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The Hong Kong Polytechnic University
Department of Applied Biology and Chemical Technology
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Course : B.Sc.(Hons.) in Chemical Technology
Year : 3 Subject: Inorganic Chemistry II
Inorganic Chemistry
Class : 12447 Subject Code: ABCT3779
12455
12047 ABCT340
Session : 2015 / 2016 , Semester 2
Date : 7 May 2016
Time Allowed: 3 hours Time: 08:45 – 11:45
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This question paper has THIRTEEN pages. (including this page)
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Instructions to Candidates :
There are TEN questions in TWO Sections. Answer all the questions in BOTH Sections.
Use separate answer scripts for Section A (WHITE) and B (GREEN)
Available from Invigilators: Table of Fundamental Constants, Valence Orbital Ionization
Energies (eV) (from H to Kr), Periodic Table, Character Tables and Tanabe-Sugano Diagrams
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DO NOT TURN OVER THIS PAGE UNTIL YOU ARE TOLD TO DO SO.
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Section A (50 marks + optional bonus 8 marks): Q1 – 6 (use the WHITE answer
book for Section A)
1. Compare the relative reactivity of CCl2 and SnCl2 with respect to the reaction with chlorine.
Explain briefly your answer.
(8 marks)
2. (a) Carbene (CH2) is a V-shaped molecule. By means of Molecular Orbital Theory, prepare the
orbital potential energy diagram of the carbene.
(7 marks)
(b) Consider the following gas-phase reaction of carbene (CH2),
CH2 + H+ → [CH3]+ proton affinity of methyl anion = HPA
Derive an expression of HPA for carbene in terms of ionization energy of hydrogen [IE(H)],
ionization energy of CH2 [IE(carbene)] and bond dissociation energy of methyl cation [BDE (C-
H)].
Given that the BDE (C-H) = 439 kJ mol-1, estimate the HPA for the carbene.
(5 marks)
3. (a) Discuss briefly whether the reaction of dinitrogen (N2) and H2 would occur spontaneously in a
concerted manner to form diazene (HN=NH) at room temperature and pressure.
(5 marks)
4. Despite the fact that benzene is more acidic (pKa ~ 45) than methane (pKa ~ 60). Explain why the
proton-transfer reaction from benzene to MeLi is kinetically slow.
(optional bonus: 8 marks)
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5. The reaction of W(CO)5(MeCN) with an excess of phenylacetylene (PhCCH) in chloroform gives
a red product A. The spectroscopic and analytical data of A are summarized as follows:
IR, cm-1
3010(s), 2980(vs), 1988(vs), 1932(s), 1770(s),1650(s)
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H NMR, (, ppm)
7.54 (m, I = 5)
3.88 (m, I = 1)
Analytical data (%) C: 45.08, H: 2.41
Propose a structural formula of A and account for as much of the spectroscopic data as possible.
(15 marks)
6. (a) Discuss the oxygen uptake properties of myoglobin and hemoglobin at different partial pressure
of oxygen.
(b) Discuss briefly the essential requirements for a clinically useful Gadolinium (Gd) based MRI
contrast agent.
(10 marks)
End of Section A
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Section B (50 marks + optional bonus 5 marks): Q7 – 10 (use the GREEN answer
book for this section)
7.a) For any three of the objects below, state the point groups and the associated symmetry elements.
Identify the point group of molecule 5 for bonus ( 3 marks)
(9 marks)
1 2 3 4 5 (Bonus-3 marks)
Give an example of a molecule from each of your selected choice of point groups from 7a)
(3 marks)
(Total 12 marks)
8. a) Consider a molecule with the general formula ML5,
What is the coordination geometry of this compound and its point group?
(2 marks)
b) Fe(CO)5 is an example of a molecule with this point group -name it according to the IUPAC system.
This compound also exist in another geometry- it is a typical fluxional compound. What is the
coordination geometry, draw the structure and explain which form is more stable
(3 marks)
c) For the complex ion [Cu(CN)2Cl3] 3-. The geometry around the central atom is a trigonal bipyramid.
The two cyanide ions may be in the axial positions, or in the trigonal plane. If the molecule with the
cyanide ionsare in the axial positions-What is the point group?
(1 mark)
d) Use the correct character table and find the symmetry species for the cyanide stretches only
(6 marks)
e) Determine whether any of the modes are IR and/or Raman active.
(2 marks)
(Total 14 marks)
9. Square planar compounds are a common geometry for transition metals. These compounds can form
isomers if there is more than one ligand in the structure. For a square planar complex - Cl2H6N2Pt – one
of these isomers is a chemotherapy drug.
a). What is the class of isomer they form? Draw out the possible isomers and label them.
(3 marks)
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Which isomer is the chemotherapy drug?
For the synthesis of the non active isomer- (the one with the weaker pharmacological effect)- one of
the synthetic route uses K2[PtCl4] which is first converted to Cl2[Pt(NH3)4] – what is the reagent used
to form this?
(Bonus 2 mark)
b) In general, low-spin d8 metal ions are all square planar. Show the nature of distortion expected. Use
an energy level diagram to explain.
(5 marks)
c) What is the coordination geometry of [Ni(CN)4]2- complex? Explain briefly your answer.
(4 marks)
(Total 12 marks)
10) The design of coordination compounds is very important in contrast agents for use in diagnostic
imaging.
Besides the use of transition metals- d elements, some inner transition metals are also used- f- elements
as well as radioactive metals.
a) Name the 7 f-electron element that is used to form the coordination compound that is used in
diagnostic imaging. (1 mark)
b) A certain type of ligand is used to form this complex- name the type of ligand and explain why it
is used? (2 marks)
c) What type of imaging modality is this coordination complex used for? (1 mark)
d) What is the saturated coordination number of this complex? (1 mark)
e) In this f-element complex -where does the toxicity comes from and explain (2 marks)
f) In PET imaging, a transition coordination complex is used. What is this transition metal and what
biological metal does it resembles? (2 marks)
g) In SPECT imaging, a radioisotope complex is used- pertecnetate. What is the geometry of this
structure? What biological anion does it resemble and where does this accumulate in the body?
(3 marks)
(Total 12 marks)
End of Section B
<END OF PAPER>
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Fundamental Constants
Name Symbol Value
Atomic mass unit u 1.660 54 10-27 kg
Avogadro’s constant NA 6.022 14 1023 mol-1
Boltzmann’s constant k 1.380 66 10-23 J K-1
Fundamental charge e 1.602 18 10-19 C
Faraday’s constant F = eNA 9.648 53 104 C mol-1
Gas Constant R = kNA 8.314 47 J K-1 mol-1
Mass of electron me 9.109 39 10-31 kg
Mass of neutron mn 1.674 93 10-27 kg
Mass of proton mp 1.672 62 10-27 kg
Planck’s constant h 6.626 08 10-34 J s
ћ = h / 2 1.054 57 10-34 J s
Rydberg constant RH 3.289 84 1015 Hz
2.179 874 10-18 J
13.605 698 eV
Speed of light in vacuum c 2.997 92 108 m s-1
Standard acceleration of free fall g 9.806 65 m s-2
Vacuum permittivity 8.854 19 10-12 J-1 C2 m-1
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Valence Orbital Ionization Energies (eV) (from H to Kr)
Atomic
Number Element
1 H -13.61
2 He -24.59
3 Li -5.39
4 Be -9.32
5 B -14.05 -8.30
6 C -19.43 -10.66
7 N -25.56 -13.18
8 O -32.38 -15.85
9 F -40.17 -18.65
10 Ne -48.47 -21.59
11 Na -5.14
12 Mg -7.65
13 Al -11.32 -5.98
14 Si -15.89 -7.78
15 P -18.84 -9.65
16 S -22.71 -11.62
17 Cl -25.23 -13.67
18 Ar -29.24 -15.82
19 K -4.34
20 Ca -6.11
30 Zn -9.39
31 Ga -12.61 -5.93
32 Ge -16.05 -7.54
33 As -18.94 -9.17
34 Se -21.37 -10.82
35 Br -24.37 -12.49
36 Kr -27.51 -14.22
Source: G. L. Miessler and D. A. Tarr (2004), Inorganic Chemistry, 3rd ed., Table 5-1, p.134
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Periodic Table of Chemical Elements
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C 2 E C2
A 1 1 z, Rz x2, y2, z2, xy
B 1 -1 x, y, Rx, Ry yz,xz
C2h E C2 i σh
Ag 1 1 1 1 Rz x2, y2, z2
Bg 1 -1 1 -1 Rx, Ry xz, yz
Au 1 1 -1 -1 z
Bu 1 -1 -1 1 x,y
C3h E C3 C32 σh S3 S35 ε=exp (2πi/3)
A' 1 1 1 1 1 1 Rz x2+y2, z2
E' 1 ε ε* 1 ε ε* (x,y) (x2-y2, xy)
1 ε* ε 1 ε* ε
A" 1 1 1 -1 -1 -1 z
E" 1 ε ε* -1 -ε -ε* (Rx, Ry) (xz, yz)
1 ε* ε -1 -ε* -ε
C3 E C3 C32 ε=exp (2πi/3)
A 1 1 1 z, Rz x2+y2, z2
E 1 ε ε* (x,y) (Rx,Ry) (x2-y2, xy), (xz, yz)
1 ε* ε
C4 E C4 C2 C43
A 1 1 1 1 z, Rz x2+y2, z2
B 1 -1 1 -1 x2-y2, xy
E 1 i -1 -i (x,y) (Rx,Ry) (xz, yz)
1 -i -1 i
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S4 E S4 C2 S43
A 1 1 1 1 Rz x2+y2, z2
B 1 -1 1 -1 z x2-y2, xy
E 1 i -1 -i (x, y); (Rx, Ry) (xz, yz)
1 -i -1 i
S6 E C3 C32 i S65 S6
Ag 1 1 1 1 1 1 Rz x2+y2, z2
Eg 1 ε ε* 1 ε ε* (Rx, Ry) (x2-y2, xy)(xz, yz)
1 ε* ε 1 ε* ε
Au 1 1 1 -1 -1 -1 z
Eu 1 ε ε* -1 -ε -ε* (x, y)
1 ε* ε -1 -ε* -ε
D2d E 2S4 C2 2C2' 2σd
A1 1 1 1 1 1 x2+y2, z2
A2 1 1 1 -1 -1 Rz
B1 1 -1 1 1 -1 x2-y2
B2 1 -1 1 -1 1 z xy
E 2 0 -2 0 0 (x, y)(Rx, Ry) (xz, yz)
D3d E 2C3 3C2 i 2S6 3σd
A1g 1 1 1 1 1 1 x2+y2, z2
A2g 1 1 -1 1 1 -1 Rz
Eg 2 -1 0 2 -1 0 (Rx, Ry) (x2-y2, xy),(xz, yz)
A1u 1 1 1 -1 -1 -1
A2u 1 1 -1 -1 -1 1 z
Eu 2 -1 0 -2 1 0 (x, y)
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Tanabe-Sugano Diagrams for d2 to
d8 Electron Configurations in
Octahedral Ligand Field. All terms
have g symmetry
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