Name : ____________________
40
Meridian Junior College
2007 JC 2 Chemical Bonding Time trial (Group A)
1
Time allocation : 12 minutes
Draw dot-and-cross diagram to show the electronic structure of the following species.
Draw and name the shape of the species. Indicate the polarity for molecules only.
Species
Dot-and-cross
Shape of Molecule
Polarity
(a) ClF5
Polar
[½]
[½]
[½]
Square pyramidal
(b)
C2O42-
NA
[½]
Trigonal planar about
each C atom
[½]
(c) ClF4NA
square planar
[½]
[½]
1
Species
Dot-and-cross
Shape of Molecule
Polarity
(d) NO3NA
Trigonal planar [½]
[½]
(e)
SbF52NA
square pyramidal [½]
[½]
(f) ClO2
Polar
[½]
[½]
[½]
bent
(g)
SO42NA
[½]
[½]
tetrahedral
[8]
2
HCJC Prelim : 10 mins
2
N2O can be made in the laboratory by heating molten ammonium nitrate. The reaction
is thought to occur by the following steps. ( NJC Prelim 2002 modified)
NH4NO3
NH3 + HNO3
2HNO3
NO2+ + H2O + NO3NH3 + NO2+ 
[H3N - NO2]+

+
H3O
+ NO3
 HNO3 + H2O
(a)
N2O + H3O+
Write a balanced equation for the overall equation.
Method : Add all the 4 equations together and cancel common intermediates
NH4NO3  N2O + 2H2O
(b)
[1]
Draw a displayed formulae showing the bonds and the bond angles in
(i)
NO2+
2 marks each:
(ii)
N2O
(iii)
[H3N - NO2]+
- 1 mark for
correct bonds
- 1 mark for
correct bond
angles
[7]
3
Time : 10 mins
3
The following lists the boiling points of fluorine and some fluoride compounds. By
reference to their chemical structures and types of bonding, explain as fully as you can
the differences in their boiling points.
Boiling point / oC
-50
50
100
Element / compound
Hydrogen sulphide
Hydrogen fluoride
Methanoic acid with apparent
Mr doubled
Aluminium oxide
Alanine
Format :
1
2
3
1257
150
State type of structure -- LOOK AT THE MP/BP and infer !
Identify bonds / intermolecular forces broken
State energy requirement
Al2O3 and Alanine

Aluminium oxide, Al2O3 has a giant ionic structure [1/2] while alanine
(H2NCH(CH3)COOH) is an amino acid and exists as a zwitterions. [1/2]

Boiling these 2 compounds involves breaking the strong electrostatic forces of
attraction between oppositely-charged ions [1]

A large amount of energy [½] is required, hence high melting points.
Methanoic acid and hydrogen fluoride

Methanoic acid (HCOOH), hydrogen sulphide and hydrogen fluoride all are simple
molecular [1/2] structures.

For methanoic acid hydrogen fluoride, boiling involves breaking the weaker (when
compared to ionic bond but stronger when compared to VDW’s )
intermolecular hydrogen bonds [½]

Methanoic acid exists as a dimer [½]and forms 2 hydrogen bonds per molecule
[½]while hydrogen fluoride can only form 1 hydrogen bond per molecule [1/2]

Extent of intermolecular hydrogen bonds: methanoic acid
fluoride [½]

Energy required to break intermolecular hydrogen bonds: methanoic acid >
hydrogen fluoride [½]
4
> hydrogen
H2S

For hydrogen sulphide, boiling involves the weak intermolecular Van der Waals’
forces, [1] hence require least amount of energy, [½] thus lowest boiling point.
[8]
RJC Prelim : 15 mins
4
Explain the following observations in terms of structure and bonding
(a)
There are two N2F2 with different boiling points
CONCEPT : N2F2 exists as 2 isomers; cis and trans isomers
[½]
[½]
Cis isomer
Trans isomer

Dipole moment cancels out hence
no net dipole moment

There is a net dipole moment

Trans isomer is non-polar [½]

Cis isomer is polar [½]

Less energy is needed to overcome
to weaker temporary dipole-dipole
attraction [½] between molecules

More energy is needed t overcome to
stronger permanent dipole-dipole
attraction [½] between molecules
Note : This is an example of a situation where you cannot state weak
VDW’s forces – but must explicitly compare strength of temporary
and permanent dipole attractions
5
(b)
Both beryllium difluoride and boron trifluoride can react with arsenic chloride but the
mole ratios are different.
Reaction of BeF2 with AsCl3
Reaction of BF3 with AsCl3
[½]
[½]
(Shape: tetrahedral about Be and As atom)
(Shape: tetrahedral about B and As atom)

BF3 needs to react with only 1 AsCl3
molecule in order for B to achieve
stable octet configuration [½]

Mole ratio: BF3  AsCl3 [1/2]
6

BeF2 needs to react with only 2 AsCl3
molecule in order for Be to achieve
stable octet configuration[½]

Mole ratio: BeF3  2AsCl3 [1/2]
(c)
I3- exists but not F3I3-

I is in period 5 hence it can expand beyond octet configuration [1/2] due to
availability of energetically accessible d orbitals to accommodate extra
electrons [1]
F3-
(d)

F is in period 2, hence it has no energetically accessibly d orbitals [½] are
available to accommodate more than 8 electrons

Hence F3- does not exist.
Liquid IF5 can conduct electricity
Concept: To conduct electricity, IF5 needs to exists as free mobile ions

IF5 undergoes self-ionisation [½] to form ions:
+
IF5

(e)
IF5
IF4 + + IF6 -
[½]
Can conduct electricity due to presence of free mobile ions, IF4 + and IF6 - . [1]
Chlorofluorocarbons can be used as aerosol propellants. Draw the displayed formulae
of the three isomers of C2H2FCl. Predict which isomer has the smallest overall dipole
moment . (JJC Prelim 2002)
Cl
H
H
C
C
C
H
F
Trans
A
H
[½]
F
C
Cl
H
C
F
[½]
C
Cl
H
[½]
Cis
B
C
 Structure A has overall smallest dipole [½]
[12]
7
Time allocation: 8 mins
5 Phosphine, PH3, is a poisonous gas which is used to kill insects and vermin in grain stores.
It can be made by treating aluminium phosphide, AlP, with dilute sulphuric acid.
(a)
Write a balanced equation for the action of dilute sulphuric acid on aluminium phosphide.
2AlP + 3H2SO4  2PH3 + Al2(SO4)3
[1]
(b)
Aluminium phosphide has a similar structure to diamond. Draw a section of the structure
of aluminium phosphide, indicating in the compound the approximate bond angles.
Structure of AlP
Al
P
[1/2 for shape
½ for correct bond angle]
(c)
Suggest, with reasoning, whether aluminium phosphide conducts electricity.
AlP does not conduct electricity [½] due to absence of delocalised electrons and
free mobile ions [½].
(d)
Phosphine is less soluble in water than ammonia. Explain the .difference in their
solubilities in water.

Hydrogen bonding is formed between NH3 and H2O molecules. [1]
Hence ammonia is soluble in water.

The hydrogen bonding between water molecules is stronger than the Van der
Waals forces of attraction between PH3 molecules [½] hence cannot displace the
hydrogen bonding [½] and is thus less soluble in water.

Hence, phosphine is less soluble in water than ammonia.
[5]
8
Additional Question
HCJC Prelim 2002 modified : 8 mins
6
Oxygen-oxygen bond lengths in some molecules are given below :
Oxygen
Ozone
Hydrogen peroxide
(a)
0.121 nm
0.128 nm
0.149 nm
Draw the shape and show the approximate bond angle in a hydrogen peroxide
molecule, oxygen molecule and ozone molecule.
[½]
Bent [½]
(b)
[½]
Bent about each O atom [½]
Trigonal planar [½]
Predict and explain the relative stability of these three molecules.
Concept: Stability of molecule  bond strength 
(c)
[½]
1
bond length

Oxygen-oxygen bond length in O2 < O3 < H2O2 [½]

Oxygen-oxygen bond strength in O2 > O3 > H2O2 [½]

Energy required to break oxygen-oxygen bond in O2 > O3 > H2O2 [½]

Relative stability of O2 > O3 > H2O2 [½]
Hydrogen peroxide and chlorine dioxide are often used as bleaches. Explain why
hydrogen peroxide is a liquid but chlorine dioxide is a gas at room temperature.
H2O2 and ClO2 have simple molecular structures. [½].
At room temperature,

there is insufficient energy [½] to overcome the strong hydrogen bonding
between H2O2 molecules [½] and hence H2O2 is a liquid.

there is sufficient energy to overcome the weaker Van der Waal’s forces between
ClO2 molecules [½] and hence ClO2 is a gas.
9