9701_m20_qp_42
A paper of Chemistry, 9701
Questions:
6
Year:
2020
Paper:
4
Variant:
2
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1
Iron is a transition element in the fourth period. Iron forms compounds containing the ions Fe2+ and Fe3+. Define the term transition element. Compare the melting point and density of iron with those of calcium, an s-block element in the fourth period. melting point density  Complete the electronic configuration of an isolated gaseous Fe2+ ion. 1s2 Aqueous Fe3+ ions form coloured complexes. Explain the origin of the colour in transition element complexes. When an excess of CN–ions is added to green [Fe(H2O)6]2+ions, yellow [Fe(CN)6]4– complex ions are formed. Heating [Fe(CN)6]4– with dilute nitricacid and then neutralising the product with Na2CO3produces red crystals, containing the [Fe(CN)5NO]2– complex ion. NO is a neutral, monodentate ligand. State the shape of the [Fe(H2O)6]2+complex ion. Write the equation for the reaction between [Fe(H2O)6]2+ions and an excess of CN–ions. Deduce the oxidation states of iron in: [Fe(CN)6]4– [Fe(CN)5NO]2–.  Define the term monodentate ligand. Complete the diagram to show the three-dimensional structure of the [Fe(CN)5NO]2– complex ion. Fe  The two complex ions [Fe(CN)6]4– and [Fe(CN)5NO]2– are different colours. Explain why the colours of the two complex ions are different. E is a complex ion, [Fe(C2O4)2Cl 2]4–, containing Fe2+ with a coordination number of 6. Define the term coordination number. E shows both optical isomerism and cis-trans isomerism. One isomer of E is shown. The C2O4 2– ion is represented as ox . In the boxes, draw three-dimensional diagrams to show: ● ● the trans isomer of E ● ● the optical isomer of E. Fe trans isomer 4– Fe optical isomer 4– Fe cis isomer E Cl Cl 4– ox ox  [Fe(C2O4)2Cl 2]4– contains ligands which are anions of ethanedioic acid, HO2CCO2H. Complete the table to show any observations for the reactions of HO2CCO2H with the named reagents. Where no change is observed, write ‘none’. reagent observations with HO2CCO2H warm acidified manganate(2,4-dinitrophenylhydrazine warm Tollens’ reagent  
1. Atomic structure  →  1.3. Electrons, energy levels and atomic orbitals
2. Atoms, molecules and stoichiometry  →  2.3. Formulas
11. Group 17  →  11.3. Some reactions of the halide ions
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Group2 metals form stable carbonates and sulfates. State and explain the trend in the thermal stability of the Group2 carbonates down the group. The sulfates of Group2 elements become less soluble down the group. Explain this trend. Aluminium is extracted from Al 2O3 by electrolysis. Al 2O3 is dissolved in cryolite in this process. The half-equation for the reaction at the anode is shown. O2– + C → CO + 2e– Use this half-equation to write the ionic equation for the electrolysis of Al 2O3. Aluminiumoxide is electrolysed for 3.0hours using carbon electrodes and a current of 3.5×105A. Calculate the mass of aluminium that is formed.  mass of aluminium = g Cryolite can be made from SiF4. The first step in this conversion is the reaction of SiF4 with H2O, forming H2SiF6 and SiO2. Write an equation for this reaction. 
10. Group 2  →  10.1. Similarities and trends in the properties of the Group 2 metals, magnesium to barium, and their compounds
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Gold is an unreactive metal that can only be oxidised under specific conditions. The standard electrode potential, E o, of Au3+/ Auis +1.50 V. Define the term standard electrode potential. Draw a fully labelled diagram of the apparatus that should be used to measure the standard cell potential, , of Au3+/ Auand HNO3/ NO. Include all necessary chemicals.  Some relevant half-equations and their standard electrode potentials are given. half-equation E o / V Au3++ 3e– Au+1.50 [AuCl 4]–+ 3e– Au+ 4Cl –+1.00 NO3 –+ 4H++ 3e– NO+ 2H2O+0.96 Write an ionic equation to show the spontaneous reaction that occurs when an electric current is drawn from the cell in . Calculate the of the reaction in .  = V Gold can be oxidised by a mixture of concentrated hydrochloricacid and concentrated nitric acid, known as aqua regia. Concentrated hydrochloric acid is 12 mol dm–3. Concentrated nitricacid is 16 mol dm–3. Explain why aquaregia is able to dissolve gold. In your answer, state and explain what effect the use of concentrated hydrochloricacid and concentrated nitricacid have on the E values of half-equations 2 and 3. Aqueous gold(chloride, AuCl 3, reacts with aqueous hydrogenperoxide, H2O2, under certain conditions, forming Au, O2 and HCl. A student carries out separate experiments using different initial concentrations of AuCl 3 and H2O2. The initial rate of each reaction is measured. The table shows the results that are obtained. experiment [AuCl 3] / mol dm–3 [H2O2] / mol dm–3 rate of production of O2/ dm3 minute–1 0.05 0.50 7.66 × 10–2 0.10 0.50 1.53 × 10–1 0.15 1.00 4.60 × 10–1 Write an equation for the reaction of AuCl 3 with H2O2. Determine the rate equation of the reaction. Show your reasoning, quoting data from the table. Use the results of experiment2 to calculate the value of the rate constant, k, for this reaction. Include the units of k.  rate constant, k =  units =  Al F3 is an ionic compound. The Born–Haber cycle for the formation of Al F3 is shown. Al 3++ 3F+ 3e– Al + 3FAl + 1.5F2Al + 1.5F2Al F3Al 3++ 3F–∆H3 ∆H2 ∆H1 ∆H4 ∆H5 ∆H6 Name the enthalpy changes labelled ∆H4 and ∆H6. ∆H4 = ∆H6 =  Use the data in the table and data from the Data Booklet to calculate the lattice energy of Al F3. process enthalpy change / kJ mol–1 Al → Al +326 Al → Al 3++5137 F→ F––328 Al + 1.5F2→ Al F3–1504  lattice energy of Al F3 = kJ mol–1 Scandiumfluoride, ScF3, is an ionic compound. Use data from the Data Booklet to suggest how the lattice energy of Al F3 compares with the lattice energy of ScF3. Explain your answer. Al F3 is sparingly soluble in water. The concentration of its saturated solution at 298 K is 6.5×10–2 mol dm–3. Write an expression for the solubility product, Ksp, of Al F3. Ksp = Calculate the numerical value of Ksp for Al F3 at 298 K.  Ksp =  
9. The Periodic Table: chemical periodicity  →  9.2. Periodicity of chemical properties of the elements in Period 3
11. Group 17  →  11.4. The reactions of chlorine
11. Group 17  →  11.3. Some reactions of the halide ions
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CompoundF has been found in small quantities in some cereals and dried fruit. O O F OH Cl O N H O O A B Give the name of the functional groups labelled A and B. A B  State the number of chiral carbon atoms in one molecule of F. F can be hydrolysed by heating with an excess of dilute hydrochloricacid, as shown. Three products are formed: G and two others. O O F OH Cl O N H O O Cl excess dilute HCl + + O O G OH OH HO OH Draw the structures of the other products of the reaction in the boxes provided. CompoundH is formed in one step of a different synthesis, as shown. O O OH OH HO O O H OH Cl OH HO Cl 2 and FeCl 3 State the role of FeCl 3 in this step. Use the Data Booklet to suggest two reasons why the chlorine atom in compound H substitutes into the ring at the position shown, instead of the other positions in the ring.  CompoundJ, CxHyOz, is also found in some cereals. Part of the mass spectrum of J is shown. The M and M+1 peaks are labelled, along with their relative intensities. relative intensity m/e M, relative intensity 100.0 M+1, relative intensity 14.4 Calculate the number of carbon atoms, x, present in J.  x = The mass spectrum has a peak at m/e = 205. Suggest the identity of the fragment lost from J to form this peak. 
22. Analytical techniques  →  22.2. Mass spectrometry
13. An introduction to AS Level organic chemistry  →  13.1. Formulas, functional groups and the naming of organic compounds
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Gallic acid, C7H6O5, is a naturally occurring aromatic molecule. gallic acid HO HO HO O OH Gallicacid contains the carboxylicacid and phenol functional groups. State and explain the relative acid strength of these two functional groups. A buffer solution was prepared by dissolving 2.04 g of gallic acid in 250 cm3 of a solution containing 0.0600 mol dm–3 of gallate ions, C7H5O5 –. C7H6O5 C7H5O5 – + H+ Ka = 3.89 × 10–5 mol dm–3 at 298 K Define the term buffer solution. Calculate the pH of this buffer solution.  pH = Write two equations to show how a solution containing gallic acid, C7H6O5, and gallate ions, C7H5O5 –, acts as a buffer. CompoundK is used as the starting material in a synthesis of gallic acid. A student suggested the first two steps of the synthesis could be as shown. K COOH NH2 L COOH NO2 NH2 concentrated HNO3 and concentrated H2SO4 step 1 concentrated HNO3 and concentrated H2SO4 step 2 M COOH NO2 O2N NH2 Nitronium ions, NO2 +, are generated by the reaction between concentrated sulfuricacid and concentrated nitricacid. Construct an equation for the formation of NO2 + by this method. Complete the mechanism and draw the intermediate of step1. Include all relevant charges and curly arrows to show the movement of electron pairs. COOH NH2 NO2 L + H+ intermediate +  State the name of the mechanism in . CompoundM is converted into compoundP as shown. CH3CH2OH and H2SO4 step 3 step 4 M COOH NO2 O2N NH2 N COOC2H5 NO2 O2N NH2 P COOC2H5 NH2 H2N NH2 State the reagents and conditions for step4. P reacts with an excess of sodiumnitrite, NaNO2, and dilute HCl at 5 °C to form compoundQ, C9H7N6O2Cl 3. CompoundQ is then converted into gallicacid. Q C9H7N6O2Cl 3 P COOC2H5 NH2 H2N NH2 an excess of NaNO2 and dilute HCl at 5 °C step 5 step 6 gallic acid HO HO HO O OH  Suggest the structure of compoundQ in the box provided. State the reagents and conditions for step6. State the number of peaks that would be observed in the 13C NMR spectrum of gallicacid. gallic acid HO HO HO O OH The proton NMR spectrum of gallic acid dissolved in D2O is recorded. ● ● Predict the number of peaks observed and any expected splitting pattern. ● ● State the expected chemical shift range (δ) of each peak predicted. 
18. Carboxylic acids and derivatives  →  18.1. Carboxylic acids
11. Group 17  →  11.3. Some reactions of the halide ions
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Valinol can be synthesised by the following reactions. Reaction 1 uses valine as the starting material. C HO O C H valine valinol NH2 C H H C CH3 H H C H NH2 C H H Cl C H H C CH3 H H C H NH2 C H H HO C H H C CH3 H H reaction 1 [H] reaction 2 NaOHWrite an equation for reaction1, using [H] to represent the reducing agent. Suggest a suitable reagent for reaction1. Name the mechanism for reaction2. Valine and glycine, H2NCH2COOH, form the tripeptide Gly–Val–Gly. Draw the structure of this tripeptide. Show the peptide bonds fully displayed.  Valine exists as two stereoisomers. Draw three‑dimensional diagrams to show the two stereoisomers of valine. In your diagrams, the –CH(CH3)2 group can be represented by –R. State the type of stereoisomerism shown. type of stereoisomerism  Valine is an amino acid. Draw the zwitterion of valine.  Valinate, Val–, is the anion of valine. It takes part in a ligand substitution reaction with hexaaquanickel(ions. ComplexZ is formed. Z [Ni(H2O)6]2++ 2Val–[Ni(H2O)2(Val)2]+ 4H2OWrite an expression for Kstab for this equilibrium. Kstab =  At room temperature, the numerical value of Kstab is 2.34×105. Explain what this value indicates about the equilibrium and the stability of complexZ. Z is an octahedral complex with formula [Ni(H2O)2(Val)2]. Use this information to state the type of ligand that the valinate ion is acting as in this complex. 
11. Group 17  →  11.3. Some reactions of the halide ions
15. Halogen compounds  →  15.1. Halogenoalkanes
17. Carbonyl compounds  →  17.1. Aldehydes and ketones
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