CHEM1101
2012-N-6
November 2012
g) In the atmosphere, nitrogen oxides exist in many forms, including NO and NO2. Two
other forms are N2O and N2O4 (the dimer of NO2). Draw Lewis structures for both
N2O and N2O4. Examine your structures closely. If you can draw a second, valid,
Lewis structure, draw it underneath.
N2O structure
N2O4 structure
Second structure, if appropriate
Second structure, if appropriate
Marks
6
CHEM1101
2010-N-6
November 2010
• The structure of tetrahydrocannabinol, the active ingredient in marijuana, is shown
below.
a
OH
II
I
O
b
What is the molecular formula
of tetrahydrocannabinol?
C21H30O2
Name the functional groups indicated in the boxes a and b.
a
alkene
b
ether
What are the approximate bond angles at the carbon atoms labelled I and II?
Atom
Bond angle
I
109.5°
II
120°
Marks
5
CHEM1101
2005-J-5
June 2005
 Complete the table below showing the number of valence electrons, a Lewis structure
and the predicted shape of each of the following species.
Formula
Number of
valence electrons
e.g. H2O
8
Lewis structure
Name of molecular shape
O
Bent (angular)
H
H
Marks
5
H
H2CO
12
C
O
see week 7
Cl
see week 7
H
H
CH3Cl
14
H
C
H
Which, if either, of H2CO and CH3Cl will have a dipole moment?
both
 Using the following electronegativity data, decide which one or more of the oxides of
C, Te, Zn and Mg would be classified as containing ionic bonds. Briefly explain your
answer.
Element
Electronegativity
O
3.5
C
2.5
Te
2.1
Zn
1.4
Mg
1.2
An electronegativity (χ) difference > 2 is classified as ionic:
C: Δ χ = 3.5 – 2.5 = 1.0  not ionic
Te: Δ χ = 3.5 – 2.1 = 1.4  not ionic
Zn: Δ χ = 3.5 – 1.4 = 2.1  ionic
Mg: Δ χ = 3.5 – 1.2 = 2.3  ionic
2
CHEM1101
2005-N-6
November 2005
 Complete the table below showing the number of valence electrons, the Lewis
structure and the predicted shape of each of the following species.
Formula
Number of
valence electrons
e.g. NH3
8
Lewis structure
H N H
ClF5
42
F
Cl
Geometry of species
trigonal pyramidal
H
F
Marks
5
F
F
see week 7
F
NO2–
18
O
N
O
see week 7
Which of NH3, ClF5 and NO2– have a non-zero dipole moment?
see week 7
 What is the approximate value of the most intense wavelength emitted by the star
Proxima Centauri, which has a temperature of 2700 K?
The maximum intensity in a thermal spectrum is approximately at a transition
energy ΔE= 4.5kBT where kB is Boltzmann’s constant (1.381  10-23 J K-1 mol-1).
Using T = 2700 K, the transition energy can be calculated:
E  4.5k B T  4.5  (1.381 10 23 )  2700  1.68  10 19 J
To convert this energy into a wavelength, Planck’s relationship must be used:
hc (6.626 10 34 )(2.998 10 8 )


 1.18  10  6 m or 1180nm
 19
E
1.68  10
Answer: 1.18  106 m or 1180nm
2
CHEM1101
2004-J-5
June 2004
 Complete the table below showing the number of valence electrons, a Lewis structure
and the predicted shape of each of the following species.
Formula
Number of
valence electrons
e.g. H2O
8
Lewis structure
Name of molecular shape
O
Bent (angular)
H
H
Marks
5
Cl
SOCl2
26
S
O
see week 7
Cl
Cl
CCl4
32
Cl
C
Cl
see week 7
Cl
Which, if either, of SOCl2 and CCl4 will have a dipole moment?
see week 7
 Using the following electronegativity data, decide which one or more of the oxides of
C, Te, Zn and Mg would be classified as containing ionic bonds. Briefly explain your
answer.
Element
Electronegativity
O
3.5
C
2.5
Te
2.1
Zn
1.4
Mg
1.2
An electronegativity (χ) difference > 2 is classified as ionic:
C: Δ χ = 3.5 – 2.5 = 1.0  not ionic
Te: Δ χ = 3.5 – 2.1 = 1.4  not ionic
Zn: Δ χ = 3.5 – 1.4 = 2.1  ionic
Mg: Δ χ = 3.5 – 1.2 = 2.3  ionic
2
CHEM1101
2004-N-6
November 2004
 The structure of tetrahydrocannabinol may be drawn thus:
a
OH
#2
#1
O
b
Name the functional groups in the boxes a and b.
a
alkene
b
ether
What are the approximate bond angles at the atoms labelled #1 and #2?
C#1 ~109.5
C#2 ~120
The infrared spectrum of tetrahydrocannabinol shows strong absorption at 3600 cm–1.
Explain this in terms of the functional groups present in the molecule.
Strong absorption at 3600 cm-1 is due to excitation of the stretching vibration of
the O-H group present in tetrahydrocannabinol. This bond has a large dipole
moment and this leads to the strength of the absorption.
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY
Marks
4