell corners at (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and (1,1,1) and in the cell faces at (½,½,0), (½,1,½), (0,½,½) (½,½,1), (½,1,½), and (1,½,½). 3.3 (a) MgSO4 = 90°, = 90° (c) and (f) are not as they have neighbouring layers of the same position. 3.19 CsI < RbCl < LiF < CaO < NiO < AlN 3.20 Ba2+; solubilities decrease with increasing radius of the cation. 3.21 (a) Schottky defects (b) Frenkel defects 3.22 Solids have a greater number of defects as temperatures approaches their melting points. 3.23 The origin of the blue color involves electron transfer from cationic centres. Vanandium carbide and manganese oxide. 3.24 Shriver & Atkins: Inorganic Chemistry 5e ANSWERS TO SELF-TESTS AND EXERCISES 3.25 3.26 3.27 Yes. A semiconductor is a substance with an electrical conductivity that decreases with increasing temperature. It has a small, measurable band gap. A semimetal is a solid whose band structure has a zero density of states and no measurable band gap. Ag2S and CuBr: p-type; VO2: n-type. CHAPTER 4 S4.9 5 Identify the acids and bases? (a) FeCl3 + Cl– → [FeCl4]–, acid is FeCl3, base is Cl–. (b) I– + I2 → I3–, acid is I2, base is I–. S4.10 The difference in structure between (H3Si)3N and (H3C)3N? The N atom of (H3Si)3N is trigonal planar, whereas the N atom of (H3C)3N is trigonal pyramidal. S4.11 Draw the structure of BF3·OEt2? Self-tests S4.1 (a) HNO3 + H2O → H3O+ + NO3– HNO3, acid. Nitrate ion, conjugate base. H2O, base. H3O+, conjugate acid. (b) CO32– + H2O → HCO3– + OH– carbonate ion, base; hydrogen carbonate, or bicarbonate, conjugate acid; H2O, acid; hydroxide ion, conjugate base. Exercises 4.1 (c) NH3 + H2S → NH4+ + HS– Ammonia, base; NH4+, conjugate acid; hydrogen sulphide, acid; HS–, conjugate base. S4.2 What is the pH of a 0.10 M HF solution? pH= 2.24 S4.3 Calculate the pH of a 0.20 M tartaric acid solution? Sketch an outline of the s and p blocks of the periodic table, showing the elements that form acidic, basic, and amphoteric oxides? The elements that form basic oxides are in plain type, those forming acidic oxides are in outline type, and those forming amphoteric oxides are in boldface type. pH=1.85 S4.4 Which solvent? dimethylsulfoxide (DMSO) and ammonia. S4.5 Is aKBrF4 an acid or a base in BrF3? A base. S4.6 Arrange in order of increasing acidity? The order of increasing acidity is [Na(H2O)6]+ < [Ni(H2O)6]2+ < < [Mn(H2O)6]2+ 3+ [Sc(H2O)6] . S4.7 4.2 Predict pKa values? (a) H3PO4 pKa ≈ 3. The actual value, given in Table 4.1, is 2.1. (b) pKa(2) ≈ 8. The actual value, given in Table 4.1, is 7.4. of the (b) HSO4–? The conjugate base is SO4–. (c) CH3OH? The conjugate base is CH3O–. (c) HPO42– pKa(3) ≈ 13. The actual value, given in Table 4.1, is 12.7. (d) H2PO4–? The conjugate base is HPO42–. (e) Si(OH)4? SiO(OH)3–. What happens to Ti(IV) in aqueous solution as the pH is raised? Treatment with ammonia causes the precipitation of TiO2. Further treatment with NaOH causes the TiO2 to redissolve. bases (a) [Co(NH3)5(OH2)]3+, conjugate base is [Co(NH3)5(OH)]2+. H2PO4– S4.8 Identify the conjugate following acids? (f) HS−? 4.3 The conjugate base is of the The conjugate base is S2–. Identify the conjugate following bases? acids Shriver & Atkins: Inorganic Chemistry 5e ANSWERS TO SELF-TESTS AND EXERCISES 6 ClO4–, cannot be studied in sulfuric acid. (a) C5H5N (pyridine)? The conjugate acid is pyridinium ion, C5H6N+. 4.9 (b) HPO42–? The conjugate acid is H2PO42–. (c) O2–? The conjugate acid is OH–. electron withdrawing (d) CH3COOH? The conjugate acid is CH3C(OH)2+. (e) [Co(CO)4]–? The conjugate acid is HCo(CO)4. (f) CN–? 4.4 Calculate the [H3O+] and pH of a 0.10 M butanoic acid solution? pH=2.85 4.5 What is the Kb of ethanoic acid? 4.6 4.7 4.11 Is the pKa for HAsO42– consistent with Pauling’s rules? No. Pauling’s rules are only approximate. 4.12 What is the order of increasing acid strength for HNO2, H2SO4, HBrO3, and HClO4? The conjugate acid is HCN. Kb = 5.6 × 10 the order is HClO4 > HBrO3 > H2SO4 >
HNO2.
4.13
–10
What is the Ka for C5H5NH+?
Ka = 5.6 × 10–6
–
Predict if F will behave as an acid or a
base in water?
4.14
F will behave as a base in water.
What are the structures and the pKa values
of chloric (HClO3) and chlorous (HClO2)
acid?
O
O
the aluminum-containing species is more
acidic.
(c) Si(OH)4 or Ge(OH)4?
Cl
H
O
Which of the following is the stronger acid?
(a) [Fe(OH2)6]3+ or [Fe(OH2)6]2+? The
Fe(III) complex, [Fe(OH2)6]3+, is the stronger
acid.
(b) [Al(OH2)6]3+ or [Ga(OH2)6]3+?
O
Cl
Account for the trends in the pKa values of
the conjugate acids of SiO44–, PO43–, SO42–,
and ClO4–?
The acidity of the four conjugate acids
increases in the order HSiO43– < HPO42– < HSO4– < HClO4. - 4.8 Is the –CN group electron donating or withdrawing? H Si(OH)4, is more acidic. O (d) HClO3 or HClO4? chloric acid HClO4 is a stronger acid. chlorous acid (e) H2CrO4 or HMnO4? Chloric acid, the predicted pKa = –2; actual value = –1. HMnO4 is the stronger acid. (f) H3PO4 or H2SO4? Chlorous acid, the predicted pKa = 3; actual value = 2. 4.8 Which bases are too strong or too weak to be studied experimentally? (a) CO32– O2–, ClO4–, and NO3– in water? H2SO4 is a stronger acid. 4.15 CO32– is of directly measurable base strength. O2–, is too strong to be studied experimentally in water. ClO42– and NO3– are too weak to be studied experimentally. order of increasing basicity is Cl2O7 < SO3 < CO2 < B2O3 < Al2O3 < BaO. 4.16 (b) HSO4–, NO3–, and ClO4–, in H2SO4? HSO4–, not too experimentally. – strong to be Arrange the following in order of increasing acidity? HSO4−, H3O+, H4SiO4, CH3GeH3, NH3, and HSO3F? increasing acidity is NH3 < CH3GeH3 < H4SiO4 < HSO4– < H3O+ < HSO3F. studied NO3 is of directly measurable base strength in liquid H2SO4. Arrange the following oxides in order of increasing basicity? Al2O3, B2O3, BaO, CO2, Cl2O7, and SO3? 4.17 Which aqua ion is the stronger acid, Na+ or Ag+? Ag+(aq). Shriver & Atkins: Inorganic Chemistry 5e ANSWERS TO SELF-TESTS AND EXERCISES 4.18 (d) AsF3(g) + SbF5(g) → [AsF2][SbF6]? The very strong Lewis acid SbF5 displaces the Lewis acid [AsF2]+ from the Lewis base F–. Which of the following elements form oxide polyanions or polycations? Al, As, Cu, Mo, Si, B, Ti? polycations: Al, Cu, and Ti. (e) EtOH readily dissolves in pyridine? A Lewis acid–base complex formation reaction between EtOH (the acid) and py (the base) produces the adduct EtOH–py. polyoxoanions (oxide polyanions): Mo polyoxoanions: As, B, and Si. 4.19 The change in charge upon aqua ion polymerization? 4.26 Polycation formation reduces the average positive charge per central M atom by +1 per M. 4.20 2[Fe(OH2)6]3+ → 2-Me-py or 4-Me-py? 4.27 4.24 4.25 – 2HCO 3- Give the equations for HF in H2SO4 and HF in liquid NH3? 4.28 H2SO4 + HF ⇔ H3SO4+ + F- 4.23 1 – (b) CO2 and CaCO3? NH3 + HF Which of the following reactions have Keq >
1? (a) R3P–BBr3 + R3N–BF3→ R3P–BF3 +
R3N–BBr3? 1.
(a) H3PO4 and Na2HPO4?
Ca2+ +
B(n-Bu)3.
(b) More basic toward BMe3: NMe3 or
NEt3? NMe3.
More balanced equations?
CO 2 + CaCO 3 + H 2O
Boron trichloride.
B(n-Bu)3 or B(t-Bu)3?
[(H2O)4Fe(OH)2Fe(OH2)4+ + 2H3O+
4.21
Select the compound with the named
characteristic? (a) Strongest Lewis acid:
BF3, BCl3, or BBr3? BBr3.
BeCl2 or BCl3?
Write balanced equations for the formation
of P4O124– from PO43– and for the formation
of [(H2O) 4Fe(OH)2Fe(OH2)4]4+ from
[Fe(OH2)6]3+?
4PO43– + 8H3O+ → P4O124– + 12H2O
7
The phosphorus atom in Me2NPF2 is the
softer of the two basic sites, so it will bond
more strongly with the softer Lewis acid BH3
NH2- + H2F+
Why is H2Se a stronger acid than H2S? As
you go down a family in the periodic chart,
the acidy of the homologous hydrogen
compounds increases.
Identifying elements that form Lewis
acids? All of the p-block elements except
nitrogen, oxygen, fluorine, and the lighter
noble gases form Lewis acids in one of their
oxidation states.
The hard nitrogen atom will bond more
strongly to the hard Lewis acid BF3.
4.29
Why is trimethylamine out of line?
Trimethyl amine is sterically large enough to
fall out of line with the given enthalpies of
reaction.
4.30
Discuss relative basicities? (a) Acetone and
DMSO?
DMSO is the stronger base regardless of how
hard or how soft the Lewis acid is. The
ambiguity for DMSO is that both the oxygen
atom and sulfur atom are potential basic sites.
Identifying acids and bases: (a) SO3 + H2O
→ HSO4– + H+? The acids in this reaction
are the Lewis acids SO3 and H+ and the base
is the Lewis base OH–.
(b) Me[B12]– + Hg2+ → [B12] +
MeHg+? The Lewis acid Hg2+ displaces the
Lewis acid [B12] from the Lewis base CH3–.
(c) KCl + SnCl2 → K+ + [SnCl3]–? The
Lewis acid SnCl2 displaces the Lewis acid K+
from the Lewis base Cl–.
Choose between the two basic sites in
Me2NPF2?
(b) Me2S and DMSO? Depending on the
EA and CA values for the Lewis acid, either
base could be stronger.
4.31
Write a balanced equation
dissolution of SiO2 by HF?
for
the
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
8
SiO2 + 6HF
2H2O + H2SiF6
or
SiO2 + 4HF
C6H5COOH + HF
2H2O + SiF4
Both a Brønsted acid–base reaction and a
Lewis acid–base reaction.
4.32
The dissolution of silicates by HF? both
4.38
Are the f-block elements hard? yes.
4.39
Calculate the enthalpy change for I2 with
phenol?
ΔfH = -20.0kJ/mol
φ
Al2 O 3 + 3H 2 S
Describe solvent properties? (a) Favor
displacement of Cl– by I– from an acid
center? If you choose a solvent that
decreases the activity of chloride relative to
iodide, you can shift the following
equilibrium to the right:
acid-Cl- + I-
CHAPTER 5
Self-tests
S5.1
acid-I- + Cl-
(b) Favor basicity of R3As over
R3N? Alcohols such as methanol or ethanol
would be suitable.
+
4.35
4.36
Propose a mechanism for the acylation of
benzene? An alumina surface, such as the
partially dehydroxylated one shown below,
would provide Lewis acidic sites that could
abstract Cl–:
Why does Hg(II) occur only as
HgS? Mercury(II) is a soft Lewis acid, and
so is found in nature only combined with soft
Lewis bases, the most common of which is
S2–.
Write Brønsted acid–base reactions in
liquid HF?
(a) CH3CH2OH?
is:
CH 3CH 2OH + HF
(b) NH3? The equation is
NH3 +
HF
S5.2
Does Cu metal dissolve in dilute HCl? No.
S5.3
Can Cr2O72– be used to oxidize Fe2+, and
would Cl– oxidation be a problem? Yes.
Cl- oxidation is not a problem.
S5.4
Fuel cell emf with oxygen and hydrogen
gases at 5.0 bar?
E = 1.25 V.
S5.5
The fate of SO2 emitted into clouds? The
aqueous solution of SO42– and H+ ions
precipitates as acid rain.
S5.6
Can Fe2+ disproportionate under standard
conditions? No.
S5.7
bpy binding to Fe(III) or Fe(II)? Fe(II)
preferentially.
S5.8
Potential of AgCl/Ag,Cl– couple?
Eox= – 1.38 V
S5.9
Latimer diagram for Pu? (a) Pu(IV)
disproportionates to Pu(III) and Pu(V) in
aqueous solution; (b) Pu(V) does not
disproportionate into Pu(VI) and Pu(IV).
S5.10
Frost diagram for thallium in aqueous
acid?
The balanced equation
CH 3CH 2OH 2+ + FN H 4+ +
F-
oxidation
2MnO4− (aq) + 5Zn(s) + 16H+(aq) →
5Zn2+(aq) + 2Mn2+(aq) + 8H2O(l)
3+
(d) Promote the reaction 2FeCl3 + ZnCl2
→Zn2+ + 2[FeCl4]−? A suitable solvent is
acetonitrile, MeCN.
Half-reactions and balanced reaction for
oxidation of zinc metal by permanganate
ions?
2[MnO4− (aq) + 8H+ (aq) + 5e– → Mn2+(aq)
reduction
+ 4H2O(l)]
5 [ Zn(s) → Zn2+(aq) + 2e– ]
(c) Favor acidity of Ag over Al ? An
example of a suitable solvent is diethyl ether.
Another suitable solvent is H2O.
4.34
C6H5COO- + H2F+
4.37
Write a balanced equation to explain the
foul odor of damp Al2S3? The foul odor
suggests H2S formation.
Al2 S 3 + 3H 2 O
4.33
(c) C6H5COOH?
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
5.3
9
Write balanced equations, if a reaction
occurs, for the following species in aerated
aqueous acid? (a) Cr2+?
4Cr2+(aq) + O2(g) + 4H+(aq) → 4Cr3+(aq)
Eº = 1.65 V
+ 2H2O(l)
(b) Fe2+?
4Fe2+(aq) + O2(g) + 4H+(aq) → 4Fe3+(aq)
Eº = 0.46 V
+ 2H2O(l)
(c) Cl–? no reaction.
(d) HOCl? No reaction.
S5.11
The oxidation number of manganese?
(e) Zn(s)?
2Zn(s) + O2(g) + 4H+(aq) → 2Zn2+(aq)
Eº = 1.99 V
+2H2O(l)
Mn2+(aq)
S5.12
Compare the strength of
oxidizing agent in acidic
solution?
NO3–
and
as an
basic
A competing reaction is:
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) (Eº =
0.763 V).
Nitrate is a stronger oxidizing agent in acidic
solution than in basic solution.
S5.13
S5.14
The possibility of finding Fe(OH)3 in a
waterlogged soil? Fe(OH)3 is not stable.
5.4
Balanced equations for redox reactions?
(a) Fe2+?
(i) Fe2+ will not oxidize water.
The minimum temperature for reduction
of MgO by carbon? 1800ºC or above.
(ii) Fe2+ will not reduce water.
Exercises
(iii) Fe2+ will reduce O2 and in doing so will
be oxidized to Fe3+.
5.1
(iv) disproportionation will not occur.
Oxidation numbers?
2 NO(g) + O2(g) → 2 NO2(g)
+1 -2
0
+4 -2
(b) Ru2+? Ru2+ will not oxidize or reduce
water. Ru2+ will reduce O2 and in doing so
will be oxidized to Ru3+. Ru2+ will
disproportionate in aqueous acid to Ru3+ and
metallic ruthenium.
2Mn3+(aq) + 2H2O → MnO2 + Mn2+ + 4H+(aq)
+3
+1 -2
+4 -2
+2
+1
LiCoO2(s) + C(s) → LiC(s) + CoO2(s)
+1 +3 -2
0
+1-1
+4 -2
→ CaH2(s)
+2 -1
(c) HClO2? HClO2 will oxidize water, will
not reduce water. HClO2 will reduce O2 and
in doing so will be oxidized to ClO3–. HClO2
will disproportionate in aqueous acid to ClO3–
and HClO.
Suggest chemical reagents for redox
transformations? (a) Oxidation of HCl to
Cl2? S2O82–, H2O2, or α–PbO2 to oxidize Cl–
to Cl2.
(d) Br2? Br2 will not oxidize or reduce water.
Br2 will not
Br2 will not reduce O2.
disproportionate in aqueous acid to Br– and
HBrO.
(b) Reducing Cr3+(aq) to Cr2+(aq)? metallic
manganese, metallic zinc, or NH3OH+.
5.5
Standard potentials vary with
temperature in opposite directions? The
amino and cyano complexes must have
different equilibrium shifts with respect to
changes in temperature that results in the
opposite directions of change for the cell
potential.
Ca(s) + H2(g)
0
0
5.2
(c) Reducing Ag+(aq) to Ag(s)? The reduced
form of any couple with a reduction potential
less than 0.799 V.
(d) Reducing I2 to I–? The reduced form of
any couple with a reduction potential less than
0.535 V.
5.6
Balance redox reaction in acid solution:
MnO4– + H2SO3 → Mn2+ + HSO4–? pH
dependence?
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
10
2MnO4− (aq) + 5H2SO3(aq) + H+(aq) →
2Mn2+(aq) +5HSO3− (aq) + 3H2O(l)
The potential decreases as the pH increases.
5.7
5.13
Calculate the equilibrium constant for
Au+(aq) + 2CN–(aq) → [Au(CN)2]–(aq)?
K = 5.7 × 1038
5.14
Find the approximate potential of an
aerated lake at pH = 6, and predict the
predominant species? (a) Fe? 0.5 – 0.6 V
Write the Nernst equation for (a) The
reduction of O2?
(b) Mn? E = 0.55 V
Q = 1/(p(O2)[H+]4)
(c) S? At pH 0, 0.387 V. At pH 14, SO42–
would again predominate. HSO4– is the
predominant sulfur species at pH 6.
and
E = Eº – [(0.059V)/4][log(1/(p(O2)[H+]4)]
(b) The reduction of Fe2O3(s)?
5.15
Frost diagram and standard potential for
the HSO4−/S8(s) couple? 0.387 V
5.16
Equilibrium constant for the reaction
Pd2+(aq) + 4 Cl–(aq) ≡ [PdCl4]2–(aq) in 1 M
HCl(aq)? K = 4.37 × 1010
5.17
Reduction potential for MnO4– to MnO2(s)
at pH = 9.00? E = 0.98 V
5.18
Tendency of mercury species to act as an
oxidizing agent, a reducing agent, or to
undergo disproportionation? Hg2+ and
Hg22+ are both oxidizing agents. None of
these species are likely to be good reducing
agents. Hg22+ is not likely to undergo
disproportionation.
Thermodynamic tendency of HO2 to
undergo disproportionation? E = +1.275 V.
(is
positive),
HO2
will
undergo
disproportionation.
Dissolved carbon dioxide corrosive towards
iron? Carbon dioxide and water generate
carbonic acid which encourages the corrosion
process by lowering solution pH.
What is the maximum E for an anaerobic
environment rich in Fe2+ and H2S? –0.1 V.
+ 6
Q = 1/[H ] and E = Eº – (RT/nF)(13.8 pH)
5.8
Using Frost diagrams? (a) What happens
when Cl2 is dissolved in aqueous basic
Cl2 is thermodynamically
solution?
susceptible to disproportionation to Cl– and
ClO4– when it is dissolved in aqueous base.
The oxidation of ClO– is slow, so a solution of
Cl– and ClO– is formed when Cl2 is dissolved
in aqueous base.
(b) What happens when Cl2 is dissolved in
aqueous acid solution? Cl2 will not
disproportionate. Cl2 is thermodynamically
capable of oxidizing water.
(c) Should HClO3 disproportionate in
aqueous acid solution? Kinetic.
5.9
Write equations for the following
reactions: (a) N2O is bubbled into aqueous
NaOH solution?
5N2O(aq) + 2OH–(aq) → 2NO3–(aq) +
4N2 (g) + H2O(l)
(b) Zinc metal is added to aqueous acidic
sodium triiodide?
Zn(s) + I3 (aq) → Zn (aq) + 3I (aq)
–
2+
3I2(s) + 5ClO3– (aq) + 3H2O(l) → 6IO3–
(aq) + 5Cl– (aq) + 6H+(aq)
5.10
Electrode potential for Ni2+/Ni couple at pH
= 14? E =– 0.21 V
5.11
Will acid or base most favour the following
half-reactions? (a) Mn2+ → MnO4–? Base
(b) ClO4– → ClO3–? Acid
Base
(d) I2 → 2I–? Acid or base, no difference.
5.12
5.20
5.21
5.22
–
(c) I2 is added to excess aqueous acidic
HClO3?
(c) H2O2 → O2?
5.19
Determine the standard potential for the
reduction of ClO4– to Cl2? 1.392 V
How will edta4– complexation affect M2+ →
M0 reductions? The reduction of a M(edta)2–
complex will be more difficult than the
reduction of the analogous M2+ aqua ion.
5.23
Which of the boundaries depend on the
choice of [Fe2+]? Any boundary between a
soluble species and an insoluble species will
change as the concentration of the soluble
species changes. The boundaries between the
two soluble species, and between the two
insoluble species, will not depend on the
choice of [Fe2+].
5.24
Under what conditions will Al reduce
MgO? Above about 1400ºC.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 11
CHAPTER 6
SF6 has Oh symmetry. Analysis of the
stretching vibrations leads to:
Γstr = A1g (Raman, polarized) + Eg
Self-tests
S6.1
S6.2
Sketch the S4 axis of an NH4+ ion. How
many of these axes are there in the ion?
Three S4 axes.
(Raman) + T1u (IR).
SF5Cl has C4v symmetry.
stretching vibrations leads to:
Γstr = 3A1 (IR and Raman, polarized)
(a) BF3 point group? D3h.
2–
(b) SO4 point group? Td.
S6.3
S6.4
S6.5
S6.14 Symmetries of all the vibration modes of
[PdCl4]2-? A1g + B1g + B2g + A2u + B2u + 2Eu
A conformation of the ferrocene molecule
that lies 4 kJ mol–1 above the lowest energy
configuration is a pentagonal antiprism. Is
it polar? No.
6.1
Is the skew form of H2O2 chiral? Yes.
S6.7
Can the bending mode of N2O be Raman
active? Yes.
S6.8
Confirm that the symmetric mode is Ag?
D2h character table, which is the Ag symmetry
type.
Show that the four CO displacements in
the square-planar (D4h) [Pt(CO)4]2+ cation
transform as A1g + B1g + Eu. How many
bands would you expect in the IR and
Raman spectra for the [Pt(CO)4]2+ cation?
The reducible representation:
D4h
E
2C4
C2
2C2’
2C2″
i
2S4
σh
2 σv
2σd
Γ3N
4
0
0
2
0
0
0
4
2
0
Reduces to A1g + B1g + Eu
A1g + B1g are Raman active. Eu is IR active.
S6.10
Orbital symmetry for a tetrahedral array
of H atoms in methane? A1
S6.11
Orbital symmetry for a square-planar
array of H atoms? B2g.
S6.12
Which Pt atomic orbitals can combine with
which of these SALCs? The atomic orbitals
much have matching symmetries to
generate SALCs. 5s and 4dz2 have A1g
symmetry; the dx2-y2 has B1g symmetry; and
5px and 5py have Eu symmetry.
S6.13
+ 2B1 (Raman) + E (IR, Raman).
Symmetry species of all five d orbitals of
the central Xe atom in XeF4 (D4h, Fig. 6.3)?
dx2-y2 is B1g;
dxy is B2g;
dxz and dyz are Eg;
dz2 is A1g.
What is the maximum possible degeneracy
for an Oh molecule? 3.
S6.6
S6.9
Analysis of the
Predict how the IR and Raman spectra of
SF5Cl differ from that of SF6?
S6.15
+T1u.
SALCs for sigma bonding in O? A1g + Eg
Exercises
Symmetry elements? (a) a C3 axis and a σv
plane in the NH3 molecule?
N
H
N
H
H
H
H
H
σv
C3
(b) a C4 axis and a σh plane in the squareplanar [PtCl4]2– ion?
Cl
Cl
Cl
Cl
Cl
Cl
Cl
C4
6.2
Cl
Pt
Pt
σh
S4 or i? (a) CO2? i
(b) C2H2? i.
(c) BF3? neither.
(d) SO42–? three different S4.
6.3
Assigning point groups: (a) NH2Cl? Cs
(b) CO32–? D3h
(c) SiF4? Td
(d) HCN? C∞v.
(e) SiFClBrI? C1.
(f) BrF4–? D4h.
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
12
6.4
6.5
How many planes of symmetry does a
benzene molecule possess? What chlorosubstituted benzene has exactly four planes
of symmetry? 7,and C6H3Cl3.
The symmetry elements of orbitals? (a) An
s orbital? Infinite number of Cn axes, plus
an infinite number of mirror planes of
symmetry, plus center of inversion, i.
6.12
and irreducible representations?
A1g + A2g + B1g + B2g + Eg + 2A2u + B2u + 3Eu
6.13
(b) A p orbital? An infinite number of mirror
planes that pass through both lobes and
include the long axis of the orbital. In
addition, the long axis is a Cn axis, where n
can be any number from 1 to ∞.
(c) A dxy orbital? Center of symmetry, three
mutually perpendicular C2 axes, three
mutually perpendicular mirror planes of
symmetry, two planes that are rotated by 45º
about the z axis from the xz plane and the yz
plane.
(d) A dz2 orbital? In addition to the
symmetry elements possessed by a p orbital:
(i) a center of symmetry, (ii) a mirror plane
that is perpendicular to the C∞ axis, (iii) an
infinite number of C2 axes that pass through
the center of the orbital and are perpendicular
to the C∞ axis, and (iv) an S∞ axis.
6.6
For the 90o-twisted form of B2F4 (D2d)
The vibrations are: 3A1 (Raman, polarized) +
B1 (Raman) + 2B2 (IR and Raman ) + 3E (IR
and Raman).
6.14
(a) Take the 4 hydrogen 1s orbitals of CH4
and determine how they transform under
Td. (b) Confirm that it is possible to reduce
this representation to A1 + T2. (c) Which
atomic orbitals on C can form MOs with
H1s SALCs?
Using symmetry Td, Γ3N reduces to: A1 + T2.
The MOs would be constructed from SALCs
with H1s and 2s and 2p atomic orbitals on C.
6.15
Use the projection operator method to
construct the SALCs of A1 + T2 symmetry
that derive from the four H1s orbitals in
methane..
SO32– ion? (a) Point group? C3v
(c) Which s and p orbitals have the
maximum degeneracy? 3px and 3py orbitals
are doubly degenerate.
PF5? (a) Point group? D3h.
(b) Degenerate MOs? 2.
(c) Which p orbitals have the maximum
degeneracy? 3px and 3py atomic orbitals are
doubly degenerate
6.8
AsCl5 Raman spectrum consistent with a
trigonal bipyamidal geometry? No.
6.9
Vibrational modes of SO3? (a) In the plane
of the nuclei? 5
(b) Perpendicular to the molecular plane?
1
6.10
Vibrations that are IR and Raman active?
(a) SF6? None.
(b) BF3? The E′ modes are active in both IR
and Raman.
6.11
IR and Raman to distinguish between: (a)
planar and pyramidal forms of PF3, (b)
planar and 90o-twisted forms of B2F4 (D2h
and D2d respectively)?
(a) Planar PF3, D3h, vibrations are: A1’
(Raman, polarized) + 2E’ (IR and Raman) +
A2” (IR).
Pyramidal PF3, C3v, vibrations are: 2A1 (IR
and Raman, polarized) + 2E’ (IR and Raman)
(b) For the planar form of B2F4 (D2h):
The vibrations are:
3Ag (Raman, polarized) + 2B2g (Raman) +
B3g (Raman) + Au(inactive) + 2B1u (IR) +
B2u (IR) + 2B3u (IR).
(b) Degenerate MOs? 2
6.7
[AuCl4]− ion? Γ of all 3N displacements
Vibrations of a C6v molecule that are
neither IR nor Raman active? Any A2, B1,
or B2 vibrations of a C6v molecule will not be
observed in either the IR spectrum or the
Raman spectrum.
s = (1/2)(ϕ1 + ϕ2 + ϕ3 + ϕ3) (= A1)
px = (1/2)(ϕ1 – ϕ2 + ϕ3 – ϕ3) (= T2)
py = (1/2)(ϕ1 – ϕ2 – ϕ3 + ϕ3) (= T2)
pz = (1/2)(ϕ1 + ϕ2 – ϕ3 – ϕ3) (= T2)
SALCs for σ-bonds
(a) BF3?
(1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’)
(1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (=
E’)
(b) PF5?
(axial F atoms are ϕ4 + ϕ5)
(1/√2)(ϕ4 + ϕ5) (= A1’)
(1/√2)(ϕ4 − ϕ5) (= A2”)
(1/√3)(ϕ1 + ϕ2 + ϕ3) (= A1’)
(1/√6)(2ϕ1 – ϕ2 – ϕ3) and (1/√2)(ϕ2 – ϕ3) (=
E’)
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES 13
Exercises
CHAPTER 7
7.1
Self-tests
S7.1
Name and draw the structures of the
Nickel
complexes? (a) [Ni(CO)4]?
tetracarbonyl or tetracarbonyl nickel(0).
Give formulas corresponding to the
following names? (a) Cisdiaquadichloroplatinum(II)? cis[PtCl2(OH2)2],
CO
Ni
trans-diaquadichloroplatinum(II), trans[PtCl2(OH2)2].
OC
(b)
Diamminetetra(isothiocyanato)
chromate(III)?
[Cr(NCS)4(NH3)2] –. ,can exist as, cis[Cr(NCS)4(NH3)2] – or trans[Cr(NCS)4(NH3)2] –.
(c) Tris(ethylenediamine)rhodium
[Rh(en)3]3+.
(b) [Ni(CN)4]2–? Tetracyanonickelate (II).
2-
NC
(III)?
Ni
CN
CN
NC
(d)
Bromopentacarbonylmanganese
(I)? [MnBr (CO)5].
(c) [CoCl4]2–? Tetrachlorocobaltate (II)
(e) Chlorotris(triphenylphosphine)rhodium
(I)? [RhCl(PPh3)3].
S7.2
CO
CO
2-
Cl
What type of isomers are possible for
[Cr(NO2)2•6H2O]? The hydrate isomers
and linkage isomers of the NO2 group. Also,
[Cr(ONO)(H2O)5]NO2 •H2O.
Co
Cl
Cl
Cl
S7.3
Identifying isomers? Note that the two
phosphine ligands in the trans isomer are
related, therefore, they exhibit the same
chemical shift.
(d) [Mn(NH3)6]2+?
Hexaamminemanganesium (II)
2+
S7.4
Sketches of the mer and fac isomers of
[Co(gly)3]?
H3N
NH3
Mn
H3N
NH3
NH3
NH3
7.2
Write the formulas for the following
complexes?
(a) [CoCl(NH3)5]Cl2
(b) [Fe(OH2)6](NO3)3
(c) cis-[FeCl2(en)2]
S7.5
–
(b) trans-[CrCl2(ox)2]3 ? Not chiral.
S7.6
(d) [Cr(NH3)5μ–OH–Cr(NH3)5]Cl5
Which of the following are chiral? (a) cis[CrCl2 (ox)2]3–? Chiral .
7.3
Name the following complexes?
(c) cis-[RhH(CO)(PR3)2]? Not chiral.
(a) cis-[CrCl2(NH3)4]+? cistetra(ammine)di(chloro)chromium(III)
Calculate all of the stepwise formation
constants? Kf1 = 1 X 105. Kf2 will be 30%
less or 30000, Kf3 = 9000, Kf4 = 2700, Kf5 =
810, and finally Kf6 = 243.
(b) trans-[Cr(NCS)4(NH3)2] ? transdi(ammine)tetrakis(isothiocyanato)chromate
(III)
–
(c) [Co(C2O4)(en)2]+?
bis(ethylenediamine)oxalatocobalt(III).
Shriver & Atkins: Inorganic Chemistry 5e
ANSWERS TO SELF-TESTS AND EXERCISES
14
7.4
Four-coordinate complexes? (a) Sketch the
two observed structures?
L
M
M
L
L
L
L
L
L
L
square planar
tetrahedral
(b)
Isomers expected for MA2B2?
tetrahedral complex, no isomers, for a squareplanar complex, two isomers, cis and trans.
7.5
For five-coordinate complexes, sketch the
two observed structures?
A
A
B
B
E
M
M
E
B
E
B
A
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