All Inorganic

Periodic Properties of Elements I.

II.

Periodic Properties of Elements in the Periodic Table A.

Periodic variation in physical properties of the elements H to Ar (First 18 elements)

B.

Variation in melting point and boiling point

Periodic relationship among the oxides, chlorides and simple hydrides of the elements Li to Cl (Elements 3-17) A.

B.

C.

Hydride 1.

Bonding

2.

Hydrolysis

Chloride 1.

Bonding

2.

Hydrolysis

Oxide 1.

Formation of acidic, amphoteric and basic oxides a)

Amphoteric oxides

s-Block Elements I.

Characteristic properties of the s-block elements A. B. C.

II.

Variation in properties of the s-block elements and their compounds A.

B. C.

D.

III.

Metallic character and low electronegativity Ionic bond formation and fixed oxidation state Flame test Reaction of the elements 1. Reaction with hydrogen 2. Reaction with oxygen 3. Reaction with chlorine 4. Reaction with water Thermal stability of carbonates and hydroxides 1. Thermal stability of carbonates 2. Thermal stability of hydroxides Solubility of sulphate and hydroxide 1. Solvation energy a) Definition of solvation and solution processes b) Factors affecting the value of hydration energy c) Effect of lattice energy and hydration energy on solubility of ionic crystal 2. Solubility of some ionic crystals a) Sulphate salt of group II metal b) Hydroxide of group II metal c) Solubility of salts of other group II metals Abnormal behaviour of lithium

Uses of the compounds of the s-block elements A. B. C. D. E. F.

Manufacture of glass Baking powder Making of soap Antacid Slaked lime Fireworks

p-Block Elements I.

Halogens A. B. C. D.

E.

II.

Laboratory preparation of chlorine Characteristic properties of the halogens Properties of halogen 1. Oxidizing power of halogen 2. Disproportionation of halogen in alkalis Properties of the halogen compound 1. Reactions of halide a) with halogens b) with conc. sulphuric(VI) acid c) with conc. phosphoric(V) acid d) with silver ions 2. Acidic properties of hydrogen halide 3. Oxoacid Uses of halogens and halogen containing compounds 1. Fluoridation of water 2. Use of chlorine 3. Photographic films

Nitrogen and its compounds A. B.

Nitrogen 1. Unreactive nature of nitrogen 2. Direct combination of nitrogen and oxygen Ammonia 1. Haber process 2. Chemical properties of ammonia a) Alkaline properties (1)

b)

(1) (2) (3) (4) (5)

C.

D. E.

Laboratory preparation of ammonia

Reducing properties Combustion Catalytic oxidation With copper(II) oxide With chlorine Thermal decomposition of ammonia

Nitric(V) acid 1. Ostwald Process 2. Oxidizing properties of nitric(V) acid a) With copper b) With iron(II) ion c) With sulphur Nitrate(V) salt 1. Thermal decomposition of nitrate 2. Brown ring test Use of nitric(V) acid a) Fertilizers b) Explosives c) Nylon d) Dyestuffs

III.

Sulphur and its compounds A. B. C.

D. E. F.

Allotrope of sulphur Burning of sulphur Sulphur dioxide / sulphate(IV) salt 1. Reducing properties of sulphur dioxide a) With manganate(VII) ion b) With dichromate(VI) ion c) With bromine 2. Oxidizing properties of sulphur dioxide a) With magnesium 3. Distinguishing from carbon dioxide Concentrated sulphuric(VI) acid 1. Contact process 2. Properties of concentrated sulphuric acid Sulphate(VI) salt 1. Test for sulphate(VI) ion Use of sulphuric(VI) acid 1. Fertilizers 2. Detergents 3. Paints, Pigments and dyestuffs

d-Block Elements I.

General features of the d-block elements from Sc to Zn A. B. C.

II.

Electronic configuration of d-block elements Electronic configuration of ions of d-block elements 1. Stability of different ions Comparison of properties between d-block and s-block metals 1. Physical properties a) Ionization enthalpies and electronegativity b) Melting point and hardness c) Atomic radii of d-block elements d) Densities 2. Chemical properties a) Reaction with water

Characteristic properties of the d-block elements and their compounds: A.

B.

C. D.

E.

Variable oxidation states 1. Common oxidation states of vanadium and manganese compounds a) Vanadium compounds b) Manganese compounds Complex formation 1. Relative stability of complex ion a) Basicity of the ligand b) Chelate effect 2. Stability constant 3. Aquaton isomerism 4. Geometrical isomerism 5. Nomenclature of complexes Coloured ions 1. Origin of the colour 2. Colours of different ions Catalytic properties of transition metals and their compounds 1. Haber process 2. Reaction between peroxodisulphate(VI) and iodide ions 3. Decomposition of hydrogen peroxide Magnetic properties of transition metal and ion

Periodic Properties of Elements I.

II.

Periodic Properties of Elements in the Periodic Table A.

Periodic variation in physical properties of the elements H to Ar (First 18 elements)

B.

Variation in melting point and boiling point

Periodic relationship among the oxides, chlorides and simple hydrides of the elements Li to Cl (Elements 3-17) A.

B.

C.

Hydride 1.

Bonding

2.

Hydrolysis

Chloride 1.

Bonding

2.

Hydrolysis

Oxide 1.

Formation of acidic, amphoteric and basic oxides a)

Amphoteric oxides

Properties of compounds and bond energies Hydrolysis of Hyrides Hydride Reaction LiH(s) LiH(s) + H2O(l) → LiOH(aq) + H2(g) BeH2(s) BeH2(s) + 2H2O(l) → Be(OH)2(aq) + 2H2(g) B2H6(g) B2H6(g) + 6H2O(l) → 2H3BO3(aq) + 6H2(g) CH4(g) NH3(g) NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq) H2O(l) H2O(l) + H2O(l) d H3O+(aq) + OH-(aq) HF(g) HF(l) + H2O(l) d H3O+(aq) + F-(aq) HF(l) + H+(aq) d H2F+(aq) NaH(s) NaH(s) + H2O(l) → NaOH(aq) + H2(g) MgH2(s) MgH2(s) + 2H2O(l) → Mg(OH)2(aq) + 2H2(g) AlH3(g) AlH3(g) + 3H2O(l) → Al(OH)3(aq) + 3H2(g) SiH4(g) SiH4(g) + 4H2O(l) → Si(OH)4(s) or SiO2·nH2O(s) + 4H2(g) PH3(g) H2S(g) H2S(aq) + 2H2O(l) d H3O+(aq) + HS-(aq) + H2O(l) d 2H3O+(aq) + S2-(aq) HCl(g) HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq) Hydrolysis of Chlorides Chloride Reaction LiCl(s) BeCl(s) BeCl2(s) + 4H2O(l) → [Be(H2O)4]2+(aq) + 2Cl-(aq) [Be(H2O)4]2+(aq) + H2O(l) → [Be(H2O)3(OH)]+(aq) + H3O+(aq) BCl3(l) BCl3(l) + 3H2O(l) → H3BO3(aq) + 3HCl(aq) CCl4(l) NCl3(l) NCl3(l) + 3H2O(l) → NH3(aq) + 3HOCl(aq) Cl2O(g) Cl2O(g) + H2O(l) → 2HOCl(aq) ClO2(g) 2ClO2(g) + H2O(l) → HClO2(aq) + HClO3(aq) Cl2O6(l) Cl2O6(l) + H2O(l) → HClO3(aq) + HClO4(aq) Cl2O7(l) Cl2O7(l) + H2O(l) → 2HClO4(aq) ClF(g) ClF(g) + H2O(l) → HF(g) + HOCl(aq) NaCl(s) MgCl(s) MgCl(s) + warm H2O(l) → MgCl(OH)(s) + HCl(aq) AlCl3(s) AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl-(aq) [Al(H2O)6]3+(aq) + H2O(l) → [Al(H2O)5OH]2+(aq) + H3O+(aq) SiCl4(l) SiCl4(l) + 4H2O(l) → Si(OH)4(aq) + 4HCl(aq) or SiCl4(l) + (x+2)H2O(l) → SiO2·xH2O(aq) + 4HCl(aq) PCl3(l) PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(aq) PCl5(s) or PCl5(s) + H2O(l) → POCl3(l) + 2HCl(g) POCl3(l) + 3H2O(l) → H3PO4(aq) + 3HCl(aq) SCl2(g) SCl2(g) + H2O(l) → HSCl(aq) + HOCl(aq) 2S2Cl2(g) + 2H2O(l) → 3S(s) + SO2(g) + 4HCl(aq) Cl2(g) Cl2(g) + H2O(l) → HCl(g) + HOCl(aq)

Page 1 Remark

Insoluble in water

Slightly soluble in water

Remark Soluble in water Dissolve and hydrolyse in water

Immisicible with water

Soluble in water Dissolve and hydrolyse in warm water

In general, chlorides hydrolyse faster in alkaline medium since OH-(aq) ion is a stronger nucleophile than water. Furthermore, the presence of alkali also shifts the equilibrium position to the right.

Properties of compounds and bond energies Hydrolysis of Oxides Oxide Reaction Li2O(s) Li2O(s) + H2O(l) → 2LiOH(aq) BeO(s) B2O3(s) B2O3(s) + 3H2O(l) → B(OH)3(aq) B(OH)3(aq) + 2H2O(l) e H3O+(aq) + [B(OH)4]-(aq) CO2(g) CO2(g) + H2O(l) d H2CO3(aq) CO(g) d H+(aq) + HCO3-(aq) d 2H+(aq) + CO32-(aq) N2O5(g) N2O5(s) + H2O(l) → 2HNO3(aq) N2O4(g) N2O4(g) [or 2NO2(g)] + H2O(l) → HNO2(aq) + HNO3(aq) N2O3(g) N2O3(g) + H2O(l) → 2HNO2(aq) NO(g) N2O(g) O2(g) F2O(g) F2O(g) + H2O(l) → 2HF(aq) + O2(g) Na2O2(s) Na2O2(s) + 2H2O(l) → 2NaOH(aq) + H2O2(aq) MgO(s) MgO(s) + H2O(l) e Mg(OH)2(s) e Mg2+(aq) + 2OH-(aq) Al2O3(s) SiO2(s) P2O3(s) P2O5(s) SO2(g) SO3(g)

Cl2O(g) ClO2(g) Cl2O6(l) Cl2O7(l)

Page 2 Remark No reaction with water but it is amphoteric

Carbonic acid or carbonated water

NO(g) and N2O(g) are neutral and insoluble in water.

Insoluble in water neutral and insoluble but hydrolyses slowly. Na(s) forms peroxide instead of oxide when burns in air. MgO(s) only reacts with water very slowly due to high lattice energy. Not reaction with water but it is amphoteric Not reaction with water but reacts with strong base

P2O3(s) + 6H2O(l) → 4H3PO3(aq) P2O5(s) + 6H2O(l) → 4H3PO4(aq) SO2(g) + H2O(l) d H2SO3(aq) d H+(aq) + HSO3-(aq) d 2H+(aq) + SO32-(aq) SO3(g) + H2O(l) → H2SO4(aq) → H+(aq) + HSO4-(aq) d 2H+(aq) + SO42-(aq) Cl2O(g) + H2O(l) → 2HOCl(aq) 2ClO2(g) + H2O(l) → HClO2(aq) + HClO3(aq) Cl2O6(l) + H2O(l) → HClO3(aq) + HClO4(aq) Cl2O7(l) + H2O(l) → 2HClO4(aq)

In general, metal oxides are basic and non-metal oxides are acidic in nature. The basic oxides could be neutralized by acid while the acidic oxides could be neutralized by alkali. The exceptions are BeO(s) and Al2O3(s). They are called amphoteric because they react with both acid and alkali. In Acid Oxide BeO(s) Al2O3(s)

Reaction BeO(s) + 2H+(aq) → Be2+(aq) + H2O(l) Al2O3(s) + 6H+(aq) → 2Al3+(aq) + 3H2O(l)

Remark Amphoteric oxide Amphoteric oxide

In Alkali Oxide BeO(s) Al2O3(s)

Reaction BeO(s) + 2OH-(aq) + H2O(l) → [Be(OH)4]2-(aq) Al2O3(s) + 2OH-(aq) + 3H2O(l) → 2[Al(OH)4]-(aq)

Remark Amphoteric oxide Amphoteric oxide

Properties of compounds and bond energies Bond energy Bond energy Bond energy E(X–H) in kJmol-1 E(X–X) in kJmol-1 (with element) (with hydrogen) Li Be B C 347 413 N 391 158 (945.4 in N≡N) O 144 (498.3 in O=O) 464 F 158 568.0 Na 72 Mg Al Si 226 315 P 198 321 S 266 364 Cl 243.4 432.0 H 435.9 435.9

Page 3 Bond energy E(X–O) in kJmol-1 (with oxygen)

Bond energy E(X–Cl) in kJmol-1 (with chlorine)

358 214 144

346

466 469 464

432.0

Periodic Properties of Elements

Unit 1

Topic

Perioidic Properties of Elements

Reference Reading

8.0 Chemistry in Context, 3rd Edition ELBS pg. 37–41 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 43–45

Page 1

Unit 1

Assignment Syllabus Notes

I.

Periodic Properties of Elements in the Periodic Table

It is important to notice that the modern periodic table was developed by a Russian scientist Mendeléev in 1867 by comparing the physical and chemical properties of different elements. However, the electronic configuration of atoms were only discovered by Bohr in 1913. Indeed, the periodic table aided the discovery of electronic configuration of atom and some new elements. In the periodic table, elements are arranged according to the atomic number, not the atomic mass. For example, the atomic mass of 18Ar is 39.9 while the atomic mass of 19K is 39.1; the atomic mass of 52Te is 127.6 while the atomic mass of 53I is 126.9. With such arrangement, elements with similar chemical properties are grouped in columns called groups.

Furthermore, elements can be classified into s-block, p-block, d-block and f-block elements depending on the orbital that the vacant electrons are filling.

Glossary

Periodic Properties of Elements

Past Paper Question

Unit 1

Page 2

91 2B 4 a i 94 1A 2 f

91 2B 4 a i 4a i Briefly explain what is meant by periodicity in the chemical properties of the elements. Elements in same group have similar chemical properties. Properties recur after an interval (period). also accept “element” properties are periodic functions of atomic number

2 1 mark 1 mark

94 1A 2 f 2f The relative atomic mass of potassium is less than that of argon. State the reason why potassium follows argon in the Periodic Table. In the Periodic Table, the elements are arranged in order of atomic number / no. of proton, but not in order of atomic mass. ½ mark K has one more proton than Ar does. ½ mark

1

Periodic Properties of Elements

Unit 2

Topic

Perioidic Properties of Elements

Reference Reading

8.1 Chemistry in Context, 3rd Edition ELBS pg. 42–50, 191–201 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 4–21 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 333–336

Page 1

Unit 2

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 35–37, 42–49, 169, 176–177 Reading Syllabus Notes

A. Periodic variation in physical properties of the elements H to Ar (First 18 elements) First ionization energy, atomic radius, electronegativity, melting point and boiling points, all varies periodically across the periodic table. For the details of variation in first ionization energy, atomic radius and electronegativity, please refer to Atomic structure – Units 4 & 6.

Variation in electronegativity

1.

Variation in melting points and boiling point

Both melting point and boiling point reflects the strength of attraction among the particles. In general, boiling point serves as a better indication of the strength of attraction than melting point does. In the boiling process, the randomly arranged liquid particles are vaporized to randomly arranged gaseous particles. However, in the melting point process, the more regularly arranged particles are melted to randomly arranged liquid particles. Besides the attraction, melting point is also depending on the arrangement of particles in the structure.

Periodic Properties of Elements

Unit 2

Page 2

For example, boiling point of unbranced alkane increases smoothly with increasing molecular size. Although melting point also shows a general increasing trend, it forms an zig-zag line instead. This is because the alkane with even no. of carbon (except methane) are more symmetrical than the ones with odd no. of carbon. They are packed more regularly in the solid, thus a little bit higher melting point.

Key -

Glossary

s.m.s.

:

g.m.s.

:

g.c.s.

:

Simple molecular structure The forces among the molecules are weak van der Waals' forces, thus the m.p. and b.p. are rather low. P4 molecule is smaller than S8 molecule, therefore S has a higher m.p. and b.p. than P. Giant metallic structure The m.p. and b.p. of a metal is depending on the strength of the metallic bond. Giant covalent structure The attraction amount the atoms are strong covalent bond, thus all giant covalent structures have very high m.p. and b.p.

simple molecular structure

giant metallic structure

giant covalent structure

Periodic Properties of Elements

Past Paper Question

91 2B 4 a ii 92 2B 6 Aa ii

Unit 2

Page 3

92 2B 6 Ba ii

91 2B 4 a ii 4a ii Draw diagrams showing the variations in melting point (or boiling point) and first ionization energy, for the elements from lithium to argon. Explain the change in ionization energy between nitrogen and oxygen.

For melting point (or boiling point) Graph shows maxima for melting point at Group IV or for boiling point at either Group III or IV. Graph shows minima at Group V to Group 0 Graphs shows decreasing in melting point (or boiling point) down the group in Group III or IV (not labelling each axis -½)

For I. E. Graph shows decrease in ionization energy down the group Graph shows 2,3,3 pattern (not labelling each axis -½) Explanation: p3 for N (p4 for O not necessary) stability of half-filled subshell 92 2B 6 Aa ii 6Aa ii Describe the structure and the bonding in nitrogen, N2, and in white phosphorus, P4. N2: N≡N structure: (linear) diatomic molecule bonding: intramolecular, covalent intermolecular, van der Waals

5

1 mark ½ mark ½ mark

1 mark 1 mark ½ mark ½ mark

2 ½ mark ½ mark ½ mark ½ mark

P P

white P4: structure: bonding:

P

P

discrete tetrahedral intramolecular, single covalent intermolecular, van der Waals

92 2B 6 Ba ii 6Ba ii Describe the structure and the bonding in oxygen, O2, and in rhombic sulphur, S8.

½ mark ½ mark ½ mark ½ mark

4

Periodic Properties of Elements O2: O=O mark structure: (linear) diatomic molecule bonding: intramolecular, covalent intermolecular, van der Waals S

S S S S S S S

S8: structure: bonding:

molecular ring structure intramolecular, single covalent intermolecular, van der Waals

Unit 2

Page 4 ½ ½ mark ½ mark ½ mark ½ mark ½ mark ½ mark ½ mark

Periodic Properties of Elements

Unit 3

Page 1

Topic

Periodic Properties of Elements

Reference Reading

8.2.0 Chemistry in Context, 3rd Edition ELBS pg. 200–208 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 21–33 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 338–340

Unit 3

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 177–178 Reading Syllabus Notes

II. Periodic relationship among the oxides, chlorides and simple hydrides of the elements Li to Cl (Elements 3-17)

Besides similarity in chemical properties among the elements within the same group, some elements show similar chemical properties to another element in another group. e.g. Li and Mg; Be and Al, B and Si. This relationship is called diagonal relationship. The chemical properties of a particle is depending on its ability to attract or donate electron. i.e. electronegativity of the element and charge density of the ion. Consider lithium and magnesium, Li+ ion is smaller than Mg2+ ion but Z+ Mg2+ ion carries higher charges than Li+ ion. Consequently, the two ions share similar charge density( r ) and + polarizing power. This makes Li compound behave similarly as the Mg compound rather than other group I metal compounds. e.g. All group I carbonates are stable to heat except lithium carbonate. Lithium carbonate decomposes upon heating like magnesium carbonate. → Li2O(s) + CO2(g) Li2CO3(s) heat MgCO3(s) heat → MgO(s) + CO2(g) However, diagonal relationship does not apply to non-metals. This is because period 2 non-metals do not have low energy vacant d-orbital but period 3 non-metals can expand its octet and use the low energy vacant d-orbital to accept electrons from the others species. This explains why CCl4 is stable in water but PCl3 hydrolyse in water readily. water

→ no reaction CCl4(l)  PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(aq) Furthermore, the electronegativity difference among the non-metals are larger than the difference among the metals.

Glossary

diagonal relationship

charge density

polarizing power

electronegativity

Periodic Properties of Elements

Past Paper Question

Unit 3

Page 2

92 2B 5 a ii

92 2B 5 a ii 5a ii Explain why there is no apparent diagonal relationship in the properties of carbon and phosphorus. The question is difficult; reasonable answer might cite: C electronegativity (2.5); P (2.05) not similar Properties differ because of d-electron availability in P bonding. Charge density Polarizing power Any one; 1 mark C This question required application of knowledge to an unfamiliar situation, and was poorly answered by most candidates. Reasons such as the different electronegativities of C and P, the different charge densities, or the availability of d orbitals for bonding in P were considered reasonable.

1

Perioidic Properties of Elements

Unit 4

Topic

Perioidic Properties of Elements

Reference Reading

8.2.1 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 69–80 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 339, 352–353

Page 1

Unit 4

Assignment Reading Syllabus

Properties of hydrides

Notes

A. Hydride

1.

Bonding

s-block hydrides are ionic in nature, consisting of metal ion and hydride ions H-. p-block hydrides are covalent in nature consisting of simple molecule. For sure, the energetic stability of a compound is depending on the strength of the intramolecular forces. In general, period 2 hydrides are more stable than period 3 hydrides since the period 2 elements have smaller atomic sizes and form stronger bond with hydrogen.

Perioidic Properties of Elements 2.

Unit 4

Page 2

Hydrolysis

s-block hydrides are highly reactive toward water and the hydride ion hydrolyses readily in water to form an alkaline solution and hydrogen. δ− δ− O H δ+

-

H

H δ+

-

O H + H2

The properties of p-block hydrides are rather complicate. Even the hydrides of the elements in the same group do not show similar properties. e.g. CH4 is stable in water but SiH4 hydrolyses in water to give hydrogen. NH3 is alkaline but PH3 is neutral in water. H2S is acidic in water. Both HF and HCl are acidic in water but the acidity of HF is much lower. The reactivity of p-block hydrides towards water is depending on several factors : i. polarity of the molecule ii. availability of low energy vacant d-orbital iii. Bond strength and difference in electronegativity (bond polarity) i.

polarity of the molecule (NH3 vs PH3)

Comparatively, N is more electronegative than H but P has similar electronegativity as H. Although both of them are pyramidal in shape, PH3 is a less polar molecule. Consequently, NH3 is a much stronger Lewis base than PH3. δ+ H δ+ H N δ− H δ+

δ− δ− O H δ+ H δ+

δ− N

H

+ H N H + O H H

H H δ+ H δ+ δ+

H

P H

H

Finally NH3 will hydrolyze in water but PH3 is insoluble and has no reaction with water at all.

ii. availability of low energy vacant d-orbital (SiH4 vs CH4) C is a period 2 element which has no low energy vacant 2d-orbital to accept electrons from water. Si is capable to use its low energy vacant 3d-orbital to accept electrons from water to form a stable intermediate. Therefore, SiH4 hydrolyses very slowly in water. (It hydrolyses faster in alkali as OH- ion is a stronger nucleophile than water.) Furthermore, the electronegativity of C is similar to that of H while Si is less electronegativity than H. Therefore, the lone pair of water would be attracted by the positive charge on the Si atom. δ− H δ−H δ+Si H δ− H δ−

δ− O H δ+ H δ+

Perioidic Properties of Elements

Unit 4

Page 3

iii. Bond strength and difference in electronegativity (bond polarity) A low bond strength and high bond polarity favor the abstraction of proton from the molecule by water. This makes the solution acidic. H2O H2S HCl HF

Difference in E.N. 1.5 0.3 1.5 2.0

Bond energy (kJmol-1) + 463 + 364 + 432 + 568

Hydrolysis in water H2O(l) + H2O(l) d H3O+(aq) + OH-(aq) H2S(aq) + 2H2O(l) d 2H3O+(aq) + S2-(aq) HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq) HF(l) + H2O(l) d H3O+(aq) + F-(aq) HF(l) + H+(aq) d H2F+(aq)

However, sometimes, it is difficult to predict whether a molecule will behave as a base or an acid. As an acid As a base

H–X + H–O–H d X- + H2O–H+ H–X + H–O–H d H2X+ + OH-

NH3(aq) + H2O(l) d NH2-(aq) + H3O+(aq) NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq)

Indeed, it is mainly depending on the relative strength of H–X and O–H bonds but it is also depending on the hydration energy of the ions formed. The acidity of HF is exceptionally low comparing with other hydrogen halide due to 2 factors : i. strong H–F bond, which does not favour the dissociation of the bond. ii. the formation of hydrogen bond between HF(l) and H+(aq), this lowers the concentration of free H+(aq) in the solution, thus the acidity.

Glossary

hydride

Past Paper Question

91 1A 3 a 93 2B 4 b iii 95 1A 2 d i ii iii 99 2A 3 c ii iii

91 2B 5 b 93 2B 6 d

91 1A 3 a 3a Explain why NaH(s) is more reducing than HCl(g) whilst HCl(g) is more acidic than NaH(s). NaH(s) is an ionic compound and its reducing properties is due to the hydride anion which can undergo the reaction, H- → ½H2 + e-, i.e. releasing electrons in an overall redox reaction. 1 mark ½ mark HCl(g) does not undergo such reaction to release electron. HCl(g) is acidic because it is a proton donor in the presence of a base. e.g. HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq). 1 mark ½ mark NaH(s) is not acidic because it is not a proton donor. 91 2B 5 b 5b Explain why SiH4 is hydrolysed by alkali, whereas CH4 is not, and give an equation for the hydrolysis. Reason A: electronegativity difference 1 mark ∴ nucleophilic attack by OH- only on Si 1 mark Reason B: state any one of the X–O, X–H bond strengths 1 mark activation energy higher for breaking C-H 1 mark Reason C: C no 2d orbital, Si has 3d (or accept C has no d orbitals) 1 mark ∴ Si can form activated complex (or Si accept lone pair/nucleophile) 1 mark Any one of the 3 reasons acceptable. Answer must have the fact (1 mark) and then deduction (1 mark). If 2 facts from different reasons are mixed, then award only 1 mark Equation 1 mark SiH4 + (n+2)H2O → SiO2⋅nH2O + 4H2 (SiH4 + 2OH- → SiO32- + 2H2O + H2) or SiH4 + 2H2O → SiO2 + 4H2 (SiH4 + H2O →Si(OH)4 + H2 (need not be balanced. Accept SiO32-. Miss out H2 -½)

3

3

Perioidic Properties of Elements Unit 4 93 2B 4 b iii 4b Account for each of the following facts: iii Carbon forms stable hydrides with different structures whereas silicon forms only unstable hydrides. Carbon atom because of its small size (less shielding effect) forms stable covalent bond with other carbon atoms ∴ tends to catenate. 1 mark Energy difference between C–C, C–H and C–O is small, while for Si, Si–Si, Si–H are weak compare with Si–O, ∴ Si–Si and Si–H bond is easily oxidized to Si–O, while C–C and C–H bond is not. 1 mark C can also form π bond by sideway overlapping of p orbitals, thus formation of C=C and C≡C are possible. For Si, because of its large size, Si=Si and Si≡Si cannot be formed easily. 1 mark 93 2B 6 d 6d Give the states of NaH and HCl at room temperature and atmospheric pressure. Explain why they are different. Compare the reaction between NaH and water with that between HCl and water. NaH - solid. It is an ionic compound, a crystalline solid with the structure similar to NaCl. 1½ mark HCl - gas. It is covalent molecule where the intermolecular force is dipole-dipole attraction. 1½ mark Reaction with water NaH: H- + H2O → H2 + OH- (result in alkaline solution) HCl: HCl + H2O → H3O+ + Cl- (result in acidic solution) 1 mark

Page 4

3

4

95 1A 2 d i ii iii 2d i State whether the following hydrides would react with water under room temperature and atmospheric pressure. If 3 there is a reaction, give the balanced equation. CH4 SiH4 H2S HCl CH4 : no reaction ½ mark SiH4 : SiH4 + 4H2O → Si(OH)4 + 4H2 (or SiO2·nH2O or SiO2) (½ mark for correct product, ½ mark fro balanced reaction) H2S + 2H2O d 2H3O+ + S2H2S : H2S + H2O d HS- + H3O+ OR (½ mark for correct product, ½ mark fro balanced reaction) ½ mark HCl : HCl + H2O → H3O+ + Clii Explain the difference in reactivity with water between CH4 and SiH4. 1½ Kinetic aspect ½ mark In CH4, there is almost no polarity in the C–H bonds ∴ CH4 is not affected by polar molecules of water In SiH4, the bond polarity is Siδ+–Hδ-. ½ mark OR ½ mark There is no empty / vacant low energy d-orbital in C ∴ lone e- pair in water will not attack CH4. ½ mark OR ½ mark In Si, there is low energy empty d-orbital. ∴ it can accept lone electron pair from water ½ mark H

H δ+ Si Hδ−

H H

H

Si OH + H2

etc.

H O H H

Thermodynamic aspect C–H bond is strong when compared with Si–H bond ∴substitution of H in CH4 is energetically unfavourable. (Max. 1½ mark, 3 points must not be in the same aspect) iii Explain the difference in reactivity with water between SiH4 and H2S. In H2S, the polarity is Hδ+–Sδ-. ∴ nucleophilic attack of H2O on H2S gives H3O+ and HS-, whereas SiH4 gives an alkaline solution since the polarity is Siδ+–Hδ-.

½ mark ½ mark ½ mark 1½ ½ mark ½ mark ½ mark

Perioidic Properties of Elements Unit 4 99 2A 3 c ii iii 3c Consider the hydrides of three Period 3 elements : SiH4, PH3 and H2S ii With reference to the electronegativity values given below, compare the reactions of these hydrides with water. Element H O Si P S Electronegativity (Pauling's scale) 2.1 3.5 1.8 2.1 2.5 iii State, with explanation, whether SiH4 or H2S has a higher boiling point. (9 marks)

Page 5

Perioidic Properties of Elements

Unit 5

Page 1

Topic

Perioidic Properties of Elements

Unit 5

Reference Reading

8.2.2 Chemistry in Context, 3rd Edition ELBS pg. 202–203 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 25–29, 50-68 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 403–405 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 179–180

Assignment Advanced Practical Chemistry, John Murray (Publisher) Ltd., 144–147 Reading Syllabus Notes

B. Chloride

1.

Bonding

The bonding of chlorides is similar to that of hydrides. s-block chlorides are ionic in nature and p-block chlorides are covalent in nature. Chlorides of phosphorus are very special comparing with the others. PCl3(l) is a liquid possessing molecular structure where PCl5(s) is a solid possessing ionic structure. However, in gaseous state, PCl5(g) also possesses a molecular structure.

+ Cl P Cl

Cl Cl

P PCl3(l)

Cl

Cl Cl Cl

-

Cl

Cl

Cl Cl

P Cl

Cl Cl

PCl5(s)

P Cl

Cl Cl PCl5(g)

Perioidic Properties of Elements 2.

Unit 5

Page 2

Hydrolysis

s-block chlorides are basically ionic and only dissolve in water without going through hydrolysis. However, owing to the high charge density of the metal ion, the highly covalent BeCl2, BCl3 and AlCl3 hydrolyse readily in water. [Be(H2O)4]2+(aq) + H2O(l) → [Be(H2O)3(OH)]+(aq) + H3O+(aq) BCl3(l) + 3H2O(l) → H3BO3(aq) + 3HCl(aq) AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl-(aq) [Al(H2O)6]3+(aq) + H2O(l) d [Al(H2O)5OH]2+(aq) + H3O+(aq) [Al(H2O)5OH]2+(aq) + H2O(l) d [Al(H2O)4(OH)2]+(aq) + H3O+(aq) [Al(H2O)4(OH)2]+(aq) + H2O+(l) d [Al(H2O)3(OH)3](s) + H3O+(aq) N.B.

For simplicity, sometimes the ligand water will be omitted is the formula, the formula [Al(H2O)3(OH)3](s) would be written as Al(OH)3(s) despite of the fact that Al3+ usually co-ordinates with 6 ligands.

The reactivity of p-block chlorides towards water is similar to that of hydrides. The main difference is that Cl is much more electronegative than H. Therefore, the central atom will always carry the δ+ instead of δ-.

Glossary

ligand

Past Paper Question

90 2B 5 a 91 1A 3 b i 92 1A 3 e 93 1A 2 a iv 94 1C 2 96 1A 2 e i

91 2B 4 b

90 2B 5 a 5a Give the structures of, and indicate the types of bonding in, LiCl, AlCl3, PCl3 and K2NiCl4. Also, discuss their physical and chemical properties in terms of structure and bonding. LiCl colourless ionic solid with Li+, Cl-. Cl

AlCl3

colourless covalent solid, dimeric in structure, Cl

PCl3

covalent colourless liquid,

P Cl

Cl

Al

Cl Cl

Al

Cl Cl

, undergoes sublimation.

Cl

, can be oxidized to POCl3. Cl

Cl

2-

Ni Cl

Cl

. K2NiCl4 coloured ionic solid with complex anion, LiCl and K2NiCl4 are ionic solids whereas AlCl3 and PCl3 are covalent molecules. LiCl and K2NiCl4 do not react with water. Whereas AlCl3 and PCl3 are hygroscopic and readily hydrolysed by water. About the bonding: LiCl, ionic chloride AlCl3, covalent Cl- as ligand covalent bond between P and Cl PCl3, NiCl42-, dative covalent bond. Physical properties e.g. melting point/boiling point; sublimation; solubility in water/organic solvent; colour of complex, etc. Chemical properties e.g. hydrolysis; ease of oxidation; complex ion in equilibrium with water, etc. Marking Scheme Overall presentation - highlight on their differences (in table form -1) 2 marks Structure and bonding (1 mark each, ½ for structure and ½ for bonding) 4 marks Physical and chemical properties (1 mark each, ½ for physical and ½ for chemical) 4 marks

10

Perioidic Properties of Elements Unit 5 91 1A 3 b i 3b Write equation(s) to describe the reaction of i ZnCl2(s) with water. ZnCl(s) + 4H2O(l) → [Zn(OH2)4]2+(aq) + 2Cl-(aq) [Zn(OH2)4]2+(aq) + H2O(l) → [Zn(OH2)3(OH)]+(aq) + H3O+ or ZnCl2 + H2O → Zn(OH)Cl + HCl

Page 3

1

1 mark

91 2B 4 b 4b Give the formula of one chloride for each of the elements N, O and S. Write an equation for the reaction (if any) of each chloride with water. Draw the electronic structure of a molecule of each chloride and describe the molecular shape. formula 3 × ½ = 1½ mark equation 3 × ½ = 1½ mark (need not be balanced, must write + H2O) electronic structure 3 × ½ = 1½ mark (must show lone pair(s) on N, O and S) shape 3 × ½ = 1½ mark (draw and/or name, if either one is correct, then ½ mark) Example: NCl3 NCl3(l) + 3H2O(l) → NH3(aq) + HOCl(aq) Cl

6

N Cl Cl

pyramidal Example: Cl2O Cl2O(g) + H2O(l) → 2HOCl(aq) Cl

O

Cl

V-shaped or 6ClO2 + 3H2O → 3HClO2 + 3HClO3 or Cl2O6 + H2O → HClO4 + HClO3 ClO3 is the same as Cl2O6, which is a solid ClO2+ClO4ClO2 + H2O → stable in dark sunlight 6ClO2(g) + 3H2O(l)  → HCl(aq) + 5HClO3(aq) Lewis and Waller give: ClO2(g) + 3H2O(l) → 2ClO-(aq) + 2H3O+(aq) 2ClO2(g) + 3H2O(l) → ClO2-(aq) + ClO3-(aq) + 2H3O+(aq) Cl2O6(l) + 3H2O(l) → ClO3-(aq) + ClO4-(aq) + 2H3O+(aq) Cl2O4(l) + 3H2O(l) → 2ClO4-(aq) + 2H3O+(aq) Cl2O7(l) + H2O(l) → 2HClO4(aq) For oxygen chlorides, accept disproportionation reaction for oxidation states 2,4,6 but not -1 or 7 for Cl. For ClO2 accept eitherClO2 + H2O → HCl + HClO3 or ClO2 + H2O → stable (hydrate) ClO2 + H2O → HClO2 + HClO3 Example: S2Cl2(l) 2S2Cl2(l) + 2H2O(l) → 3S(s) + SO2(aq) + 4HCl(aq) Cl

S

S

Cl

any sensible description e.g. zigzag shape or SCl2, SCl4 Cl S Cl

Cl S Cl

Cl Cl

For S2Cl2 + H2O → S + SO2 (or SO3) don’t need HCl For SCl4, reasonable answer to give SO2 For SCl2 + H2O → HOCl + HSCl (or accept S + SO2 + HCl) 92 1A 3 e 3e Both FeCl2 and FeCl3 undergo hydrolysis in water. Write equations for these reactions and explain why FeCl3 gives the more acidic solution. Both aqueous solutions contain hydrated cations and Cl-(aq). [Fe(H2O)6]2+ + H2O → [Fe(H2O)5OH]+ + H3O+ (pKa = 9.5) 1 mark [Fe(H2O)6]3+ + H2O → [Fe(H2O)5OH]2+ + H3O+ (pKa = 2.2) 1 mark 1 mark Smaller size and greater charge of Fe3+ ion results in more acidic solution than that of Fe2+ ion.

3

Perioidic Properties of Elements Unit 5 93 1A 2 a iv 2a Answer the following questions by choosing in each case one of the species listed below, putting it in the box and giving the relevant equation(s). Al(s), AlCl3(s), AlO2-(aq), Na(s), CO32-(aq), Cu2+(aq), P4O10(s), S(s), S2O32-(aq), Zn2+(aq) iv Which species fumes in moist air? AlCl3(s) + 3H2O(l) → Al(OH)3(s) + 3HCl(g) 1 mark AlCl3(s) 96 1A 2 e i 2e Which is the stronger acid in each of the following pairs of substances ? Briefly explain your choice. i [Fe(H2O)6]2+(aq) , [Fe(H2O)6]3+(aq) [Fe(H2O)6]3+ ½ mark ½ mark The conjugate base [Fe(H2O)5OH]2+ is stabilized to a greater extent ½ mark because the anionic ligand OH- tends to stabilize the electron deficient Fe3+ more than Fe2+ / causes the eqm. to lie towards the right / increase the ease of ionization of H ½ mark Or, ½ mark Fe3+ has a higher charge to radius ratio / charge density, ½ mark greater polarizing power, and hence the H develops larger partial +ve charge.♣ ½ mark

Page 4

2

2

Perioidic Properties of Elements

Unit 6

Page 1

Topic

Perioidic Properties of Elements

Unit 6

Reference Reading

8.2.3 Chemistry in Context, 3rd Edition ELBS pg. 204–208 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 27, 29–31, 34–49 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 425–426

Assignment Advanced Practical Chemistry, John Murray (Publisher) Ltd., 151–153 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 180–184, 283–285 Reading Syllabus Notes

C. Oxide

1.

Formation of acidic, amphoteric and basic oxides

Some oxides are acidic. e.g. CO2(g), SO2(g). When reacted with water, they form acids. e.g. CO2(aq) + H2O(l) d H2CO3(aq) SO2(aq) + H2O(l) d H2SO3(aq)

H O H

H

O S

H

O

H

O O

S O

O H O

S O

Perioidic Properties of Elements

Unit 6

Page 2

Some oxides are basic. e.g. Na2O(s), K2O(s). When reacted with water, they forms alkalis. e.g. Na2O(s) + H2O(l) → 2NaOH(aq) K2O(s) + H2O(l) → 2KOH(aq)

O

2-

H O

2

O H

H In general, non-metal oxides are acidic and metal oxides are basic.

A. Amphoteric oxides There are a few oxides which reacts with both acid and alkali. e.g. ZnO(s), PbO(s) and Al2O3(s. ZnO(s) + 2H+(aq) → Zn2+(aq) + H2O(l) ZnO(s) + 2OH-(aq) + H2O(l) → [Zn(OH)4]2-(aq) white zincate ion (colourless) PbO(s) + 2H+(aq) → Pb2+(aq) + H2O(l) PbO(s) + 2OH-(aq) + H2O(l) → [Pb(OH)4]2-(aq) white plumbite ion (colourless) Al2O3(s) + 6H+(aq) → 2Al3+(aq) + 3H2O(l) Al2O3(s) + 2OH-(aq) + 3H2O(l) → 2[Al(OH)4]-(aq) (to be more accurate [Al(H2O)2(OH)4]-(aq)] white aluminate ion (colourless) The hydroxides of zinc, aluminium and lead are also amphoteric. Zn(OH)2(s) + 2H+(aq) → Zn2+(aq) + H2O(l) Zn(OH)2(s) + 2OH-(aq) → [Zn(OH)4]2-(aq) white colourless Pb(OH)2(s) + 2H+(aq) → Pb2+(aq) + H2O(l) Pb(OH)2(s) + 2OH-(aq) → [Pb(OH)4]2-(aq) white colourless Al(OH)3(s) + 3H+(aq) → Al3+(aq) + 3H2O(l) Al(OH)3(s) + OH-(aq) → [Al(OH)4]-(aq) white colourless Since Zn(OH)2(s), Pb(OH)2(s) and Al(OH)3(s) are insoluble in water. If a little alkali (e.g. NaOH(aq)) is added into the Zn2+(aq), Pb2+(aq) or Al3+(aq) solution. The hydroxides will be precipitated first. If excess alkali is added, the hydroxide precipitate will redissolve to form a clear solution. This serves as a very good test for the three ions. e.g. Zn2+(aq) + 2OH-(aq) → Zn(OH)2(s) colourless white Zn(OH)2(s) + 2OH-(aq) → [Zn(OH)4]2-(aq) white excess colourless For a similar result, Al, Pb and Zn metals also reacts directly with strong alkali besides strong acid. 2Al(s) + 2OH-(aq) + 6H2O(l) → 2[Al(OH)4]-(aq) + 3H2(g) Zn(s) + 2OH-(aq) + 2H2O(l) → [Zn(OH)4]2-(aq) + H2(g) Pb(s) + 2OH-(aq) + 2H2O(l) → [Pb(OH)4]2-(aq) + H2(g)

Perioidic Properties of Elements

Unit 6

Glossary

acidic oxides

amphoteric oxides

Past Paper Question

92 2B 5 b iii 93 1A 2 a i 95 1A 2 b ii 95 1B 4 g ii

92 2B 6 Ac ii

Page 3

basic oxides 92 2B 6 Bc ii

92 2B 5 b iii 5b iii Write equations showing the amphoteric nature of lead(II) oxide. PbO(s) + 2HCl(aq) → PbCl2(s) + H2O(l) H2O(l) + PbO(s) + NaOH(aq) → NaPb(OH)3(aq) 92 2B 6 Ac ii 6Ac ii Which is more reactive to oxidation, NH3 or PH3? Explain your answer. PH3, Greater stability of P(V) : P4O10 92 2B 6 Bc ii 6Bc ii Which is more reactive to oxidation, H2O or H2S? Explain your answer. H2S any reasonable answer, such as more higher oxidation states available for S or enthalpy of formation of H2O is strongly exothermic (∆Hf -285.9 kJmol-1) but H2S only weakly so (∆Hf -20.6 kJmol-1)

2 1 mark 1 mark

2 1 mark + 1 mark

2 1 mark

1 mark

93 1A 2 a i 2a Answer the following questions by choosing in each case one of the species listed below, putting it in the box and giving the relevant equation(s). Al(s), AlCl3(s), AlO2-(aq), Na(s), CO32-(aq), Cu2+(aq), P4O10(s), S(s), S2O32-(aq), Zn2+(aq) i Which species can react with excess dilute NaOH with effervescence, forming a solution which gives a white precipitate on addition of dilute HCl? 2Al(s) + 2OH-(aq) + 2H2O(l) → 2AlO2-(aq) + 3H2(g) Al(s) AlO2-(aq) + H+(aq) + H2O(l) → Al(OH)3(s) OR 2OH-(aq) + 2Al(s) + 6H2O(l) → 2[Al(OH)4]-(aq) + 3H2(g) 2[Al(OH)4]-(aq) + 2H+(aq) → Al(OH)3(s) + 2H2O(l) 3 marks 95 1A 2 b ii 2b ii Is the reaction below a "redox reaction" or an "acid-base reaction" ? Explain. CaO + SiO2 → CaSiO3 This is an acid-base reaction. CaO is a basic oxide and SiO2 is an acidic oxide The reaction does not involve a change in oxidation state.

3

2 ½ mark 1 mark ½ mark

95 1B 4 g ii 4g Give the observations when (ii) sodium hydroxide solution are added dropwise, until in excess, to aqueous solutions containing Pb2+ and Cu2+ ions respectively. These experiments are carried out at room temperature. (ii) sodium hydroxide solution Pb2+(aq) Cu2+(aq) (3 marks for 8 observations; deduct ½ mark for each incorrect observation) (ii) sodium hydroxide solution 2+ Pb (aq) White ppt. Soluble in excess / Colour solution Cu2+(aq) Blue ppt. Insoluble in excess

3

s-Block Elements I.

Characteristic properties of the s-block elements A. B. C.

II.

Variation in properties of the s-block elements and their compounds A.

B. C.

D.

III.

Metallic character and low electronegativity Ionic bond formation and fixed oxidation state Flame test Reaction of the elements 1. Reaction with hydrogen 2. Reaction with oxygen 3. Reaction with chlorine 4. Reaction with water Thermal stability of carbonates and hydroxides 1. Thermal stability of carbonates 2. Thermal stability of hydroxides Solubility of sulphate and hydroxide 1. Solvation energy a) Definition of solvation and solution processes b) Factors affecting the value of hydration energy c) Effect of lattice energy and hydration energy on solubility of ionic crystal 2. Solubility of some ionic crystals a) Sulphate salt of group II metal b) Hydroxide of group II metal c) Solubility of salts of other group II metals Abnormal behaviour of lithium

Uses of the compounds of the s-block elements A. B. C. D. E. F.

Manufacture of glass Baking powder Making of soap Antacid Slaked lime Fireworks

s-Block Elements

Unit 1

Page 1

Topic

s-Block Elements

Unit 1

Reference Reading

9.0–9.1 Chemistry in Context, 3rd Edition ELBS pg. Inorganic Chemistry, 4th Edition, T.M. Leung pg. 84–86 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 360–361

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 211–213, 283 Reading Syllabus Notes

I.

Characteristic properties of the s-block elements

A. Metallic character and low electronegativity All s-block elements are soft metals. Since the outermost shell has been effectively shielded from the nuclear attraction, the metallic bond would not be very strong and the electronegativity would also be very low (i.e. very high electropositivity.) B. Ionic bond formation and fixed oxidation state s-block elements lose their outermost electron readily to form positive ions. In general, they only form ionic compound. For the group 1 elements, the difference between the 1st I.E. and 2nd I.E. is very large, therefore group 1 elements only form ions with +1 oxidation state. Similarly, for group 2 elements, the difference between 1st and 2nd I.E is very small but the difference between 2nd I.E. and 3rd I.E. is very larger, group II elements only form ions with +2 oxidation state. Comparing with Fe, a transition metal, the difference between 1st and 2nd I.E. is small but the difference between 2nd I.E. and 3rd I.E. is moderate. Fe is capable to form ions with oxidation state +2 and +3. C. Flame test All s-block metal or compound containing s-block elements burning with characteristic colour due to the excitation of the electrons. The colour of the flame can be used to identify the metal or metal ion that the sample contains. Generally speaking, chloride salt is lowest melting and boiling among different salts. Therefore, it order to obtain the best result, it would be better to heat the sample with the bunsen flame in the presence of conc. HCl(aq) so the sample would vaporize more readily. Furthermore, conc. HCl(aq) is also used to clean the nichrome or platinum wire to be used in the flame test. This is done by dipping the wire in the acid and heating the wire with bunsen flame repeatedly until no colored flame is observed. Flame colour of some metals / metal salts

Glossary

Group I metals Lithium (Li) Sodium (Na) Potassium (K) Rubidium (Rb) Caesium (Cs)

Deep red Yellow Lilac (a purple colour) Bluish Red Blue

Other metals Copper (Cu)

Bluish green

nichrome wire

flame test

Group II metals Beryllium (Be) Magnesium (Mg) Calcium (Ca) Strontium (Sr) Barium (Ba)

no special colour no special colour Brick red Blood red / Crimson red Apple green

s-Block Elements

Past Paper Question

Unit 1

Page 2

94 1A 2 e 96 2B 4 b ii 99 1B 7 c i ii

94 1A 2 e 2e Why is caesium, rather than sodium, used in a photo-electric cell? (A photo-electric cell is a device which produces electrons when irradiated by photons.) Cs, having a lower I.E., can be ionized easily by absorption of visible light ∴ it is more suitable to be used in making photo-electric cell. 1 mark 96 2B 4 b ii 4b For each of the following pairs of compounds, suggest a test to distinguish one compound from the other : ii KCl(s) and MgCl2(s). Carry out a flame test, 1 mark ½ mark KCl(s) gives a lilac flame ½ mark while MgCl2(s) gives no characteristic flame colour Or 1 mark Add NaOH(aq) / Na2CO3(aq) / Na2HPO4(aq) to aqueous solution of the substance. ½ mark MgCl2 gives white precipitate ½ mark while KCl does not. (Accept any CORRECT test for the two compounds) 99 1B 7 c i ii 7c i Describe the procedure for a flame test. ii Describe how to distinguish between a sodium salt and a potassium salt of the same anion by flame test.

1

2

s-Block Elements

Unit 2

Topic

s-Block Elements

Reference Reading

9.2.0–9.2.2 Chemistry in Context, 3rd Edition ELBS pg. 242–243 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 94–98, 103–105 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 363–367

Page 1

Unit 2

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 213–214 Reading Syllabus

Reactivities of s-block elements Thermal stability of carbonates and hydroxide of s-block elements.

Notes

II. Variation in properties of the s-block elements and their compounds A. Reaction of the elements 1.

Reaction with hydrogen

s-block elements are very electropositive, even more electropositive than hydrogen. When heated with hydrogen, ionic hydride will be formed. e.g. 2Na(s) + H2(g) → 2NaH(s) 2.

Reaction with oxygen

Ordinary s-block elements are very reactive towards oxygen. They almost tarnish immediately to form metal oxide when exposed to air. e.g. 4Na(s) + O2(g) → 2Na2O(s) (white powder) 2Ca(s) + O2(g) → CaO(s) N.B.

Aluminium seems to be less reactive than the other metals in air because it is protective by a thin but dense oxide layer. Once the oxide layer have been removed by dipping in mercury(II) chloride, aluminium also reacts vigorously to form white powder of Al2O3(s) with air.

Besides the normal oxides, sodium forms peroxide, potassium and other more reactive group I metals form superoxides when burn in air. e.g. 2Na(s) + O2(g) → Na2O2(s ) (white when cold but yellow when hot) K(s) + O2(g) → KO2(s) (yellow powder) The hydrolysis of peroxide and superoxide is also different from that of normal oxide Normal oxide M2O(s) + H2O(l) → 2MOH(aq) Peroxide M2O2(s) + 2H2O(l) → 2MOH(aq) + H2O2(aq) or 2M2O2(s) + 2H2O(l) → 4MOH(aq) + O2(g) Superoxide 2MO2(s) + 2H2O(l) → 2MOH(aq) + H2O2(aq) + O2(g) or 4MO2(s) + 2H2O(l) → 4MOH(aq) + 3O2(g) 3.

Reaction with chlorine

Electropositive s-block elements also react with electronegative chlorine to form ionic chloride. Group I Group II

2M(s) + Cl2(g) → 2MCl(s) M(s) + Cl2(g) → MCl2(s)

s-Block Elements

Unit 2 4.

Page 2

Reaction with water

s-block elements, especially group I elements, reacts readily with water with a reactivity increasing down the group. They react with water to form an solution of metal hydroxide and hydrogen. 2M(s) + H2O(l) → 2MOH(aq) B. Thermal stability of carbonates and hydroxides 1.

Thermal stability of carbonates

Generally speaking, carbonate is unstable to heat. Upon heating, most carbonates decompose into oxides and carbon dioxide. → M2O(s) + CO2(g) M2CO3(s) heat heat MCO3(s)  → MO(s) + CO2(g)

-

O

O C O

C O -

+

O

2-

O

Depending on the polarizing power (i.e. charge density) of the cation, different carbonates have different stability to heat. A carbonate containing a more polarizing cation is less stable to heat. This is because the electron cloud of the carbonate ion will be more distorted. Furthermore, upon decomposition, it will form a more stable oxide with stronger ionic bond within the lattice. The evolution of CO2 gas will also increase the disorderness of the system which is favourable in nature. The temperature of a normal bunsen flame is about 1300 K. A subtance having a decomposition temperature higher than this temperature is usually regarded as stable to heat. Comparatively, group I ions are much bigger than group II ions, except Li2CO3(s), group I carbonate is regarded as stable to heat. → Li2O(s) + CO2(g) Li2CO3(s) heat

Among the group II carbonates, the decomposition temperature increases down the group as the polarizing power of the cation decreases. Group II carbonates BeCO3(s) MgCO3(s) CaCO3(s) SrCO3(s) BaCO3(s)

Decomposition temperature / ºC ~ 100 540 900 1290 1360

s-Block Elements

Unit 2 2.

Page 3

Thermal stability of hydroxides

The decomposition of hydroxides is somehow similar to the decomposition of carbonates. Upon decomposition, the large hydroxide ion is converted to small oxide ion with liberation of water. Hydroxides of group 1 (except Li) are stable to heat. Those of Group 2 and lithium are decomposed by heat to form oxides. The thermal stability of group II hydroxide is parallel to group II carbonates where the thermal stability increases down the group. → M2O(s) + H2O(l) 2MOH(s) heat M(OH)2(s) heat → MO(s) + H2O(l)

-

O H O H

-

H

O H

Glossa ry

oxide

peroxide

Past Paper Question

90 1A 3 e ii 91 2B 5 d 92 1A 3 a 93 2B 4 b i 95 2B 5 a v 96 2B 4 b i 99 2A 1 a iii

superoxide

+

O

2-

thermal stability

90 1A 3 f

95 2B 5 b i ii

90 1A 3 e ii 3e Give observation, and write balanced equations for the following: ii a small, freshly-cut piece of sodium is exposed to air for a few minutes; a similar piece of sodium is placed on filter paper on the surface of water in a glass trough. The surface of the sodium becomes dull (less shiny, more white) 4Na(s) + O2(g) → 4Na2O(s) The piece of sodium melts and burns with a yellow flame. 2Na(s) + 2H2O(l) → 2Na+(aq) + 2OH-(aq) + H2(g) 2H2(g) + O2(g) → 2H2O(l)

4 ½ mark 1 mark ½ mark 1 mark 1 mark

90 1A 3 f 3f Account for the observation that when a piece of aluminium foil is dipped into a solution of mercury(II) chloride and then withdrawn, it becomes very hot and disintegrates into a white powder. The protective layer of Al2O3 is destroyed when the piece of Al is dipped in a solution of HgCl2. 1 mark 1 mark When withdrawn from the solution, the fresh surface of Al reacts highly exothermically 1 mark with O2(air) to form white Al2O3 powder. 91 2B 5 d 5d Account for the observation that lithium carbonate decomposes at 700ºC whereas sodium carbonate is thermally stable at this temperature. Li+ is more polarizing than Na+ 1 mark because it is smaller, with same charge or higher charge density 1 mark The charge cloud of the oxy-anion is more distorted, causing the anion to break up to give O2- and CO2. O 2M+ + -

O

C

O

-

2M+O2- + O C O

Change of lattice energy from Li2CO3 to Li2O is greater than Na2CO3 to Na2O.

1 mark 2 mark

3

3

s-Block Elements Unit 2 92 1A 3 a 3a Give balanced equations for two different reactions that are common to the three elements: lithium, sodium and potassium. Use the symbol M for the elements. 1 mark 2M(s) + Cl2(g) → 2MCl(s) 2M(s) + 2H2O(l) → 2M+(aq) + 2OH-(aq) + H2(g) 1 mark

Page 4 2

93 2B 4 b i 4b Account for each of the following facts: i The decomposition temperature of MgCO3 is lower than that of CaCO3. 3 charge When compared with Ca2+, Mg2+ has higher radius ratio and is more polarizing, 1 mark 2∴ CO3 in MgCO3 is polarized to a greater extent than in CaCO3. 1 mark 1 mark Hence MgCO3 undergoes thermal decomposition to form oxide (small anion) more readily than CaCO3. 95 2B 5 a v 5a The table below lists some properties of the alkali metals. Standard electrode Melting point / ºC Element Atomic Ionic radius / nm First ionization potential / V radius / nm energy / kJmol-1 Li 0.123 0.060 520 -3.04 180 Na 0.157 0.095 495 -2.71 98 K 0.203 0.133 418 -2.92 64 Rb 0.216 0.148 403 -2.93 39 Cs 0.235 0.169 374 -2.95 29 v With reference to the standard electrode potentials and the melting points of the elements, explain why the reactivity of the alkali metals with water increases as the group is descended. All group I elements react with cold water to give OH- and H2. 1 mark OR M(s) + H2O(l) → MOH(aq) + ½H(g) The electrode potential of group I metal increases down the group with the exception of Li, therefore, the reactivity increase. 1 mark For Li, the relatively high melting point reduces the reactivity Q activation energy for the reaction is higher. 2 marks 95 2B 5 b i ii 5b Explain the following facts : i The thermal stability of sodium hydroxide is higher than that of magnesium hydroxide. Mg2+ is more polarizing than Na+ due to its smaller size and greater charge. The OH- is polarized to a greater 2 marks extent in Mg(OH)2. OR Because of its high charge / radius ratio, Mg2+ forms more stable lattice with small anions e.g. O2-, ∴ Mg(OH)2 is ready to decompose to give MgO. 2 marks ii The thermal stability of the carbonates of the alkaline earth metals increases as the group is descended. As the group is descended, the polarizing power of the cation decreases. Therefore, cations of the lighter elements distort the anion electron clouds to a larger extend. Anions with highly distorted electron clouds are more readily decomposed than those with little distortion. Hence the thermal stability increases as the group is descended. 3 marks OR The charge / radius ratio of cation increases as the group is ascended. Cations with high charge/radius ratio tend to form stable lattice with small anions. 3 marks Therefore, carbonates of Be, Mg tend to decompose to form the more stable oxides. 96 2B 4 b i 4b For each of the following pairs of compounds, suggest a test to distinguish one compound from the other : i Na2CO3(s) and NaHCO3(s) NaHCO3 decomposes on heating to give CO2 /a gas ½ + ½ mark ½ + ½ mark which can turn lime water milky but Na2CO3 does not decompose. Or ½ + ½ mark dissolve the solid in water, add MgSO4(aq) to the solutions ½ + ½ mark Na2CO3(aq) gives a white ppt.; but NaHCO3(aq) does not. 99 2A 1 a iii 1a iii What would be observed when a small piece of rubidium is added to 2-methylpropan-2-ol ? Write the balanced

4

2

3

2

s-Block Elements equation for the reaction.

Unit 2

Page 5

s-Block elements

Unit 3

Topic

s-Block Elements

Reference Reading

9.2.3.0 Chemistry in Context, 3rd Edition ELBS pg. 142–143, 183–184, 243–244 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 105–112 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 365–366

Page 1

Unit 3

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 127–128, 162–163, 215, 284 Reading Syllabus

Solvation process Solubility of ionic crystal

Notes

C. Solubility of sulphate and hydroxide The feasibility of a change is depending on the relative stability of the reactants and the products. Similarly, whether a substance would be soluble is depending on the stability of the solvent and solute comparing with the stability of the solution. The change in stability of a system is measured by standard change in Gibb's free energy ∆G which is defined as ∆G = ∆H - T∆S. Dissolution is a process accompanied with an increase in disorderness i.e. ∆S is positive. Therefore, an dissolution process is energetically feasible if ∆H of solution is negative, zero or slightly positive. The solute would be insoluble if ∆H is highly positive. Or if we disregard the change in disorderness in the dissolution process, ∆G ≈ ∆H. The solute would be soluble if the ∆H of solution is negative. According to Hess's law, the enthalpy change of a process is independent of the pathway of the process actually taken. In the determination of ∆H of solution of an ionic compound, the process can be imagined as vaporization of the substance followed by solvation of the gaseous ion to form the solution. 1.

Solvation energy

a) Definition of solvation and solution processes Solvation process – Solution process –

The process in which the particle of the solute in gaseous state is dissolved in a solvent to form a solution. The process in which a solute in any physical state is dissolved in a solvent to form a solution.

Solvation energy (Hydration energy) – Enthalpy change when 1 mole of formula units of gaseous particles are solvated to form a infinitely diluted solution. If the solvent is water, then the enthalpy change is called hydration energy.

s-Block elements

Unit 3

Page 2

Considering the process of solution of NaCl(s),

NaCl(s) → Na+(aq) + Cl-(aq)

∆Hosol’n = +5 kJmol-1

The process can be divided into two steps : ∆Ho = - ∆Holat i. NaCl(s) → Na+(g) + Cl-(g) For an ionic solid to dissolve, the crystal lattice must be broken up and the cations and anions have to be separated from each other. If the ions are separated to infinite distance apart, the energy required is numerically equal to ∆Holat. ∆Hohyd or ∆Hosolv ii. Na+(g) + Cl-(g) → Na+(aq) + Cl-(aq) This process is called either hydration or solvation and the energy involved is called hydration energy or solvation energy. By Hess’s Law, the heat of solution ∆Hosol’n = -∆Holat + ∆Hohyd Whether the heat of solution ∆Hosol’n is exothermic or endothermic depending on the relative magnitudes of ∆Holat and ∆Hohyd. Heat is always liberated in the hydration process. If the heat of hydration is high enough, the energy released will be high enough to overcome the attractions between the ions. This makes the crystal soluble. If ∆Hohyd is more negative than ∆Holat If ∆Hohyd is less negative than ∆Holat

⇒ the solute would be more soluble ⇒ the solute would be less insoluble

s-Block elements

Unit 3

Page 3

b) Factors affecting the value of hydration energy The hydration energy is depending on the attraction between the gaseous ion and the solvent molecules. If an ion has high charge density, when it is solvated, the attraction forces developed between the solvent particles and the ion will be very large and the hydration energy with be very exothermic. Since the cation and anion are solvated independently, the total hydration energy can be expressed as : 1 1 Hydration energy ∝ ( r + r ) + -

r +, r - :

radii of cation and anion

On top of charge density, the magnitude of lattice energy is also depending on the packing efficiency of the ions in the crystal. If the cations and anions in the crystal lattice have similar size, the packing will be more efficient and gives a higher lattice energy. Furthermore, lattice energy is depending on the attraction between the cation and anion, therefore the lattice energy can be expressed as : 1 Lattice energy ∝ r + r + -

c)

r +, r - :

radii of cation and anion

Effect of lattice energy and hydration energy on solubility of ionic crystal

Solubility of a solute is depending on the interaction between the solute and solvent particle. A solute will only be soluble if the attraction between the solute and solvent particles is greater than the attractions between the solute-solute particles and solventsolvent particles.

Sizes of the cations and anions

Both of them are small Both of them are large They have different sizes

Hydration energy 1 1 (∝ ( r + r )) + High Low High

Lattice energy 1 (∝ r + r ) + High Low Low

Solubility

Low to moderate Low to moderate High to moderate

In general, if the sizes of the ions in the crystal are different, the crystal will be soluble. And, if the sizes of the ions in the crystal are similar, the crystal will be insoluble. This assumes that the crystal is purely ionic. e.g. This is not applicable to the case of AgCl which is not purely ionic.

s-Block elements

Unit 3 2.

Page 4

Solubility of some ionic crystals

a) Sulphate salt of group II metal Compound MgSO4 CaSO4 SrSO4 BaSO4

Relative solubility Soluble Sparingly soluble Insoluble Insoluble

1 Lattice energy ∝ r + r + -

Solubility mol/100g of water at 298 K 1.83 × 10-1 4.66 × 10-3 7.11 × 10-5 9.43 × 10-7

1 1 Hydration energy ∝ ( r + r ) + -

On descending the group, magnitudes of both hydration energy and lattice energy decrease as the size of the cation increases. SO42- ion is a polyatomic ion with a much larger size than group II ions. This makes the magnitude of lattice energy rather constant and decreases slower than the magnitude of hydration energy on moving down the group. ∆Hosol’n = -∆Holat + ∆Hohyd As a result, on moving down the group, ∆Hosol’n becomes less negative and the solubility decreases. b) Hydroxide of group II metal Compound Mg(OH)2 Ca(OH)2 Sr(OH)2 Ba(OH)2

Relative solubility Insoluble Sparingly soluble Slightly soluble Slightly soluble

1 Lattice energy ∝ r + r + -

Solubility mol/100g of water at 298 K 2.00 × 10-5 1.53 × 10-3 3.37 × 10-3 1.50 × 10-2

1 1 Hydration energy ∝ ( r + r ) + -

Once again, on moving down the group, magnitudes of both hydration energy and lattice energy decrease as the size of the cation increases. OH- ion is a small anion. The hydration energy is very large, and rather constant on moving down the group since it is almost dominated by the small size of OH- ion. ∆Hosol’n = -∆Holat + ∆Hohyd As a result, on moving down the group, ∆Hosol’n becomes more negative and the solubility increases. N.B.

In general, a substance with a solubility lower than 10-2 mol/100g of water is considered to be not very soluble.

s-Block elements

Unit 3 c)

Page 5

Solubility of salts of other group II metals

Results obtained when 0.1M solution Mg2+, Ca2+, Sr2+ and Ba2+ are treated with various solutions.

Difference in solubility is a useful tool in qualitative analysis. For example Ca2+(aq) and Sr2+(aq) solutions can be distinguished by adding saturated CaSO4(aq) solution. N.B.

Although OH- ion is a polyatomic ion, it is considered as a small ion.

Glossary

solvation

Past Paper Question

90 2B 4 e 91 2A 1 b iii 93 2B 4 b ii 96 2B 4 c 98 1A 3 b iii 99 1A 1 b ii

hydration

solution

91 2B 5 c

90 2B 4 e 4 Account for the following observations. 4e Lithium salts generally have high solubilities in water. Li+ has very high charge density because of the small ionic radius. there formation of Li(H2O)n+ involve high hydration energy. Li salts - lattice energy is low because of the crystal packing is not efficient.

3 1 mark 1 mark 1 mark

91 2A 1 b iii iii When solid sodium chloride is dissolved in water, the process is endothermic. Explain this observation. The dissolving of a salt in water is a case of an enthalpy change accompanying a chemical process. The first process requires an input of energy to break the lattice. The second process involves the release of energy when new bonds are made between ions and water. The enthalpy of solution is a measure of the difference between 2 marks these two processes. In case of NaCl, the lattice energy is larger than the enthalpy of hydration.

2

s-Block elements Unit 3 Page 6 91 2B 5 c 5c Explain why the solubility of group(II) salts of large anions decreases on descending the group, whereas the 4 solubility of group(II) salts of small anions shows the reverse trend. 1 mark ∆Hsoln = -∆HLE + ∆HHE ∆HLE: Lattice energy - energy released in the formation of the solid crystal from the gaseous ions ∆HHE: Hydration enthalpy - energy released when the solute particles in gaseous state become associated with water molecule to form a solution. ½ mark More negative the ∆Hsoln, more soluble would be the salt. 1 ½ mark Lattice energy ∝ r+ + r− 1 1 Hydration energy ∝ ( + ) ½ mark r+ r− On descending the group, magnitude of both hydration energy and lattice energy decrease. ½ mark For large anions, the lattice energy is rather constant. The magnitude of hydration energy decreases more than that of lattice energy as the size of the cation increases. ½ mark ½ mark The ∆Hsoln becomes less negative on descending the group and solubilities decrease. For small anions with small cation, the packing efficiency is high result in higher lattice energy. The magnitude of hydration energy decreases less than that of lattice energy as the size of the cation increases. The ∆Hsoln becomes more negative on descending the group and solubilities increase. 93 2B 4 b ii 4b Account for each of the following facts: ii MgCrO4 is more soluble in water than BaCrO4. 3 2CrO4 is a large anion 1 mark The solubility of an ionic compound with large anion is affected more by the hydration energies of its ions than its lattice energy. 1 mark Mg2+ is smaller than Ba2+, it has much higher hydration energy than Ba2+, ∴ MgCrO4 is more soluble than 1 mark BaCrO4. 96 2B 4 c 4c Briefly explain why magnesium sulphate(VI) is soluble in water, whereas barium sulphate(VI) is almost insoluble. 2 Hydration energy of Mg2+ is more exothermic than that of Ba2+ 1 mark due to the very small size of Mg2+ cation. Therefore, the lattice of MgSO4 can easily be broken down by water. 1 mark 98 1A 3 b iii 3b Sketch the trends for the properties mentioned in (i), (ii), and (iii) below, and account for the trend in each case. iii solubility in water of the Group II sulphates(VI), MgSO4, CaSO4, SrSO4 and BaSO4

99 1A 1 b ii 1b Account for each of the following : ii The solubility in water of magnesium hydroxide is less than that of barium hydroxide.

s-Block elements

Unit 4

Topic

s-Block Elements

Reference Reading

9.2.4 Chemistry in Context, 3rd Edition ELBS pg. 243 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 104 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 374

Page 1

Unit 4

Assignment Reading Syllabus

Anomaly of lithium

Notes

D. Abnormal behaviour of lithium Lithium exhibits anomalous behaviour comparing with other group I metal. The properties of lithium resembles that of magnesium more than that of other group I metal. This is an example of diagonal relationship. The anomaly is mainly due to the exceptional high charge density and polarizing power of lithium ion comparing with other group I metal ions. Properties Burning in air Heating of carbonate Heating of hydroxide Heating of nitrate(V) Formation of hydrogencarbonate Formation of organometallic compound Solubility of salts Stability of compound

Glossary

anomaly

Lithium Formation of normal oxide Decompose into oxide and carbon dioxide Decompose into oxide and water Decompose into oxide, nitrogen dioxide and oxygen Do not form stable solid hydrogencarbonate Form organometallic compound e.g. RLi

Magnesium Formation of normal oxide Decompose into oxide and carbon dioxide Decompose into oxide and water Decompose into oxide, nitrogen dioxide and oxygen Do not form stable solid hydrogencarbonate Form organometallic compound e.g. MgR2

Insoluble phosphate, fluoride and carbonate Form stable nitride

Insoluble phosphate, fluoride and carbonate Form stable nitride

organometallic compound

Sodium Formation of peroxide on top of normal oxide Stable to heat Stable to heat Decompose into nitrate(III) and oxygen only Form stable solid hydrogencarbonate Do not form any stable organometallic compound at all Soluble phosphate, fluoride and carbonate Do not form stable nitride,

s-Block elements

Past Paper Question

Unit 4

Page 2

93 2B 4 a 96 2B 4 a

93 2B 4 a 4a The first member of a Group in Periodic Table usually exhibits anomalous behaviour. For lithium in Group I, give two examples of anomalous behaviour and explain how this behaviour arises. Any two of the following examples: 1. Lithium carbonate, nitrate, hydroxide decomposes upon heating, while carbonate of other Group I elements do not.

3

Li2CO3  → Li2O + CO2 heat

4LiNO3  → 2Li2O + 4NO2 + O2 heat

2LiOH  → Li2O + H2O 2. Lithium do not form any superoxide and peroxide. 3. Lithium forms a number of organic compound e.g. RLi 4. Slow reaction with water and not explosive with acid. 5. Do not form solid hydrogencarbonate. 2 marks 6. Li+2S2- is insoluble in H2O Q Li has a small size, high charge density, high polarizing power and more electronegative(less stable in forming ionic compound) - forms more stable compounds with small anions e.g. O2-, H-, N3- due to high lattice energy. ∴ Li2CO3, LiNO3 tend to decompose on heating - forms more stable covalent bond with C when compare with other element in Group I. 1 mark heat

96 2B 4 a 4a Give TWO reactions to illustrate the following statement : 'Lithium compounds differ in their chemical properties from compounds of other Group I elements .' Lithium carbonate decomposes on heating to give (lithium oxide) and carbon dioxide. ½ + ½ mark Li2CO3 → Li2O + CO2 (1 mark) 1 mark But other alkali metal carbonates do not. ½ mark Lithium nitrate(V) decomposes on heating to give (lithium oxide), ½ mark nitrogen dioxide / a brown gas and oxygen. 2LiNO3 → Li2O + 2NO2 + ½O2 (1 mark) 1 mark But other alkali metal nitrate(V) give nitrate(III) and oxygen. 2MNO3 → 2MNO2 + O2 (1 mark)

4

s-Block Elements

Unit 5

Topic

s-Block Elements

Reference Reading

9.3 Chemistry in Context, 3rd Edition ELBS pg. Inorganic Chemistry, 4th Edition, T.M. Leung pg. 113–115 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 369

Page 1

Unit 5

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 219–221 Reading Syllabus Notes

III. Uses of the compounds of the s-block elements A. Manufacture of glass Sodium carbonate is used in the manufacture of glass. Soda glass is a mixture of sodium silicate and calcium silicate, which is made by fusing the carbonates with silica SiO2(s) (from sand) at 1500ºC. Na2CO3(s) + SiO2(s) → Na2SiO3(s) + CO2(g) CaCO3(s) + SiO2(s) → CaSiO3(s) + CO2(g) B. Baking powder The main ingredient of baking powder is sodium hydrogencarbonate which is not very stable to heat. Upon heating, it decomposes into sodium carbonate, carbon dioxide and water. The carbon dioxide evolved will help to raise the bread in the baking process inside the oven. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l) N.B.

HCO3- is a large anion with high polarizability. It only forms stable solid hydrogencarbonate with big group I metal ions. Group II and other metal hydrogencarbonate do not exist at all.

C. Making of soap Aqueous sodium hydroxide is used in the hydrolysis of fat (saponification) in the preparation of soap.

D. Antacid Mg(OH)2(s) is commonly used as the antacid to neutralize the excess acid in the stomach.

s-Block Elements

Unit 5

Page 2

E. Slaked lime Limestone (CaCO3(s)) is heated to give quicklime (CaO(s)) which is 'slaked' with water to give slaked lime (Ca(OH)2(s)). When dissolved in water, the aqueous solution of slaked lime is called limewater. → CaO(s) + CO2(g) CaCO3(s) heat CaO(s) + H2O(l) → Ca(OH)2(s)

Slaked lime (Ca(OH)2(s)) is used in the removal the acidity of the soil, making of mortar (slaked lime + sand + water), bleaching powder and manufacture of calcium hydrogensulphite which is used in paper industry. F. Fireworks Salts of s-block metals are mixed with gunpowder in fireworks to give different colour to the fireworks.

Glossary

silica

baking powder

Past Paper Question

94 2B 6 a 97 1A 3 a iv

slaked lime

quicklime

94 2B 6 a 6a Self-raising flour for baking cakes and bread contains sodium hydrogencarbonate. Explain briefly, with the aid of 3 chemical equation(s), the role of sodium hydrogencarbonate in the baking process. Sodium hydrogencarbonate in the self-raising flour undergoes thermal decomposition to give CO2 gas. 1 mark 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) 1 mark 1 mark The CO2 gas produced causes the cakes to rise. 97 1A 3 a iv 3a For each of the following materials, identify ONE inorganic compound which is an active ingredient: iv antacid

3

p-Block Elements I.

Halogens A. B. C. D.

E.

II.

Laboratory preparation of chlorine Characteristic properties of the halogens Properties of halogen 1. Oxidizing power of halogen 2. Disproportionation of halogen in alkalis Properties of the halogen compound 1. Reactions of halide a) with halogens b) with conc. sulphuric(VI) acid c) with conc. phosphoric(V) acid d) with silver ions 2. Acidic properties of hydrogen halide 3. Oxoacid Uses of halogens and halogen containing compounds 1. Fluoridation of water 2. Use of chlorine 3. Photographic films

Nitrogen and its compounds A. B.

Nitrogen 1. Unreactive nature of nitrogen 2. Direct combination of nitrogen and oxygen Ammonia 1. Haber process 2. Chemical properties of ammonia a) Alkaline properties (1)

b)

(1) (2) (3) (4) (5)

C.

D. E.

Laboratory preparation of ammonia

Reducing properties Combustion Catalytic oxidation With copper(II) oxide With chlorine Thermal decomposition of ammonia

Nitric(V) acid 1. Ostwald Process 2. Oxidizing properties of nitric(V) acid a) With copper b) With iron(II) ion c) With sulphur Nitrate(V) salt 1. Thermal decomposition of nitrate 2. Brown ring test Use of nitric(V) acid a) Fertilizers b) Explosives c) Nylon d) Dyestuffs

III.

Sulphur and its compounds A. B. C.

D. E. F.

Allotrope of sulphur Burning of sulphur Sulphur dioxide / sulphate(IV) salt 1. Reducing properties of sulphur dioxide a) With manganate(VII) ion b) With dichromate(VI) ion c) With bromine 2. Oxidizing properties of sulphur dioxide a) With magnesium 3. Distinguishing from carbon dioxide Concentrated sulphuric(VI) acid 1. Contact process 2. Properties of concentrated sulphuric acid Sulphate(VI) salt 1. Test for sulphate(VI) ion Use of sulphuric(VI) acid 1. Fertilizers 2. Detergents 3. Paints, Pigments and dyestuffs

I. Halgoens

Unit 1

Page 1

Topic

I. Halogens

Unit 1

Reference Reading

10.1.0–10.1.2 Chemistry in Context, 3rd Edition ELBS pg. 265, 270–276 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 122–129 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 396, 398–399

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 224–233 Reading Syllabus Notes

I.

Halogens

A. Laboratory preparation of chlorine Cl2(g) is a strong oxidizing agent. In order to prepare Cl2(g) from Cl- ion, an oxidizing agent stronger than Cl2(g) must be used. KMnO4(s) and MnO2(s) are two commonly used oxidizing agents. Since the oxidizing power of KMnO4(s) is very strong, heating is not essential. However, if MnO2(s) is used, heating is mandatory. The source of Cl- ions may come from NaCl(s) or conc. HCl(aq). If NaCl(s) is used, H2SO4(aq) must be added to provide the H+(aq) ions required in the reaction. MnO2(s) + 4HCl(aq)  → MnCl2(aq) + 2H2O(l) + 2Cl2(g) heat

10HCl(aq) + 2MnO4-(aq) + 6H+(aq) → 5Cl2(g) + 8H2O(l) + 2Mn2+(aq)

I. Halgoens

Unit 1

Page 2

B. Characteristic properties of the halogens All halogens have high electronegative and accept electrons readily. The word "halogen" means salt maker. It forms salt readilty with metal. C. Properties of halogen 1.

Oxidizing power of halogen

All halogens are strong oxidizing agents. Furthermore, free halogen atom also has high electron affinity. They all accepts electron readily. Comparatively, the oxidizing power of fluorine is even stronger than potassium manganate(VII). Reduction half ionic equation I2(s) + 2e- d 2I-(aq) Br2(g) + 2e- d 2Br-(aq) MnO2(s) + 4H+(aq) + 2e- d Mn2+(aq) + 2H2O(l) Cl2(g) + 2e- d 2Cl-(aq) MnO4-(aq) + 8H+(aq) + 5e- d Mn2+(aq) + 4H2O(l) F2(g) + 2e- d 2F-(aq)

Standard reduction potential + 0.54 V + 1.07 V + 1.23 V + 1.36 V + 1.51 V + 2.87 V

Reaction with sodium Cl2(g) + 2Na(s) → 2NaCl(s) Br2(l) + 2Na(s) → 2NaBr(s) I2(s) + 2Na(s) → 2NaI(s) Reaction with iron(II) ion Cl2(g) + 2Fe2+(aq) → 2Cl-(aq) + 2Fe3+(aq) Br2(l) + 2Fe2+(aq) → 2Br-(aq) + 2Fe3+(aq) I2(l) + 2Fe2+(aq) → 2I-(aq) + 2Fe3+(aq) Reaction with phosphorus Most phosphorus form two kinds of halides PX3 (X = F, Cl, Br, I) and PX5 (X = F, Cl, Br). P(s) + F2(g) → PF3(g) (colourless gas) P(s) + F2(g) → PF5(g) (colourless gas) P(s) + Cl2(g) → PCl3(l) (colourless liquid) P(s) + Cl2(g) → PCl5(s) (white solid) P(s) + Br2(l) → PBr3(l) (colourless liquid) P(s) + Br2(l) → PBr5(s) (yellow solid) P(s) + I2(s) → PI3(s)

I. Halgoens

Unit 1 2.

Page 3

Disproportionation of halogen in alkalis

Except fluorine, all halogens disproportionate in alkali. However, different halogens disproportionate differently in alkali medium at different temperature. Reaction of fluorine with alkali 2F2(g) + 2OH-(aq) → OF2(aq) + 2F-(aq) + H2O(l) 0 -1 -1

at room temp. (not a disproportionation reaction)

2F2(g) + 4OH-(aq) → O2(g) + 4F-(aq) + H2O(l) 0 -1

at about 70ºC

F2(g) is extremely electronegative, it only behaves as an oxidizing agent but not a reducing agent. Disproportionation of chlorine in alkali Cl2(g) + 2OH-(aq) → ClO-(aq) + Cl-(aq) + H2O(l) 0 +1 -1

at room temp.

3Cl2(g) + 6OH-(aq) → ClO3-(aq) + 5Cl-(aq) + 3H2O(l) 0 +5 -1

at about 70ºC

This may be considered as thermal decomposition of ClO-(aq). 3ClO-(aq) → 2Cl-(aq) + ClO3-(aq) +1 -1 +5 Disproportionation of bromine in alkali 3Br2(l) + 6OH-(aq) → BrO3-(aq) + 5Br-(aq) + 3H2O(l) 0 +5 -1

at room temp.

Disproportionation of iodine in alkali 3I2(s) + 6OH-(aq) → IO3-(aq) + 5I-(aq) + 3H2O(l) 0 +5 -1

at room temp.

The difference in mode of disproportionation can be explained by the difference in electronegativity of halogen atom. As the most electronegative element, F tends to attain an oxidation state of -1 only. With decreasing electronegativity on moving down the group, the element disproportionates and gets a more positive oxidation state more readily.

Glossary

disproportionation

Past Paper Question

90 1A 3 c 91 2B 4 c ii 92 1A 3 f iii 93 2B 6 a i 94 1A 2 a ii 96 1A 2 c d I ii 97 1B 7 a 98 2B 8 d i ii

93 2B 6 a ii 94 2B 4 a

I. Halgoens Unit 1 90 1A 3 c 3c By means of balanced equations, illustrate the reactions of the halogens F2(g), Cl2(g) and Br2(l) with OH-(aq) at room temperature. 1 mark 2F2(g) + 2OH-(aq) → OF2(aq) + 2F-(aq) + H2O(l) Cl2(g) + 2OH-(aq) → OCl-(aq) + Cl-(aq) + H2O(l) 1 mark or 3Cl2(g) + 6OH-(aq) → ClO3-(aq) + 5Cl-(aq) + 3H2O(l) 3Br2(l) + 6OH-(aq) → BrO3-(aq) + 5Br-(aq) + 3H2O(l) 1 mark 91 2B 4 c ii 4c ii Explain what is meant by ‘disproportionation’, and write an equation involving a compound of chlorine to illustrate your answer. Definition : Reaction in which a single species/ion/substance is simultaneously oxidized and reduced. (no mark for atom/element -½) 1 mark H2O + Cl2 → HCl + HOCl 2NaOH + Cl2 → NaCl + NaOCl + H2O 4KClO3 → 3KClO4 + KCl 3HOCl + 3H2O → 3H3O+ + ClO3- + 2Cl3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O 1 mark (½ mark for unbalanced equation, 1 mark for balanced) 92 1A 3 f iii 3f Complete and balance the following equations: temperature iii dil NaOH + Cl room  → H O + + 2

93 2B 6 a i ii 6a Compare the reaction of aqueous alkali with each of fluorine, chlorine, bromine and iodine i at 295 K, At 295 K, Cl2, Br2 and I2 disproportionate according to the equation X2 + 2OH-(aq) → OX-(aq) + X-(aq) + H2O(l) OBr- and OI- are unstable and will undergo further disproportionation. X = Br or I 3OX- d XO3- + 2XF2 is the strongest [O] agent, it will not disproportionate, instead, it will oxidize the OH- ion. 2F2 + 2OH- → OF2(g) + 2F- + H2O ii at 350 K. At 350 K, Cl2, Br2 and I2 all disproportionate as follows 3X2 + 6OH-(aq) → XO3-(aq) + 5X-(aq) + 3H2O(l) F2 oxidizes the alkali, F2 + 4OH- → O2(g) + 4F- + 2H2O

3 1 mark 1 mark 1 mark 2 1 mark 1 mark

2KOH(aq) + Cl2(g)  → KCl(aq) + KOCl(aq) + H2O(l) cold dil.

2OH-(aq) + Cl2(g)  → Cl-(aq) + OCl-(aq) + H2O(l) cold dil.

or 6KOH(aq) + 3Cl2(g)   → 5KCl(aq) + KClO3(aq) + 3H2O(l) hot conc.

hot conc.

2

1 mark

94 1A 2 a ii 2a ii Write a balance equation, involving the reaction of chlorine with KOH(aq), to illustrate a disproportionation.

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

3

1

2

temperature 2NaOH (aq, dilute) + Cl 2(g) room  → NaOCl(aq) + NaCl (aq) + H 2 O (l)

Page 4

1 mark

1

I. Halgoens Unit 1 94 2B 4 a 4a Describe how a sample of dry chlorine is prepared in the laboratory. State the safety precaution(s) that is/are required. Heating conc. HCl in the presence of manganese(IV) oxide MnO2(s) + 4HCl(aq)  → MnCl2(aq) + 2H2O(l) + 2Cl2(g) OR Add conc. HCl to solid KMnO4 10HCl(aq) + 2MnO4-(aq) + 6H+(aq) → 5Cl2(g) + 8H2O(l) + 2Mn2+(aq) heat

Page 5 5

2 marks

Pass the Cl2 produced through cold water to remove HCl, dry the gas by passing through conc. H2SO4 and collect 2 marks by downward delivery. The experiment should be carried out in a fume cupboard. 1 mark 96 1A 2 c d i ii 2c What is the oxidation state of the central atom in each of the following compounds ? HClO4 HClO3 H2CrO4 HMnO4 HClO3 +5 or V HClO4 +7 or VII H2CrO4 +6 or VI HMnO4 +7 or VII (DO NOT accept a number preceding the charge e.g. 5+) 2d i Explain why among the four compounds in (c), only HClO3 can undergo disproportionation. The central atoms of the other three compounds are in their highest oxidation states whereas in HClO3, Cl is in an intermediate oxidation state. ii Write a balanced equation for the disproportionation of HClO3 to give HClO4 and ClO2. 3HClO3 → HClO4 + H2O + 2ClO2

2 4 × ½ mark 1 ½ mark ½ mark 1 1 mark

97 1B 7 a 7a Briefly describe how a sample of dry chlorine gas can be prepared in the laboratory. Draw a labelled diagram of the laboratory set-up and state the safety precaution(s) that is/are required. 98 2B 8 d i ii 8d i Write the balanced equation for the reaction that occurs when chlorine gas is passed into an aqueous sodium hydroxide solution at (I) 298 K (II) 343 K. ii At each of the above temperatures, iodine solid instead of chlorine gas is added to the aqueous sodium hydroxide solution. Would you expect the reaction of iodine with sodium hydroxide to be similar to the reaction of chlorine with sodium hydroxide at each temperature ? If not, explain why a different reaction takes place.

6

4

I. Halogens

Unit 2

Topic

I. Halogens

Reference Reading

10.1.3.1 Chemistry in Context, 3rd Edition ELBS pg. 276–278 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 129–135 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 402–404

Page 1

Unit 2

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 233–234 Reading Syllabus Notes

D. Properties of the halogen compound 1.

Reactions of halide

a) with halogens The oxidizing power of halogens show a decreasing trend on moving down the group. Therefore, F2 is the halogen with the strongest oxidizing power. A more reactive halogen is capable to displace a less reactive halogen from its salt. e.g.

Cl2 + 2Br- → 2Cl- + Br2

Aqueous halogens have similar colours with different intensity ranging from yellow, orange to brown. The halogen displaced could be identified more accurately by extracting with an organic solvent. Chlorine is pale green in organic layer while bromine is orange and iodine is purple. b) with conc. sulphuric(VI) acid Conc. sulphuric(VI) acid (b.p. 338ºC) is an involatile acid and a strong oxidizing agent. However, its oxidizing power is not strong enough to oxidize F- ion or Cl- ions. It can only displace HF(g) (b.p. 20ºC) and HCl(g) (b.p. 85ºC) from the salts since HF(g) and HCl(g) are very volatile. NaF(s) + H2SO4(l) → HF(g) + NaHSO4(s) NaCl(s) + H2SO4(l) → HCl(g) + NaHSO4(s) The oxidizing powers of Br2 and I2 are weaker than conc. H2SO4(l). Therefore, conc. H2SO4(l) is capable to oxidze Br- ion and I- ion besides displacing HBr(g) (b.p. -67ºC) and HI(g) (b.p. -35ºC) from the salts. Moreover, I- ion is a stronger reducing agent than Br- ion. I- ion is capable to reduce conc. H2SO4(l) to H2S(g) while Br- ion is only capable to reduce conc. H2SO4(l) to SO2(g). NaBr(s) + H2SO4(l) → HBr(g) + NaHSO4(s) 2HBr(g) + H2SO4(l) → SO2(g) + Br2(g) + 2H2O(l) NaI(s) + H2SO4(l) → HI(g) + NaHSO4(s) 8HI(g) + H2SO4(l) → 4I2(g) + H2S(g) + 4H2O(l) c)

with conc. phosphoric(V) acid

Similar to conc. H2SO4(l), conc. H3PO4(l) (b.p. 213ºC) is also involatile but it is non-oxidizing. Conc. H3PO4(l) is also a viscous liquid due to the formation of extensive hydrogen bonds among the molecules. It is capable to displace hydrogen halides form the solid halide salts without oxidizing the halide ion to halogen. NaF(s) + H3PO4(l) → HF(g) + NaH2PO4(s) NaCl(s) + H3PO4(l) → HCl(g) + NaH2PO4(s) NaBr(s) + H3PO4(l) → HBr(g) + NaH2PO4(s) NaI(s) + H3PO4(l) → HI(g) + NaH2PO4(s)

I. Halogens

Unit 2

Page 2

d) with silver ions The presence of halide ion in an aqueous solution can be identified by using acidified silver nitrate solution. Silver nitrate must be used in acidic medium because silver(I) ion will form silver(I) oxide in the presence of hydroxide. 2Ag+(aq) + OH-(aq) → Ag2O(s) + H2O(l) Aqueous silver(I) ion forms silver halide precipitate with different colour and stability under light. Ag+(aq) + Cl-(aq) → AgCl(s) white ppt. soluble in NH3(aq) and deocmposes under light to purple grey ppt. Ag+(aq) + Br-(aq) → AgBr(s) yellow ppt. slightly soluble in NH3(aq) and decomposes under light to yellow grey ppt. Ag+(aq) + I-(aq) → AgI(s) yellow ppt. insoluble in NH3(aq) and stable to light N.B.

Solubility and formation of ammonia complex AgBr(s) and AgI(s) are not very soluble in NH3(aq) solution because they are highly covalent. AgCl(s) is soluble in aqueous ammonia because it has a larger share of ionic character and Ag+ ion is capable to form a soluble complex with NH3(aq) molecules. AgCl(s) + 2NH3(aq) d Ag(NH3)2+(aq) + Cl-(aq) Besides Ag+(aq), Cu2+(aq) and Zn2+(aq) are also capable to form soluble complex with NH3(aq) molecules. At first, the three ions will form insoluble precipitate of oxide and hydroxides upon addition of a little aqueous ammonia since ammonia is a weak alkali. 2Ag+(aq) + 2OH-(aq) → Ag2O(s) + H2O(l) colourless brown Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s) blue pale blue Zn2+(aq) + 2OH-(aq) → Zn(OH)2(s) colourless white If excess ammonia is added, the three precipitates will redissolve at high NH3(aq) concentration. Ag2O(s) + 2NH3(aq) + H2O(l) d Ag(NH3)2+(aq) + OH-(aq) brown colourless (diamminesilver(I) ion) Cu(OH)2(s) + 4NH3(aq) d Cu(NH3)42+(aq) + 2OH-(aq) pale blue deep blue (tetraamminecopper(II) ion) Zn(OH)2(s) + 4NH3(aq) d Zn(NH3)42+(aq) + 2OH-(aq) white colourless (tetraamminezinc(II) ion)

Glossary

displace involatile tetraamminecopper(II) ion

Past Paper Question

90 2B 6 c 92 1A 3 f i ii 93 2B 6 c 94 2B 4 d i ii 95 1B 4 a 96 2B 5 c i ii iii iv 98 2B 8 b ii 99 1A 3 d ii

soluble complex redissolve tetraamminezinc(II) ion

diamminesilver(I) ion

I. Halogens Unit 2 90 2B 6 c 6c Explain why copper(II) chloride is a stable compound whereas copper(II) iodide does not exist. Cl is much more electronegative than I / the size of I- is larger than ClCl- is more resistant to oxidation than Ior Cu2+ is an oxidizing agent strong enough to oxidize I- to I2 or I- is a stronger reducing agent I- ion is polarized by Cu2+ CuI2 disproportionates, CuI2 → CuI + ½I2 92 1A 3 f i ii 3f Complete and balance the following equations: temperature i conc H 2 SO 4 + KBr room  → H 2 O + + ii

conc H 3 PO 4 + KBr   →

3 1 mark 1 mark

1 mark

1

+

temperature 2KBr(s) + 3H 2 SO 4 (conc) room  → 2KHSO 4(aq) + 2H 2 O (l) + SO 2(g) + Br2(g) heat

Page 3

1 mark 1

+

KBr(s) + H 3 PO 4(conc) heat  → HBr(g) + KH 2 PO 4(aq)

93 2B 6 c 6c Write the equation in each case, for the reaction of concentrated H3PO4 with NaCl and NaBr. Will concentrate H2SO4 give a similar reaction in each case? Explain any difference(s) in reactivity between concentrated H3PO4 and concentrated H2SO4. Reaction between NaX and H3PO4 (X = Cl or Br) H3PO4 + NaX → HX + NaH2PO4 With H2SO4, NaCl gives similar reaction NaBr will react to give also Br2 and SO2, 2HBr + H2SO4 → 2H2O + SO2 + Br2 Q conc. H2SO4 is an oxidizing agent while conc. H3PO4 is not. 94 2B 4 d i ii 4d Write balanced equation(s) and comment on the difference for the reaction between i concentrated H2SO4 and NaCl, and NaCl + H2SO4 → NaHSO4 + HCl ii concentrated H2SO4 and NaI. NaI + H2SO4 → HI + NaHSO4 8HI(g) + H2SO4(l) → 4I2(s) + H2S(g) + 4H2O(l) HI / I-, being a stronger reducing agent than HCl, will be oxidized by conc. H2SO4 to give I2.

1 mark

3

1 mark 1 mark 1 mark

1 1 mark 2 1 mark 1 mark

95 1B 4 a 4a Devise an experiment to distinguish between KBr(s) and a mixture containing approximately 40% KCl(s) and 60% KBr(s) by mass. Prepare conc. aqueous solutions by dissolving the solids in water. 1 mark ½ mark Add AgNO3(aq) to solution in test tube ½ mark Add excess NH3(aq) to the precipitate 1 mark Let the precipitate settle, decant the solution and compare the height of the precipitate. (Note : Br- is insoluble in dil.NH3 but soluble in conc. NH3)

3

I. Halogens Unit 2 Page 4 96 2B 5 c i ii iii iv 5c Hydrogen chloride can be prepared by the action of concentrated sulphuric(VI) acid on potassium chloride. i Write a balanced equation for the reaction of concentrated sulphuric(VI) acid and potassium chloride. 1 1 mark KCl + H2SO4 → KHSO4 + HCl ii Briefly describe how a sample of dry hydrogen chloride can be prepared in the laboratory using the above 6 reaction. Draw a labelled diagram of the laboratory set-up and state the safety precaution(s) that is/are required. Hydrogen chloride can be prepared by dropping (slow addition of) c. H2SO4 onto KCl ½ mark ½ mark and warming the reaction mixture. ½ The HCl produced is tried by passing through anhydrous CaCl2 /concentrated H2SO4 mark ½ mark and then collected by downward delivery / in a gas syringe. 1 mark Precaution: carry out experiment in fume cupboard. [2 marks for description of the expt; 1 mark for safety precaution.]

marks for the diagram for the set-up 3 marks (½ mark for a tap funnel; ½ marks for labelling the reactants 1 mark for correct set-up for the drying process; 1 mark for the collection of HCl) (Deduct ½ mark for each minor mistake) iii Briefly explain why hydrogen bromide cannot be prepared by the action of concentrated sulphuric(VI) acid on potassium bromide. Write a balanced equation to represent the reaction that occurs. c. H2SO4 can oxidize HBr produced to Br2 1 mark 2HBr + H2SO4 → 2H2O + SO2 + Br2 1 mark Or (1 2KBr + 3H2SO4 → 2KHSO4 + 2H2O + SO2 + Br2 mark) iv Suggest how a sample of hydrogen bromide can be prepared in the laboratory. HBr can be prepared by heating KBr with cone. H3PO4. 1 mark Or (1 mark) Add Br2(l) to moist red phosphorus. 98 2B 8 b ii 8b For each of the following, state the expected observation and write the relevant balanced equation(s). ii Ammonia solution is added dropwise, until in excess, to a solution of copper(II) sulphate(VI). 99 1A 3 d ii 3d For each of the following, state the expected observation and write the relevant balanced equation(s). ii KBr(s) is heated with concentrated H2SO4.

2

1

6

I. Halogens

Unit 3

Page 1

Topic

I. Halogens

Unit 3

Reference Reading

10.1.3.2–10.1.3.3 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 135–139 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 404–405

Assignment Reading Syllabus Notes

2.

Acidic properties of hydrogen halide

All hydrogen halides are acidic in water. The acidity of hydrogen halide decreases on moving down the group. HA(aq) + H2O(l) d A-(aq) + H3O+(aq) pKa E(H–X) ∆ EN

HF 3.2 568 3.9

HCl -7 432 0.9

HBr -9 366 0.7

HI -10 298 0.4

This can be explained the decrease in bond energy despite of the increase in bond polarity. Furthermore, from the experimental data, it can be concluded that the solvation of the halide ion is not a determining factor affecting the strength of the acid. HF shows exceptionally low acidity because of the very strong H–F bond and the capability of the molecule to form hydrogen bond with the free proton in the aqueous medium. This reduces the amount of the hydroxonium ions in the solution and lowers the acidity of HF(aq). HF(aq) + H+(aq) d H2F+(aq) 3.

Oxoacid

An acid containing oxygen is called an oxoacid. They have the forms HXO, HXO2, HXO3, H2XO3, H2XO4 etc. The salts formed by them are called oxo-salts. e.g.

O

H

O

O

O

O

S

H O Cl

O H

chloric(I) acid (HOCl)

O P

H

sulphuric(VI) acid (H2SO4)

H

O H

phosphoric(V) acid (H3PO4)

The acidity of an oxoacid is depending on the no. of oxygen attaching to the atom X. Since O is an electronegative atom, it imposes a negative inductive effect on the central atom X. This will make the δ-O–Hδ+ bond more polar. The attraction between the water molecules and Hδ+ will be strengthened and this favours the dissociation of the proton. Therefore, the molecule will become more acidic. Moreover, the oxoacid with more no. of O atom would also be more acidic because the anions formed would be more stable. This is because they are capable to form more identical resonance structures which imply a more evenly distributed electron cloud and more stable structure.

Glossary

oxoacid

oxo-salt

I. Halogens

Unit 3

Page 2

91 1A 3 c 94 1A 2 c iii iv 94 2B 5 a v 96 2B 5 b 99 1A 3 d i

Past Paper Question

91 1A 3 c 3c Give the formulae of two chlorine-containing salts in which the oxidation number of Cl are +1 and +5 respectively. Suggest one common chemical properties shown by both compounds. NaOCl and KClO3 ½ mark each 1 mark both are strong oxidizing agents 94 1A 2 c iii iv 2c iii Write the formulae of three oxoacids of chlorine, and arrange them in order of increasing acid strength. HOCl < HClO2 < HClO3 < HClO4 1 mark for 3 formulae, 1 mark for order of acidity iv Explain why the acidity of 0.1 M HF(aq) is weaker than that of 0.1 M HCl(aq) HF is extensively hydrogen-bonded in water. (H–F)n + H2O(l) d H3O+(aq) + F-(aq) the equilibrium lies to the L.H.S. 1 mark The Cl- ion, having a larger size, is highly polarizable, it is stable in water. Hence the equilibrium 1 mark HCl(aq) + H2O(l) d H3O+(aq) + Cl-(aq) lies to the right, so HCl is a strong acid. OR HX bond strength is greater in H–F than in H–Cl 1 mark 1 mark this outweighs the reduced hydration energy from F- to ClOR 1 mark HF slightly ionizes in H2O & HCl completely ionizes in H2O 94 2B 5 a v 5a v Draw the three-dimensional structure of the manganate(VII) ion. Mn O

2

1

-

OO

2

O

tetrahedral

1 mark

96 2B 5 b 5b What are the highest and lowest oxidation states of bromine ? Give one compound in which bromine is in its highest oxidation state and another compound in which bromine is in its lowest oxidation state. highest oxidation state = +7 / VII ½ mark 1 mark example: BrO4- / HBrO4 lowest oxidation state = -1 / -I ½ mark 1 mark example: Br- / HBr (Accept any correct example of Br containing compound) 99 1A 3 d i 3d For each of the following, state the expected observation and write the relevant balanced equation(s). i KIO3(aq) is added to acidified KI(aq).

3

I. Halogens

Unit 4

Page 1

Topic

I. Halogens

Unit 4

Reference Reading

10.1.4 Chemistry in Context, 3rd Edition ELBS pg. 278–279 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 140–141 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 408–410

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 235–236 Reading Syllabus Notes

E. Uses of halogens and halogen containing compounds 1.

Fluoridation of water

Fluoridation is the addition of a chemical called fluoride to water supplies to help teeth resist decay. In the 1930's, researchers discovered that people who grew up where water naturally contained fluoride had up to two-thirds fewer cavities than people living in areas without fluoride in the water. Today, about half the people of the United States drink fluoridated water. The use of fluoride tablets and toothpastes and the application of strong fluoride solutions to the teeth by dentists can also help prevent tooth decay. Local governments or the people of a community often must decide whether the water supply should be fluoridated. This means balancing the benefits, risks, and costs of fluoridation, as well as moral questions about its widespread use. People disagree over these issues, and fluoridation has always been controversial. Benefits, risks, and costs. Many studies have shown that fluoridation reduces tooth decay substantially. However, rates of tooth decay also have declined in areas without fluoridated water, perhaps chiefly because of the widespread use of fluoride toothpastes. 2.

Use of chlorine

Manufacture of poly(chloroethene) Industrially, vinyl chloride is prepared by heating 1,2-dichloroethane at 500 ºC which is prepared from addition reaction between chlorine and ethene. H

H C C

H

H

H H H C

H H

Cl2 Electrophilic addition

H C C H Cl Cl

H

Heat 500 ° C

C H

H C

Cl

Cl Cl

C H

Eventually, poly(vinyl chloride) / poly(chloroethene) is prepared through a radical addition mechanism. H H H

H C

R

Cl

H H R C C Cl H

R C C

C

Cl H

H

H

H C

Cl

C H

H H H H R C C

C

C

Cl H Cl H

I. Halogens

Unit 4

Page 2

Manufacture of bleach and disinfectant Bleaching solution and bleaching powder are made by mixing chlorine with sodium hydroxide solution and calcium hydroxide respectively. Cl2(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O(l) Cl2(g) + Ca(OH)2(s) → Ca(ClO)2(s) + CaCl2(s) + H2O(l) Besides using as a bleaching agent, chlorine bleach can also be used as a very effective disinfectant. In is also used in water and sewage treatment. 3.

Photographic films

Black and white film is made of emulation of silver bromide crystal in gelatin coated on a plastic base. Upon exposure to light, silver bromide will decompose to black silver crystal. In the developing process, the unexposed silver bromide will be washed away and a black image will be left behind to form a negative image.

Glossary

fluoridation

Past Paper Question

96 1A 2 a 97 1A 3 a i iii

bleach

disinfectant

photographic film

96 1A 2 a 2a State the method by which sodium hydroxide is produced industrially. Give TWO other products obtained by this industrial process. NaOH(aq) is produced by electrolysis of brine / concentrated NaCl(aq) 1 mark (Deduct ½ mark for not mentioning ‘concentrated’.) ½ + ½ mark Other products: H2 / Cl2 / NaOCl solution (bleach solution) (any TWO) 97 1A 3 a i iii 3a For each of the following materials, identify ONE inorganic compound which is an active ingredient: i photographic film iii bleaching powder

2

3

II. Nitrogen and its compounds

Unit 1

Topic

II. Nitrogen and its compounds

Reference Reading

10.2.0–10.2.2

Page 1

Unit 1

Assignment Chemistry in Context, 3rd Edition ELBS pg. 362–365 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 443–448 Reading

A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 142–159 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 312–315

Syllabus Notes

II. Nitrogen and its compounds A. Nitrogen 1.

Unreactive nature of nitrogen

Comparing with other p-block elements, nitrogen is quite unreactive because of the strong N≡N triple bond. A very high temperature is required to break the bond. Therefore, the reaction with N2(g) involves a very high activation energy.

When Mg(s) is burning in air, the temperature is high enough to break the N≡N bond. This leads to formation of yellow magnesium nitride. Mg(s) + N2(g) → Mg3N2(s) N–N N=N N≡N N–H H–H

Bond energy (kJmol-1) + 158 + 410 + 945 + 391 + 436

C–C C=C C≡C C–H

Bond energy (kJmol-1) + 347 + 612 + 838 + 435

II. Nitrogen and its compounds 2.

Unit 1

Direct combination of nitrogen and oxygen

Normally, nitrogen doesn't react with oxygen. But at a very high temperature, e.g. during lightning or in the internal combustion engine, N2(g) combines with O2(g) to produce colourless NO(g). In air, NO(g) is further oxidized to NO2(g). N2(g) + O2(g) → NO(g) (involves high activation energy) NO(g) + O2(g) → NO2(g) (spontaneous at room temperature) In laboratory, this can be done by passing a high tension across an electric arc. The test tube of air will turn brown gradually because of the formation of NO2(g).

NO2(g) is an acid anhydride which form nitric(V) and nitric(III) acid when dissolved in water. NO2(g) + H2O(l) → HNO3(aq) + HNO2(aq) HNO2(aq) is a weak acid which is vulnerable to air oxidation to strong HNO3(aq) acid. 2HNO2(aq) + O2(g) → 2HNO3(aq) This is why the area with heavy traffic usually has more serious acid rain problem. Overall equation

4NO2(g) + 2H2O(l) + O2(g) → 4HNO3(aq)

Page 2

II. Nitrogen and its compounds

Unit 1

Page 3

B. Ammonia 1.

Haber process

Haber Process – manufacturing of ammonia gas N2(g) + 3H2(g) d 2NH3(g)

∆H = -92 kJ mol-1

Fe(s) catalyst, 200 atm, 500ºC

This method was invented by a German scientist called Haber.

Effect of temperature and pressure on the yield and rate of Haber process N2(g) + 3H2(g) d 2NH3(g)

∆H = -92 kJ mol-1

According to Le Chatelier’s principle, the reaction is favored by low temperature and high pressure. On another hand, an increase in pressure or temperature both increases the rate of reaction. ↑Pressure ↑Temperature

yield ↑ ↓

rate of reaction ↑ ↑

Therefore, a high pressure is ideal for achieving both high yield and high rate of reaction. However, manufacturing at high pressure requires the use of a strong reacting vessel which is costly to build and operate. Therefore, only 200 atm was chosen. 500ºC is chosen because if the temperature is too high, the yield will be too low. Moreover, if the temperature is too low, the rate of reaction will be too slow and a very long time will be required for the reactants to react.

The percentage of ammonia in the equilibrium mixture obtained from a 1:1 mixture of N2 and H2 at different temperature and pressure

The conditions chosen is actually a compromise between yield and rate of reaction. At such conditions, the production of ammonia would be most economical.

II. Nitrogen and its compounds 2.

Unit 1

Page 4

Chemical properties of ammonia

a) Alkaline properties Ammonia is a weak alkali in water. Kb of ammonia at 25ºC is 1.8 × 10-5 moldm-3. NH3(aq) + H2O(l) d NH4+(aq) + OH-(aq) (1) Laboratory preparation of ammonia For this reason, ammonia gas is usually prepared by heating an ammonium salt will a strong alkali in laboratory. e.g. soda lime (a mixture of Ca(OH)2(s) and NaOH(s)) 2NH4Cl(s) + Ca(OH)2(s) → NH3(g) + CaCl2(aq) + H2O(l) Since water is produced, it has to be removed by a drying agent. Anhydrous calcium oxide is the only suitable common drying agent. This is because other common drying agents e.g conc. H2SO4(l) and anhydrous CaCl2(s), react with ammonia. Sulphuric acid neutralizes ammonia. 2NH3(g) + H2SO4(aq) → (NH4)2SO4(aq) Anhydrous calcium chloride forms complex with ammonia. 4NH3(g) + CaCl2(s) → CaCl2·4NH3(s) In the experimental setup, the reacting flask containing ammonium chloride and strong alkali have to be inclined downwards to prevent the water condensed at the mouth from cracking the flask. Moreover, ammonia is lighter than air and very soluble in water, it is collected by the method of upward delivery (downward displacement of air) or using gas syringe.

b) Reducing properties The nitrogen atom in ammonia has the lowest possible oxidation state of nitrogen, -3. Therefore, ammmonia can only behave as a reducing agent in all kinds of redox reaction. Indeed, it is a fairly strong reducing agent. (1) Combustion Ammonia doesn’t burn in ordinary air but it burns with a yellow flame in pure oxygen to give nitrogen. 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l)

II. Nitrogen and its compounds

Unit 1

Page 5

(2) Catalytic oxidation (Ostwald process in laboratory scale) In the presence of platinum catalyst, ammonia can be oxidized further to NO(g) at high temperature. In the experimental setup, oxygen is bubbled through concentrated ammonia solution and mixed with ammonia vapour. A red hot coil of platinum catalyst is then lowered into the flask. Pt catalyst

→ 4NO(g) + 6H2O(l) 4NH3(g) + 5O2(g) 

∆H = -907 kJmol-1

Since the reaction is highly exothermic, the platinum wire will continue to glow once the reaction has started. The initial product NO(g) is colourless which reacts with atmospheric oxygen to form brown NO2(g). NO(g) + O2(g) → NO2(g) Brown fume of NO2(g) will be observed at the mouth of the flask. (3) With copper(II) oxide Ammonia can also reduce oxides of unreactive metal to the metals with heating. e.g. CuO(s) → Cu(s) 3CuO(s) + 2NH3(g) → 3Cu(s) + N2(g) + 3H2O(g)

(4) With chlorine Chlorine is a strong oxidizing agent. Ammonia burns spontaneously in chlorine. 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g) If ammonia is in excess, a dense white fume of NH4Cl(s) will also be observed. NH3(g) + HCl(g) → NH4Cl(s)

II. Nitrogen and its compounds

Unit 1

Page 6

(5) Thermal decomposition of ammonia Thermal decomposition of ammonia is only the reverse of Haber process. The decomposition is almost complete at 1 atm in the presence of Fe catalyst with heating. This is done by passing ammonia back and forth the heated iron wool.

2NH3(g) → N2(g) + 3H2(g) Because 2 moles of ammonia will decompose into 4 moles of gaseous products. The volume of the gas will be doubled if all ammonia is decomposed.

Glossary

high tension electric arc Haber process Le Chatelier’s principle catalytic oxidation (Ostwald process) thermal decomposition of ammonia

II. Nitrogen and its compounds

Past Paper Question

90 2B 5 b 92 2B 6 Ac iii 97 2B 8 b ii iii 99 2A 1 c i

Unit 1

Page 7

97 2B 8 c i ii 99 2A 4 a iii

90 2B 5 b 5b Outline an industrial preparation of ammonia. Your answer should include the physico-chemical principles involved and also the role of any catalyst. Ammonia N2 + 3H2 d 2NH3 ∆H < 0 / exothermic 1 mark Conditions of reaction: 200 atm, 500ºC, finely divided iron catalyst 1 mark Effect of temperature on the equilibrium constant 1 mark Effect of pressure 1 mark } any 2 1 mark 2 marks Effect of concentration of N2 and H2 1 mark Catalyst - required because of the inertness of N2 and H2 molecules 1 mark NH3 is removed from the reaction mixture Overall presentation 1 mark 92 2B 6 Ac iii 6Ac iii Ammonia is synthesized by the Haber Process. With reference to the availability and preparation of raw materials, discuss the major economic aspects of this process. N2(g) atmosphere fractional distillation H2(g) petroleum industry cracking /steam reforming locate near oil refinery to save transport costs recycle of raw material use of catalyst 1 mark each 97 2B 8 b ii iii 8b The synthesis of ammonia using the Haber Process involves the following: N2(g) + 3H2(g) d 2NH3(g) ∆Ho = -92 kJ mol-1 ii Name a catalyst for the process and state the effect of the catalyst on the reaction. iii State how ammonia is isolated from unreacted nitrogen and hydrogen in the process. 97 2B 8 c i ii 8c The reaction of chlorine with excess ammonia in the gas phase to give ammonium chloride and nitrogen involves several steps. i Write a balanced equation for the overall reaction. ii Briefly explain how ammonia acts as a reducing agent and as a base in the reaction. 99 2A 1 c i 1c For each of the following pairs of substances, suggest a chemical test to distinguish one from the other. For each test, give the reagent(s) used and the expected observation. i CsCl(s) and NH4Cl(s) 99 2A 4 a iii 4a In the Haber process, ammonia is synthesized by the exothermic reaction of nitrogen and hydrogen at around 723K. N2(g) + 3H2(g) d 2NH3(g) In a simulation of the process, a mixture of nitrogen and hydrogen was placed in a closed container. The initial concentrations of nitrogen and hydrogen were 0.50 mol dm-3 and 1.50 mol dm-3 respectively. When the equilibrium was attained at 723 K, 25.0% of the original nitrogen was consumed. iii (I) State, with explanation, the effect of temperature on Kc for the reaction. (II) Explain why the Haber process is not operated at temperatures much higher or much lower than 723 K.

7

4

7

4

II. Nitrogen and its compounds

Unit 2

Topic

II. Nitrogen and its compounds

Reference Reading

10.2.3

Page 1

Unit 2

Assignment Chemistry in Context, 3rd Edition ELBS pg. 365–366 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 159–162 Reading

A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 453–457 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 315

Syllabus Notes

C. Nitric(V) acid 1.

Ostwald Process

Nitric(V) acid is manufactured from NH3(g) industrially through the NO(g) and NO2(g) intermediates. The process is called Ostwald process. Although NO(g) can be prepared by direct oxidation of N2(g) molecule, this is not economical because the activation energy involved is very high.. NH3(g) serves as an intermediate to break the whole process into steps with lower activation energies. N2 → NH3(g) → NO2(g) → HNO3(aq) Ostwald process a.

NH3(g) is first oxidized by O2(g) in air to NO(g) in the presence of Pt(s) catalyst at 900 ºC and 8 atm. Pt 4NH3(g) + 5O2(g)  → 4NO(g) + 6H2O(l) ∆H = - 950 kJmol-1 Similar to Haber process, the conditions chosen in Ostwald process is also a compromise between yield and rate of reaction. Since the reaction is highly exothermic, the heat generated is diverted to heat up the raw material in the heat exchanger.

b.

Cooled NO(g) is further oxidized by O2(g) to NO2(g) 2NO(g) + O2(g) → 2NO2(g)

c.

Nitrogen dioxide is dissolved in water in the presence of O2(g) to form HNO3(aq) 4NO2(g) + O2(g) + 2H2O(l) → 4HNO3(aq) The final solution contains about 60% by mass of nitric(V) acid. It can be concentrated to concentrated nitric(V) acid by distillation. Concentrated nitric(V) acid is an azeotrope of 68% nitric(V) acid. It can further be concentrated to pure nitric(V) acid by distilling with conc. H2SO4(l) or P2O5(s).

II. Nitrogen and its compounds 2.

Unit 2

Page 2

Oxidizing properties of nitric(V) acid

In contrast to ammonia, nitrogen in nitric(V) acid has the highest possible oxidation state, +5. It can only behaves as an oxidizing agent. Depending on the concentration, nitric(V) acid shows different degree of oxidizing power and gives different products. Concentrated nitric(V) acid (16M) Dilute nitric(V) acid (2M) Very dilute nitric(V) acid (0.1M)

NO3-(aq) + 2H+(aq) + e- d NO2(g) NO3-(aq) + 4H+(aq) + 3e- d NO(g) 2H+(aq) + 2e- d H2(g)

Very dilute nitric(V) acid behaves similarly to other aqueous acids. It has only very weak oxidizing power offered by the H+(aq) ions. Indeed, nitric(V) acid can also be reduced to other products with different oxidation states depending on the condition. NO3-(aq) + 5H+(aq) + 4e- d 2½H2O + ½N2O NO3-(aq) + 8H+(aq) + 6e- d 2H2O + H2NOH NO3-(aq) + 10H+(aq) + 8e- d 3H2O + NH4+ a) With copper Concentrated nitric(V) acid is a very strong oxidizing. It even oxidizes less reactive metal like copper. with very dilute nitric(V) acid with dilute nitric(V) acid (with heating) with concentrated nitric(V) acid (cold)

no reaction 3Cu(s) + 2NO3-(aq) + 8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l) Cu(s) + 2NO3-(aq) + 4H+(aq) → Cu2+(aq) + 2NO2(g) + 2H2O(l)

Surprisingly, concentrated HNO3(aq) has no reaction with Fe(s) due to the formation of an impermeable oxide layer. Thus, Fe(s) is said to be passive to concentrated HNO3(aq). Therefore, concentrated HNO3(aq) is usually stored in steel tanks. b) With iron(II) ion 3Fe2+(aq) + NO3-(aq) + 4H+(aq) → 3Fe3+(aq) + NO(g) + 2H2O(l) Concentrated nitric(V) acid oxidizes iron(II) salt to iron(III) salt. The NO(g) produced will be further oxidized to brown NO2(g) in air. Furthermore, the NO molecule is capable to form a brown complex [Fe(H2O)5NO]2+(aq) with Fe2+(aq). The Fe2+(aq) solution may turn the solution brown eventually.

II. Nitrogen and its compounds c)

Unit 2

Page 3

With sulphur Sulphur has no reaction with cold conc. HNO3(aq). But upon heating, conc. HNO3(aq) can oxidize S(s) to sulphuric acid. S(s) + 6HNO3(aq) → H2SO4(aq) + 6NO2(g) + 2H2O(l)

Glossary

Ostwald process

Past Paper Question

92 2B 6 Ab ii 93 2B 5 Ia 97 2B 8 a ii

passive

92 2B 6 Ab ii What are the products of the reactions between ii cold, concentrated nitric(V) acid and hydrogen sulphide gas? H2S(g) + 2HNO3(conc.) → S(s) + 2H2O(l) + 2NO2(g) note: on boiling, H2SO4(aq) formed

6Ab

1 1 mark

93 2B 5 Ia 5I For the industrial preparation of nitric(V) acid from nitrogen , give the chemical reactions and the conditions a under which they take place. Pass purified N2 and H2 (ratio 1:3) over iron catalyst at 450ºC and 200 atm pressure. 2 marks N2 + 3H2 d 2NH3 Ammonia is mixed with O2 (excess air) and passed over a platinum catalyst at 900ºC at 8 atm. 4NH3 + 5O2 → 4NO + 6H2O 2NO + O2 → 2NO2 3 marks Resulting gaseous mixture is passed through water. 1 mark 4NO2 + O2 + 2H2O → 4HNO3 97 2B 8 a ii 8a Suggest how the following nitrogen oxides can be prepared in the laboratory. In each case, state the reactant(s) used and the reaction conditions, and write balanced equation(s) for the reaction(s) involved. ii nitrogen monoxide, NO

6

6

II. Nitrogen and its compounds

Unit 3

Topic

II. Nitrogen and its compounds

Reference Reading

10.2.4

Page 1

Unit 3

Assignment Chemistry in Context, 3rd Edition ELBS pg. Inorganic Chemistry, 4th Edition, T.M. Leung pg. 162–169 Reading

A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 457–458

Syllabus Notes

D. Nitrate(V) salt 1.

Thermal decomposition of nitrate

Upon heating, different nitrate(V) salts will decompose to give different products depending on the reactivity of the metal in the salt. Reactivity series of metal K, Na, Ca, Mg, Al, Zn, Fe, Pb, Cu, Hg, Ag, Au i.

For reactive metal nitrate(V) e.g. KNO3(s) and NaNO3(s), they will decompose into nitrate(III) and oxygen. thermal decomposition

→ Reactive metal nitrate(III) + Oxygen Reactive metal nitrate(V)   e.g. 2NaNO3(s) → 2NaNO2(s) + O2(g) For this reason, KNO3 is an ingredient of gunpowder. Upon heating, it releases oxygen which supports the burning of gunpowder. ii. For less reactive metal nitrate(V) e.g. LiNO3(s), from Ca(NO3)2(s) to Cu(NO3)2(s), it will decompose into oxide, nitrogen dioxide and oxygen. thermal decomposition

→ Less reactive metal oxide + Nitrogen dioxide + Oxygen Less reactive metal nitrate(V)   e.g 2Ca(NO3)2(s) → 2CaO(s) + 4NO2(g) + O2(g) iii. For non-reactive metal nitrate(V) e.g. HgNO3(s) and AgNO3(s), they will decompose into metal, nitrogen dioxide and oxygen. (N.B. Au does not form stable solid nitrate at all.) thermal decomposition

→ Non-reactive metal + Nitrogen dioxide + Oxygen Non-reactive metal nitrate(V)   e.g. 2AgNO3(s) → 2Ag(s) + 2NO2(g) + O2(g) This may be considered as a 2 steps decomposition. 4AgNO3(s) → 2Ag2O(s) + 4NO2(g) + O2(g) 2Ag2O(s) → 4Ag(s) + O2(g) The stability of a nitrate(V) salt serves as an important clue of identifying the cation.

II. Nitrogen and its compounds

Unit 3

Page 2

iv. Ammonium nitrate(V) NH4NO3(s) will decompose to dinitrogen oxide and water leaving no residue behind. thermal decomposition

NH4NO3(s)   → N2O(g) + 2H2O(l) In this reaction, a solid reagent expands suddenly to gaseous products, therefore, NH4NO3(s) is potentially explosive. It should not be heated too strongly. N.B.

0 − 5° C

Recall the action of nitric(III) acid on amino group, R–NH2 + HNO2(aq) → N2(g) + ROH NH4NO2(s) may be considered as a mixture of HNO2 and NH3 NH4NO2 d HNO2 + NH3 Therefore, NH4NO2(s) decomposes upon heating to give N2(g) and H2O(l) only. thermal decomposition

NH4NO2 (s)   → N2(g) + 2H2O(l)

2.

Brown ring test

The presence of nitrate(V) ion can be detected by brown ring test. In brown ring test, the salt to be test is dissolved in a freshly prepared iron(II) sulphate solution. Concentrated sulphuric acid is then added slowly and carefully into the resulting mixture without stirring it up. Because concentrated sulphuric acid is denser in water, it will sink to the bottom. A brown ring will form at the interface between the concentrated sulphuric acid and the aqueous layer if nitrate(V) ion is present.

The reactions occurring at the interface are : NO3-(aq) + H2SO4(l) → HNO3(aq) + HSO4-(aq)f HNO3(aq) + 3Fe2+(aq) + 3H+(aq) → 2H2O(l) + NO(g) + 3Fe3+(aq) FeSO4(aq) + NO(g) → FeSO4·NO(aq) a simplified formula of the brown complex or [Fe(H2O)5NO]2+](aq)

Glossary

thermal decomposition of ntirate

brown ring test

II. Nitrogen and its compounds

Past Paper Question

91 1A 3 h 97 1A 3 a ii

Unit 3

Page 3

91 1B 4 a 97 2B 8 a i

91 1A 3 h 3h When solid ammonium nitrate(V) is transported, the vehicle must carry a sign clearly showing : AMMONIUM NITRATE ; DANGEROUS GOODS ; NO SMOKING. Give chemical reasons why this is necessary. NH4NO3(s) is unstable and readily/easily decomposes to give large volumes of gaseous products (N2O and H2O) 1 mark and is therefore explosive. 1 mark No smoking because N2O is a supporter of combustion at elevated temperature 91 1B 4 a 4a You are give two solid samples, one of which is sodium nitrate(V) and the other sodium bromide. Outline laboratory tests to identify each anion positively, giving the reagents, reaction conditions and the expected observations (Equations are not required.) For NaNO3(s), add conc. H2SO4 and heat / add conc. H2SO4 + Cu and heat to give brown fumes of NO2 having no action on fluorescein (C20H12O5 yellowish to red in colour) impregnated paper. Add freshly prepared FeSO4 and conc. H2SO4 slowly / dropwise to give brown ring at the liquid junction.

2

4

For NaBr(s), add conc. H2SO4 to give acid vapour of HBr which gives white fumes with NH3 and reddish brown vapours of Br2 which turns fluorescein impregnated paper red. or Add acidified AgNO3 solution to give pale yellow ppt which is slightly / partially / sparingly soluble in excess NH3 or which darkens in colour on exposure to sunlight. or Add HOCl solution to give orange-red colour of bromine. When the solution is shaken with CS2 / CHCl3 / CCl4, ½ mark for each point the Br2 dissolves in the organic solvent leaving a colourless aqueous layer. 97 1A 3 a ii 3a For each of the following materials, identify ONE inorganic compound which is an active ingredient: ii gun powder 97 2B 8 a i 8a Suggest how the following nitrogen oxides can be prepared in the laboratory. In each case, state the reactant(s) used and the reaction conditions, and write balanced equation(s) for the reaction(s) involved. i dinitrogen oxide, N2O

3

6

II. Nitrogen and its compounds

Unit 4

Topic

II. Nitrogen and its compounds

Reference Reading

10.2.5

Page 1

Unit 4

Assignment A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 456 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 355–357 Reading

Modern Organic Chemistry (4th ed.), Bell & Hyman, 154–155, 320 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 316–317

Syllabus

Use of nitrc(V) acid

Notes

E. Use of nitric(V) acid Majority of nitric acid produced is used to make fertilizers and explosives. They accounts for 75% and 15% of the total consumption. a) Fertilizers Nitric acid is used in making of fertilizer such as NH4NO3 by neutralization with ammonia. b) Explosives Nitroglycerin, a constituent of several explosives is manufactured by nitration of propane-1,2,3-triol using concentrated nitric acid. CH2 OH

CH2 O NO2

CH OH + 3 HNO3

CH O NO2

CH2 OH

CH2 O NO2

propane-1,2,3-triol (glycerol)

propane-1,2,3-triyl trinitrate (nitroglycerine)

CH2 O NO2 12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)

CH O NO2 CH2 O NO2 propane-1,2,3-triyl trinitrate (nitroglycerine)

Nitroglycerine is a colourless, oily liquid which is violently exploded on slight shock. Oxygen is present in the molecule, and carbon dioxide, water vapour and nitrogen are liberated to produce a very large pressure. Dynamite, invented by the Swedish chemist, Nobel, is made by allowing kieselguhr (a kind of clay from seashell) to absorb nitroglycerin. Although it retains its explosive properties, the nitroglycerin is less sensitive to shock. Besides nitroglycerin, many other explosives like trinitrotoluene (T.N.T., the yellow explosive), gun-cotton are also prepared by nitration using concentrated nitric acid. CH3 O 2N

CH2ONO2

NO2

O

H H ONO2

H

H

ONO2

NO2 methyl-2,4,6-trinitrobenzene (trinitrotoluene, T.N.T.)

H

n cellulose nitrate (nitrocellulose, gun cotton)

II. Nitrogen and its compounds c)

Unit 4

Page 2

Nylon

In industrial production of nylon, hexanedioic acid is manufactured by oxidation of a mixture of cyclohexanol and cyclohexanone using moderately concentrated nitric acid and a copper salt as catalyst. OH

O 60% HNO3, 60 º C, cat.

+

HOOC

CH2CH2CH2CH2 COOH

Part of the hexanedioic acid is converted to hexane-1,6-diamine through hexanedinitrile. Hexanedioic acid and hexane-1,6-diamine are then used to make nylon-6.6. d) Dyestuffs Azo dye is prepared by coupling diazonium ion with phenolic compound.

+ N N

H N

OH

diazonium ion

+ H2O

N

OH

OH

+

naphthalen-2-ol

N

-

OH

N H

azo dye (an orange red ppt.)

For example, the food colour Orange II is made by diazotising the sodium salt of 4-aminobenzenesulphonic acid and coupling the product to naphthalen-2-ol : -

+

SO2O Na

+-

Na OO2S

O

+ N N +

-

N N OH Orange II

Nitric acid is used to prepare nitrobenzene through which diazonium ion is prepared. NO2 HNO3 , H2SO4 50 - 55 蚓

+ N N

NH2 Sn HCl

NaNO2 HCl, 0-5 蚓 benzenamine (an aromatic primary amine)

diazonium ion

II. Nitrogen and its compounds

Glossary

nitroglycerine

Past Paper Question

97 2B 8 b iv

Unit 4 trinitrotoluene (T.N.T.)

97 2B 8 b iv 8b iv Give TWO major uses of ammonia in industry.

Page 3 dynamite

azo dye

diazonium ion

III. Sulphur and its compounds

Unit 1

Page 1

Topic

III. Sulphur and its compounds

Unit 1

Reference Reading

10.3.0–10.3.2 Advanced Practical Chemistry, John Murray (Publisher) Ltd., 192–194

Assignment A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 417–418, 426–427 Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 348–351 Reading Syllabus Notes

III. Sulphur and its compounds A. Allotrope of sulphur Normally, sulphur molecule consists of 8 sulphur atoms joined together, S8. S8 molecule has the shape like a crown. The atomicity of sulphur is said to be 8. Atomicity means the number of atoms in a molecule of an element.

Sulphur burns in oxygen with a blue flame and sulphur dioxide is formed.

or simply

S8(s) + 8 O2(g) → 8 SO2(g) S(s) + O2(g) → SO2(g)

Heating of sulphur

Characteristics of sulphur dioxide 1. Smell – SO2(g) has characteristic smell. 2. Acidity – SO2(g) forms sulphurous acid, H2SO3(aq) in water which turns blue litmus paper red. 3. Reducing properties – SO2(g) is a reducing agent. It turns acidified K2Cr2O7(aq) from orange to green. Sulphur crystal

Different forms of the same element in the same physical state are called allotropes of the element. e.g. diatomic oxygen O2(g) and ozone O3(g) are two allotropes of oxygen There are several different sulphur allotropes. e.g. rhombic sulphur, monoclinic sulphur and plastic sulphur At room temperature, sulphur exists in 2 forms of crystal called rhombic sulphur and monoclinic sulphur. They melts at 113ºC and 119ºC respectively. Thermodynamically, rhombic sulphur is more energetically stable at room temperature while monoclinic sulphur is more stable at high temperature. Rhombic sulphur can be prepared by evaporating a solution of sulphur in carbon disulphide at room temperature while monoclinic sulphur can be prepared by cooling molten sulphur slowly at an elevated temperature.

III. Sulphur and its compounds

Unit 1

Page 2

Both rhombic and monoclinic sulphur consists of S8 molecules. Upon heating, S8 molecules agitate and the solid starts to melt. It turns into a light yellow liquid (a → b). If the heating is continued, the ring will open to form short chains (b → c). The short chains join to form long chains. Each chains has up to 106 atoms. At this stage, the liquid becomes dark red and viscous due to the large van der Waals’ forces. (c → d → e). If the sulphur is heated continuously, it will boil at 444ºC and the long chain will start to break up. However, if the boiling sulphur is cooled rapidly, the long chain structure will be freezed and forming plastic sulphur. It is stretchable like plastic (e → f). On standing, it changes back to rhombic sulphur gradually.

III. Sulphur and its compounds

Unit 1

Page 3

B. Burning of sulphur Sulphur burns in air with a blue flame to form sulphur dioxide. S(s) + O2(g) → SO2(g) This can be used to prepared sulphur dioxide industrially for the contact process in the manufacturing of sulphuric(VI) acid.

C. Sulphur dioxide / sulphate(IV) salt Laboratory preparation of sulphur dioxide However, sulphur dioxide is seldom produced by burning of sulphur in laboratory due to the complicate experimental setup required. Depending on the amount of sulphur dioxide to be prepared, a redox reaction or a displacment reaction may be used. Hot conc. H2SO4(l) is capable to oxidize copper to copper(II) ion while the acid will be reduced to SO2(g). → CuSO4(aq) + SO2(g) + 2H2O(l) Cu(s) + 2H2SO4(l) heat

Besides SO2(g), water is also produced which has to be removed by passing through a washing bottle conc. H2SO4(l). Eventually, the SO2(g) can be collected by downward delivery or a gas syringe because it is heavier than air and quite soluble in water. If only a small amount of SO2(g) is required, a sulphate(IV) salt would be used instead. Upon the addition of a strong acid e.g. HCl(aq), the weak acid, sulphuric(IV) acid would be displaced from the sulphate(IV) salt. Eventually, SO2(g) could be prepared. SO32-(aq) + 2H+(aq) → H2SO3(aq) → SO2(g) + H2O(l)

III. Sulphur and its compounds 1.

Unit 1

Page 4

Reducing properties of sulphur dioxide

Sulphur dioxide is a soluble gas, though not as soluble as HCl(g) or NH3(g). In aqueous state, SO2(g) weakly dissociates to HSO3-(aq) or SO32-(aq). SO2(g) + H2O(l) d H3O+(aq) + HSO3-(aq)

Ka1 = 1.5 × 10-2 mol dm-3

HSO3-(aq) + H2O(l) d H3O+(aq) + SO32-(aq)

Ka2 = 6.2 × 10-8 mol dm-3

Since the dissociation is rather weak, H2SO3(aq) can also be written as SO2(aq) and H2O(l). And it is observed that the redox properties of SO2(aq) and SO32-(aq) are similar. 4H+(aq) + SO42-(aq) + 2e- d H2SO3(aq) + H2O(l)

Eo = + 0.17 V

Because SO2(aq) is a mild reducing agent it reacts with strong oxidizing agent like MnO4-(aq) and Cr2O72-(aq). a) With manganate(VII) ion H2SO3(aq) + H2O(l) d 4H+(aq) + SO42-(aq) + 2e- × 5 MnO4-(aq) + 8H+(aq) + 5e- d Mn2+(aq) + 4H2O(l) × 2 Overall 5SO32-(aq) + 2MnO4-(aq) + 6H+(aq) d 5SO42-(aq) + 2Mn2+(aq) + 3H2O(l)

Eo = - 0.17 V Eo = + 1.51 V Eo = + 1.34 V

The purple colour of MnO4-(aq) will be decolorized by sulphur dioxide. b) With dichromate(VI) ion H2SO3(aq) + H2O(l) d 4H+(aq) + SO42-(aq) + 2e- × 3 Cr2O72-(aq) + 14H+(aq) + 6e- d 2Cr3+(aq) + 7H2O(l) Overall 3SO32-(aq) + Cr2O72-(aq) + 8H+(aq) d 3SO42-(aq) + 2Cr3+(aq) + 4H2O(l)

Eo = - 0.17 V Eo = + 1.33 V Eo = + 1.16 V

Orange Cr2O72-(aq) will be reduced to green Cr3+(aq) by sulphur dioxide. c)

With bromine H2SO3(aq) + H2O(l) d 4H+(aq) + SO42-(aq) + 2eBr2(aq) + 2e- d 2Br-(aq)

Overall SO32-(aq) + Br2(aq) + H2O(l) d SO42-(aq) + 2Br-(aq) + 2H+(aq)

Eo = - 0.17 V Eo = + 1.09 V Eo = + 0.92 V

The orange colour of Br2(aq) will be decolorized. 2.

Oxidizing properties of sulphur dioxide

Normally, sulphur dioxide doesn’t behave as a oxidizing agent. But in the presence of a strong reducing agent. e.g. Mg(s), SO2(g) can also be reduced to S(s). a) With magnesium 2Mg(s) + SO2(g) → 2MgO(s) + S(s) Magnesium burns continuously in a gas jar of sulphur dioxide to from white magnesium oxide and yellow sulphur powder.

III. Sulphur and its compounds 3.

Unit 1

Page 5

Distinguishing from carbon dioxide

Comparatively, sulphur dioxde and carbon dioxide are quite similar in chemical properties. Both form white precipitate with limewater, silver nitrate and barium chloride solutions. SO32-(aq) + Ca2+(aq) → CaSO3(s) CO32-(aq) + Ca2+(aq) → CaCO3(s) SO32-(aq) + 2Ag+(aq) → Ag2SO3(s) CO32-(aq) + 2Ag+(aq) → Ag2CO3(s) SO32-(aq) + Ba2+(aq) → BaSO3(s) CO32-(aq) + Ba2+(aq) → BaCO3(s) However, they also show certain differences. Sulphur dioxide is much more soluble in water and is pungent in smell. Moreover, the main difference between the two is that sulphur dioxide is reducing while carbon dioxide is non-reducing. Therefore, they are be distinguished by using an oxidizing agent e.g. KMnO4/H+(aq), K2Cr2O7/H+(aq).

Glossary

atomicity

Past Paper Question

93 2B 5 IIc

rhombic sulphur

monoclinic sulphur

plastic sulphur

93 2B 5 IIc II SO2 can act as an oxidant or a reductant. Give one reaction in each case to illustrate this. c SO2 as reducing agent : reaction with acidified K2Cr2O7 / KMnO4 Cr2O72- + 3SO2 + 2H+ → 2Cr3+ + 3SO42- + 5H2O or MnO4- + 5SO2 + 2H2O → 2Mn2+ + 5SO42- + 5H2O + 4H+ SO2 as oxidising agent : H2S + SO2 → S(s) + H2O (any 2 appropriate examples)

2

2 marks

III. Sulphur and its compounds

Unit 2

Topic

III. Sulphur and its compounds

Reference Reading

10.3.3–10.3.4

Page 1

Unit 2

Assignment Chemistry in Context, 3rd Edition ELBS pg. 359–362 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 428–434 Reading

Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 348–350 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 310–311

Syllabus Notes

D. Concentrated sulphuric(VI) acid 1.

Contact process

Sulphur dioxide used in contact process in obtained by 1. Burning sulphur S(s) + O2(g) → SO2(g) 2. Roasting sulphur containing ores in air 4FeS2(s) + 11O2(g) → 2Fe2O3(s) + 8SO2(g) 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

1.

Vanadium(V) oxide (V2O5(s)) catalyst, 450 ºC, 1 atm SO2(g) + O2(g) d SO3(g) ∆H = -197 kJ mol-1 During the reaction, the gaseous reactant is in contact with the catalyst, so the process is called contact process. Because the reaction is exothermic, the hot emerging SO3(g) is diverted to the heat exchanger to heat up the incoming gas. This saves the fuel required to heat up the gas. Indeed, the conversion is done in several stages. The gaseous mixture is directed through the catalytic converter and the heat exchanger for several times until the conversion reaches 98%.

2.

Sulphur trioxide, SO3(g) is dissolved in conc. H2SO4(l) to form oleum, H2S2O7(l) (also called fuming sulphuric acid). SO3(g) + H2SO4(l) → H2S2O7(l)

3.

H2S2O7(l) is added to the correct amount of water to form conc. H2SO4(l). H2S2O7(l) + H2O(l) → 2H2SO4(l)

Note :

SO3(g) is not added to the water directly because the reaction is highly exothermic. This will vaporize the sulphuric acid formed and cause difficulty in collection of acid.

III. Sulphur and its compounds

Unit 2

Page 2

Effect of temperature and pressure on the yield 2SO2(g) + O2(g) d 2SO3(g)

∆H = -197 kJ mol-1

According to Le Chatelier’s principle the reaction is favored by a low temperature and a high pressure. However, if the temperature is too low, the rate of reaction will also be lowered and a longer time will be required to reach the equilibrium. Furthermore, though a high pressure gives a higher yield, the yield at 1 atm is already 98% after several stages of conversion. It is not worthy to use a pressure higher than 1 atm. This requires the use of a stronger reacting vessel which is more costly to build. ↑ Pressure ↑ Temperature

yield ↑ ↓

rate of reaction ↑ ↑

Actually, the condition, 1 atm and 450ºC with the catalyst of V2O5(s) is a compromise between yield and rate of reaction. At this condition, the production of SO3(g) is most economical. 2.

Properties of concentrated sulphuric acid

i.

Acid

Effect of temperature on the equilibrium constant and percentage of conversion

Conc. H2SO4(l) shows acidic properties when water is added. ii. Oxidizing properties with Cu(s) Normally, Cu(s) does not react will dilute sulphuric acid. It reacts with hot conc. H2SO4(l) because of its strong oxidizing properties. Oxidation half equation Reduction half equation

Cu(s) → Cu2+(aq) + 2eSO42-(aq) + 4H+(aq) + 2e- → SO2(g) + 2H2O(l)

Ionic equation

Cu(s) + SO42-(aq) + 4H+(aq) → Cu2+(aq) + SO2(g) + 2H2O(l)

Full equation

Cu(s) + 2H2SO4(l) → CuSO4(aq) + SO2(g) + 2H2O(l)

When water is added into the reacting mixture, it turns blue because of the presence of Cu2+(aq) ion.

III. Sulphur and its compounds

Unit 2

Page 3

iii. High affinity for water - Hygroscopic and dehydrating properties Conc. H2SO4(l) is hygroscopic and dehydrating. Hygroscopic is the properties of absorbing water vapour from air. Dehydrating is the properties of removing water from a stable structure and this implies a structural change. Hygroscopic properties Because of the hygroscopic properties, conc. H2SO4(l) can be used as a drying agent. By bubbling the wet gas through conc. H2SO4(l), the gas can be dried. However, since conc. H2SO4(l) is also acidic and oxidizing, it cannot be used to dry alkaline or reducing gases. e.g. Alkaline gas - ammonia NH3(g) Reducing gas - ethene C2H4(g)

Dehydrating properties with copper(II) sulphate-5-water Conc. H2SO4(l) removes the water of crystallization from CuSO4·5H2O(l). This turns the blue crystal to white powder. conc . H SO

2 4(l ) → CuSO4·5H2O(s)   blue crystal

CuSO4(s) + 5H2O(l) white powder

with paper The main composition of paper is cellulose which is a carbohydrate. It has the general formula Cx(H2O)y. Conc. H2SO4(l) is capable to extract the water from the structure of cellulose and char the paper. conc . H SO

2 4(l ) → xC(s) + yH2O(l) Cx(H2O)y(s)  

According to this principle, dil. H2SO4(aq) can be used as a magic ink. The word written on the paper can be developed by heating the paper to evaporate the water. Once the sulphuric acid is concentrated, it will char the paper and the word will appear. with sugar Similarly, sugar is another carbohydrate, it is also charred by conc. H2SO4(l). Upon heating, conc. H2SO4(l) is also capable to oxidize the C(s) to CO2(g) and raise the spongy mass. Oxidation half equation Reduction half equation

C(s) + 2H2O(l) → CO2(g) + 4H+(aq) + 4eSO42-(aq) + 4H+(aq) + 2e- → SO2(g) + 2H2O(l)

Ionic equation

C(s) + 2H2O(l) + 2SO42-(aq) + 8H+(aq) → CO2(g) + 4H+(aq) + 2SO2(g) + 4H2O(l) C(s) + 2SO42-(aq) + 4H+(aq) → CO2(g) + 2SO2(g) + 2H2O(l)

Full equation

C(s) + 2H2SO4(l) → CO2(g) + 2SO2(g) + 2H2O(l)

III. Sulphur and its compounds

Unit 2

Page 4

iv. Non-volatile acid conc. H2SO4(l) is a non-volatile acid which boils at 380ºC. This makes it capable to displace the more volatile. Displacement of hydrogen chloride Boiling point of conc. H2SO4(l) : 380 ºC Boiling point of hydrogen chloride : -85 ºC Room temperature : 25 ºC H2SO4(l) + NaCl(s) → NaHSO4(s) + HCl(g) Even at room temperature, hydrogen chloride is boiling. Conc. H2SO4(l) is capable to displace HCl(g) from NaCl(s). Displacement of nitric acid Boiling point of nitric acid : 83 ºC Similarly, pure HNO3(l) can be displaced from NaNO3(s). However, the boiling point of nitric acid is higher than room temperature. Gentle heating is required. → NaHSO4(s) + HNO3(g) H2SO4(l) + NaNO3(s) heat

E. Sulphate(VI) salt 1.

Test for sulphate(VI) ion

SO42-(aq) is usually tested with BaCl2(aq) solution. In the presence of BaCl2(aq), SO42-(aq) ion will be precipitated as insoluble BaSO4(s) and the solution will turn into a white suspension. Ba2+(aq) + SO42-(aq) → BaSO4(s) However, SO32-(aq) and CO32-(aq) also form white precipitate with Ba2+(aq). They are distinguished from SO42-(aq) by addition of acid. In acidic medium, BaSO3(s) and BaCO3(s) will redissolve. BaSO3(s) + 2HCl(aq) → BaCl2(aq) + H2O(l) + SO2(g) BaCO3(s) + 2HCl(aq) → BaCl2(aq) + H2O(l) + CO2(g) BaSO4(s) + HCl(aq) → no reaction (the ppt. remains insoluble)

Glossary

contact process barium chloride

vanadium(V) oxide

oleum / fuming sulphuric acid

hygroscopic

dehydrating

III. Sulphur and its compounds

Past Paper Question

Unit 2

90 2B 5 b 92 2B 6 Bb ii 93 2B 5 IIa 98 1B 7 a iii 99 2A 1 c ii

Page 5

98 2B 8 c i ii

90 2B 5 b 5b Outline an industrial preparation of sulphuric acid. Your answer should include the physico-chemical principles involved and also the role of any catalyst. Sulphuric acid S + O2 → SO2 2SO2 + O2 d 2SO3; ∆H < 0 very slow in the absence of catalyst H O SO3 + H2SO4 → H2S2O7  → 2H2SO4 3 marks Steps involved: Production of SO2, conversion to SO3, removal of SO3 Conditions of conversion: 1 atm, 450ºC, vanadium(V) oxide catalyst 1 mark 1 mark Effect of [O2], removal of SO3 Effect of pressure 1 mark } any 1 Effect of temperature on the equilibrium constant 1 mark 1 mark 1 mark Catalyst - lower the activation energy for the SO2 + O2 reaction as O2 is kinetically inert Overall presentation 1 mark

7

2

92 2B 6 Bb ii 6Bb What are the products of the reactions between ii concentrated sulphuric(VI) acid and hydrogen sulphide gas? H2SO4(conc.) + H2S(g) → S(s) + 2H2O(l) + SO2(g)

1 1 mark

93 2B 5 IIa II For the industrial preparation of sulphuric(VI) acid from sulphur, give the chemical reactions and the conditions a under which they take place. Sulphur is burnt in air. S + O2 → SO2 1 mark A mixture of SO2 + O2 (slightly excess air) is passed over V2O5 catalyst at 450ºC and 1 atm. pressure. 2 mark 2SO2 + O2 → 2SO3 1 mark SO3 gas is passed into conc. H2SO4 and then diluted with water SO3 + H2SO4 → H2S2O7 1 mark H2S2O7 + H2O → 2H2SO4 1 mark 98 1B 7 a iii 7a Briefly describe how you would test for the presence of the following chemical species in a sample of iron alum, (NH4)2SO4·Fe2(SO4)3·24H2O. iii sulphate(VI) ions 98 2B 8 c i ii 8c The conversion of sulphur dioxide to sulphur trioxide in the Contact Process involves the following : ∆Ho = -98 kJmol-1 SO2(g) + ½O2(g) d SO3(g) with V2O5(s) catalyst i State, with explanation, the effect of each of the following changes on the position of the equilibrium. (I) increasing the total pressure (II) increasing the temperature ii Briefly describe how sulphur trioxide is converted to sulphuric(VI) acid in this process. 99 2A 1 c ii 1c For each of the following pairs of substances, suggest a chemical test to distinguish one from the other. For each test, give the reagent(s) used and the expected observation. ii Na2SO3(aq) and Na2S2O3(aq) (4 marks)

6

1

5

III. Sulphur and its compounds

Unit 3

Page 1

Topic

III. Sulphur and its compounds

Reference Reading

10.3.5

Unit 3

Assignment Chemistry in Context (4th ed.), Thomas Nelson and Sons Ltd., 350–351 Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 312 Reading Syllabus

Use of sulphuric(VI) acid

Notes F. Use of sulphuric(VI) acid a) Fertilizers (1) By reacting with concentrated sulphuric(VI) acid, the water-insoluble calcium phosphate(V) (found in phosphate ores) can be converted to the more soluble calcium dihydrogenphosphate(V) which can be used as a phosphorus fertilizer. Ca3(PO4)2(s) + 2H2SO4(l) → Ca(H2PO4)2(s) + 2CaSO4(s) (2) Ammonium sulphate(VI) is produced by the reaction of ammonia with sulphuric(VI) acid. 2NH3(g) + H2SO4(aq) → (NH4)2SO4(aq) crystallization

→ (NH4)2SO4(s) (NH4)2SO4(aq)  b) Detergents Soapless detergents can be made by treating hydrocarbons from petroleum industry with concentrated sulphuric(VI) acid followed by sodium hydroxide. For example,

CH3(CH2)11

c)

(i) conc. H2SO4 (ii) NaOH

CH3(CH2)11

-

+

SO3 Na

Paints, Pigments and Dyestuffs (1) The white pigment titanyl sulphate, TiOSO4, is produced by dissolving titanium(IV) oxide, TiO2, in hot concentrated sulphuric(VI) acid. Besides, barium sulphate(VI) and calcium sulphate(VI) are used as paint additives. (2) Concentrated sulphuric(VI) acid is used with concentrated nitric(V) acid in the preparation of nitrobenzene which is used for making azo dyes. (3) By reacting the dye with concentrated sulphuric(VI) acid, sulphonic group –SO3H can be introduced to the dye molecule. This group makes the dye more soluble in water and easily attached to fibre.

III. Sulphur and its compounds

Unit 3

Page 2

Uses of sulphuric acid

Glossary

fertilizer calcium phosphate(V) calcium dihydrogenphosphate(V) ammonium sulphate(VI) detergent paint pigment dyestuff titanyl sulphate titanium(IV) oxide barium sulphate(VI) calcium sulphate(VI) azo dye sulphonic group

Past Paper Question

92 2B 6 Bc iii

92 2B 6 Bc iii iii Describe the part that H2SO4 plays in any TWO important industrial processes. not Contact Process Production of 'superphosphate' fertilizer 1 mark 2 marks Ca3(PO4)2 + H2SO4 → Ca(H2PO4)2 + 2CaSO4 acts as acid 1 mark or Production of fertilizer 1 mark 2 marks 2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq) acts as acid 1 mark or Manufacture of detergents or Manufacture of dyestuffs or Manufacture of drugs or Manufacture of explosives or Manufacture of rayon (artificial silk) 1 mark for function and 1 mark for process or Cleaning of metal before plating/galvanizing 2 processes required

6Bc

4

Topic

Nitrogen and Sulphur

Reference Reading

10.0 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.

Unit 1

Assignment Reading Syllabus Notes Glossary Past Paper Question

90 2B 5 b 81 2B 8 a i ii iii d 82 2B 10 b 84 2B 10 a i ii b 88 2B 6 d 89 2B 6 b

90 2B 5 b 5b Outline an industrial preparation of EITHER ammonia OR sulphuric acid. Your answer should include the physico-chemical principles involved and also the role of any catalyst. Ammonia N2 + 3H2 d 2NH3 ∆H < 0 / exothermic 1 mark Conditions of reaction: 200 atm, 500ºC, finely divided iron catalyst 1 mark Effect of temperature on the equilibrium constant 1 mark Effect of pressure 1 mark } any 2 1 mark 2 marks Effect of concentration of N2 and H2 1 mark Catalyst - required because of the inertness of N2 and H2 molecules 1 mark NH3 is removed from the reaction mixture Overall presentation 1 mark Sulphuric acid S + O2 → SO2 2SO2 + O2 d 2SO3; ∆H < 0 very slow in the absence of catalyst H O → 2H2SO4 SO3 + H2SO4 → H2S2O7  Steps involved: Production of SO2, conversion to SO3, removal of SO3 Conditions of conversion: 1 atm, 450ºC, vanadium(V) oxide catalyst 1 mark Effect of [O2], removal of SO3 Effect of pressure 1 mark } any 1 Effect of temperature on the equilibrium constant 1 mark Catalyst - lower the activation energy for the SO2 + O2 reaction as O2 is kinetically inert Overall presentation 2

3 marks 1 mark 1 mark 1 mark 1 mark

7

81 2B 8 a i ii iii d 8a Describe

I

II

8d

EITHER the process for the conversion of ammonia gas into (68%) nitric(V) acid, 12 OR the process for the conversion of sulphur dioxide gas into (98%) sulphuric acid. Give the conditions under which the various steps take place and briefly discuss which of the following will involve the highest costs: i energy, ii capital investment in plant, and iii handling. Concentrated nitric(V) acid 1 mark Ammonia(g) is mixed with air. passed over a Pt gauze or (Pt/Rh) 1 mark catalyst which is maintained at 800ºC and 8 atm. 1 mark 1 mark the reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) 1 mark takes place with the evolution of heat. The gases must be pure and avoid poisoning the catalyst. 1 mark The NO(g) is then oxidised by further air to NO2(g) 2NO(g) + O2(g) → 2NO2(g) 1 mark which is then absorbed in water. 1 mark 2NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) the NO(g) is recycled. 1 mark Absorption may be carried out up to 68%(conc.) HNO3 which is the constant boiling mixture. (i) energy – the reaction condition of 800ºC is not difficult and expensive to maintain because the reaction is exothermic. 1 mark (ii) The catalyst is expensive and it will lose and need to be refilled from time to time so it costs a lot. ( The lost of catalyst is due to the highly exothermic reaction produce energy which may vaporize the catalyst, Pt). 1 mark (iii) Because NO, NO2 and HNO3 are highly corrosive and toxic, so special chemical resistant materials have to be chosen for handling. 1 mark Concentrated sulphuric(VI) acid 1 mark SO2(g) is mixed with air 1 mark and the mixture is passed over a catalyst of Pt or V2O5. 1 mark Usually V2O5 is used because Pt is expensive and easily to be poisoned by impurities. e.g. As2O3. The operating temperature is about 450ºC at 1 atm. 1 mark 1 mark 2SO2(g) + O2(g) → 2SO3(g) where the reaction is exothermic. 1 mark The SO3(g) produced is then dissolved in 98% H2SO4 and fuming sulphuric acid (oleum) is formed. SO3(g) + H2SO4(l) → H2S2O7(l) 1 mark 1 mark Oleum is then diluted with calculated amount of water so as to give back 98% H2SO4(l) SO3(g) is not dissolved directly in water to form sulphuric acid because the reaction between SO3(g) and water is violent and a mist of acid droplets forms instead of a solution. 1 mark (i) energy – the reaction condition of 450ºC is not difficult and expensive to maintain because the reaction is exothermic. 1 mark (ii) The catalyst is expensive and it will lose and need to be refilled from time to time so it costs a lot. ( The lost of catalyst is due to the highly exothermic reaction produce energy which may vaporize the catalyst, Pt). 1 mark (iii) Because SO2, H2SO4 and H2S2O7 are highly corrosive and toxic, so special chemical resistant materials have to be chosen for handling. . 1 mark Write on the present-day uses of EITHER concentrated nitric(V) acid, 4 OR concentrated sulphuric(VI) acid. Uses : 4 uses of conc. HNO3 or H2SO4. 1 mark each conc. HNO3 Manufacture of explosive such as T.N.T. Manufacture of fertilizers such as NH4NO3 Manufacture of dyestuffs Manufacture of hexane-1,6-diamine which is a raw material for making nylon. Manufacture of soapless detergents conc. H2SO4 Manufacture of fertilizers such as (NH4)2SO4 Manufacture of explosives Manufacture of pigments

84 2B 10 a i ii b 10a i Give the formulae of the oxides of EITHER nitrogen OR sulphur

2

ii

10b

having oxidation numbers (O.N.) of +2, +4 and +6. N: NO(+2), NO2/N2O4(+4), NO3(+6) 2 marks S: SO/S2O2(+2), SO2(+4), SO3(+6) Describe what you would observe and give a balanced equation for the reaction the oxide of EITHER nitrogen (O.N. +2) OR sulphur (O.N. +4) (A) with acidified potassium manganate(VII) solution, (B) with an appropriate iron sulphate solution, (C) with rain, and (D) to form the element. Nitrogen (A) 5NO + 3MnO4- + 4H+ → 5NO3- + 3Mn2+ + 2H2O (Purple colour decolourized / colourless solution) (B) FeSO4 + NO → FeSO4·NO (Brown solution formed) (C) 2NO + O2 → NO2; 2NO2 + H2O → HNO3 + HNO2; 2HNO2 + O2 → 2HNO3 (no special observation) (D) Electric: 2NO d N2 + O2 at 3000ºC (no special observation) OR Burning white phosphorus: 10NO + 4P → 2P2O5 + 5N2 (white fume) Sulphur (A) 5SO2 + 2MnO4- + 2H2O → 5SO42- + 2Mn2+ + 4H+ (Purple colour decolourized / colourless solution) (B) 2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+ (Solution change from yellow brown to pale green / colourless) (C) SO2 + H2O → H2SO3; 2H2SO3 + O2 → 2H2SO4 (no special observation) (D) Burning magnesium: SO2 + 2Mg → 2MgO + S (yellow / white ppt. formed) OR Mix with hydrogen sulphide: SO2 + 2H2S → 3S + 2H2O (yellow / white ppt. formed) 2 marks each Describe the industrial preparation of EITHER nitric(V) acid from nitrogen OR sulphuric(VI) acid from sulphur. Nitrogen 1. Conversion of nitrogen to ammonia, N2 + 3H2 d 2NH3 450-500ºC; 200-1000 atmosphere; iron or iron oxide catalyst 3 marks 2. Conversion of ammonia to nitrogen(II) oxide, 4NH3 + 5O2 d 4NO + 6H2O 600ºC; 8 atmosphere; platinium catalyst 3 marks 3. Conversion of nitrogen(II) oxide to nitric(V) acid, 4NO + 3O2 + 2H2O → 4HNO3 Room temperature; 1 atmosphere; add water and air; no catalyst required. 3 marks Sulphur 1. Conversion of sulphur to sulphur dioxide, S + O2 → SO2 Burning sulphur, 1 atmosphere, no catalyst required. 2 marks 2. Conversion of sulphur(IV) oxide to sulphur(VI) oxide, 2SO2 + O2 d 2SO3 450ºC; 1 atmosphere; excess oxygen or air; platinium or vanadium(V) oxide catalyst. 4 marks O 3. Conversion of sulphur(VI) oxide to sulphuric(VI) acid, SO3 + H2SO4 → H2S2O7 H  → 2H2SO4 Add oxide to acid then dilute with water 3 marks

8

9

2

88 2B 6 d 6d Give examples of compounds of EITHER sulphur OR nitrogen to display the principal oxidation states and also give an account of their redox reactions. Redox chemistry of sulphur overall presentation 1 mark Principal oxidation states with examples 2 marks e.g. -2(H2S), 0(S), +4(SO2) and +6(SO3 or SO42-) Any sulphur containing compounds, upon oxidation, give SO2 or SO42SO2 – through combustion of the sulphur compounds in air 1 mark SO42- – oxidation in homogeneous solution. Upon reduction, S or H2S are usually produced H2S is a mild reducing agent but SO42- is usually stable towards redox reactions in the absence of H+. 1 mark Intermediate oxidation states, for example, S2O32- or S4O62-, will undergo disproportionation in water. 1 mark Redox chemistry of nitrogen overall presentation 1 mark Principal oxidation states with examples 2 marks e.g. 0(N2), -3(NH3), +4(NO2), +5(NO3-) The highest oxidation state is +5, and NO3- is usually the final product upon oxidation of any nitrogen containing compounds under drastic conditions. 1 mark NO3- is stable, but in the presence of H+, it is a strong oxidizing agent, upon reduction, give N2 or NH3. 1 mark

6

N2 is the most stable form, usually chemically insert in normal condition. NH3 – obtained by reduction of any nitrogen-containing compounds under drastic conditions. Compounds in intermediate oxidation states such NO2 undergo disproportionation 2NO2 + H2O d HNO3 + HNO2

1 mark 1 mark 1 mark

89 2B 6 b 6b Nitric(V) acid and sulphuric(VI) acid can be reduced to form products having a variety of oxidation states of N or S. For EITHER nitric(V) acid OR sulphuric(VI) acid, illustrate using balanced half equations, how four possible reduction products, having different oxidation states of N or S, may be obtained. Explain why an aqueous solution of either Na2SO4 or NaNO3 is not a strong oxidizing agent. Depends on the nature of the reducing agents and the reaction conditions employed: Nitric(V) acid NO3- + 2H+ + e- d NO2 + H2O 1 mark 1 mark 2NO3- + 12H+ + 10e- d N2 + 6H2O 1 mark NO3- + 9H+ + 8e- d NH3 + 3H2O 1 mark NO3- + 4H+ + 3e- d NO + 2H2O other possible product: N2O, H2NNH2, NO21 mark H+ is required if NO3- acts as an oxidizing agent Sulphuric(VI) acid SO42- + 4H+ + 2e- d SO2 + 2H2O 1 mark 1 mark SO42- + 8H+ + 6e- d S + 4H2O 1 mark SO42- + 10H+ + 8e- d H2S + 4H2O 1 mark 2SO42- d S2O32- or S4O62-, etc. 1 mark H+ is required if SO42- acts as an oxidizing agent

5

Nitrogen 10.2.2.1 90 2B 5 b 5b Outline an industrial preparation of ammonia. Your answer should include the physico-chemical principles involved and also the role of any catalyst. Ammonia N2 + 3H2 d 2NH3 ∆H < 0 / exothermic 1 mark Conditions of reaction: 200 atm, 500ºC, finely divided iron catalyst 1 mark Effect of temperature on the equilibrium constant 1 mark Effect of pressure 1 mark } any 2 1 mark 2 marks Effect of concentration of N2 and H2 1 mark Catalyst - required because of the inertness of N2 and H2 molecules 1 mark NH3 is removed from the reaction mixture Overall presentation 1 mark

7

10.2.3.1 81 2B 8 a i ii iii 8a Describe the process for the conversion of ammonia gas into (68%) nitric(V) acid. 12 Give the conditions under which the various steps take place and briefly discuss which of the following will involve the highest costs: i energy, ii capital investment in plant, and iii handling. I Concentrated nitric(V) acid 1 mark Ammonia(g) is mixed with air. passed over a Pt gauze or (Pt/Rh) 1 mark catalyst which is maintained at 800ºC and 8 atm. 1 mark 1 mark the reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) 1 mark takes place with the evolution of heat. The gases must be pure and avoid poisoning the catalyst. 1 mark The NO(g) is then oxidised by further air to NO2(g) 2NO(g) + O2(g) → 2NO2(g) 1 mark which is then absorbed in water. 1 mark 2NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) the NO(g) is recycled. 1 mark Absorption may be carried out up to 68%(conc.) HNO3 which is the constant boiling mixture. (i) energy – the reaction condition of 800ºC is not difficult and expensive to maintain because the reaction is exothermic. 1 mark (ii) The catalyst is expensive and it will lose and need to be refilled from time to time so it costs a lot. ( The lost of catalyst is due to the highly exothermic reaction produce energy which may vaporize the catalyst, Pt). 1 mark (iii) Because NO, NO2 and HNO3 are highly corrosive and toxic, so special chemical resistant materials have to be chosen for handling. 1 mark 10.2.5.0 81 2B 8 d 8d Write on the present-day uses of concentrated nitric(V) acid. Uses : 4 uses of conc. HNO3. 1 mark each Manufacture of explosive such as T.N.T. conc. HNO3 Manufacture of fertilizers such as NH4NO3 Manufacture of dyestuffs Manufacture of hexane-1,6-diamine which is a raw material for making nylon. 10.2.4.0 84 2B 10 a i ii 10a i Give the formulae of the oxides of nitrogen having oxidation numbers (O.N.) of +2, +4 and +6. N: NO(+2), NO2/N2O4(+4), NO3(+6) 2 marks ii Describe what you would observe and give a balanced equation for the reaction the oxide of nitrogen (O.N. +2) (A) with acidified potassium manganate(VII) solution, (B) with an appropriate iron sulphate solution, (C) with rain, and (D) to form the element. Nitrogen (A) 5NO + 3MnO4- + 4H+ → 5NO3- + 3Mn2+ + 2H2O (Purple colour decolourized / colourless solution) (B) FeSO4 + NO → FeSO4·NO (Brown solution formed) (C) 2NO + O2 → NO2; 2NO2 + H2O → HNO3 + HNO2; 2HNO2 + O2 → 2HNO3 (no special observation) (D) Electric: 2NO d N2 + O2 at 3000ºC (no special observation) OR Burning white phosphorus: 10NO + 4P → 2P2O5 + 5N2 (white fume) 2 marks each 10.2.3.1 84 2B 10 b 10b Describe the industrial preparation of nitric(V) acid from nitrogen. Nitrogen 1. Conversion of nitrogen to ammonia, N2 + 3H2 d 2NH3 450-500ºC; 200-1000 atmosphere; iron or iron oxide catalyst 2. Conversion of ammonia to nitrogen(II) oxide, 4NH3 + 5O2 d 4NO + 6H2O 600ºC; 8 atmosphere; platinium catalyst

4

2 8

9

3 marks 3 marks

3.

Conversion of nitrogen(II) oxide to nitric(V) acid, 4NO + 3O2 + 2H2O → 4HNO3 Room temperature; 1 atmosphere; add water and air; no catalyst required.

3 marks

10.2.4.0 88 2B 6 d 6d Give examples of compounds of nitrogen to display the principal oxidation states and also give an account of their 6 redox reactions. Redox chemistry of nitrogen overall presentation 1 mark Principal oxidation states with examples 2 marks e.g. 0(N2), -3(NH3), +4(NO2), +5(NO3-) The highest oxidation state is +5, and NO3- is usually the final product upon oxidation of any nitrogen containing compounds under drastic conditions. 1 mark NO3- is stable, but in the presence of H+, it is a strong oxidizing agent, upon reduction, give N2 or NH3. 1 mark 1 mark N2 is the most stable form, usually chemically insert in normal condition. 1 mark NH3 – obtained by reduction of any nitrogen-containing compounds under drastic conditions. Compounds in intermediate oxidation states such NO2 undergo disproportionation 1 mark 2NO2 + H2O d HNO3 + HNO2 10.2.3.2 89 2B 6 b 6b Nitric(V) acid can be reduced to form products having a variety of oxidation states of N. For nitric(V) acid, illustrate using balanced half equations, how four possible reduction products, having different oxidation states of N, may be obtained. Explain why an aqueous solution of NaNO3 is not a strong oxidizing agent. Depends on the nature of the reducing agents and the reaction conditions employed: Nitric(V) acid NO3- + 2H+ + e- d NO2 + H2O 1 mark + + 12H + 10e d N + 6H O 1 mark 2NO3 2 2 + 1 mark NO3 + 9H + 8e d NH3 + 3H2O 1 mark NO3- + 4H+ + 3e- d NO + 2H2O other possible product: N2O, H2NNH2, NO21 mark H+ is required if NO3- acts as an oxidizing agent

5

Sulphur 10.3.3.1 90 2B 5 b 5b Outline an industrial preparation of sulphuric acid. Your answer should include the physico-chemical principles involved and also the role of any catalyst. Sulphuric acid S + O2 → SO2 2SO2 + O2 d 2SO3; ∆H < 0 very slow in the absence of catalyst H O SO3 + H2SO4 → H2S2O7  → 2H2SO4 3 marks Steps involved: Production of SO2, conversion to SO3, removal of SO3 Conditions of conversion: 1 atm, 450ºC, vanadium(V) oxide catalyst 1 mark 1 mark Effect of [O2], removal of SO3 Effect of pressure 1 mark } any 1 Effect of temperature on the equilibrium constant 1 mark 1 mark 1 mark Catalyst - lower the activation energy for the SO2 + O2 reaction as O2 is kinetically inert Overall presentation 1 mark 2

7

10.3.3.1 81 2B 8 a i ii iii 8a Describe the process for the conversion of sulphur dioxide gas into (98%) sulphuric acid. Give the conditions under which the various steps take place and briefly discuss which of the following will involve the highest costs: i energy, ii capital investment in plant, and iii handling. Concentrated sulphuric(VI) acid 1 mark SO2(g) is mixed with air 1 mark and the mixture is passed over a catalyst of Pt or V2O5. 1 mark Usually V2O5 is used because Pt is expensive and easily to be poisoned by impurities. e.g. As2O3. The operating temperature is about 450ºC at 1 atm. 1 mark 1 mark 2SO2(g) + O2(g) → 2SO3(g) where the reaction is exothermic. 1 mark The SO3(g) produced is then dissolved in 98% H2SO4 and fuming sulphuric acid (oleum) is formed. SO3(g) + H2SO4(l) → H2S2O7(l) 1 mark 1 mark Oleum is then diluted with calculated amount of water so as to give back 98% H2SO4(l) SO3(g) is not dissolved directly in water to form sulphuric acid because the reaction between SO3(g) and water is violent and a mist of acid droplets forms instead of a solution. 1 mark (i) energy – the reaction condition of 450ºC is not difficult and expensive to maintain because the reaction is exothermic. 1 mark (ii) The catalyst is expensive and it will lose and need to be refilled from time to time so it costs a lot. ( The lost of catalyst is due to the highly exothermic reaction produce energy which may vaporize the catalyst, Pt). 1 mark (iii) Because SO2, H2SO4 and H2S2O7 are highly corrosive and toxic, so special chemical resistant materials have to be chosen for handling. 1 mark 10.3.5.0 81 2B 8 d 8d Write on the present-day uses of concentrated sulphuric(VI) acid. Uses : 4 uses of conc. H2SO4. Manufacture of soapless detergents conc. H2SO4 Manufacture of fertilizers such as (NH4)2SO4 Manufacture of explosives Manufacture of pigments

4 1 mark each

10.3.2.0 84 2B 10 a i ii 10a i Give the formulae of the oxides of sulphur having oxidation numbers (O.N.) of +2, +4 and +6. S: SO/S2O2(+2), SO2(+4), SO3(+6) 2 marks ii Describe what you would observe and give a balanced equation for the reaction the oxide of sulphur (O.N. +4) (A) with acidified potassium manganate(VII) solution, (B) with an appropriate iron sulphate solution, (C) with rain, and (D) to form the element. Sulphur (A) 5SO2 + 2MnO4- + 2H2O → 5SO42- + 2Mn2+ + 4H+ (Purple colour decolourized / colourless solution) (B) 2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+ (Solution change from yellow brown to pale green / colourless) (C) SO2 + H2O → H2SO3; 2H2SO3 + O2 → 2H2SO4 (no special observation) (D) Burning magnesium: SO2 + 2Mg → 2MgO + S (yellow / white ppt. formed) OR Mix with hydrogen sulphide: SO2 + 2H2S → 3S + 2H2O (yellow / white ppt. formed) 2 marks each 10.3.3.1 84 2B 10 b 10b Describe the industrial preparation of sulphuric(VI) acid from sulphur. Sulphur 1. Conversion of sulphur to sulphur dioxide, S + O2 → SO2 Burning sulphur, 1 atmosphere, no catalyst required. 2. Conversion of sulphur(IV) oxide to sulphur(VI) oxide, 2SO2 + O2 d 2SO3 450ºC; 1 atmosphere; excess oxygen or air; platinium or vanadium(V) oxide catalyst.

12

2

8

9

2 marks 4 marks

3.

O  → 2H2SO4 Conversion of sulphur(VI) oxide to sulphuric(VI) acid, SO3 + H2SO4 → H2S2O7 H Add oxide to acid then dilute with water marks 2

3

10.3.3.2 88 2B 6 d 6d Give examples of compounds of sulphur to display the principal oxidation states and also give an account of their redox reactions. Redox chemistry of sulphur overall presentation 1 mark Principal oxidation states with examples 2 marks e.g. -2(H2S), 0(S), +4(SO2) and +6(SO3 or SO42-) Any sulphur containing compounds, upon oxidation, give SO2 or SO42SO2 – through combustion of the sulphur compounds in air 1 mark SO42- – oxidation in homogeneous solution. Upon reduction, S or H2S are usually produced H2S is a mild reducing agent but SO42- is usually stable towards redox reactions in the absence of H+. 1 mark Intermediate oxidation states, for example, S2O32- or S4O62-, will undergo disproportionation in water. 1 mark 10.3.3.2 89 2B 6 b 6b Sulphuric(VI) acid can be reduced to form products having a variety of oxidation states of S. For Sulphuric(VI) acid, illustrate using balanced half equations, how four possible reduction products, having different oxidation states of S, may be obtained. Explain why an aqueous solution of either Na2SO4 is not a strong oxidizing agent. Depends on the nature of the reducing agents and the reaction conditions employed: Sulphuric(VI) acid SO42- + 4H+ + 2e- d SO2 + 2H2O 1 mark 1 mark SO42- + 8H+ + 6e- d S + 4H2O 1 mark SO42- + 10H+ + 8e- d H2S + 4H2O 1 mark 2SO42- d S2O32- or S4O62-, etc. 1 mark H+ is required if SO42- acts as an oxidizing agent

6

5

d-Block Elements I.

General features of the d-block elements from Sc to Zn A. B. C.

II.

Electronic configuration of d-block elements Electronic configuration of ions of d-block elements 1. Stability of different ions Comparison of properties between d-block and s-block metals 1. Physical properties a) Ionization enthalpies and electronegativity b) Melting point and hardness c) Atomic radii of d-block elements d) Densities 2. Chemical properties a) Reaction with water

Characteristic properties of the d-block elements and their compounds: A.

B.

C. D.

E.

Variable oxidation states 1. Common oxidation states of vanadium and manganese compounds a) Vanadium compounds b) Manganese compounds Complex formation 1. Relative stability of complex ion a) Basicity of the ligand b) Chelate effect 2. Stability constant 3. Aquaton isomerism 4. Geometrical isomerism 5. Nomenclature of complexes Coloured ions 1. Origin of the colour 2. Colours of different ions Catalytic properties of transition metals and their compounds 1. Haber process 2. Reaction between peroxodisulphate(VI) and iodide ions 3. Decomposition of hydrogen peroxide Magnetic properties of transition metal and ion

d-Block Elements

Unit 1

Page 1

Topic

d-block Elements

Unit 1

Reference Reading

11.1.0–11.1.1 Chemistry in Context, 3rd Edition ELBS pg. 292–295 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 192–196 A-Level Chemistry, 3rd Edition, E.N. Ramsden pg. 490

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 249–251 Reading Syllabus Notes

I.

General features of the d-block elements from Sc to Zn

A. Electronic configuration of d-block elements d-block elements are usually defined as those with partially filled d-orbitals or forms compound with partially filled d-orbitals. Or they may be defined as those situated between s-block and p-block elements. However, according to the first definition, Sc and Zn would be excluded since they only form Sc3+ and Zn2+ ions which do not have partially filled d-orbitals. In certain extent, Sc and Zn resemble the properties of s-block metals more than other dblock metals. 3d 4s 21Sc

[Ar] 3d14s2

[Ar]

2

2

[Ar]

23V

3

[Ar] 3d 4s

2

[Ar]

24Cr

[Ar] 3d54s1

[Ar]

22Ti

[Ar] 3d 4s

5

25Mn

[Ar] 3d 4s

2

[Ar]

26Fe

[Ar] 3d64s2

[Ar]

7

27Co

[Ar] 3d 4s

2

[Ar]

28Ni

[Ar] 3d84s2

[Ar]

10

29Cu

[Ar] 3d 4s

1

[Ar]

30Zn

[Ar] 3d104s2

[Ar]

10000 11000 11100 11111 11111 21111 22111 22211 22222 22222

2 2 2 1 2 2 2 2 1 2

half-filled 3d and 4s orbitals

full-filled 3d and half-filled 4s orbitals

B. Electronic configuration of ions of d-block elements When 3d orbital is empty, the energy of 4s orbital is lower than that of 3d orbital due to the penetration effect of 4s orbital despite of the further position of 4s orbital from the nucleus. However, if 3d orbital is occupied, the electron in the 3d orbital will offer a primary shielding effect on the 4s electrons and makes them more energetic. Therefore, when an atom from Sc to Zn is going to lose its electron, 4s electrons will be lost first instead of the 3d electron.

Element [Ar] 3d14s2 21Sc [Ar] 3d24s2 22Ti [Ar] 3d34s2 23V [Ar] 3d54s1 24Cr [Ar] 3d54s2 25Mn [Ar] 3d64s2 26Fe [Ar] 3d74s2 27Co [Ar] 3d84s2 28Ni [Ar] 3d104s1 29Cu [Ar] 3d104s2 30Zn

Oxidation State +2

+1

3+

Cu+

[Ar] 3d10

Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+

[Ar] 3d5 [Ar] 3d6 [Ar] 3d7 [Ar] 3d8 [Ar] 3d9 [Ar] 3d10

Sc Ti3+ V3+ Cr3+ Mn3+ Fe3+ Co3+

+3 [Ar] [Ar] 3d1 [Ar] 3d2 [Ar] 3d3 [Ar] 3d4 [Ar] 3d5 [Ar] 3d6

d-Block Elements

Unit 1

Page 2

Except Sc and Zn, other transition metals are capable to form compounds with more than 1 oxidation state. The ions listed above are only common ones. For an oxidation state higher than 3, metals never exist as simple ions due to the high charge density. Probably, they will form complex with ligands or covalently bonded to other atoms.

1.

Stability of different ions

Comparing Mn2+ [Ar] 3d5 and Mn3+ [Ar] 3d4, it is found that Mn2+ is the more stable one in natural environment. This can be explained by the presence of half filled d-orbital of Mn2+. The same logical also applies to Fe2+ and Fe3+ where Fe3+ is the more stable one. However, considering Cu+ [Ar] 3d10 and Cu2+ [Ar] 3d9, Cu2+ is the more abundant one in nature despite of the fullfilled d-orbital of Cu+. This seems to be contradictory. When we discussed the stability of ion, we only considered the stability of the electronic configuration of an isolated ion only. In the nature, ions are not isolated. They may be solvated or forming compounds with the other ions. In aqueous state, two Cu+(aq) ions are capable to undergo disproportionation to give a Cu(s) atom and a Cu2+(aq) ion. Cu+(aq) + e- d Cu(s) Cu2+(aq) + e- d Cu+(aq)

Eo = +0.52 V Eo = +0.15 V

2Cu+(aq) d Cu2+(aq) + Cu(s)

Eo = (+0.52V) + (-0.15V) = +0.37V (energetically feasible)

Therefore, aqueous Cu+(aq) does not exist. Cu(I) compound only exists in solid state or if complexed.

Glossary

d-block elements

Past Paper Question

91 2B 6 a i 92 2B 4 a i ii iii 94 2B 5 a i ii iii 95 1A 2 f i ii iii 96 2B 6 a 99 1A 3 a

half-filled orbital

full-filled orbital

91 2B 6 a i 6a i What is the meaning of ‘d-block element’? Element which has incompletely filled d orbital when combined in compounds (or complete in case of Zn) between s and p block

1 1 mark ½ mark

d-Block Elements Unit 1 92 2B 4 a i ii iii 4a i Using the ‘electrons-in-boxes’ representation, give the electronic structure outside the argon core of Cu and Cr3+. 3d 4s 3d 4s

11100 0 1 mark each Which of the two ions, copper(I) or copper(II), has the more stable electronic structure? Give an explanation. Cu+ d10 is expected to be more stable than Cu2+ d9 since it has completely filled d-orbital. 1 mark iii Why is the copper(I) ion unstable in aqueous solution? 29Cu

22222 1

0

24Cr

2Cu

2+ ( aq )

  → Cu disproportionation

2

3+

ii

+ ( aq )

Page 3

+ Cu( s )

1 1

½ mark for disproportionation and ½ mark for correct product

94 2B 5 a i ii iii 5a Manganese and iron are d-block elements. i What do you understand by the term “d-block element”? 1 Elements with partially filled d-orbitals. 1 mark 2 ii Using the “electron-in-boxes” representation, write the electronic structure outside the argon core for Mn2+ and Fe2+ respectively. Mn2+ [Ar] 11111 1 mark 2+ 1 mark Fe [Ar] 21111 iii Explain briefly why Fe2+ ions are easily oxidized to Fe3+ ions, but Mn2+ ions are not readily oxidized to Mn3+ ions. 3 Mn2+ with half-filled 3d orbitals is relatively stable. Removing e- disturbs this stable configuration; relatively high 2 marks energy is required. Therefore, Mn2+ is not readily oxidized to Mn2+. Fe2+ with an electronic configuration of d6, after removal of an e- from the outermost shell will attain the relatively stable e- configuration with half-filled 3d orbitals. Therefore, the Fe2+ ions are easily oxidized to Fe3+. 1 mark 95 1A 2 f i ii iii 2f i Show the electronic configuration of a ground state Cr2+ ion by filling in the appropriate boxes below. 1s 2s 2p 3s 3p 3d 4s 4p 4d

1

0 0 000 0 000 00000 0 000 00000

1s

2s

2p

3s

3p

3d

4s

4p

4d

2 2 222 2 222 11110 0 000 00000 What is the highest possible oxidation state of chromium ? +6 / VI iii Give a chromium compound in which chromium is in its highest oxidation state. CrO42- / Cr2O72- / HCrO4- / CrO2Cl2 / CrO3 etc. or name (any chromium(VI) compound)

1 mark

ii

1 1 1 mark

96 2B 6 a 6a Using the "electrons-in-boxes" representation, write the ground state electronic structure outside the argon core for 2 V2+ and Cu+ respectively. 3d 1 mark [Ar] 11100 V2+ + Cu [Ar] 22222 1 mark 99 1A 3 a 3a Deduce the number of unpaired electrons for Fe3+(g) at its ground state.

d-Block Elements

Unit 2

Topic

d-Block Elements

Reference Reading

11.1.2 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.

Page 1

Unit 2

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 251–254 Reading Syllabus Notes

C. Comparison of properties between d-block and s-block metals 1.

Physical properties

Main group elements (s-block and p-block) show a considerable change in properties on moving across a period. i.e. from metallic properties to non-metallic properties. Comparing with main group elements, all elements in the first transition period share similar properties. They are all hard, dense and comparatively less reactive. These properties can be explained by the weak shielding effect of diffuse d-electrons and a rather constant effective nuclear charge experienced by the valence electron on moving across the period. a) Ionization enthalpies and electronegativity On moving across a transition period (e.g. from Sc to Cu), comparing with the main group elements, the 1st I.E. of d-block elements are rather constant. This can be interpreted as a rather constant effective nuclear charge experienced by the outermost electron.

On moving across the first transition period, electrons are added to the 3d oribitals which will impose a primary shielding effect on the 4s electrons (valence electron). This is different from the situation in the main group, where an electron is added to s or p orbital on moving across the period and the shielding effect is only secondary in nature. As a result of the stronger primary shielding effect, the ionization energies do not raise sharply across the transition period. However, unlike the change from Ne to Na, where the shielding effect is primary in nature, the I.E. doesn't drop sharply either. The primary shielding effect of the d-electrons must not be very strong. The weak shielding effect can be explained by the diffuse nature of d-electrons. As a result, the increase in nuclear charge has been effectively shielded and the attraction between the nucleus and the 4s electrons (valence electrons) would remain rather constant on moving across the period.

d-Block Elements

Unit 2

Page 2

Furthermore, the weak shielding effect also accounts for the high I.E of d-block elements (631-906 kJmol-1) comparing with the lower 1st I.E. of s-block elements (419-516 kJmol-1). Therefore, the primary shielding effect of the diffuse d-electrons must be not as effective as the primary shielding effect found in s-block elements. For the same reason, the electronegativities of the d-block elements are rather constant and only shows a gradual increase across the period. Nevertheless, comparing with s-block element, d-block elements are more electronegative, i.e. less electropositive, thus less metallic.

b) Melting point and hardness Both melting point and hardness are indications of the strength of the metallic bond. It is found that the melting point and hardness of dblock elements are much higher than that of s-block elements. The difference in strength of a metallic bond can be explained by the effective nuclear charge and the no. of delocalized electrons available. First of all, the effective nuclear charge experienced by the valence electrons of d-block metal is higher than that of s-block metal because of the poor shielding effect of diffuse d-electrons. Furthermore, the availability of the delocalized electrons are also higher. d-block metals can use its 3d electrons on top of the 4s electrons in the formation of metallic bond. It is observed that on moving across the first transition period, the melting point increases gradually and drops when approaching the end of the period with a depression in the middle. The increase is due to the general increase in effective nuclear charge experienced by the valence electrons and availability of delocalized electron. Any further increase in effective nuclear charge will lower the availability of delocalized electron and make the melting point drop. When we look at the electronic configuration of Mn, it will not be surprised that it has a melting point much lower than the expected. Since Mn has a half-filled 3d orbital and full-filled 4s orbital, the availability of the delocalized electrons will be lowered. 3d 21Sc

1

2

2

2

[Ar] 3d 4s

22Ti

[Ar] 3d 4s

23V

[Ar] 3d34s2

24Cr

[Ar] 3d54s1 5

2

25Mn

[Ar] 3d 4s

26Fe

[Ar] 3d64s2

27Co

[Ar] 3d74s2

28Ni

[Ar] 3d84s2

29Cu

[Ar] 3d104s1

30Zn [Ar]

3d104s2

4s

10000 [Ar] 11000 [Ar] 11100 [Ar] 11111 [Ar] 11111 [Ar] 21111 [Ar] 22111 [Ar] 22211 [Ar] 22222 [Ar] 22222 2 [Ar]

2 2 2 1 2 2 2 2 1

half-filled 3d and 4s orbitals

full-filled 3d and half-filled 4s orbitals

d-Block Elements c)

Unit 2

Page 3

Atomic radii of d-block elements

For similar reasons, the atomic radii of d-block elements are quite small and rather constant across the period, comparing with the s-block metals. However, a small variation is observed. On moving across the period, the atomic size decreases gradually due to the small increase in effective nuclear change and remain constant thereafter. On reaching the end of the transition period, the size of the atom gets larger as the repulsion among the larger no. of electrons becomes significant.

d) Densities Owing to the small atomic sizes, strong metallic bond and close crystal packing, the d-block metals are much denser than the s-block metals.

structure density g/cm3

19K

20Ca

21Sc

22Ti

23V

24Cr

25Mn

26Fe

27Co

28Ni

29Cu

30Zn

BCC 0.86

FCC 1.55

HCP 3.0

HCP 4.5

BCC 6.1

BCC 7.2

BCC 7.4

BCC 7.9

FCC 8.9

FCC 8.9

FCC 8.9

HCP 7.1

BCC

: body centered cubic (not a close packing)

FCC

: face centered cubic (a close packing)

HCP

: hexagonal close packing (a close packing)

d-Block Elements 2.

Unit 2 Chemical properties

a) Reaction with water In general, d-block metals are much less reactive towards water than s-block metals do. This is parallel to low electropositivity of d-block elements.

Glossary Past Paper Question

Page 4

d-Block Elements

Unit 3

Topic

d-Block Elements

Reference Reading

11.2.1 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.

Page 1

Unit 3

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 254–257 Reading Syllabus

Variable oxidation states

Notes

II. Characteristic properties of the d-block elements and their compounds: A. Variable oxidation states

Unlike s-block metals, many transition metals are capable to form compounds with more than one oxidation state. The no. of possible oxidation states increases from Sc and reaches the maximum at Mn due to the increase in the number of d-electrons. Thereafter, the no. of possible oxidation states decreases from Fe to Zn as the nuclear charge and ionization energies increase. Concerning the energetic of formation of ion, if the difference in successive ionization energy is small, the formation of the highest oxidation state would be preferable. e.g. Ca only forms +2 ion but not +1 ion. If the difference in successive ionization energy is large, the formation of the lowest oxidation state would be preferable. e.g. K only forms +1 but not +2 ion. Therefore, formation of compounds with variable oxidation state must be due to a moderate and gradual increase in successive ionization energies. One of the explanations to this phenomenon is the small difference in energy between the (n-1)d and ns electrons. As the energy of (n-1)d and ns electrons are similar, once a ns electron is removed, the energy of (n-1)d electron will decrease moderately as the p to e- ratio increase. Consequently, this will cause a moderate increase in successive ionization energies.

d-Block Elements

Unit 3

kJmol-1 K Ca Mn

1st I.E. 419 590 717

2nd I.E. (∆ I.E.) 3051 (2632) 1145 (555) 1509 (792)

Page 2 3rd I.E. (∆ I.E.) 4412 (1361) 4912 (3767) 3249 (1740)

4th I.E. (∆ I.E.) 5887 (1475) 6474 (1562) 4940 (1691)

5th I.E. (∆ I.E.) 7975 (2088) 8144 (1670) 6985 (2045)

First 5 successive ionization energies On moving across the transition period, the gap between first I.E. and second I.E. are rather constant while the gap between second I.E. and third I.E. widens. As a result, the elements at the beginning of the transition period favours the formation of +3 oxidation state. When approaching the end of the transition period, the formation of +2 oxidation state is more favorable. This reflects the increasing difficulty of removing a 3d electron as the nuclear charge increases. Although many transition metals are capable to form compounds with high oxidation state e.g. Mn(VII), ions with more than 3+ charge do not exist freely at all due to the high charge density of the ion. They are either covalently bonded (e.g. Mn2O7) or complexed (e.g. MnO4-). 1.

Common oxidation states of vanadium and manganese compounds

a) Vanadium compound +2, +3, +4 and +5 are the common oxidation states of vanadium compound. Their existence can be demonstrated by shaking a solution of ammonium vanadate(V) (NH4VO3) with zinc power in dilute sulphuric acid. VO3-(aq) + 2H+(aq) → VO2+(aq) + H2O(l) VO2+(aq) + 2H+(aq) + e- → VO2+(aq) + H2O(l) VO2+(aq) + 2H+(aq) + e- → V3+(aq) + H2O(l) V3+(aq) + e- → V2+(aq)

this is not a redox reaction, V(V) white → V(V) yellow V(V) yellow → V(IV) blue V(IV) blue → V(III) green V(III) green → V(II) violet

Upon shaking, the solution will turn from yellow, blue, green and eventually violet since zinc is a very strong reducing agent which is capable to reduce V(V) to V(II) stepwise.

d-Block Elements

Unit 3

Page 3

The following is an electrode potential - oxidation state chart which can be used to predict the energetic feasibility of a redox reaction. Please note that the scales of both x and y axes are reversed so that the half-equations will be in line with the definition of reduction potential and the order of the electrochemical series.

Electrode potential - oxidation state chart Regarding the use of the chart, there is a rule called anticlockwise rule. For any two half-equations, the combination with an anticlockwise direction as shown on the diagram on the right would be energetically feasible. e.g. Zn(s) + Fe2+(aq) → Zn2+(aq) + Fe(s) Zn2+(aq) + Fe(s) → Zn(s) + Fe2+(aq)

energetically feasible energetically not feasible

Example Fe2+(aq) can be used to reduce VO2+(aq) to VO2+(aq) Fe2+(aq) + VO2+(aq) + H2O(l) → Fe3+(aq) + VO2+(aq) + 2H+(aq)

Eo = (-0.77V) + (1.00V) = +0.23V

Since the overall Eo is positive, the change is energetically feasible. However, the value of Eo doesn't tell anything about the kinetic feasibility of the reaction. Similarly, I-(aq) is capable to reduce VO2+(aq) to VO2+(aq) but neither Fe2+(aq) nor I-(aq) is capable to reduce VO2+(aq) further to V3+(aq) according to the chart.

d-Block Elements

Unit 3

Page 4

2) Manganese compounds +2, +4 and +7 are the three most common oxidation states of manganese compound. Common manganese containing species +2 +3 +4 +6 +7

Mn2+(aq) colourless (or very pale pink) Mn2O3(s) brown, Mn3+(aq) red MnO2(s) brown MnO42-(aq) green MnO4-(aq) purple

The energetic feasibility of redox reaction involving manganese can also be predicted by using the standard electrode potential as mentioned above. O2(g) + 2H2O(l) + 2e- d 4OH-(aq) Cu+(aq) + e- d Cu(s) I2(aq) + 2e- d 2I-(aq) MnO42-(aq) + 2H2O(l) + 2e- d MnO2(s) + 4OH-(aq) 2H+(aq) + O2(g) + 2e- d H2O2(aq) MnO2(s) + 4H+(aq) + 2e- d Mn2+(aq) + 2H2O(l) Mn3+(aq) + e- d Mn2+(aq) MnO4-(aq) + 8H+(aq) + 5e- d Mn2+(aq) + 4H2O(l) MnO4-(aq) + 4H+(aq) + 3e- d MnO2(s) + 2H2O(l) H2O2(aq) + 2H+(aq) + 2e- d 2H2O(l)

Eo / V + 0.40 + 0.52 + 0.54 + 0.59 + 0.68 + 1.23 + 1.49 + 1.51 + 1.70 + 1.77

KMnO4(s) is one of the most popular strong oxidizing used in the laboratory. The oxidizing power of MnO4-(aq) is so strong that it is capable to oxidize water and give O2(g) and MnO2(s) slowly. This reaction can be accelerated by light therefore KMnO4(aq) have to be stored in a brown bottle. 4MnO4-(aq) + 4H+(aq) → 4MnO2(s) + 3O2(g) + 2H2O(l) In acidic medium, when MnO4-(aq) is reduced. It will be converted to colourless Mn2+(aq). Eo = + 1.51 V MnO4-(aq) + 8H+(aq) + 5e- d Mn2+(aq) + 4H2O(l) In less acidic or alkaline medium, MnO4-(aq) will be converted to MnO2(s) instead. From the value of Eo, MnO4-(aq) shows a even stronger oxidizing power in less acidic medium with the formation of the brown MnO2(s) ppt. Eo = + 1.70 V MnO4-(aq) + 4H+(aq) + 3e- d MnO2(s) + 2H2O(l) Therefore, when MnO4-(aq) is used as a testing agent or an ordinary oxidizing agent, acidic medium will be used. If MnO4-(aq) is used as a very strong oxidizing agent, alkaline medium may be used.

Glossary

variable oxidation state

d-Block Elements

Past Paper Question

Unit 3

Page 5

91 2B 6 a iv 94 2B 5 b i

91 2B 6 a iv 6a iv State the reagents you would employ to obtain (I) vanadium(IV) and (II) vanadium (II) from vanadium(V) in an acidic solution of ammonium metavanadate, NH4VO3, and give the colours of the products. (I) by mild reductants e.g. H2S, SO2, Fe2+, oxalate, Sn2+, I1 mark Blue ½ mark (II) by Zn or Zn/HCl 1 mark Violet ½ mark 94 2B 5 b i 5b Account for each of the following: i KMnO4 solutions are best stored in brown bottles. MnO4- is unstable in the presence of sunlight. it will decompose to form brown MnO2

3

2 1 mark 1 mark

d-Block Elements

Unit 4

Topic

d-Block Elements

Reference Reading

11.2.2 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.

Page 1

Unit 4

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 201–207, 257–258, 269–270 Reading Syllabus Notes

B. Complex formation When an ion is placed in a polar solvent e.g. water, it will be hydrated due to the electrostatic attraction between the ion and water molecules. Moreover, if the ion possesses a very high charge density and a low lying energy vacant orbital, the ion may be capable to form a dative covalent bond with the lone pair from the solvent molecule. The dative covalent bond formed in this way would be much strong than the ordinary ion-dipole attraction. For example, when anhydrous NaCl(s) and anhydrous CoCl2(s) are dissolved in water, both of them will dissolve and the ions will be hydrated. However, upon crystallization, anhydrous NaCl(s) crystal will be obtained from the NaCl(aq) solution while only [Co(H2O)6]Cl2(s) will be obtained from the CoCl2(aq) solution. Furthermore, the comparatively higher electronegativity of d-block elements also favours the formation of dative covalent bond. The molecules which form dative covalent bond with the ion (or an atom) are called ligands. The species with the ion (or the atom) joining together with the ligands is called a complex. Because the ion is coordinated with ligands, a complex is also called a coordinated compound and the dative covalent bond formed is called coordinate bond.

NC

Ag

CN

+

NC

Cu

2+

3-

CN

H3N CN

H3N

Pt

NH3 NH3

CN

2+

H2O H2O H2O

Co

OH2 OH2

H2O

linear complex

tetrahedral complex

square planar complex

octahedral complex

d-Block Elements 1.

Unit 4

Page 2

Relative stability of complex ion

a) Basicity of the ligand Basically, stability of a complex ion is depending on the strength of the dative bond. For a given central ion, the strength of the coordinate bond is depending on the electron donating ability of the ligand i.e. basicity (or nucleophilicity) of the ligand. Consider the ligands Cl-, H2O, NH3 and CNparent acid pKa conjugate base

H3O+ -1.74 H2O

HCl -7 Cl-

NH4+ 9.2 NH3

HCN 9.3 CN-

basicity of the conjugate base increase

→ NH3 molecule is a strong base than H2O molecule, NH3 molecule is also a stronger ligand than H2O molecule. At the same concentration, the strong ligand is capable to displace the weak ligand from a complex. If the concentrations are not the same, the equilibrium position may be shifted back. Iron(II)

[Fe(Cl)4]2-(aq)

Iron(III)

[Fe(Cl)4]-(aq) yellow

Cobalt(II)

H2O CN[Fe(CN)6]3-(aq) [Fe(H2O)6]3+(aq) d d colourless/yellow red

[Co(Cl)4]2-(aq) blue

Copper(II)

[Cu(Cl)4]2-(aq) yellow

N.B.

H2O CN[Fe(CN)6]4-(aq) [Fe(H2O)6]2+(aq) d d green yellow

H2O NH3 CN[Co(NH3)6]2+(aq) [Co(CN)6]4-(aq) [Co(H2O)6]2+(aq) d d d pink deep red H2O NH3 CN[Cu(NH3)4(H2O)2]2+(aq) [Cu(CN)4]2-(aq) [Cu(H2O)6]2+(aq) d d d blue deep blue

Upon addition of aqueous NH3(aq), Fe2+(aq) and Fe3+(aq) do not form soluble complex with ammonia molecules. Indeed, they will form insoluble ppt. of Fe(OH)2(s) and Fe(OH)3(s) with the hydroxide ions in the ammonia solution and remain undissolved.

Sometimes, the complexes of Cu2+ are written as [Cu(H2O)4]2+(aq) instead of [Cu(H2O)6]2+(aq) and [Cu(NH3)4]2+(aq) instead of [Cu(NH3)4(H2O)2]2+(aq). This is because the two water molecule ligands at axial positions are only loosely held. The complexes are sometimes regarded as square planar instead of octahedral. b) Chelate effect Besides the basicity, the strength of a ligand is also depending on the no. of dative covalent bond that the ligand can form. A ligand which is capable to form 1 dative covalent bond is called a monodentate ligand; a ligand which is capable to form 2 dative covalent bonds is called a bidentative ligand etc. In general, the larger the no. of bond a ligands can form, the stronger will be the ligand.

d-Block Elements

Unit 4

Page 3

Consider the following example [M(H2O)6]2+(aq) + bi → [M(bi)(H2O)4]2+(aq) + 2H2O(l) n+

bi

H2O H2O

M

H2O

OH2 OH2

+ bi

H2O

H2O

M

H2O

OH2

H2N

n+

CH2 CH2 NH2

+ 2H2O

H2O

a bidentate ligand

bi : a bidentate ligand In the above reaction, one bidentate ligand e.g. ethane-1,2-diamine (H2NCH2CH2NH2 or ethylenediamine or en) displaces two water molecules from the complex. There is an overall change in the no. of particles in this reaction. At the reactant side, there are only 2 particles but at the product side, there are 3 particles. Besides the displacement of ligand, there is an overall increase in the no. of molecules and disorderness of the system. i.e. ∆S increase. The extra strength of a polydentate ligand due to increase in disorderness is called chelate effect. 2.

Stability constant

The strength of a ligand can be expressed as an equilibrium constant called stability constant, Kst. Consider the formation of a complex in aqueous medium Cu2+(aq) + 4NH3(aq) d [Cu(NH3)4]2+(aq) Cu2+(aq) ≡ [Cu(H2O)6]2+(aq) ≡ [Cu(H2O)4]2+(aq)

[Cu(NH3)4]2+(aq) ≡ [Cu(NH3)4(H2O)2]2+(aq)

[[Cu(NH3)4]2+(aq)] Kst = [Cu2+ ][NH ]4 (aq) 3(aq) The larger the value of Kst, the stronger would be the complexing ability of the ligand. The displacement of the ligands may also be expressed in consecutive steps Cu2+(aq) + NH3(aq) d [Cu(NH3)]2+(aq)

[[Cu(NH3)]2+(aq)] K1 = [Cu2+ ][NH ] (aq) 3(aq)

[Cu(NH3)]2+(aq) + NH3(aq) d [Cu(NH3)2]2+(aq)

[[Cu(NH3)2]2+(aq)] K2 = [Cu(NH )]2+ ][NH ] (aq) 3 3(aq)

[Cu(NH3)2]2+(aq) + NH3(aq) d [Cu(NH3)3]2+(aq)

[[Cu(NH3)3]2+(aq)] K3 = [Cu(NH ) ]2+ ][NH ] (aq) 3 2 3(aq)

[Cu(NH3)3]2+(aq) + NH3(aq) d [Cu(NH3)4]2+(aq)

[[Cu(NH3)4]2+(aq)] K4 = [Cu(NH ) ]2+ ][NH ] (aq) 3 3 3(aq)

K 1 × K2 × K3 × K4 =

[[Cu(NH3)2]2+(aq)] [[Cu(NH3)3]2+(aq)] [[Cu(NH3)4]2+(aq)] [[Cu(NH3)]2+(aq)] × × × 2+ 2+ 2+ [Cu (aq)][NH3(aq)] [Cu(NH3)] (aq)][NH3(aq)] [Cu(NH3)2] (aq)][NH3(aq)] [Cu(NH3)3]2+(aq)][NH3(aq)]

=

[[Cu(NH3)4]2+(aq)] [Cu2+(aq)][NH3(aq)]4

=

Kst

d-Block Elements

Unit 4

3.

Page 4

Aquation isomerism

Some coordinated compounds share the same empirical formula but when dissolved in water, they give different number of ions. This phenomenon is called aquation isomerism. Cr(H2O)6Cl3 is one of the examples. Structural formula [Cr(H2O)6]Cl3 [CrCl(H2O)5]Cl2·H2O [CrCl2(H2O)4]Cl·2H2O 4.

Ions present in aqueous solution [Cr(H2O)6]3+(aq), 3Cl-(aq) [CrCl(H2O)5]2+(aq), 2Cl-(aq) [CrCl2(H2O)4]+(aq), Cl-(aq)

Colour Violet Blue-green Green

Geometrical isomerism

Tetra-coordinated complex Square planar complex [MA2B2]n+ n+ B

n+

A

A

B

A

M A

B M B

cis-isomer

trans-isomer Hexa-coordinated complex Octahedral complex [MA2B4]n+ n+

A B

[MA3B3]n+

B

B

M B

B

A

N.B. 5.

B

B

cis-isomer

B

B

n+

B B

M A

B

n+

A

M B

trans-isomer

n+

A

A M

A

B

A

meridional isomer

A A

facial isomer

The exact geometry of a complex cannot be predicted by the Valence Shell Electron Pair Repulsion Theory that have been studied.

Nomenclature of complexes

Naming of a complex is based on the name of the ligand, the no. of ligands and overall charge of the complex. 1.

Like all ionic compounds, cations are named first and anions second.

2.

Ligand name of a negative ligand will end with -o, e.g. CN- cyano, Cl- chloro. Ligand name of a neutral ligand is not changed e.g. H2NCH2CH2NH2 is just called ethane-1,2-diamine Exceptions : H2O is called aqua and NH3 is called ammine.

3.

The number of each ligand in the complex is indicated by the prefixes di, tri, tetra etc.

4.

The oxidation state of the metal is indicated in parentheses. Furthermore, If the complex is cationic or neutral, the name of the metal is used. e.g. [Fe(H2O)6]3+ hexaaquairon(III) ion If the complex is anionic, the ending -ate is added to the name of the metal, in certain cases, Latin name will be used. e.g. [Fe(CN)6]3- hexacyanoferrate(III) ion

5.

The negative ligands are placed before the neutral ligands in the name. (Not in alphabetical order.) e.g. [CoCl3(NH3)3] trichlorotriamminecobalt(III)

6.

Geometrical isomers are designated by cis- or trans- and mer- or fac-.

d-Block Elements

Unit 4

Other examples K4[Fe(CN)6] [Fe(Cl)4][Cr(Cl)4][Fe(H2O)6]3+ [Co(Cl)4]2[Cu(CN)4]2Ligand Cl-

Fe2+ [Fe(Cl)4]2tetrachloroferrate(II) ion

H2O

potassium hexacyanoferrate(II) tetrachloroferrate(III) ion tetrachlorochromate(III) ion hexaaquairon(III) ion tetrachlorocobaltate(II) ion tetracyanocuprate(II) ion Fe3+ [Fe(Cl)4]tetrachloroferrate(III) ion [Fe(H2O)6]3+ hexaaquairon(III) ion

Co2+ [Co(Cl)4]2tetrachlorocobaltate(II) ion

Cu2+

Cr3+ [Cr(Cl)4]tetrachlorochromate(III) ion

[Cu(NH3)4]2+ tetraamminecopper(II) ion [Cu(CN)4]2tetracyanocuprate(II) ion

NH3 CN-

Page 5

[Fe(CN)6]2+ hexacyanoferrate(II) ion

Glossary

Complex ligand ion-dipole attraction coordinated compound coordinate bond tetrahedral square planar octahedral chelate effect monodentate ligand bidentate ligand polydentate ligand stability constant aquation isomerism meridional isomer facial isomer

Past Paper Question

90 2B 6 a iv 91 1A 3 b i 92 1A 3 e 93 1A 2 b iv 94 1A 2 b i 95 1B 4 g 96 1A 2 e i 97 1B 7 b 99 1A 3 c

90 2B 6 b 91 1A 3 f 92 2B 4 a vi vii

91 2B 6 a ii iii

94 2B 5 b iii 96 1A 2 f i

96 2B 6 b i ii

90 2B 6 a iv 6a The following species are either impossible to prepare or very unstable. Explain, in each case, why this is so. iv [NaI4]3Because there is no low energy vacant orbitals for bonding / Na+ has too low charge density for complex ion formation. 1½ mark 90 2B 6 b 6b Account for the following observations. “An aqueous solution of FeSO4 readily gives a rusty brown precipitate on standing in air, whereas an aqueous solution of K4[Fe(CN)6] under similar conditions does not.” nH 2 O

linear

detail explanation FeSO4 → Fe(H2O)n2+/Fe2+(aq) Fe(H2O)n2+ + O2 → Fe(H2O)n3+ air oxidation Fe(H2O)n3+ → Fe2O3 or Fe(OH)3 + H+ product Fe(CN)64- is stable because CN- is a better ligand than H2O / CN- ligands stabilize Fe2+ (no mark for Fe(CN)64- is a stable complex) The redox potential of Fe(CN)64- is different from Fe(H2O)n2+

1½

5

1 mark 1 mark 1 mark 1 mark 1 mark

91 1A 3 f 3 3f Write equations for stepwise reactions of [Cu(H2O)4]2+(aq) with aqueous ammonia. Describe what changes you would observe during these reactions. 1 mark [Cu(H2O)6]2+(aq) + 2NH3(aq) → [Cu(H2O)4]2+·Cu(OH)2(s) + 2NH4+(aq) or Cu(OH)2(s) [Cu(H2O)4]2+·Cu(OH)2(s) + excess NH3(aq) → [Cu(NH3)4]2+ 1 mark When NH3(aq) is added slowly, at first insoluble pale blue basic salt or hydroxide is formed, but these dissolve 1 mark readily in excess NH3(aq) to give deep blue solution containing tetraaminecopper(II) ion.

d-Block Elements Unit 4 91 2B 6 a ii iii 6a ii Give the name and formula, and draw the 3-dimensional structure of a hexa-coordinated complex of a d-block element. Name (must be named correctly) e.g. hexaamminecobalt(III) ion ½ mark ½ mark Formula e.g. [Co(NH3)6)]3+ Shape e.g. Co

NH3 NH3

NH3

1 mark

iii Why do V4+ ions react to give VO2+ ions in aqueous solution? V4+ is small and highly charged (or has high charge density) V4+ ion polarizes H2O molecule so strongly that it attracts the electrons shared between H and O V4+ + O

2

3+

NH3 NH3 NH3

Page 6

2 1 mark 1 mark

H 2H+ + VO2+ H

92 2B 4 a vi vii 4a vi State two reasons why transition metal ions readily form complexes whereas Group I metal ions do not. 2 Atoms lose electrons to form ions, whose polarizing power increases with number of electrons lost. Atoms have partially-filled d-orbitals that can interact with ligand orbitals by transfer of electron density Complex ion formation is favoured by high charge density on cation High electronegativity of cation Available orbitals of right energy to interact with ligand orbitals 1 mark each vii Write the chemical equation on which the overall stability constant for the reaction between the cation Cu(H2O)42+ 1 and the ligand NH3 in aqueous medium can be based. Cu(H2O)42+ + 4NH3 d Cu(NH3)42+ + 4H2O 1 mark 93 1A 2 b iv 2b iv Write equations for the reactions described below. The addition of an aqueous solution of KSCN to a solution containing Fe(III) ions gives a complex with a deep red colour. The deep red colour fades when excess NaF solution is added. On the addition of dilute alkali to the resulting solution, a brown precipitate forms. 1 mark Fe3+(aq) + SCN-(aq) → Fe(SCN)2+(aq) [deep red] Fe(SCN)2+(aq) + 6F-(aq) → [FeF6]3-(aq) + SCN-(aq) 1 mark [FeF6]3-(aq) + 3OH-(aq) → Fe(OH)3(s) [brown ppt.] + 6F-(aq) 1 mark

3

94 1A 2 b i 2 2b i Explain why Ni2+ forms more complexes than K+ does. 2+ + Ni has higher charge density / more polarizing than K / greater ability to form covalent bond (dative bond) with ligand Any 2, 1 mark each Ni2+ has low lying , vacant d orbital. 94 2B 5 b iii 5b Account for each of the following: iii The addition of 2M NaOH to an aqueous solution of Fe3+ ions gives a brown precipitate, whereas the addition of 2M NaOH to an aqueous solution of Fe3+ ions in the presence of excess CN- ions gives no precipitate. In aqueous solution, Fe3+ exists as aquo-complex. Addition of OH- causes the formation of a brown ppt. [Fe(H2O)6]3+(aq) + 3OH-(aq) → Fe(OH)3(s) + 6H2O(l) 1 mark brown ppt. In the presence of excess CN-, Fe3+ forms a very stable complex [Fe(CN)6]3-, concentration of free Fe3+ ions is lowered ∴ addition of 2M NaOH cannot cause [Fe3+][OH-] to exceed the Ksp of Fe(OH)3. 2 marks

3

d-Block Elements Unit 4 Page 7 95 1B 4 g 4g Give the observations when (i) ammonia solution and (ii) sodium hydroxide solution are added dropwise, until in 3 excess, to aqueous solutions containing Pb2+ and Cu2+ ions respectively. These experiments are carried out at room temperature. (i) ammonia solution (ii) sodium hydroxide solution Pb2+(aq) Cu2+(aq) (3 marks for 8 observations; deduct ½ mark for each incorrect observation) (i) ammonia solution (ii) sodium hydroxide solution Pb2+(aq) White ppt. White ppt. Insoluble in excess Soluble in excess / Colour solution Cu2+(aq) Blue ppt. Blue ppt. Soluble in excess / Deep blue solution Insoluble in excess 96 1A 2 f i 2f The complex ion [Co(H2NCH2CH2NH2)2Cl2]+ has THREE isomers. These isomers belong to TWO types of isomerism. i Draw a three-dimensional structure for each isomer.

+

Cl en

Co Cl

+

en en

Co

Cl Cl

en Cl Cl

en

2

+

Co en

1 + ½ + ½ mark

(en represents H2NCH2CH2NH2) (1 mark for the trans-isomer; ½ mark for each cis-isomer ) 96 2B 6 b i ii 6b What do you understand by the following terms ? In each case, illustrate your answer with an example involving copper(II). i complex ion A complex ion is formed by combination of a central metal ion ½ mark ½ mark with ligands / complexing agents. The ligands/complexing agents are either neutral molecules / negatively charged ions carrying lone pair(s) of electrons 1 mark ½ mark which can be donated to the metal ion / can form dative bond with the metal ion. ½ mark e.g. [Cu(NH3)4]2+ (accept any complex of Cu(II) ii stability constant Stability constant Kst : e.g. [Cu(NH3)4]2+ Kst is the equilibrium constant for the following process 1 or 0 mark Cu2+(aq) + 4NH3(aq) d [Cu(NH3)4]2+ [Cu(NH3)42+] i.e. Kst = [Cu2+][NH ]4 1 mark 3 (Accept expressions for stability constant of other copper(II) complexes and answers for stepwise and overall stability constants) 97 1B 7 b 7b An aqueous solution contained 0.40 g of a complex salt Cr(NH3)5Cl3. The free chloride ions present in the solution required 21.90 cm3 of 0.150 M silver nitrate(V) solution for complete precipitation. Deduce the structural formula of the complex ion present in the salt. 99 1A 3 c 3c Draw the structures of all isomers of the complex [Co(NH3)4Cl2]Cl and give the systematic name of each isomer.

3

2

d-Block Elements

Unit 5

Topic

d-Block Elements

Reference Reading

11.2.3 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.

Page 1

Unit 5

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 259–261 Reading Syllabus

Coloured ions

Notes

C. Coloured ions 1.

Origin of the colour

The origin of the colour of an transition metal ion is similar to that of emission spectrum. Instead of emitting radiation while the excited atom is falling back to the lower energy state, radiation is absorbed from the illuminating radiation. The energy absorbed is eventually dissipated in form of heat. The atom returns to the lower energy state. The colour of hydrated ion (e.g. Cu2+(aq)) is the result of absorption of light. Most pure substances in nature are colourless or white because they do no absorb light from the visible region. However, most substances in nature are coloured because most of them are mixtures containing a very small percentage of coloured substance. Whether a substance would be coloured or not is solely a coincidence. Many transition metal complexes, including aqueous complex, are coloured because under the influence of the ligands, the 5 degenerate d-orbitals will be split into different levels with small gaps between them. The degenerate energy levels split because the five 5 dorbitals are not equally arranged in space and are affected by the ligands differently.

Eventually, if the d-orbitals are partially filled (not completely filled nor completely empty), the electron transition will produce a colour. The electron transition is called d-d transition. This explains why anhydrous copper(II) sulphate is white while hydrated copper(II) sulphate is blue. Sc(III) and Zn(II) compounds are not coloured while the other transition metal compounds are usually coloured.

d-Block Elements 2.

Unit 5

Page 2

Colours of different ions

[Fe(Cl)4]2-(aq)

[Fe(Cl)4]-(aq) yellow

H2O CN[Fe(CN)6]4-(aq) [Fe(H2O)6]2+(aq) d d green yellow

H2O CN[Fe(CN)6]3-(aq) [Fe(H2O)6]3+(aq) d d colourless/yellow red

[Co(Cl)4]4-(aq) blue [Cu(Cl)4]4-(aq) yellow

Glossary

d-d transition

Past Paper Question

93 1A 2 b i 94 1B 4 d iii 99 2A 2 d i ii

H2O NH3 CN[Co(NH3)6]2+(aq) [Co(CN)6]4-(aq) [Co(H2O)6]2+(aq) d d d pink deep red H2O NH3 CN[Cu(NH3)4(H2O)2]2+(aq) [Cu(CN)4]2-(aq) [Cu(H2O)6]2+(aq) d d d blue deep blue

94 2B 5 b ii

93 1A 2 b i 2b i Explain, in terms of electronic arrangement, why most transition metal ions are coloured. 1. Absorption of photon in the visible region 2. Unpaired electron or partly filled orbital 3. Small energy difference between d orbitals under the influence of the ligands 4. Transition of electron 94 1B 4 d iii 4d This question refers to reactions which occur between the following pairs of substances. boxes in the following table. Substances Reaction method Expected observation(s) Example: CH3COOH(l) and Warm / reflux with miscible, fruity odour C2H5OH(l) conc H2SO4 (iii) CuSO4(aq) and 2M NH3 pale blue precipitate, deep blue solution Substances Reaction method Expected observation(s) Example: CH3COOH(l) and Warm / reflux with miscible, fruity odour C2H5OH(l) conc H2SO4 (iii) CuSO4(aq) and 2M NH3 Add NH3(aq) dropwise pale blue precipitate, (1) deep blue solution 94 2B 5 b ii 5b Account for each of the following: ii Cobalt(II) chloride can be used to test for the presence of water. Cobalt(II) chloride, in anhydrous state is blue in colour. After absorption of water / forming complex ion with H2O, it changes colour to pink.

2

2 marks

Complete the empty

6

Main product(s) CH3COOC2H5(l) Cu(OH)2(s) [Cu(NH3)4]2+(aq) Main product(s) CH3COOC2H5(l) Cu(OH)2(s) [Cu(NH3)4]2+(aq)

2 1 mark 1 mark

99 2A 2 d i ii 2d For each of the following experiments, state the expected observation and write the relevant balanced equation(s). i Concentrated hydrochloric acid is added dropwise to an aqueous solution of copper(II) sulphate(VI) until no further change is observed. ii Aqueous ammonia is added dropwise, until in excess, to the resulting solution in (i).

d-Block Elements

Unit 6

Topic

d-Block Elements

Reference Reading

11.2.4–11.2.5 11.2.0 Modern Physical Chemistry ELBS pg. Chemistry in Context, 3rd Edition ELBS pg. Physical Chemistry, Fillans pg.

Page 1

Unit 6

Assignment Chemistry in Context, 5th Edition, Thomas Nelson and Sons Ltd. pg. 261–262 Reading Syllabus

Catalytic properties Magnetic properties

Notes

D. Catalytic properties of transition metals and their compounds Many transition metals or compounds of transition metal exhibit catalytic behavior. This is because, basically, all chemical reactions involve electron transfer or redistribution of electron. Many transition metal has low lying energy vacant d-orbital to accept the electron and possesses variable oxidation state which aids the electron transfer process and lowers the activation energy of redox reaction. 1. 2. 3.

Haber process Reaction between peroxodisulphate(VI) and iodide ions Decomposition of hydrogen peroxide

For details please refer to Rate of reaction – Catalysis E. Magnetic properties of transition metal and ion When an electron spins, a magnetic moment will be induced. Normally, electrons will pair up in an atom and the magnetic moments will be canceled out. Therefore, most substance are non-magnetic i.e. diamagnetic. In certain substance containing unpaired electrons, e.g. Mn2+ [Ar] 3d5, the magnetic moments are not canceled out completely and they are weakly magnetic i.e. paramagnetic. A very few no. of substances are strongly magnetic e.g. Fe, due to the spinning of the protons. They are ferromagnetic.

Glossary

diamagnetic

paramagnetic

ferromagnetic

d-Block Elements

Past Paper Question

Unit 6

Page 2

90 1A 3 d 92 1A 3 h 94 2B 5 a iv 96 2B 6 c

90 1A 3 d 3d Illustrate, with examples using nickel, two characteristic properties of transition metals. Catalytic properties. e.g. the use of a nickel catalyst in the hydrogenation of oils to make margarine Complex ion formation. e.g. [Ni(H2O)6]2+ Accept also ‘colours’, ‘paramagnetic’ or other properties which can be illustrated.

2 1 mark 1 mark

92 1A 3 h 3h Which two properties of transition metal ions make them suitable for use as catalyst? I Vacant orbitals may act as electron acceptors from reactant species; full orbitals can act as electron donors to reactant species and weak bonds (chemisorption) are formed between catalyst and reactant particles. 1 mark, either II Variable oxidation states enable oxidation-reduction cycle of transition metal catalyst. 1 mark 94 2B 5 a iv 5a iv Explain briefly why d-block elements and their compounds are frequently used as catalysts in redox reactions. Give one example of such a catalytic reaction. Variable oxidation states of d-block elements allow electron transfer between reactants and products by means of 2 marks catalyst changing between two oxidation states. Any one example of transition metal or its compound as catalyst V O

2 5 → SO3 (redox reaction) e.g. 2SO2 + O2  

2

3

1 mark

96 2B 6 c 6c State THREE characteristic properties of d-block elements, apart from complex ion formation. In each case, illustrate your answer with an example involving copper or vanadium. Any THREE of the following: ½+½ exhibition of variable oxidation states e.g. Cu(I) & Cu(II) / V2+, V3+, VO2+ and VO2+ mark ½ + ½ mark formation of coloured compounds e.g. Cu2+(aq) is blue / VO2+ is yellow ½ + ½ mark exhibition of catalytic properties e.g. V2O5 in contact process ½ + ½ mark exhibition of paramagnetic properties e.g. Cu2+ / V2+ are paramagnetic ½ + ½ mark formation of non-stoichiometric compounds e.g. sulphide of vanadium (½ marks for each properties; ½ marks for a correct example)

3