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Inorganic chemistry revision package solutions


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ER for 08 inorganic chem

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Inorganic chemistry revision package solutions

  1. 1. GROUP IIAJC/P2/3CJC/P31 (a) (i) MgO (s) + 2HNO3 (aq) → Mg(NO3)2 (aq) + H2O (l) 0.500 Amount of MgO = = 0.0124 mol 24.3 + 16.0 50 Amount of HNO3 = × 0.100 = 0.00500 mol (limiting reagent) 1000 0.005 Amount of Mg(NO3)2 produced = = 0.00250 mol 2 (a) (ii) Mg(OH)2 + 2NH4NO3 → Mg(NO3)2 + 2NH3 + 2H2O 1.5 Amount of Mg(NO3)2 = = 0.0101 mol 24.3 + 28.0 + 96.0 Volume of NH3 produced = 0.0101 × 2 × 24 = 0.485 dm3 (iii) 2Mg(NO3)2 → 2MgO + 4NO2 + O2 Brown gas of NO2 will be observed. (iv) Ba(NO3)2 is more thermally stable than Mg(NO3)2, hence it will not decompose at 330°C. Ba2+ is less polarising than Mg2+ as it has a larger size, hence the NO3- is polarised to a smaller extent in Ba(NO3)2.YJC/ JC2/ 2008 1
  2. 2. HCI/P2 ORYJC/ JC2/ 2008 2
  3. 3. RJC/P2/Q3 heat 1c(i) M(NO3)2(s)  MO(s) + 2NO2(g) + → O2(g) 2 (ii) Amt of nitrate = 1/2 x Amt of NO2 = ½ x 1.84/(14.0 + 2(16.0)) = 0.0200 mol Mr of nitrate = 2.96/0.0200 = 148.0 g mol─1 Mr of M = 148.0 – 2(14.0 + 3 x 16.0) = 24.0 g mol─1 Hence, M is magnesium.(d) Lattice energy cannot be used to determine relative thermal stability of Group II nitrates. Barium nitrate has a higher decomposition temperature because Ba2+ is a larger ion or Ba2+ has a lower charge density, therefore Ba2+ has lower polarising power. Hence, Ba2+ polarises the electron cloud of the nitrate ion less strongly and weakens the N–O bonds to a lesser extent in barium nitrate.IJC/P3/Q3YJC/P2/3ai MagnesiumMg(s) + H2O(l) MgO(s) + H2(g)Mg reacts with slowly with cold water to give H2(g). Y, Y(s) + H2O(l) Y(OH)2(s) + H2(g), Y reacts vigorously with cold water to give hydrogen gasii. The more negative the reduction potential, the greater the reducing power as the valence electrons are more readily lost.b. Down the gp, size of cation increases and the charge density decreases, polarising power of Y2+ decreases [1m] and C-O bond is less polarised [1m] and require more energy to break, therefore it has a high decomposition By losing 2 electrons, Be forms a stable duplet structure ii Be2+ has a highest charge density (high charge and smallest size). It polarises the oxide ion significantly which leads to covalency . Hence, the oxide is ionic with covalent character and is amphoteric.YJC/ JC2/ 2008 3
  4. 4. TJC/P32a) • Down Group II, the thermal stability of the nitrates increases. • Down the group, since the charge density of the M2+ decreases, the M2+ becomes less polarising. Thus the covalent bonds within the NO3− anion is the least polarised with Ba2+ and hence would require greater heat energy to decompose into simpler substances. • 2 M(NO3)2 → 2 MO + 4 NO2 + O2b) • Mass of MO residue obtained = (100 − 65.8) × 1.00 100 = 0.342 g Number of moles of MO obtained = 0.342 ÷ (Ar M + 16.0) • Since M(NO3)2 ≡ MO, 1.00 ÷ (Ar M + 124.0) = 0.342 ÷ (Ar M + 16.0) • Ar M = 40.1 Hence the metal is • BaO + H2SO4 → BaSO4 + H2O • BaO2 + H2SO4 → BaSO4 + H2O2ii) Using the Data Booklet, H2O2 ≡ 2I− ≡ I2 • Number of moles of H2O2 produced = 1.18 × 10−3 mol Hence, number of moles of BaSO4 produced from BaO2 = 1.18 × 10−3 mol • Mass of BaO2 present in 1.00 g solid = 1.18 × 10−3 × 169.0 = 0.199 g [Mr BaO2 = 169.0] • Mass of BaO present in 1.00 g solid = 1.00 − 0.199 = 0.801 gVJC/P2/Q2/bYJC/ JC2/ 2008 4
  5. 5. IJC/P3MJC/P2/Q2biii.iiiYJC/ JC2/ 2008 5
  6. 6. GROUP VIICJC/P22 (a)(i) HCl is a weaker acid than HI because the H-Cl bond is stronger than H-I bond . Hence HCl dissociate to give H+ to a smaller extent than HI. (Or vice versa) HX + H2O H3O+ + X- (or) HX H+ + X- (ii) Cl- + H2SO4 HCl + HSO4- I- + H2SO4 HI + HSO4- 8 HI + H2SO4 4I2 + 4H2O + H2S Cl- is a weaker reducing agent than I-. con. sulphuric acid is strong enough to oxidise HI to I2AJC/P3NJC/P2/YJC/ JC2/ 2008 6
  7. 7. SRJC/P3/3 ai White / steamy fumes of HCl will be observed HCl is not oxidized by concentrated H2SO4, no Cl2 obtained. KCl (s) + H2SO4(l) KHSO4 (aq) + HCl (g) ii Concentrated H2SO4 reacts with KI to form HI. KI(s) + H2SO4 (aq) KHSO4 (aq) + HI (g) HI is then oxidized by concentrated H2SO4 to give the violet fumes, I2 and the pungent gas, SO2. 2HI (g) + H2SO4 (aq) I2 (g) + SO2 (g) + 2H2O (l) iii Test the fumes with moist starch paper.Starch paper should turn black/blue/blue-black. bi Bottle C is Cl2 (aq) which oxidized / displaced Br- (aq) to Br2 (aq) and I- (aq) to I2 (aq) which account for the brown mixture obtained. Cl2 (aq) + 2Br- (aq) 2Cl- (aq) + Br2 (aq) Cl2 (aq) + 2I- (aq) 2Cl- (aq) + I2 (aq) ii. Add hexane to Bottle A and B after reaction with Cl2 (aq). Shake and allow to stand The bottle with reddish brown organic layer contains KBr (aq) initially. The bottle with violet / purple organic layer contains NaI (aq) initially.YJC/ JC2/ 2008 7
  8. 8. IJC/P3/4 RJC/P2/3a i. Add hot, aqueous sodium hydroxide to aqueous chlorine, which will then disproportionate to form aqueous sodium chloride and aqueous sodium chlorate(V). ii. Ethanedioic acid is functioning as the reducing agent because the oxidation state of chlorine decreased from +5 in ClO3– to +4 in ClO2. iii. 2ClO2(g) + H2O(l)  ClO2–(aq) + ClO3–(aq) + 2H+(aq) E → ££¢¡  = –0.20 V Since E ££¢¡      = E R – E L +0.95 V – (+1.15 V) 0 V, the disproportionation of ClO2(aq) to ClO2–(aq) and ClO3–(aq) is energetically not feasible and would not occur under standard conditions. YJC/P3/33 (a) (i) Cl2 + 2OH- ClO- + Cl- + H2O [2] 3Cl2 + 6OH- ClO3- + 5Cl- + 3H2O [1 m for each equation] (ii) [1m] for multiplying 2 to amount of X- and SO42- [3] S2O32- : X- : SO42- : H+ 1 : 8 : 2 : 10 [1m] 4X2 + S2O32- + 5H2O 2SO42- + 8X- + 10H+ (1m) (iii) Y2 + 2S2O32- 2Y- + S4O62- [1m] [2] The change in oxidation state of S is smaller in (ii) compared to (iii). Hence X2 has stronger oxidising power. [1m] YJC/ JC2/ 2008 8
  9. 9. TPJ/P26(a) Boiling points increase as you go down group Electron cloud / size of molecule / atomic radius increases going down group Id-id VDW intermolecular forces increases going down group More energy required to overcome intermolecular forces (b) (i) H2 + Cl2 2HCl white fumes are evolved (ii) Going down group, Electron affinity decreases Strength of H-X bond decreases Tendency of halogen to be reduced decreases OR Oxidation strength of halogen decreases (c) (i) NaCl + H2SO4 NaHSO4 + HCl (ii) NaI + H2SO4 NaHSO4 + HI 8HI + H2SO4 4I2 + H2S + 4H2O Going down group, Reducing power of halogen increases HI is oxidized to iodine, hence pure HI cannot be obtained. HCI/P3/Q2YJC/ JC2/ 2008 9
  10. 10. TRANSITION METALSAJC/P3/2HCI/P3/Q3YJC/ JC2/ 2008 10
  11. 11. HCI/P3/4YJC/ JC2/ 2008 11
  12. 12. MJC/P2bibiiciiididiidiiiYJC/ JC2/ 2008 12
  13. 13. PJC/P3abiSAJC/P32(d) The five 3d orbitals of d-block free metal ions are degenerate (at same energylevel). When they are bonded to ligands, the electronic repulsion between theligands and the metal ion causes the 3d orbitals to undergo splitting, results in 2groups of non-degenerate orbitals. called the d-d splitting. By absorption ofenergy, electrons can be promoted from a lower energy d-orbital to a higherenergy d-orbital. This is called the d-d transition. This energy is related to thewavelength of the light absorbed, and the light not absorbed is thus seen as thecolour of the complexSc and Zn do not form coloured complexes because Sc3+ [Ar]3d0, does not have3d electrons and Zn2+ [Ar]3d10, has completely- filled 3d-subshell . Hence d-dtransition is not possible.2(e)(i) Copper(II) sulphate salts dissolve in water to give a blue colour solution,which is attributed to [Cu(H2O)6]2+ (aq)CuSO4 (s) + 6H2O(l) Cu(H2O)6]2+ (aq) + SO42- (aq) Blue solutionWhen dilute aqueous NH3 is gradually added, a blue precipitate of Cu(OH)2 is firstformed.NH3 + H2O ⇔ NH4+ + OH-[Cu(H2O)6]2+(aq) + 2OH-(aq) Cu(OH)2(s) + 6H2OYJC/ JC2/ 2008 13
  14. 14. blue ppt.When excess aqueous NH3 is added, the precipitate dissolves to give a deep bluesolution, due to the formation of the soluble complex ion, [Cu(NH3)]2+.Cu(OH)2 + 4NH3 + 2H2O ⇔ [Cu(NH3)4(H2O)2]2+ + 2OH- deep blue soln.NH3 is a stronger ligand than H2O and so, displaces water from [Cu(H2O)6]2+ toform [Cu(NH3)4(H2O)2]2+.NB: The identification of the coloured compounds is important.2(f) The uncatalysed reaction is slow due to high activation energy since 2negatively ions S2O82- and I- are involved and they repel each other.The Co3+ acts as a homogenous catalyst. The catalysed pathway involves twosteps:Step 1: 2Co3+ + 2I- 2Co2+ + I2 E(cell)= +1.28V 2+ 2-Step 2: 2Co + S2O8 2Co3+ + 2SO42- E(cell) = +0.19VOverall: S2O82- + 2I- 2SO42- + I2Both steps in the catalysed reaction involve the interaction of oppositely chargedions, which attract one another strongly; hence the activation energy is lower andenhances the rate of reaction.TJC/P31ai) • Colour of Fe2+ (aq): green Colour of Fe3+ (aq): yellowii) • For transition metals ions e.g. Fe3+ (aq), In the presence of water ligands, the degeneracy of the 5 3d orbitals is raised and split into two energy levels due to the repulsion of the electrons in the metal ion and the lone pairs on the water ligands. • When an electron is promoted from the d-orbital of lower energy to one of higher energy (d-d transition), an amount of energy, , which happens to be in the visible region of the electromagnetic spectrum, has to be absorbed. • The colour observed is complement to the light energy • Oxidation state of iron in Na2FeO4 = +6 8OH- + Fe3+ → FeO42- + 4H2O + 3e • 2e + H2O + OCl- → Cl- + 2OH- • Overall: 10OH- + 2Fe3+ + 3OCl- → 2FeO42- + 5H2O + 3Cl-YJC/ JC2/ 2008 14
  15. 15. VJC/P32ai.iibcTPJC/P3/1(d) Cu2+ ion has the electronic configuration of [Ar] 3d9. When CuSO4(s) is dissolved in water, Cu2+ ions are hydrated by water ligands. This causes d- orbital splitting and enables electrons from the orbital of lower energy to absorb light energy and jump to the partially-filled orbital of higher energy. Light is absorbed form red spectrum. Hence, a blue solution is seen.YJC/ JC2/ 2008 15
  16. 16. RJC/P3/3(a)(i) Electronic configurations: 2+ 5 2+ 25Mn : [Ar]3d ; 26Fe : [Ar]3d6 ; 27Co 2+ : [Ar]3d7 Third ionisation energy of iron is lower than 3rd ionisation energy of Co since Co2+ has one more proton than Fe2+ and hence greater effective nuclear charge than Fe2+ (shielding being approximately the same). Despite its greater effective nuclear charge, Fe has lower 3rd IE than Mn. This is because the electron to be removed in Fe2+ is from a paired orbital and is aided by electron−electron repulsion. (ii) With reference to the Data Booklet, 2− − 2− O S2O8 + 2e− 2SO4 − E = +2.01 V 3+ 2+ O Fe + e− Fe E = +0.77 V O I2 + 2e− 2I− E = +0.54 V 2− The reaction between S2O8 − and I− is kinetically slow as it involves reaction between 2 species which are negatively charged. 3+ 3+ 2+ On addition of Fe , Fe oxidises I− to I2 and itself is reduced to Fe . The reaction is spontaneous (E cell = +0.77 − 0.54 = +0.23 V0) and kinetically favourable as it involves species which are oppositely charged. 2+ 2− 2− 3+ The Fe formed then reduces S2O8 − to SO4 − and itself is oxidised back to Fe . The reaction is spontaneous (E cell = +2.01 − 0.77 = +1.24 V0) and kinetically favourable as well as it also involves species which are oppositely charged. 3+ Since Fe is regenerated, it is chemically unchanged and only a small amount is needed for it to catalyse the reaction. 2+ 3+ 2+ (iii) Both aqueous Fe and Fe exist as aqua complexes with formulae [Fe(H2O)6] 3+ and [Fe(H2O)6] respectively. 3+ 3+ 3+ Due to the higher charge density of Fe , water molecules coordinated to Fe in [Fe(H2O)6] is + more polarised so that more H3O are produced due to greater hydrolysis: 3+ 2+ + [Fe(H2O)6] (aq) + H2O (l) [Fe(H2O)5(OH)] (aq) + H3O (aq) Hence, for the same concentration, pH of aq FeCl3 is lower than pH of aq FeCl2 as + concentration of H3O in aq FeCl3 is higher.NJC/P3/4a.b.YJC/ JC2/ 2008 16
  17. 17. ciciiYJC/ JC2/ 2008 17