WAEC SSCE - Chemistry - 2003

Bayanin Amsa

Substance Q reacts with cold water to form a solution which turns red litmus blue. This indicates that the substance is basic or alkaline in nature. Out of the given options, sodium is the only substance that is basic in nature and reacts with cold water to form a solution which turns red litmus blue. Therefore, substance Q could be sodium.

Bayanin Amsa

The correct statement is that "the pH value of the CH₃COOH was the highest." pH is a measure of how acidic or basic a solution is, and it is based on the concentration of hydrogen ions (H+) present in the solution. A lower pH indicates a higher concentration of H+ ions and a more acidic solution, while a higher pH indicates a lower concentration of H+ ions and a more basic solution. For the given solutions, HCl is a strong acid that completely dissociates in water to give H+ and Cl- ions. H₂SO₄ is also a strong acid that dissociates into two H+ ions and one SO₄^2- ion. CH₃COOH is a weak acid that only partially dissociates into H+ and CH₃COO- ions. The pH of HCl and H₂SO₄ solutions will be lower (more acidic) than the pH of the CH₃COOH solution because they are strong acids, and their concentration of H+ ions is higher. However, we cannot determine if they have the same pH values or not from the given information. The sum of pH values of these solutions will not be equal to 14, as this value is the pH of a neutral solution (where the concentration of H+ ions is equal to the concentration of OH- ions). The solutions given in the question are acidic, and their pH values will be less than 7. Therefore, the correct statement is that the pH value of the CH₃COOH was the highest because it is a weak acid that only partially dissociates, resulting in a lower concentration of H+ ions and a higher pH value.

Bayanin Amsa

To determine the formula of the compound formed between elements X and Y, we need to find out the number of electrons that each element can gain or lose to form ions. Element X has 7 valence electrons and needs to gain one electron to achieve the stable configuration of the nearest noble gas, which is 3s₂3p₆. Thus, element X can form an anion with a charge of -1, which is denoted by X^-. Element Y has 5 valence electrons and needs to gain three electrons to achieve the stable configuration of the nearest noble gas, which is 3s₂3p₆. Thus, element Y can form an anion with a charge of -3, which is denoted by Y^3-. To form a compound, the total positive charge must balance the total negative charge. The least common multiple of 1 and 3 is 3, so we need three X^- ions to balance two Y^3- ions. Therefore, the formula of the compound formed is X₃Y₂. So, the correct option is: XY₂.

Bayanin Amsa

The main impurity in haematite is typically SiO₃, also known as silica. Haematite is an iron ore consisting primarily of iron oxide (Fe₂O₃) and is a common source of iron. However, impurities can be present in the ore, and silica is one of the most common impurities. Silica can affect the quality of the iron produced from the ore and can cause problems during the smelting process. Therefore, it is important to remove as much silica as possible during the processing of haematite.

Bayanin Amsa

When water is boiled, it becomes more receptive to lathering with soap. This indicates that the sample contains calcium hydrogentrioxocarbonate (IV) (Ca(HCO₃)₂). Hard water contains calcium and magnesium ions which reacts with soap to form an insoluble precipitate (soap scum), which reduces the effectiveness of soap. Boiling hard water removes temporary hardness (calcium hydrogen-carbonate) by causing the bicarbonate ions to decompose and produce insoluble carbonate, which can be removed by filtration, leaving behind soft water that lathers more easily with soap.

Bayanin Amsa

The substance responsible for the banana taste of a food flavour is Pentylethanoate. This is because Pentylethanoate, also known as amyl acetate, is a chemical compound that naturally occurs in bananas, as well as in other fruits such as apples and pears. It is commonly used as a flavoring agent in the food industry to replicate the taste and smell of these fruits, and is known for its distinct banana-like flavor and aroma. Therefore, Pentylethanoate is the substance that could be responsible for the banana taste of a food flavor.

Bayanin Amsa

Bayanin Amsa

The compound used for removing impurities from bauxite is NaOH (sodium hydroxide). This process is called the Bayer's process. Bauxite contains impurities like iron oxide (Fe2O3), silicon dioxide (SiO2), and titanium dioxide (TiO2), which must be removed before the extraction of aluminum from bauxite. In the Bayer's process, crushed bauxite is treated with a hot, concentrated solution of NaOH (sodium hydroxide), which dissolves the aluminum oxide (Al2O3) present in bauxite, forming sodium aluminate (NaAlO2). The impurities such as iron oxide, silicon dioxide, and titanium dioxide are not soluble in NaOH and are filtered out. The filtered solution of sodium aluminate is then treated with carbon dioxide (CO2) to form aluminum hydroxide (Al(OH)3) which is then calcined to obtain pure aluminum oxide, which is used for the extraction of aluminum metal.

Bayanin Amsa

The process that involves neutralization is liming of soil. When soil is acidic, it can be treated with a base such as calcium carbonate (CaCO3) to raise its pH and reduce its acidity. This process is known as liming of soil and involves a neutralization reaction between the acidic components of the soil and the basic components of the calcium carbonate. The result is a more neutral soil pH which is more suitable for plant growth. In the other options, hardening of oils involves hydrogenation, souring of milk involves fermentation, and charring of sugar involves thermal decomposition.

Bayanin Amsa

The equation n(C6H10O5) + nH2O → nC6H12O6 represents the hydrolysis of starch. Hydrolysis is a chemical process that uses water to break the bonds between molecules. In this case, the bond between the glucose molecules in starch is broken down by the addition of a water molecule, resulting in the formation of glucose molecules. Therefore, "Hydrolysis of starch" is the correct answer.

Bayanin Amsa

The equation for the reaction at the anode during the electrolysis of concentrated sodium chloride solution is: 2CI^- _(aq) → CI_2(g) + 2e^- During the electrolysis of a concentrated sodium chloride solution, the chloride ions (CI^-) are oxidized at the anode to form chlorine gas (CI₂) and electrons. The resulting electrons flow to the cathode, where they reduce sodium ions (Na⁺) to form sodium metal (Na). Therefore, the overall reaction is: 2NaCl _(aq) → 2Na_(s) + CI_2(g) Note that this equation does not include the reaction at the cathode.

Bayanin Amsa

The term used to describe an oxide whose aqueous solution turns blue litmus red is an acid anhydride. Acid anhydrides are oxides that react with water to form acidic solutions. They are usually non-metal oxides that react with water to form an acidic solution. When an acid anhydride dissolves in water, it reacts with the water molecules to produce an acid. The acid then turns blue litmus red, indicating the presence of an acidic solution. Therefore, option B, acid anhydride, is the correct answer.

Bayanin Amsa

Group O, also known as the noble gases, are unreactive because their highest occupied energy level is full, which means they have a stable electron configuration. They have no tendency to gain, lose or share electrons, as they already have the full complement of electrons in their outermost shell. This makes them extremely unreactive and inert, which is why they are often referred to as noble or inert gases. Therefore, the correct answer is "their highest occupied energy level is full".

Bayanin Amsa

The electronic configuration of an element is 1s²2s²2p⁶3s²3p³ indicates that the element has two electrons in its 1s orbital, two electrons in its 2s orbital, six electrons in its 2p orbital, two electrons in its 3s orbital, and three electrons in its 3p orbital. Based on this configuration, we can locate the element on the periodic table by looking at the highest energy level filled with electrons, which is the 3rd energy level or period, and the number of valence electrons, which is 5. Thus, the element is located in Group V and Period 3 of the periodic table. Therefore, the correct option is "Group V, period 3".

Bayanin Amsa

The quantity that is the same for one mole of Br_2(l) and one mole of He_(g) is the number of molecules. One mole of any substance contains the same number of particles, known as Avogadro's number (6.022 x 10²³ particles per mole). Therefore, one mole of Br_2(l) and one mole of He_(g) both contain the same number of molecules. The number of atoms and total mass of particles in one mole of Br_2(l) and one mole of He_(g) are different, as Br_2(l) is a molecular substance and contains two bromine atoms per molecule, while He_(g) is an element and contains only one helium atom per particle. The volume occupied at s.t.p. by one mole of Br_2(l) and one mole of He_(g) is also different, as the volume depends on the size and shape of the particles, and the conditions of temperature and pressure.

Bayanin Amsa

When excess ethene is shaken with acidified KMnO₄ solution, the product obtained is ethane-1,2-diol. This is because acidified KMnO₄ solution is a strong oxidizing agent and can oxidize alkenes such as ethene. When ethene is reacted with KMnO₄, it undergoes oxidative cleavage where the double bond is broken and two hydroxyl groups are added to the carbons of the double bond. The resulting product is a diol, specifically ethane-1,2-diol.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Discharge of industrial effluents into waterways can pollute water. When industries release their effluents, which are waste materials such as chemicals, heavy metals, or other pollutants, into waterways, they can contaminate the water and harm aquatic life. The pollutants can also make the water unfit for human consumption, agriculture or recreational purposes. Therefore, it is important to properly manage and treat industrial waste before discharging it into the environment to prevent water pollution. The other options do not directly cause water pollution: exposure of water to ultraviolet rays can help to disinfect the water, passage of river water through a sand bed can help to filter the water, and addition of a measured quantity of chlorine can help to disinfect the water as well.

Bayanin Amsa

The electrochemical cell is composed of two half-cells: Zn_(s)/Zn²⁺_(aq) and Cu²⁺_(aq)/Cu_(s). The anode (negative electrode) is the Zn_(s)/Zn²⁺_(aq) half-cell, where oxidation occurs, and the cathode (positive electrode) is the Cu²⁺_(aq)/Cu_(s) half-cell, where reduction occurs. During the oxidation process at the anode, solid zinc (Zn) is oxidized to form Zn²⁺ ions, which are released into the solution. This is represented by the half-equation: Zn_(s) → Zn²⁺_(aq) + 2e^- During the reduction process at the cathode, Cu²⁺ ions in the solution are reduced to form solid copper (Cu). This is represented by the half-equation: Cu²⁺_(aq) + 2e^- → Cu_(s) The overall equation for the cell reaction can be obtained by adding the two half-equations and canceling out the electrons: Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s) Therefore, the correct equation that describes the reaction occurring in the cell is Zn_(s) + Cu²⁺_(aq) → Zn²⁺_(aq) + Cu_(s). Answer (C).

Bayanin Amsa

The reaction of C₂H₅OH with excess acidified K₂Cr₂O_7(aq) is an oxidation reaction. In this reaction, the ethanol (C₂H₅OH) is oxidized to ethanoic acid (CH₃COOH). Therefore, the organic product of this reaction is CH₃COOH, which is acetic acid or ethanoic acid.

Bayanin Amsa

Calcium oxide (CaO) can be prepared from sea shells which mainly contain calcium carbonate (CaCO₃) by heating at a high temperature. This process is called thermal decomposition, where heat is applied to the sea shells causing the CaCO₃ to break down into its constituent elements, calcium oxide (CaO) and carbon dioxide (CO₂). The reaction can be represented as: CaCO₃ → CaO + CO₂ Therefore, heating the sea shells at a high temperature will yield calcium oxide (CaO).

Bayanin Amsa

The electrolysis of a fused chloride involves the use of electricity to extract the metal from its molten chloride. The process involves the reduction of the metal ion to the metal at the cathode while the chloride ions are oxidized to chlorine gas at the anode. From the reactivity series, metals that are high up are more reactive than those below them. Therefore, the most reactive metal, V, is the most likely to be extracted by the electrolysis of its fused chloride. This is because it requires more energy to extract a metal that is higher up in the reactivity series compared to one that is lower down. Hence, V requires more energy in the form of electricity to extract it from its fused chloride compared to the other metals, making it the most likely to be extracted through this method.

Bayanin Amsa

An exothermic reaction is a type of chemical reaction in which energy is released from the reaction system into the surroundings. Hence, the correct statement about an exothermic reaction is that the products have less heat content than reactants. This means that the energy released during the reaction is given off to the surroundings, and the enthalpy change for the reaction is negative. In contrast, an endothermic reaction is a reaction that absorbs energy from the surroundings, and has a positive enthalpy change.

Bayanin Amsa

Based on the given order of reactivity, any metal placed before Y in the sequence can displace Y from its salt, but any metal after Y cannot. Therefore, the answer is Z because it is placed after Y in the sequence of reactivity. Thus, Z cannot displace Y from its aqueous salt solution.

Bayanin Amsa

Bayanin Amsa

The correct answer is that the knocking tendency of the fuel is low. The octane number of a fuel is a measure of its ability to resist "knocking" or "pinging" during combustion, caused by the air/fuel mixture detonating prematurely in the engine. A fuel with a higher octane number is more resistant to knocking and is generally considered to be of higher quality. An octane number of 100 indicates that the fuel has excellent resistance to knocking, and is therefore considered a high-quality fuel.

Bayanin Amsa

The octane number is a measure of the ability of a fuel to resist knocking, which is the uncontrolled and explosive combustion of fuel in the engine's combustion chamber. A higher octane number means that the fuel has a greater ability to resist knocking. Therefore, if a fuel has an octane number of 100, it indicates that the knocking tendency of the fuel is low, and it can resist knocking well. Thus, the correct option is "the knocking tendency of the fuel is low."

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The first step is to calculate the number of moles of NaOH present in 5.0 g of the solution using the formula: Number of moles = mass ÷ molar mass The molar mass of NaOH is 40g mol^-1. Therefore: Number of moles = 5.0 g ÷ 40g mol^-1 = 0.125 mol Next, we use the formula: Number of moles = concentration x volume to calculate the volume of the NaOH solution needed to contain 0.125 mol of NaOH. Rearranging the formula gives: Volume = number of moles ÷ concentration Volume = 0.125 mol ÷ 0.20 mol dm^-3 = 0.625 dm³ = 625 cm³ Therefore, the volume of 0.20 mol dm^-3 NaOH solution needed to yield 5.0 g of NaOH on evaporation to dryness is 625 cm³. The correct option is (b) 625 cm³.

Bayanin Amsa

Bayanin Amsa

When bees sting, they inject an acidic substance into the skin, which causes pain and irritation. To treat the sting, one needs a base that can neutralize the acid. Among the options given, NaHCO3 (sodium bicarbonate) is a base that can effectively neutralize the acid and alleviate the pain and irritation caused by the sting. CH3COOH (acetic acid) is also acidic and cannot neutralize the acid. C2H5OH (ethanol) and H2O2 (hydrogen peroxide) are not bases and cannot neutralize the acid either.

Bayanin Amsa

Gin is a distilled alcoholic beverage that typically contains about 40% ethanol. To produce gin from wine, the wine must undergo a process called distillation. Distillation involves heating the wine to vaporize the ethanol, which has a lower boiling point than water, and then condensing the vapor to collect the ethanol. This results in a much higher concentration of ethanol in the final product. Fermentation, which is the process by which wine is produced, can only produce ethanol concentrations of up to about 17%. Therefore, distillation is necessary to increase the ethanol content of the wine to the level required for gin. Evaporation and oxidation are not used to produce gin from wine.

Bayanin Amsa

Barium chloride solution (ll) will give a precipitate with dilute tetraoxosulphate (V) acid. This is because barium chloride is a salt of barium and chloride ions, while tetraoxosulphate (V) acid is a solution of sulphate ions and hydrogen ions. When barium chloride solution is added to dilute tetraoxosulphate (V) acid, the barium ions react with the sulphate ions to form a white insoluble solid called barium sulphate. This is the precipitate that is formed. Lead (ll) trioxonitrate (V) solution and sodium chloride solution will not give a precipitate with dilute tetraoxosulphate (V) acid.

Bayanin Amsa

The reaction of vegetable oil with a solution of wood ash is called saponification. This is a chemical process that involves the conversion of a fat or oil into soap and glycerol. When vegetable oil is mixed with a solution of wood ash, the ash provides the hydroxide ions which react with the oil to form soap molecules and glycerol. The soap molecules are composed of a hydrophilic (water-loving) head and a hydrophobic (water-repelling) tail. This allows them to interact with both water and oil, making them effective at cleaning. Saponification is an important industrial process for producing soap, as well as for the production of biodiesel.

Bayanin Amsa

The statement that "each participating atom contributes one electron" is not correct. In dative bonding, also known as coordinate bonding, one atom donates a pair of electrons to another atom. The atom that donates the electron pair is called the donor, and the atom that accepts the electron pair is called the acceptor. The donor atom contributes both electrons in the bond, while the acceptor atom does not contribute any electrons to the bond. Therefore, each participating atom does not contribute one electron.

Bayanin Amsa

(a)(i) Three characteristic properties of transition metals are:

1. They exhibit variable oxidation states.
2. They form coloured ions or compounds.
3. They form complex ions.

(ii)

• In \(\mathrm{MnCl_2}\): \(x+2(-1)=0\), hence \(x=+2\).
• In \(\mathrm{MnO_2}\): \(x+2(-2)=0\), hence \(x=+4\).
• In \(\mathrm{MnO_4^-}\): \(x+4(-2)=-1\), hence \(x=+7\).

(iii) Manganese is metallic and contains mobile, delocalised electrons which carry electric current in the solid state. Manganese(II) chloride is ionic. In the solid state its ions are fixed in a lattice, but when molten or dissolved in water, the ions are free to move and conduct electricity.

(b)(i) A collision produces products only when:

1. the colliding particles possess energy equal to or greater than the activation energy; and
2. the particles collide with the correct orientation.

(ii) The energy profile below is for an exothermic reaction. \(E_A\) is the energy difference between the reactants and the activated complex, while \(\Delta H\) is the energy difference between products and reactants.

For this exothermic reaction, the products are at a lower energy than the reactants; therefore, \(\Delta H\) is negative.

(c)(i) Addition of \(\mathrm{KSCN(aq)}\) increases the concentration of \(\mathrm{SCN^-}\) ions. The equilibrium shifts to the right to use up some of the added \(\mathrm{SCN^-}\):

\[\mathrm{Fe^{3+}(aq)+3SCN^-(aq)\rightleftharpoons Fe(SCN)_3(aq)}\]

More deep-red \(\mathrm{Fe(SCN)_3}\) is formed, so the pale red solution becomes deeper red.

(ii) The ion that forms an insoluble hydroxide is \(\mathrm{Fe^{3+}}\).

\[\mathrm{Fe^{3+}(aq)+3OH^-(aq)\rightarrow Fe(OH)_3(s)}\]

(iii) On adding \(\mathrm{NaOH(aq)}\):

1. a reddish-brown precipitate of \(\mathrm{Fe(OH)_3}\) is formed; and
2. the red colour fades or disappears, leaving a colourless solution above the precipitate.

Bayanin Amsa

(a) Empirical and molecular formula of X (40% C, 6.67% H, O = 100 - 40 - 6.67 = 53.33%)

(i) Empirical formula = CH₂O (mass = 30).
(ii) \(\dfrac{60}{30}=2\), so molecular formula = C₂H₄O₂.

(b) Alkanoic acid Y, M_r = 74. For C_nH_2n+1COOH, propanoic acid C₂H₅COOH has M_r = 74.

(i) Functional group: carboxyl group, -COOH. (ii) Converting Y to an alkanoate (salt) is a neutralization reaction. (iii) Structural formula of Y: CH₃CH₂COOH. (iv) 2CH₃CH₂COOH + 2Na → 2CH₃CH₂COONa + H₂.

(v) X (C₂H₄O₂) is ethanoic acid; Y is propanoic acid, in the same homologous series. Y has a larger relative molecular mass (74 > 60) and longer chain, so stronger van der Waals forces; its boiling point will be higher than 118°C.

(c)(i) Fermentation is the slow chemical breakdown of a complex organic substance (such as sugar) into simpler substances (ethanol and carbon(IV) oxide) by the action of enzymes from micro-organisms such as yeast. (ii) C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. (iii) Yeast (which supplies the enzyme zymase) must be added. (iv) The palm wine keeps fermenting and produces carbon(IV) oxide gas; because the bottle is full to the brim and tightly corked, the gas cannot escape, so the pressure builds up until it exceeds the strength of the bottle and it shatters.

Bayanin Amsa

(a) (i) Used in bleaching: sulphur(IV) oxide, SO₂. (ii) Reddish-brown gas: nitrogen(IV) oxide, NO₂. (iii) Amphoteric (reacts with NaOH and HCl): aluminium oxide, Al₂O₃ (or ZnO). (iv) Dissolves in water to give pH greater than 7: sodium oxide, Na₂O (or CaO). (v) Oxidizes hot concentrated HCl to chlorine: manganese(IV) oxide, MnO₂.

(b)(i) Three methods to remove hardness due to Ca(HCO₃)₂ (temporary hardness): boiling; adding slaked lime (Clark's method); adding sodium trioxocarbonate(IV) (washing soda).

(ii) Sulphur(IV) oxide pollution forms acid with rain water (SO₂ + H₂O → H₂SO₃, and this is oxidized to H₂SO₄). This acid attacks limestone (CaCO₃), wearing the building away: CaCO₃ + H₂SO₃ → CaSO₃ + H₂O + CO₂. The limestone is eroded, so it is unwise to build with it there.

(c)(i) Two potassium compounds that yield oxygen on strong heating: potassium trioxochlorate(V), KClO₃, and potassium trioxonitrate(V), KNO₃ (or KMnO₄).

(ii) Amount of gas in 250 cm³ at s.t.p.:

(iii) 250 cm³ at s.t.p. (273 K) heated to 27°C (300 K) at constant pressure:

(iv) On heating at constant pressure the molecules gain kinetic energy and move faster, colliding more frequently and more forcefully; the gas expands (volume increases) so that the number of collisions per unit area of the walls, and hence the pressure, stays constant.

Bayanin Amsa

(a) In the treatment of river water for town supply:

(i) Sand bed: used for filtration to remove suspended solid particles.

(ii) Alum: used as a coagulant to cause fine suspended and colloidal particles to clump together and settle.

(iii) Chlorine: used for sterilization; it kills harmful bacteria and other micro-organisms.

(b)(i) Three major uses of H₂SO₄:

• Manufacture of fertilizers.
• As an electrolyte in lead-acid batteries.
• Manufacture of detergents, pigments and plastics.

(ii) Concentrated H₂SO₄ is a strong dehydrating agent. It removes water from the blue litmus paper, leaving carbon behind; hence, the paper becomes black and charred. Dilute H₂SO₄ produces H₃O⁺ ions in water, which turn blue litmus red. It does not char the paper because it is not sufficiently dehydrating.

(iii)

Zn + 2H₂SO₄(conc.) → ZnSO₄ + SO₂ + 2H₂O

(c)(i) Two gaseous pollutants generated by burning coal are sulphur(IV) oxide, SO₂, and carbon(II) oxide, CO.

(ii) When coal is broken into smaller pieces, it has a larger surface area exposed to oxygen in the air. This increases the rate of combustion, so it burns more easily than coal in lump form.

(iii) Methane, CH₄ (firedamp), is responsible for most explosions in coal mines.

(iv) The non-volatile residue left after the destructive distillation of coal is coke.

(d)(i) Allotropy is the existence of an element in two or more different forms in the same physical state.

(ii) The two crystalline allotropes of carbon are diamond and graphite.

(iii) Two elements apart from carbon that exhibit allotropy are sulphur and phosphorus.

(iv)

Mass of one mole of C₆₀ = 60 × 12 g = 720 g mol⁻¹.

Bayanin Amsa

(a)(i) Two reasons aluminium is preferred to copper for overhead cables: aluminium is much lighter (lower density), so the cables sag less and need fewer supports; aluminium is cheaper and more abundant than copper (while still being a good conductor).

(ii) Electrolytic extraction of aluminium: Purified alumina (Al₂O₃) is dissolved in molten cryolite (Na₃AlF₆) to lower the melting point and improve conductivity. The molten mixture is electrolyzed in a steel tank lined with graphite (the cathode), with carbon anodes.
Cathode: Al³⁺ + 3e^- → Al (molten aluminium collects at the bottom and is tapped off).
Anode: 2O^2- → O₂ + 4e^-; the oxygen burns the carbon anodes to CO₂, so they are replaced from time to time.

(b) In copper purification the impure copper is the anode and pure copper is the cathode, in copper(II) tetraoxosulphate(VI) solution.
(i) The cathode (pure copper) increases in mass, because Cu²⁺ ions are discharged and deposited on it (Cu²⁺ + 2e^- → Cu).
(ii) Oxidation occurs at the anode, because there copper atoms lose electrons to form ions (Cu → Cu²⁺ + 2e^-).
(iii) III is the electrolyte, copper(II) sulphate solution. Its blue colour does not fade because the rate at which Cu²⁺ ions leave the solution at the cathode equals the rate at which Cu²⁺ ions enter the solution from the dissolving anode, so [Cu²⁺] stays constant.

(c) Deposit 8.00 g Ca in 1 h 15 min = 4500 s, Ca²⁺ + 2e^- → Ca.

Bayanin Amsa

(a) Identifying the elements from configuration (not required, but shown for clarity)

• P: 1s\(^2\)2s\(^2\)2p\(^2\) (Z = 6, group IV)
• Q: 1s\(^2\)2s\(^2\)2p\(^4\) (Z = 8, group VI)
• R: 1s\(^2\)2s\(^2\)2p\(^6\) (Z = 10, group VIII, noble gas)
• S: 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\) (Z = 12, group II)
• T: 1s\(^2\)2s\(^2\)2p\(^6\)3s\(^2\)3p\(^5\) (Z = 17, group VII)

(i) Belongs to group VI: Q (six outermost electrons, 2s\(^2\)2p\(^4\)).

(ii) Strongly metallic: S (a group II element that readily loses electrons).

(iii) Readily ionises by gaining one electron: T (needs one electron to complete its octet, forming a 1\(^-\) ion).

(iv) Two unpaired electrons in the ground-state atom: P and Q. In 2p\(^2\) the two electrons occupy separate orbitals (Hund's rule); in 2p\(^4\) one orbital is doubly filled leaving two singly-occupied orbitals - both give two unpaired electrons.

(v) Readily loses two electrons in bonding: S (3s\(^2\) outer electrons lost to form a 2\(^+\) ion).

(vi) Does not take part in chemical reactions: R (a stable, filled-shell noble gas).

(vii) An s-block element: S (its outermost electrons enter the 3s sub-shell).

(b)(i) Completed table

Neutrons = mass number - atomic number. For Al\(^{3+}\), three electrons are lost (13 - 3 = 10). For the oxide ion, two electrons are gained (8 + 2 = 10).

(b)(ii) Trend in atomic radius

Down a group the number of occupied electron shells increases and inner-shell shielding increases, so the outermost electrons lie farther from the nucleus and the radius increases. Across a period electrons are added to the same shell while the nuclear charge rises; the greater effective nuclear attraction pulls the shell inward, so the radius decreases.

(c)(i) Half-life

The half-life of a radioactive element is the time taken for half the atoms (nuclei) in a given sample to decay (disintegrate).

(c)(ii) Alpha decay of polonium-210

(d) Type of bonding

• (i) Chlorine as discrete molecules: covalent bonding (shared electron pair within each Cl\(_2\) molecule).
• (ii) Sodium chloride dissolving readily in water: ionic (electrovalent) bonding (the polar water molecules hydrate the Na\(^+\) and Cl\(^-\) ions).
• (iii) Deep-blue complex ion with excess ammonia: co-ordinate (dative) covalent bonding (lone pairs on N donate into the Cu\(^{2+}\) ion).