Name:
CG:
Date:
SERANGOON JUNIOR COLLEGE
H2 9746 H1 8872 CHEMISTRY
2010 JC1
Chemical Bonding
Tutorial 5
Compulsory Questions
The compulsory questions are divided into three parts. Part A contains questions from TYS
and will be discussed during tutorial. Questions in Part B are for self-practice. Questions in
Part C are for submission.
Part A
[To be discussed during tutorial. Do all questions on foolscap paper. Show working for
MCQ.]
Comparison between the basic types of bonding
Qn 1 Electric cable, as used in houses, is made of copper wire surrounded by
poly(chloroethene), also called polyvinylchloride or PVC.
(a) (i) Describe the bonding in copper and explain how it conducts electricity.
 definition of metallic bonds
Describe:
Give definition
of bonds,
not just the name
Copper has strong electrostatic forces of attractions between a
giant metallic lattice structure of cations/Cu2+ and sea of electrons.
Copper contains mobile electrons to conduct electricity.
(ii) Describe the bonding in PVC and hence explain why it acts as an
insulator.
 definition of covalent bonds
PVC is polyvinyl chloride, a polymer with molecular structure with
strong electrostatic forces of attraction between shared pair of
electrons and nuclei
PVC does not contain mobile ions or electrons to conduct electricity.
9647 H2 8872 H1Chemistry
Page 1 of 31
(b)
However, electric cable used in fire alarm systems has copper wire
surrounded by magnesium oxide which acts as an insulator. The whole
cable is encased in thin copper tubing.
(i)
Describe the bonding in magnesium oxide and explain why it acts
as an insulator.
 definition of ionic bonds
Strong electrostatic forces of attraction between oppositely
charged Mg2+ and O2- ions.
Ions can only vibrate about fixed position and hence are not mobile
to conduct electricity.
(ii)
Suggest two reasons why magnesium oxide is preferred to PVC
as an insulator.
 MgO will not burn to produce toxic gases
 MgO has higher melting point to withstand high temperature in
event of fire.
(iii)
Suggest a reason why copper is suitable for encasing the
magnesium oxide.
Copper is
 High tensile strength provides support for brittle MgO
(because MgO is giant ionic)
 Or Resistant to corrosion
Note: Tensile strength is the resistance of a material to tear apart,
measured as the maximum tension the material can withstand.
9647 H2 8872 H1Chemistry
Page 2 of 31
Qn 2 (a) Boron trifluoride, BF3 and aluminium fluoride, AlF3 differ markedly in their
physical properties.
Compound
Melting point / °C
BF3
-144
AlF3
1291
Deduce the type of bonding present in each of these compounds and draw
dot and cross diagrams to illustrate this bonding.
Deduce/Predict: need to give reason with inference from the given data
Analysis: Formula suggested AlF3 and BF3 should have similar type of structure.
But, melting point differ actual type of structure  actual type of bonding
Answer to include: structure and type of intramolecular bonding
Low melting point indicates BF3 exist as simple molecular structure with
covalent bonds between B and F atoms
High melting point indicates AlF3 exist as giant ionic structure with
ionic bonds between oppositely charged Al3+ and F- ions
F B F
F
3+
Al
a)
3
Al forms ionic compounds
with F, O and N;
F
but covalent compounds with
other non-metals, eg. Cl, Br, I
Boron trifluoride forms a compound with ammonia. Describe the type of bond
that is formed during this reaction. Draw a diagram to illustrate the shape and
bonding in the product.
N atom in NH3 donates the lone pair of electrons to B with vacant
orbital/incomplete octet forming a co-ordinate (or dative covalent) bond.
9647 H2 8872 H1Chemistry
Page 3 of 31
Shapes and Polarity
Qn 3 Draw the dot-and-cross diagram and Lewis structure (if applicable) for each
substance and hence explain the following observations:
(a)
(a)
CO2 and ICl2- are both linear, but have different number of electron pairs
CO2
OCO
X
X
X
X
O C O
 There are 2 bond pairs and 0 lone pair around C
 To minimise repulsion, the 2 electron pairs are directed to opposite
sides of each other
 CO2 is linear
Cl
I
Cl
x
x
Cl
x
x
x
x
x
ICl2-
-
I
Lone pair should
be represented by
dots
Cl
 There are 2 bond pairs and 3 lone pairs around I
 To minimise repulsion, the 5 electron pairs are directed to the corners of
a trigonal bipyramid
 Since lone pair - lone pair repulsion > lone pair – bond pair
repulsion > bond pair-bond pair repulsion
 ICl2- is also linear
Features of linear shape

2 bond pairs as indicated by the 2 neighbouring atoms

number of lone pairs is either 0 or 3.
9647 H2 8872 H1Chemistry
Page 4 of 31
Qn 3 Draw the dot-and-cross diagram and Lewis structure for each substance and
hence explain the following observations:
(b)
SO2 and SCl2 are both bent, but show different bond angles.
(b) SO2
OSO
X
X
XX
X
X
O
S
O
 There are 2 bond pairs and 1 lone pair around S
 To minimise repulsion, the 3 electron pairs are directed to the corners of
an equilateral triangle
 Since lone pair - bond pair repulsion > bond pair-bond pair repulsion
 SO2 is bent with bond angle of <1200C.
SCl2
x
x
x
Cl
x x
S
S
x
x
x
x
Cl
x
x
x
x
x
Cl
Cl
 There are 2 bond pairs and 2 lone pairs around S
 To minimise repulsion, the 4 electron pairs are directed to the corners of
a tetrahedron
 Since lone pair – lone pair repulsion > lone pair - bond pair
repulsion > bond pair-bond pair repulsion
 SCl2 is bent with bond angle of 104.50C
Features of bent or non-linear shape

2 bond pairs as indicated by the 2 neighbouring atoms

number of lone pairs is either 1 or 2

bond angle: 2 lone pairs (104.50) < 1 lone pair (<1200)
due to additional lone pair-lone pair repulsion
9647 H2 8872 H1Chemistry
Page 5 of 31
Qn 3 Draw the dot-and-cross diagram and Lewis structure for each substance and
hence explain the following observations:
(c)
(b)
NO3- and SO3 are both trigonal planar, but show different bond types.
Charge of -1  add 1 electron to more electronegative O
such that 1 O atom has 7electrons form a single bond
NO3-
-
xx
x
xO
x
x
x
O
N
x
xx
O
x
x
x
xx
-
O
x
x
O
N
O
SO3
xx
x
x
O
x
x
x
xO x
xx
S
O
x
x
x
S
x
x
x Ox
xx
O
O
 Both have 3 bond pairs and 0 lone pair around N and S respectively
 To minimise repulsion, the 3 electron pairs are directed to the corners
of a trigonal pyramid
 Both are trigonal planar
 N, being an element in period 2, has no d orbitals in valence shell
(n=2) to accommodate electrons and cannot expand beyond octet
structure. Hence, dative bond is formed instead of double bond.
Period 3 onwards elements
Period 2 elements
expand beyond octet (>8 electrons)
cannot expand beyond octet
double bonds are preferred
dative bonds as last resort
(lone pair from central atom to neighbour)
9647 H2 8872 H1Chemistry
Page 6 of 31
Qn 3 Draw the dot-and-cross diagram and Lewis structure for each substance and
hence explain the following observations:
(d)
CH4 and NH3 have the same number of electron pairs around the central atom,
but show different bond angles.
(d) Both have 4 electron pairs directed to the corners of a tetrahedron
CH4
H
H C H
H
H
x
x
x
x
H
C
H
H

There are 4 bond pairs and 0 lone pair around C

CH4 is tetrahedral with bond angle of 109.5
NH3
HN H
H
x
x
x

H
N
H
H
There are 3 bond pairs and 1 lone pair around N
 Since lone pair - bond pair repulsion > bond pair-bond pair repulsion,
 NH3 is trigonal pyramidal in shape with bond angle compressed to 107
Bond angle decreases with increasing number of lone pairs
9647 H2 8872 H1Chemistry
Page 7 of 31
Qn 3 Draw the dot-and-cross diagram and Lewis structure for each substance and
hence explain the following observations:
(e) The bond angle in water increases slightly when the ion, H3O+, is formed
(e)
H2O
O
OH
H
x
x
H
H

There are 2 bond pairs and 2 lone pairs around O

To minimise repulsion, the 4 electron pairs are directed to corners of a
tetrahedron

Since lone pair-lone pair repulsion > lone pair-bond pair repulsion >
bond pair-bond pair repulsion,
H2O is bent with bond angle of 104.50

Charge of +1  subtract 1 electron from less electronegative H
form a dative bond (lone pair of O donates to H+)
H3O+
+
H OH
H
x
x
H
O
+
H H

O donates 1 lone pair to H+ with vacant orbital forming a dative covalent
bond such that there are now 3 bond pairs and 1 lone pair around O

Since lone pair-bond pair repulsion >
bond pair-bond pair
repulsion,
H3O+ is trigonal pyramidal with bond angle of 1070

2
structural requirements for dative bond:
(a) lone pair of electrons; (b) atom with incomplete octet/duplet for H
 Bond
angle increases by 2.50 when 1 lone pair is changed to 1 bond pair
9647 H2 8872 H1Chemistry
Page 8 of 31
Qn 3 Draw the dot-and-cross diagram for each substance and hence explain the
following observations:
H
Cl
Cl
C C
(f)
H
Cl
Cl
C C
H is polar.
is non-polar, but H
(f)
H
x
C
x
x
x
x xx
x
Cl x
x x
Cx
H
Cl
Cl x
C
x
H
x xx
x
Cl xx
x
x
x Cx
H
Both isomers have 2 polar C-Cl bonds with no lone pair around each C
 The C-Cl bonds in 1st (trans form) isomer are opposite each other across
the C=C bond such that dipole moments associated with the polar bonds
cancel out exactly
H
Cl
C C

H
Cl
is non-polar
 The C-Cl bonds in 2nd (cis form) isomer are on the same side of C=C
bond such that dipole moments associated with the polar bonds do not
cancel out exactly
Cl
Cl
C C

H
H is polar
Factors affecting polarity of molecule:
 polar bond,
δ+
X-Yδ- and lone pair(s)
 orientation of polar bond(s) in molecule ~ idea of “tug of war”/“symmetrical shape”
9647 H2 8872 H1Chemistry
Page 9 of 31
Comparison between the types of structure/bonding
Qn 4 Complete the table below for the following substances in terms of structure
and bonding.
Substance
Chemical structure
E.g S8
Simple molecular
(a) CH3CH2OH
(e) CH3Cl
Substance
CH3CH2OH
Chemical bond to be overcome during
melting / boiling
(b) SiC
(f) SiCl4
Van der Waals’ forces of attraction
between S8 molecules
(c) CaSO4
(g) CH3NH2
(d) Ca
(h) Ne
Chemical
Chemical bond to be overcome during
structure
melting / boiling
simple molecular
weak intermolecular hydrogen bonds
giant molecular
strong covalent bonds
(alcohol)
SiC
between Si and C atoms
CaSO4
giant ionic lattice
strong ionic bonds between
oppositely charged ions (ie cations & anions)
Ca
giant metallic lattice
strong metallic bonds between
cations and sea of electrons
CH3Cl
simple molecular
weak Van der Waals’ forces of attraction
due to permanent dipoles (polar molecules)
SiCl4
simple molecular
weak Van der Waals’ forces of attraction
due to temporary dipoles (non-polar
molecules)
CH3NH2
simple molecular
weak intermolecular hydrogen bonds
simple molecular
weak Van der Waals’ forces of attraction
(amine)
Ne
(noble gas)
9647 H2 8872 H1Chemistry
between atoms
Page 10 of 31
Qn 5 In 1886, Henri Moissan succeeded in obtaining fluorine by the electrolysis of
molten potassium hydrogen-difluoride, KHF2, which is an ionic compound
containing one cation and one anion.
(i)
Write the formula of the ions present in KHF2.
Cation: K+
Anion: [HF2]-
(ii)
Suggest a structure for the anion and state what types of bonding occur
within it.
For HF2- :
H: 1 electron (central atom)
1 F: 7 electrons, forms single bond with H (covalent bonding)
1F: 8 electrons, does not form dative bond with H (H does not have
empty orbital to accept)  forms hydrogen bond with H
covalent bond
F HF
Label:
- δ+ on H and δ- on 2 F
- lone pair of electrons on F
hydrogen bond
9647 H2 8872 H1Chemistry
Page 11 of 31
Comparison of the various physical properties
Melting / Boiling Point
Qn 6 Explain the following observations:
(a)
To include:
- Type of structure
- Relative strength of
intermolecular forces
of attraction
- Amount of energy
Carbon dioxide is a gas whereas silicon dioxide (IV) oxide is a solid of high
melting point.

CO2 has a simple molecular structure.

Smaller amount of energy is needed to overcome the weaker
intermolecular Van der Waals’ forces of attractions.

Hence, CO2 has a low boiling point and is a gas.

SiO2 has a giant molecular structure.

Larger amount of energy is needed to overcome the stronger covalent
bonds between Si and O atoms.

(b)
Hence, SiO2 is a solid with high melting point.
The boiling points of the halogens show a trend.
Element
boiling point / °C
Cl2
-35
Br2
+59
I2
+184

Halogens have simple molecular structures

No of electrons of halogens increases down the group.

Larger amount of energy is required to overcome the more extensive
or stronger
intermolecular Van der Waals’ forces of attraction down the group

Hence, boiling point increases down the group.
9647 H2 8872 H1Chemistry
Page 12 of 31
(c)
Butane, CH3CH2CH2CH3, has a higher boiling point than 2-methylpropane,
CH(CH3)3.

Both butane and 2-methylpropane have simple molecular structures
with the same number of electrons.

Butane is a unbranched (or more elongated) with greater surface area
than 2-methylpropane which is branched (or more spherical)

More energy required to overcome the more extensive intermolecular
Van der Waals’ forces of attraction in butane.
avoid using “stronger’ as
same number of electrons

Hence butane (CH3CH2CH2CH3) has a higher boiling point than
2-methylpropane (CH(CH3)3)
(d)
butane has a lower boiling point than butanol (CH3CH2CH2CH2OH)

Both butane and butanol have simple molecular structures.

Less energy is required to overcome the weaker intermolecular Van
der Waals’ forces of attraction in butane than the stronger
intermolecular hydrogen bonds in butanol.

Hence butane has lower boiling point than butanol.
9647 H2 8872 H1Chemistry
Page 13 of 31
Solubility
Qn 7 The solubilities in water of the three gases are given in the table below:
gas
Solubility in
water / mol dm-3
NH3
18
HCl
23
CO2
0.033
Each of these gases is soluble in water because it interacts with the solvent.
(i)
(ii)
Write equations for any chemical reactions that occur.
NH3 (g) + H2O (l)
NH4+ + OH- (aq)
CO2 (g) + H2O (l)
H2CO3 (aq)
HCl (g) + H2O (l) →
H3O+ + Cl- (aq)
Expected to include H2O as reactant
Suggest reasons for the much higher solubilities of ammonia and
hydrogen chloride, compared to that of carbon dioxide.
 NH3 can form favourable intermolecular hydrogen bonds with water
molecules
 HCl dissociates in water to produce ions, hence the ions forms
favourable ion-dipole interaction with water.
Hence, both ammonia and hydrogen chloride are water-soluble.
 Weaker intermolecular Van der Waals’ forces of attractions in CO2
are not able to displace the stronger intermolecular hydrogen
bonds in water for hydration.
Hence carbon dioxide is less soluble in water.
9647 H2 8872 H1Chemistry
Page 14 of 31
Electrical Conductivity and others
Qn 8
(a) Graphite conducts electricity but diamond does not.
 Graphite is a good conductor of electricity parallel to the layers as nonbonding valence electrons of the carbon atoms are mobile along layers to
conduct electricity.

In diamond, the electrons are localised in the covalent bonds and not
mobile to conduct electricity.
(b)

Potassium is good electrical conductor at any states while potassium chloride
conducts in molten state or aqueous solution, not solid state.
In potassium, presence of mobile electrons to conduct electricity.

In potassium chloride, ions in solid state can only vibrate about fixed
positions and hence are not mobile to conduct electricity, whereas ions in
molten or aqueous are mobile to conduct electricity.
(c)

Copper is ductile while copper(II) sulfate is brittle.
In copper, stress applied on a metallic lattice causes sliding of layers of
cations without breaking the metallic structure as the sea of electrons are
still holding the cations together. Hence, copper is ductile.

In CuSO4, stress applied on the ionic lattice with regular pattern allows causes
sliding of layers resulting in ions of similar charges coming together.
The resultant repulsion shatters the ionic structure. Hence, CuSO4 is brittle.
(d)


Ethanoic acid, C2H4O2 in the gas phase just above its boiling point has an
apparent Mr of 120.
Ethanoic acid (CH3CO2H, Mr of 60.0) has an apparent Mr of 120 which
doubles the expected value.
Each acid molecule pairs up to form a cyclic (closed ring) dimer via
intermolecular hydrogen bonds.

 
O
Labels:
- δ+ on H and δ- on 2 O atoms
- lone pair of electrons on O
9647 H2 8872 H1Chemistry
CH3
C
H O
C CH3
O H
 
O

Page 15 of 31
Combined physical properties
Qn 9
Lead, lead(II) chloride and lead(IV) chloride have melting points of 327oC, 498oC
and -15oC respectively.
(a)
Describe the bonding of these three substances and explain how it accounts
for the above melting points.

PbCl2 has a giant ionic lattice structure with strong electrostatic forces of
attraction between oppositely charged ions.
Largest amount of energy is required to overcome the strong ionic bonds.
Hence, PbCl2 has the highest melting point.

Pb has a giant metallic lattice structure with strong electrostatic force of
attraction between cations and sea of electrons..
Large amount of energy is required to overcome the strong metallic bonds.
Hence, Pb has a high melting point.

PbCl4 has a simple molecular structure with weak intermolecular Van der
Waals’ forces of attraction.
Smallest amount of energy is required to overcome the weaker Van der
Waals’ forces. Hence, PbCl4 has the lowest melting point.
9647 H2 8872 H1Chemistry
Page 16 of 31
(b)
State and explain with reasoning the relative solubility of PbCl2 and PbCl4 in
polar solvents such as water.
 PbCl2 can form favourable ion-dipole interactions with water molecules that
results in the release of energy to break the giant ionic lattice structure for
hydration to occur.
Hence, PbCl2 is more soluble in water.
 No favourable interactions between PbCl4 and water molecules can be
formed as the weak intermolecular Van der Waals’ forces of attraction in
PbCl4 are not able to displace the stronger intermolecular hydrogen
bonds of water.
Hence, PbCl4 is insoluble in water.
(c)
State and explain the relative electrical conductivity of these chlorides
PbCl2
PbCl4
 Good conductor in molten/aqueous  Non-electrical conductor
state but a non-conductor in solid
state
 In molten/aqueous state, ions are
mobile to conduct electricity
 In solid state, ions can only vibrate
about fixed positions and not
 Electrons are localised
in covalent bonds and
not mobile to conduct
electricity
mobile
9647 H2 8872 H1Chemistry
Page 17 of 31
Integrated Question
Qn 10 Modified TJC Prelim 2008/1/2a,b
(a)
Chemical
Bonding
Some data on three nitrogen-containing compounds are given in the table
below:
Compound
Molecular
Boiling Point / 0C
Formula
dinitrogen pentoxide
N2O5
Decomposes
nitric acid
HNO3
83
nitrosyl chloride
NOCl
-6.4
(i) Draw Lewis structures to illustrate the shapes of N2O5 (a symmetrical
molecule), HNO3 (with N and O as centres) and NOCl. Indicate the relevant
bond angles in each case.
O
N
N
O
O
O
0
N
O
O
N104.50N
O
N
H
O
120
O
O
O
O
0
120
0
104.5
O
Cl
O
O
H
O
N
N
1200
O
trigonal planar (about each N) / bent (about central O)
O
Cl
<120
0
bent (about N)
(ii) Nitrosyl chloride is a yellow gas most commonly encountered as a
decomposition product of aqua regia, a mixture of hydrochloric and nitric acid
Nitric acid, also known as aqua fortis, is a common bench reagent used in
chemical laboratories.
Explain the difference in boiling points of nitric acid and nitrosyl chloride in
terms of structure and bonding.
Both nitric acid and nitrosyl chloride have simple molecular structures.
More energy is required to overcome the stronger intermolecular hydrogen
bonds in nitric acid than the weaker intermolecular Van Der Waals’ forces
of attractions in nitrosyl chloride
Hence nitric acid has a higher boiling point than nitrosyl chloride.
9647 H2 8872 H1Chemistry
Page 18 of 31
(b) The dissociation of nitrosyl chloride into nitric oxide and chlorine is an
endothermic process which takes place according to the equation:
2NOCl (g)
∆H > 0
2NO (g) + Cl2 (g)
In an evacuated 20 dm3 vessel at 400 K, 0.5 mol of NOCl is injected and the
equilibrium pressure is 101 kPa.
(i) Calculate the total number of moles of gas at equilibrium, assuming the
gases behave ideally.
(i)
PV = nRT
n=
Ideal gas
equation
PV
101  10 3  20  10 3
=
RT
8.31  400
n = 0.06077 = 0.608 mol (3 sf)
(i) Hence calculate the percentage of the nitrosyl chloride that has dissociated.
(ii)
Let x be equilibrium amount of Cl2 in mol.
2NOCl (g)
Initial /mol
2NO (g) + Cl2(g)
0.5
0
Change /mol
- 2x
+ 2x
+x
Equilibrium/mol
0.5 – 2x
+ 2x
+ x
Chemical
Equilibria
0
Total equilibrium number of mol = 0.5 – 2x + 2x + x = 0.6077
x = 0.1077
% of NOCl dissociated =
9647 H2 8872 H1Chemistry
2(0.1077)
100 % = 43.1 %
0.5
Page 19 of 31
Part B
[Drilling questions. These questions will not be discussed in class, please check ASPIRE II
for more information
Qn 1
Describe the structure & type of bonding in potassium, calcium oxide and hydrogen
iodide. Draw dot-and-cross diagrams to illustrate the arrangement of valence
electrons in calcium oxide and hydrogen iodide.
Potassium has a giant metallic lattice structure with strong electrostatic forces
of attraction between cations (K+ ) and sea of electrons.
Calcium oxide has a giant ionic lattice structure with strong electrostatic forces
of attraction between the oppositely charged (Ca2+ and O2-) ions.
2+
Ca
2xx
Ox
x
xx
Hl has a simple molecular structure with stronger electrostatic forces of
attraction between the nuclei and the shared pair of electron and
weaker intermolecular van der waals’ forces
xx
H xxxI xx
9647 H2 8872 H1Chemistry
Page 20 of 31
Qn 2
Complete the table below for the following molecules / ions.
Dot-andcross
diagram
Molecule
/ion
Central atom
Lewis structure
Neutral substance
Shape/
Bond angle
Polarity
(polar/nonpolar)
Cations
Anions
BH2+
AlCl4-
By Group No.
(a) Gp II/III
BeCl2
MgO
Mg3N2
(b) Gp IV: C
C2H6
C2H4
CO
CH2F+
CNO-
(c) Gp V: N
NO2
N2O4
N2H4
NO2+
NO2-
(d) Gp V
AsBr3
SbF5
POCl2+
PCl6-
(e) Gp VI: O
O3
HOBr
H3O2+
OH-
(f) Gp VI
H2SO3 H2SO4 (2 centres)
SF3+
SO32-
SO42-
(g) Gp VII
ClFO2 BrFO3 IF3O
IF4+
ClF4-
ClO3-
(h) Gp 0
XeF4
XeOF4
XeF3+
XeO2F2
-
(a) Group II/III
Molecule
/ion
BeCl2
Dot-and-cross
diagram
Lewis
structure
Cl Be Cl
xx
xx
x
x
x
x
Cl
Be
Cl
x
x
xx
xx
MgO
2+
Polarity
Non-polar
Note: Group II/III metals form
2xx
Ox
x
xx
Mg
Shape / Bond
Angle
Linear / 1800
predominantly ionic bonds with the top
most electronegative atoms: F, O and N
2+
Mg3N2
3
Mg
3xx
2
Nx
xx
+
BH2+
x B
H
AlCl4-
H
H
xx
x
x
x Cl x
x
xx
x xx
x x Al Cl x
x
xCl
x x x x xx
x Cl x
x xx x
9647 H2 8872 H1Chemistry
+
-
B
Cl
Al
Cl
__
Tetrahedral/
109.5°
__
H
-
Cl
Linear / 1800
Cl
Page 21 of 31
(b) Group IV: C
Molecule
/ion
CO
C2H6
C2H4
CH2F+
Dot-and-cross
diagram
x
O
xCx
x
Lewis
structure
x
xC
O
Hx Hx
xC x H
HxC
x
x
H H
H
H
x
x
x
x
xN
x
H
H
H
H
xx
x
Non-polar
Trigonal planar/
1200
+
C
H
C Ox
Non-polar
H
F
-
Polar
Trigonal planar
(each C) / 120°
H
C C
+
Fx
HxC H
CNO-
C C
Polarity
Tetrahedral
(about each C) /
109.5°
H
H
Hx Hx
H x C xx C x H
Shape /
Bond Angle
Linear / 1800
H
-
__
Linear /
C O
N
__
1800
xx
(c) Group V: N
Molecule
/ion
NO2
Dot-and-cross
diagram
x
N
O
x
x
x
x
x
xx
N2O4
N2H4
Lewis structure
N
O
xx
Oxx
O
O
O
O
xx
xx
x
O x N x N xx O xx
x
x
xO x
xx
N N
O
O
xx
HNNH
HH
x
x
x
x
x
NO2+
ON O
x
x
x
x
NO2-
xx
ON O
x
H
+
N
H
N
H
O N O
-
O
x
x
H
-
+
Shape/
Bond angle
Slightly bent /
<1200
Polarity
Trigonal planar
(about each N) /
1200
Non-polar
Trigonal
pyramidal
(each N) / 1070
Polar
Linear / 1800
__
Bent / <1200
__
Slightly
polar
N
O
9647 H2 8872 H1Chemistry
Page 22 of 31
(d) Group V
Molecule
/ion
AsBr3
Dot-and-cross
diagram
x
x
xx
Lewis structure
xx
x
x
Br x As xBr
xx
xx
x
x
x
As
Br
Br
Br
x
Br x
xx
SbF5
xx
x xx
x
x x x
x
x
x x
x
x
xx
F
F
F
Sbx
x
x x x
x
xFx x F x
xx
xx
x
x
xx
POCl2+
x
x
xx
x
x
x
xx x x xx
x
x
x
xx
xx
Cl P O
Cl
F
Cl
Sb
F
-
P
Cl
P
Polarity
Trigonal
bipyramidal /
1200 (equatorial)
& 900 (axial)
Non-polar
Trigonal planar/
120°
Polar
__
Cl
-
Cl
Cl
x xCl
x
x Px
x Cl
+
O
Cl
Cl
F
F
+
PCl6Cl
F
Shape/
Bond angle
Trigonal
pyramidal / 1070
Cl
__
Octahedral /
900
Cl
Cl
Cl
Molecule
/ion
HOBr
Dot-and-cross
diagram
Lewis structure
O3
O xxO xx
xx
x
x
xO x
(e) Group VI: O
O
O
xx
O
Shape/
Bond angle
Bent /
< 104.5°
Bent /
< 120°
OHOx H
9647 H2 8872 H1Chemistry
-
O H
Polar
Polar
1st O: Bent/104.50
2nd O:Trigonal
pyramidal/
1070
H3O2+
Polarity
__
__
Linear /
1800
Page 23 of 31
(f) Group VI:
Molecule
/ion
H2SO3
Dot-and-cross
diagram
x S x xx
x
x
x
Oxx x x O
x
x Ox
H xx H
xx
H2SO4
Lewis structure
Shape/
Bond angle
Polarity
Polar
H
S: Trigonal
pyramidal /1070
Each O: bent /
104.50
Polar
O
S: Tetrahedral /
1090
Each O: bent /
104.50
Trigonal
pyramidal / 1070
__
Trigonal
pyramidal / 1070
__
2-
Tetrahedral /
109.5°
__
Lewis structure
Shape/
Bond angle
Trigonal
pyramidal /
1070
Polarity
Tetrahedral /
109.5°
Polar
See saw /
<1200 (equatorial)
& 900 (axial)
Polar
S
O O O
H
O
O
xx
O
S
xS
x
x
x
Ox
O H
H
O
x
O
H
H
SF3+
SO32-
2--
SO42-
2-
S
x S x xx
x
xx x
xx x
Oxx xx x O
x
O
x
xx
O O O
2-
O
O
xx
O
xS
x
x
x
S
O
O
O
O
O
(g) Group VII
Molecule
/ion
ClFO2
Dot-and-cross
diagram
BrFO3
IF3O
O
I
F
F F
9647 H2 8872 H1Chemistry
Polar
Page 24 of 31
Molecule
/ion
IF4+
Dot-and-cross
diagram
Lewis structure
+
F
F
I
Shape/
Bond angle
Polarity
See saw /
<1200 (equatorial)
& 900 (axial)
__
Square planar /
90o
__
Trigonal
pyramidal /
107°
__
F F
ClF4x
x xx
xx x
F
-
x
xx x
x
x x
F
F
x Cl x x
x xx
x
xFxx
xFxx
x
x
ClO3x
x
xx
xx
x
x
x x x x x xx
x
x
x
x
O Cl O
O
F
Cl
F
-
F
-
-
Cl
O
O
O
(h) Group 0
Molecule
/ion
XeF4
Dot-and-cross
diagram
x
x
xx
F
xx x
xx
x
x
xF
xx
xx
Xe
XeOF4
x
x
xF
xx
x
xx
F xx
xx
Lewis structure
F
F
Xe
F
F
F
F
XeO2F2
O
F
Xe
O
F
F
Xe
O
XeF3+
F
+
F
Xe
Shape/
Bond angle
Square
planar /
90o
Polarity
Nonpolar
Square
pyramidal /
90o
Polar
See saw /
<1200
(equatorial)
& 900 (axial)
Polar
T-shaped /
90o
__
F
F
9647 H2 8872 H1Chemistry
Page 25 of 31
Qn 3 N2006//III/2a
Boron forms simple trihalides of formula BX3 with all the halogens. BF3 and BCl3 are
the most common. Both find uses as Friedel-Crafts catalysts since they readily react
with electron pair donors.
(i)
Describe and explain the shape of the BF3 molecule.
There are 3 bond pairs and 0 lone pairs around B.
To minimise repulsion, the 3 electron pairs are directed to the corners of an
equilateral triangle.
Hence BF3 is trigonal planar
(ii)
BF3 and trimethylamine, (CH3)3N, react in a 1:1 ratio to give a white crystalline
solid. Draw a diagram to show the bonding within a molecule of this solid,
explaining the type of bonding involved.
[5]
N atom in N(CH3)3 donates the lone pair of electrons to B with vacant
orbital/incomplete octet forming a co-ordinate (or dative covalent) bond.
F
H3C
H3C
H3C
F
N B
F
Qn 4 Cyanamide, NH2CN, is used for organic synthesis in a stable commercial form
and it is formed from the acidification of calcium cyanamide, CaNCN.
Cyanamide is also capable of forming an addition product with boron
trifluoride, BF3.
(a)
Write down an equation representing the formation of cyanamide from
calcium cyanamide.
CaNCN + 2H+
NH2CN + Ca2+
(b)
Comment on the difference in melting points of cyanamide and calcium
cyanamide with respect to their structures and bonding.
NH2CN is a simple molecular structure and
CaNCN has a giant ionic lattice structure.
Smaller amount of energy is required to overcome the weaker
intermolecular hydrogen bonds in NH2CN than the stronger ionic
bonds in CaNCN.
Hence melting point of NH2CN is lower than CaNCN.
(c)
State the likely shapes of NH2CN and BF3, hence draw a diagram to
illustrate the likely shape of the addition product.
H
N
H
F
C
N
+
F
B
F
NC
H
H
F
F
N B
F
NH2CN is trigonal pyramidal, BF3 is trigonal planar
Addition product: tetrahedral with respect to each N and B centre
9647 H2 8872 H1Chemistry
Page 26 of 31
Qn 5 N2001/II/2a
A newly discovered source of frozen fuel is “methane ice”, also called ‘methane
hydrate’. This is methane trapped in ice about 500 m to 2000 m below the ocean
surface. Deposits have been detected off Norway, North Carolina and in the Pacific
Ocean off Japan, Indonesia and New Zealand.
The ice in methane hydrate has a more open structure than ordinary ice and
contains spaces large enough to contain methane molecules.
(i)
Draw a diagram of a water molecule and explain why its bond angle is about
105o
O
H
H
There are 2 bond pairs and 2 lone pairs around O. To minimise repulsion,
the 4 electron pairs are directed to the corners of a tetrahedron.
Since lone pair-lone pair repulsion > lone pair-bond pair repulsion > bond
pair-bond pair repulsion, bond angle is compressed to 104.50 (≈105°)
(ii)
The diagram below shows part of the structure of ordinary ice.
Explain why the bond angles in ice are 109o.
In ice, each water molecule forms two hydrogen bonds with its neighbouring
water molecules such that each oxygen atom is surrounded by four hydrogen
bonded atoms.
There are a total of 4 bond pairs and 0 lone pairs around O,
hence shape is tetrahedral about O and bond angle is 109°.
9647 H2 8872 H1Chemistry
Page 27 of 31
Qn 6 Nitrogen and boron combine to form boron nitride, with empirical formula BN,
which has a graphite-like structure.
(a)
By considering the electron distribution of your above structure, suggest with
reasoning the likely electrical conductivity of boron nitride.
Boron nitride:

non-conductor of electricity when perpendicular to layers: electrons are
localised in the covalent bonds and not mobile to conduct electricity

non-conductor of electricity when parallel to layers. Lone pairs of
electrons on N atoms are held closely to the highly electronegative N
atom and are not mobile to conduct electricity.
(b)
Suggest one probable industrial application of boron nitride.

It could be used as a machinery lubricant.
Qn 7 NaN3 hydrolyses slowly in water to form hydrazoic acid, HN3.
(a)
Draw the dot and cross diagram for ionic compound, NaN3.
-
+
Na
(b)
N N N
Explain why NaN3 has low solubility in hexane, an organic solvent in
terms of chemical structure and bonding.

NaN3 has a giant ionic lattice structure with strong ionic bonds.

Hexane has a simple molecular structure with weak intermolecular
Van der Waals’ forces of attraction.

Hence, no favourable ion-dipole interactions between ions and nonpolar hexane molecules can be formed to break down the giant ionic
lattice structure.
9647 H2 8872 H1Chemistry
Page 28 of 31
Qn 8 A student is given oxides of elements W, X, Y, Z and their properties are
given below:
Formula
Melting point / oC
Appearance at
Conductivity
of oxide
r.t.p.
WO2
1700
White solid
None
X2O
0
Colourless liquid
Poor
YO
2850
White solid
Good in molten state
ZO2
-73
Colourless gas
None
She was told that the four samples could be water, silicon dioxide, carbon dioxide or
magnesium oxide.
(a) Identify the four oxides.
Formula of oxide
Identity
WO2
Silicon dioxide
X2O
Water
YO
Magnesium oxide
ZO2
Carbon dioxide
(b) Explain your reasoning for the identification of X2O, based on the physical
properties given (i.e. its melting point and conductivity in molten state).
X2O Simple molecular structure

Smaller amount of energy is needed to break the relatively weaker
intermolecular hydrogen bonds, hence low melting point.

Poor electrical conductor because electrons are localised in the
covalent bonds and not mobile to conduct electricity
(c) Explain why the melting point of YO is higher than ZO2.
YO
Giant ionic lattice structure
Larger amount of energy is needed to break the stronger ionic
bonds
ZO2
Simple molecular structure
Small amount of energy is needed to break the weaker
intermolecular Van Der Waals’ forces of attraction
(d) Explain why YO can conduct electricity in molten state but not WO2.
YO
Good electrical conductor in molten state: ions are mobile to conduct
electricity
WO2 Non electrical conductor: electrons are localised in the covalent
bonds in the giant molecular structure,
hence, not mobile to conduct electricity
9647 H2 8872 H1Chemistry
Page 29 of 31
Part C
These questions are for submission and answer will be posted on ASPIRE.
Deadline for submission: ________
Qn 1 Total 10 marks (Standard A level)
Ethene, C2H4, and hydrazine, N2H4, are hydrides of elements which are adjacent in
the Periodic Table. Data about ethene and hydrazine are given in the table below.
C2H4
N2H4
o
melting point/ C
-169
+2
o
boiling point/ C
-104
+114
solubility in water
Insoluble
high
solubility in ethanol
high
high
(a) Ethene and hydrazine have a similar arrangement of atoms but differently
shaped molecules.
(i) What is the H-C-H bond angle in ethene? Bond angle: 1200 [1]
(ii) Draw a ‘dot-and-cross’ diagram for hydrazine.
xx
HNNH
H H [1]
x
x
x
x
x
(iii) What is the H-N-H bond angle in hydrazine? Bond angle: 1070 [1]
(iv) State and explain whether hydrazine is polar or non-polar.
N-H bond is polar and there is 1 lone pair around each N
N2H4 is trigonal pyramidal about each N such that the dipole moments
[5]
associated with the polar bonds and lone pair do not cancel out exactly. [1]
N2H4 is polar [1]
(b) The melting and boiling points of hydrazine are much higher than those of
ethene. Suggest reasons for these differences in terms of the intermolecular
forces each compound possesses.
[2]
Both have simple molecular structures.
More energy is required to overcome the stronger intermolecular hydrogen
bonding in hydrazine than the weaker intermolecular Van der Waals’ forces
of attraction in ethene. Hence higher melting & boiling of hydrazine than ethene.
(c) Explain, with the aid of a diagram showing lone pairs of electrons and dipoles,
why hydrazine is very soluble in ethanol.
[3]
1m show/state hydrogen bond
1m draw N2H4 and C2H5OH
1m show lone pairs and partial charges
9647 H2 8872 H1Chemistry

O 
C2 H 5 H

N H 
H
N
H

H
O
hydrogen

bonds
C2 H5 H
Page 30 of 31
Qn 2 Total 10 marks (More Challenging)
(a)
Cyanogen, a highly toxic gas can be represented by the formula, (CN)2. When
subjected to extremely high pressure, a non-conducting solid is formed.
Research done suggested that the electrical conductivity of the solid could be
increased by inserting certain atoms into its structure for example, caesium.
(i)
Draw the dot and cross diagram of a cyanogen molecule.
[1]
N CC N
[1 or 0]
(ii)
Determine with reasoning the shape of the cyanogen molecule.
[2]
There are 2 bond pairs and 0 lone pairs around C .
To minimise repulsion, the 2 electron pairs are directed to opposite
sides of each other [1]
Cyanogen is linear [1].
(iii)
(b)
Explain how the introduction of caesium enhances the electrical
conductivity of cyanogen.
[1]
Caesium has a giant metallic lattice structure with mobile electrons
to enhance the electrical conductivity of cyanogen. [1]
Cyanide poisoning occurs when a living organism is exposed to cyanide ions.
Common life threatening chemicals include cyanide salts such as potassium
cyanide.
(i)
Draw the dot and cross diagram for cyanide ion. Given that cyanide ion
has a linear shape, predict its Lewis structure.
[2]
-
C
N
[1 or 0]
(ii)
-
C
N
[1 ecf]
Predict and explain how you would expect the boiling point of
potassium cyanide to compare with hydrogen fluoride.
[4]
Potassium cyanide has a giant ionic lattice structure while
Hydrogen fluoride has a simple molecular structure. [1]
Larger amount of energy is required to overcome the stronger ionic
bonds [1] in KCN than the weaker intermolecular hydrogen bonds
in HCN [1].
Potassium cyanide will have a higher [1] boiling point compared to
hydrogen fluoride.
Live Chemistry, Dream Chemistr, Breathe Chemistry, Eat Chemistry, Love Chemistry
9647 H2 8872 H1Chemistry
Page 31 of 31