CHEM0030 page 1 of 13
UNIVERSITY COLLEGE LONDON
B.SC. DEGREE 2020 M.SCI. DEGREE 2020
CHEM0030: ADVANCED TOPICS IN INORGANIC CHEMISTRY
Credit value: 15
CHEM0030 page 2 of 13
Candidates should attempt ALL questions. Each question is marked out of 25 and the numbers in square brackets in the right-hand margin indicate the provisional allocation of marks to the subsections of a question.
1. Answer ALL parts.
Sketch two graphs showing size variation with atomic number of (i) lanthanide atoms and (ii) lanthanide 3+ cations. Briefly discuss the similarities and differences between the two graphs. [4]
(ii) [2] For both Ln and Ln3+, there is a contraction with increasing Z and the f-electrons
are increasingly pulled in towards the core due to poor shielding. The cations all have the electronic configuration [Xe]4f n with the f electrons pulled into the core
and so a smooth curve is obtained. [1] By contrast, for the metals, both the 6s2 and to some extent the 4f-electrons, participate in metal-metal bonding. However, but the half-filled and fully-filled
4f shells of Eu and Yb, respectively, result in no contribution to metal-metal bonding from the 4f and so the bonding is weaker and the radii are slightly larger
for the atoms in these two metals. [1]
(i) Derive, with justification, the spectroscopic ground-state term symbol for
any Ln3+ cation of your choice, excluding the middle (Gd3+) and end (La3+, Lu3+) members of the series. [4]
[Part marks]
A model answer is given below for neodymium:
Nd3+ is [Xe]4f 3 [1]
L = 6 [1⁄2] S = 11⁄2 (so 2S + 1 = 4) [1⁄2] J = 6 − 11⁄2 = 41⁄2 as less than half-full shell [1⁄2] Ground state term is: 4I41⁄2 [1⁄2]
Other answers are 2F2.5 (Ce), 3H4 (Pr), 4I4.5 (Nd), 5I4 (Pm), 6H2.5 (Eu), 7F6 (Tb), 6H7.5 (Dy), 5I8 (Ho), 4I7.5 (Er), 3H6 Tm, 2F3.5 (Yb)
(ii) State the equation for the calculation of the magnetic moment of a lanthanide cation in dilute solution. Hence determine the magnetic moment for the Ln3+ cation chosen in (i) in dilute solution given that the Landé factor gJ is given by:
gJ = 3/2 + [S (S + 1) − L (L + 1)] / [2J (J + 1)]
What assumption is made in this calculation? [4]
A model answer is given below for neodymium:
μeff = gJ [J (J + 1)]1⁄2 [1] Students need to calculate gJ first using the S, L, and J values derived:
gJ = 8/11 [1] μeff = 3.62 Bohr magnetons [1] The assumption made is that there are no cations in excited states as these would exhibit a different magnetic moment as the excited states have different L, S, and
J quantum numbers. There are no low–lying excited states for Nd3+ so the value matches the experiment. [1]
(iii) Briefly discuss the application of Nd3+ in solid-state lasers. [3]
Exploits metastable excited state in Nd3+ cations, which are doped into a matrix
of Y based cubic oxide such as a garnet Y3Al5O12 (YAG) or a vanadate YVO4. [1]
Student could show a diagram like the above or explain the diagram in words. [1]
Start with a di-iodosalt so that Sm is in (II) oxidation state and use a dry solvent such as THF in an inert atmosphere:
SmI2 + 2NaC5Me5 +2THF= Sm(C5Me5)2(THF)2 [1] The product is then heated to remove the solvent that is bound to the product: Sm(C5Me5)2(THF)2 = Sm(C5Me5)2 + 2THF [1]
(ii) Explain why a similar synthesis of Sm(C5H5)2 is more problematic. [1]
Lanthanides have a very strong binding affinity towards oxygen and so anhydrous oxygen containing solvents such as ethers are problematic. They can
be removed by heating but this only works if steric effects from the other ligands help drive off the solvent. This is not the case for an unsubstituted cyclopentadienyl ring which is not very bulky and so Sm(C5H5)2(THF)2 does not decompose to the desired product on heating. [1]
(iii) Discuss differences in reactivity of anhydrous FeCl2 with lanthanide and actinide cyclopentadienyl compounds. [2]
Cyclopentadienyl complexes such as LnCp3 have ionic bonding of the Cp rings to
the lanthanide cation and so will exchange with the Cl− in FeCl2 to form ferrocene. [1] By contrast, the bonding in actinide complexes such as Th(C5H5)4 have a large degree of covalent bonding due to the 5f orbitals not being full pulled into the core and so the Cp rings will not exchange with the Cl− in FeCl2 to form ferrocene.
The transition is in the infrared region of the electromagnetic spectrum but visible light in the green region can be obtained using a frequency doubling crystals such
as KTP (KTiOPO4) = Potassium Titanyl Phosphate. [1]
(c) (i) Give a reaction scheme for the synthesis of Sm(C5Me5)2 starting from a halide salt dissolved in an appropriate solvent. [2]
CS Help, Email: tutorcs@163.com
(d) Marie Curie’s research work opened up the field of chemistry on the radioactive
elements. Discuss the impact of her work (maximum 300 words). [5]
This is an open-ended question but candidates need to pick out 5 impact points and argue
how they impact modern society. (Candidates could pick out fewer, but then need to argue in more depth.) The following list of impact areas is not exhaustive:
Marie Curie coined the term “radioactivity”, which is still in use today, i.e. she had an impact on the language we use.
With colleagues, Marie Curie discovered the first of the actinide elements (Actinium) that led to the discovery of the 2nd row of f-block elements as well as polonium and radium using radioactivity as an analytical tool to monitor them.
Work on radium led to the foundation of the Radium Institute in Paris for the radiotherapy treatment of cancer, a technique still used to this day, though not with radium. (It also led to a lot of “quack” remedies with radium too.)
A minor impact was in the use of radium for the development of radio- luminescent paints for watch dials and, more importantly, for aircraft instrument dials for night-time flying. However, the consequential occupational hazard from pointing the paint brushes with ones tongue was a factor in the establishment of occupational-disease labour laws in the USA.
Impact on her daughter’s career who went on to discover the trans-uranium elements by neutron bombardment (for which a Nobel prize was awarded), which in turn led to the synthesis of 239Pu for atomic weapons and 238Pu for nuclear batteries used in space exploration, plus 241Am in smoke alarms to prevent deaths from household fires.
Marie Curie’s experience with X-rays was exploited in the use of X-rays in medical field hospitals in WWI (though the exposure to the X-rays was to subsequently kill her).
The impact of her work led to several subsequent Nobel prizes.
Role model for women in the STEM subjects. Scholarships etc. are now named
after her to encourage younger scientists, especially female ones.
appreciation of the impact of her work.
are not given for simply reciting her life history… but for demonstrating an
2. Answer ALL parts.
(a) The catalytic cycle below shows the rhodium-catalysed hydroformylation of an
(i) Give the valence electron count and the oxidation state of Rh for compounds A to F.
A 16 e-; Rh(I)
B 18 e-; Rh(I)
C 16 e-; Rh(I)
D 18 e-; Rh(I)
E 16 e-; Rh(I)
F 18 e-; Rh(III) 1
(ii) Explain the mechanisms of the reactions A B, B C, D E, E F
and F A. Justify your answer. [5] A B alkene coordination; electron count increases by 2. 1
B C 1,2-alkene insertion into Rh-H bond; no change in oxidation state; electron count reduces by 2. 1
D E CO insertion into Rh-C bond; no change in oxidation state; electron count reduces by 2. 1 E F Oxidative addition of H2; oxidation state increases by 2 1 F A Reductive elimination; oxidation state decreases by 2 1
(iii) Describe how the alkene ligand coordinates to the Rh centre in B. [4]
Dewar-Chatt Duncanson model – σ-donation from the C=C π orbital with concomitant π- backbonding into an empty π* orbital on the alkene. There is synergistic effect because the greater is the σ-donation to the metal, the greater the π-backbonding to the alkene. The greater the electron density back donated into the π* orbitals on the alkene, the weaker the C=C bond and the greater is its length. It is also possible to say that the hybridisation of the alkene carbons changes from sp2 to sp3 as back donation increases. (2)
(b) Describe each of the reaction types taking place in the reactions below and identify each of the six platinum compounds G-M.
G = [Pt(PPh3)2Cl]+[BF4]- (halide abstraction) (AgCl) 1
H = Pt(PPh3)2Cl2MeI (oxidative addition) 2
I = Pt(PPh3)2PhCl (salt elimination) (ClMgBr) 2
J = Pt(PPh3)2(COPh)Cl (CO insertion) 2
E = Pt(PPh3)2MeCl (salt elimination) (ClMgBr) 1
F = Pt(PPh3)2{C(H)MeCH2Me}Cl (insertion) also credit if square pyramidal intermediate alkene complex given. 2
3. Answer ALL parts.
(a) Explain why, in an n-type doped semiconductor, the effective mass of the
conduction band minimum plays a large role in determining the magnitude of the
conductivity. [3]
Effective mass, m*, stems from large orbital overlap – linked to high mobility [1 mark] μ = eτ/m* — The smaller m* is the higher the mobility. [1 mark]
σ = nμe [1 mark]
(b) Describe how the conductivity of a metal and a semiconductor evolve with temperature. [6]
σ = ne2τ/m*
In a metal: the carrier concentration, n, changes very slowly with temperature (1 mark).
τ is inversely proportional to temperature (τ α 1/T), due to scattering by lattice vibrations (phonons). Therefore, a plot of σ vs. 1/T (or ρ vs. T) is essentially linear. (1 mark) Conductivity goes down as temperature increases (1 mark).
In a semiconductor: The carrier concentration increases as temperature goes up, due to excitations across the band gap, Eg. n is proportional to exp{-Eg/2kT}. (1 mark)
τ is inversely proportional to temperature (1 mark)
The exponential dependence of n dominates, therefore, a plot of ln σ vs. 1/T is essentially linear. Conductivity increases as temperature increases (1 mark)
(c) The table below presents the undoped optical band gap (Egopt, undoped), the ionization potential (IP), and the electron affinity (EA) for some direct band gap oxides when undoped, and the optical band gap (Egopt,doped) when doped with 1 × 1020 cm−3 of electrons.
Egopt,undoped (eV) Egopt,doped (eV)
IP (eV) EA (eV)
Material A Material B 2.8 2.0 3.2 3.2 7.6 7.2 4.8 5.2
Material C 3.0 3.1 7.0 4.0
Material D 1.3
(i) Which material is most likely to be Justify your answer.
both p-type and n-type dopable?
Material D has a low IP of 5.6 eV, meaning it is the most likely to be also p-type dopable (1 mark).
It also shows a BM shift of 0.3 eV, indicating it is a degenerate n-type semiconductor (1 mark).
(ii) Given the equation below, which material is likely to possess the highest mobility for electrons? Justify you answer.
𝐸𝐵𝑀 = ħ2 (3𝜋2𝑛𝑒)2/3 2𝑚∗
(iii) Which material would be most suited for use as a potential solar absorber? Justify you answer. [2]
Material D, as it has a band gap of 1.3 eV, which is near the maximum of the Shockley Quisser limit. (1 mark)
Material D also has a direct band gap, for strong absorption, suitable for thin film PV. (1 mark)
(d) Write out the equation for the dimensionless figure of merit (ZT) of a thermoelectric. Define all the factors in the equation, and explain using equations why these factors are interconnected. [5]
The equation for the dimensionless figure of merit is:
𝑍𝑇 = 𝛼2𝜎𝑇 = 𝛼2𝑇 𝜅 𝜌𝜅
Where α is the Seebeck coefficient, σ the electrical conductivity, ρ the electrical resistivity, and κ the total thermal conductivity (κ = κelec + κlatt; the electronic and lattice contributions, respectively) (1⁄2 mark for correct equation, 1⁄2 mark for explanation of all the terms, 1 mark for knowing that K is broken into Kelec and Klatt)
𝛥 𝑉 2 𝑘 𝐵2 𝑚 ∗ 𝑇 𝜋 23 𝛼= 𝛥𝑇= 3𝑒ħ2 (3𝑛)
Seebeck coefficient (α) is maximized by a high m* (flat bands), and a low carrier concentration.
It favours poor conductors. (1⁄2 mark for equation, 1⁄2 mark for explanation of all
terms, 1 mark for knowing that it favours poor conductors)
Material B has the largest BM shift (1.2 eV). (1 mark)
As n is the same for all systems, then m* has to be the smallest for material B to achieve such a big BM shift – highest mobility. (1 mark)
conductivity is maximised by high carrier concentration (n) and a low effective mass (High μ). (1⁄2 mark for equation, 1⁄2 mark for explanation that you need high mobility and high carrier concentration)
𝑘𝑒𝑙𝑒𝑐 = 𝐿0𝜎𝑇
Electronic thermal conductivity is proportional to conductivity. So as you increase the conductivity, you increase kelec, lowering the ZT. (1⁄2 mark for equation, 1⁄2 mark for understanding.)
Briefly outline why the design of a high performance p-type transparent
conducting oxide remains on the “to do” list for the research community. [5]
P-type TCOs needed for transparent transistors and “invisible electronics” (0.5 marks) Current TCOs only transparent electrodes, not semiconductors (0.5 marks)
All industry TCOs are n-type(0.5 marks)
Doping n-type TCOs p-type impossible due to large IPs and localized oxgen holes(0.5 marks)
Strategies to make p-type materials include CMVB – Cu –O initially(0.5 marks) However, poor conductivities, indirect band gaps, deep defect levels(0.5 marks)
Extend to Cu- Ch – challenges are band gap, but better mobility and conductivity. (0.5 marks)
D6 spinels also limited success(0.5 marks)
Lone pair oxides limited SnO. (0.5 marks)
Cr based oxides also limited. (0.5 marks)
Programming Help, Add QQ: 749389476
4. Answer ALL parts.
(a) Determine the number of d-electrons and assign the nature of the principal
transition that leads to the following complexes being visibly coloured: [CrO4]3−, [CoCl4]2−, [Cu(His)2]2+, and [Au(bipy)2]+. When charge transfer is the principal cause, indicate whether it originates from the metal or the ligand. [4]
Answer) [CrO4]3- = d1, d-d transition [CoCl4]2-= d7, d-d transition
[Cu(His)2]2+ = d9, d-d transition [Au(bipy)2]+ = d10, MLCT charge transfer
(b) The UV-visible spectrum for a d3 octahedral complex is shown below together with the corresponding Tanabe-Sugano diagram:
(i) Specify the term symbols that give rise to the two major bands in the spectrum. [4]
Answer) 4A2g 4T2g (4F) & 4A2g 4T1g (4P)
(ii) Using the Tanabe-Sugano diagram, calculate the position of a 3rd band
that might be observed if the spectrum was acquired down to 200 nm. [4]
Answer) Candidates will need to firstly work out from the UV-Vis diagram the band positions; these are at ~ 450 and ~ 600 nm. From this they should be able to work out that this corresponds to (E)v1B and (E)v2B values of 30 and 40 respectively, leading to a 0/B value ~ 30. Using the 600 nm (16667 cm-1) band to calculate the 0 (~ 16680) & B (556), allows one to work out that the (E)v3B value should be found at 65. Multiplying 65 * 556 gives a band position at 36140 cm-1 which corresponds to a wavelength of 276 nm.
程序代写 CS代考 加QQ: 749389476
(iii) Propose the position of at least one additional band if the spectrum were
to be extended to 900 nm. Why is this transition so weak with respect to
the other transitions observed? [5]
Answer) This will be the 4A2g 2T1g/4A2g 2Eg transitions and can be calculated using the Tanabe Sugano diagram yielding an (E)vB value of 21 and therefore a band position of 21 * 556 = 11676 cm-1 or 856 nm. The transition is weak since it is disallowed according to spin and orbit selection rules.
(iv) Explain why no weak shoulders are observed in the spectrum. [2]
Answer) These arise due to the presence of a Jahn-Teller distortion but since the complex is not distorted this is not observed.
(c) Give TWO examples of how optical spectroscopy can be used to obtain information on the evolving coordination and/or oxidation state of a transition metal ion in a functional material. Propose how this information can be used to design better performing materials (maximum 300 words). [6]
Marks were awarded as follows:
1 mark for each ‘optical spectroscopy’ technique identified. Candidates needed to have identified that optical spectroscopy refers to techniques such as IR, UV-Vis or Raman spectroscopy (not X-rays). Maximum 2 marks.
1 mark for each description of how an optical spectroscopy technique can be used to differentiate between a change in transition metal coordination/oxidation state. Ideally this would have been illustrated using exemplars from the lecture notes/literature. Maximum 2 marks.
If the candidates had used the lecture notes/literature for the exemplars they could have extrapolated how this information could be used to design better performing materials. Marks were given for simple observations pertaining to the transition metal behaviour such as understanding/improving adsorption/desorption, redox state, coordination state and more general comments regarding understanding reaction mechanism, effecting lower temperature of reactivity, improved stability etc. Maximum 2 marks.