Periodic Table SQ Ans

7/23/2019 Periodic Table SQ Ans

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Structured Questions Answers

1

Elements Li Be B C N O F Ne

Nature of

bonding

Strong metallic

bond between

positive ions and

delocalized

electrons

Strong covalent

bond between

nuclei and shared

electrons of the

bonded atoms

Weak van der Waals’ forces between -molecules or atoms

Structure Giant metallic Giant covalent Simple molecular

Each correct answer [2

1]

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)

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)

M e l t i n g p o i n t

Na Mg Al Si P S Cl Ar




Correct labelling [1]

(b)



(c)



R e l a t i v

e e l e c t r i c a l

c o n d u c

t i v i t y






In Period 2, only Li and Be can conduct electricity. [1] There are three elements in Period 3 which

can conduct electricity. [1]

5(a) Element with a giant covalent structure (i.e. Si) has the highest melting point in the period. [

2

1]

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,

Li Be B C N O F Ne

R e l a t i v e e l e c t r i c a l

c o n d u c t i v i t y



Cl and Ar), [2

1] the absence of delocalized electrons makes them insulators of electricity. [

2

1]

6 (a)

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 athree-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)

P S Cl ArSiMg Al

Na



Li Be B C N O F Ne





(b)



(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)



ArP S

ClSiMg Al Na


R e l a t i v e e l e c t r i c a l

c o n d u c t i v i t y

P S Cl ArSiMg Al Na



(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 (P4) 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)

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]

P S Cl ArSiMg Al Na




13 (a)

Element Electronic

configuration

Physical state

at 25

C

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]

Semi-metal: X [2

1]

Non-metal: Y, Z [1]

(c) X is a semi-metal and it has a giant covalent structure. [2

1] Atoms of X are held together by strong

covalent bonds. A large amount of energy is needed to overcome the strong covalent bonds before

melting can occur, [2

1] so it has the highest melting point.

W is a metal. It has a giant metallic structure [

2

1] and the atoms are held by strong metallic bonds.

However, the metallic bonds do not have to be substantially broken to form a liquid, [2

1] so its

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

weak van der Waals’ forces, [2

1] so they have lower melting points than X and W . Since Y has a

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]

Y and Z are non-metals. Their electrons are not free to move, [2

1] so they cannot conduct

electricity.

(e) X :

[1]

Y :

[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:

[1]

Graphite:



[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] Themetallic 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 conductelectricity. [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) Na2O. [1] The difference in electronegative values between Na and O is larger than that between

Al and O. [1]

(c) Na2O2 [1]

O.N. of O = 1 [1]

(d) 6Na(s) + 2O2(g) 2Na2O(s) + Na2O2(s) [1]

19

Oxide Na2O MgO Al2O3 SiO2 P4O10 SO2 Cl2O

Colour and

state at

25

C

White solidColourless

gas

Orange

gas

Structure Giant ionicGiant

covalentSimple molecular

Acid-base

propertyBasic 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) Na2O reacts vigorously with water to form sodium hydroxide solution. [1]

Na2O(s) + H2O(l) 2NaOH(aq) [1]

Al2O3 is insoluble in water. [1]

Cl2O reacts readily with water to form hypochlorous acid. [2

1] Hypochlorous acid ionizes slightly

in water to form hydrogen ions and hypochlorite ions. [2

1]

Cl2O(g) + H2O(l) 2HOCl(aq) [2

1]

HOCl(aq) H+(aq) + OCl(aq) [2

1]



22 (a) Na2O, MgO, Al2O3, SiO2, P4O10, SO2 [3]

(DO NOT accept Na2O2 as one of the answers.)

(b) P4O10(s) + 6H2O(l) 4H3PO4(aq) [1]

(c) 4Al(s) + 3O2(g) 2Al2O3(s) [1]

(d) SiO2(s) + 2OH

(aq) SiO32(aq) + H2O(l) [1]

(e) Cl2O(g) + H2O(l) 2HOCl(aq) [1]

23 (a) B. [1] It has the lowest melting point. [1]

(b) Giant covalent structure [1]

(c) SiO2 [1]

(d) Sodium oxide [1]

Giant ionic structure [1]

24 (a) (i) Amphoteric oxide [1]

(ii) Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l) [1]

Al2O3(s) + 2KOH(aq) + 3H2O(l) 2KAl(OH)4(aq) [1]

(iii) BeO(s) + H2SO4(aq) BeSO4(aq) + H2O(l) [1]

BeO(s) + 2NaOH(aq) + 3H2O(l) Na2Be(OH)4 [1]

(b) (i)

[1]

(ii) P4O6(s) + 6H2O(l) 4H3PO3(aq) [1]P4O10(s) + 6H2O(l) 4H3PO4(aq) [1]

25 (a) Glass contains mainly silicon dioxide, [1] which is acidic and can react with sodium hydroxide. [1]

SiO2(s) + 2NaOH(aq) Na2SiO3(aq) + H2O(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) 2SO2(g) + O2(g) 2SO3(g) [1]

Vanadium(V) oxide/V2O5 [1]

(iii) Colourless [1]



(b) (i) [1]

(ii) Orange [1]

27 (a) Na2O:

The compound is soluble in water. [1]

Na2O(s) + H2O(l) 2NaOH(aq) [1]

Al2O3:

The compound is insoluble in water. [1]

SiO2:

The compound is insoluble in water. [1]

(b) The solution made with Na2O is alkaline. [1]

The liquid obtained by mixing Al2O3 with water is neutral. [1]

The liquid obtained by mixing SiO2 with water is neutral. [1]

(c) Al2O3(s) + 6HNO3(aq) 2Al(NO3)3(aq) + 3H2O(l) [1]

Al2O3(s) + 2NaOH(aq) + 3H2O(l) 2NaAl(OH)4(aq) [1]

(d) Amphoteric [1]

28 (a) Phosphorus pentoxide:

[1]

Sulphur dioxide:

[1]

(b) MgO: MgO(s) + H2O(l) Mg(OH)2(s) [1]

P4O10: P4O10(s) + 6H2O(l) 4H3PO4(aq) [1]

SO2: SO2(g) + H2O(l) H2SO3(aq) [1]

(c) P4O10 [1]29 (a)

Correct ratio of elements [1]



Correct arrangement of elements [1]

(b) Statement (1) is correct. [1] SiO2 is acidic, so it will react with alkalis to form salts. [1]

Statement (2) is incorrect [1] since SiO2 will not react with acids. [1]

30 (a) The universal indicator turns red. [1] P4O10 reacts with water to form phosphoric acid. [1]

P4O10(s) + 6H2O(l) 4H3PO4(aq) [1]

(b) As a desiccant/dehydrating agent. [1]

(c) The student is incorrect. [1] An intact molecule of P4O10 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]

Semi-metal oxide: C [2

1]

Non-metal oxide: A [2

1]

(b) Acidic oxides: A, C [1]

Amphoteric oxide: D [21 ]

Basic oxide: B [2

1]

(c) C : Silicon dioxide/SiO2 [2

1]

D: Aluminium oxide/Al2O3 [2

1]

For oxides of the elements across Period 3, only Al2O3 and SiO2 are insoluble in water. [1] Al 2O3

is amphoteric and it reacts with both dilute acids and alkalis. [1] SiO2 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) Al2O3 [1]

Al2O3 reacts with dilute acids to form salts.

Al2O3(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2O(l) [1]

Al2O3 reacts with dilute alkalis to form aluminate ions.

Al2O3(s) + 2NaOH(aq) + 3H2O(l) 2NaAl(OH)4(aq) [1]

(c) SO2 and Cl2O [2]

(d) (i) P4O10(s) + 6H2O(l) 4H3PO4(aq) [1]

(ii) Molecules of H3PO4 are held together by hydrogen bonds. [1]

[1]

(iii) It is a tribasic acid [1] because 3 hydrogen atoms from the three hydroxyl groups are capable

of ionization. [1]/H3PO4(aq) 3H(aq) + PO43(aq) [1]

34 (a) X = Magnesium [1]

Y = Aluminium [1]

Z = Silicon [1]

(b) (i)

[1]

(ii) MgO(s) + H2O(l) Mg(OH)2(s) [1]

Mg(OH)2(s) + water Mg2+(aq) + 2OH

(aq) [1]

(c) An amphoteric substance can act both as an acid and as a base. [1]

Al2O3(s) + 6H+(aq) 2Al3+(aq) + 3H2O(l) [1]

Al2O3(s) + 2OH(aq) + 3H2O(l) 2Al(OH)4(aq) [1]

(d) SiO2 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]

Aluminium oxide is insoluble in water. [

2

1]

hydrogen bond



pH of the solution does not change./pH = 7 [2

1]

Sodium oxide dissolves in water to give an alkaline solution. [2

1]

Na2O(s) + H2O(l) 2NaOH(aq) [ 2

1

]

Phosphorus pentoxide dissolves in water to give an acidic solution. [2

1]

P4O10(s) + 6H2O(l) 4H3PO4(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

moles (mol) 7.69

4.74 mol = 1.07 mol0.16

6.25 mol = 1.60 mol [1]

Mole ratio 2 3 [1]

Empirical formula of gallium oxide is Ga2O3. [2

1]

Assume that there are 100 g of arsenic oxide,

Element As O

Mass (g) 65.2 34.8

Number of

moles (mol) 9.742.65 mol = 0.870 mol

0.168.34 mol = 2.18 mol [1]

Mole ratio 2 5 [1]

Empirical formula of arsenic oxide is As2O5. [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

der Waals’ forces. [2

1]

37 (a) (i) P4O10 reacts with water to give phosphorus acid. [1]

P4O10(s) + 6H2O(l) 4H3PO4(aq) [1]

SO2 reacts with water to give sulphurous acid. [1]

SO2(g) + H2O(l) H2SO3(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]

SiO2(s) + 2OH

(aq) SiO32(aq) + H2O(l) [1]

(b) SiO2 has a giant covalent structure. [2

1] The atoms are held together by strong covalent bonds.

[2

1]

SO2 has a simple molecular structure. [2

1] The molecules are held together by weak van der

Waals’ forces. [2

1]

(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/GeO2/SnO2/PbO2 (Any ONE) [1]

(c) (i) 2Al(OH)3(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 6H2O(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) 2V3+(aq) + Zn(s) 2V2+(aq) + Zn2+(aq) [1]

(ii) The solution changes from green to violet. [1]

(c) (i) Vanadium(V) oxide/V2O5 [1]

(ii) The oxidation state of vanadium in V2O5 is +5. [1]



(iii) Contact Process [1]

41 (a) Fe2+ and Fe3+. [1]

(b) Fe2+ [1]

(c) The reaction mixture changes from green to yellow. [1]

(d) 2Fe2+(aq) + Cl2(aq) 2Fe3+(aq) + 2Cl(aq) [1]

42 (a) (i) Cr 3+: +3 [1]

Cr 2O72: +6 [1]

(ii) Cr 3+: green [1]

Cr 2O72: orange [1]

(b) (i) Cr 2O72(aq) + 14H+(aq) + 6e 2Cr 3+(aq) + 7H2O(l) [1]

(ii) The solution changes from orange to green at the end point. [1]

(iii) K 2Cr 2O7(aq) + 4H2SO4(aq) + 3H2S(g)

Cr 2(SO4)3(aq) + K 2SO4(aq) + 7H2O(l) + 3S(s) [1]

K 2Cr 2O7(aq) + H2SO4(aq) + 3SO2(g)

Cr 2(SO4)3(aq) + K 2SO4(aq) + H2O(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 Cu2+ ion [2]

Iron: Fe2+ ion and Fe3+ ion [2]

(c) Scandium and zinc [2]

44 (a) The orange dichromate ions (Cr 2O72(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 2O72(aq) is +6. [1]

45 (a) The oxidation number of vanadium in VO2+ is +5. [1]

The oxidation number of vanadium in VO2+ is +4. [1]

(b) V2+(aq) ion is violet [1] and V3+(aq) ion is green. [1]

(c) (i) 2SO2(g) + O2(g) 2SO3(g) [1](ii) The oxidation number of vanadium in V2O5 is +5. [1]

(d) (i) Stage 2: 2VO2(s) +2

1O2(g) V2O5(s) [1]

(ii) V2O5 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) MnO2(s) + 4HCl(aq) MnCl2(aq) + 2H2O(l) + Cl2(g) [1]

(ii) The colour of the resultant solution is very pale pink [1] due to the presence of Mn2+(aq) ions.

[1]

(c) (i) Purple [1]

(ii) MnO4

(aq) + 8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5Fe3+(aq) [1]



47 (a) N2(g) + 3H2(g) 2NH3(g) [1]

Finely divided iron [1]

(b) 2SO2(g) + O2(g) 2SO3(g) [1]

Vanadium(V) oxide/V2O5 [1]

(c) 2NO(g) + 2CO(g) N2(g) + 2CO2(g) [1]

Platinum/Pt and rhodium/Rh [1]

(d) 2H2O2(aq) 2H2O(l) + O2(g) [1]

Manganese(IV) oxide/MnO2 [1]

(e)

[1]

Titanium(IV) chloride/TiCl4 [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/VO2 [1]

(ii) It oxidizes X (VO2) to V2O5, [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: Fe2+ [1]

B: Fe3+ [1]

(b) Fe2+, I

and SO42. [3]

(c) Firstly, S2O82

ions are reduced by A.2Fe2+(aq) + S2O8

2(aq) 2Fe3+(aq) + 2SO42(aq) [1]

Secondly, I

ions are oxidized by B.

2Fe3+(aq) + I(aq) 2Fe2+(aq) + I2(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) + 2nH2O(l) + 3O2(g) 2Fe2O3 • nH2O(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) FeCl2(aq) + H2(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) N2(g) + 3H2(g) 2NH3(g) [1]

(d) (i) It oxidizes I to I2. [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) + S2O82(aq) I2(aq) + 2SO4

2(aq) [1]

52 (a) The oxidation state of manganese in MnSO4 is +2. [1]

The oxidation state of manganese in MnO2 is +4. [1]

The oxidation state of manganese in KMnO4 is +7. [1]

(b) (i) Manganese(IV) oxide [1]

(ii) Decomposition of hydrogen peroxide [1]

53 (a) Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g) [1]

2Fe(s) + 3Cl2(g) 2FeCl3(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]

2Fe3+(aq) + 2I

(aq) 2Fe2+(aq) + I2(aq) [1]Iron(II) ions reduce peroxodisulphate ions to give sulphate ions. Iron(II) ions are oxidized to

iron(III) ions. [1]

2Fe2+(aq) + S2O82(aq) 2Fe3+(aq) + 2SO4

2(aq) [1]

54 (a) 2I(aq) + S2O82(aq) I2(aq) + 2SO4

2(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]

2Fe3+(aq) + I(aq) 2Fe2+(aq) + I2(aq) [1]

Secondly, iron(II) ions reduce peroxodisulphate ions to give sulphate ions. Iron(II) ions areoxidized to iron(III) ions. Thus, iron(III) ions are regenerated. [1]

2Fe2+(aq) + S2O82(aq) 2Fe3+(aq) + 2SO4

2(aq) [1]

(c) It is essential for the synthesis of haemoglobin. [1]

55 (a) (i) Displacement reaction [1]

(ii) Zn(s) + CuSO4(aq) ZnSO4(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) N2(g) + 2CO2(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]

Periodic Table SQ Ans

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