Structured Questions Answers 1 Elements Li Be B C N O F Ne ...

Structured Questions Answers
1
Elements Li Be B C N O F Ne
Strong metallic Strong covalent
bond between bond between
Nature of
Weak van der Waals’ forces between -
positive ions and nuclei and shared
bonding
molecules or atoms
delocalized electrons of the
electrons
bonded atoms
Structure Giant metallic Giant covalent Simple molecular
1
Each correct answer [ ] 2
2 (a) Periodicity is the recurrence of similar patterns at regular intervals [1] when elements are arranged
in increasing atomic number. [1]
(b) Across a period, the number of protons increases and new electrons enter the same shell. The
nuclear attraction to the new electrons increases. [1] The tendency for the outermost shell electrons
in elements to delocalize decreases, so metallic character decreases. [1]
(c) Down a group, the number of protons increases and the new electrons enter a new shell. The size
of atoms increases. [1] The tendency for the outermost shell electrons in elements to delocalize
increases, so metallic character increases. [1]
3 (a)
Melting point
Na Mg Al Si P S Cl Ar
Correct sketching [1]
Correct labelling of axes [1]
(b) The melting point increases from Group I to Group III. [1] From Group I to III, the number of
delocalized electrons increases from one to three. [1] Thus, the strength of metallic bonds
increases. [1] Energy required to break the metallic bonds increases. [1]
(c) Silicon. [1] It has a giant covalent structure. [1]
4 (a)

Melting point
Na Mg Al Si P S Cl Ar
(b)
Correct sketching [1]
Correct labelling [1]
Relative electrical
conductivity
Na Mg Al Si P S Cl Ar
(c)
Correct sketching [1]
Correct labelling [1]

Relative electrical
conductivity
Li Be B C N O F Ne
5
Correct sketching [1]
Correct labelling [1]
In Period 2, only Li and Be can conduct electricity. [1] There are three elements in Period 3 which
can conduct electricity. [1]
1
(a) Element with a giant covalent structure (i.e. Si) has the highest melting point in the period. [ ] 2
Before melting can occur, most of the covalent bonds have to be broken [ 2
1 ] and a large amount of
energy is needed to overcome the strong covalent bonds. [ 2
1 ]
For elements with giant metallic structures (i.e. Na, Mg and Al), the metallic bonds do not have to
be substantially broken to form a liquid. [ 2
1 ] Thus, the melting points of metals are not as high as
that of Si. [ 2
1 ]
In melting simple molecular substances (i.e. P, S, Cl and Ar), much less energy is needed to
overcome the weak intermolecular forces. [ 2
1 ] Therefore, they have much lower melting points.
[ 2
1 ]
(b) Metals in this period (i.e. Na, Mg and Al) contain delocalized electrons, [ 2
1 ] so they can conduct
electricity. [ 2
1 ] With the increasing number of valence electrons, [ 2
1 ] electrical conductivities
increase from Na to Al. [ 2
1 ]
For elements with a giant covalent structure (i.e. Si) [ 2
1 ] or a simple molecular structure (i.e. P, S,

Cl and Ar), [ 2
1 ] the absence of delocalized electrons makes them insulators of electricity. [ 2
1 ]
6 (a)
Melting point
Na
Mg
Al
Si
P S Cl Ar
Correct trend of melting points from Na to Si [1]
Correct trend of melting points from P to Ar [1]
(b) Carbon has a giant covalent structure. The atoms are held by strong covalent bonds. [1] In melting,
most of the strong covalent bonds have to be broken. [1] A large amount of energy is needed, [1]
so the melting point is very high.
7 (a) Graphite. [1]
There are delocalized electrons in each layer of graphite. [1]
(b) In diamond, each carbon atom is covalently bonded to four other atoms, forming a
three-dimensional giant network. [1] To break the structure, numerous strong covalent bonds
between carbon atoms must be broken. [1] This makes diamond extremely hard.
In graphite, each carbon atom is covalently bonded to three other atoms. The carbon atoms are
arranged in flat, parallel layers. [1] Only weak van der Waals’ forces exist between adjacent layers.
[1] This makes graphite crystal easy to cleave.
(c) Giant covalent structure [1]
8 (a)
Melting point
Li Be B C N O F Ne

(b)
Correct sketching [1]
Correct labelling [1]
Melting point
Na
Mg
Al
Si
P S Cl
Ar
Correct sketching [1]
Correct labelling [1]
(c) Each lithium and sodium atom contributes one delocalized electron into the ‘electron sea’. [1] Due
to the smaller size, lithium ion has a stronger attraction on the delocalized electrons than sodium.
[1] Therefore, the metallic bonds in lithium are stronger. Lithium has a higher melting point. [1]
(d) Fluorine and chlorine have simple molecular structures. [1] Chlorine has a larger molecular size
than fluorine, so chlorine has stronger van der Waals’ forces between molecules. [1] Therefore,
chlorine has a higher melting point. [1]
(e) Carbon and silicon have giant covalent structures. [1] Since a CC bond is stronger than a SiSi
bond, [1] more energy is needed to break the covalent bonds between carbon atoms. Therefore,
carbon has a higher melting point. [1]
9 (a)
Relative electrical
conductivity
Na
Mg
Al
Si
P S Cl Ar
Correct sketching [1]
Correct labelling [1]

(b) Electrical conductivity depends on the number of delocalized electrons per atom. [1] Since
magnesium has two outermost shell electrons while sodium has only one, [1] magnesium has a
higher electrical conductivity. [1]
10 (a) x = 6 [1]
(b) The one with atomic number of x + 4. [1]
(c) Si has a giant covalent structure [1] in which electrons are not free to move. [1]
(d) Al has a larger number of delocalized electrons than Mg. [1] Therefore, the electrical conductivity
of Al is higher.
11 (a) Period 2 [1]
(b) No. [1] All the three elements have simple molecular structures and they have similar molecular
sizes. [1]
(c) The student is incorrect. [1] Although phosphorus (P 4 ) also has a simple molecular structure, the
molecular size of phosphorus is significantly larger [1] than the above three elements. The van der
Waals’ forces between phosphorus molecules are much stronger, so the melting point of
phosphorus is much higher. [1]
12 (a) Na, Mg, Al: giant metallic [1]
Si: giant covalent [1]
P, S, Cl, Ar: simple molecular [1]
(b) (i) The melting point of a substance is the temperature at which it changes from a solid to a
liquid. [1]
(ii)
Melting point
Na
Mg
Al
Si
P S Cl Ar
Correct sketching [1] , Correct labelling [1]
(iii) Si has a giant covalent structure and it has the highest melting point across the period. In
melting, most of the covalent bonds have to be broken. [1] A large amount of energy is
needed to overcome the strong covalent bonds. [1]
From Na to Al, they have giant metallic structures and the metallic bonds do not have to be
substantially broken to form a liquid. [1] Thus, the melting points of the metals are lower than
Si. [1] Due to the increase in number of delocalized electrons, melting point increases from
Na to Al. [1]

13 (a)
Element Electronic
configuration
Physical state
at 25C
Melting point
(1-highest, 4-lowest)
Electrical
conductivity
W 2, 1 Solid 2 conductor
X 2, 3 Solid 1 depends on
temperature
Y 2, 8, 6 Solid 3 insulator
Z 2, 8, 8 Gas 4 insulator
Each correct answer [ 2
1 ]
(b) (i) W: Lithium [ 2
1 ]
X: Boron [ 2
1 ]
Y: Sulphur [ 2
1 ]
Z: Argon [ 2
1 ]
(ii) Metal: W [ 2
1 ]
1
Semi-metal: X [ ] 2
Non-metal: Y, Z [1]
1
(c) X is a semi-metal and it has a giant covalent structure. [ ] Atoms of X are held together by strong
2
covalent bonds. A large amount of energy is needed to overcome the strong covalent bonds before
1
melting can occur, [ ] so it has the highest melting point.
2
W is a metal. It has a giant metallic structure [ 2
1 ] and the atoms are held by strong metallic bonds.
1
However, the metallic bonds do not have to be substantially broken to form a liquid, [ ] so its 2
melting point is not as high as X.
For Y and Z, they have simple molecular structures. Much less energy is needed to overcome the
1
weak van der Waals’ forces, [ ] so they have lower melting points than X and W. Since Y has a
2
larger molecular size than Z, [ 2
1 ] Y has a higher melting point than Z.
(d) W is a metal, so it contains delocalized electrons to conduct electricity. [ 2
1 ]

X is a semi-metal. Its electrons are not free to move at room temperature, [ 2
1 ] so it is an insulator
at room temperature. However, at higher temperatures, it conducts electricity as the electrons are
free to move. [ 2
1 ]
(e) X:
Y and Z are non-metals. Their electrons are not free to move, [ 2
1 ] so they cannot conduct
electricity.
Y:
[1]
[1]
14 (a) Li: alkali metal [1]
Be: alkaline earth metal [1]
F: halogen [1]
Ne: noble gas/inert gas [1]
(b) (i) Carbon can exist as diamond and graphite. For diamond, it has a giant covalent structure and
all four outermost shell electrons in each carbon atom are used for formation of covalent
bonds. [1] There are no delocalized electrons in its structure, so diamond is an insulator of
electricity. [1] On the other hand, graphite also has a giant covalent structure. Three of the
outermost shell electrons in each carbon atom are used for formation of covalent bonds. [1]
There is one electron left which is free to move and conduct electricity. [1] Therefore,
graphite is an conductor of electricity.
(ii) Diamond:
Graphite:
[1]

[1]
(c) (i) Be has a larger number of delocalized electrons [1] than Li. The metallic bonds in Be are
stronger and more energy is needed to break the bonds, [1] so Be has a higher melting point
than Li.
(ii) F and Ne have simple molecular structures. [1] In melting the solid, much less energy is
needed to overcome the weak intermolecular forces. [1] Therefore, they have relatively lower
melting points.
15 (a) Both have giant metallic structures. [1] Beryllium has two outermost shell electrons while lithium
has only one. [1] Metallic bonds in beryllium are stronger than those in lithium. [1] Therefore,
beryllium has a higher melting point than lithium.
(b) Carbon has a giant covalent structure while nitrogen has a simple molecular structure. [1] The
covalent bonds between carbon atoms are much stronger than the weak van der Waals’ forces
between nitrogen molecules. [1] Much more energy is needed to break the strong covalent bonds
between carbon atoms, [1] so carbon has a higher melting point.
(c) Lithium has a giant metallic structure while fluorine has a simple molecular structure. [1] The
metallic bonds in lithium are much stronger than the weak van der Waals’ forces between fluorine
molecules. [1] Much more energy is needed to break the strong metallic bonds in lithium, [1] so
lithium has a higher melting point.
16 (a) Both of them have giant metallic structures. [1] They contain delocalized electrons to conduct
electricity. [1] Going from sodium to aluminium, the number of delocalized electrons increases. [1]
Therefore, the electrical conductivities also increase.
(b) Both have giant covalent structures. [1] Graphite has a layered structure in which each carbon
atom is bonded to three other atoms and delocalized electrons are present in each layer to conduct
electricity. [1] Diamond has a giant three-dimensional network in which each carbon atom is
bonded to four other atoms, so there are no delocalized electrons in diamond. [1]
(c) Boron has a giant covalent structure and it does not have delocalized electrons. [1] Aluminium has
a giant metallic structure and it contains delocalized electrons to conduct electricity. [1]
17 The student is partly correct. [1]
Metals contain delocalized electrons which are free to move [1]. Therefore, metals are conductors of
electricity.
For graphite, although it is a non-metal, [1] it contains delocalized electrons in each layer of its
structure. [1] These electrons are free to move along the layers. [1] Therefore, graphite is also an
electrical conductor.

18 (a) Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. [1]
(b) Na 2 O. [1] The difference in electronegative values between Na and O is larger than that between
Al and O. [1]
(c) Na 2 O 2 [1]
O.N. of O = 1 [1]
(d) 6Na(s) + 2O 2 (g) 2Na 2 O(s) + Na 2 O 2 (s) [1]
19
Oxide Na 2 O MgO Al 2 O 3 SiO 2 P 4 O 10 SO 2 Cl 2 O
Colour and
Colourless Orange
state at
White solid
gas gas
25C
Giant
Structure
Giant ionic
Simple molecular
covalent
Acid-base
property
Basic Amphoteric Acidic
Each correct blank [1]
20 (a) Across Period 3, the electronegativities of elements change from low values on the left to high
values on the right. [1]
(b) The difference in electonegativity values between the element and oxygen decreases across a
period, [1] so the bonding of the oxides becomes less ionic and more covalent. [1] Therefore, the
structure of the oxides changes from giant ionic to giant covalent, and then simple molecular. [1]
(c) Basic oxides react with water to form hydroxides and with dilute acids to form salts. [1] Example:
sodium oxide. [1]
Acidic oxides react with water to give acids and with dilute alkalis to form salts. [1] Example:
Sulphur dioxide. [1]
21 (a) X: Na [1]
Y: Al [1]
Z: Cl [1]
(b) Na 2 O reacts vigorously with water to form sodium hydroxide solution. [1]
Na 2 O(s) + H 2 O(l) 2NaOH(aq) [1]
Al 2 O 3 is insoluble in water. [1]
1
Cl 2 O reacts readily with water to form hypochlorous acid. [ ] Hypochlorous acid ionizes slightly
2
in water to form hydrogen ions and hypochlorite ions. [ 2
1 ]
Cl 2 O(g) + H 2 O(l) 2HOCl(aq) [ 2
1 ]
HOCl(aq) H + (aq) + OCl (aq) [ 2
1 ]

22 (a) Na 2 O, MgO, Al 2 O 3 , SiO 2 , P 4 O 10 , SO 2 [3]
(DO NOT accept Na 2 O 2 as one of the answers.)
(b) P 4 O 10 (s) + 6H 2 O(l) 4H 3 PO 4 (aq) [1]
(c) 4Al(s) + 3O 2 (g) 2Al 2 O 3 (s) [1]
(d) SiO 2 (s) + 2OH (aq) SiO 2 3 (aq) + H 2 O(l) [1]
(e) Cl 2 O(g) + H 2 O(l) 2HOCl(aq) [1]
23 (a) B. [1] It has the lowest melting point. [1]
(b) Giant covalent structure [1]
(c) SiO 2 [1]
(d) Sodium oxide [1]
Giant ionic structure [1]
24 (a) (i) Amphoteric oxide [1]
(ii) Al 2 O 3 (s) + 6HCl(aq) 2AlCl 3 (aq) + 3H 2 O(l) [1]
Al 2 O 3 (s) + 2KOH(aq) + 3H 2 O(l) 2KAl(OH) 4 (aq) [1]
(iii) BeO(s) + H 2 SO 4 (aq) BeSO 4 (aq) + H 2 O(l) [1]
BeO(s) + 2NaOH(aq) + 3H 2 O(l) Na 2 Be(OH) 4 [1]
(b) (i)
[1]
(ii) P 4 O 6 (s) + 6H 2 O(l) 4H 3 PO 3 (aq) [1]
P 4 O 10 (s) + 6H 2 O(l) 4H 3 PO 4 (aq) [1]
25 (a) Glass contains mainly silicon dioxide, [1] which is acidic and can react with sodium hydroxide. [1]
SiO 2 (s) + 2NaOH(aq) Na 2 SiO 3 (aq) + H 2 O(l) [1]
(b) Sodium silicate [1]
(c) Use dilute alkalis as titrant. [1]
Rinse the burette throughout after experiment. [1]
(d) Sodium hydroxide pellets absorb water from the air, [1] so it is not possible to accurately weigh
them. [1]
26 (a) (i) Sulphur dioxide:
[1]
Sulphur trioxide:
[1]
(ii) 2SO 2 (g) + O 2 (g) 2SO 3 (g) [1]
Vanadium(V) oxide/V 2 O 5 [1]
(iii) Colourless [1]

(b) (i) [1]
(ii) Orange [1]
27 (a) Na 2 O:
The compound is soluble in water. [1]
Na 2 O(s) + H 2 O(l) 2NaOH(aq) [1]
Al 2 O 3 :
The compound is insoluble in water. [1]
SiO 2 :
The compound is insoluble in water. [1]
(b) The solution made with Na 2 O is alkaline. [1]
The liquid obtained by mixing Al 2 O 3 with water is neutral. [1]
The liquid obtained by mixing SiO 2 with water is neutral. [1]
(c) Al 2 O 3 (s) + 6HNO 3 (aq) 2Al(NO 3 ) 3 (aq) + 3H 2 O(l) [1]
Al 2 O 3 (s) + 2NaOH(aq) + 3H 2 O(l) 2NaAl(OH) 4 (aq) [1]
(d) Amphoteric [1]
28 (a) Phosphorus pentoxide:
Sulphur dioxide:
[1]
[1]
(b) MgO: MgO(s) + H 2 O(l) Mg(OH) 2 (s) [1]
P 4 O 10 : P 4 O 10 (s) + 6H 2 O(l) 4H 3 PO 4 (aq) [1]
SO 2 : SO 2 (g) + H 2 O(l) H 2 SO 3 (aq) [1]
(c) P 4 O 10 [1]
29 (a)
Correct ratio of elements [1]

Correct arrangement of elements [1]
(b) Statement (1) is correct. [1] SiO 2 is acidic, so it will react with alkalis to form salts. [1]
Statement (2) is incorrect [1] since SiO 2 will not react with acids. [1]
30 (a) The universal indicator turns red. [1] P 4 O 10 reacts with water to form phosphoric acid. [1]
P 4 O 10 (s) + 6H 2 O(l) 4H 3 PO 4 (aq) [1]
(b) As a desiccant/dehydrating agent. [1]
(c) The student is incorrect. [1] An intact molecule of P 4 O 10 consists of 4 phosphorus atoms and 10
oxygen atoms. [1]
[1]
31 (a) (i) Silicon dioxide [1]
(ii) Covalent [1]
(b) (i) Orange [1]
(ii) This indicates that the solution is acidic. [1] Glass contains mainly silicon dioxide, which
react with the alkaline solution. The reaction mixture becomes acidic since some OH (aq)
ions are removed. [1]
(iii) Metal oxides [1]
(c) (i) This is because sodium hydroxide will react with silicon oxide to corrode the glass bottle. [1]
(ii) Plastic bottle [1]
32 (a) Metal oxides: B, D [1]
1
Semi-metal oxide: C [ ] 2
Non-metal oxide: A [ 2
1 ]
(b) Acidic oxides: A, C [1]
Amphoteric oxide: D [ 2
1 ]
Basic oxide: B [ 2
1 ]
(c) C: Silicon dioxide/SiO 2 [ 2
1 ]
D: Aluminium oxide/Al 2 O 3 [ 2
1 ]
For oxides of the elements across Period 3, only Al 2 O 3 and SiO 2 are insoluble in water. [1] Al 2 O 3
is amphoteric and it reacts with both dilute acids and alkalis. [1] SiO 2 is acidic and it reacts with
dilute alkalis. [1]

(d) A has the lowest melting point. [1] A is a non-metal oxide and it has a simple molecular structure.
[1] Its discrete molecules are held together by weak van der Waals’ forces only. In melting the
solid, small amount of energy is needed to overcome the weak intermolecular forces, [1] so it has a
relatively low melting point.
33 (a) The oxides changes from basic, through amphoteric, to acidic. [1]
(b) Al 2 O 3 [1]
Al 2 O 3 reacts with dilute acids to form salts.
Al 2 O 3 (s) + 3H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (aq) + 3H 2 O(l) [1]
Al 2 O 3 reacts with dilute alkalis to form aluminate ions.
Al 2 O 3 (s) + 2NaOH(aq) + 3H 2 O(l) 2NaAl(OH) 4 (aq) [1]
(c) SO 2 and Cl 2 O [2]
(d) (i) P 4 O 10 (s) + 6H 2 O(l) 4H 3 PO 4 (aq) [1]
(ii) Molecules of H 3 PO 4 are held together by hydrogen bonds. [1]
hydrogen bond
[1]
(iii) It is a tribasic acid [1] because 3 hydrogen atoms from the three hydroxyl groups are capable
of ionization. [1]/H 3 PO 4 (aq) 3H (aq) + PO 3 4 (aq) [1]
34 (a) X = Magnesium [1]
Y = Aluminium [1]
Z = Silicon [1]
(b) (i)
[1]
(ii) MgO(s) + H 2 O(l) Mg(OH) 2 (s) [1]
Mg(OH) 2 (s) + water Mg 2+ (aq) + 2OH (aq) [1]
(c) An amphoteric substance can act both as an acid and as a base. [1]
Al 2 O 3 (s) + 6H + (aq) 2Al 3+ (aq) + 3H 2 O(l) [1]
Al 2 O 3 (s) + 2OH (aq) + 3H 2 O(l) 2Al(OH) 4 (aq) [1]
(d) SiO 2 has a giant covalent structure. [1]
35 (a) Put the three solids into distilled water [1] and measure the pH of the solutions by pH meter. [1]
1
Aluminium oxide is insoluble in water. [ ] 2

pH of the solution does not change./pH = 7 [ 2
1 ]
Sodium oxide dissolves in water to give an alkaline solution. [ 2
1 ]
Na 2 O(s) + H 2 O(l) 2NaOH(aq) [ 2
1 ]
Phosphorus pentoxide dissolves in water to give an acidic solution. [ 2
1 ]
P 4 O 10 (s) + 6H 2 O(l) 4H 3 PO 4 (aq) [ 2
1 ]
(b) Aluminium oxide and sodium oxide have giant ionic structures. [ 2
1 ] Their ions are held together
by strong ionic bonds. [ 2
1 ]
Phosphorus pentoxide has a simple molecular structure. [ 2
1 ] The molecules are held together by
weak van der Waals’ forces. [ 2
1 ]
36 (a) Assume that there are 100 g of gallium oxide,
Element Ga O
Mass (g) 74.4 25.6
Number of 74.4
moles (mol) 69. 7
mol = 1.07 mol
25.6
16.0
mol = 1.60 mol [1]
Mole ratio 2 3 [1]
Empirical formula of gallium oxide is Ga 2 O 3 . [ 2
1 ]
Assume that there are 100 g of arsenic oxide,
Element As O
Mass (g) 65.2 34.8
Number of 65.2
moles (mol) 74. 9
mol = 0.870 mol
34.8
16.0
mol = 2.18 mol [1]
Mole ratio 2 5 [1]
Empirical formula of arsenic oxide is As 2 O 5 . [ 2
1 ]
(b) Gallium oxide has a giant ionic structure. [ 2
1 ] The ions are held together by strong ionic bonds.
[ 2
1 ]

Arsenic oxide has a simple molecular structure. [ 2
1 ] The molecules are held together by weak van
1
der Waals’ forces. [ ] 2
37 (a) (i) P 4 O 10 reacts with water to give phosphorus acid. [1]
P 4 O 10 (s) + 6H 2 O(l) 4H 3 PO 4 (aq) [1]
SO 2 reacts with water to give sulphurous acid. [1]
SO 2 (g) + H 2 O(l) H 2 SO 3 (aq) [1]
(ii) Silicon dioxide has a giant covalent structure, so it cannot dissolve in water. [1] But it reacts
with dilute alkalis to form salts. [1]
SiO 2 (s) + 2OH (aq) SiO 2 3 (aq) + H 2 O(l) [1]
1
(b) SiO 2 has a giant covalent structure. [ ] The atoms are held together by strong covalent bonds.
2
[ 2
1 ]
SO 2 has a simple molecular structure. [ 2
1 ] The molecules are held together by weak van der
1
Waals’ forces. [ ] 2
(c) (i) Dehydrating agent [1]
(ii) Manufacture of sulphuric acid/Preservative [1]
38 (a) A substance is described as ‘amphoteric’ if it can act both as an acid and as a base. [1]
(b) BeO/GeO 2 /SnO 2 /PbO 2 (Any ONE) [1]
(c) (i) 2Al(OH) 3 (s) + 3H 2 SO 4 (aq) Al 2 (SO 4 ) 3 (aq) + 6H 2 O(l) [1]
(ii) Al(OH) 3 (s) + NaOH(aq) NaAl(OH) 4 (aq) [1]
(d) Electrolysis of molten aluminium oxide [1]
39 (a) Generally, the outermost shells of the atoms of transition metals have either one or two electrons.
[1] Due to this similar outer electron arrangement, [1] transition metals across a period have
similar chemical properties.
(b) Sodium atoms only have one outermost shell electron that contributes to the ‘sea of electrons’. [1]
On the other hand, electrons in both the outermost shell and the next inner shell of copper atoms
contribute to the ‘sea of electrons’. [1] Therefore, the strength of metallic bonds in copper is much
stronger than that in sodium. [1] This makes copper much harder than sodium.
(c) Iron(III) ions [1]
(d) White [1]
40 (a) Green [1]
(b) (i) 2V 3+ (aq) + Zn(s) 2V 2+ (aq) + Zn 2+ (aq) [1]
(ii) The solution changes from green to violet. [1]
(c) (i) Vanadium(V) oxide/V 2 O 5 [1]
(ii) The oxidation state of vanadium in V 2 O 5 is +5. [1]

(iii) Contact Process [1]
41 (a) Fe 2+ and Fe 3+ . [1]
(b) Fe 2+ [1]
(c) The reaction mixture changes from green to yellow. [1]
(d) 2Fe 2+ (aq) + Cl 2 (aq) 2Fe 3+ (aq) + 2Cl (aq) [1]
42 (a) (i) Cr 3+ : +3 [1]
Cr 2 O 2 7 : +6 [1]
(ii) Cr 3+ : green [1]
Cr 2 O 2 7 : orange [1]
(b) (i) Cr 2 O 2 7 (aq) + 14H + (aq) + 6e 2Cr 3+ (aq) + 7H 2 O(l) [1]
(ii) The solution changes from orange to green at the end point. [1]
(iii) K 2 Cr 2 O 7 (aq) + 4H 2 SO 4 (aq) + 3H 2 S(g)
Cr 2 (SO 4 ) 3 (aq) + K 2 SO 4 (aq) + 7H 2 O(l) + 3S(s) [1]
K 2 Cr 2 O 7 (aq) + H 2 SO 4 (aq) + 3SO 2 (g)
Cr 2 (SO 4 ) 3 (aq) + K 2 SO 4 (aq) + H 2 O(l) [1]
(iv) Iodide ion is oxidized to iodine which is brown in aqueous solution. [1] This makes the end
point difficult to be detected. [1]
43 (a) It is because a variable number of electrons can be removed from the outermost shell and the next
inner shell of the atoms of transition metals. [1]
(b) (i) Copper and iron [2]
(ii) Copper: Cu + ion and Cu 2+ ion [2]
Iron: Fe 2+ ion and Fe 3+ ion [2]
(c) Scandium and zinc [2]
44 (a) The orange dichromate ions (Cr 2 O 2 7 (aq)) are reduced by ethanol to give chromium(III) ions
(Cr 3+ (aq)) which are green in colour. [2]
(b) The oxidation number of chromium in Cr 3+ (aq) is +3 [1] and that in Cr 2 O 2 7 (aq) is +6. [1]
45 (a) The oxidation number of vanadium in VO + 2 is +5. [1]
The oxidation number of vanadium in VO 2+ is +4. [1]
(b) V 2+ (aq) ion is violet [1] and V 3+ (aq) ion is green. [1]
(c) (i) 2SO 2 (g) + O 2 (g) 2SO 3 (g) [1]
(ii) The oxidation number of vanadium in V 2 O 5 is +5. [1]
(d) (i)
1
Stage 2: 2VO 2 (s) + O2 (g) V 2 O 5 (s) [1]
2
(ii) V 2 O 5 is considered to be a catalyst because it is regenerated at the end of the reactions. [1]
46 (a) Decomposition of hydrogen peroxide [1]
(b) (i) MnO 2 (s) + 4HCl(aq) MnCl 2 (aq) + 2H 2 O(l) + Cl 2 (g) [1]
(ii) The colour of the resultant solution is very pale pink [1] due to the presence of Mn 2+ (aq) ions.
[1]
(c) (i) Purple [1]
(ii) MnO 4 (aq) + 8H + (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + 4H 2 O(l) + 5Fe 3+ (aq) [1]

47 (a) N 2 (g) + 3H 2 (g) 2NH 3 (g) [1]
Finely divided iron [1]
(b) 2SO 2 (g) + O 2 (g) 2SO 3 (g) [1]
Vanadium(V) oxide/V 2 O 5 [1]
(c) 2NO(g) + 2CO(g) N 2 (g) + 2CO 2 (g) [1]
Platinum/Pt and rhodium/Rh [1]
(d) 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) [1]
Manganese(IV) oxide/MnO 2 [1]
(e)
[1]
Titanium(IV) chloride/TiCl 4 [1]
48 (a) A catalyst alters the rate of a chemical reaction [1] without being used up in the process. [1]
(b) (i) Vanadium(IV) oxide/VO 2 [1]
(ii) It oxidizes X (VO 2 ) to V 2 O 5 , [1] so the catalyst is regenerated and is available for further
catalysis. [1]
(c) A catalyst provides an alternative pathway [1] with lower energy barrier/activation energy, [1] so
more reactant particles can possess enough energy to have effective collisions. [1]
49 (a) A: Fe 2+ [1]
B: Fe 3+ [1]
(b) Fe 2+ , I and SO 2 4 . [3]
(c) Firstly, S 2 O 2 8 ions are reduced by A.
2Fe 2+ (aq) + S 2 O 2 8 (aq) 2Fe 3+ (aq) + 2SO 2 4 (aq) [1]
Secondly, I ions are oxidized by B.
2Fe 3+ (aq) + I (aq) 2Fe 2+ (aq) + I 2 (aq) [1]
50 (a) (i) A variable number of electrons [1] can be removed from the outermost shell and the next
inner shell of the atoms of transition metals. [1]
(ii) Electrons in both outermost shell and the next inner shell [1] of most transition metal atoms
contribute to the ‘sea of electrons’. [1]
(b) (i) Copper [1]
(ii) Iron [1]
(iii) Zinc [1]
(iv) Titanium [1]
51 (a) (i) 4Fe(s) + 2nH 2 O(l) + 3O 2 (g) 2Fe 2 O 3 • nH 2 O(s) [1]
(ii) From 0 to +3 [1]
(b) (i) The solution changes from colourless to green. [1]
Gas bubbles are evolved. [1]
(ii) Fe(s) + 2HCl(aq) FeCl 2 (aq) + H 2 (g) [1]
(iii) It is a redox reaction. [1]
The oxidation number of iron changes from 0 to +2. The oxidation number of hydrogen

changes from +1 to 0. [1]
(c) (i) A catalyst alters the rate of a chemical reaction [1] without being used up in the process. [1]
(ii) N 2 (g) + 3H 2 (g) 2NH 3 (g) [1]
(d) (i) It oxidizes I to I 2 . [1]
(ii) Iron(II) ions reduce peroxodisulphate ions to sulphate ions. [1] Then the iron(III) ions formed
are reduced to iron(II) ions by iodide ions. [1]
(iii) 2I (aq) + S 2 O 2 8 (aq) I 2 (aq) + 2SO 2 4 (aq) [1]
52 (a) The oxidation state of manganese in MnSO 4 is +2. [1]
The oxidation state of manganese in MnO 2 is +4. [1]
The oxidation state of manganese in KMnO 4 is +7. [1]
(b) (i) Manganese(IV) oxide [1]
(ii) Decomposition of hydrogen peroxide [1]
53 (a) Fe(s) + 2HCl(aq) FeCl 2 (aq) + H 2 (g) [1]
2Fe(s) + 3Cl 2 (g) 2FeCl 3 (aq) [1]
(b) Adding sodium hydroxide solution to the test tubes containing iron(II) chloride solution and
iron(III) chloride solution respectively. [1]
A green precipitate can be observed in the test tube containing iron(II) chloride solution. [1]
A yellow/brown precipitate can be observed in the test tube containing iron(III) chloride solution.
[1]
(c) Iodide ions reduce iron(III) ions to give iron(II) ions as the intermediate. Iodide ions are oxidized
to iodine as one of the products. [1]
2Fe 3+ (aq) + 2I (aq) 2Fe 2+ (aq) + I 2 (aq) [1]
Iron(II) ions reduce peroxodisulphate ions to give sulphate ions. Iron(II) ions are oxidized to
iron(III) ions. [1]
2Fe 2+ (aq) + S 2 O 2 8 (aq) 2Fe 3+ (aq) + 2SO 2 4 (aq) [1]
54 (a) 2I (aq) + S 2 O 2 8 (aq) I 2 (aq) + 2SO 2 4 (aq) [1]
(b) Firstly, iodide ions reduce iron(III) ions to give iron(II) ions as the intermediate. Iodide ions are
oxidized to iodine as one of the products. [1]
2Fe 3+ (aq) + I (aq) 2Fe 2+ (aq) + I 2 (aq) [1]
Secondly, iron(II) ions reduce peroxodisulphate ions to give sulphate ions. Iron(II) ions are
oxidized to iron(III) ions. Thus, iron(III) ions are regenerated. [1]
2Fe 2+ (aq) + S 2 O 2 8 (aq) 2Fe 3+ (aq) + 2SO 2 4 (aq) [1]
(c) It is essential for the synthesis of haemoglobin. [1]
55 (a) (i) Displacement reaction [1]
(ii) Zn(s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu(s) [1]
(b) Zinc ribbons dissolve. [1]
Reddish brown solid is formed. [1]
Blue colour of the solution fades. [1]
(c) Zinc: making dry batteries/production of brass/as anti-corrosion coating for iron (Any ONE) [1]
Copper: making electrical cables and wirings/making water taps and pipes/production of alloys

(Any ONE) [1]
(Accept other reasonable answers.)
56 (a) Platinum [1] and rhodium. [1]
(b) 2NO(g) + 2CO(g) N 2 (g) + 2CO 2 (g) [1]
(c) Iron is essential for the synthesis of haemoglobin in human body. [1]
(d) High mechanical strength/low density/resistant to corrosion/can withstand extreme temperatures
(Any TWO) [2]