0% found this document useful (0 votes)
8 views26 pages

24 Merged

This document is an examination paper for the B.Sc.(Hons.) in Chemical Technology course at The Hong Kong Polytechnic University, specifically for the Inorganic Chemistry II subject. It consists of two sections with a total of ten questions covering various topics in inorganic chemistry, including molecular orbital theory, reaction kinetics, and coordination compounds. The paper also includes fundamental constants and ionization energies for reference.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
0% found this document useful (0 votes)
8 views26 pages

24 Merged

This document is an examination paper for the B.Sc.(Hons.) in Chemical Technology course at The Hong Kong Polytechnic University, specifically for the Inorganic Chemistry II subject. It consists of two sections with a total of ten questions covering various topics in inorganic chemistry, including molecular orbital theory, reaction kinetics, and coordination compounds. The paper also includes fundamental constants and ionization energies for reference.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 26

© The Hong Kong Polytechnic University

The Hong Kong Polytechnic University

Department of Applied Biology and Chemical Technology

____________________________________________________________________________

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


____________________________________________________________________________

This question paper has THIRTEEN pages. (including this page)

____________________________________________________________________________

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
____________________________________________________________________________

DO NOT TURN OVER THIS PAGE UNTIL YOU ARE TOLD TO DO SO.

1
© The Hong Kong Polytechnic University

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)

2
© The Hong Kong Polytechnic University

5. The reaction of W(CO)5(MeCN) with an excess of phenylacetylene (PhCCH) 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)
1
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

3
© The Hong Kong Polytechnic University

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)
4
© The Hong Kong Polytechnic University

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>

5
© The Hong Kong Polytechnic University

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

6
© The Hong Kong Polytechnic University

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

7
© The Hong Kong Polytechnic University

Periodic Table of Chemical Elements

8
© The Hong Kong Polytechnic University

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

9
© The Hong Kong Polytechnic University

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)

10
© The Hong Kong Polytechnic University

11
© The Hong Kong Polytechnic University

12
© The Hong Kong Polytechnic University

Tanabe-Sugano Diagrams for d2 to


d8 Electron Configurations in
Octahedral Ligand Field. All terms
have g symmetry

13
© The Hong Kong Polytechnic University

The Hong Kong Polytechnic University

Department of Applied Biology and Chemical Technology

____________________________________________________________________________

Course : B.Sc.(Hons.) in Chemical Technology

Year : 3 Subject: Inorganic Chemistry II


Inorganic Chemistry

Class : 12447 / 12455 Subject Code: ABCT3779


12047 ABCT340

Session : 2014/2015 , Semester 2

Date : 7 May 2015

Time Allowed: 3 hours Time: 08:45 – 11:45


____________________________________________________________________________

This question paper has THIRTEEN pages. (including this page)

____________________________________________________________________________

Instructions to Candidates :

There are TEN questions in TWO Sections. Answer all the questions in BOTH Sections.

Use separate answer scripts for Section A (BLUE) and B (PINK)

Available from Invigilators: Table of Fundamental Constants, Valence Orbital Ionization


Energies (eV) (from H to Kr), Periodic Table, Character Tables and Tanabe-Sugano Diagrams
____________________________________________________________________________

DO NOT TURN OVER THIS PAGE UNTIL YOU ARE TOLD TO DO SO.

1
© The Hong Kong Polytechnic University

Section A (50 marks + optional bonus 8 marks): Q1 – 6 (use the BLUE answer
book for Section A)

1. The ionization energy of NO (9.26 eV) is considerably smaller than that of CO (14.01 eV).
Explain this finding on the basis of electronic structures of the molecules.
(5 marks)

2. Consider the following gas-phase reaction,

Y + H+ → [B-H+] (Y = proton acceptor) proton affinity = HPA

Derive an expression of HPA in terms of ionization energy of hydrogen [IE(H)], ionization


energy of Y [IE(Y)] and bond dissociation energy (BDE) of [B-H+].
(5 marks)

3. With reference to their electronic structures, discuss why ammonia (NH3) has higher proton
affinity than methane (CH4). [proton affinities (kJ mol-1): 543 (methane); 854 (ammonia)]

(10 marks)

4. (a) Acetylene (HCCH) is a linear molecule. Prepare the molecular orbital energy diagram of
acetylene.

(10 marks)

(b) Discuss briefly whether acetylene and H2 would react spontaneously to form ethylene at
room temperature and pressure.

(optional bonus: 8 marks)

5. [C5H5Cr(CO)3]2 reacts with tetramethylthiuramdisulfide (A) in refluxing benzene to give a


chromium-containing compound (B) having the following spectroscopic features:

Structural formula A Spectroscopic features of B


IR: 1950, 1860 cm-1
H 3C S S CH3
1
H NMR: δ5.48 (I = 5), δ3.21 (I = 6)
N C C N

H 3C S S CH3
MS: m/e = 293 (A pattern similar to
the Cr isotope pattern is observed)
(A)

Suggest a structural formula for B. (10 marks)

2
© The Hong Kong Polytechnic University

6. (a) Briefly describe the mode of binding of O2 to the iron metal centre in a haem unit of
haemoglobin.

(b) What are “picket fence” porphyrins? Stretch one example of this.

(c) Why are these porphyrins used in model studies of O2 binding to haemoglobin or myoglobin?

(d) The binding of O2 to haemoglobin exhibits “cooperativity” effect. Explain this statement.

(10 marks)

End of Section A

3
© The Hong Kong Polytechnic University

Section B (50 marks + optional bonus 6 marks): Q7 – 10 (use the PINK answer
book for this section)

7. For ANY three of the four molecules below, state the point groups and the associated
symmetry elements. (9 marks)
(Optional Bonus 3 marks for the fourth molecule)

(Total 12 marks)

1 2 3 4

N
H H
S Cl N
Cl Co
O Ru
Cl N
H3C CH3 Cl
N

8. [Ni(H2O)6]2+ forms a green complex and would give a blue solution upon addition of aqueous
ammonia. The blue solution would then change to violet when adding “en” ligand.

For ALL the three Nickel Complexes:


(a) Draw the structural formulae of the complexes in their correct geometry; identify if there
are any possible isomers and state what type of ligand “en” is compared to water and
ammonia.
(5 marks)

(b) Name the complexes (3 marks)

(c) List the ligands in order of their increasing energies in terms of CF/LF theory and explain
your answer (5 marks)

(d) Give the complementary colours of the complexes (Optional Bonus: 3 marks)

(Total 16 marks)

9. (a) Ni2+ and Co2+ can form high-spin complexes in octahedral and tetrahedral geometry. Which
species is more likely to form a tetrahedral complex? Explain your answer.
(5 marks)

(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 14 marks)

4
© The Hong Kong Polytechnic University

10. N2F2 molecule has two isomers.


(a) Draw the structures of the two isomers and label them (2 marks)

(b) For the isomer that contains an inversion centre, what is the point group of that molecule?
(1 mark)

(c) Use the correct character table and find the symmetry species for its normal modes of
vibration
(8 marks)

(d) Determine whether any of the modes are IR and/or Raman active.
(3 marks)

(Total 14 marks)

End of Section B

<END OF PAPER>

5
© The Hong Kong Polytechnic University

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

6
© The Hong Kong Polytechnic University

Valence Orbital Ionization Energies (eV) (from H to Kr)

Atomic Number Element 1s 2s 2p 3s 3p 4s 4p


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

7
© The Hong Kong Polytechnic University

Periodic Table of Chemical Elements

8
© The Hong Kong Polytechnic University

C2 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

9
© The Hong Kong Polytechnic University

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)

10
© The Hong Kong Polytechnic University

11
© The Hong Kong Polytechnic University

12
© The Hong Kong Polytechnic University

Tanabe-Sugano Diagrams for d2 to


d8 Electron Configurations in
Octahedral Ligand Field. All terms
have g symmetry

13

You might also like