H2 REVISION PACKAGE FOR BT1– 2012

SOLUTIONS TO INORGANIC CHEMISTRY QUESTIONS

MCQs – Section A

1 C 21 D 41 A 61 D 81 B2 D 22 D 42 C 62 B 82 A3 B 23 C 43 C 63 C 83 D4 C 24 A 44 A 64 A 84 B5 D 25 B 45 C 65 C 85 C6 B 26 C 46 B 66 B 86 A7 B 27 D 47 C 67 D 87 D8 A 28 C 48 C 68 D 88 C9 C 29 A 49 C 69 C 89 D10 A 30 B 50 D 70 B 90 D11 A 31 D 51 C 71 C 91 B12 D 32 B 52 D 72 D 92 C13 C 33 B 53 C 73 D14 A 34 A 54 C 74 C15 C 35 B 55 B 75 B16 B 36 C 56 D 76 C17 A 37 A 57 C 77 C18 A 38 D 58 C 78 B19 D 39 B 59 B 79 B20 A 40 D 60 C 80 D

MCQs – Section B

1 B 11 C 21 A2 B 12 A 22 A3 B 13 D 23 D4 C 14 A 24 B5 C 15 D 25 B6 C 16 B7 B 17 B8 B 18 A9 B 19 C10 C 20 B

Structured/Free-Response

1 (a) Sodium, magnesium and aluminium are metals. They have metallic structures

consisting of a regular arrangement of cations in a sea of delocalised electrons which

allows them to conduct electricity.

Conductivity increases from sodium to aluminum due to an increase in the number of

valence electrons.

Silicon is a metalloid with a low conductivity.

Phosphorous, sulphur, chlorine and argon are non-conductors of electricity because

they are non-metals with no mobile electrons.

(b) Na2O is an ionic oxide which is basic in nature. It has no reaction with NaOH but

reacts with HCl to form a salt.

Na2O + 2 HCl 2 NaCl + H2O

Due to the high charge density on Al3+ ion and the large extent of polarization in

Al2O3 , Al2O3 is an ionic compound with covalent character. Hence it is amphoteric

and can react with both HCl and NaOH.

Al2O3 + HCl 2 AlCl3 + 3 H2O

Al2O3 + 2 NaOH + 3 H2O 2 Na+ + 2 [Al(OH)4]-

SiO2 is a covalent oxide which is acidic in nature. Hence it has no reaction with HCl

but only NaOH to form a salt.

SiO2 + 2 NaOH Na2SiO3 + H2O

(c) (i) Na2O2 + CO2 Na2CO3 + H2O

(ii) Total volume of CO2 exhaled = 8 x 600 = 4800 dm3

Amount of CO2 exhaled = 4800 / 24 = 200 mol

Amount of Na2O2 needed = 200 mol

Mass of Na2O2 needed per day = 200 x (23.1 x 2 + 16.0x 2)

= 15 600 g

2 (a) (ii) Due to the high charge density of Al3+ ion, it undergoes hydrolysis readily in

water to form a strongly acidic solution.

(iii) AlCl3 + 6 H2O [Al(H2O)5(OH)] Cl2 + HCl

PCl5 + 4 H2O H3PO4 + 5 HCl

(b) (i) Mg or Al

Only oxides of Mg, Al and Si are insoluble in water. However, only oxides of Mg

and Al are able to react with acids.

(ii) Add aqueous sodium hydroxide.

Only Al2O3 will dissolve in aqueous NaOH to give a colourless solution. MgO will

not dissolve.

3 (a) X: Na Y: Al

A neutral solution is formed which does not react with Na2CO3 when the chloride of X

is dissolved in water.

Vigorous reaction of X with water implies that X could be sodium.

Na + H2O NaOH + ½ H2

NaCl Na+ + Cl-

From the reaction shown, the compound has undergone electrophilic substitution of

the benzene ring, hence a halogen carrier is needed. Therefore, Y is Al.

Al is unreactive towards water due to the layer of the protective oxide layer.

(b) Si has a giant molecular structure, where Si and O atoms are bonded by strong

covalent bonds. Large amount of energy is needed to break the covalent bonds,

hence it has the highest melting point.

P4, S8, Cl2 and Ar are non-polar, simple covalent molecules with weak van der waals’

forces existing between the molecules. Hence, little amount of energy is needed to

overcome the weak van der Waals’ forces and so, they have low melting points.

The strength of van der Waals’ forces depends on the number of electrons in the

molecule. The greater the number of electrons, the more polarisable the molecule

becomes, the stronger the van der waals’ forces. Since the number of electrons of S8

is larger than P4, Cl2 and Ar, it has the highest melting point, followed by P4, Cl2 and

Ar.

(c) SiO2 is insoluble in water because it has a giant covalent structure with strong

covalent bonds.

Na2O reacts with water to form an alkaline solution:

Na2O + H2O 2 NaOH

SiO2 is an acidic oxide which will react with the alkaline solution to form a salt and

water:

SiO2 + 2 NaOH Na2SiO3 + H2O

4 (a) Aluminium chloride has a simple molecular structure. Less energy is needed to

overcome the weak van der Waals’ forces between the molecules.

Chromium (III) chloride has a giant ionic structure with strong ionic bonding between

the oppositely charged ions. Large amount of energy is needed to overcome the ionic

bonding and hence, it has a much higher melting point.

(b) Al3+ ions have high charge density due to its small size and high charge.

It polarizes the water molecule strongly, weakening the O—H bond to form H+.

Hence, it undergoes hydrolysis.

[Al(H2O)6]3+ [Al(H2O)5(OH)]2+ + H+

(c) (i) Al2O3 is able to react with both acids and alkalis.

(ii) Al2O3 + 6 H+ 2 Al3+ + 3 H2O

Al2O3 + 2 OH- + 3 H2O 2 Al(OH)4 -

5 (a) (i) Both Na2O and Al2O3 have giant ionic structures with strong ionic bonds

between the oppositely charged ions which require a lot of energy to break them

resulting in high melting points.

Al3+ ions have a higher charge and smaller ionic radius than Na+, hence the

lattice energy for Al2O3 is higher than Na2O. This accounts for the higher melting

point of Al2O3 than Na2O.

SO3 has a simple molecular structure with weak van der Waals’ forces between

the molecules. Small amount of energy is needed to break them resulting in a

low melting point.

(ii) Na2O dissolves readily in water to give a strongly alkaline solution of pH 13 (or

14).

Na2O + H2O 2 NaOH

Al2O3 is insoluble in water due to its extremely high lattice energy.

SO3 dissolves readily in water to give a strongly acidic solution with pH 1 or 2.

SO3 + H2O H2SO4

(b) Mr of AgCl = 143.5

No. of moles of AgCl precipitated = 0.574 / 143.5 = 4.00 x 10-3 mol

Mr of Si2OCl6 = 285.2

No. of moles of Cl in Si2OCl6 = (0.2 / 285.2) x 6 = 4.21 x 10-3 mol

Mr of Si3O2Cl8 = 400.3

No. of moles of Cl in Si3O2Cl8 = (0.2 / 400.3) x 8 = 4.00 x 10-3 mol

Since the number of moles of Cl in AgCl = No. of moles of Cl in Si3O2Cl8

therefore, the identity of the oxochloride = Si3O2Cl8

6 (a) Most d-block compounds are coloured due to the splitting of the d-orbitals in the

metal ion. When they are bonded to ligands, the 3d orbitals undergo splitting,

resulting in 2 groups of non-degenerate orbitals. This is called the d-d splitting.

By absorption of energy, electrons can be promoted from a lower energy d-orbital to a

higher energy d-orbital. This is called the d-d transition. The light not absorbed is thus

seen as the colour of the complex.

Sc and zinc compounds are colourless because such d-d transitions cannot occur in

ions with empty d-orbitals or completely filled d-orbitals.

(b) Copper(II) sulphate salts dissolve in water to give a blue colour solution, which is attributed to [Cu(H2O)6]2+ (aq)

CuSO4 (s) +aq [Cu(H2O)6]2+ (aq) + SO42- (aq)

Blue solution

When dilute aqueous NH3 is gradually added, a blue precipitate of Cu(OH)2 is first formed.

NH3 + H2O NH4+ + OH-

[Cu(H2O)6]2+(aq) + 2OH-(aq) Cu(OH)2(s) + 6H2O blue solution blue precipitate

When excess aqueous NH3 is added, the precipitate dissolves to give a deep blue solution, due to the formation of the soluble complex ion, [Cu(NH3)]2+.

Cu(OH)2 + 4NH3 + 2H2O [Cu(NH3)4(H2O)2]2+ + 2OH-

deep blue solution

NH3 is a stronger ligand than H2O, so it displaces water from [Cu(H2O)6]2+ to form [Cu(NH3)4(H2O)2]2+.

Overall reaction: [Cu(H2O)6]2+ + 4NH3 [Cu(NH3)4(H2O)2]2+ + 4H2O blue solution deep blue solution

(c) The uncatalysed reaction is slow due to high activation energy since 2 negatively ions S2O8

2- and I- are involved and they repel each other

The Co3+ acts as a homogenous catalyst in this redox reaction of I- and S2O82-. The

catalysed pathway involves two steps:

Step 1: 2Co3+ + 2I- 2Co2+ + I2

Reduction: Co3+ + e Co2+ Ered° = +1.82 VOxidation: 2 I- I2 + 2 e Eoxd° = - 0.54 V

Ecell= 1.82 + (-0.54) = + 1.28 V

Step 2: 2Co2+ + S2O82- 2Co3+ + 2SO4

2-

Reduction: S2O82- + 2e 2SO4

2- Ered° = +2.01 VOxidation: Co2+ Co3+ + e Eoxd° = -1.82 V Ecell = 2.01 + (-1.82) = +0.19V Overall: S2O8

2- + 2I- 2SO42- + I2

Both steps in the catalysed reaction involve the interaction of oppositely charged ions, which attract one another strongly; hence the activation energy is lower and enhances the rate of reaction.

7 Ti 4+ : [Ar]3d0 and Ti 3+ : [Ar] 3d1

Ti3+ has one d electron. In [Ti(H2O)6]3+, the presence of the H2O ligands causes the 3d

orbitals to split into 2 different energy levels which are close in energy. When energy from

light is absorbed, the d electron can be promoted from the lower energy to higher energy d-

orbital. The light energy not absorbed will be seen as the colour of the complex.

Ti4+ has no d electron and d-d transition cannot occur. Hence, it appears colourless.

8 (a) Transition elements are able to exhibit a variety of oxidation states because of the

close similarity in energy of the 4s and 3d electrons. Hence, transition metals are able

to lose all of the electrons in the 3d and 4s subshells.

(b) No. of moles of KMnO4 used = (20.0 / 1000) x 0.100 = 2.00 x 10-3 mol

No. of moles of electrons gained by MnO4- = 2.00 x 10-3 x 5 =1.00 x 10-2 mol

Mole ratio of VOClx : electrons

0.01 : 1.00 x 10-2

1 : 1

Hence, the oxidation number of vanadium increases by 1.

Therefore, the original oxidation state of V in VOClx = +4

x = 2

9 (a) In the complex, the degenerate 3d orbitals are split into 2 sets of different energy

levels, separated by a small energy gap. Electrons absorb energy from light and are

promoted from a lower to a higher energy 3d orbitals. The wavelength of light not

absorbed will be seen as the colour of the complex.

(b) No. of moles of KMnO4 = (19.60 / 1000) x 0.02 = 3.92 x 10-4 mol

No. of moles of Fe = no. of moles of Fe2+

= (3.92 x 10-4) x 5

= 1.96 x 10-3 mol

Percentage of iron in the meteorite = [(1.96 x 10-3 x 55.8) / 0.20] x 100%

= 54.7 %

10 (a) Due to the close proximity in energies of the 4s and 3d orbitals, transition metals can

exhibit variable oxidation states by losing different number of electrons in the 3d and

4s orbitals.

(b) In the presence of ligands, the degenerate 3d orbitals in copper (II) ions are split into

2 sets with different energies with a small energy gap between them.

When a d-electron from the lower energy group is promoted to the higher energy

group (d-d transition), light energy is absorbed. The light energy not absorbed will be

seen as the colour of the complex.

11 (a) (i) Across the period from Na to Cl,

nuclear charge increases;

but increase in shielding effect is negligible as electron is added to the

same quantum shell;

Hence, the effective nuclear charge increases OR the electrostatic forces

of attraction between the nucleus and the outermost/ valence electrons

increases across the period.

Therefore, the atomic radius decreases across the period from Na to Cl.

(ii) Na+ has one electron less than Na or Na+ has one quantum shell less than Na.

Nuclear charge remains the same. The electrostatic forces of attraction

between the nucleus and the remaining electrons are greater for Na+.

(b) (i)

(ii) The anionic radii of P3–, S2– and Cl– are larger than their respective atoms.

This is due to electrons are being added to the same valence shell, resulting

in greater inter-electronic repulsions between the valence electrons.

Across the period from P3–, S2– and Cl–, the anionic radii decrease. The

nuclear charge increases across the period with negligible increase in

shielding effect resulting in the corresponding increase in the effective nuclear

charge OR electrostatic forces of attraction between the nucleus and the

valence electrons.

12 (a) (i) Al2O3 has a giant ionic lattice structure with strong ionic bonds / high lattice

energy / partial covalent character, hence it is insoluble.

SiO2 has a giant molecular structure with strong covalent bonds requiring

large amounts of energy to break, hence it is insoluble.

(ii) AlCl3 dissolves to give [Al(H2O)6]3+(aq) which undergoes hydrolysis due to the

high charge density on Al3+:

[Al(H2O)6]3+ [Al(H2O)5(OH)]2+ + H+

OR 2AlCl3 + 6H2O A⇌ l2O3 + 6HCl

OR AlCl3 + 3H2O A⇌ l(OH)3 + 3HCl

SiCl4 hydrolyses in water to give HCl, a strong acid / H+ is produced:

SiCl4 + 2H2O SiO2 + 4HCl

OR SiCl4 + 4H2O Si(OH)4 + 4HCl

13 (i) AB(CO3)2 → AO + BO + 2CO2

(ii) Assume: AB(CO3)2 → AO(s) + BO (aq) + 2CO2

Masses given: 0.400 0.057 (0.275-0.057) x

Therefore, x = 0.400 – 0.275 = 0.125 g

(iii) Soln: Let Ar of A be a and that of B be b

AB(CO3)2 → AO(s) + BO (aq) + 2CO2

Mass/g: 0.400 0.057 0.218 0.125

Determining amts and using mol ratios

Amt of CO2 = 0.125/44 = 0.00284 mol

Amt of AO = 0.057/(a + 16) = 0.00284/2 mol

Giving a = 24.1, Therefore A is Mg

Amt of BO = 0.218/(b + 16) = 0.00284/2 MOL

Giving b = 137.5, Therefore B is Ba

14 (a) The alkaline earth metals exhibit +2 oxidation state in their compounds because they

have 2 valence electrons which are lost to achieve stable octet.

They do not exhibit +1 oxidation state due to the low lattice energy of the compounds

formed.

+3 oxidation state is also not favorable due to the very high 3rd ionization energy

required to remove the third electron from the inner quantum shell.

15 (a) (i) CaCO3(s) CaO(s) + CO2(g)

(ii) From Data Booklet: Zn2+ : 0.074 nm; Ca2+ : 0.099 nm

Since Zn2+ has a larger cationic radius than Ca2+, the charge density of Zn2+ is

larger and it has a higher polarizing power than Ca2+. This leads to a greater

distortion of CO32- electron cloud, thus there is greater weakening of bonds in

CO32-. thermal stability of ZnCO3 < CaCO3.

(iii)

16 (b) (i)

ease of breaking the H−X bond: HCl < HBr < HI

thermal stability: HCl > HBr > HI

(ii) Hydrogen bonding exist between HF molecules

17 (b) (i) x = 3

2ICl3 + 6I- → 4I2 + 6Cl-

HCl HBr HI

Bond Energy

(kJmol−1)431 336 229

18 (a)

19 (a) Iodine oxidises thiosulfate to tetrathionate resulting in a change of average

oxidation number from +2 to +2.5:

I2 + 2S2O32 2I + S4O6

2

Chlorine oxidises thiosulfate to sulfate(VI) resulting in a change of average

oxidation number from +2 to +6:

4Cl2 + S2O32 + 5H2O 8Cl + 2SO4

2 +10H+

Since chlorine is able to increase the oxidation state of S in thiosulfate more greatly

than iodine, chlorine has higher oxidising power than iodine.

(b) (i) NaCl(s) + H2SO4(l) NaHSO4(s) + HCl(g)

(ii) Hydrogen iodide formed is oxidised by conc H2SO4 to iodine since HI is easily

oxidised.

(iii) Concentrated phosphoric(V) acid.

20 (a) Any two properties:

Fe has a higher melting/boiling point than Al.

Fe has stronger metallic bonds as both the 3d and 4s electrons can be used in

metallic bonding due to their proximity in energies, hence more energy is required to

overcome the stronger bonds.

Fe has a greater density than Al.

Fe has a greater atomic mass but its atomic radius is smaller. Hence atomic volume

is smaller. Since density = mass/volume, density of Fe is greater than Al.

Fe has better conductivity than Al as both the 3d and 4s electrons are available in the

mobile sea of electrons due to the proximity in energy of 3d and 4s orbitals.

Fe is harder than Al due to stronger metallic bonding in Fe as Fe can use both its 3d

and 4s electrons for metallic bonding due to the proximity in energy of 3d and 4s

electrons.

(b) 26Fe 1s2 2s2 2p6 3s2 3p6 3d6 4s2 13Al 1s2 2s2 2p6 3s2 3p1

26Fe3+ 1s2 2s2 2p6 3s2 3p6 3d5 13Al3+ 1s2 2s2 2p6

Although Fe has one more quantum shell than Al and higher nuclear charge, the

increase in shielding effect due to the d electrons is proportionately less than the

increase in nuclear charge, hence Fe experiences higher effective nuclear charge

and is smaller.

Fe3+ has one more quantum shell than Al3+ and the additional shielding is due to s and

p electrons which are more effective in shielding compared to d electrons.

21 (a) Fe(H2O)63+ Fe(H2O)5(OH)2+ + H+

(b) (i) 1s22s22p63s23p63d10

(ii) Colourless

because Cu + does not have unfilled/ partially filled d-orbital

22 (a)

(b)

23 (i) K2Cr2O7 + 2 HBr 2 KCrO3Br + H2O

(ii) No, oxidation no. of Cr doesn’t change, remains at +6

(iii) Eocell = +1.33 – (+1.07) = +0.26 V >0

Products should be Cr3+(aq) and Br2(l)

(iv) Reaction didn’t take place at standard condition.

OR The reaction in step I took place under cooled conditions.

(v) AgBr

(vi) Cr is reduced from +6 (orange solution of Cr2O72-) to +3 Cr3+ (deep green

solution)

24

25

26 (a) X = Mg

Y = Ca or Sr or Ba

Reason: Reducing power increases down the group.

(b) High melting point of MgO

(c) (i) SiO2 does not dissolve in water.

pH 7

Na2O dissolves readily in water to give a colourless solution.

Na2O + H2O 2NaOH

pH 13-14

CaO dissolves sparingly in water.

CaO + H2O Ca(OH)2

pH 9-12

(ii) The basic metal oxide will react / undergo neutralisation with the acidic SO2.

OR

Acid-base reaction

(d) (i) Br2 has a smaller electron cloud size than I2.

electron cloud of Br2 is less polarisable

weaker Van der Waals’ forces between the molecules

More energy is required to overcome the interaction between I2 molecules,

leading to a higher boiling point.

(ii) Cl- has one more electron; hence its p/e ratio is smaller

attraction of the nucleus for the outer electrons is weaker. Therefore it has a

bigger radius.

OR electron-electron repulsion

(e) Disproportionation

3Cl2(g) + 6OH-(aq) 5Cl-(aq) + ClO3-(aq) + 3H2O(l)

Smell of Cl2 disappears OR

greenish-yellow/pale green/green colour of Cl2 disappears (hard to observe)

27 Li+ is much smaller than Cs+

Charge density of Li+ is higher than that of Cs+

I3- is polarized to a greater extent by Li+

CsI3 is more stable than LiI3.

28 (i) F : [Fe(CN)6]4-

G : [Fe(CN)6]3-

(ii) Ligands are molecules or ions that possess at least one lone pair electrons that

can form dative covalent bonds to the central metal ion.

Transition elements have strong tendency to form complexes as their cations

are small and highly charged.

Central metal ion have empty low-lying 3d orbitals to accommodate the lone

pair electrons from the ligands.

(iii) Fe3[Fe(CN)6]2 / Fe4[Fe(CN)6]3

(iv) Hexadentate ligand EDTA can form a very stable complex with Fe2+, thus

removing the catalytic effect of Fe2+

29 (a) (i) SiO2 is insoluble in water due to the high bond dissociation energy needed to

break the strong Si—O covalent bonds.

SiCl4 (l) + 4H2O (l) Si(OH)4 (s) + 4HCl (aq)

(ii) Green

Red

30 (i) Co(OH)2

(ii) [Co(NH3)6]2+

(iii) [Co(NH3)6]3+

Oxidation by O2(g) in the air

(iv) 2Co3+ + 2I- → 2Co2+ + I2

31 (i) G is hydrogen peroxide

BaO2 + 2H2O Ba(OH)2 + H2O2

Hydrogen peroxide which in acidified condition, acts as a reducing agent to

reduce MnO4- ions to Mn2+ ions.

(ii) Ca2+ being the smaller cation than Ba2+, has greater charge density and has

better ability to polarise/distort the large electron cloud of the nitrate ion. As a

result, one N-O bond is weakened and the nitrate ion is more easily

decomposed to form an oxide anion. Conversely, Ba2+ ions are large and have

the lower charge density, resulting in poorer polarising power. Thus, the Ba2+

cation does not have the ability to distort the electron cloud of the nitrate ions.

Hence, Ba(NO3)2 is more stable to heating than Ca(NO3)2 .

32 (i) T: I2

U: Br2

(ii) 4Br2 + S2O32- + 5H2O 8 Br─ + 2SO4

2- + 10H+

(iii) Having the highest Eθ= +1.36V, Cl2 is a strongly oxidizing, it is able to oxidise sulphur in +2 oxidation state in S2O3

2- to +6 oxidation state in SO42-.

4Cl2 + S2O32- + 5H2O 2SO4

2- + 10H+ + 8Cl-

Br2 is also quite oxidising, Eθ=+1.07 V. It is able to oxidize S2O32- to SO4

2-.

4Br2 + S2O32- + 5H2O 2SO4

2- + 10H+ + 8 Br─

Being less oxidizing (Eθ=+0.54V), I2 is only able to oxidize sulphur in +2 oxidation state in S2O3

2- to + 2.5 oxidation state in S4O62-.

I2 + 2S2O32- S4O6

2- + 2I-

This trend is supported by the decreasing oxidising power down the group as illustrated by E0 values.

(iv) Aqueous iron(II) sulfate, followed by aqueous sodium hydroxide

33 (i) Transition metal ions have empty 3d orbitals, thus they can act as Lewis acids

(electron pair acceptors) and form dative bonds with ligands to form complexes.

(ii) Copper (II) salts dissolve in water to give a blue colour solution, which is due to

the complex ion [Cu(H2O)6] 2+ .

When dilute aqueous NH3 is gradually added, a blue precipitate of Cu(OH)2 is

first formed.

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

[Cu(H2O)6]2+(aq) + 2OH-(aq) Cu(OH)2(s) + 6H2O (l) blue solution blue precipitate

When excess aqueous NH3 is added, the precipitate dissolves to give a deep blue solution due to the formation of the soluble complex ion, [Cu(NH3)] 2+ .

Cu(OH)2(s) + 4NH3(aq) + 2H2O(l) [Cu(NH3)4(H2O)2]2+(aq) + 2OH-(aq) deep blue solution

NH3 is a stronger ligand than H2O and so, displaces water from [Cu(H2O)6]2+ to

form [Cu(NH3)4(H2O)2]2+, which is deep blue in colour.

Overall reaction:

[Cu(H2O)6]2+(aq) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 4 H2O(l)

blue solution deep blue solution

On heating, the precipitate forms copper(II) oxide.

Cu(OH)2(s) → CuO(s) + H2O(l)

34 (i) A = Na

NaO and NaCl are giant ionic solids .

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

NaCI (s) Na+ (aq) + CI- (aq)

(ii) B = P

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

Or P4O6 (s) + 6H2O (l) 4H3PO3 (aq)

PCI3 (l) + 3H2O (l) H3PO3 (aq) + 3HCI (aq)

Or PCI5 (l) + 4H2O (l) H3PO4 (aq) + 5HCI (aq)

(iii) C= Si

SiO2 has giant molecular structure with strong covalent bonds between atoms, thus insoluble in water.

35 Solubility of Li2O is lower than that of BaO.

Therefore, the basicity of Li2O is also lower than that of BaO.

36 (a) (i) X is Na/ Mg

W is Si

(ii) SiCl4 + 2H2O SiO2 + 4HCl

(b) To 2 separate samples of the powder, add in dilute HCl and dilute NaOH respectively.

If the powder dissolves in both dilute HCl and dilute NaOH to form colourless solution,

it is Al2O3.

If the powder dissolves only in dilute HCl to form colourless solution, it is MgO.

If the powder does not dissolve in both dilute HCl and dilute NaOH, it is SiO2.

37 (a) (i) To increase the yield of PCl5 since reaction is exothermic. To cool the

exothermic reaction for safety.

(ii) PCl5 will react/hydrolyse in water to give an acidic solution.

PCl5(s) + H2O(l) → POCl3(aq) + 2HCl(aq) OR

PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq)

(b) A white precipitate of AgCl is obtained when sodium chloride is mixed with silver

nitrate. Precipitate dissolves in aqueous ammonia to give a colourless solution of

Ag(NH3)2+.

Ag+(aq) + Cl–(aq) → AgCl(s)

Ag+(aq) + 2NH3(aq) → Ag(NH3)2+(aq)

When NH3 is added, the formation of the complex ion decreases the concentration of

Ag+, causing equilibrium position of the equilibrium AgCl Ag+ + Cl– to shift to the

right such that ionic product < Ksp(AgCl).

(c) NaAt(s) + H2SO4(l) → HAt(g) + NaHSO4(s)

2HAt(g) + H2SO4(l) → At2(g) + SO2(g) + 2H2O(l)

8HAt(g) + H2SO4(l) → 4At2(g) + H2S(g) + 4H2O(l)

38 (d) (i) E is MgCO3

F is CO2

G is Mg(NO3)2.6H2O or Mg(NO3)2. xH2O

(ii) MgCO3 → MgO + CO2

MgCO3 + 2HNO3 → Mg(NO3)2 + H2O + CO2

2Mg(NO3)2. xH2O → 2MgO + 4NO2 + O2 + xH2O

(iii) Mg2+ has a smaller radius than Ba2+.

Mg2+ has a higher charge density than Ba2+ and hence stronger polarising

power to polarise the electron cloud of the large NO3‾ anion.

39 (a) (ii) In the presence of water, AlCl3 undergoes hydrolysis to form an acidic solution

of pH≈ 3.

AlCl3 (s) + aq [Al(H2O)6]3+ (aq) + 3Cl (aq)

[Al(H2O)6]3+ (aq) + H2O (l) [Al(H2O)5(OH)]2+ (aq) + H3O+ (aq)

The acidic nature of the solution is due to the relatively small but highly charged

Al3+ ion which polarizes the surrounding water molecules to give up H+ ions.

(iii) SiCl4 (l) + 4H2O (l) Si(OH)4 (s) + 4HCl (aq)

PCl5 (s) + 4H2O (l) H3PO4 (aq) + 5HCl (aq)

40 (a) (i) dehydration of alcohol to form alkene:

OR esterification:

nitration of benzene:

OR esterification:

OR hydration of alkene:

(ii) H2SO4 (l) + NaBr (s) → HBr (g) + NaHSO4 (aq)

H2SO4 (l) + 2HBr (g) → Br2 (g) + SO2 (g) + 2H2O (l)

(b) (i) Tetrachloromethane / hexane (or any other non-polar organic solvent)

(ii) The halide is iodide.

2I–(aq) + Cl2(g) → 2Cl–(aq) + I2(aq)

41 (a) (i) A: [Cr(H2O)6]3+

B: [Cr(H2O)6]2+

C: Cr(OH)3 or Cr(H2O)3(OH)3

D: CrO42– or Na2CrO4

(ii) E:

H2C CH2 + H2OCH3CH2OHconc. H2SO4

CH2=CH2 + H2O CH3CH2OHconc. H2SO4

+ conc. HNO3

conc. H2SO4

NO2

+ H2O

RCO2H + R'OH RCO2R' + H2Oconc. H2SO4

RCO2H + R'OH RCO2R' + H2Oconc. H2SO4

(iii) Cr3+ has a high charge density. Hence [Cr(H2O)6]3+ can undergo hydrolysis in

water to produce H+ ions, forming CO2 with carbonate ions.

(iv) Ligand exchange

(b) Cr3+ has d3 electronic configuration.

In an octahedral ligand field, the 6 NH3 ligands will split the five degenerate 3d orbitals

into 2 groups of different energy levels.

The difference in the two energy levels, E, falls within the visible region of the

electromagnetic spectrum. An electron in a lower d orbital energy level can absorb

radiation in the visible spectrum and be promoted into the higher d orbital energy

level.

This dd electron transition gives rise to the colour as the complement of the

absorbed colour.

42 (i) Both Cl2 and I2 exist as diatomic molecules and have dispersion forces existing

between molecules.

I, being lower down in the group than Cl, has a bigger electron cloud size and

hence, stronger dispersion forces. Therefore I2 is less volatile than Cl2.

(ii) 4Cl2 (aq) + S2O32– (aq) + 5H2O (l) 8Cl – (aq) + 2SO4

2– (aq) + 10H+ (aq)

I 2 (aq) + 2S2O32– (aq) 2I – (aq) + S4O6

2– (aq)

E / V

Cl2 + 2e 2Cl – +1

I2 + 2e 2I –36 +0.54

ECl2/Cl– is more positive than E

I2/I–, indicating that Cl2 is a stronger oxidizing

agent than I2. Hence, Cl2 can oxidize S2O32– to SO4

2– while I2 can only oxidize

S2O32– to S4O6

2–.

43 (a)

Na Mg Al Si P Si Cl Ar

M.p. /oC

High melting point from Na to Al as they exist as giant metallic structures with strong

metallic bonds of increasing strengths due to smaller cationic radius and increased

number of delocalized electrons.

Very high melting point for Si as it exists as a giant covalent structure with an

extensive network of strong covalent bonds.

These strong bonds require a lot of energy to break before melting can occur.

Low melting point from P to Ar as they exist as simple molecular structures,

consisting of discrete molecules with weak dispersions forces between the molecules.

Melting point decreases from S8 > P4 > Cl2 > Ar because the size of the electron

S

Melting point /oC

clouds decreases from S8 > P4 > Cl2 > Ar, such that dispersion forces are weaker.

(b) Mg(OH)2 is basic, and reacts only with acids:

Mg(OH)2(s) + 2HCl (aq) MgCl2(aq) + 2H2O(l)

Be(OH)2 is amphoteric, and reacts with both acids and bases:

Be(OH)2(s) + 2HCl (aq) BeCl2(aq) + 2H2O(l)

Be(OH)2(s) + 2NaOH(aq) Na2Be(OH)4(aq)

44 (a) (i) A = [Fe(H2O)6]2+

B = [Fe(H2O)6]3+

(ii) X = K2Cr2O7 or KMnO4 or Cl2 or H2O2 (or in words)

Y = potassium thiocyanate or KSCN (or NH4CN)

(iii)

(b) (i) This is due to the presence of partially-filled 3d subshells that can accept

electrons from lone pairs or form temporary bonds with reactant molecules

during adsorption, thus weakening the bonds in the reactant molecules and

lowering the activation energy of the reaction.

(ii)

45 (a) (i) Al2O3 + 6H+ 2Al3+ + 3H2O

Al2O3 + 2OH + 3H2O 2[Al(OH)4]-

(b) (i)

N2 + H2

NH3

H

Ea (uncatalysed)

Ea (catalysed)Ener

gy /

kJ m

ol–1

Reaction progress

Ionic Radius

Na Mg Al Si P S Cl

pH

Na2O MgO Al2O3 SiO2 P4O10 SO3

7

(ii) Ionic Radius:

There are two isoelectronic series in period 3.

Within each series, an increasing number of protons results in

increasingly stronger nuclear attraction experienced by the valence

electrons.

Valence electrons are pulled closer to nucleus, leading to a decreasing

trend in ionic radius.

Anionic series has a larger size than cationic series as the valence

electrons are in a higher principal quantum shell, experience weaker

nuclear attraction and are less tightly held.

pH:

Na2O and MgO are ionic oxides that dissolve completely in water to form

basic solution.

Al2O3 and SiO2 are insoluble in water pH of solution is that of water.

P4O10 and SO3 are covalent oxides that hydrolyse in water to give acidic

solution.

(c) (i) [Mg(H2O)6]2+ + H2O [Mg(H2O)5(OH)]+ + H3O+

46 (a)

CaIx x

x

x

xxx

OO

Ox

2+

2

(b) (i) Calcium iodate is white as the electronic transitions are between different

principal quantum shells.

In the presence of ligands, the degenerate d-orbitals of Fe ions undergo d-

splitting. An electron in the lower energy d-orbital undergoes d-d transition by

absorbing visible light.

(ii) Fe has more protons than Ca, thus has a higher nuclear charge. The inner 3d

electrons of Fe are poor shielding compared to the inner 3s and 3p electrons of

Ca. Thus, the valence electrons of Fe experience greater attraction to the

nucleus and require more energy to remove compared to Ca.

(c) (i) 2 Ca(IO3)2 (s) 2 CaO (s) + 2 I2 (g) + 5 O2 (g)

(ii)The charge density of the ions given by decreases down the group.

Hence, polarizing power of the ions decreases and the electron cloud of the

iodate ion is polarized to a smaller extent.

The iodine-oxygen bonds in the iodate ion becomes less weakened and more

energy is required to break the bonds. Thus, thermal stability of the group II

iodates increase.

47 (b) (i) e- config of cation if Y : 1s2 2s2 2p6 3s2 3p6 3d9

Identity of X : NH3 or ammonia

(ii) (Y represents Cu)

Cu(NO3)2 + 6H2O [Cu(H2O)6]2+ + 2NO3-

Blue

[Cu(H2O)6]2+ + 2OH- Cu(OH)2(s) + 6H2O

Light blue ppt.

[Cu(H2O)6]2+ + 4NH3 [Cu(NH3)4(H2O)2]2+ + 4H2O

Excess dark blue filtrate

Mg(H2O)62+ + 2OH- Mg(OH)2 + 6H2O

White residue, insoluble in excess NH3

48 (b) (i) A nickel (II) sulphate solution contains [Ni(H2O)6]2+ ions.

In the presence of H2O ligands, the 3d orbitals are split into two groups with a

small energy gap between them.

Electrons in the lower energy 3d orbital absorb energy as they move to a

vacant and higher energy 3d orbital. The energy absorbed corresponds to

some wavelengths of visible light.

The unabsorbed complementary colours are then emitted.

Such d-d electrons transitions are responsible for the colour of the

compounds.

(ii) The NH3 ligands split the d orbitals of the Ni2+ ion into two sets of slightly

different energies to a different extent from that of the H2O ligands. (size gap

is different)

Hence, visible radiation of a different wavelength is absorbed and emitted for

d-d transitions, giving rise to a different colour.

(iii) A bidentate ligand has 2 groups with lone pairs of electrons and hence can

form 2 dative bonds to the central metal atom/ion.

49 (c) (i)

(ii)

(d) (i)

(ii)

50 (a) (i)

(ii)

51 (a) (i) CaCO3

CaCO3 + SO2 CaSO3 + CO2

(ii)Mass of S in coal = 1 x 109 x = 2.5 x 107kg = 2.5 x 1010 g

Amt of S = 2.5 x 1010 32.1 = 7.788 x 108 mol

Amt of CaCO3 = Amt of S = 7.788 x 108 mol

Mass of CaCO3 = 7.788 x 108 x 100.1 = 7.80 x 1010 g

(b) Solution F is calcium chloride solution which is effectively neutral.

For BeCl2, it becomes the aqua complex, [Be(H2O)4]Cl2 upon contact with water. The

small, highly polarising Be2+ weakens the O–H bonds of the water molecules in

its surrounding sphere of coordination and results in the release of H+ (or H3O+)

ions in solution, thus giving rise to an acidic solution.

BeCl2(s) + 4H2O(l) → [Be(H2O)4]2+(aq) + 2Cl– (aq)

[Be(H2O)4]2+(aq) + H2O(l) [Be(OH)(H2O)3]+(aq) + H3O+(aq)

(c) (i) CaCO3 CaO + CO2

Amt of CaO = 1.00/100.1 x 56.1 = 0.560 g

(ii)

(iii) From Data Booklet

time

1.00

0

Mass/ g

Decomposition curve of CaCO3

Decomposition curve of MgCO3

x

ion Mg2+ Ca2+

Ionic radius 0.065 0.099

The thermal stability depends on the charge density of the cation. The greater

the charge density, the thermally less stable the carbonate.

Since charge density and hence polarsing power of Mg2+ is higher than that of

Ca2+, the distortion of the electron cloud of the carbonate anion, thus

weakening effect of the carbonoxygen bonds in the magnesium carbonate

occurs to a greater extent.

Hence MgCO3 is more unstable than CaCO3 and should decompose to MgO

and CO2 at a faster rate (since its decomposition temperature is lower than that

of CaCO3.)

The mass of MgO obtained is 0.478 g, lower than x (i.e. 0.560 g)

52 (a) From data booklet, atomic size: F 0.072 nm ;

Cl: 0.099 nm ; Br : 0.114 nm ; I : 0.133 nm

As size of halogen atom increases down the group, the bond length of H – X

increases/effectiveness of overlap of orbitals decreases. Bond energy of H – X

decreases and ease of decomposition of HX increases.

Or

From data booklet: Bond energy values: H – F : 562

kJ mol-1; H – Cl : 431 kJ mol-1 ; Br : 366 kJ mol-1 and H – I : 299 kJ mol-1

As the bond energy of H – X decreases down the group, the ease of decomposition

of HX increases. The bond energy decreases down the group as the atomic size of

halogen atoms increases and H – X bond length increases.

(b) (i) 2 HBr + H2SO4 SO2 + Br2 + 2H2O

(ii) HI is a stronger reducing agent than HBr

It is able to reduce S from oxidation state +6 (in H2SO4) to -2 (in H2S).

HBr is only able to reduce S to +4 oxidation state (in SO2)

53 Since the oxide of Y can react with both NaOH and HCl, it is an amphoteric oxide.

Among the possible elements, only the oxide of aluminium is amphoteric in nature:

Compound Y is Al.

Al2O3 (s) + 6HCl (aq) 2AlCl3 (aq) + 3H2O (l)

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

In the presence of water, AlCl3 undergoes hydrolysis to form an acidic solution, which

reacts with the NaOH.

AlCl3 (s) + aq [Al(H2O)6]3+ (aq) + 3Cl (aq)

[Al(H2O)6]3+ (aq) + H2O (l) [Al(H2O)5(OH)]2+ (aq ) + H3O+ (aq)

Phosphorus chloride and phosphorus oxide will react with water but silicon oxide is

insoluble in water.

Compound Z is P.

PCl3(l) +3H2O(l) H3PO3 (aq)+3HCl (aq)

or

PCl5 (s)+ 4H2O(l) H3PO4(aq) + 4HCl (aq)

P4O6 (s) + 6H2O (l) 4H3PO3 (aq)

or

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

54 (i) 2Ba(NO3)2 → 2BaO + 4NO2 + O2

(ii) The temperature of the thermal decomposition increases down the group.

This is because down the group,

- the size of the cations increase

- hence polarizing power of cations decrease

- ability of cations to distort the large anion decrease

nitrates are more stable to heat

(iii) L.E is proportionate to ionic charge but inversely proportionate to ionic

radius

or L.E.

- Ionic size of nitrate ion bigger than that of chloride

- Hence, LE for barium nitrate is SMALLER than barium chloride.

55 (a) (i) MgO

has a giant ionic lattice structure

large amount of energy is needed to

overcome the strong ionic bonds

OR electrostatic forces of attraction between oppositely charged ions.

has a high boiling point

P4O6

has a simple molecular structure

small amount of energy is needed to

overcome the weak intermolecular Van der Waals’ forces of attraction

has a low boiling point

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

(iii) SO2 (g) + 2NaOH (aq) Na2SO3 (aq) + H2O (l)

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

(iv) Al2O3 behaves as an a mphoteric oxide or

Al2O3 is able to react with both acid and base

(b) (i) Mg(NO3)2(s) MgO(s) + 2NO2 (g)+ ½ O2 (g)

(ii) Down the group,

Ionic radius of cation: Mg2+ < Sr2+.

Charge density of cation: Mg 2+ >Sr 2+ / decreases

Polarising power of cation: Mg 2+ > Sr 2+ / decreases

or

Polarising effect on NO3- anion: Mg 2+ > Sr 2+ / decreases

Mg(NO3)2 is less stable than Sr(NO3)2, hence it decomposes at a lower

temperature

Thermal stability of Group II nitrates increases down the group; hence barium

nitrate would decompose at a higher temperature compared to strontium nitrate.

56 (b) (i) Transition metals possess variable oxidation states due to the small energy

difference between the 3d and 4s electrons. Thus different number of 3d and 4s

electrons may be lost to form stable ions or compounds of different oxidation

states.

(ii) NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Cu2+(aq) + 2OH-(aq) Cu(OH)2 (s) ----- (1)

When NH3(aq) is added gradually, [OH-] will increase

Ionic product of Cu(OH)2 > Ksp of Cu(OH)2

Pale blue ppt, Cu(OH)2 is formed

[Cu(H2O)6]2+(aq) + 4NH3(aq) [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) -----(2)

deep blue

When excess NH3 is added,

NH3 ligands replaces the H2O ligands, forming a more stable deep blue

[Cu(NH3)4(H2O)2]2+ complex with Cu2+(aq).

[Cu2+] decreases as it is being used to form the complex, equilibrium position

in (1) shifts left to increase [Cu2+]

Pale blue ppt dissolves.

(iii) The d orbitals of Cu2+ (aq) are split into two different energy level due to

presence of H2O ligands.

The d electron undergoes d-d transition and is promoted to a higher energy d

orbital.

During the process, red wavelength of light energy from the visible region of the

electromagnetic spectrum is absorbed and blue wavelength is transmitted which

appears as the colour observed.

57 (d) In haemoglobin molecule, Fe(II) ion is octahedrally bonded to five nitrogen atoms

and a water molecule.

H2O ligand may be replaced by an oxygen (O2) ligand in a reversible reaction

allowing the haemoglobin to carry oxygen from one part of body to another.

Stronger ligands like CO (or CN-) bond irreversibly with haemoglobin and this

prevents the haemoglobin from carrying oxygen.

58 (d)

59 (a)

(b) (i)

(ii)

(c) (i)

(ii)

60 (a) (i) NaCl: giant ionic crystal lattice, strong electrostatic forces of attraction between

oppositely charged Na+ & Cl- ions. Large amount of energy required to break the

strong ionic bonds hence high melting point.

SiCl4: non-polar simple covalent molecule, weak induced dipole-induced dipole

interactions between molecules. Small amount of energy required to break the

weak induced dipole-induced dipole interactions hence low melting point.

(ii) Refer to Chemical Periodicity notes

61 (a) (i) 23V 1s2 2s2 2p6 3s2 3p6 4s2

23V2+ 1s2 2s2 2p6 3s2 3p6 3d3

(ii) In the presence of ligands, the 3d orbitals are split into two groups of

slightly different energy levels. V2+ has a partially filled d orbitals, transition of

electrons between the two different levels take place with absorption of energy

in the visible spectrum. The wavelengths that are transmitted correspond to

the complementary colour observed.

In Ca2+, there is no d-d transition due to absence of 3d electrons (or absence

of partially filled 3d orbitals or electron transition takes place outside the

visible region), hence it appears colourless.

(b) The first ionization energy of the elements from scandium to copper remains relatively

invariant as it involves the removal of 4s electrons. The significant screening effect

of 3d electrons in the penultimate shell nullifies the increase in nuclear charge.

The transition elements have higher 1st ionization energy than Ca since they have

higher effective nuclear charge than Ca due to the fact that d electrons are poorly

shielded.

The increasing trend in third ionization energy is due to the increase in the effective

nuclear charge and relatively constant shielding effect as 3d electrons are now in

valence shell.

Third ionization energy of calcium is much higher than that of the other transition

elements as the electron to be removed is from inner 3p orbital which is closer to the

nucleus and has lower energy, held more strongly by the nucleus (or electron is

removed from a full octet to form Ca3+). Therefore, larger amount of energy is

needed to remove it.

The 3rd electron from Fe2+ is removed from an orbital containing a pair of electrons;

less energy is expected due to inter-electronic repulsion.

62 (b) (i) Pb(OH)2.PbCO3(s) 2PbO(s) + CO2(g) +H2O(g)

(ii) Mr of Pb(OH)2.PbCO3 = 508

% loss in mass = [(44.0 + 18.0)/508] x 100 = 12.2

(iii) The decomposition temperature for Pb(OH)2.PbCO3 is expected to be higher.

Pb2+ ion is larger in size than Ca2+ hence charge density is lower resulting in

lower polarizing power which leads to less distortion on anionic charge

clouds, hence white lead is more stable.

63 (a) Add excess NaOH(aq).

Only aluminum oxide will dissolve to form a soluble complex:

Al2O3 + 2NaOH + 3H2O 2NaAl(OH)4

The mixture is then filtered and the MgO residue is washed with distilled water.

64 (a) (i) NaCl – electrostatic attraction exist between oppositely charged ions in an ionic

lattice

PCl3 – covalent bonds exist between the P and Cl atoms to form discrete

molecules.

Large difference in electronegativity between Na and Cl, hence, complete

transfer of electrons – ionic compound; small difference in electronegativity

between P and Cl, hence, atoms share electrons – covalent compound.

Large amount of energy required to overcome the strong ionic bonds between

oppositely charged ions in NaCl, while much lesser energy is required to

overcome weak intermolecular forces of attraction / vdw forces between PCl3

molecules.

(ii) Electrical conductivity

NaCl – able to conduct electricity in molten state as the ions are free to carry the

charge, while PCl3 does not contain mobile charge carriers in molten state

(iii) [Mg(H2O)6]2+ [Mg(H2O)5(OH)]+ + H+

(iv) Observations: white fumes of HCl is produced

(at the beginning when water was first added)

Reaction equation: PCl3 + 3H2O H3PO3 + 3HCl

Or

PCl3 hydrolyses completely in water to form HCl and H3PO3.

pH between 0 to 2