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Question 1

Topic: 9.2

Vanadium, niobium and tantalum are metals in the same group of the Periodic Table.
(a) The shorthand electronic configuration of vanadium in the ground state is \([Ar]3d^34s^2\).
(i) State what is meant by the term ground state.
(ii) Show the electronic configuration of vanadium using electrons in boxes notation.

Electronic Configuration Notation

(iii) Deduce the total number of electrons in the p sub-shells of a vanadium atom.
(b) Pelopium was the suggested name for a new element discovered in a mineral. Pelopium was later found to be a mixture of niobium, Nb, and tantalum, Ta. Only one naturally occurring isotope exists for each of Nb and Ta.
(i) Complete Table 1.1.

Table 1.1

(ii) Define relative isotopic mass.

(iii) A sample of pelopium contains 90.9% by mass \(^{93}_{41}Nb\) and 9.1% by mass \(^{181}_{73}Ta\). Calculate the theoretical relative atomic mass of pelopium based on these data and Table 1.1. Give your answer to two decimal places. Show your working.

▶️ Answer/Explanation
Solution

(a)(i) The ground state refers to the lowest energy state of an atom where all electrons occupy the lowest available orbitals.

(a)(ii) The electronic configuration of vanadium in boxes notation is:

Electronic Configuration Boxes

(a)(iii) Vanadium has 12 electrons in the p sub-shells (from the argon core \([Ar]\), which includes \(2p^6\) and \(3p^6\)).

(b)(i) The completed Table 1.1 is:

Completed Table 1.1

(b)(ii) Relative isotopic mass is the mass of an isotope relative to \(\frac{1}{12}\) the mass of a carbon-12 atom.

(b)(iii) The relative atomic mass of pelopium is calculated as:
\( (92.91 \times 0.909) + (180.95 \times 0.091) = 84.46 + 16.46 = 100.92 \).
Final Answer: \(\boxed{100.92}\)

Question 2

Topic: 11.4

Oxygen is a Group 16 element.

(a) (i) Write equations for the following reactions.
• sodium and oxygen
• sulfur and oxygen
(ii) Draw a dot-and-cross diagram to show the species present in Al₂O₃. Draw outer electrons only.

(iii) The maximum oxidation state of the Period 3 elements in their oxides varies across the period. State and explain the variation.

(b) \(H_2O\) reacts with both inorganic and organic compounds.
(i) Complete Table 2.1 to give details of the reactions of some Period 3 oxides with \(H_2O\).

(ii) Write an equation for the reaction of CH₃CN with H₂O in acidic conditions.

(iii) Draw the structures of the two alcohols formed in the reaction shown in equation 1.

(iv) Explain why alcohols are less acidic than water.

(c) Fig. 2.1 shows the boiling points of \(H_2O\) and other Group 16 hydrides.

(i) Explain the trend in the boiling points of the Group 16 hydrides H₂S to H₂Te.
(ii) Explain why the boiling point of \(H_2O\) is much higher than that of \(H_2S\).

▶️ Answer/Explanation
Solution

(a)(i)

• Sodium and oxygen: \(4Na + O_2 \rightarrow 2Na_2O\)
• Sulfur and oxygen: \(S + O_2 \rightarrow SO_2\)

Explanation: Sodium reacts with oxygen to form sodium oxide (\(Na_2O\)), while sulfur burns in oxygen to produce sulfur dioxide (\(SO_2\)).

(a)(ii)

Explanation: The dot-and-cross diagram for \(Al_2O_3\) shows aluminum donating 3 electrons and oxygen accepting 2 electrons, forming an ionic lattice.

(a)(iii)

The maximum oxidation state increases across Period 3.

Explanation: As the number of valence electrons increases, elements can lose, share, or donate more electrons, leading to higher oxidation states in their oxides.

(b)(i)

Explanation: The table is completed with the correct reactions of Period 3 oxides with water, including their pH and product types.

(b)(ii)

\(CH_3CN + 2H_2O \rightarrow CH_3COOH + NH_3\) (in acidic conditions)

Explanation: Acetonitrile (\(CH_3CN\)) hydrolyzes in acidic water to form acetic acid (\(CH_3COOH\)) and ammonia (\(NH_3\)).

(b)(iii)

Structures: Methanol (\(CH_3OH\)) and Ethanol (\(C_2H_5OH\)).

Explanation: The reaction produces two alcohols, methanol and ethanol, as shown in the given equation.

(b)(iv)

Alcohols are less acidic than water because alkyl groups are electron-donating, strengthening the O—H bond and making \(H^+\) less likely to be donated.

Explanation: The inductive effect of alkyl groups reduces the acidity of alcohols compared to water.

(c)(i)

The boiling points increase from \(H_2S\) to \(H_2Te\) due to stronger London dispersion forces as molecular size and electron count increase.

Explanation: Larger molecules have more electrons, leading to stronger intermolecular forces and higher boiling points.

(c)(ii)

\(H_2O\) has a much higher boiling point than \(H_2S\) due to hydrogen bonding, which is stronger than London dispersion forces.

Explanation: The high electronegativity of oxygen enables hydrogen bonding in \(H_2O\), requiring more energy to break these bonds.

Question 3

Topic: 11.2

Nitrogen and phosphorus are elements in Group 15 of the Periodic Table.

(a) Nitrogen is found in inorganic compounds such as nitrogen oxides (\(NO_x\)), nitrates and nitric acid.
(i) Identify one natural and one man-made occurrence of nitrogen oxides in the atmosphere.
(ii) Write an equation to describe the role of NO₂ in the direct formation of acid rain.
(iii) Peroxyacetyl nitrate, PAN, is a component of photochemical smog. Describe how PAN forms from NO₂.
(iv) Nitric acid reacts with basic oxides to form nitrates. Write an equation for the reaction of nitric acid with calcium oxide.
(v) Describe what is seen when solid calcium nitrate is heated strongly.
(b) A common test for nitrates is the reaction with NaOH and Al. Equation 1 shows the reaction.

(i) Deduce the oxidation state of nitrogen in \(NO_3^–\).
(ii) Identify the species that is oxidised in equation 1.
(iii) NH3 is a basic gas. Describe how NH₃ is able to act as a base.

(iv) Suggest the shape of the [Al(OH)₄]⁻ ion.

(c) Fig. 3.1 shows a sketch of some of the ionisation energies of phosphorus, P.

(i) Construct an equation to represent the third ionisation energy of P.
(ii) Complete the graph in Fig. 3.1 to show the third to sixth ionisation energies of P.

(d) Complete Table 3.1 to show the properties of nitrogen and phosphorus in their standard states.

(e) A form of solid nitrogen has a lattice structure similar to solid iodine. Identify the type of lattice structure of solid nitrogen.
(f) At very high temperatures, phosphorus can form \(P_2\) molecules. \(P_2\) contains a triple bond, \(P \equiv P\).
(i) Describe the formation of the \(P \equiv P\) bond in terms of orbital overlap.
(ii) The bond energy of \(P \equiv P\) is 485 kJ mol⁻¹. The bond energy of \(N \equiv N\) is 944 kJ mol⁻¹. Compare the reactivity of \(P_2\) and \(N_2\). Explain your answer.

▶️ Answer/Explanation
Solution

(a)(i) Natural: Lightning discharges. Man-made: Combustion of fossil fuels.

Explanation: Nitrogen oxides form naturally during lightning strikes due to high temperatures, while man-made sources include vehicle emissions and industrial processes.

(a)(ii) \(2NO_2 + H_2O \rightarrow HNO_3 + HNO_2\)

Explanation: NO₂ reacts with water to form nitric acid (HNO₃) and nitrous acid (HNO₂), contributing to acid rain.

(a)(iii) PAN forms when NO₂ reacts with volatile organic compounds (VOCs) in sunlight.

Explanation: Photochemical smog involves NO₂ and hydrocarbons undergoing reactions under sunlight to produce PAN.

(a)(iv) \(2HNO_3 + CaO \rightarrow Ca(NO_3)_2 + H_2O\)

Explanation: Nitric acid neutralizes calcium oxide, forming calcium nitrate and water.

(a)(v) Brown fumes of NO₂ gas are observed.

Explanation: Heating calcium nitrate decomposes it into calcium oxide, nitrogen dioxide (NO₂), and oxygen.

(b)(i) +5

Explanation: The oxidation state of N in \(NO_3^-\) is calculated as \(x + 3(-2) = -1\), giving \(x = +5\).

(b)(ii) Aluminium (Al)

Explanation: Al is oxidized from 0 to +3, losing electrons in the reaction.

(b)(iii) NH₃ donates a lone pair of electrons to protons (H⁺).

Explanation: Ammonia acts as a base by accepting protons using its lone pair on nitrogen.

(b)(iv) Tetrahedral

Explanation: The [Al(OH)₄]⁻ ion has four bonding pairs around Al, adopting a tetrahedral geometry.

(c)(i) \(P^{2+}(g) \rightarrow P^{3+}(g) + e^-\)

Explanation: The third ionisation energy removes an electron from \(P^{2+}\) to form \(P^{3+}\).

(c)(ii) Graph shows increasing ionisation energies with a large jump after the 5th ionisation.

Explanation: The graph reflects the removal of electrons from inner shells, requiring more energy.

(d) See image for completed table.

Explanation: Nitrogen is a diatomic gas (N₂), while phosphorus exists as P₄ molecules in a solid state.

(e) Molecular lattice

Explanation: Solid nitrogen, like iodine, forms a molecular lattice held by weak van der Waals forces.

(f)(i) Overlap of three p-orbitals from each P atom forms a triple bond.

Explanation: The \(P \equiv P\) bond involves one sigma and two pi bonds from p-orbital overlap.

(f)(ii) \(P_2\) is more reactive than \(N_2\) due to its weaker triple bond (485 vs. 944 kJ mol⁻¹).

Explanation: The lower bond energy of \(P \equiv P\) makes it easier to break, increasing reactivity compared to \(N \equiv N\).

Question 4

Topic: 16.1

Bromoalkanes are used widely in industry, although there is increasing concern about their environmental impact.

(a) Complete Fig. 4.2 to show the mechanism for the formation of 1,2-dibromoethane in reaction 1. Include charges, dipoles, lone pairs of electrons and curly arrows as appropriate.

(b) The enthalpy change of reaction 1, \(\Delta H_r\) = –90.0kJmol⁻¹.

The enthalpy change of formation of ethene, \(ΔH_f\) = +52.2 kJ mol⁻¹. Calculate the enthalpy change of formation of 1,2-dibromoethane.

(c) (i) Complete Fig. 4.1 to:
• draw the structure of compound A
• name compound B.
(ii) Draw the structure of one repeat unit of polymer C in the box.

(iii) In reaction 5, compound B reacts with an excess of NaOH dissolved in ethanol. The products are HBr, \(H_2O\), and an unsaturated hydrocarbon D. Suggest the identity of D.

(d) Compound E is the only isomer of 1,2-dibromoethane. Alkaline hydrolysis of E gives compound F.

(i) Identify the type of isomerism shown by E and 1,2-dibromoethane.
(ii) Name the homologous series that F belongs to.
(iii) Complete Table 4.1 to state what is observed when F reacts with the reagents listed.

(e) Compound F reacts with reagent G to form compound H.

The infrared spectrum of H is shown in Fig. 4.3.

H also shows a molecular ion peak at m/e = 60 in its mass spectrum.
(i) Use the information in (e), Fig. 4.3 and Table 4.2 to deduce the structure of H. Explain your answer fully.

(ii) Suggest the role of reagent G.

▶️ Answer/Explanation
Solution

(a)

Explanation: The mechanism involves electrophilic addition where the π bond in ethene attacks Br₂, forming a bromonium ion intermediate. The Br⁻ then attacks from the opposite side, resulting in 1,2-dibromoethane.

(b) \(\Delta H_f\) of 1,2-dibromoethane = –142.2 kJ mol⁻¹.

Explanation: Using Hess’s Law, \(\Delta H_r = \Delta H_f(\text{products}) – \Delta H_f(\text{reactants})\). Given \(\Delta H_r = -90.0\) kJ mol⁻¹ and \(\Delta H_f(\text{ethene}) = +52.2\) kJ mol⁻¹, solving gives \(\Delta H_f(\text{1,2-dibromoethane}) = -90.0 – 52.2 = -142.2\) kJ mol⁻¹.

(c)(i) A: \(CH_2BrCH_2Br\), B: 1,2-dibromoethane.

(ii) Polymer C repeat unit: \(-CH_2-CH_2-\)

(iii) D: \(C_2H_2\) (ethyne).

Explanation: Elimination of HBr from 1,2-dibromoethane in the presence of NaOH/ethanol produces ethyne.

(d)(i) Structural/positional isomerism.

(ii) Aldehyde.

(iii)

Explanation: F (ethanal) forms a silver mirror with Tollens’ reagent and a red precipitate with Fehling’s solution due to its aldehyde group.

(e)(i) H: \(CH_3COOH\) (ethanoic acid).

Explanation: The IR spectrum shows a broad O-H stretch (~3000 cm⁻¹) and a C=O stretch (~1700 cm⁻¹), consistent with a carboxylic acid. The mass spectrum peak at m/e = 60 matches ethanoic acid’s molecular weight.

(ii) Oxidising agent.

Explanation: Reagent G (e.g., acidified KMnO₄ or K₂Cr₂O₇) oxidises the aldehyde (F) to a carboxylic acid (H).

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