Senior Six Chemistry Notes 2023-2024 PDF

SENIOR SIX CHEMISTRY UNITS’SYLLABUS CHEMISTRY FOR SENIOR SIX COURSE OUTLINE Unit 1: Properties and uses of transition metals Unit 2: Extraction of metals Unit 3: NPK as components of fertilizers Unit 4: Benzene Unit 5: Derivatives of benzene Unit 6: Polymers and polymerization Unit 7: Solven...

SENIOR SIX CHEMISTRY UNITS’SYLLABUS CHEMISTRY FOR SENIOR SIX COURSE OUTLINE Unit 1: Properties and uses of transition metals Unit 2: Extraction of metals Unit 3: NPK as components of fertilizers Unit 4: Benzene Unit 5: Derivatives of benzene Unit 6: Polymers and polymerization Unit 7: Solvent extraction and colligative properties Unit 8: Quantitative chemical equilibrium Unit 9: PH of acidic and alkaline solutions Unit 10: Indicators and titration curves Unit 11: Solubility of sparingly soluble salts Unit 12: Electrochemical cell and its applications Unit 13: Factors that affect the rate of reactions Unit 14: Rate laws and measurements Unit 15: Radioactivity KEY COMPETENCES: By the end of senior six of Advanced level of all chemistry combinations, students should be able to:  Explain the properties and uses of transition metals  Relate the properties of metals to their methods of extraction and uses and suggest preventive measures to dangers associated with their extraction.  Analyze the components of quality fertilizers and their benefits, effects of misuse and dangers associated with standard fertilizers.  Relate the chemistry and uses of benzene and its derivatives to their nature and structures.  Relate the types of polymers to their structural properties and uses.  Apply partition and Raoult‟s laws to separate mixtures and determine the molecular formula and mass of compounds using colligative properties.  Write expressions and calculate the values of equilibrium constant, interpret the values of Kc in relation to the yield of the products in reversible reactions.  Prepare solutions, measure the pH and calculate the pH of acidic and alkaline solutions. Prepared by AIMABLE DUKUZE Page 1 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS  Explain the concept of buffer solutions, hydrolysis of salts and discuss its applications in manufacturing industry and biological processes.  Relate titration curves of acid-base titration properly choose and use indicators in acid-base titrations.  Calculate the solubility product constant of sparingly soluble salts and deduce the applications of common-ion effect in industry.  Explain the working and industrial applications of electrochemical and electrolytic cells.  Measure the rates of reactions and formulate simple rate equations using experimental results.  Explain the factors that affect the rate of chemical reaction and use Arrhenius equation to calculate the ratio of rate constant and activation energy with change in temperature.  Explain the importance and dangers of radioisotopes in everyday life. TOPIC AREA 3: THE PERIODIC TABLE SUB-TOPIC AREA 3.2: TRANSITION METALS UNIT 1: PROPERTIES AND USES OF TRANSITION METALS Key unit competence:  Explain the properties and uses of transition metals 1.1. Definition and electronic configuration of transition metals. A transition metal is a metal which forms at least one stable ion with partially filled d-orbitals. It is a d-block element that forms at least one ion with an incomplete d sub-shell. According to IUPAC system, a transition metal is "an element whose atom has a partially filled d sub-shell, or which can give rise to cations with an incomplete d sub-shell". Transition metals are located between groups 1& 2 (s-block) and group 13 (p-block) on the periodic table. The elements are also called d-block elements because their valence electrons are in d-orbitals. Since zinc and scandium do not share this property, they are not transition metals, and indeed do not show many of the properties generally attributed to transition metals. They are however still classified as d-block elements. The first row d-block elements which are also transition metals are therefore titanium, vanadium, chromium, manganese, iron, cobalt, nickel and copper. The outer electronic configuration of elements in the first row of the d-block is as follows: 4s 3d Sc [Ar] ↑↓ ↑ Ti [Ar] ↑↓ ↑ ↑ V [Ar] ↑↓ ↑ ↑ ↑ Prepared by AIMABLE DUKUZE Page 2 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Cr [Ar] ↑ ↑ ↑ ↑ ↑ ↑ Mn [Ar] ↑↓ ↑ ↑ ↑ ↑ ↑ Fe [Ar] ↑↓ ↑↓ ↑ ↑ ↑ ↑ Co [Ar] ↑↓ ↑↓ ↑↓ ↑ ↑ ↑ Ni [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑ Cu [Ar] ↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ Zn [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ Table 1.1: Electronic configuration of series 1 transition metals Note: The unusual structures of chromium and copper. The 4s and 3d subshells are very similar in energy and therefore it is easy to promote electrons from the 4s into the 3d orbitals. In chromium the 4s13d5 structure is adopted because the repulsion between two paired electrons in the 4s orbital is more than the energy difference between the 4s and 3d subshells. It is thus more stable to have unpaired electrons in the higher energy 3d orbital than paired electrons in the lower energy 4s orbital. In copper and zinc the 3d subshell is actually lower in energy than the 4s subshell. The 3d orbitals are thus filled before the 4s orbital. Thus copper adopts a 4s13d10 configuration. In all ions of d-block elements, the 3d subshell is lower in energy than the 4s subshell so the 4s electrons are always removed first. 3d electrons are only removed after all 4s electrons have been removed. ACTIVITY 1: 1. What is the difference between a d-block and transition elements? 2. Write the electronic configuration of; Fe2+, Ti and Cu2+ 1.2. Properties of transition metals (a) Oxidation states and ionization energies The oxidation state formed by an element in its compounds is determined by the maximum number of electrons it can lose without requiring so much energy to remove the electrons that the energy cannot be recovered in bonding. s-block elements only form one stable oxidation state in their compounds. They lose all their valence electrons easily but cannot lose any more electrons since there is a large amount of energy required to remove the electrons from the inner shell. This jump in energy is best shown graphically: Eg Na Eg Mg Prepared by AIMABLE DUKUZE Page 3 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS First eight ionisation energies of sodium First eight ionisation energies of magnesium 5 5 4.5 4.5 4 4 log (ionisation energy log (ionisation energy 3.5 3.5 3 3 2.5 Series2 2.5 Series2 2 2 1.5 1.5 1 1 0.5 0.5 0 0 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 number of electrons removed number of electrons removed Na always adopts the +1 oxidation state in its compounds because there is a large jump between the first and the second ionization energies. Mg always adopts the +2 oxidation state in its compounds because there is a small jump between the first and the second ionization energies but a very large jump between the second and third ionization energies. In the d-block elements, however, there are often a large number of valence electrons and removing them all would require so much energy that it would be unfeasible. It is usually only possible to remove some of the valence electrons. All d-block elements can give up their 4s electrons fairly easily but the d-electrons are harder to remove. Moreover, since the successive ionization energies of d-electrons increase steadily, it is difficult to predict how many can be lost. This effect can be shown graphically by considering the successive ionization energies of an element such as manganese: First eight ionisation energies of manganese 4.5 4 3.5 log (ionisation energy 3 2.5 2 1.5 1 0.5 0 1 2 3 4 5 6 7 8 number of electrons removed The ionization energies increase steadily after the removal of the 4s electrons. It turns out that the energy required to remove the 3d electrons is sometimes recovered in bonding, but not always. The number of 3d electrons removed thus varies from compound to compound.d-block metals are thus able to adopt a variety of oxidation states. The oxidation states most commonly formed by the first-row d-block elements are as follows: Sc: +3 only (d0)Ti: +3 (d1), +4 (d0) Prepared by AIMABLE DUKUZE Page 4 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS V: +2 (d3), +3 (d2), +4 (d1), +5 (d0)Cr: +3 (d3), +6 (d0) Mn: +2 (d5), +3 (d4), +4 (d3), +6 (d1), +7 (d0)Fe: +2 (d6), +3 (d5) Co: +2 (d7), +3 (d6)Ni: +2 (d8) Cu: +1 (d10), +2 (d9)Zn: +2 only (d10) It is sometimes possible to remove all 3d electrons to form a do configuration. However for elements beyond manganese it is unusual to find an oxidation state containing fewer than 5 3d electrons. The d5 configuration is quite stable because paired d-electrons can be removed to reach d5 but unpaired electrons must be removed to reach d4. Unpaired electrons are harder to remove because there is less repulsion between the electrons. Note that scandium only forms +3 ions (d0). This is because the low effective nuclear charge on scandium enables all three valence electrons to be removed fairly easily. Note also that zinc only forms +2 ions (d10). This is because the high effective nuclear charge on zinc prevents any 3d electrons from being removed. All the other elements form at least one stable ion with partially filled d-orbitals, and it is this property which defines a transition metal. (b) Coloured compounds When white light passes through solution containing metal ions, some of the wavelengths of visible light are absorbed. The colour that we observe is a mixture of the wavelengths of light that have not been absorbed. Many coloured inorganic compounds contain transition metal ions. In fact, it is quite difficult to find coloured inorganic compounds that do not contain transition metal ions. Colour in inorganic chemistry is linked to the partially filled d-orbital of transition metal ions. (c) Catalytic action A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative route for the reaction to follow- the alternative route has a lower activation energy. The ability of transition metals to form more than one stable oxidation state means that they can accept and lose electrons easily. This enables them to catalyse certain redox reactions. They can be readily oxidised and reduced again, or reduced and then oxidised again, as a consequence of having a number of different oxidation states of similar stability. They can behave either as homogeneous catalysts or as heterogeneous catalysts. The reasons for transition metals to work as catalysts:  Presence of empty d orbitals which enable transition metal ions (or atoms) to form temporary bonds with reactant molecules at the surface of a catalyst and weakens the bond in the reactant molecules  Variable oxidation states which allow them to work as catalysts in the reactions involving the transfer of electrons How does a catalyst work? A catalyst is a substance which alters the rate of a reaction without itself being chemically unchanged at the end.A catalyst lowers the activation energy for the reaction by providing an alternative route. An extra step is introduced. Prepared by AIMABLE DUKUZE Page 5 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS e n uncatalysed reaction e (high activatio energy) r g reactants y catalysed reaction (low activation energy) products reaction pathway Because the activation energy is lower, more particles have enough energy to react when they collide and so the fraction of successful collisions is higher. Examples of catalysts:  In Born-Haber process, industrial manufacture of ammonia, iron metal is used as a catalyst. It is used to lower the temperature at which the reaction takes place(activation energy), hence the rate of reaction increases.  In contact process, industrial manufacture of sulphuric acid, vanadium (V) oxide, V2O5, is used.  In hydrogenation of ethane, nickel metal, Ni, is used. (d) Transition elements have variable oxidation states Oxidation state is a number assigned to an element in chemical combination which represents the number of electrons lost or gained. If the number is negative this means that the atom has gained electrons; and if the number is positive, this means that the atom has lost electrons; this number shows the number of electrons gained or lost respectively.Transitions metals show variable oxidation states because of the availability of 3d electrons. The common oxidation states shown by the first transition series are: Prepared by AIMABLE DUKUZE Page 6 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Table 1.2: The oxidation states shown by the transition metals (series) Sc Ti V Cr Mn Fe Co Ni Cu Zn +3 +2 +2 +2 +2 +2 +2 +2 +1 +2 +3 +3 +3 +3 +3 +3 +3 +2 +4 +4 +4 +4 +4 +4 +4 +5 +5 +5 +6 +6 +6 +7 (e) Densities and atomic/metallic radii The transition elements are much denser than the s-block elements and show in generala gradual increase in density from left to right in a period as you can see below fromscandium to copper. This trend in density can be explained by a decrease in metallic radii coupled with the relative increase in atomic mass. Table 1.3: Density/g cm -3of the first transition series Sc Ti V Cr Mn Fe Co Ni Cu Zn 3.43 4.1 6.07 7.19 7.21 7.8 8.7 8.9 8.9 7.1 Table1.4: Metallic radii of the first transition series Sc Ti V Cr Mn Fe Co Ni Cu Zn 164 147 135 129 137 126 125 125 128 137 (f) Formation of complex ions A complex or coordination compound is a chemical species made of a central metal (cation or neutral) bonded to other chemical species called ligands by coordination or dative bonds. A complex may be neutral, positively or negatively charged. Transition metal form complexes because of:  Their small and highly charged ions,  The presence of vacant(empty) d-orbitals which can accommodate lone pair of electrons donated by other groups (ligands) The general formula of a complex is the following: y [MLn] Where:  M-metal ion or atom  L-Ligand  n-the number of ligands surrounding the metal  y-the charge of the complex; [MLn] indicates a neutral complex. ► Coordination number of a complex: is the number of coordinate bonds on the central metal in a complex. ► Ligand: It is a species (anion or a molecule) that is bonded to a central metal ion or atom in a complex. A ligand should have at least one lone pair of electrons to form a coordinate bond. (g) Many transition metal ions and their compounds are coloured Formation of colored ions by transition elements is associated with presence of incompletely filled 3d orbitals. Prepared by AIMABLE DUKUZE Page 7 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Ion (aq) Colour Outer 3d electrons Sc3+ Colourless 3d0 Ti3+ Purple 3d1 V3+ Green 3d2 Cr3+ Violet 3d3 Mn2+ Pink 3d5 Fe2+ Green 3d6 Co2+ Pink 3d7 Ni2+ Green 3d8 Cu2+ Blue 3d9 Zn2+ Colourless 3d10 TABLE 1.5: Transition metals ions and their respective colours This property has its origin in the excitation of d electrons from lower energy d-orbitals to higher energy. In fact, when the central metal is surrounded by ligands, these cause d orbitals to be split into groups of higher and lower energy orbitals. When electrons fill d-orbitals, they fill first of all the lower energy orbitals; if there is free space in higher energy d-orbitals, an electron can be excited from lower energy d-orbitals to higher energy d-orbitals by absorbing a portion of light corresponding to a given colour, the remaining color light is the white light minus the absorbed colour. Figure 1.4: d-orbital split into two groups of different energies When a coloured object is hit by white light, the object absorbs some colour and the colour transmitted or reflected by the object is the colour which has notbeen absorbed. The observed colour is called complementary colour. (h) Formation of alloys An alloy is a mixture (solid solution) made by combining two or more elements where at least one is a metal. Importance of alloying:  Increase of the strength of a metal,  Resistance to corrosion,  Gives to the metal a good appearance Table 1.6: The properties and uses of some common alloys formed by transition metals (first series) ALLOY Colour COMPOSITION PROPERTIES USES To build statues Hard and strong and monuments Bronze Chocolate 90 % copper Does not corrode In the making of brown 10 % tin easily medals, swords and Prepared by AIMABLE DUKUZE Page 8 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Has shiny surface artistic materials. In the making of Brass Yellow 70 % copper Harder than copper musical instruments 30 % zinc and kitchenware. Steel Blue, grey, etc 99 % iron Hard and strong In the construction 1 % carbon of building and bridges. In the building of the body of cars and railway tracks. Ranges of To make cutlery colors, and surgical including blue, 74 % iron Shiny instruments. Stainless steel black, bronze, 8 % carbon Strong gold, green, 18 % chromium Does not rust red/violet 93 % aluminium Light To make the body Duralumin Metallic colour 3 % copper Strong of airplanes and 3 % magnesium bullet trains. 1 % manganese Pewter Grey 96 % tin Luster In the making of 3 % copper Shiny souvenirs. 1 % antimony Strong (i) Most transition metal ions are paramagnetic Paramagnetism is a property of substances to be attracted in a magnetic field. Substances which are not attracted (i.e slightly repelled) in a magnetic field are said to be diamagnetic.Transition metal ions show paramagnetism because of the presence of unpaired electrons in their 3d sub-shells. The greater the number of unpaired electrons, the stronger the paramagnetism; that is the reason why:  Fe3+ is more paramagnetic than Fe because Fe has five unpaired electrons while Fe2+ has four 2+ 3+ unpaired.  Sc and Zn have no paramagnetic effect because they do not have unpaired electrons. 3+ 2+ 1.3. Complex ions 1.3.1. Definition A complex ion is an ion comprising one or more ligand attached to a central metal cation by means of a dative covalent bond.A ligand is a species which can use its lone pair of electrons to form a dative covalent bond with a transition metal. Examples of ligand are H2O, NH3, Cl-, OH-, CN-,cations which form complex ions must have two features:  They must have a high charge density, and thus be able to attract electrons from ligands.  They must have empty orbitals of low energy, so that they can accept the lone pair of electrons from the ligands. Prepared by AIMABLE DUKUZE Page 9 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Cations of d-block metals are small, have a high charge and have available empty 3d and 4s orbitals of low energy. They thus form complex ions readily. The number of lone pairs of electrons which a cation can accept is known as the coordination number of the cation. It depends on the size and electronic configuration of that cation, and also on the size and charge of the ligand. 6 is the most common coordination number, although 4 and 2 are also known. Examples of complex ions are [Fe(H2O)6]2+, [CoCl4]2-, [Cu(NH3)4(H2O)2]2+. Note that the formula of the ion is always written inside square brackets with the overall charge written outside the brackets. 1.3.2. Shapes of complex ions  6-coordinate complexes are all octahedral, and are formed with small ligands such as H2O and NH3. They can thus be drawn as follows: [Fe(H2O)6]2+ [Cr(NH3)6]3+ [Cu(NH3)4(H2O)2]2+ 2+ 3+ 2+ H2.O..NH. 3 H2.O. H2O: :OH2 H3N: : NH3 H3N: : NH3 Fe Cr Cu H2O: :OH2 H3N: :NH3 H3N: :NH3...... H2O NH3 H2O  4-coordinate complexes are generally tetrahedral, and are formed with larger ligands such as Cl-. Larger ligands cannot fit around the transition metal so easily and hence form smaller complexes. They can be drawn as follows: [CoCl4]2- 2-.Cl. Co.. Cl.. Cl.. Cl  2-coordinate complexes are in general linear, and are formed with Ag+ ions. They can be drawn as follows: [Ag(NH3)2]+ + H3N: Ag :NH3 The rules covering the likely coordination number of transition metal complexes are: a) Silver ions form linear complexes with a coordination number of 2; Prepared by AIMABLE DUKUZE Page 10 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS b) Chloride ions and other large ions form tetrahedral complexes with a coordination number of 4; c) Most other transition metal complexes are octahedral with a coordination number of 6. 1.3.3. Representation of metal salts When metal ions are in solution, they are usually represented as the simple ion, such as Fe2+(aq), Co2+(aq), Cr3+(aq) or Fe2+(aq). This is, however, a simplified representation as all d-block cations and many other cations with high polarising power exist as the hexaaqua complex, eg [Fe(H2O)6]2+.  FeSO4(aq) consists of [Fe(H2O)6]2+ and SO42- ions  Fe2(SO4)3(aq) consists of [Fe(H2O)6]3+ and SO42- ions  CuCl2(aq) consists of [Cu(H2O)6]2+ and Cl- ions Many complex ions exist in the solid state. In these cases the ligands are often written after the rest of the compound, seperated by a dot:  [Fe(H2O)6]SO4(s) is generally written FeSO4.6H2O(s)  [Cr(H2O)6]Cl3(s) is generally written CrCl3.6H2O(s) The dot does not always, however, mean that there are ligands present; some salts contain water of crystallisation, where the water is not a ligand but a link between the ions.  In Na2CO3.10H2O, the water molecules are not behaving as ligands, but as water molecules of crystallization.  In CuSO4.5H2O, four of the water molecules are behaving as ligands, and the fifth as a water molecule of crystallization: [Cu(H2O)4]SO4.H2O 1.3.4. Naming of complex ions and isomerism in compounds of transition elements 1.3.4.1. Naming of complex ions Naming molecules requires the knowledge of certain rules, such as how to name cations, anions, where to start from when both a cation and an anion are combined in an ionic molecule or when two non metals are combined in a covalent molecule. Like other compounds, complex compounds/ions are named by following a set of rules. You are familiar with some of them and the new ones can be understood and applied easily. 1. In simple metal compounds, the metal is named first then the anion. Example: CaCl2: calcium chloride 2. In naming the complex: (a) Name the ligands first, in alphabetical order, then the metal atom or cation, followed by its oxidation state written between brackets as Roman number, though the metal atom or cation is written before the ligands in the chemical formula. Example: [CuBr4]2-: Tetrabromocuprate (II) ion The names of some common ligands are listed in the table below: Table 1.10: Common ligands Anionic Ligands Names Neutral Ligands Names Br- bromo NH3 ammine Prepared by AIMABLE DUKUZE Page 11 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS F- fluoro H2O Aqua/aquo O2- oxo NO Nitrosyl OH- Hydroxo CO Carbonyl CN- cyano O2 dioxygen C2O42- oxalato N2 dinitrogen CO32- carbonato C5H5N pyridine CH3COO- Acetato H2NCH2CH2NH2 ethylenediamine (b) Greek prefixes are used to indicate the number of each type of ligand in the complex: The numerical prefixes are listed in the following table: Table 1.11: Numerical prefixes Number Prefix Number Prefix Number Prefix 1 Mono 5 penta (pentakis) 9 nona 2 di (bis) 6 hexa (hexakis) 10 deca 3 tri (tris) 7 hepta 11 undeca 4 tetra (tetrakis) 8 octa 12 dodeca (c) After naming the ligands, name the central metal.  If the complex bears a positive charge (cationic complex), the metal is named by its usual name. Example: Cu: Copper Pt: Platinum  If the complex bears a negative charge (anionic complex), the name of the metal ends with the suffix –ate Example: Co in a complex anion is called cobaltate and Pt is called platinate. For some metals, the Latin names are used in the complex anions e.g. Fe is called ferrate (not ironate). See table below: Table 1.12: Latin names of some transition metals in anionic complexes Name of Metal Name in an Anionic Complex Iron Ferrate Copper Cuprate Lead Plumbate Silver Argentate Gold Aurate Tin Stannate 3. For historic reasons, some coordination compounds are called by their common names. Example: Fe(CN)63- and Fe(CN)64- are named ferricyanide and ferrocyanide respectively, and Fe(CO)5 is called iron carbonyl. 4. To name a neutral complex molecule, follow the rules of naming a complex cation. Example: [Cr(NH3)3Cl3]: triamminetrichlorochromium (III) You can have a compound where both the cation and the anion are complex ions. Notice how the name of the metal differs even though they are the same metal ions.Remember: Name the (possibly complex) cation BEFORE the (possibly complex) anion. Example: [Ag(NH3)2][Ag(CN)2] is diamminesilver(I)dicyanoargentate(I) Note that: - The names are written as a one word: Tetraamminecupper(II), not Tetraammine copper (II). Prepared by AIMABLE DUKUZE Page 12 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS - Complex ions formula is written between square brackets and the charge of the ion as superscript outside the brackets: [Cu(NH3)4]2+. When oppositely charged ions approach the complex ion, a neutral molecule can be obtained: [Cu(NH3)4]2+2Cl- or simply, [Cu(NH3)4]Cl2: tetraamminecopper(II)chloride. The ions outside the square brackets are known as “counter ions”. 1.3.4.2. Isomerism in complexes Isomers are different chemical species that have the same chemical formula. Isomerism among transition metal complexes arises as a result of different arrangements of their constituent ligands around the metal. The categories of isomerism exhibited by complexesare provided below: In this unit, we are specifically concerned with „stereoisomerism‟ which gives rise to isomers known as “stereoisomers”. Stereoisomers have the same structural formulae but different arrangements of ligandsin space. They are classified in two categories: geometrical isomers and optical isomers. 1) Geometrical isomers Coordination complexes, with two different ligands in the cis and trans positions from a ligand of interest, form isomers. For example, the square planar, diammine dichloroplatinum (II) Pt(NH3)2Cl2),can be presented as follows: The octahedral [Co(NH3)4Cl2]+ ion can also have geometrical isomers. Prepared by AIMABLE DUKUZE Page 13 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Different geometrical isomers are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of [Co(NH3)4Cl2]NO3 differ in color; the cis form is violet, and the trans form is green. Furthermore, these isomers have different dipole moments, solubilities and reactivities. 2) Optical isomers (enantiomers) In optical isomers, or enantiomers; two objects are exact mirror images of each other but cannot be superimposable. This means that optical isomers are non-superimposable mirror images of each other. A classic example of this is your two hands (left and right); hold them face-to-face: one is the mirror image of the other. Now try to superimpose them one over another: they are non-superimposable (only the middle fingers superimpose one over the other. Chemical compounds that behave like the hands are called “chiral”, in reference to the Greek word for hands. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, optical isomers have identicalphysical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as [M(en)3]n+ [in which Mn+ is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in figure below.These two isomers will react differently with other optical isomers. For example, DNA helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of [M(en)3]n+ and not the other. 1.3.4. Chemical properties of complex ions (a) Polarizing power The effect of ligands in a complex ion is to significantly stabilise the metal cation by increasing its size and hence reducing its polarizing power. Many compounds which are covalent in the anhydrous state are actually ionic in the hydrated state:  FeCl3(s) is covalent but FeCl3.6H2O(s) is ionic.  AlCl3(s) is covalent but AlCl3.6H2O(s) is ionic. (b) Precipitation reactions The properties of anions are also changed if they are behaving as ligands in a complex ion. In particular, they are much less readily precipitated by cations. Prepared by AIMABLE DUKUZE Page 14 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS For example, AgNO3(aq) will form a precipitate of AgCl if added to a solution of copper (II) chloride, [Cu(H2O)4]Cl2 but not from sodium tetrachlorocobaltate (II), Na2[CoCl4]. BaCl2(aq) will form a precipitate of BaSO4 if added to a solution of [Cr(H2O)5Cl]SO4 but not with [Cr(H2O)5SO4]Cl. (c) Heating The properties of molecules also changes if they are behaving as ligands. In particular, they are much less readily removed by heating. For example, Na2CO3.10H2O(s) can be dehydrated to Na2CO3(s) by gentle heating at 50oC. FeSO4.6H2O(s), however, can only be dehydrated to FeSO4(s) by strong heating at 150oC, since the water is strongly bonded to the Fe2+ as a ligand. CuSO4.5H2O(s) can be dehydrated to CuSO4.4H2O(s) by gentle heating at 50oC, and to CuSO4 by strong heating at 100oC. One of the water molecules is a molecule of crystallisation but the others are ligands. (d) Polydentate ligands Some ligands are capable of forming more than one dative covalent bond per ligand. Examples are ethanedioate (C2O42-) and 1,2-diaminoethane (H2NCH2CH2NH2), both of which donate 2 lone pairs per ligand and are said to be bidentate. One unusual ligand, known as edta4-, can form 6 dative covalent bonds per ligand. It is thus said to be hexadentate. Ligands such as H2O, Cl- and CN- form only one dative covalent bond per ligand and are said to be unidentate. Ligands are classified depending on the number of sites at which one molecule of a ligand is coordinated to the central metallic atom; the ligands are classified as monodentate (or unidenate) and polydentate (or multidentate) ligands. With polydentate ligands it often appears that the coordination number of the complex is 1 or 3, when in fact the coordination number is 6 as normal - it is just that each ligand is bonded twice. Some examples of these complex ions are shown below: [Fe(C2O4)3]3- [Cr(H2NCH2CH2NH2)3]3+ [Cu(edta)]2- 3- 2- O CH2 3+ CH2 O C O C C O.. O CH2.NH. 2.N. CH2 O CH2 :O O: C O: :O C H2N: :NH2 CH2 O C Cu CH2 Fe C Cr O: :N CH2 O: :O O.... CH2 C O H2N: :NH2 O CH2 O C.. CH2 NH2 C O CH2 O (e) Formation of coloured ions When a cation forms a complex ion, the incoming ligands repel the electrons in the atom, and thus they are raised in energy. Some of the d-orbitals, however, are repelled more than others and the result Prepared by AIMABLE DUKUZE Page 15 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS is that the d-orbitals are split into 2 groups of orbitals, with three orbitals being slightly lower in energy than the other two. The difference in energy between these groups of orbitals is similar to the energy of visible light. The energy of the light is related to the frequency of the light by the equation E = hf. If these d-orbitals are partially filled, some of the electrons in the lower energy orbitals are excited into the higher energy orbitals, and in doing so absorb the light that corresponds to that frequency. The resultant light is deficient in the light of that frequency and thus appears coloured. Transition metal ions are coloured because d-electrons can absorb light and get excited into higher energy d- orbitals. The resultant light is thus missing certain frequencies and is hence coloured. Note that two criteria must be satisfied if the ion is to be coloured: - There must be a splitting of the d-orbitals. This only happens in the presence of ligands and thus only complex ions are coloured. Anhydrous ions do not have split d-orbitals and so cannot aborb light in the visible spectrum and are thus white. Eg anydrous CuSO4 (d9) is white but hydrated CuSO4.5H2O is blue. - The d-orbitals must be partially filled. If the d-orbitals are empty (Eg Sc3+, Al3+) then there are no electrons which can be excited into the higher energy d-orbitals and the ions will be colourless. If the d-orbitals are full (Eg Cu+, Zn2+) then there are no empty orbitals into which the electrons can be excited and the ions will be colourless. The colour of a complex ion depends on: - the ligand - the coordination number - the oxidation state of the metal - the identity of the metal 1.3.5. Other applications of transition metal complexes i) Haemoglobin Another important complex ion involving multidentate ligands is haemoglobin. Haem is a complex ion consisting Fe2+ and a complex tetradentate ligand called porphyrin. The complex is generally found with a protein called globin, which provides the fifth coordinate bond, and a molecule of oxygen which forms the sixth bond. The complete six coordinate complex is called haemoglobin. This structure is responsible for carrying oxygen in the blood throughout the human body. Fe2+ + porphyrin  haem and haem + globin + O2 haemoglobin Carbon monoxide is a similar size and shape to oxygen and forms a much stronger bond with the iron. It thus diplaces the oxygen from the complex and reduces the blood‟s ability to carry oxygen. It is thus a very poisonous gas. ii) Cisplatin The square planar complex cisplatin has the following structure: Cl NH3 Pt Cl NH3 Prepared by AIMABLE DUKUZE Page 16 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS It kills cancerous cells and now widely used in cancer treatment. iii) Tests for aldehydes and halide ions Diammine silver (I) has the formula [Ag(NH3)2]+ and is the active ion in Tollen‟s reagent. It is reduced to silver by reducing sugars and aldehydes (but not ketones). It produces a characteristic silver mirror on the side of the test-tube and this is used as the basis of the test for aldehydes. It is also formed in the test for halide ions in solution. The silver halides are insoluble in water but AgCl and AgBr will dissolve in ammonia due to the formation of the diammine silver complex: AgCl(s) + 2NH3(dilute)  [Ag(NH3)2]+(aq) + Cl-(aq) AgBr(s) + 2NH3(conc)  [Ag(NH3)2]+(aq) + Br-(aq) iv) Photography Silver bromide, AgBr, is the substance on photographic film. In the presence of light, it decomposes into silver and bromine: 2AgBr  2Ag(s) + Br2(l). The unreacted AgBr is removed when sodium thiosulpate is added. This forms a complex with the AgBr and washes it off the film, leaving only the silver metal on the film. The result is the negative image. AgBr(s) + 2S2O32-(aq)  [Ag(S2O3)2]3-(aq) + Br-(aq) v) Electroplating Metals such as silver and gold generally occur native, but in very impure form. They can be extracted using cyanide ions which form stable complexes with silver and gold. The Ag is oxidized to Ag+ and then complexed as [Ag(CN)2]-. This complex is widely used in electroplating. To coat another object with silver, place the metal object to be coated at the cathode and use [Ag(CN)2]- as the electrolyte. The complex breaks up, Ag+ ions move to the cathode and the object is coated with a layer of silver. 1.4. IDENTIFICATION TESTS OF CATIONS Introduction This mainly deals with identification of ions (cations and anions) present in different inorganic compounds. When carrying out qualitative analysis on inorganic substances, a rough guide of what a substance consistsisgiven by preliminary tests such as: Prepared by AIMABLE DUKUZE Page 17 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS  Physical appearance of solid salts Colour DEDUCTION 1 White (whether dry or moist) Non-transition cation present e.g. Mg2+, Ba2+, Ca2+, Pb2+, Al3+, Zn2+ or NH4+ 2 Green Fe2+, Cu+, Cr3+ or Ni2+ 3 Blue or Green Cu2+, Cr3+or Ni2+(but bright green) 4 Yellow or reddish-brown Fe3+ 5 Red or Pink or purple Mn2+ or Co2+ 6 Black Sulphide or oxide  Characteristic colours of aqueous solutions of cations COLOUR OF SOLUTION DEDUCTION FORMED 1 Colourless Mg2+, Ca2+, Ba2+, Pb2+, Al3+, Ag+ Zn2+Mn2+or NH4+, Sc3+, Ti4+ 2 Green Cu+, Fe2+, Cr3+, or Ni2+ 3 Blue Cu2+, or Co2+(anhydrous) 4 Yellow or brown Fe3+, Cr6+ (from CrO42-) 5 Pale pink or red Mn2+, Co2+ hydrated 6 Milky Sn2+ 7 Orange Cr6+ (from Cr2O72-) 8 Purple Mn7+ 9 Black Sulphides of Cu2+, Ni2+, Co2+, Pb2+ or Fe2+ Copper(I)oxide,Cu2O  Smell / Odour SMELL SUSPECTED SOLID Smell of ammonia NH4+ salt Smell of H2S Sulphide(s2-) Smell of sulphur dioxide Sulphite(SO32-)  Solubility of inorganic salts in water  All K+, Na+ and NH4+ salts are soluble  All nitrates are soluble  All carbonates are insoluble except carbonates of K+, Na+, and NH4+  All sulphites aresoluble except sulphites of Ca2+, Ba2+, and Pb2+.  All sulphates are soluble except CaSO4, Ag2SO4 that are moderately soluble; PbSO4 and BaSO4 are insoluble.  All hydroxides are insoluble except hydroxides of NH4+, Na+, K+; Ca(OH)2 is moderately or sparingly soluble in cold water.  All chlorides are soluble except PbCl2 that is sparingly soluble in hot water and AgCl insoluble.  All chromates are soluble except; CaCrO4 is moderately soluble and PbCrO4, BaCrO4 and Ag2CrO4 are insoluble.  All oxalates and phosphates are insoluble except oxalates or phosphates of NH4+, Na+, Mg2+ and K+  Effect of heat on inorganic solid salts To a spatula of unknown solid in a dry test tube, heat first gently and then strongly until no further change occurs. During the test, note the following changes: (i) Nature and colour of the sublimate if any Prepared by AIMABLE DUKUZE Page 18 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS (ii) The nature and colour changes of the residue when hot and then on cooling (iii) Test and identify all the gases or vapours evolved COLOUR DEDUCTION A Colourless liquid water or water vapour ( use anhydrous copper(II) sulphate or cobalt (II)Chloride paper that turns from Hydrated salts white to blue or from blue to pink respectively). B Carbon dioxide gas evolved (use calcium hydroxide solution CO2 from CO32-, HCO32-, (lime water) that turns milky. C2O42- C Ammonia gas (smell or test with red litmus paper, which turns NH3 gas from NH4+ blue, and/or HCl gas from concentrated hydrochloric acid, which forms white fumes of NH4Cl). D Oxygen (colourless but re-lights a glowing splint) and nitrogen O2 and NO2 gas from NO3- dioxide gas (acidic by use of blue litmus paper, reddish-brown with pungent smell). E Reddish-brown gas evolved NO2 from NO3- or bromine from Br- F Sulphur dioxide gas evolved (smells, acidic, turns purple acidified SO2 gas from SO32- or SO42- potassium permanganate solution colourless or turns potassium dichromate solution from orange to green) G Hydrogen chloride gas (smells, acidic by use of litmus paper, Hydrated Cl- except those of fumes with ammonia). group (I) elements and barium chloride H Sublimate NH4+ I Decrepitation/Cracking sound Common for crystals Note: During heating the permanent changes in colour may occur to the solid residue OBSERVATION DEDUCTION A Pink Blue black Co2+ B Yellow Red(black) Fe3+ C Violet Green Cr3+ D Green Yellow Red (black) Fe2+ E Blue White Black Cu2+ During heating the following temporary colour changes may also occur: OBSERVATION DEDUCTION Hot Cold A Yellow White Zn2+ B Red (orange) Yellow Pb2+ C Black Red Fe3+  Effect of aqueous sodium hydroxide and aqueous ammonia on solutions containing transition metal ions The hydroxides of transition metals are precipitated from solutions of the metal ions by the addition of hydroxide ionsor ammonia. The colour of the precipitate can often be used to identify the metal present. The precipitates formed are gelatinous and often coloured and some form soluble complex ions with excessammonia. Prepared by AIMABLE DUKUZE Page 19 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS a) To about 1cm3 of the solution containing the positive ion (cation), add 2M aqueous sodium hydroxide dropwise until in excess A white precipitate formed, soluble in excess to Zn2+ probably present form a colorless solution Zn2+(aq) + 2OH-(aq) Zn(OH)2(s) Zn(OH)2(s) + 2OH-(aq) [Zn(OH)4]2-(aq) 2+ A pale blue precipitate formed, insoluble in excess Cu present Cu2+(aq) + 2OH-(aq) Cu(OH)2(s) 2+ A dirty green precipitate formed, insoluble in excess Fe present and turns brown on standing due to aerial oxidation Fe2+(aq) + 2OH-(aq) Fe(OH)2(s) 4Fe(OH)2(s) + O2(g) + 2H2O(l) 4Fe(OH)3(s) 3+ A rust-brown precipitate formed, insoluble in excess Fe present Fe3+(aq) + 3OH-(aq) Fe(OH)3(s) A white precipitate formed, insoluble in excess and Mn2+ present rapidly turns brown on standing due to aerial Mn2+(aq) + 2OH-(aq) Mn(OH)2(s) oxidation 2Mn(OH)2(s) + O2(g) 2MnO2xH2O(s) A green precipitate formed, soluble in excess to give Cr3+ present a green solution 3+ - Cr(OH)3(s) Cr (aq) + 3OH (aq) Cr(OH)3(s) + OH-(aq) [Cr(OH)4]-(aq) A green precipitate formed, insoluble in excess Ni2+ present Ni2+(aq) + 2OH-(aq) Ni(OH)2(s) 2+ A blue precipitate formed, insoluble in excess and Co present turns pink on standing Co2+(aq) + 2OH-(aq) Co(OH)2(s) b) To about 1cm3 of the solution containing the positive ion (cation), add 2M aqueous ammonia dropwise until in excess A white precipitate formed, soluble in excess Zn2+ present to form a colorless solution Zn2+(aq) + 2OH-(aq) Zn(OH)2(s) Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]2+(aq) + 2OH-(aq) A blue precipitate formed, soluble in excess Cu2+ present to form a deep blue solution Cu2+(aq) + 2OH-(aq) Cu(OH)2(s) Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH-(aq) A dirty green precipitate formed, insoluble in Fe2+ present excess and turns brown on standing due to Fe2+(aq) + 2OH-(aq) Fe(OH)2(s) aerial oxidation 4Fe(OH)2(s) + O2(g) + 2H2O(l) 4Fe(OH)3(s) A rust-brown precipitate formed, insoluble in Fe3+ present excess Fe3+(aq) + 3OH-(aq) Fe(OH)3(s) A white precipitate formed, insoluble in Mn2+ present excess and rapidly turns brown on standing Mn2+(aq) + 2OH-(aq) Mn(OH)2(s) due to aerial oxidation 2Mn(OH)2(s) + O2(g) 2MnO2xH2O(s) 3+ A grey-green precipitate formed, soluble in Cr present excess to give a green solution Cr3+(aq) + 3OH-(aq) Cr(OH)3(s) Cr(OH)3(s) + 6NH3(aq) [Cr(NH3)6]3+(aq) + 3OH-(aq) 2+ A green precipitate formed, soluble in excess Ni present to form a blue solution Ni2+(aq) + 2OH-(aq) Ni(OH)2(s) Prepared by AIMABLE DUKUZE Page 20 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Ni(OH)2(s) + 6NH3(aq) [Ni(NH3)6]2+(aq) + 2OH-(aq) 2+ A blue precipitate is formed Co present Co2+(aq) + 2OH-(aq) Co(OH)2(s)  Confirmatory tests for some transition metal ions Confirmatory tests are the tests required to confirm the analysis. Generally, a confirmatory test is used only after other reactions have been to isolate/identify the ion. The presence some ions cannot be confirmed by using only sodium hydroxide and ammonia solution. Such ions have specific reagents that are used to confirm them. (a) Zinc ions (i) Addition of little solid ammonium chloride followed by disodium hydrogen phosphate solution to a solution of zinc ions gives a white precipitate. The precipitate dissolves in excess ammonia or dilute mineral acids. (ii) Addition of potassium ferrocyanide solution to a solution of zinc ions gives a white precipitate. Zn2+(aq) + [Zn(CN)6]-(aq) Zn2[Fe(CN)6](s) (b) Chromium ions To a solution of chromium (III) ions, add excess aqueous sodium hydroxide followed by little hydrogen peroxide and boil the resultant mixture. A yellow solution of a chromate is formed. 2Cr3+(aq) + 10OH-(aq) + 3H2O2(aq) 2CrO42-(aq) + 8H2O(l) Treatment of the yellow solution with: (i)Lead(II) ethanoate or Lead(II)nitrate solution gives a yellow precipitate of Lead(II) chromate. Pb2+(aq) + CrO42-(aq) PbCrO4(s) (ii) Barium nitrate (or chloride) solution gives a yellow precipitate of barium chromate. Ba2+(aq) + CrO42-(aq) BaCrO4(s) (iii) Silver nitrate gives a brick red precipitate of silver chromate 2Ag+(aq) + CrO42-(aq) Ag2CrO4(s) (iv) A little alcohol (for example, butan-1-ol) and dilute sulphuricacid , a blue color is formed in the alcohol layer. The blue color is due to unstable CrO5. (c) Manganese (II) ions To the solution of manganese(II) ions, add little concentrated nitric acid followed by little solid lead(IV) oxide or solid sodium bismuthate(V) and boil the mixture. A purple solution is formed due to MnO4- ion. 2Mn2+(aq) + 5BiO -(aq) + 14H+(aq) 2MnO4-(aq) + 5Bi3+(aq) + 7H2O(l) 3 2Mn2+(aq) + 5PbO2(s) + 4H+(aq) 2MnO4-(aq) + 5Pb2+(aq) + 2H2O(l) Prepared by AIMABLE DUKUZE Page 21 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS (d) Iron (II) ions (i) Addition of potassium hexacyanoferrate(III) solution to a solution of iron(II) ions gives a dark blue precipitate. (ii) Addition of few drops of concentrated nitric acid to a solution of iron(II) ions gives a yellowish solution due to iron(III) ions formed. The solution gives positive test for iron(III) ions. (e) Iron(III) ions (i) Addition of potassium hexacyanoferrate(II) solution to a solution of iron(III) ions gives a dark blue precipitate (ii)Addition of potassium thiocyanate or ammonium thiocyanate solution to a solution of iron(III) ions gives a blood red coloration. (f) Cobalt (II) ions Addition of potassium thiocyanate or ammonium thiocyanate solution to a solution of cobalt(II) ions gives a blue colored product of potassium cobalt(II) tetrathiocyanate. Co2+(aq) + 4KSCN(aq) KCo(SCN)4(aq) + 2K+(aq) g) Nickel (II) ions (i) Addition of potassium cyanide solution gives a yellow-green precipitate of Nickel(II) cyanide. The precipitate dissolves in excess reagent to form a dark yellow solution tetracyanonickel (II) ion. Ni2+(aq) + 2CN-(aq) Ni(CN)2(s) Ni(CN)2 (s) + 2CN-(aq) [Ni(CN)4]2-(aq) (ii) Addition of aqueous ammonia followed by 2 to 3 drops of dimethylglyoxime solution to a solution of nickel (II) ions gives a red precipitate. The formation of this precipitate may sometimes require that the solution mixture would be warmed. h) Copper (II) ions In addition to use of aqueous ammonia, the copper(II) ions can be confirmed by addition of the following reagents to an aqueous solution of copper(II) ions: (i) Potassium iodide solution: A white precipitate of copper (I) iodide stained brown with free iodine. 2Cu2+(aq) + 4I-(aq) Cu2I2(s) + I2(aq) The brown color fades on addition of sodium thiosulphate solution due to the reaction below: I2 (aq) + 2S2O32-(aq) 2I-(aq) + S4O62-(aq) (ii) Potassium hexacyanoferrate (II) solution: A brown precipitate is formed. 2Cu2+(aq) + [Fe(CN)6]4-(aq) Cu2Fe(CN)6(s) (iii) Potassium (or ammonium) cyanide solution: A yellow precipitate is formed. The precipitate rapidly turns white. Cu2+(aq) + 2CN-(aq) Cu(CN)2(s) Cu(CN)2(s) CuCN(s) + C2N2(g) Prepared by AIMABLE DUKUZE Page 22 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Identification of Mg2+ Tests Observations Deductions (a) To 2 spatula endfuls of a substance in A colourless solution Mg2+, Ca2+, Ba2+, Pb2+, Al3+, a test tube, add about 6 cm3 of formed Zn2+ or NH4+ suspected water.Shake and divide the solution into four portions (i) To the first portion, add sodium A white precipitate Ca2+, Mg2+, Ba2+ suspected hydroxide solution drop wise until in insoluble in excess excess sodium hydroxide solution formed. (ii) To the second portion, add ammonia A white precipitate Pb2+, Al3+, Mg2+ suspected solution dropwise until in excess, insoluble in excess then warm the mixture sodium hydroxide solution formed (iii) To the third portion, add few drops No observable change Ca2+, Pb2+, Ba2+ absent of dil. H2SO4 (iv) To the forth portion, add dilute A blue precipitate Mg2+ present hydrochloric acid followed by magneson reagent. Finally, add 3 drops of sodium hydroxide solution to the resultant solution. Identification of Cu2+ Tests Observations Deductions I. To 2 spatula endfuls of a substance A blue or green solution Cu2+, Fe2+, Ni2+, Co2+,Cr3+ in a test tube, add about 6 cm3 of suspected water. Shake and divide it into three parts II. To the first portion, add sodium A pale blue precipitate Cu2+present hydroxide solution drop wise until in insoluble in excess sodium excess. Heat gently hydroxide solution which turns black on heating III. To the second portion, add White precipitate of Cu2I2, Cu2+ present potassium iodide solution stained brown with free iodine IV. To the third portion, add Blue precipitate soluble in Cu2+confirmed ammonia solution dropwise until in excess ammonia solution to excess. form a deep blue solution V. To the fourth portion, add few A reddish-brown gelatinous Cu2+confirmed drops of potassium ppt of Cu2 [Fe(CN)6] formed ferrocyanide,K4Fe(CN)6 solution Identification of Ba2+ Tests Observations Deductions To 3 spatula endfuls of a substance in a test A colourless Mg2+, Ca2+, Ba2+, Pb2+, Al3+, Zn2+ or tube, add about 6 cm3 of water.Shake and solution formed NH4+ suspected divide it into three portions (i)To the first portion, add sodium hydroxide A white Mg2+, Ca2+,Ba2+ suspected Prepared by AIMABLE DUKUZE Page 23 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS solution drop wise until in excess precipitate (Ba2+ and Ca2+ precipitate when insoluble in excess the test solution is concentrated) sodium hydroxide solution (ii)To the second portion, add ammonia No observable Pb2+, Al3+, Zn2+ absent solution drop wise until in excess change (iii)To the third portion add dilute sulphuric A white acid precipitate formed Ba2+ present (iv)To the forth portion, add potassium A yellow Ba2+ confirmed chromate(VI) solution precipitate is formed (v) Add saturated CuSO4 solution White ppt of Ba2+ confirmed BaSO4 formed Identification of Al3+ Tests Observations Deductions i. To 2 spatula endfuls A colourless solution formed Mg2+, Ca2+, Ba2+, Pb2+, Al3+, Zn2+ or NH4+ susp of a substance in a test tube, add about 6 cm3 of water. Shake and divide it into five parts ii. To the first portion, A white precipitate soluble in excess Pb2+, Al3+ or Zn2+ suspected add sodium NaOH formed hydroxide solution dropwise until in excess iii. To the second A white precipitate insoluble in excess Pb2+, Al3+ suspected portion, add ammonia formed ammonia solution drop wise until in excess iv. To the thirdth The dye is taken up by the precipitate Al3+confirmed solution, add litmus which is coloured solution and then blue (blue lake) aq.NH3until alkaline Identification of Pb2+ Tests Observations 1. To 2 spatula endfuls of a substance A colourless solution formed in a test tube, add about 6 cm3 of water. Shake and divide it into sixparts 2. To the first portion add sodium A white precipitate soluble in excess sodium hydroxide solution gives a colourless hydroxide solution drop wise until in solution. excess Prepared by AIMABLE DUKUZE Page 24 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS 3. To the second portion add A white precipitate insoluble in excess reagent formed. ammonia solution drop wise until in Pb(OH)2 excess 4. To the third portion add dilute A white precipitate PbSO4 formed. sulphuric acid 5. To the fourth portion add potassium A yellow precipitate PbI2 formed. iodide solution 6. To the fifth portion, add potassium A bright yellow precipitatePbCrO4 formed chromate solution Identification of NH4+ Tests Observations Deductions Note the appearance of the White crystalline solid Mg2+, Ca2+, Ba2+, Pb2+, Al3+, solid Zn2+ or NH4+ suspected To 2 spatula endfuls of a A colourless solution Mg2+, Ca2+, Ba2+, Pb2+, Al3+, substance in a test tube, add about formed Zn2+ or NH4+ suspected 6 cm3 of water. To 1 spatula endful of the A gas that turns red litmus NH4+ suspected substance in a test tube, add paper blue is evolved The gas given off is sodium hydroxide solution and ammonia then heat. Identification of Mn2+ Tests Observations Deductions To 2 spatula endfuls of a Pale pink solution formed A transition cation suspected substance in a test tube, add about 6 cm3 of water. Shake and divide it into three parts (i)To the first part add sodium A white precipitate insoluble in Mn2+ present hydroxide solution dropwise excess sodium hydroxide solution until in excess formed which turns brown on standing (ii)To the second portion, add A white precipitate insoluble in Mn2+ present ammonia solution dropwise until excess sodium hydroxide solution in excess formed which turns brown on standing (iii)To the third portion, add few Purple solution formed Mn2+ confirmed drops of conc. HNO3 followed by a small quantity of solid sodium bismuthate and boil Identification of Zn2+ To 2 spatula endful of a substance in a A colourless solution formed Mg2+, Ca2+, Ba2+, Pb2+, Al3+, 3 test tube, add about 6 cm of Zn2+ or NH4+ suspected water.Shake and divide it into three portions (i)To the first part add sodium A white precipitate soluble in Zn2+,Pb2+, or Al3+suspected hydroxide solution drop wise until in excess sodium hydroxide solution Prepared by AIMABLE DUKUZE Page 25 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS excess gives a colourless solution. (ii)To the second part add ammonia A white precipitate soluble in Zn2+ confirmed solution drop wise until in excess excess ammonia solution to give a colourless solution. (iii) To the third part add NH4Cl or White (dirty white) ppt of ZnS Zn2+ confirmed NH3 then pass H2S or add Na2S formed solution NB: Confirmatory test the white precipitate dissolves in excess of sodium hydroxide and ammonia solution Identification of Fe2+ Tests Observations Deductions To 2 spatula endfuls of a substance in A light green solution formed Fe2+, Cu2+, Ni2+, Cr3+ suspected a test tube, add about 6 cm3 of water.Shake and divide it into four portions. (i)To the first portion, add sodium A dirty green precipitate Fe2+ present hydroxide solution dropwise until in insoluble in excess sodium excess hydroxide solution formed but turns brown on standing (ii)To the second portion, add A dirty green precipitate Fe2+ present ammonia solution drop wise until in insoluble in excess ammonia excess solution formed but turns brown on standing (iii)To the third portion, add few Deep (dark) blue precipitate Fe2+ confirmed drops of potassium hexacyanoferrate (III) To the four part add 1 cm3 of H2O2 Green solution turns yellow Fe2+ confirmed warm then add 3 drops of dil NaOH brown Yellow brown ppt of Fe(OH)2 formed Identification of Fe3+ Tests Observations Deductions To 2 spatula endful of M in a test A yellowish solution formed Fe3+, CrO42- suspected tube, add about 6 cm3 of water, shake and divide into four portions (i)To the first portion, add sodium A reddish-brown precipitate Fe3+, Mn3+, Ag+suspected hydroxide solution drop wise until in insoluble in excess sodium excess hydroxide solution formed Fe(OH)3 (ii)To the second portion, add A reddish-brown precipitate Fe3+ present ammonia solution drop wise insoluble in excess ammonia until in excess, then warm the solution formed Fe(OH)3 mixture (iii) (a)To the third portion, add few Red blood colouration drops of potassium thiocyanate Fe3+ confirmed 2+ solution Fe[(H2O)5SCN] (b) To the fourth portion, add 2-3 drops of potassium hexacyanoferrate Dark blue precipitate formed Fe3+ confirmed Prepared by AIMABLE DUKUZE Page 26 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS (II) To test solution add a few drops of Deep red solution formed Fe3+ confirmed H2O2 followed by dil NaOH Red brown ppt of Fe(OH)3 formed Identification of Cr 3+ Tests Observations Deductions A green solution formed Fe2+, Cu2+, Ni2+, Cr3+ suspected To 2 spatula endfuls of a substance in a test tube, add about 6 cm3 of water.Shake and divide it into three portions (i)To the first portion, add sodium A bluish-green precipitate which Cr3+, Cu2+ suspected hydroxide solution drop wise until dissolves in excess sodium in excess hydroxide to give a green solution (ii)To the second portion, add A grey-green precipitate slightly Cr3+ present ammonia solution drop wise until soluble in excess to give a pink in excess solution (iii)To the third portion, add A yellow solutionformed which Cr3+ confirmed NaOH(aq) and hydrogen peroxide precipitates on addition of lead solution followed by lead nitrate nitratesolution Summary flowchart to identification of cations Ca2+ , Mg2+ ,Ba2+ probably Na+, K+,NH4+ white ppt insoluble no ppt formed Heat in excess no observable Add NaOH Solution until the A gas which turns change: Na+ + excess red litmus blue is evolved or K + NH4 present white ppt Colored precipitate which dissolve in excess Al3+, Zn2+ ,Pb2+ Cu2+, Fe2+,Fe3+,Co3+,Ni2+,Mn2+ Green pppBrown ppt Blue ppp is Fe2+ is Fe3+ Add NH3 until is Cu2+ the excess white ppt white ppt insoluble in excess A yellow ppt confirms dissolve in excess Pb2+ Add a solution Zn2+ Al3+ , Pb2+ of KI No observable change confirms Al3+ Prepared by AIMABLE DUKUZE Page 27 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS 1.5. IDENTIFICATION TESTS OF ANIONS Summary flowchart to identification of anions SO42- , SO 2- , 3 NO3-, CO32-,HCO3- Cl-, I-,Br- Add BaCl2 followed by dilute acid(e.g HCl) White precipitate No observable change insoluble in dilute acid White ppt soluble in dilute acid NO3-,Cl-,I-,Br- SO42- CO32-,HCO3-,SO32- Add AgNO3 If effervescence If SO2evolves CO2 occurs pale ye pale ye No observable CO32-,HCO3- 2- llow ppt White ppt llow ppt SO3 change inssoluble formed soluble in ammonia in Heat the solution ammonia NO3- Cl- Br- I- CO2 formed and no change a white ppt may be formed CO32- HCO3- Note: CO32- + 2H+ CO2 + H2O Efferversence SO32- + 2H + SO2 + H2O gas which turns orange dichromate green S2- + 2H+ smell of rotten eggs H2S S2O32- +2H+ S(s) + SO2 + H2O yellow ppt of Sulfur Identification of CO32- Tests Observations Deductions a) Heat one spatula endful of a A colourless gas given off HCO3-, SO42-, C2O42-, substance in a dry test tube, first which turns a blue litmus CO32-, SO32- suspected gently and then strongly until no paper pink further change. Test the gas evolved by moist blue litmus paper CO2 and SO2 suspected Prepared by AIMABLE DUKUZE Page 28 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS b) To about 6 cm3 of water,add 2 A colourless solution formed NO3-, HCO3-, SO42-, spatula endfuls of a substance. Shake CO32-, C2O42-, SO32- of and divide the solution into 3 portion Group I elements or NH4+or HCO32-of Group(II)saltssuspecte d (i) To the first portion, add dil. A colourless gas evolved CO32-, HCO3-, C2O42- H2SO4 and test the gas evolved which forms a milky present usinglimewater and blue litmus suspension with lime water. paper The gas turns moist blue litmus paper pink (ii) To the second portion, add A colourless gas evolved CO32-, HCO3-, C2O42- dilute HCl solution and test the which forms a milky present gas evolved using lime water suspension with lime water. solution and moist blue litmus The gas turns moist blue paper litmus paper pink (iii) To the third portion, add dilute The purple colour of C2O42- absent nitric acid then boil , and add potassium permanganate dilute sulphuric acid immediately persists. followed by 2-3 drops of potassium permanganate solution. (iv) To the fourth portion, add A white precipitate CO32- confirmed magnesium sulphate solution formed (v) Add lead(II) nitrate or lead(II) PbCO3 ppt insoluble on CO32- confirmed oxalate solution, heat then add heating but soluble in nitric HNO3 acid Identification ofSO42- Tests Observations Deductions (a) Heat a spatula endful of a substance A colourless gas evolved. It SO32-, SO42- suspected strongly in a dry test-tube and test the turns yellow potassium gas evolved using potassium dichromate solution green dichromate solution and blue litmus and moist blue litmus paper paper red (b) To 2ml of a solution of the substance A white precipitate of SO42- confirmed in a test tube, add barium chloride BaSO4 formed solution followed by dilute hydrochloric acid and warm (c) To 2ml of a solution of the substance A white precipitate of in a test tube, add nitric acid solution BaSO4 formed followed by BaCl2(aq) Identification of NO3- Tests Observations Deductions Prepared by AIMABLE DUKUZE Page 29 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS (a) To 2 spatula endfuls of a A colourless HCO3-, SO42-, CO32-, solution substance, add 6cm3 of water. formed C2O42-, SO32- of Group I Shake and divide it into two elements or NH4+or portions HCO32-or all nitratesof Group(II)saltssuspected (b) To the first portion, add freshly A brown ring forms at NO3-, NO2- suspected prepared iron (II) sulphate the interface of two followed by conc. Sulphuric acid layers. carefully down the side of the test tube. (c) To the second portion, addcopper Reddish-brown fumes NO3- confirmed turnings followed by concentrated evolved, which turn blue sulphuric acid. litmus paper red and a greenish-blue solution formed. Identification of Br- Tests Observations Deductions (a) Place 2 spatula endfuls of a Brown vapours evolved. I-, Br-, NO3- solidsubstanceinto a test tube and heat suspected gently. (b) To 1 spatula endful of the solid substance, A colourless gas evolved I-, Br-, NO3- add conc. Sulphuric acid. and red vapours formed suspected on warming. (c) To a solution of the substance, add dilute A pale yellow Br- confirmed nitric acid followed by silver nitrate precipitate formed solution. Then, add ammonia solution. which is slightly soluble in ammonia solution. Identification of I- Tests Observations Deductions (a) Place 2 spatula endfuls of a substance into Violet vapours evolved. I- suspected a test tube and heat gently. (b) To 1 spatula endful of the solid substance, violet vapours formed I- present add conc. Sulphuric acid. (c) To a solution of the substance, add dilute A pale yellow precipitate I- present nitric acid followed by silver nitrate formed which is solution. Then, add ammonia solution. insoluble in ammonia solution. (d) To a solution of the substance, add lead Bright- yellow precipitate I- confirmed nitrate solution formed Identification of Cl- Tests Observations Deductions (a)To 2 spatula endfuls of the substance in a A colourless solution Cl- , HCO3-, SO42-, Prepared by AIMABLE DUKUZE Page 30 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS test tube, add about 6 cm3 of water. Shake formed CO32-, C2O42-, SO32- of and divide it into five parts Group I elements or NH4+or HCO32-or all nitratesof Group(II)salts suspected (i)To the first portion, add lead (II) A white precipitate Cl- suspected nitrate solution and heat formed which disappears on heating (ii) To the first portion, add dilute nitric A white precipitate Cl- present acid then silver nitrate solution formed which dissolves followed by ammonia solution in ammonia solution (b)Heat 1 spatula endful of a substance in a A greenish-yellow gas Cl- confirmed dry test-tube and test the gas evolved evolved Worked example 1: You are provided with substance Q which contains 2 cations and one anion. You are required to identify the cations and anion in Q Carry out the following tests and record your observations and deductions in the table below. Identify any gas evolves. Tests Observations Deductions 1. Heat one spatula end-full of Q -Colorless vapor condenses on cooler -Water vapor from in a dry test tube until there is no part of test tube hydrated salt further change -A colorless gas with pungent smell ,turns - Ammonia gas red litmus blue forms white fumes with HCl -colorless gas with pungent smell turns orange dichromate paper green -sulfur dioxide gas - light green crystals turns into white powder ,then yellow, then reddish brown NH4+, Fe2+ suspected 2. Put 2 spatula end-fulls of Q in a Colorless solution - test tube ,add about 5cm3of water and shake. Divide the solution into five portions 3. To the first solution add dilute Green ppt insoluble in excess, on heating -Fe2+ is suspected NaOH, drop wise until the excess gas turns red litmus blue and form white fumes with Conc.HCl. ppt turns brown Ammonia gas ,NH4+ is on heating confirmed. Ion (III) oxide from Prepared by AIMABLE DUKUZE Page 31 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS Fe2+which is decomposed. 4.To second, add ammonia Green ppt insoluble in excess Fe2+ c suspected solution drop wise until in excess 5.To the 3rd solution add 2-3 A deep blue ppt Fe2+ confirmed drops of potassium hexacyano ferrate (III) solution 6. To the 4th add Lead (II) nitrate White ppt SO42-, Cl- suspected solution followed by dilute nitric acid 7. To the 5th add Barium chloride White precipitate insoluble in dil.HCl SO42- confirmed. solution followed by dilute hydrochloric acid (HCl) Cations are: NH4+ and Fe2+ Anion is : SO42- Worked example 2: Analysis of a substance A which contains two cations and two anions Tests Observations Deductions Heat two spatula ends full of A gently Vapour of water condenses to give a Hydrated salt present then later strongly colorless liquid which turns anhydrous copperII salt blue Colorless ,odorless gas turns blue Carbon dioxide gas, and litmus red and limewater milky hence CO32-or HCO3- or C2O42-present Residue yellow then green Ni2+or Cr2+ suspected Shake two spatula –ends full of A with Filtrate is a green solution Cu2+,Fe2+, Ni2+or Cr2+ about 6cm3of water ,filter and retain suspected both filtrate and residue Residue is green solid Cu2+,Fe2+, Ni2+or Cr2+ suspected Divide the filtrate into 2 parts To the A bluish green ppt soluble in excess Cr3+ suspected first part add dilute ammonia solution to give a violet solution dropwise until in excess To the second portion, add little 1- A blue color appears in the1- Cr3+ confirmed butanol followed by dilute sulfuric Prepared by AIMABLE DUKUZE Page 32 of 197  SENIOR SIX CHEMISTRY UNITS’SYLLABUS acid. butanol layer To the 3rd part, add lead ethanoate White ppt formed and persists on CO32-or HCO3- or C2O42- solution and heat heating or SO4 2-suspected To the 4th part, add little dilute nitric White ppt SO42- confirmed acid followed by a few drops of barium nitrate solution Wash the residue with some water Green solid dissolves with CO32- confirmed and dissolve it in about 5cm3of dilute effervescence of a colorless odorless nitric acid gas which turns blue litmus red and lime water milky C