WAEC SSCE - Chemistry - 2008 (Essay)
Tambaya 1 Rahoto
(a) Consider the following reaction sequence.
(i) What process leads/tathe formation of K?
(ii) Write the formula of K.
(iii) Write the structural formula of L and name L.
(iv) Name A\(_n\)
(v) Write the structure of M and name M.
(b)(i) What are carbohydrates?
(ii) Give one example each of a I. monosaccharide; II. disaccharide; III. polysaccharide.
(c) Consider the following structure of a simple sugar :
(i) Which functional group makes the compound a reducing agent?
(ii) State what would C = O be observed when I. the compound is mixed with Fehling's solution and boiled;
I. few drops of concentrated H\(_2\)SO\(_4\) is added to the sample of the compound. H?C?OH
(iii) Write an equation for the reaction in (c)(ii)(II).
(d) A hydrocarbon Z with molecular mass 78 on combustion gave 3.385 g of CO\(_2\) and and 0.692 g of H\(_2\)O. Determine the molecular formula of Z. [ H = 1, C = 12, O = 16]
(a) Reaction sequence. The diagram shows: C2H5OH treated with conc. H2SO4 at 180°C gives K; K with KMnO4/H2O gives L; K with a catalyst and heat gives An; K with Cl2 gives M.
(a)(i) Process forming K: dehydration of ethanol (removal of water). Ethanol loses a molecule of water to form ethene.
\[ \text{C}_2\text{H}_5\text{OH} \xrightarrow[180^{\circ}\text{C}]{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O} \]
(a)(ii) Formula of K: C2H4 (ethene).
(a)(iii) L. Ethene with cold dilute KMnO4/H2O is oxidised (hydroxylated) to a diol. Structural formula: CH2(OH)–CH2(OH). Name: ethane-1,2-diol (ethylene glycol).
(a)(iv) An. With a catalyst and heat, ethene molecules join by addition polymerization to form poly(ethene). An is polyethene (polythene), (C2H4)n.
\[ n\,\text{CH}_2{=}\text{CH}_2 \rightarrow \ [\!-\text{CH}_2-\text{CH}_2-\!]_n \]
(a)(v) M. Ethene adds chlorine across its double bond. Structure: CH2Cl–CH2Cl. Name: 1,2-dichloroethane.
\[ \text{CH}_2{=}\text{CH}_2 + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}{-}\text{CH}_2\text{Cl} \]
(b)(i) Carbohydrates are organic compounds made up of carbon, hydrogen and oxygen (the hydrogen and oxygen present in the ratio 2:1 as in water), having the general formula Cx(H2O)y; they are polyhydroxy aldehydes or ketones.
(b)(ii) Examples:
- I. Monosaccharide: glucose (or fructose).
- II. Disaccharide: sucrose (or maltose).
- III. Polysaccharide: starch (or cellulose).
(c) Simple sugar (open-chain glucose, containing a terminal C=O and several –C(H)(OH)– groups).
(c)(i) Functional group responsible for its reducing action: the aldehyde group, –CHO (the terminal C=O).
(c)(ii) Observations:
- I. With Fehling's solution, warmed: the blue solution gives a brick-red precipitate of copper(I) oxide, Cu2O, because the aldehyde group reduces Cu2+ to Cu+.
- II. With a few drops of concentrated H2SO4: the sugar is dehydrated and turns black (charring), leaving a black spongy mass of carbon as water is removed.
(c)(iii) Equation for (c)(ii)II (dehydration of glucose by concentrated sulphuric acid):
\[ \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{conc. H}_2\text{SO}_4} 6\text{C} + 6\text{H}_2\text{O} \]
(d) Molecular formula of hydrocarbon Z (molar mass 78).
Amount of carbon (from CO2):
\[ n(\text{CO}_2) = \frac{3.385}{44} = 0.0769\ \text{mol} \Rightarrow n(\text{C}) = 0.0769\ \text{mol} \]
Amount of hydrogen (from H2O):
\[ n(\text{H}_2\text{O}) = \frac{0.692}{18} = 0.0384\ \text{mol} \Rightarrow n(\text{H}) = 2 \times 0.0384 = 0.0769\ \text{mol} \]
Mole ratio C : H
\[ \text{C} : \text{H} = 0.0769 : 0.0769 = 1 : 1 \]
Empirical formula = CH, of mass \(12 + 1 = 13\).
\[ n = \frac{78}{13} = 6 \]
Therefore the molecular formula of Z is C6H6 (benzene).
Bayanin Amsa
(a) Reaction sequence. The diagram shows: C2H5OH treated with conc. H2SO4 at 180°C gives K; K with KMnO4/H2O gives L; K with a catalyst and heat gives An; K with Cl2 gives M.
(a)(i) Process forming K: dehydration of ethanol (removal of water). Ethanol loses a molecule of water to form ethene.
\[ \text{C}_2\text{H}_5\text{OH} \xrightarrow[180^{\circ}\text{C}]{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O} \]
(a)(ii) Formula of K: C2H4 (ethene).
(a)(iii) L. Ethene with cold dilute KMnO4/H2O is oxidised (hydroxylated) to a diol. Structural formula: CH2(OH)–CH2(OH). Name: ethane-1,2-diol (ethylene glycol).
(a)(iv) An. With a catalyst and heat, ethene molecules join by addition polymerization to form poly(ethene). An is polyethene (polythene), (C2H4)n.
\[ n\,\text{CH}_2{=}\text{CH}_2 \rightarrow \ [\!-\text{CH}_2-\text{CH}_2-\!]_n \]
(a)(v) M. Ethene adds chlorine across its double bond. Structure: CH2Cl–CH2Cl. Name: 1,2-dichloroethane.
\[ \text{CH}_2{=}\text{CH}_2 + \text{Cl}_2 \rightarrow \text{CH}_2\text{Cl}{-}\text{CH}_2\text{Cl} \]
(b)(i) Carbohydrates are organic compounds made up of carbon, hydrogen and oxygen (the hydrogen and oxygen present in the ratio 2:1 as in water), having the general formula Cx(H2O)y; they are polyhydroxy aldehydes or ketones.
(b)(ii) Examples:
- I. Monosaccharide: glucose (or fructose).
- II. Disaccharide: sucrose (or maltose).
- III. Polysaccharide: starch (or cellulose).
(c) Simple sugar (open-chain glucose, containing a terminal C=O and several –C(H)(OH)– groups).
(c)(i) Functional group responsible for its reducing action: the aldehyde group, –CHO (the terminal C=O).
(c)(ii) Observations:
- I. With Fehling's solution, warmed: the blue solution gives a brick-red precipitate of copper(I) oxide, Cu2O, because the aldehyde group reduces Cu2+ to Cu+.
- II. With a few drops of concentrated H2SO4: the sugar is dehydrated and turns black (charring), leaving a black spongy mass of carbon as water is removed.
(c)(iii) Equation for (c)(ii)II (dehydration of glucose by concentrated sulphuric acid):
\[ \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{conc. H}_2\text{SO}_4} 6\text{C} + 6\text{H}_2\text{O} \]
(d) Molecular formula of hydrocarbon Z (molar mass 78).
Amount of carbon (from CO2):
\[ n(\text{CO}_2) = \frac{3.385}{44} = 0.0769\ \text{mol} \Rightarrow n(\text{C}) = 0.0769\ \text{mol} \]
Amount of hydrogen (from H2O):
\[ n(\text{H}_2\text{O}) = \frac{0.692}{18} = 0.0384\ \text{mol} \Rightarrow n(\text{H}) = 2 \times 0.0384 = 0.0769\ \text{mol} \]
Mole ratio C : H
\[ \text{C} : \text{H} = 0.0769 : 0.0769 = 1 : 1 \]
Empirical formula = CH, of mass \(12 + 1 = 13\).
\[ n = \frac{78}{13} = 6 \]
Therefore the molecular formula of Z is C6H6 (benzene).
Tambaya 2 Rahoto
(a)(1) Define covalent bond.
(ii) Give two properties of covalent compounds
(ii) With the aid of a diagram, show how ammonia molecule is formed
(iv) Illustrate with a diagram the formation of ammonium ion?
(v) What type of bond(s) exist(s) in I. ammonia, H. ammonium ion? (\(_1\)H\(_7\)N)
(b)(i) Write three subatomic particles with their corresponding relative masses. CH\(_2\)OH
(ii) Name the possible states in which water can exist.
(c) (i) State Graham's law of diffusion
(ii) Arrange the following gases, He, CH\(_4\) and N\(_2\) in order of increasing rates of diffusion. Give a reason for the order. [ H = 1, He = 4, C = 12, N = 14 ]
(d) Draw the structures of the following compounds:
(i) 2,3-dimethylbutane;
(ii) 1,4-dibromocyclohexane.
(a) Covalent bonding
(i) A covalent bond is a bond formed when two atoms contribute and share one or more pairs of electrons.
(ii) Properties of covalent compounds include:
- They are generally non-conductors of electricity.
- They usually have low melting and boiling points.
- They are usually insoluble in water but soluble in non-polar organic solvents.
(iii) and (iv) The dot-and-cross diagrams below show the formation of ammonia and ammonium ion. Dots represent electrons from nitrogen, while crosses represent electrons from hydrogen.
(v)
- In ammonia, NH3, the N-H bonds are ordinary covalent bonds.
- In ammonium ion, NH4+, there are covalent bonds and one co-ordinate (dative covalent) bond formed by donation of the lone pair on nitrogen to H+. After formation, the four N-H bonds are equivalent.
(b)
(i) The subatomic particles and their relative masses are:
| Subatomic particle | Relative mass |
|---|---|
| Proton | 1 |
| Neutron | 1 |
| Electron | \(\dfrac{1}{1840}\) |
(ii) Water exists as a solid (ice), liquid (water), and gas (steam or water vapour).
(c) Diffusion
(i) Graham's law states that, at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass or vapour density.
\[r \propto \frac{1}{\sqrt{M}}\]
(ii) \(M_r(\mathrm{He})=4\), \(M_r(\mathrm{CH_4})=16\), and \(M_r(\mathrm{N_2})=28\). Hence, the order of increasing rate of diffusion is:
\[\boxed{\mathrm{N_2\; <\; CH_4\; <\; He}}\]
This is because gases with smaller molar masses diffuse faster.
(d) Structural formulae
Bayanin Amsa
(a) Covalent bonding
(i) A covalent bond is a bond formed when two atoms contribute and share one or more pairs of electrons.
(ii) Properties of covalent compounds include:
- They are generally non-conductors of electricity.
- They usually have low melting and boiling points.
- They are usually insoluble in water but soluble in non-polar organic solvents.
(iii) and (iv) The dot-and-cross diagrams below show the formation of ammonia and ammonium ion. Dots represent electrons from nitrogen, while crosses represent electrons from hydrogen.
(v)
- In ammonia, NH3, the N-H bonds are ordinary covalent bonds.
- In ammonium ion, NH4+, there are covalent bonds and one co-ordinate (dative covalent) bond formed by donation of the lone pair on nitrogen to H+. After formation, the four N-H bonds are equivalent.
(b)
(i) The subatomic particles and their relative masses are:
| Subatomic particle | Relative mass |
|---|---|
| Proton | 1 |
| Neutron | 1 |
| Electron | \(\dfrac{1}{1840}\) |
(ii) Water exists as a solid (ice), liquid (water), and gas (steam or water vapour).
(c) Diffusion
(i) Graham's law states that, at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass or vapour density.
\[r \propto \frac{1}{\sqrt{M}}\]
(ii) \(M_r(\mathrm{He})=4\), \(M_r(\mathrm{CH_4})=16\), and \(M_r(\mathrm{N_2})=28\). Hence, the order of increasing rate of diffusion is:
\[\boxed{\mathrm{N_2\; <\; CH_4\; <\; He}}\]
This is because gases with smaller molar masses diffuse faster.
(d) Structural formulae
Tambaya 3 Rahoto
(a)(i) Define hard water
(ii) Name two substances responsible for hardness in water
(iii) Give two methods for the removal of hardness in water.
(b)(i) What are the raw materials require for the manufacture of tetraoxosulphate (VI) acid by the contact process?
(ii) Write an equation for tr reaction that requires a catalyst in the contact process (iii) State the catalyst used in (b)(ii).
(c)(i) Give two uses of sodium tetraoxosulphate (VI) Consider the reaction represented by the following equation:
Na\(_2\)SO\(_4\).10H\(_2\)O\(_{(s)}\) \(\to\) Na\(_2\)SO\(_4\).H\(_2\)O + 9H\(_2\)O\(_{(g)}\)
(ii) What name is given to this type of reaction?
(iii) Calculate the solubility of Na\(_2\)CO\(_3\) at 25°C. if 30.0 cm\(^3\) of its saturated solution at that temperature gave 1.80 g of the anhydrous salt. [C = 12, 0 = 16, Na = 23]
(d)(i) Define the term activated complex
(ii) State one reason why a collision may not produce a chemical reaction
(iii) The formation of water gas is represented by the following equation:
C\(_{(s)}\) + H\(_2\)O\(_{(g)}\) \(\to\) CO\(_{(g)}\) + H\(_{2(0)}\) \(\Delta\)H= + 131 kJmol\(^{-1}\)
Draw an energy profile diagram for the reaction showing the I. activated complex, II. enthalpy of reactants.
(a)(i) Definition of hard water
Hard water is water that does not readily form a lather with soap.
(a)(ii) Substances responsible for hardness
- Calcium hydrogencarbonate, \( \mathrm{Ca(HCO_3)_2} \)
- Magnesium sulphate, \( \mathrm{MgSO_4} \)
Other acceptable substances include calcium sulphate, magnesium hydrogencarbonate and magnesium chloride. Calcium and magnesium ions cause hardness.
(a)(iii) Removal of hardness
- Boiling the water, which removes temporary hardness caused by hydrogencarbonates.
- Adding washing soda, sodium trioxocarbonate(IV), \( \mathrm{Na_2CO_3} \), which removes calcium and magnesium ions as insoluble carbonates.
Ion-exchange resins (permutit) and distillation are also acceptable methods.
(b)(i) Raw materials for the Contact process
Sulphur and air (or oxygen) are the main raw materials. Sulphur is burned in air to form sulphur dioxide.
(b)(ii) Catalysed reaction in the Contact process
\[ 2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)} \]
(b)(iii) Catalyst
Vanadium(V) oxide, \( \mathrm{V_2O_5} \).
(c)(i) Uses of sodium tetraoxosulphate(VI), \( \mathrm{Na_2SO_4} \)
- It is used in the manufacture of glass or paper.
- It is used in the manufacture of detergents.
(c)(ii) Type of reaction
\[ \mathrm{Na_2SO_4 \cdot 10H_2O(s) \rightarrow Na_2SO_4 \cdot H_2O(s) + 9H_2O(g)} \]
This reaction is called efflorescence. The hydrated salt loses some of its water of crystallisation.
(c)(iii) Solubility of \( \mathrm{Na_2CO_3} \) at \(25^\circ\mathrm{C}\)
Mass of anhydrous \( \mathrm{Na_2CO_3} \) in \(30.0\ \mathrm{cm^3}\) of saturated solution \(=1.80\ \mathrm{g}\).
\[ \text{Mass in }1000\ \mathrm{cm^3} = \frac{1000 \times 1.80}{30.0} = 60.0\ \mathrm{g} \]
\[ M_r(\mathrm{Na_2CO_3}) = 2(23)+12+3(16)=106 \]
\[ \text{Solubility} = \frac{60.0}{106}=0.566\ \mathrm{mol\,dm^{-3}} \]
Therefore, the solubility is \(0.566\ \mathrm{mol\,dm^{-3}}\) (or \(60.0\ \mathrm{g\,dm^{-3}}\)).
(d)(i) Activated complex
An activated complex is a temporary, unstable, high-energy arrangement of atoms formed when reacting particles collide effectively, before products are formed.
(d)(ii) Reason a collision may not cause a reaction
The colliding particles may have energy less than the activation energy. Therefore, they cannot form the activated complex. A collision may also fail if the particles have an unsuitable orientation.
(d)(iii) Energy profile diagram for the formation of water gas
The reaction has \( \Delta H=+131\ \mathrm{kJ\,mol^{-1}} \), so it is endothermic: the products are at a higher enthalpy level than the reactants.
The key feature is that the products are higher than the reactants, because a positive \( \Delta H \) means energy is absorbed overall.
Bayanin Amsa
(a)(i) Definition of hard water
Hard water is water that does not readily form a lather with soap.
(a)(ii) Substances responsible for hardness
- Calcium hydrogencarbonate, \( \mathrm{Ca(HCO_3)_2} \)
- Magnesium sulphate, \( \mathrm{MgSO_4} \)
Other acceptable substances include calcium sulphate, magnesium hydrogencarbonate and magnesium chloride. Calcium and magnesium ions cause hardness.
(a)(iii) Removal of hardness
- Boiling the water, which removes temporary hardness caused by hydrogencarbonates.
- Adding washing soda, sodium trioxocarbonate(IV), \( \mathrm{Na_2CO_3} \), which removes calcium and magnesium ions as insoluble carbonates.
Ion-exchange resins (permutit) and distillation are also acceptable methods.
(b)(i) Raw materials for the Contact process
Sulphur and air (or oxygen) are the main raw materials. Sulphur is burned in air to form sulphur dioxide.
(b)(ii) Catalysed reaction in the Contact process
\[ 2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)} \]
(b)(iii) Catalyst
Vanadium(V) oxide, \( \mathrm{V_2O_5} \).
(c)(i) Uses of sodium tetraoxosulphate(VI), \( \mathrm{Na_2SO_4} \)
- It is used in the manufacture of glass or paper.
- It is used in the manufacture of detergents.
(c)(ii) Type of reaction
\[ \mathrm{Na_2SO_4 \cdot 10H_2O(s) \rightarrow Na_2SO_4 \cdot H_2O(s) + 9H_2O(g)} \]
This reaction is called efflorescence. The hydrated salt loses some of its water of crystallisation.
(c)(iii) Solubility of \( \mathrm{Na_2CO_3} \) at \(25^\circ\mathrm{C}\)
Mass of anhydrous \( \mathrm{Na_2CO_3} \) in \(30.0\ \mathrm{cm^3}\) of saturated solution \(=1.80\ \mathrm{g}\).
\[ \text{Mass in }1000\ \mathrm{cm^3} = \frac{1000 \times 1.80}{30.0} = 60.0\ \mathrm{g} \]
\[ M_r(\mathrm{Na_2CO_3}) = 2(23)+12+3(16)=106 \]
\[ \text{Solubility} = \frac{60.0}{106}=0.566\ \mathrm{mol\,dm^{-3}} \]
Therefore, the solubility is \(0.566\ \mathrm{mol\,dm^{-3}}\) (or \(60.0\ \mathrm{g\,dm^{-3}}\)).
(d)(i) Activated complex
An activated complex is a temporary, unstable, high-energy arrangement of atoms formed when reacting particles collide effectively, before products are formed.
(d)(ii) Reason a collision may not cause a reaction
The colliding particles may have energy less than the activation energy. Therefore, they cannot form the activated complex. A collision may also fail if the particles have an unsuitable orientation.
(d)(iii) Energy profile diagram for the formation of water gas
The reaction has \( \Delta H=+131\ \mathrm{kJ\,mol^{-1}} \), so it is endothermic: the products are at a higher enthalpy level than the reactants.
The key feature is that the products are higher than the reactants, because a positive \( \Delta H \) means energy is absorbed overall.
Tambaya 4 Rahoto
(a) State (i) Pauli's Excusion principle;
(ii) Hund's rule of maximum multiplicity.
(b)(i) Write the electronic configuration of each of the following ions of copper: I. Cu\(_+\) II. Cu\(_{(2+)}\) [\(_{29}Cu\)]
(ii) Give the number of unpaired-electrons in each of the ions in (b)(i) above.
(iii) State the type of reaction represented by the following equation:
2Cu\(^+_{(aq)}\) \(\to\) Cu\(^{2+}_{(aq)}\) + Cu\(_{(s)}\)
(iv) Write the formula of one compound of Cu+.
(c)(i) Name the type of radiation that will I. penetrate lead block; II. be stopped by thin paper
(ii) Give the charge on each of the radiations mentioned in I(c)(i) above.
(iii) What term is used to describe each of the following nuclear processes?
I. Combination of two lighter nuclei to form a heavy nucleus II. Splitting of a heavy nucleus into two or more lighter nuclei
III. Time required for one-half of the atoms of a radioactive substance to decay.
(d) Arrange the following ions in order of increasing size. Give a reason for your answer in each case. I. Li\(^{+}\), K\(^{+}\), Na\(^{+}\); II. O\(^{2-}\), F\(^{-}\), N\(^{3-}\)
(e) Determine the percentage composition of phosphorus and oxygen in phosphorus (V) oxide [ P = 31, O = 16 ]
(a) Statements of the rules
- (i) Pauli's Exclusion Principle: No two electrons in the same atom can have the same set of four quantum numbers. In effect, an orbital can hold a maximum of two electrons, and these two must have opposite (paired) spins.
- (ii) Hund's Rule of Maximum Multiplicity: When electrons are being placed in orbitals of equal energy (degenerate orbitals such as the three p or five d orbitals), each orbital is singly occupied first, all with parallel spins, before any orbital receives a second electron. This gives the maximum number of unpaired electrons and the lowest energy arrangement.
(b) Copper ions (Cu has atomic number 29; atom = \([Ar]3d^{10}4s^{1}\)). Electrons are lost from the 4s orbital first, then from 3d.
- (i) I. Cu\(^{+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\) i.e. \([Ar]3d^{10}\)
- I. Cu\(^{2+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{9}\) i.e. \([Ar]3d^{9}\)
- (ii) Unpaired electrons: Cu\(^{+}\) (\(3d^{10}\)) has 0 unpaired electrons (all paired); Cu\(^{2+}\) (\(3d^{9}\)) has 1 unpaired electron.
- (iii) The reaction \(2Cu^{+} \to Cu^{2+} + Cu\) is a disproportionation reaction (a redox reaction in which the same species, Cu\(^{+}\), is simultaneously oxidized to Cu\(^{2+}\) and reduced to Cu).
- (iv) A compound of Cu\(^{+}\): copper(I) oxide, Cu\(_2\)O (or copper(I) chloride, CuCl).
(c) Radioactivity
- (i) I. Radiation that penetrates a lead block: gamma (\(\gamma\)) radiation. II. Radiation stopped by thin paper: alpha (\(\alpha\)) radiation.
- (ii) Charges: gamma radiation carries no charge (0); alpha radiation carries a charge of +2 (it is a helium nucleus, \(^{4}_{2}He^{2+}\)).
- (iii) I. Nuclear fusion. II. Nuclear fission. III. Half-life.
(d) Order of increasing ionic size
- I. Li\(^{+}\) < Na\(^{+}\) < K\(^{+}\). Reason: these are ions of the same group with the same charge; going down the group the number of occupied electron shells increases, so the ionic radius increases.
- II. F\(^{-}\) < O\(^{2-}\) < N\(^{3-}\). Reason: these are isoelectronic ions (each has 10 electrons). As the nuclear charge (number of protons) decreases from F (9) to O (8) to N (7), the pull on the fixed number of electrons weakens, so the electron cloud expands and the ionic radius increases.
(e) Percentage composition of P\(_2\)O\(_5\) (phosphorus(V) oxide; P = 31, O = 16)
Molar mass \(= (2\times31) + (5\times16) = 62 + 80 = 142\ \text{g mol}^{-1}\).
\[\%\,P = \frac{62}{142}\times100 = 43.66\%\]
\[\%\,O = \frac{80}{142}\times100 = 56.34\%\]
So P\(_2\)O\(_5\) is about 43.7% phosphorus and 56.3% oxygen.
Bayanin Amsa
(a) Statements of the rules
- (i) Pauli's Exclusion Principle: No two electrons in the same atom can have the same set of four quantum numbers. In effect, an orbital can hold a maximum of two electrons, and these two must have opposite (paired) spins.
- (ii) Hund's Rule of Maximum Multiplicity: When electrons are being placed in orbitals of equal energy (degenerate orbitals such as the three p or five d orbitals), each orbital is singly occupied first, all with parallel spins, before any orbital receives a second electron. This gives the maximum number of unpaired electrons and the lowest energy arrangement.
(b) Copper ions (Cu has atomic number 29; atom = \([Ar]3d^{10}4s^{1}\)). Electrons are lost from the 4s orbital first, then from 3d.
- (i) I. Cu\(^{+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}\) i.e. \([Ar]3d^{10}\)
- I. Cu\(^{2+}\): \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{9}\) i.e. \([Ar]3d^{9}\)
- (ii) Unpaired electrons: Cu\(^{+}\) (\(3d^{10}\)) has 0 unpaired electrons (all paired); Cu\(^{2+}\) (\(3d^{9}\)) has 1 unpaired electron.
- (iii) The reaction \(2Cu^{+} \to Cu^{2+} + Cu\) is a disproportionation reaction (a redox reaction in which the same species, Cu\(^{+}\), is simultaneously oxidized to Cu\(^{2+}\) and reduced to Cu).
- (iv) A compound of Cu\(^{+}\): copper(I) oxide, Cu\(_2\)O (or copper(I) chloride, CuCl).
(c) Radioactivity
- (i) I. Radiation that penetrates a lead block: gamma (\(\gamma\)) radiation. II. Radiation stopped by thin paper: alpha (\(\alpha\)) radiation.
- (ii) Charges: gamma radiation carries no charge (0); alpha radiation carries a charge of +2 (it is a helium nucleus, \(^{4}_{2}He^{2+}\)).
- (iii) I. Nuclear fusion. II. Nuclear fission. III. Half-life.
(d) Order of increasing ionic size
- I. Li\(^{+}\) < Na\(^{+}\) < K\(^{+}\). Reason: these are ions of the same group with the same charge; going down the group the number of occupied electron shells increases, so the ionic radius increases.
- II. F\(^{-}\) < O\(^{2-}\) < N\(^{3-}\). Reason: these are isoelectronic ions (each has 10 electrons). As the nuclear charge (number of protons) decreases from F (9) to O (8) to N (7), the pull on the fixed number of electrons weakens, so the electron cloud expands and the ionic radius increases.
(e) Percentage composition of P\(_2\)O\(_5\) (phosphorus(V) oxide; P = 31, O = 16)
Molar mass \(= (2\times31) + (5\times16) = 62 + 80 = 142\ \text{g mol}^{-1}\).
\[\%\,P = \frac{62}{142}\times100 = 43.66\%\]
\[\%\,O = \frac{80}{142}\times100 = 56.34\%\]
So P\(_2\)O\(_5\) is about 43.7% phosphorus and 56.3% oxygen.
Tambaya 5 Rahoto
(a)(i) Define in terms of electron transfer I. oxidizing agent; II. reducing agent.
(ii) Write a balanced equation to show that carbon in a reducing agent.
(iii) State the change in oxidation number of the specie that reacted with carbon in (a)((ii).
(b) A gas X has a vapour density of 32. It reacts with sodium hydroxide solution to form salt and water only. It decolourizes acidified potassium tetraoxomanganate (VII) solution and reacts with H\(_2\)S to form sulphur. Using the information provided:
(i) identify gas X; (ii) state two properties exhibited by X;
(iii) give two uses of X.
(c) Consider the following substances: sodium; lead (II) iodide; hydrogen; magnesium; oxygen. Which of the substances
(i) conducts electricity?
(ii) is produced at the cathode during electrolysis of H\(_2\)SO\(_{4(aq)}\)?
(iii) corresponds to the molecular formula A\(_2\)?
(iv) is an alkaline earth metal?
d)(i) Define the term salt.
(ii) Mention two types of salt
(iii) Give an example of each of the salts mentioned in (d)(ii) above.
(e) In a neutralization reaction, dilute tetraoxosulphate (VI) acid completely reacted with sodium hydroxide solution.
(i) Write a balanced equation for the reaction
(ii) How many moles of sodium hydroxide would be required for the complete neutralization of 0.50 moles of tetraoxosulphate (VI) acid?
(a) Redox in terms of electron transfer
- (i) I. Oxidizing agent: a substance that accepts (gains) electrons from another species during a reaction, and is itself reduced.
- II. Reducing agent: a substance that donates (loses) electrons to another species during a reaction, and is itself oxidized.
- (ii) Carbon acting as a reducing agent: \[2CuO + C \to 2Cu + CO_2\] Here carbon removes oxygen from (reduces) copper(II) oxide.
- (iii) The species that reacted with carbon is copper in CuO. Its oxidation number changes from +2 to 0 (a decrease of 2, i.e. it is reduced).
(b) Gas X (vapour density 32, so molar mass \(= 2\times32 = 64\ \text{g mol}^{-1}\)). It reacts with NaOH to give salt and water only (acidic oxide), decolourizes acidified KMnO\(_4\) (it is a reducing agent) and reacts with H\(_2\)S to give sulphur.
- (i) Gas X is sulphur(IV) oxide, SO\(_2\) (molar mass 64).
- (ii) Two properties: it is a colourless gas with a choking, irritating smell; it is acidic and a strong reducing agent (bleaching action; turns moist blue litmus red).
- (iii) Two uses: as a bleaching agent (for straw, paper and wool); as a food preservative/fumigant and in the manufacture of sulphuric acid (contact process).
(c) From sodium, lead(II) iodide, hydrogen, magnesium, oxygen
- (i) Conducts electricity: sodium and magnesium (metals conduct in the solid state).
- (ii) Produced at the cathode during electrolysis of dilute H\(_2\)SO\(_4\): hydrogen.
- (iii) Corresponds to formula A\(_2\): hydrogen (H\(_2\)) and oxygen (O\(_2\)) are diatomic molecules.
- (iv) Alkaline earth metal: magnesium (Group 2).
(d) Salts
- (i) A salt is a compound formed when the ionisable hydrogen ion(s) of an acid are wholly or partly replaced by a metallic ion or the ammonium ion.
- (ii) Two types: normal (neutral) salt and acid salt (also basic salt, double salt, complex salt).
- (iii) Examples: normal salt, e.g. Na\(_2\)SO\(_4\) (sodium sulphate); acid salt, e.g. NaHSO\(_4\) (sodium hydrogen sulphate).
(e) Neutralization of H\(_2\)SO\(_4\) with NaOH
- (i) \[H_2SO_4 + 2NaOH \to Na_2SO_4 + 2H_2O\]
- (ii) From the equation, 1 mole of H\(_2\)SO\(_4\) needs 2 moles of NaOH. Therefore for 0.50 mole of acid: \[0.50 \times 2 = 1.0\ \text{mole of NaOH}\]
Bayanin Amsa
(a) Redox in terms of electron transfer
- (i) I. Oxidizing agent: a substance that accepts (gains) electrons from another species during a reaction, and is itself reduced.
- II. Reducing agent: a substance that donates (loses) electrons to another species during a reaction, and is itself oxidized.
- (ii) Carbon acting as a reducing agent: \[2CuO + C \to 2Cu + CO_2\] Here carbon removes oxygen from (reduces) copper(II) oxide.
- (iii) The species that reacted with carbon is copper in CuO. Its oxidation number changes from +2 to 0 (a decrease of 2, i.e. it is reduced).
(b) Gas X (vapour density 32, so molar mass \(= 2\times32 = 64\ \text{g mol}^{-1}\)). It reacts with NaOH to give salt and water only (acidic oxide), decolourizes acidified KMnO\(_4\) (it is a reducing agent) and reacts with H\(_2\)S to give sulphur.
- (i) Gas X is sulphur(IV) oxide, SO\(_2\) (molar mass 64).
- (ii) Two properties: it is a colourless gas with a choking, irritating smell; it is acidic and a strong reducing agent (bleaching action; turns moist blue litmus red).
- (iii) Two uses: as a bleaching agent (for straw, paper and wool); as a food preservative/fumigant and in the manufacture of sulphuric acid (contact process).
(c) From sodium, lead(II) iodide, hydrogen, magnesium, oxygen
- (i) Conducts electricity: sodium and magnesium (metals conduct in the solid state).
- (ii) Produced at the cathode during electrolysis of dilute H\(_2\)SO\(_4\): hydrogen.
- (iii) Corresponds to formula A\(_2\): hydrogen (H\(_2\)) and oxygen (O\(_2\)) are diatomic molecules.
- (iv) Alkaline earth metal: magnesium (Group 2).
(d) Salts
- (i) A salt is a compound formed when the ionisable hydrogen ion(s) of an acid are wholly or partly replaced by a metallic ion or the ammonium ion.
- (ii) Two types: normal (neutral) salt and acid salt (also basic salt, double salt, complex salt).
- (iii) Examples: normal salt, e.g. Na\(_2\)SO\(_4\) (sodium sulphate); acid salt, e.g. NaHSO\(_4\) (sodium hydrogen sulphate).
(e) Neutralization of H\(_2\)SO\(_4\) with NaOH
- (i) \[H_2SO_4 + 2NaOH \to Na_2SO_4 + 2H_2O\]
- (ii) From the equation, 1 mole of H\(_2\)SO\(_4\) needs 2 moles of NaOH. Therefore for 0.50 mole of acid: \[0.50 \times 2 = 1.0\ \text{mole of NaOH}\]
Tambaya 6 Rahoto
(a)(i) What are acidic oxides?
(ii) Give one example of each of the following oxides: I. acidic oxide; II. basic oxide; III. amphoteric oxide; IV. neutral oxide
(b)(i) Define each of the following terms; I. Heat II. Heat of neutralization
(ii) Weite the above an equation to illustrate each of the terms in (b)(i) above
(ii) Given that the standard heat of combustion of butane (C\(_4\)H\(_{(10)}\) is + 5877 kJmol\(^{-1}\), calculate the heat of 14.5 g butane. [ H = 1, = 12 ]
(c) (i) Name two allotropes of sulphur
(ii) State one difference between the two allotropes.
(d)(i) Give two characteristics of noble eases
(ii) State one use each of I. He; II. Ar.
(e) State what is observed on warming ammonium trioxonitrate (V) with sodium hydroxide.
(a) Oxides
- (i) Acidic oxides are oxides (usually of non-metals) that react with water to form acids, and react with bases/alkalis to form a salt and water only.
- (ii) Examples: I. acidic oxide CO\(_2\) (or SO\(_2\), SO\(_3\)); II. basic oxide Na\(_2\)O (or CaO, MgO); III. amphoteric oxide Al\(_2\)O\(_3\) (or ZnO, PbO); IV. neutral oxide CO (or NO, H\(_2\)O, N\(_2\)O).
(b) Heat and heat of neutralization
- (i) I. Heat: a form of energy that flows from a body at a higher temperature to a body at a lower temperature. II. Heat of neutralization: the heat change when one mole of water is formed from the neutralization of an acid by a base (alkali) under standard conditions.
- (ii) Illustrative equations: heat change accompanying a reaction, e.g. \[C + O_2 \to CO_2,\ \Delta H = -393\ \text{kJ mol}^{-1}\] heat of neutralization, e.g. \[HCl_{(aq)} + NaOH_{(aq)} \to NaCl_{(aq)} + H_2O_{(l)},\ \Delta H = -57.3\ \text{kJ mol}^{-1}\]
- (iii) Heat released by 14.5 g of butane (C\(_4\)H\(_{10}\), molar mass \(= 48+10 = 58\ \text{g mol}^{-1}\); heat of combustion given as 5877 kJ per mole). Amount of butane \(= \dfrac{14.5}{58} = 0.25\ \text{mol}\). Heat evolved \(= 0.25 \times 5877 = 1469.25\ \text{kJ}\). So about 1469 kJ of heat is released (combustion is exothermic).
(c) Sulphur
- (i) Two allotropes: rhombic (alpha) sulphur and monoclinic (beta) sulphur.
- (ii) Difference: rhombic sulphur is the stable form below 96 degrees Celsius, while monoclinic sulphur is stable between 96 degrees Celsius and its melting point (about 119 degrees Celsius); they also differ in crystal shape (rhombic is octahedral, monoclinic is needle-shaped).
(d) Noble gases
- (i) Two characteristics: they are chemically inert/unreactive (stable, filled outer shells); they are monatomic, colourless gases with very low melting and boiling points.
- (ii) Uses: I. He used to fill balloons and airships, and (with oxygen) in breathing mixtures for divers. II. Ar used to provide an inert atmosphere in electric (filament) light bulbs and in welding.
(e) On warming ammonium trioxonitrate(V) (ammonium nitrate) with sodium hydroxide, a colourless gas with a pungent, choking smell (ammonia) is evolved which turns moist red litmus paper blue: \[NH_4NO_3 + NaOH \to NaNO_3 + H_2O + NH_3\uparrow\]
Bayanin Amsa
(a) Oxides
- (i) Acidic oxides are oxides (usually of non-metals) that react with water to form acids, and react with bases/alkalis to form a salt and water only.
- (ii) Examples: I. acidic oxide CO\(_2\) (or SO\(_2\), SO\(_3\)); II. basic oxide Na\(_2\)O (or CaO, MgO); III. amphoteric oxide Al\(_2\)O\(_3\) (or ZnO, PbO); IV. neutral oxide CO (or NO, H\(_2\)O, N\(_2\)O).
(b) Heat and heat of neutralization
- (i) I. Heat: a form of energy that flows from a body at a higher temperature to a body at a lower temperature. II. Heat of neutralization: the heat change when one mole of water is formed from the neutralization of an acid by a base (alkali) under standard conditions.
- (ii) Illustrative equations: heat change accompanying a reaction, e.g. \[C + O_2 \to CO_2,\ \Delta H = -393\ \text{kJ mol}^{-1}\] heat of neutralization, e.g. \[HCl_{(aq)} + NaOH_{(aq)} \to NaCl_{(aq)} + H_2O_{(l)},\ \Delta H = -57.3\ \text{kJ mol}^{-1}\]
- (iii) Heat released by 14.5 g of butane (C\(_4\)H\(_{10}\), molar mass \(= 48+10 = 58\ \text{g mol}^{-1}\); heat of combustion given as 5877 kJ per mole). Amount of butane \(= \dfrac{14.5}{58} = 0.25\ \text{mol}\). Heat evolved \(= 0.25 \times 5877 = 1469.25\ \text{kJ}\). So about 1469 kJ of heat is released (combustion is exothermic).
(c) Sulphur
- (i) Two allotropes: rhombic (alpha) sulphur and monoclinic (beta) sulphur.
- (ii) Difference: rhombic sulphur is the stable form below 96 degrees Celsius, while monoclinic sulphur is stable between 96 degrees Celsius and its melting point (about 119 degrees Celsius); they also differ in crystal shape (rhombic is octahedral, monoclinic is needle-shaped).
(d) Noble gases
- (i) Two characteristics: they are chemically inert/unreactive (stable, filled outer shells); they are monatomic, colourless gases with very low melting and boiling points.
- (ii) Uses: I. He used to fill balloons and airships, and (with oxygen) in breathing mixtures for divers. II. Ar used to provide an inert atmosphere in electric (filament) light bulbs and in welding.
(e) On warming ammonium trioxonitrate(V) (ammonium nitrate) with sodium hydroxide, a colourless gas with a pungent, choking smell (ammonia) is evolved which turns moist red litmus paper blue: \[NH_4NO_3 + NaOH \to NaNO_3 + H_2O + NH_3\uparrow\]
