All-Inorganic

Periodic Properties of Elements I. 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

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

A. Hydride

1. Bonding

2. Hydrolysis

B. Chloride

1. Bonding

2. Hydrolysis

C. Oxide

1. Formation of acidic, amphoteric and basic oxides

a) Amphoteric oxides

s-Block Elements I. Characteristic properties of the s-block elements

A. Metallic character and low electronegativity B. Ionic bond formation and fixed oxidation state C. Flame test

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

1. Reaction with hydrogen 2. Reaction with oxygen 3. Reaction with chlorine 4. Reaction with water

B. Thermal stability of carbonates and hydroxides 1. Thermal stability of carbonates 2. Thermal stability of hydroxides

C. 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

D. Abnormal behaviour of lithium

III. Uses of the compounds of the s-block elements A. Manufacture of glass B. Baking powder C. Making of soap D. Antacid E. Slaked lime F. Fireworks

p-Block Elements I. Halogens

A. Laboratory preparation of chlorine B. Characteristic properties of the halogens C. Properties of halogen

1. Oxidizing power of halogen 2. Disproportionation of halogen in alkalis

D. 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

E. Uses of halogens and halogen containing compounds 1. Fluoridation of water 2. Use of chlorine 3. Photographic films

II. Nitrogen and its compounds

A. Nitrogen 1. Unreactive nature of nitrogen 2. Direct combination of nitrogen and oxygen

B. Ammonia 1. Haber process 2. Chemical properties of ammonia

a) Alkaline properties (1) Laboratory preparation of ammonia

b) Reducing properties (1) Combustion (2) Catalytic oxidation (3) With copper(II) oxide (4) With chlorine (5) Thermal decomposition of ammonia

C. Nitric(V) acid 1. Ostwald Process 2. Oxidizing properties of nitric(V) acid

a) With copper b) With iron(II) ion c) With sulphur

D. Nitrate(V) salt 1. Thermal decomposition of nitrate 2. Brown ring test

E. Use of nitric(V) acid a) Fertilizers b) Explosives c) Nylon d) Dyestuffs

III. Sulphur and its compounds A. Allotrope of sulphur B. Burning of sulphur C. 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 D. Concentrated sulphuric(VI) acid

1. Contact process 2. Properties of concentrated sulphuric acid

E. Sulphate(VI) salt 1. Test for sulphate(VI) ion

F. 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. Electronic configuration of d-block elements B. Electronic configuration of ions of d-block elements

1. Stability of different ions C. 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

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

1. Common oxidation states of vanadium and manganese compounds a) Vanadium compounds b) Manganese compounds

B. 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

C. Coloured ions 1. Origin of the colour 2. Colours of different ions

D. Catalytic properties of transition metals and their compounds 1. Haber process 2. Reaction between peroxodisulphate(VI) and iodide ions 3. Decomposition of hydrogen peroxide

E. Magnetic properties of transition metal and ion

Periodic Properties of Elements I. 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

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

A. Hydride

1. Bonding

2. Hydrolysis

B. Chloride

1. Bonding

2. Hydrolysis

C. Oxide


a) Amphoteric oxides

Properties of compounds and bond energies Page 1 Hydrolysis of Hyrides Hydride Reaction Remark 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) Insoluble in water 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) Slightly soluble in water 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 Remark LiCl(s) Soluble in water 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)

Dissolve and hydrolyse in water

BCl3(l) BCl3(l) + 3H2O(l) → H3BO3(aq) + 3HCl(aq) CCl4(l) Immisicible with water NCl3(l) NCl3(l) + 3H2O(l) → NH3(aq) + 3HOCl(aq) Cl2O(g) ClO2(g) Cl2O6(l) Cl2O7(l)

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)

ClF(g) ClF(g) + H2O(l) → HF(g) + HOCl(aq) NaCl(s) Soluble in water MgCl(s) MgCl(s) + warm H2O(l) → MgCl(OH)(s) + HCl(aq) Dissolve and hydrolyse in warm water 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) PCl5(s)

PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(aq) 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) 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 Page 2 Hydrolysis of Oxides Oxide Reaction Remark Li2O(s) Li2O(s) + H2O(l) → 2LiOH(aq) BeO(s) No reaction with water but it is amphoteric 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) CO(g)

CO2(g) + H2O(l) d H2CO3(aq) d H+

(aq) + HCO3-(aq)

d 2H+(aq) + CO3

2-(aq)

Carbonic acid or carbonated water

N2O5(g) N2O4(g) N2O3(g) NO(g) N2O(g)

N2O5(s) + H2O(l) → 2HNO3(aq) N2O4(g) [or 2NO2(g)] + H2O(l) → HNO2(aq) + HNO3(aq) N2O3(g) + H2O(l) → 2HNO2(aq)

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

O2(g) Insoluble in water F2O(g) F2O(g) + H2O(l) → 2HF(aq) + O2(g) neutral and insoluble but hydrolyses slowly. Na2O2(s) Na2O2(s) + 2H2O(l) → 2NaOH(aq) + H2O2(aq) Na(s) forms peroxide instead of oxide when burns in air. MgO(s) MgO(s) + H2O(l) e Mg(OH)2(s) e Mg2+

(aq) + 2OH-(aq) MgO(s) only reacts with water very slowly due to high

lattice energy. Al2O3(s) Not reaction with water but it is amphoteric SiO2(s) Not reaction with water but reacts with strong base P2O3(s) P2O5(s)

P2O3(s) + 6H2O(l) → 4H3PO3(aq) P2O5(s) + 6H2O(l) → 4H3PO4(aq)

SO2(g) SO3(g)

SO2(g) + H2O(l) d H2SO3(aq) d H+

(aq) + HSO3-(aq)

d 2H+(aq) + SO3

2-(aq)

SO3(g) + H2O(l) → H2SO4(aq) → H+

(aq) + HSO4-(aq)

d 2H+(aq) + SO4

2-(aq)

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

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 Reaction Remark BeO(s) BeO(s) + 2H+

(aq) → Be2+(aq) + H2O(l) Amphoteric oxide

Al2O3(s) Al2O3(s) + 6H+(aq) → 2Al3+

(aq) + 3H2O(l) Amphoteric oxide In Alkali Oxide Reaction Remark BeO(s) BeO(s) + 2OH-

(aq) + H2O(l) → [Be(OH)4]2-(aq) Amphoteric oxide

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

(aq) Amphoteric oxide

Properties of compounds and bond energies Page 3 Bond energy Bond energy

E(X–X) in kJmol-1 (with element)

Bond energy E(X–H) in kJmol-1 (with hydrogen)

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

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

Li Be B C 347 413 358 346 N 158 (945.4 in N≡N) 391 214 O 144 (498.3 in O=O) 464 144 F 158 568.0 Na 72 Mg Al Si 226 315 466 P 198 321 S 266 364 469 Cl 243.4 432.0 H 435.9 435.9 464 432.0

Periodic Properties of Elements Unit 1 Page 1

Topic Perioidic Properties of Elements Unit 1

Reference Reading

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

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 Unit 1 Page 2 Past Paper Question

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. 2 Elements in same group have similar chemical properties. 1 mark

Properties recur after an interval (period). 1 mark also accept “element” properties are periodic functions of atomic number

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. 1

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



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

Assignment Reading

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

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.


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 - s.m.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.

g.m.s. : Giant metallic structure The m.p. and b.p. of a metal is depending on the strength of the metallic bond. g.c.s. : 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.

Glossary simple molecular structure giant metallic structure giant covalent structure


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

5

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

For I. E. Graph shows decrease in ionization energy down the group 1 mark Graph shows 2,3,3 pattern 1 mark (not labelling each axis -½) Explanation: p3 for N (p4 for O not necessary) ½ mark stability of half-filled subshell ½ mark

92 2B 6 Aa ii 6Aa ii Describe the structure and the bonding in nitrogen, N2, and in white phosphorus, P4. 2 N2: N≡N ½ mark

structure: (linear) diatomic molecule ½ mark bonding: intramolecular, covalent ½ mark intermolecular, van der Waals ½ mark

white P4:

P

P PP

½ mark structure: discrete tetrahedral ½ mark bonding: intramolecular, single covalent ½ mark intermolecular, van der Waals ½ mark

92 2B 6 Ba ii 6Ba ii Describe the structure and the bonding in oxygen, O2, and in rhombic sulphur, S8. 4

Periodic Properties of Elements Unit 2 Page 4 O2: O=O ½

mark structure: (linear) diatomic molecule ½ mark bonding: intramolecular, covalent ½ mark intermolecular, van der Waals ½ mark

S8: SS

S

SSS S

S

½ mark structure: molecular ring structure ½ mark bonding: intramolecular, single covalent ½ mark intermolecular, van der Waals ½ mark


Topic Periodic Properties of Elements Unit 3

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

Assignment Reading

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

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

Mg2+ ion carries higher charges than Li+ ion. Consequently, the two ions share similar charge density( Z+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. Li2CO3(s) heat → Li2O(s) + CO2(g) 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. CCl4(l)

water → no reaction 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


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. 1 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.

Perioidic Properties of Elements Unit 4 Page 1


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

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 Unit 4 Page 2 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.

H2HO-

+H- HO

H

δ−

δ+

δ+δ−

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.

N

H

H

H

δ+

δ+

δ+

δ− + HO-HO

H

δ−

δ+

δ+δ−N

H

H

H

H+

P

HH HN

HH H

δ−

δ+δ+

δ+

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.

Si

H

H H

H

δ−

δ+δ−

δ−

δ− δ+ O H

Hδ+

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. Difference in E.N. Bond energy (kJmol-1) Hydrolysis in water H2O 1.5 + 463 H2O(l) + H2O(l) d H3O+

(aq) + OH-(aq)

H2S 0.3 + 364 H2S(aq) + 2H2O(l) d 2H3O+(aq) + S2-

(aq) HCl 1.5 + 432 HCl(aq) + H2O(l) → H3O+

(aq) + Cl-(aq)

HF 2.0 + 568 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 H–X + H–O–H d X- + H2O–H+ NH3(aq) + H2O(l) d NH2

-(aq) + H3O+

(aq)

As a base H–X + H–O–H d H2X+ + OH- 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 91 2B 5 b 93 2B 4 b iii 93 2B 6 d 95 1A 2 d i ii iii 99 2A 3 c ii iii

91 1A 3 a 3a Explain why NaH(s) is more reducing than HCl(g) whilst HCl(g) is more acidic than NaH(s). 3 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 HCl(g) does not undergo such reaction to release electron. ½ mark 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

NaH(s) is not acidic because it is not a proton donor. ½ mark

91 2B 5 b 5b Explain why SiH4 is hydrolysed by alkali, whereas CH4 is not, and give an equation for the hydrolysis. 3 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 SiH4 + (n+2)H2O → SiO2⋅nH2O + 4H2 (SiH4 + 2OH- → SiO3

2- + 2H2O + H2) 1 mark or SiH4 + 2H2O → SiO2 + 4H2 (SiH4 + H2O →Si(OH)4 + H2 (need not be balanced. Accept SiO3

2-. Miss out H2 -½)

Perioidic Properties of Elements Unit 4 Page 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. 3 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. 4

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

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

there is a reaction, give the balanced equation. CH4 SiH4 H2S HCl

3

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 : H2S + H2O d HS- + H3O+ OR H2S + 2H2O d 2H3O+ + S2- (½ mark for correct product, ½ mark fro balanced reaction) HCl : HCl + H2O → H3O+ + Cl- ½ mark

ii Explain the difference in reactivity with water between CH4 and SiH4. 1½ Kinetic aspect

In CH4, there is almost no polarity in the C–H bonds ½ mark ∴ CH4 is not affected by polar molecules of water In SiH4, the bond polarity is Siδ+–Hδ-. ½ mark OR There is no empty / vacant low energy d-orbital in C ½ mark ∴ lone e- pair in water will not attack CH4. ½ mark OR In Si, there is low energy empty d-orbital. ½ mark ∴ it can accept lone electron pair from water ½ mark

SiH

H

H

OH + H2 etc.Si

H

H

H

Hδ−δ+

O

H

H

½ mark Thermodynamic aspect C–H bond is strong when compared with Si–H bond ½ mark ∴substitution of H in CH4 is energetically unfavourable. ½ mark (Max. 1½ mark, 3 points must not be in the same aspect)

iii Explain the difference in reactivity with water between SiH4 and H2S. 1½ In H2S, the polarity is Hδ+–Sδ-. ½ mark

∴ nucleophilic attack of H2O on H2S gives H3O+ and HS-, ½ mark whereas SiH4 gives an alkaline solution since the polarity is Siδ+–Hδ-. ½ mark

Perioidic Properties of Elements Unit 4 Page 5 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)



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 Reading

Advanced Practical Chemistry, John Murray (Publisher) Ltd., 144–147

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.

PCl

ClCl

PCl3(l)

+

PCl

Cl

Cl

Cl

-

P

Cl

ClCl Cl

Cl Cl

PCl5(s)

P

Cl

Cl

Cl ClCl

PCl5(g)

Perioidic Properties of Elements Unit 5 Page 2 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 91 2B 4 b 92 1A 3 e 93 1A 2 a iv 94 1C 2 96 1A 2 e i

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. 10

LiCl colourless ionic solid with Li+, Cl-.

AlCl3 colourless covalent solid, dimeric in structure, Al

ClCl

AlClCl

ClCl , undergoes sublimation.

PCl3 covalent colourless liquid, P

Cl ClCl , can be oxidized to POCl3.

K2NiCl4 coloured ionic solid with complex anion, Ni

Cl

Cl Cl

Cl2-

. 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 PCl3, covalent bond between P and Cl NiCl4

2-, 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

Perioidic Properties of Elements Unit 5 Page 3 91 1A 3 b i 3b Write equation(s) to describe the reaction of i ZnCl2(s) with water. 1 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 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.

6

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)

NCl Cl

Cl

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

OCl Cl V-shaped

or 6ClO2 + 3H2O → 3HClO2 + 3HClO3 or Cl2O6 + H2O → HClO4 + HClO3 ClO3 is the same as Cl2O6, which is a solid ClO2

+ClO4-

ClO2 + H2O → stable in dark 6ClO2(g) + 3H2O(l)

sunlight → 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 either ClO2 + 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)

S SCl

Cl

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

SCl

Cl S

Cl

ClClCl

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. 3

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 Smaller size and greater charge of Fe3+ ion results in more acidic solution than that of Fe2+ ion. 1 mark

Perioidic Properties of Elements Unit 5 Page 4 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), CO3

2-(aq), Cu2+

(aq), P4O10(s), S(s), S2O32-

(aq), Zn2+(aq)

iv Which species fumes in moist air? 2 AlCl3(s) AlCl3(s) + 3H2O(l) → Al(OH)3(s) + 3HCl(g) 1 mark 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) 2

[Fe(H2O)6]3+ ½ mark The conjugate base [Fe(H2O)5OH]2+ is stabilized to a greater extent ½ mark because the anionic ligand OH- tends to ½ mark 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, Fe3+ has a higher charge to radius ratio / charge density, ½ mark greater polarizing power, ½ mark and hence the H develops larger partial +ve charge.♣ ½ mark



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 Reading

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

Syllabus

Notes C. Oxide


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)

S

O

O

OH

H

OH

HS

O

OS

O

O

OH

H

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 OH

H

2- O H2

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 Page 3 Glossary acidic oxides amphoteric oxides basic oxides

Past Paper Question

92 2B 5 b iii 92 2B 6 Ac ii 92 2B 6 Bc ii 93 1A 2 a i 95 1A 2 b ii 95 1B 4 g ii

92 2B 5 b iii 5b iii Write equations showing the amphoteric nature of lead(II) oxide. 2 PbO(s) + 2HCl(aq) → PbCl2(s) + H2O(l) 1 mark

H2O(l) + PbO(s) + NaOH(aq) → NaPb(OH)3(aq) 1 mark

92 2B 6 Ac ii 6Ac ii Which is more reactive to oxidation, NH3 or PH3? Explain your answer. 2 PH3, Greater stability of P(V) : P4O10 1 mark + 1 mark 92 2B 6 Bc ii 6Bc ii Which is more reactive to oxidation, H2O or H2S? Explain your answer. 2 H2S 1 mark

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) 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), CO3

2-(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?

3

Al(s) 2Al(s) + 2OH-(aq) + 2H2O(l) → 2AlO2

-(aq) + 3H2(g)

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 2

This is an acid-base reaction. ½ mark CaO is a basic oxide and SiO2 is an acidic oxide 1 mark The reaction does not involve a change in oxidation state. 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. 3

(ii) sodium hydroxide solution Pb2+

(aq) Cu2+

(aq) (3 marks for 8 observations; deduct ½ mark for each incorrect observation) (ii) sodium hydroxide solution Pb2+

(aq) White ppt. Soluble in excess / Colour solution

Cu2+(aq) Blue ppt.

Insoluble in excess

s-Block Elements I. Characteristic properties of the s-block elements

A. Metallic character and low electronegativity B. Ionic bond formation and fixed oxidation state C. Flame test

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

1. Reaction with hydrogen 2. Reaction with oxygen 3. Reaction with chlorine 4. Reaction with water

B. Thermal stability of carbonates and hydroxides 1. Thermal stability of carbonates 2. Thermal stability of hydroxides

C. 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

D. Abnormal behaviour of lithium

III. Uses of the compounds of the s-block elements A. Manufacture of glass B. Baking powder C. Making of soap D. Antacid E. Slaked lime F. 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 Reading

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

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 Group I metals Group II metals Lithium (Li) Deep red Beryllium (Be) no special colour Sodium (Na) Yellow Magnesium (Mg) no special colour Potassium (K) Lilac (a purple colour) Calcium (Ca) Brick red Rubidium (Rb) Bluish Red Strontium (Sr) Blood red / Crimson red Caesium (Cs) Blue Barium (Ba) Apple green Other metals Copper (Cu) Bluish green

Glossary nichrome wire flame test

s-Block Elements Unit 1 Page 2 Past Paper Question

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.) 1

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). 2 Carry out a flame test, 1 mark

KCl(s) gives a lilac flame ½ mark while MgCl2(s) gives no characteristic flame colour ½ mark Or Add NaOH(aq) / Na2CO3(aq) / Na2HPO4(aq) to aqueous solution of the substance. 1 mark MgCl2 gives white precipitate ½ mark while KCl does not. ½ mark (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.



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

Assignment 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 2M(s) + Cl2(g) → 2MCl(s) Group II M(s) + Cl2(g) → MCl2(s)

s-Block Elements Unit 2 Page 2 4. 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. M2CO3(s) heat → M2O(s) + CO2(g) MCO3(s) heat → MO(s) + CO2(g)

CO O2-

O+C OO

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. Li2CO3(s) heat → Li2O(s) + CO2(g) Among the group II carbonates, the decomposition temperature increases down the group as the polarizing power of the cation decreases. Group II carbonates Decomposition temperature / ºC BeCO3(s) ~ 100 MgCO3(s) 540 CaCO3(s) 900 SrCO3(s) 1290 BaCO3(s) 1360

s-Block Elements Unit 2 Page 3 2. 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. 2MOH(s) heat → M2O(s) + H2O(l) M(OH)2(s) heat → MO(s) + H2O(l)

2-O+

O H-

O H-

O H

H

Glossa ry oxide peroxide superoxide thermal stability

Past Paper Question

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

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. 4

The surface of the sodium becomes dull (less shiny, more white) ½ mark 4Na(s) + O2(g) → 4Na2O(s) 1 mark The piece of sodium melts and burns with a yellow flame. ½ mark 2Na(s) + 2H2O(l) → 2Na+

(aq) + 2OH-(aq) + H2(g) 1 mark

2H2(g) + O2(g) → 2H2O(l) 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. 3

The protective layer of Al2O3 is destroyed when the piece of Al is dipped in a solution of HgCl2. 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. 1 mark

91 2B 5 d 5d Account for the observation that lithium carbonate decomposes at 700ºC whereas sodium carbonate is thermally

stable at this temperature. 3

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

CO O

--2M+ + 2M+O2- + C OO

1 mark Change of lattice energy from Li2CO3 to Li2O is greater than Na2CO3 to Na2O. 2 mark

s-Block Elements Unit 2 Page 4 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. 2

2M(s) + Cl2(g) → 2MCl(s) 1 mark 2M(s) + 2H2O(l) → 2M+

(aq) + 2OH-(aq) + H2(g) 1 mark

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

When compared with Ca2+, Mg2+ has higher chargeradius ratio and is more polarizing, 1 mark

∴ CO32- in MgCO3 is polarized to a greater extent than in CaCO3. 1 mark

Hence MgCO3 undergoes thermal decomposition to form oxide (small anion) more readily than CaCO3. 1 mark

95 2B 5 a v 5a The table below lists some properties of the alkali metals. Element Atomic

radius / nm Ionic radius / nm First ionization

energy / kJmol-1 Standard electrode

potential / V Melting point / ºC

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. 4

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. 2 Mg2+ is more polarizing than Na+ due to its smaller size and greater charge. The OH- is polarized to a greater

extent in Mg(OH)2. 2 marks 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. 3 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. Therefore, carbonates of Be, Mg tend to decompose to form the more stable oxides. 3 marks

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) 2 NaHCO3 decomposes on heating to give CO2 /a gas ½ + ½ mark

which can turn lime water milky but Na2CO3 does not decompose. ½ + ½ mark Or 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. ½ + ½ mark

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

s-Block Elements Unit 2 Page 5 equation for the reaction.

s-Block elements Unit 3 Page 1


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

Assignment Reading

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

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 – The process in which the particle of the solute in gaseous state is dissolved in a solvent to

form a solution. Solution process – 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 : i. NaCl(s) → Na+

(g) + Cl-(g) ∆Ho = - ∆Holat

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. ii. Na+

(g) + Cl-(g) → Na+

(aq) + Cl-(aq) ∆Hohyd or ∆Hosolv

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 ⇒ the solute would be more soluble If ∆Hohyd is less negative than ∆Holat ⇒ 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 :

Hydration energy ∝ ( 1r+

+ 1r-

) 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 :

Lattice energy ∝ 1

r+ + r- r+, r- : radii of cation and anion

c) 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 solvent-solvent particles.

Sizes of the cations and anions Hydration energy Lattice energy Solubility

(∝ ( 1r+

+ 1r-

)) (∝ 1

r+ + r- )

Both of them are small High High Low to moderate Both of them are large Low Low Low to moderate They have different sizes High Low 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 Page 4 2. Solubility of some ionic crystals

a) Sulphate salt of group II metal Compound Relative solubility Solubility mol/100g of water at 298 K MgSO4 Soluble 1.83 × 10-1 CaSO4 Sparingly soluble 4.66 × 10-3 SrSO4 Insoluble 7.11 × 10-5 BaSO4 Insoluble 9.43 × 10-7


r+ + r- Hydration energy ∝ (

1r+

+ 1r-

)

On descending the group, magnitudes of both hydration energy and lattice energy decrease as the size of the cation increases. SO4

2- 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 Relative solubility Solubility mol/100g of water at 298 K Mg(OH)2 Insoluble 2.00 × 10-5 Ca(OH)2 Sparingly soluble 1.53 × 10-3 Sr(OH)2 Slightly soluble 3.37 × 10-3 Ba(OH)2 Slightly soluble 1.50 × 10-2


r+ + r- Hydration energy ∝ (

1r+

+ 1r-

)

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 Page 5 c) 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 hydration solution

Past Paper Question

90 2B 4 e 91 2A 1 b iii 91 2B 5 c 93 2B 4 b ii 96 2B 4 c 98 1A 3 b iii 99 1A 1 b ii

90 2B 4 e 4 Account for the following observations. 4e Lithium salts generally have high solubilities in water. 3 Li+ has very high charge density because of the small ionic radius. 1 mark

there formation of Li(H2O)n+ involve high hydration energy. 1 mark

Li salts - lattice energy is low because of the crystal packing is not efficient. 1 mark

91 2A 1 b iii iii When solid sodium chloride is dissolved in water, the process is endothermic. Explain this observation. 2 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 these two processes. In case of NaCl, the lattice energy is larger than the enthalpy of hydration. 2 marks

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

solubility of group(II) salts of small anions shows the reverse trend. 4

∆Hsoln = -∆HLE + ∆HHE 1 mark ∆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. More negative the ∆Hsoln, more soluble would be the salt. ½ mark

Lattice energy ∝ 1r r+ −+

½ mark

Hydration energy ∝ ( )1 1r r+ −

+ ½ mark

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 The ∆Hsoln becomes less negative on descending the group and solubilities decrease. ½ mark 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 CrO4

2- 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 BaCrO4. 1 mark

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 Page 1


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

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 Lithium Magnesium Sodium

Burning in air Formation of normal oxide

Formation of normal oxide

Formation of peroxide on top of normal oxide

Heating of carbonate Decompose into oxide and carbon dioxide

Decompose into oxide and carbon dioxide

Stable to heat

Heating of hydroxide Decompose into oxide and water

Decompose into oxide and water

Stable to heat

Heating of nitrate(V) Decompose into oxide, nitrogen dioxide and oxygen

Decompose into oxide, nitrogen dioxide and oxygen

Decompose into nitrate(III) and oxygen only

Formation of hydrogencarbonate

Do not form stable solid hydrogencarbonate

Do not form stable solid hydrogencarbonate

Form stable solid hydrogencarbonate

Formation of organometallic compound

Form organometallic compound e.g. RLi

Form organometallic compound e.g. MgR2

Do not form any stable organometallic compound at all

Solubility of salts Insoluble phosphate, fluoride and carbonate

Insoluble phosphate, fluoride and carbonate

Soluble phosphate, fluoride and carbonate

Stability of compound Form stable nitride Form stable nitride Do not form stable nitride,

Glossary

anomaly organometallic compound

s-Block elements Unit 4 Page 2 Past Paper Question

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. 3

Any two of the following examples: 1. Lithium carbonate, nitrate, hydroxide decomposes upon heating, while carbonate of other Group I elements

do not. Li2CO3

heat → Li2O + CO2 4LiNO3

heat → 2Li2O + 4NO2 + O2 2LiOH heat → 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. 6. Li+

2S2- is insoluble in H2O 2 marks 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

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 .' 4

Lithium carbonate decomposes on heating to give (lithium oxide) and carbon dioxide. ½ + ½ mark Li2CO3 → Li2O + CO2 (1 mark) But other alkali metal carbonates do not. 1 mark Lithium nitrate(V) decomposes on heating to give (lithium oxide), ½ mark nitrogen dioxide / a brown gas and oxygen. ½ mark 2LiNO3 → Li2O + 2NO2 + ½O2 (1 mark) But other alkali metal nitrate(V) give nitrate(III) and oxygen. 1 mark 2MNO3 → 2MNO2 + O2 (1 mark)



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

Assignment 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. CaCO3(s) heat → CaO(s) + CO2(g) 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 slaked lime quicklime

Past Paper Question

94 2B 6 a 97 1A 3 a iv

94 2B 6 a 6a Self-raising flour for baking cakes and bread contains sodium hydrogencarbonate. Explain briefly, with the aid of

chemical equation(s), the role of sodium hydrogencarbonate in the baking process. 3

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 The CO2 gas produced causes the cakes to rise. 1 mark

97 1A 3 a iv 3a For each of the following materials, identify ONE inorganic compound which is an active ingredient: 3 iv antacid

p-Block Elements I. Halogens

A. Laboratory preparation of chlorine B. Characteristic properties of the halogens C. Properties of halogen

1. Oxidizing power of halogen 2. Disproportionation of halogen in alkalis

D. 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

E. Uses of halogens and halogen containing compounds 1. Fluoridation of water 2. Use of chlorine 3. Photographic films

II. Nitrogen and its compounds

A. Nitrogen 1. Unreactive nature of nitrogen 2. Direct combination of nitrogen and oxygen

B. Ammonia 1. Haber process 2. Chemical properties of ammonia

a) Alkaline properties (1) Laboratory preparation of ammonia

b) Reducing properties (1) Combustion (2) Catalytic oxidation (3) With copper(II) oxide (4) With chlorine (5) Thermal decomposition of ammonia

C. Nitric(V) acid 1. Ostwald Process 2. Oxidizing properties of nitric(V) acid

a) With copper b) With iron(II) ion c) With sulphur

D. Nitrate(V) salt 1. Thermal decomposition of nitrate 2. Brown ring test

E. Use of nitric(V) acid a) Fertilizers b) Explosives c) Nylon d) Dyestuffs

III. Sulphur and its compounds A. Allotrope of sulphur B. Burning of sulphur C. 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 D. Concentrated sulphuric(VI) acid

1. Contact process 2. Properties of concentrated sulphuric acid

E. Sulphate(VI) salt 1. Test for sulphate(VI) ion

F. 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 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) heat → MnCl2(aq) + 2H2O(l) + 2Cl2(g)

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 Standard reduction potential I2(s) + 2e- d 2I-

(aq) + 0.54 V Br2(g) + 2e- d 2Br-

(aq) + 1.07 V MnO2(s) + 4H+

(aq) + 2e- d Mn2+(aq) + 2H2O(l) + 1.23 V

Cl2(g) + 2e- d 2Cl-(aq) + 1.36 V

MnO4-(aq) + 8H+

(aq) + 5e- d Mn2+(aq) + 4H2O(l) + 1.51 V

F2(g) + 2e- d 2F-(aq) + 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 Page 3 2. 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) at room temp. (not a disproportionation reaction)

0 -1 -1 2F2(g) + 4OH-

(aq) → O2(g) + 4F-(aq) + H2O(l) at about 70ºC

0 -1 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) at room temp. 0 +1 -1 3Cl2(g) + 6OH-

(aq) → ClO3-(aq) + 5Cl-

(aq) + 3H2O(l) at about 70ºC 0 +5 -1 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) at room temp. 0 +5 -1 Disproportionation of iodine in alkali 3I2(s) + 6OH-

(aq) → IO3-(aq) + 5I-

(aq) + 3H2O(l) at room temp. 0 +5 -1 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 93 2B 6 a ii 94 1A 2 a ii 94 2B 4 a 96 1A 2 c d I ii 97 1B 7 a 98 2B 8 d i ii

I. Halgoens Unit 1 Page 4 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.

3

2F2(g) + 2OH-(aq) → OF2(aq) + 2F-

(aq) + H2O(l) 1 mark 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. 2

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

- + 2Cl- 3Cl2 + 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: iii dil NaOH Cl H O 2

room temperature2+ → + + 1

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

(aq) 2 (l)+ → 1 mark

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, 3 At 295 K, Cl2, Br2 and I2 disproportionate according to the equation

X2 + 2OH-(aq) → OX-

(aq) + X-(aq) + H2O(l) 1 mark

OBr- and OI- are unstable and will undergo further disproportionation. 3OX- d XO3

- + 2X- X = Br or I 1 mark F2 is the strongest [O] agent, it will not disproportionate, instead, it will oxidize the OH- ion. 2F2 + 2OH- → OF2(g) + 2F- + H2O 1 mark

ii at 350 K. 2 At 350 K, Cl2, Br2 and I2 all disproportionate as follows

3X2 + 6OH-(aq) → XO3

-(aq) + 5X-

(aq) + 3H2O(l) 1 mark F2 oxidizes the alkali, F2 + 4OH- → O2(g) + 4F- + 2H2O 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. 1 2KOH(aq) + Cl2(g)

cold dil . → KCl(aq) + KOCl(aq) + H2O(l) 2OH-

(aq) + Cl2(g) cold dil . → Cl-

(aq) + OCl-(aq) + H2O(l)

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

6OH-(aq) + 3Cl2(g)

hot conc. → 5Cl-(aq) + ClO3

-(aq) + 3H2O(l) 1 mark

I. Halgoens Unit 1 Page 5 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. 5

Heating conc. HCl in the presence of manganese(IV) oxide MnO2(s) + 4HCl(aq)

heat → MnCl2(aq) + 2H2O(l) + 2Cl2(g) 2 marks OR Add conc. HCl to solid KMnO4 10HCl(aq) + 2MnO4

-(aq) + 6H+

(aq) → 5Cl2(g) + 8H2O(l) + 2Mn2+(aq)

Pass the Cl2 produced through cold water to remove HCl, dry the gas by passing through conc. H2SO4 and collect by downward delivery. 2 marks 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 ?

HClO3 HClO4 HMnO4 H2CrO4

HClO3 +5 or V HClO4 +7 or VII HMnO4 +7 or VII H2CrO4 +6 or VI (DO NOT accept a number preceding the charge e.g. 5+) 4 × ½ mark

2

2d i Explain why among the four compounds in (c), only HClO3 can undergo disproportionation. 1 The central atoms of the other three compounds are in their highest oxidation states ½ mark

whereas in HClO3, Cl is in an intermediate oxidation state. ½ mark

ii Write a balanced equation for the disproportionation of HClO3 to give HClO4 and ClO2. 1 3HClO3 → HClO4 + H2O + 2ClO2 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. 6

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.

4

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.

I. Halogens Unit 2 Page 1


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

Assignment 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)4

2+(aq) + 2OH-

(aq) pale blue deep blue (tetraamminecopper(II) ion) Zn(OH)2(s) + 4NH3(aq) d Zn(NH3)4

2+(aq) + 2OH-

(aq) white colourless (tetraamminezinc(II) ion)

Glossary displace involatile soluble complex redissolve diamminesilver(I) ion tetraamminecopper(II) ion tetraamminezinc(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

I. Halogens Unit 2 Page 3 90 2B 6 c 6c Explain why copper(II) chloride is a stable compound whereas copper(II) iodide does not exist. 3 Cl is much more electronegative than I / the size of I- is larger than Cl- 1 mark

Cl- is more resistant to oxidation than I- 1 mark Cu2+ is an oxidizing agent strong enough to oxidize I- to I2 or I- is a stronger reducing agent or I- ion is polarized by Cu2+ CuI2 disproportionates, CuI2 → CuI + ½I2 1 mark

92 1A 3 f i ii 3f Complete and balance the following equations: i conc H SO KBr H O + + 2 4

room temperature2+ → + 1

2KBr 3H SO 2KHSO 2H O SO + Br(s) 2 4 (conc)room temperature

4(aq) 2 (l) 2(g) 2(g)+ → + + 1 mark

ii conc H PO KBr 3 4heat+ → + 1

KBr H PO HBr KH PO (s) 3 4(conc)heat

(g) 2 4(aq)+ → + 1 mark

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.

3

Reaction between NaX and H3PO4 (X = Cl or Br) H3PO4 + NaX → HX + NaH2PO4 1 mark With H2SO4, NaCl gives similar reaction NaBr will react to give also Br2 and SO2, 2HBr + H2SO4 → 2H2O + SO2 + Br2 1 mark Q conc. H2SO4 is an oxidizing agent while conc. H3PO4 is not. 1 mark

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 1 NaCl + H2SO4 → NaHSO4 + HCl 1 mark ii concentrated H2SO4 and NaI. 2 NaI + H2SO4 → HI + NaHSO4

8HI(g) + H2SO4(l) → 4I2(s) + H2S(g) + 4H2O(l) 1 mark HI / I-, being a stronger reducing agent than HCl, will be oxidized by conc. H2SO4 to give I2. 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. 3

Prepare conc. aqueous solutions by dissolving the solids in water. 1 mark Add AgNO3(aq) to solution in test tube ½ mark Add excess NH3(aq) to the precipitate ½ mark Let the precipitate settle, decant the solution and compare the height of the precipitate. 1 mark (Note : Br- is insoluble in dil.NH3 but soluble in conc. NH3)

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 KCl + H2SO4 → KHSO4 + HCl 1 mark ii Briefly describe how a sample of dry hydrogen chloride can be prepared in the laboratory using the above

reaction. Draw a labelled diagram of the laboratory set-up and state the safety precaution(s) that is/are required.

6

Hydrogen chloride can be prepared by dropping (slow addition of) c. H2SO4 onto KCl ½ mark and warming the reaction mixture. ½ mark The HCl produced is tried by passing through anhydrous CaCl2 /concentrated H2SO4 ½ mark and then collected by downward delivery / in a gas syringe. ½ mark Precaution: carry out experiment in fume cupboard. 1 mark [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.

2

c. H2SO4 can oxidize HBr produced to Br2 1 mark 2HBr + H2SO4 → 2H2O + SO2 + Br2 1 mark Or 2KBr + 3H2SO4 → 2KHSO4 + 2H2O + SO2 + Br2 (1 mark)

iv Suggest how a sample of hydrogen bromide can be prepared in the laboratory. 1 HBr can be prepared by heating KBr with cone. H3PO4. 1 mark

Or Add Br2(l) to moist red phosphorus. (1 mark)

98 2B 8 b ii 8b For each of the following, state the expected observation and write the relevant balanced equation(s). 6 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.



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) HF HCl HBr HI pKa 3.2 -7 -9 -10 E(H–X) 568 432 366 298 ∆ EN 3.9 0.9 0.7 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.

ClOH

SO O

O OH H

PO O

O OH H

H

chloric(I) acid (HOCl) sulphuric(VI) acid (H2SO4) 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 Past Paper Question

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

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 both are strong oxidizing agents 1 mark

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. 2 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) 2 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 HCl(aq) + H2O(l) d H3O+

(aq) + Cl-(aq) lies to the right, so HCl is a strong acid. 1 mark

OR HX bond strength is greater in H–F than in H–Cl 1 mark this outweighs the reduced hydration energy from F- to Cl- 1 mark OR HF slightly ionizes in H2O & HCl completely ionizes in H2O 1 mark

94 2B 5 a v 5a v Draw the three-dimensional structure of the manganate(VII) ion. 1

Mn

OO O

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.

3

highest oxidation state = +7 / VII ½ mark example: BrO4

- / HBrO4 1 mark lowest oxidation state = -1 / -I ½ mark example: Br- / HBr 1 mark (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).



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 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.

C CH H

H HC C

H

H

H

H

Cl Cl

Cl2Electrophilicaddition

C C

H

H

H

H

Cl Cl

C CH H

Cl H

Heat 500 °C

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

C CH H

Cl HR

R C C

H H

HCl

R C C

H H

HCl

C C

Cl

H H

H

R C C

H H

HClC C

H 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 bleach disinfectant photographic film

Past Paper Question

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

96 1A 2 a 2a State the method by which sodium hydroxide is produced industrially. Give TWO other products obtained by this

industrial process. 2

NaOH(aq) is produced by electrolysis of brine / concentrated NaCl(aq) 1 mark (Deduct ½ mark for not mentioning ‘concentrated’.) Other products: H2 / Cl2 / NaOCl solution (bleach solution) (any TWO) ½ + ½ mark

97 1A 3 a i iii 3a For each of the following materials, identify ONE inorganic compound which is an active ingredient: 3 i photographic film iii bleaching powder

II. Nitrogen and its compounds Unit 1 Page 1

Topic II. Nitrogen and its compounds Unit 1

Reference Reading

10.2.0–10.2.2

Assignment Reading

Chemistry in Context, 3rd Edition ELBS pg. 362–365 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 443–448 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) Bond energy (kJmol-1) Bond energy (kJmol-1) N–N + 158 C–C + 347 N=N + 410 C=C + 612 N≡N + 945 C≡C + 838 N–H + 391 C–H + 435 H–H + 436

II. Nitrogen and its compounds Unit 1 Page 2 2. 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)

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. yield rate of reaction ↑Pressure ↑ ↑ ↑Temperature ↓ ↑ 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 Unit 1 Page 4 2. 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. 4NH3(g) + 5O2(g)

Pt catalyst → 4NO(g) + 6H2O(l) ∆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 Unit 1 Page 7 Past Paper Question

90 2B 5 b 92 2B 6 Ac iii 97 2B 8 b ii iii 97 2B 8 c i ii 99 2A 1 c i 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. 7

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 Effect of concentration of N2 and H2 1 mark 2 marks Catalyst - required because of the inertness of N2 and H2 molecules 1 mark NH3 is removed from the reaction mixture 1 mark 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. 4

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 7

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. 4

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.



Reference Reading

10.2.3

Assignment Reading

Chemistry in Context, 3rd Edition ELBS pg. 365–366 Inorganic Chemistry, 4th Edition, T.M. Leung pg. 159–162 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. 4NH3(g) + 5O2(g) Pt → 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 Unit 2 Page 2 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) NO3

-(aq) + 2H+

(aq) + e- d NO2(g) Dilute nitric(V) acid (2M) NO3

-(aq) + 4H+

(aq) + 3e- d NO(g) Very dilute nitric(V) acid (0.1M) 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 no reaction with dilute nitric(V) acid (with heating) 3Cu(s) + 2NO3

-(aq) + 8H+

(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)

with concentrated nitric(V) acid (cold) 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 Unit 2 Page 3 c) 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 passive

Past Paper Question

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

92 2B 6 Ab ii 6Ab What are the products of the reactions between ii cold, concentrated nitric(V) acid and hydrogen sulphide gas? 1 H2S(g) + 2HNO3(conc.) → S(s) + 2H2O(l) + 2NO2(g) 1 mark

note: on boiling, H2SO4(aq) formed

93 2B 5 Ia 5Ia

For the industrial preparation of nitric(V) acid from nitrogen , give the chemical reactions and the conditions under which they take place.

6

Pass purified N2 and H2 (ratio 1:3) over iron catalyst at 450ºC and 200 atm pressure. N2 + 3H2 d 2NH3 2 marks 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. 4NO2 + O2 + 2H2O → 4HNO3 1 mark

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. 6

ii nitrogen monoxide, NO



Reference Reading

10.2.4

Assignment Reading

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

Reactive metal nitrate(V) thermal decomposition → Reactive metal nitrate(III) + Oxygen 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. Less reactive metal nitrate(V) thermal decomposition → Less reactive metal oxide + Nitrogen dioxide + Oxygen 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.)

Non-reactive metal nitrate(V) thermal decomposition → Non-reactive metal + Nitrogen dioxide + Oxygen 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.

NH4NO3(s)

thermal decomposition → 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. Recall the action of nitric(III) acid on amino group, R–NH2 + HNO2(aq)

0 5− ° →C 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. NH4NO2 (s)

thermal decomposition → 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 Unit 3 Page 3 Past Paper Question

91 1A 3 h 91 1B 4 a 97 1A 3 a ii 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.

2

NH4NO3(s) is unstable and readily/easily decomposes to give large volumes of gaseous products (N2O and H2O) and is therefore explosive. 1 mark No smoking because N2O is a supporter of combustion at elevated temperature 1 mark

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.)

4

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. 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, the Br2 dissolves in the organic solvent leaving a colourless aqueous layer. ½ mark for each point

97 1A 3 a ii 3a For each of the following materials, identify ONE inorganic compound which is an active ingredient: 3 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. 6

i dinitrogen oxide, N2O



Reference Reading

10.2.5

Assignment Reading

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

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

CH

CH2

CH2

O

O

O

NO2

NO2

NO2

CH

CH2

CH2

OH

OH

OH3 HNO3+

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

CH

CH2

CH2

O

O

O

NO2

NO2

NO2

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

12 CO2(g) + 10 H2O(g) + 6 N2(g) + O2(g)

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

NO2

NO2

O2N

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

HH

H

HH

CH2ONO2

ONO2

ONO2

O

ncellulose nitrate (nitrocellulose, gun cotton)

II. Nitrogen and its compounds Unit 4 Page 2 c) 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

+ CH2CH2CH2CH2 COOHHOOC60% HNO3, 60 º C, cat.

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

OHH

naphthalen-2-ol

azo dye(an orange red ppt.)

H2O+N

N

OH

OH-

N

N

OH

H

+diazonium ion

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 :

O-

OH

N N

SO2O-Na+

N NNa+-OO2S+

+

Orange II

Nitric acid is used to prepare nitrobenzene through which diazonium ion is prepared.

NO2

diazonium ion

NaNO2HCl, 0-5 蚓

NH2

benzenamine(an aromatic primary amine)

N N+

SnHCl

HNO3 , H2SO4 50 - 55 蚓

II. Nitrogen and its compounds Unit 4 Page 3 Glossary nitroglycerine trinitrotoluene (T.N.T.) dynamite azo dye diazonium ion

Past Paper Question

97 2B 8 b iv

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

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 Reading

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

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. S8(s) + 8 O2(g) → 8 SO2(g) or simply S(s) + O2(g) → SO2(g)

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. Heating of sulphur

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). Cu(s) + 2H2SO4(l) heat → CuSO4(aq) + SO2(g) + 2H2O(l) 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. SO3

2-(aq) + 2H+

(aq) → H2SO3(aq) → SO2(g) + H2O(l)

III. Sulphur and its compounds Unit 1 Page 4 1. 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 SO3

2-(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 SO3

2-(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 Cr2O7

2-(aq).

a) With manganate(VII) ion H2SO3(aq) + H2O(l) d 4H+

(aq) + SO42-

(aq) + 2e- × 5 Eo = - 0.17 V MnO4

-(aq) + 8H+

(aq) + 5e- d Mn2+(aq) + 4H2O(l) × 2 Eo = + 1.51 V

Overall 5SO3

2-(aq) + 2MnO4

-(aq) + 6H+

(aq) d 5SO42-

(aq) + 2Mn2+(aq) + 3H2O(l) 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 Eo = - 0.17 V Cr2O7

2-(aq) + 14H+

(aq) + 6e- d 2Cr3+(aq) + 7H2O(l) Eo = + 1.33 V

Overall 3SO3

2-(aq) + Cr2O7

2-(aq) + 8H+

(aq) d 3SO42-

(aq) + 2Cr3+(aq) + 4H2O(l) Eo = + 1.16 V

Orange Cr2O7

2-(aq) will be reduced to green Cr3+

(aq) by sulphur dioxide. c) With bromine H2SO3(aq) + H2O(l) d 4H+

(aq) + SO42-

(aq) + 2e- Eo = - 0.17 V Br2(aq) + 2e- d 2Br-

(aq) Eo = + 1.09 V Overall SO3

2-(aq) + Br2(aq) + H2O(l) d SO4

2-(aq) + 2Br-

(aq) + 2H+(aq) 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 Unit 1 Page 5 3. 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. SO3

2-(aq) + Ca2+

(aq) → CaSO3(s) CO3

2-(aq) + Ca2+

(aq) → CaCO3(s) SO3

2-(aq) + 2Ag+

(aq) → Ag2SO3(s) CO3

2-(aq) + 2Ag+

(aq) → Ag2CO3(s) SO3

2-(aq) + Ba2+

(aq) → BaSO3(s) CO3

2-(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 rhombic sulphur monoclinic sulphur plastic sulphur

Past Paper Question

93 2B 5 IIc

93 2B 5 IIc IIc

SO2 can act as an oxidant or a reductant. Give one reaction in each case to illustrate this. 2

SO2 as reducing agent : reaction with acidified K2Cr2O7 / KMnO4 Cr2O7

2- + 3SO2 + 2H+ → 2Cr3+ + 3SO42- + 5H2O

or MnO4- + 5SO2 + 2H2O → 2Mn2+ + 5SO4

2- + 5H2O + 4H+ SO2 as oxidising agent : H2S + SO2 → S(s) + H2O (any 2 appropriate examples) 2 marks



Reference Reading

10.3.3–10.3.4

Assignment Reading

Chemistry in Context, 3rd Edition ELBS pg. 359–362 A-Level Chemistry (3rd ed.), Stanley Thornes (Publisher) Ltd., 428–434 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. SO2(g) + O2(g) d SO3(g) ∆H = -197 kJ mol-1 Vanadium(V) oxide (V2O5(s)) catalyst, 450 ºC, 1 atm 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. yield rate of reaction ↑ Pressure ↑ ↑ ↑ Temperature ↓ ↑ 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.

Effect of temperature on the equilibrium constant and percentage

of conversion

2. Properties of concentrated sulphuric acid

i. Acid

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 Cu(s) → Cu2+

(aq) + 2e- Reduction half equation SO4

2-(aq) + 4H+

(aq) + 2e- → SO2(g) + 2H2O(l) Ionic equation Cu(s) + SO4

2-(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. CuSO4·5H2O(s)

conc H SO l. ( )2 4 → CuSO4(s) + 5H2O(l) blue crystal 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. Cx(H2O)y(s)

conc H SO l. ( )2 4 → xC(s) + yH2O(l) 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 C(s) + 2H2O(l) → CO2(g) + 4H+

(aq) + 4e- Reduction half equation SO4

2-(aq) + 4H+

(aq) + 2e- → SO2(g) + 2H2O(l) Ionic equation C(s) + 2H2O(l) + 2SO4

2-(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. H2SO4(l) + NaNO3(s) heat → NaHSO4(s) + HNO3(g)

E. Sulphate(VI) salt

1. Test for sulphate(VI) ion

SO4

2-(aq) is usually tested with BaCl2(aq) solution. In the presence of BaCl2(aq), SO4

2-(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, SO3

2-(aq) and CO3

2-(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 vanadium(V) oxide oleum / fuming sulphuric acid hygroscopic dehydrating barium chloride

III. Sulphur and its compounds Unit 2 Page 5 Past Paper Question

90 2B 5 b 92 2B 6 Bb ii 93 2B 5 IIa 98 1B 7 a iii 98 2B 8 c i ii 99 2A 1 c ii

90 2B 5 b 5b Outline an industrial preparation of sulphuric acid.


Sulphuric acid S + O2 → SO2 2SO2 + O2 d 2SO3; ∆H < 0 very slow in the absence of catalyst SO3 + H2SO4 → H2S2O7

H O2 → 2H2SO4 Steps involved: Production of SO2, conversion to SO3, removal of SO3 3 marks Conditions of conversion: 1 atm, 450ºC, vanadium(V) oxide catalyst 1 mark Effect of [O2], removal of SO3 1 mark Effect of pressure 1 mark } any 1 Effect of temperature on the equilibrium constant 1 mark 1 mark Catalyst - lower the activation energy for the SO2 + O2 reaction as O2 is kinetically inert 1 mark Overall presentation 1 mark

92 2B 6 Bb ii 6Bb What are the products of the reactions between ii concentrated sulphuric(VI) acid and hydrogen sulphide gas? 1 H2SO4(conc.) + H2S(g) → S(s) + 2H2O(l) + SO2(g) 1 mark 93 2B 5 IIa IIa

For the industrial preparation of sulphuric(VI) acid from sulphur, give the chemical reactions and the conditions under which they take place.

6

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 1 98 2B 8 c i ii 8c The conversion of sulphur dioxide to sulphur trioxide in the Contact Process involves the following :

SO2(g) + ½O2(g) d SO3(g) with V2O5(s) catalyst ∆Ho = -98 kJmol-1 5

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)



Reference Reading

10.3.5

Assignment Reading

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

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) (NH4)2SO4(aq)

crystallization → (NH4)2SO4(s) 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 CH3(CH2)11 SO3-Na+(i) conc. H2SO4

(ii) NaOH

c) 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.


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 6Bc iii Describe the part that H2SO4 plays in any TWO important industrial processes. 4 not Contact Process

Production of 'superphosphate' fertilizer 1 mark Ca3(PO4)2 + H2SO4 → Ca(H2PO4)2 + 2CaSO4 2 marks acts as acid 1 mark or Production of fertilizer 1 mark 2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq) 2 marks 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

Topic Nitrogen and Sulphur Unit 1

Reference Reading

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

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.


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 Effect of concentration of N2 and H2 1 mark 2 marks Catalyst - required because of the inertness of N2 and H2 molecules 1 mark NH3 is removed from the reaction mixture 1 mark Overall presentation 1 mark Sulphuric acid S + O2 → SO2 2SO2 + O2 d 2SO3; ∆H < 0 very slow in the absence of catalyst SO3 + H2SO4 → H2S2O7


81 2B 8 a i ii iii d 8a Describe EITHER the process for the conversion of ammonia gas into (68%) nitric(V) acid,

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.

12

I Concentrated nitric(V) acid Ammonia(g) is mixed with air. 1 mark passed over a Pt gauze or (Pt/Rh) 1 mark catalyst which is maintained at 800ºC and 8 atm. 1 mark the reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) 1 mark takes place with the evolution of heat. 1 mark 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. 2NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) 1 mark the NO(g) is recycled. Absorption may be carried out up to 68%(conc.) HNO3 which is the constant boiling mixture. 1 mark (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

II Concentrated sulphuric(VI) acid 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. 1 mark The operating temperature is about 450ºC at 1 atm. 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. 1 mark SO3(g) + H2SO4(l) → H2S2O7(l) 1 mark Oleum is then diluted with calculated amount of water so as to give back 98% H2SO4(l) 1 mark 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


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

8d Write on the present-day uses of EITHER concentrated nitric(V) acid, OR concentrated sulphuric(VI) acid.

4

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. conc. H2SO4 Manufacture of soapless detergents 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

having oxidation numbers (O.N.) of +2, +4 and +6.

N: NO(+2), NO2/N2O4(+4), NO3(+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 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.

8

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

10b Describe the industrial preparation of EITHER nitric(V) acid from nitrogen OR sulphuric(VI) acid from sulphur.

9

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 3. Conversion of sulphur(VI) oxide to sulphuric(VI) acid, SO3 + H2SO4 → H2S2O7 H O2 → 2H2SO4

Add oxide to acid then dilute with water 3 marks

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. 6

Redox chemistry of sulphur overall presentation 1 mark Principal oxidation states with examples e.g. -2(H2S), 0(S), +4(SO2) and +6(SO3 or SO4

2-) 2 marks Any sulphur containing compounds, upon oxidation, give SO2 or SO4

2- SO2 – through combustion of the sulphur compounds in air SO4

2- – oxidation in homogeneous solution. 1 mark Upon reduction, S or H2S are usually produced H2S is a mild reducing agent but SO4

2- is usually stable towards redox reactions in the absence of H+. 1 mark Intermediate oxidation states, for example, S2O3

2- or S4O62-, will undergo disproportionation in water. 1 mark

Redox chemistry of nitrogen overall presentation 1 mark Principal oxidation states with examples e.g. 0(N2), -3(NH3), +4(NO2), +5(NO3

-) 2 marks 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

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. 1 mark Compounds in intermediate oxidation states such NO2 undergo disproportionation 2NO2 + H2O d HNO3 + HNO2 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.

5

Depends on the nature of the reducing agents and the reaction conditions employed: Nitric(V) acid NO3

- + 2H+ + e- d NO2 + H2O 1 mark 2NO3

- + 12H+ + 10e- d N2 + 6H2O 1 mark NO3

- + 9H+ + 8e- d NH3 + 3H2O 1 mark NO3

- + 4H+ + 3e- d NO + 2H2O 1 mark other possible product: N2O, H2NNH2, NO2

- H+ is required if NO3

- acts as an oxidizing agent 1 mark Sulphuric(VI) acid SO4

2- + 4H+ + 2e- d SO2 + 2H2O 1 mark SO4

2- + 8H+ + 6e- d S + 4H2O 1 mark SO4

2- + 10H+ + 8e- d H2S + 4H2O 1 mark 2SO4

2- d S2O32- or S4O6

2-, etc. 1 mark H+ is required if SO4

2- acts as an oxidizing agent 1 mark

Nitrogen 10.2.2.1 90 2B 5 b 5b Outline an industrial preparation of ammonia.


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 Effect of concentration of N2 and H2 1 mark 2 marks Catalyst - required because of the inertness of N2 and H2 molecules 1 mark NH3 is removed from the reaction mixture 1 mark Overall presentation 1 mark

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.

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.

12

I Concentrated nitric(V) acid Ammonia(g) is mixed with air. 1 mark passed over a Pt gauze or (Pt/Rh) 1 mark catalyst which is maintained at 800ºC and 8 atm. 1 mark the reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) 1 mark takes place with the evolution of heat. 1 mark 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. 2NO2(g) + H2O(l) → 2HNO3(aq) + NO(g) 1 mark the NO(g) is recycled. Absorption may be carried out up to 68%(conc.) HNO3 which is the constant boiling mixture. 1 mark (i) energy – the reaction condition of 800ºC is not difficult and expensive to maintain because the reaction is


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. 4 Uses : 4 uses of conc. HNO3. 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.

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. 2 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.

8

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. 9 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 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

redox reactions. 6

Redox chemistry of nitrogen overall presentation 1 mark Principal oxidation states with examples e.g. 0(N2), -3(NH3), +4(NO2), +5(NO3

-) 2 marks 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 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. 1 mark Compounds in intermediate oxidation states such NO2 undergo disproportionation 2NO2 + H2O d HNO3 + HNO2 1 mark

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.

5

Depends on the nature of the reducing agents and the reaction conditions employed: Nitric(V) acid NO3

- + 2H+ + e- d NO2 + H2O 1 mark 2NO3

- + 12H+ + 10e- d N2 + 6H2O 1 mark NO3

- + 9H+ + 8e- d NH3 + 3H2O 1 mark NO3

- + 4H+ + 3e- d NO + 2H2O 1 mark other possible product: N2O, H2NNH2, NO2

- H+ is required if NO3

- acts as an oxidizing agent 1 mark

Sulphur 10.3.3.1 90 2B 5 b 5b Outline an industrial preparation of sulphuric acid.


Sulphuric acid S + O2 → SO2 2SO2 + O2 d 2SO3; ∆H < 0 very slow in the absence of catalyst SO3 + H2SO4 → H2S2O7


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.

12

Concentrated sulphuric(VI) acid 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. 1 mark The operating temperature is about 450ºC at 1 atm. 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. 1 mark SO3(g) + H2SO4(l) → H2S2O7(l) 1 mark Oleum is then diluted with calculated amount of water so as to give back 98% H2SO4(l) 1 mark 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


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. 4 Uses : 4 uses of conc. H2SO4. 1 mark each

conc. H2SO4 Manufacture of soapless detergents Manufacture of fertilizers such as (NH4)2SO4 Manufacture of explosives Manufacture of pigments

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. 2

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.

8

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. 9 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

3. Conversion of sulphur(VI) oxide to sulphuric(VI) acid, SO3 + H2SO4 → H2S2O7 H O2 → 2H2SO4

Add oxide to acid then dilute with water 3 marks

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. 6

Redox chemistry of sulphur overall presentation 1 mark Principal oxidation states with examples e.g. -2(H2S), 0(S), +4(SO2) and +6(SO3 or SO4

2-) 2 marks Any sulphur containing compounds, upon oxidation, give SO2 or SO4

2- SO2 – through combustion of the sulphur compounds in air SO4

2- – oxidation in homogeneous solution. 1 mark Upon reduction, S or H2S are usually produced H2S is a mild reducing agent but SO4

2- is usually stable towards redox reactions in the absence of H+. 1 mark Intermediate oxidation states, for example, S2O3

2- 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.

5

Depends on the nature of the reducing agents and the reaction conditions employed: Sulphuric(VI) acid SO4

2- + 4H+ + 2e- d SO2 + 2H2O 1 mark SO4

2- + 8H+ + 6e- d S + 4H2O 1 mark SO4

2- + 10H+ + 8e- d H2S + 4H2O 1 mark 2SO4

2- d S2O32- or S4O6

2-, etc. 1 mark H+ is required if SO4

2- acts as an oxidizing agent 1 mark

d-Block Elements I. General features of the d-block elements from Sc to Zn

A. Electronic configuration of d-block elements B. Electronic configuration of ions of d-block elements

1. Stability of different ions C. 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

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

1. Common oxidation states of vanadium and manganese compounds a) Vanadium compounds b) Manganese compounds

B. 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

C. Coloured ions 1. Origin of the colour 2. Colours of different ions

D. Catalytic properties of transition metals and their compounds 1. Haber process 2. Reaction between peroxodisulphate(VI) and iodide ions 3. Decomposition of hydrogen peroxide

E. 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 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 d-block metals. 3d 4s

21Sc [Ar] 3d14s2 [Ar] 10000 2

22Ti [Ar] 3d24s2 [Ar] 11000 2

23V [Ar] 3d34s2 [Ar] 11100 2

24Cr [Ar] 3d54s1 [Ar] 11111 1 half-filled 3d and 4s orbitals

25Mn [Ar] 3d54s2 [Ar] 11111 2

26Fe [Ar] 3d64s2 [Ar] 21111 2

27Co [Ar] 3d74s2 [Ar] 22111 2

28Ni [Ar] 3d84s2 [Ar] 22211 2

29Cu [Ar] 3d104s1 [Ar] 22222 1 full-filled 3d and half-filled 4s orbitals

30Zn [Ar] 3d104s2 [Ar] 22222 2 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.

Oxidation State

Element +1 +2 +3

21Sc [Ar] 3d14s2 Sc3+ [Ar]

22Ti [Ar] 3d24s2 Ti3+ [Ar] 3d1

23V [Ar] 3d34s2 V3+ [Ar] 3d2

24Cr [Ar] 3d54s1 Cr3+ [Ar] 3d3

25Mn [Ar] 3d54s2 Mn2+ [Ar] 3d5 Mn3+ [Ar] 3d4

26Fe [Ar] 3d64s2 Fe2+ [Ar] 3d6 Fe3+ [Ar] 3d5

27Co [Ar] 3d74s2 Co2+ [Ar] 3d7 Co3+ [Ar] 3d6

28Ni [Ar] 3d84s2 Ni2+ [Ar] 3d8

29Cu [Ar] 3d104s1 Cu+ [Ar] 3d10 Cu2+ [Ar] 3d9

30Zn [Ar] 3d104s2 Zn2+ [Ar] 3d10


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 full-filled 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) Eo = +0.52 V Cu2+

(aq) + e- d Cu+(aq) 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 half-filled orbital full-filled orbital

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

91 2B 6 a i 6a i What is the meaning of ‘d-block element’? 1 Element which has incompletely filled d orbital when combined in compounds 1 mark

(or complete in case of Zn) between s and p block ½ mark

d-Block Elements Unit 1 Page 3 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+. 2 3d 4s 3d 4s

29Cu0 22222 1 24Cr3+ 11100 0 1 mark each

ii Which of the two ions, copper(I) or copper(II), has the more stable electronic structure? Give an explanation. 1 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? 1 2 2Cu Cu Cuaq

disproportionationaq s( ) ( ) ( )

+ + → + 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 ii Using the “electron-in-boxes” representation, write the electronic structure outside the argon core for Mn2+ and

Fe2+ respectively. 2

Mn2+ [Ar] 11111 1 mark Fe2+ [Ar] 21111 1 mark

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

energy is required. Therefore, Mn2+ is not readily oxidized to Mn2+. 2 marks 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

0 0 000 0 000 00000 0 000 00000

1

1s 2s 2p 3s 3p 3d 4s 4p 4d

2 2 222 2 222 11110 0 000 00000 1 mark

ii What is the highest possible oxidation state of chromium ? 1 +6 / VI iii Give a chromium compound in which chromium is in its highest oxidation state. 1 CrO4

2- / Cr2O72- / HCrO4

- / CrO2Cl2 / CrO3 etc. or name (any chromium(VI) compound) 1 mark

96 2B 6 a 6a Using the "electrons-in-boxes" representation, write the ground state electronic structure outside the argon core for

V2+ and Cu+ respectively. 2

3d V2+ [Ar] 11100 1 mark Cu+ [Ar] 22222 1 mark

99 1A 3 a 3a Deduce the number of unpaired electrons for Fe3+

(g) at its ground state.


Topic d-Block Elements Unit 2

Reference Reading

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

Assignment 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 d-block 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 4s

21Sc [Ar] 3d14s2 [Ar] 10000 2

22Ti [Ar] 3d24s2 [Ar] 11000 2

23V [Ar] 3d34s2 [Ar] 11100 2

24Cr [Ar] 3d54s1 [Ar] 11111 1 half-filled 3d and 4s orbitals

25Mn [Ar] 3d54s2 [Ar] 11111 2

26Fe [Ar] 3d64s2 [Ar] 21111 2

27Co [Ar] 3d74s2 [Ar] 22111 2

28Ni [Ar] 3d84s2 [Ar] 22211 2

29Cu [Ar] 3d104s1 [Ar] 22222 1 full-filled 3d and half-filled 4s orbitals

30Zn [Ar] 3d104s2 [Ar] 22222 2

d-Block Elements Unit 2 Page 3 c) 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.

19K 20Ca 21Sc 22Ti 23V 24Cr 25Mn 26Fe 27Co 28Ni 29Cu

30Zn

structure BCC FCC HCP HCP BCC BCC BCC BCC FCC FCC FCC HCP density

g/cm3 0.86 1.55 3.0 4.5 6.1 7.2 7.4 7.9 8.9 8.9 8.9 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 Unit 2 Page 4 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



Reference Reading


Assignment 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 Page 2 kJmol-1 1st I.E. 2nd I.E. (∆ I.E.) 3rd I.E. (∆ I.E.) 4th I.E. (∆ I.E.) 5th I.E. (∆ I.E.) K 419 3051 (2632) 4412 (1361) 5887 (1475) 7975 (2088)

Ca 590 1145 (555) 4912 (3767) 6474 (1562) 8144 (1670)

Mn 717 1509 (792) 3249 (1740) 4940 (1691) 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) this is not a redox reaction, V(V) white → V(V) yellow VO2

+(aq) + 2H+

(aq) + e- → VO2+(aq) + H2O(l) V(V) yellow → V(IV) blue

VO2+(aq) + 2H+

(aq) + e- → V3+(aq) + H2O(l) V(IV) blue → V(III) green

V3+(aq) + e- → V2+

(aq) 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) energetically feasible

Zn2+(aq) + Fe(s) → Zn(s) + Fe2+

(aq) 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 Mn2+

(aq) colourless (or very pale pink) +3 Mn2O3(s) brown, Mn3+

(aq) red +4 MnO2(s) brown +6 MnO4

2-(aq) green

+7 MnO4-(aq) purple

The energetic feasibility of redox reaction involving manganese can also be predicted by using the standard electrode potential as mentioned above. Eo / V O2(g) + 2H2O(l) + 2e- d 4OH-

(aq) + 0.40 Cu+

(aq) + e- d Cu(s) + 0.52 I2(aq) + 2e- d 2I-

(aq) + 0.54 MnO4

2-(aq) + 2H2O(l) + 2e- d MnO2(s) + 4OH-

(aq) + 0.59 2H+

(aq) + O2(g) + 2e- d H2O2(aq) + 0.68 MnO2(s) + 4H+

(aq) + 2e- d Mn2+(aq) + 2H2O(l) + 1.23

Mn3+(aq) + e- d Mn2+

(aq) + 1.49 MnO4

-(aq) + 8H+

(aq) + 5e- d Mn2+(aq) + 4H2O(l) + 1.51

MnO4-(aq) + 4H+

(aq) + 3e- d MnO2(s) + 2H2O(l) + 1.70 H2O2(aq) + 2H+

(aq) + 2e- d 2H2O(l) + 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). MnO4

-(aq) + 8H+

(aq) + 5e- d Mn2+(aq) + 4H2O(l) Eo = + 1.51 V

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. MnO4

-(aq) + 4H+

(aq) + 3e- d MnO2(s) + 2H2O(l) Eo = + 1.70 V 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 Unit 3 Page 5 Past Paper Question

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.

3

(I) by mild reductants e.g. H2S, SO2, Fe2+, oxalate, Sn2+, I- 1 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. 2 MnO4

- is unstable in the presence of sunlight. 1 mark it will decompose to form brown MnO2 1 mark



Reference Reading


Assignment Reading

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

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.

Ag CNNC+

Cu

CN

NCCN

CN

3-

PtNH3

NH3H3N

H3N2+

Co

H2O

H2O

OH2

OH2H2O

H2O

2+

linear complex tetrahedral complex square planar complex octahedral complex

d-Block Elements Unit 4 Page 2 1. 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 CN- parent acid HCl H3O+ NH4

+ HCN pKa -7 -1.74 9.2 9.3 conjugate base Cl- H2O NH3 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)

H2Od

[Fe(H2O)6]2+(aq)

CN-

d [Fe(CN)6]4-

(aq) green yellow

Iron(III) [Fe(Cl)4]-(aq)

H2Od

[Fe(H2O)6]3+(aq)

CN-

d [Fe(CN)6]3-

(aq) yellow colourless/yellow red

Cobalt(II) [Co(Cl)4]2-(aq)

H2Od

[Co(H2O)6]2+(aq)

NH3

d [Co(NH3)6]2+

(aq) CN-

d [Co(CN)6]4-

(aq) blue pink deep red

Copper(II) [Cu(Cl)4]2-(aq)

H2Od

[Cu(H2O)6]2+(aq)

NH3

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

(aq) CN-

d [Cu(CN)4]2-

(aq) yellow blue deep blue N.B. 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)

M

H2O

H2O

OH2

OH2H2O

H2OM

H2OOH2H2O

H2O

bi

+ bi + 2H2O

n+n+

H2N CH2 CH2 NH2

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)

Kst = [[Cu(NH3)4]2+

(aq)][Cu2+

(aq)][NH3(aq)]4

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) K1 = [[Cu(NH3)]2+

(aq)] [Cu2+

(aq)][NH3(aq)]

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

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

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

(aq)][NH3(aq)]

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

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

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

(aq)][NH3(aq)]

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

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

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

(aq)][NH3(aq)]

K1 × K2 × K3 × K4

= [[Cu(NH3)]2+

(aq)] [Cu2+

(aq)][NH3(aq)] × [[Cu(NH3)2]2+

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

(aq)][NH3(aq)] × [[Cu(NH3)3]2+

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

(aq)][NH3(aq)] × [[Cu(NH3)4]2+

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

(aq)][NH3(aq)]

= [[Cu(NH3)4]2+

(aq)][Cu2+

(aq)][NH3(aq)]4

= Kst

d-Block Elements Unit 4 Page 4 3. 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 Ions present in aqueous solution Colour [Cr(H2O)6]Cl3 [Cr(H2O)6]3+

(aq), 3Cl-(aq) Violet

[CrCl(H2O)5]Cl2·H2O [CrCl(H2O)5]2+(aq), 2Cl-

(aq) Blue-green [CrCl2(H2O)4]Cl·2H2O [CrCl2(H2O)4]+

(aq), Cl-(aq) Green

4. Geometrical isomerism

Tetra-coordinated complex

Square planar complex [MA2B2]n+

MA

BA

Bn+n+

trans-isomer

MB

BA

An+

cis-isomer

Hexa-coordinated complex

Octahedral complex [MA2B4]n+ [MA3B3]n+

M

A

A

B

BB

B

n+

M

A

B

B

AB

B

n+

M

A

A

B

AB

B

n+

M

B

A

A

AB

B

n+

trans-isomer cis-isomer meridional isomer facial isomer

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

5. 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 Page 5 Other examples

K4[Fe(CN)6] potassium hexacyanoferrate(II) [Fe(Cl)4]- tetrachloroferrate(III) ion [Cr(Cl)4]- tetrachlorochromate(III) ion [Fe(H2O)6]3+ hexaaquairon(III) ion [Co(Cl)4]2- tetrachlorocobaltate(II) ion [Cu(CN)4]2- tetracyanocuprate(II) ion

Ligand Fe2+ Fe3+ Co2+ Cu2+ Cr3+ Cl- [Fe(Cl)4]2-

tetrachloroferrate(II) ion

[Fe(Cl)4]- tetrachloroferrate(III) ion

[Co(Cl)4]2- tetrachlorocobaltate(II) ion

[Cr(Cl)4]- tetrachlorochromate(III) ion

H2O [Fe(H2O)6]3+ hexaaquairon(III) ion

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

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

[Cu(CN)4]2-

tetracyanocuprate(II) ion

Glossary Complex ligand ion-dipole attraction coordinated compound coordinate bond linear

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 90 2B 6 b 91 1A 3 b i 91 1A 3 f 91 2B 6 a ii iii 92 1A 3 e 92 2B 4 a vi vii 93 1A 2 b iv 94 1A 2 b i 94 2B 5 b iii 95 1B 4 g 96 1A 2 e i 96 1A 2 f i 96 2B 6 b i ii 97 1B 7 b 99 1A 3 c

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]3- 1½ Because 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.”

5

FeSO4nH O2 → Fe(H2O)n

2+/Fe2+(aq) detail explanation 1 mark

Fe(H2O)n2+ + O2 → Fe(H2O)n

3+ air oxidation 1 mark Fe(H2O)n

3+ → Fe2O3 or Fe(OH)3 + H+ product 1 mark Fe(CN)6

4- is stable because CN- is a better ligand than H2O / CN- ligands stabilize Fe2+ 1 mark (no mark for Fe(CN)6

4- is a stable complex) The redox potential of Fe(CN)6

4- is different from Fe(H2O)n2+ 1 mark

91 1A 3 f 3f Write equations for stepwise reactions of [Cu(H2O)4]2+

(aq) with aqueous ammonia. Describe what changes you would observe during these reactions.

3

[Cu(H2O)6]2+(aq) + 2NH3(aq) → [Cu(H2O)4]2+·Cu(OH)2(s) + 2NH4

+(aq) or Cu(OH)2(s) 1 mark

[Cu(H2O)4]2+·Cu(OH)2(s) + excess NH3(aq) → [Cu(NH3)4]2+ 1 markWhen NH3(aq) is added slowly, at first insoluble pale blue basic salt or hydroxide is formed, but these dissolve readily in excess NH3(aq) to give deep blue solution containing tetraaminecopper(II) ion. 1 mark

d-Block Elements Unit 4 Page 6 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. 2

Name (must be named correctly) e.g. hexaamminecobalt(III) ion ½ mark Formula e.g. [Co(NH3)6)]3+ ½ mark Shape e.g.

Co

NH3

NH3

NH3NH3

NH3NH3

3+

1 mark

iii Why do V4+ ions react to give VO2+ ions in aqueous solution? 2 V4+ is small and highly charged (or has high charge density) 1 mark

V4+ ion polarizes H2O molecule so strongly that it attracts the electrons shared between H and O 1 mark

V4+ + OH

H2H+ VO2++

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+

and the ligand NH3 in aqueous medium can be based. 1

Cu(H2O)42+ + 4NH3 d Cu(NH3)4

2+ + 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.

3

Fe3+(aq) + SCN-

(aq) → Fe(SCN)2+(aq) [deep red] 1 mark

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

94 1A 2 b i 2b i Explain why Ni2+ forms more complexes than K+ does. 2 Ni2+ has higher charge density / more polarizing than K+ / greater ability to form covalent bond (dative bond) with

ligand Ni2+ has low lying , vacant d orbital. Any 2, 1 mark each

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. 3

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

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

excess, to aqueous solutions containing Pb2+ and Cu2+ ions respectively. These experiments are carried out at room temperature.

3

(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. Insoluble in excess

White ppt. Soluble in excess / Colour solution

Cu2+(aq) Blue ppt.

Soluble in excess / Deep blue solution Blue ppt.

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. 2

Co

Cl

Cl

enen

+

Co ClCl

en

en+

CoClCl

en

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 3 A complex ion is formed by combination of a central metal ion ½ mark

with ligands / complexing agents. ½ mark The ligands/complexing agents are either neutral molecules / negatively charged ions carrying lone pair(s) of electrons 1 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) ½ mark

ii stability constant 2 Stability constant Kst : e.g. [Cu(NH3)4]2+

Kst is the equilibrium constant for the following process Cu2+

(aq) + 4NH3(aq) d [Cu(NH3)4]2+ 1 or 0 mark

i.e. Kst = [Cu(NH3)4

2+][Cu2+][NH3]4 1 mark

(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.



Reference Reading


Assignment 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 d-orbitals 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 Unit 5 Page 2 2. Colours of different ions

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

H2Od

[Fe(H2O)6]2+(aq)

CN-

d [Fe(CN)6]4-

(aq) green yellow

[Fe(Cl)4]-(aq)

H2Od

[Fe(H2O)6]3+(aq)

CN-

d [Fe(CN)6]3-

(aq) yellow colourless/yellow red

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

H2Od

[Co(H2O)6]2+(aq)

NH3

d [Co(NH3)6]2+

(aq) CN-

d [Co(CN)6]4-

(aq) blue pink deep red

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

H2Od

[Cu(H2O)6]2+(aq)

NH3

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

(aq) CN-

d [Cu(CN)4]2-

(aq) yellow blue deep blue

Glossary d-d transition

Past Paper Question

93 1A 2 b i 94 1B 4 d iii 94 2B 5 b ii 99 2A 2 d i ii

93 1A 2 b i 2b i Explain, in terms of electronic arrangement, why most transition metal ions are coloured. 2 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 2 marks

94 1B 4 d iii 4d This question refers to reactions which occur between the following pairs of substances. Complete the empty

boxes in the following table. 6

Substances Reaction method Expected observation(s) Main product(s) Example: CH3COOH(l) and

C2H5OH(l)

Warm / reflux with conc H2SO4

miscible, fruity odour CH3COOC2H5(l)

(iii) CuSO4(aq) and 2M NH3 pale blue precipitate, deep blue solution

Cu(OH)2(s) [Cu(NH3)4]2+

(aq)

Substances Reaction method Expected observation(s) Main product(s) Example: CH3COOH(l) and

C2H5OH(l)

Warm / reflux with conc H2SO4

miscible, fruity odour CH3COOC2H5(l)

(iii) CuSO4(aq) and 2M NH3 Add NH3(aq) dropwise (1)

pale blue precipitate, deep blue solution

Cu(OH)2(s) [Cu(NH3)4]2+

(aq)

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. 2 Cobalt(II) chloride, in anhydrous state is blue in colour.

After absorption of water / forming complex ion with H2O, 1 mark it changes colour to pink. 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).



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.

Assignment 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. Haber process 2. Reaction between peroxodisulphate(VI) and iodide ions 3. 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 Unit 6 Page 2 Past Paper Question

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. 2 Catalytic properties. e.g. the use of a nickel catalyst in the hydrogenation of oils to make margarine 1 mark

Complex ion formation. e.g. [Ni(H2O)6]2+ 1 mark Accept also ‘colours’, ‘paramagnetic’ or other properties which can be illustrated.

92 1A 3 h 3h Which two properties of transition metal ions make them suitable for use as catalyst? 2 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. 3

Variable oxidation states of d-block elements allow electron transfer between reactants and products by means of catalyst changing between two oxidation states. 2 marks Any one example of transition metal or its compound as catalyst e.g. 2SO2 + O2

V O2 5 → SO3 (redox reaction) 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. 3

Any THREE of the following: exhibition of variable oxidation states e.g. Cu(I) & Cu(II) / V2+, V3+, VO2+ and VO2

+ 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 + ½ mark (½ marks for each properties; ½ marks for a correct example)

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