Chemistry: Unit-wise Subjective Question Bank

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Chemistry: Unit-wise Subjective Question Bank

Groups B and C of the Grade 11 Chemistry paper carry sixty-four of the seventy-five marks: eight questions of five marks and three of eight. This chapter gives worked practice for both, arranged by the four areas of the CDC syllabus — General and Physical, Inorganic, Organic, and Applied Chemistry. Every model answer is written the way you should write it in the hall: balanced equations with conditions over the arrow, numericals with the formula first, and structures named in full.

How to use this chapter
  • Attempt first. Cover the answer, write your own against the clock (12 minutes per 5-marker, 19 per 8-marker), then mark yourself.
  • Count your points. A 5-mark theory answer wants about five markable statements; an 8-mark answer wants a structure: definition, equations or diagram, body, example or numerical.
  • Balance and label. An unbalanced equation or a missing condition (catalyst, temperature, solvent) usually costs the mark.
  • Atomic masses used: H = 1, C = 12, N = 14, O = 16, Na = 23, Cl = 35.5, Ca = 40, Cu = 63.5.
About years and sources

Every question below is an original practice question written to the CDC pattern and carries no past-paper year. The syllabus outline asks for "years each appeared", but Grade 11 papers are not published by NEB, so no year is claimed. Instead the section headings give each area's teaching hours (General and Physical 48, Inorganic 35, Organic 30, Applied 15 of 128), the public measure of how much of the paper to expect.

6.1 General and Physical Chemistry (units 1–8, 48 hours)

Group B — Short answer (5 marks each)

  1. A compound contains 40.0\ \% carbon, 6.7\ \% hydrogen and 53.3\ \% oxygen, and its molar mass is 180\ \text{g mol}^{-1}. Find its empirical and molecular formulae. [5]
  2. 6\ \text{g} of hydrogen and 56\ \text{g} of nitrogen are allowed to react to form ammonia. Identify the limiting reactant and find the mass of ammonia formed and of the reactant left over. [5]
  3. On heating 25\ \text{g} of \text{CaCO}_3, 10\ \text{g} of \text{CaO} is obtained. Find the theoretical and percentage yield. [5]
  4. State Bohr's postulates. Calculate the wavelength of the photon emitted when the electron of a hydrogen atom falls from n = 3 to n = 2 (E_n = -13.6/n^{2}\ \text{eV}; hc = 1240\ \text{eV nm}). [5]
  5. Write the four quantum numbers of the last electron of chlorine (Z = 17) and state why n = 2, l = 2 is not allowed. [5]
  6. Explain why the first ionization energy of nitrogen is higher than that of oxygen, and why that of magnesium is higher than that of aluminium. [5]
  7. Using VSEPR theory, predict and explain the shapes of \text{NH}_3 and \text{H}_2\text{O}, and account for their bond angles. [5]
  8. Balance, by the oxidation number method, \text{MnO}_4^{-} + \text{Fe}^{2+} + \text{H}^{+} \to \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_2\text{O}. [5]
  9. State Faraday's laws of electrolysis. Find the mass of copper deposited by a current of 2\ \text{A} in 30 minutes from \text{CuSO}_4 solution (1\ \text{F} = 96\,500\ \text{C}). [5]
  10. A gas occupies 500\ \text{cm}^{3} at 27^{\circ}\text{C} and 1\ \text{atm}. Find its volume at STP, and state the law used. [5]
  11. In the reaction \text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2 at equilibrium, [\text{N}_2\text{O}_4] = 0.04\ \text{mol L}^{-1} and [\text{NO}_2] = 0.12\ \text{mol L}^{-1}. Find K_c and K_p at 300\ \text{K} (R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}). [5]
Answers — Group B, General and Physical

1. Moles in 100\ \text{g}: C = 40/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33. Divide by 3.33: 1 : 2 : 1. Empirical formula \text{CH}_2\text{O} (mass 30). n = 180/30 = 6. Molecular formula \text{C}_6\text{H}_{12}\text{O}_6.

2. \text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3. Moles: \text{N}_2 = 56/28 = 2; \text{H}_2 = 6/2 = 3. Two moles of \text{N}_2 would need 6 mol \text{H}_2, so hydrogen is limiting. 3 mol \text{H}_2 gives 2 mol \text{NH}_3 = 34\ \text{g}. \text{N}_2 used = 1 mol, left = 1 mol = 28\ \text{g}.

3. \text{CaCO}_3 \to \text{CaO} + \text{CO}_2; 25\ \text{g} = 0.25 mol, so theoretical \text{CaO} = 0.25\times56 = 14\ \text{g}. Percentage yield = \dfrac{10}{14}\times100 = 71.4\ \%.

4. Postulates: (i) electrons revolve in fixed circular orbits of definite energy without radiating; (ii) the angular momentum is quantized, mvr = \dfrac{nh}{2\pi}; (iii) energy is absorbed or emitted only when an electron jumps between orbits, \Delta E = h\nu. \Delta E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 13.6\times\dfrac{5}{36} = 1.89\ \text{eV}. \lambda = \dfrac{1240}{1.89} = 656\ \text{nm} (the red \text{H}_\alpha line of the Balmer series).

5. Cl: 1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}; the last electron is in 3p: n = 3, l = 1. Filling the three p orbitals (m_l = -1, 0, +1) with \uparrow first and then pairing, the fifth electron has m_l = 0 and m_s = -\tfrac{1}{2}. For n = 2, l may be only 0 or 1 (l = 0 to n - 1), so l = 2 is not allowed.

6. N: 2p^{3} is half-filled and extra stable; O: 2p^{4} has a paired electron, and pair repulsion makes it easier to remove, so \text{IE}_{N} > \text{IE}_{O}. Mg: 3s^{2} is a full subshell; Al: the outer electron is in the higher-energy, better shielded 3p, so \text{IE}_{Mg} > \text{IE}_{Al}.

7. Electron pairs around the central atom repel and arrange as far apart as possible. \text{NH}_3: four pairs (three bond, one lone), tetrahedral arrangement, trigonal pyramidal shape, angle 107^{\circ}. \text{H}_2\text{O}: four pairs (two bond, two lone), tetrahedral arrangement, bent shape, angle 104.5^{\circ}. Repulsion order: lone–lone > lone–bond > bond–bond, so lone pairs squeeze the angle below 109.5^{\circ} (more lone pairs, smaller angle).

8. Oxidation numbers: Mn +7 \to +2 (gain of 5e^{-}); Fe +2 \to +3 (loss of 1e^{-}). Multiply Fe by 5: \text{MnO}_4^{-} + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+}. Balance O with 4\text{H}_2\text{O} and H with 8\text{H}^{+}: \text{MnO}_4^{-} + 5\text{Fe}^{2+} + 8\text{H}^{+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}. Check charge: -1 + 10 + 8 = +17 and 2 + 15 = +17 ✓.

9. First law: the mass deposited is proportional to the charge passed, m = ZIt. Second law: for the same charge, masses deposited are proportional to equivalent masses. Q = It = 2\times1800 = 3600\ \text{C}. \text{Cu}^{2+} + 2e^{-} \to \text{Cu}: 63.5\ \text{g} needs 193\,000\ \text{C}. m = \dfrac{63.5\times3600}{193\,000} = 1.18\ \text{g}.

10. Charles' law with Boyle's law (combined gas law): \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}. V_2 = \dfrac{500\times1\times273}{300\times1} = 455\ \text{cm}^{3}.

11. K_c = \dfrac{[\text{NO}_2]^{2}}{[\text{N}_2\text{O}_4]} = \dfrac{(0.12)^{2}}{0.04} = 0.36\ \text{mol L}^{-1}. \Delta n = 2 - 1 = 1, so K_p = K_c(RT)^{\Delta n} = 0.36\times0.0821\times300 = 8.87\ \text{atm}.

Group C — Long answer (8 marks each)

  1. (a) State Avogadro's law and deduce the relation between molecular mass and vapour density. [3](b) Propane burns as \text{C}_3\text{H}_8 + 5\text{O}_2 \to 3\text{CO}_2 + 4\text{H}_2\text{O}. For 11.2\ \text{L} of propane at STP, find the volume of oxygen needed, the volume of \text{CO}_2 formed, and the mass of water formed. [5] [8]
  2. Describe the shapes of \text{BeF}_2, \text{BF}_3, \text{CH}_4, \text{PCl}_5 and \text{SF}_6 using VSEPR theory and the hybridization of the central atom. [8]
  3. (a) Derive the ideal gas equation from Boyle's, Charles' and Avogadro's laws. [4](b) Find the pressure exerted by 4.4\ \text{g} of \text{CO}_2 in a 10\ \text{L} vessel at 27^{\circ}\text{C}. [2](c) Why do real gases deviate from ideal behaviour? [2] [8]
  4. (a) State the law of mass action. [1](b) Derive the relation K_p = K_c(RT)^{\Delta n}. [4](c) State the effect of pressure, temperature and a catalyst on \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 (\Delta H < 0). [3] [8]
Answers — Group C, General and Physical

1. (a) Equal volumes of all gases under the same conditions of temperature and pressure contain equal numbers of molecules. Vapour density is the ratio of the mass of a volume of gas to the mass of the same volume of hydrogen, so \text{V.D.} = \dfrac{M_{gas}}{M_{\text{H}_2}} (equal volumes contain equal molecules). The hydrogen molecule has mass 2 (two atoms), so \text{V.D.} = \dfrac{M}{2} and M = 2\times\text{V.D.} (b) 11.2\ \text{L} = 0.5 mol propane. Moles of \text{O}_2 = 5\times0.5 = 2.5 mol, volume = 2.5\times22.4 = 56\ \text{L}. Moles of \text{CO}_2 = 3\times0.5 = 1.5 mol, volume = 33.6\ \text{L}. Moles of \text{H}_2\text{O} = 4\times0.5 = 2.0 mol, mass = 2\times18 = 36\ \text{g}.

2. The central atom's electron pairs spread out as far as possible; hybridization is the mixing of orbitals to describe the geometry.

Molecule Pairs on central atom Hybridization Shape Angle
\text{BeF}_2 2 bond sp linear 180^{\circ}
\text{BF}_3 3 bond sp^{2} trigonal planar 120^{\circ}
\text{CH}_4 4 bond sp^{3} tetrahedral 109.5^{\circ}
\text{PCl}_5 5 bond sp^{3}d trigonal bipyramidal 90^{\circ} and 120^{\circ}
\text{SF}_6 6 bond sp^{3}d^{2} octahedral 90^{\circ}

In \text{PCl}_5 the two axial bonds are longer than the three equatorial bonds because each axial bond has three nearest neighbours at 90^{\circ} against two for an equatorial bond. All the molecules above have no lone pairs, so the shape follows the pair arrangement. 3. (a) Boyle: V \propto \dfrac{1}{P} (n, T constant). Charles: V \propto T (n, P constant). Avogadro: V \propto n (P, T constant). Combining, V \propto \dfrac{nT}{P}, so V = R\dfrac{nT}{P}, i.e. PV = nRT, where R is the universal gas constant (8.314\ \text{J mol}^{-1}\text{K}^{-1}). (b) n = 4.4/44 = 0.1 mol, T = 300\ \text{K}: P = \dfrac{nRT}{V} = \dfrac{0.1\times0.0821\times300}{10} = 0.246\ \text{atm}. (c) The ideal gas assumes that molecules have no volume and no attraction. At high pressure the molecular volume is not negligible; at low temperature intermolecular attractions lower the pressure. So real gases deviate at high pressure and low temperature. 4. (a) The rate of a chemical reaction is directly proportional to the product of the active masses (molar concentrations) of the reactants, each raised to the power of its coefficient. (b) For aA + bB \rightleftharpoons cC + dD: K_p = \dfrac{p_C^{c}p_D^{d}}{p_A^{a}p_B^{b}} and K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}. For a gas p = \dfrac{n}{V}RT = [\,\,]RT, so p_i = [i]RT. Substituting: K_p = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}(RT)^{(c + d) - (a + b)} = K_c(RT)^{\Delta n}. (c) Pressure: raising it shifts the equilibrium to the right (4 mol of gas become 2). Temperature: raising it shifts it to the left because the forward reaction is exothermic; lowering it favours ammonia but slows the rate. Catalyst: no change in the position or K; it only reaches equilibrium faster.

6.2 Inorganic Chemistry (units 9–11, 35 hours)

Group B — Short answer (5 marks each)

  1. Describe the preparation of ozone from oxygen, its structure, and the part played by chlorofluorocarbons in ozone layer depletion. [5]
  2. Describe the laboratory preparation of ammonia, with the equation, the drying agent and the test for the gas. [5]
  3. Write the equations for the reaction of copper with(a) dilute and(b) concentrated nitric acid, and state the role of the acid. [5]
  4. Describe the brown ring test for a nitrate, with equations. [5]
  5. Describe the laboratory preparation of chlorine and show with equations that it is a stronger oxidizing agent than bromine and iodine. [5]
  6. Describe the preparation of hydrogen sulphide in Kipp's apparatus and give three of its reactions. [5]
  7. Distinguish between calcination and roasting with an example and equation of each, and state what slag is. [5]
  8. Explain the trend in the thermal stability of the carbonates of Group 2 elements and the solubility of their sulphates. [5]
  9. Describe the preparation and setting of plaster of Paris and give two uses. [5]
  10. Name the metal ions concerned in nerve impulses, in chlorophyll and in haemoglobin, and name two toxic metals with one effect each. [5]
Answers — Group B, Inorganic

1. Preparation: dry oxygen is passed through a silent electric discharge in an ozonizer: 3\text{O}_2 \rightleftharpoons 2\text{O}_3, \Delta H = +285\ \text{kJ} (about 10 % conversion). Structure: a bent resonance hybrid, \text{O}-\text{O} bond length 128\ \text{pm}, angle 117^{\circ}. Depletion: UV breaks a CFC (\text{CF}_2\text{Cl}_2 \to \text{CF}_2\text{Cl}\cdot + \text{Cl}\cdot); then \text{Cl}\cdot + \text{O}_3 \to \text{ClO}\cdot + \text{O}_2 and \text{ClO}\cdot + \text{O} \to \text{Cl}\cdot + \text{O}_2. Chlorine is regenerated, so one atom destroys thousands of ozone molecules.

2. Heat a mixture of ammonium chloride and slaked lime: 2\text{NH}_4\text{Cl} + \text{Ca(OH)}_2 \to \text{CaCl}_2 + 2\text{NH}_3\uparrow + 2\text{H}_2\text{O}. The gas is dried with quicklime (CaO) — not conc. \text{H}_2\text{SO}_4 or \text{CaCl}_2, which react with it. It is collected by downward displacement of air (lighter than air). Tests: it turns moist red litmus blue; a glass rod dipped in conc. HCl gives dense white fumes of \text{NH}_4\text{Cl}.

3. (a) Dilute: 3\text{Cu} + 8\text{HNO}_3 \to 3\text{Cu(NO}_3)_2 + 2\text{NO}\uparrow + 4\text{H}_2\text{O}. (b) Concentrated: \text{Cu} + 4\text{HNO}_3 \to \text{Cu(NO}_3)_2 + 2\text{NO}_2\uparrow + 2\text{H}_2\text{O} (brown fumes). Nitric acid acts as an oxidizing agent: the nitrogen is reduced from +5 to +2 (dilute) or +4 (concentrated), while copper is oxidized from 0 to +2.

4. Add freshly prepared \text{FeSO}_4 solution to the nitrate solution, then pour conc. \text{H}_2\text{SO}_4 slowly down the side of the tube. Nitrate is reduced: \text{NO}_3^{-} + 3\text{Fe}^{2+} + 4\text{H}^{+} \to \text{NO} + 3\text{Fe}^{3+} + 2\text{H}_2\text{O}. The NO combines with excess \text{Fe}^{2+}: [\text{Fe(H}_2\text{O)}_6]^{2+} + \text{NO} \to [\text{Fe(H}_2\text{O)}_5\text{NO}]^{2+} + \text{H}_2\text{O}. A brown ring appears at the junction of the two liquids.

5. \text{MnO}_2 + 4\text{HCl} \to \text{MnCl}_2 + \text{Cl}_2\uparrow + 2\text{H}_2\text{O} (warm; the gas is washed with water and dried with conc. \text{H}_2\text{SO}_4, collected by upward displacement). Displacement: \text{Cl}_2 + 2\text{KBr} \to 2\text{KCl} + \text{Br}_2 (orange); \text{Cl}_2 + 2\text{KI} \to 2\text{KCl} + \text{I}_2 (violet in \text{CCl}_4). Chlorine displaces both, and \text{Br}_2 displaces \text{I}_2 but not \text{Cl}^{-}, so the order of oxidizing power is \text{Cl}_2 > \text{Br}_2 > \text{I}_2.

6. In Kipp's apparatus iron(II) sulphide reacts with dilute acid: \text{FeS} + \text{H}_2\text{SO}_4 \to \text{FeSO}_4 + \text{H}_2\text{S}\uparrow. The gas has a rotten-egg smell. Reactions: (a) with copper sulphate: \text{CuSO}_4 + \text{H}_2\text{S} \to \text{CuS}\downarrow + \text{H}_2\text{SO}_4 (black); (b) reducing action: \text{H}_2\text{S} + \text{Cl}_2 \to 2\text{HCl} + \text{S}; (c) burning: 2\text{H}_2\text{S} + 3\text{O}_2 \to 2\text{SO}_2 + 2\text{H}_2\text{O}.

7. Calcination: heating the ore in limited or no air to remove moisture and volatile matter, e.g. \text{ZnCO}_3 \to \text{ZnO} + \text{CO}_2. Roasting: heating the ore in excess air, e.g. 2\text{ZnS} + 3\text{O}_2 \to 2\text{ZnO} + 2\text{SO}_2. Slag: the fusible product of the flux and gangue, e.g. \text{CaO} + \text{SiO}_2 \to \text{CaSiO}_3, which floats on the molten metal and is removed.

8. Carbonates decompose on heating, \text{MCO}_3 \to \text{MO} + \text{CO}_2, and the stability increases down the group (\text{BeCO}_3 < \text{MgCO}_3 < \text{CaCO}_3 < \text{SrCO}_3 < \text{BaCO}_3) because the larger cation polarizes the carbonate ion less. Sulphate solubility decreases down the group: \text{BeSO}_4 and \text{MgSO}_4 dissolve freely, \text{CaSO}_4 is sparingly soluble, and \text{BaSO}_4 is insoluble, because the lattice energy falls less quickly than the hydration energy.

9. Gypsum is heated to about 120^{\circ}\text{C}: \text{CaSO}_4\cdot2\text{H}_2\text{O} \to \text{CaSO}_4\cdot\tfrac{1}{2}\text{H}_2\text{O} + 1\tfrac{1}{2}\text{H}_2\text{O}. On mixing with water it sets to a hard mass as it re-forms gypsum: \text{CaSO}_4\cdot\tfrac{1}{2}\text{H}_2\text{O} + 1\tfrac{1}{2}\text{H}_2\text{O} \to \text{CaSO}_4\cdot2\text{H}_2\text{O} with slight expansion. Uses: plaster for fractures, moulds and casts, wall plaster.

10. Nerve impulses: \text{Na}^{+} and \text{K}^{+} (sodium–potassium pump). Chlorophyll: \text{Mg}^{2+}. Haemoglobin: \text{Fe}^{2+}. Toxic metals: mercury (Minamata disease, nerve damage); lead (damage to the brain and kidney, anaemia); arsenic (skin lesions and cancer); cadmium (Itai-itai disease, kidney damage).

Group C — Long answer (8 marks each)

  1. Describe the extraction of sodium by Down's process, with the reactions at each electrode. [8]
  2. Describe the chemical properties of concentrated sulphuric acid as a dehydrating agent and an oxidizing agent, with equations. [8]
  3. Compare the halogens chlorine, bromine and iodine in physical state, preparation from their salts, and displacement reactions, and describe the tests for chloride, bromide and iodide ions. [8]
Answers — Group C, Inorganic

1. Raw material: fused sodium chloride (m.p. 801^{\circ}\text{C}) mixed with calcium chloride, which lowers the melting point to about 600^{\circ}\text{C}. Cell: a steel container lined with firebrick; a graphite anode at the centre; a circular iron cathode; a nickel gauze diaphragm between them to keep sodium and chlorine apart; an inverted trough above the cathode to collect sodium; a dome over the anode to collect chlorine. Reactions: anode: 2\text{Cl}^{-} \to \text{Cl}_2 + 2e^{-}; cathode: \text{Na}^{+} + e^{-} \to \text{Na}. Overall: 2\text{NaCl} \to 2\text{Na} + \text{Cl}_2. Molten sodium floats on the electrolyte and rises into the collector. By-product: chlorine. Why not aqueous NaCl: hydrogen would be discharged at the cathode instead of sodium.

2. Dehydrating agent: it removes the elements of water. Sugar: \text{C}_{12}\text{H}_{22}\text{O}_{11} \rightarrow\ [\text{conc. H}_2\text{SO}_4] 12\text{C} + 11\text{H}_2\text{O} (black carbon). Blue vitriol turns white: \text{CuSO}_4\cdot5\text{H}_2\text{O} \to \text{CuSO}_4 + 5\text{H}_2\text{O}. Ethanol gives ethene at 170^{\circ}\text{C}. Oxidizing agent (hot, concentrated): \text{Cu} + 2\text{H}_2\text{SO}_4 \to \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}; \text{C} + 2\text{H}_2\text{SO}_4 \to \text{CO}_2 + 2\text{SO}_2 + 2\text{H}_2\text{O}; \text{S} + 2\text{H}_2\text{SO}_4 \to 3\text{SO}_2 + 2\text{H}_2\text{O}. Sulphur is reduced from +6 to +4. Non-volatile acid: \text{NaCl} + \text{H}_2\text{SO}_4 \to \text{NaHSO}_4 + \text{HCl}, used to make hydrogen chloride.

3. State: \text{Cl}_2 greenish-yellow gas; \text{Br}_2 red-brown liquid; \text{I}_2 violet-black solid (sublimes). Preparation: \text{MnO}_2 + 4\text{HCl} \to \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}; 2\text{NaBr} + 3\text{H}_2\text{SO}_4 + \text{MnO}_2 \to 2\text{NaHSO}_4 + \text{MnSO}_4 + 2\text{H}_2\text{O} + \text{Br}_2; similarly iodide with \text{MnO}_2 and conc. \text{H}_2\text{SO}_4 gives \text{I}_2. Displacement: \text{Cl}_2 displaces \text{Br}^{-} and \text{I}^{-}; \text{Br}_2 displaces \text{I}^{-} only. Tests (with \text{AgNO}_3 then ammonia): \text{Cl}^{-} white \text{AgCl}, soluble in dilute \text{NH}_3; \text{Br}^{-} pale-yellow \text{AgBr}, soluble in conc. \text{NH}_3; \text{I}^{-} yellow \text{AgI}, insoluble in \text{NH}_3. Also: free \text{I}_2 turns starch blue-black.

6.3 Organic Chemistry (units 12–15, 30 hours)

Group B — Short answer (5 marks each)

  1. Give the IUPAC names of \text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3, \text{CH}_3\text{CH}_2\text{CH(OH)CH}_3, \text{CH}_3\text{CH}_2\text{CHO}, \text{CH}_3\text{CH}{=}\text{CHCH}_3 and \text{CH}_3\text{C}{\equiv}\text{CCH}_3. [5]
  2. Describe the Lassaigne test for nitrogen, sulphur and chlorine in an organic compound. [5]
  3. Write all the structural isomers of \text{C}_4\text{H}_{10}\text{O} and name the type of isomerism between the alcohols and the ethers. [5]
  4. Explain the inductive effect with examples of +I and -I groups, and use it to compare the acid strengths of acetic acid and chloroacetic acid. [5]
  5. Write the equations for four methods of preparing an alkane (ethane or methane). [5]
  6. State Markovnikov's rule and the peroxide effect, and give the products of the addition of HBr to propene in each case. [5]
  7. Explain the acidic nature of ethyne, with two reactions, and give a test to distinguish ethyne from ethene. [5]
  8. Describe Kekulé's structure of benzene and give the evidence for the resonance hybrid. [5]
  9. Explain the mechanism of the nitration of benzene. [5]
  10. Write the equations for the Friedel–Crafts alkylation and acylation of benzene. [5]
Answers — Group B, Organic

1. 2-methylbutane; butan-2-ol; propanal; but-2-ene; but-2-yne.

2. The organic compound is fused with a piece of sodium, the hot tube is plunged into distilled water, boiled and filtered; this is the sodium extract, containing \text{NaCN}, \text{Na}_2\text{S} and \text{NaCl}. Nitrogen: boil with \text{FeSO}_4, acidify with dilute \text{H}_2\text{SO}_4; a Prussian blue colour or precipitate shows N. Sulphur: add lead acetate solution; a black precipitate of PbS shows S. Chlorine: acidify with dilute \text{HNO}_3, add \text{AgNO}_3; a curdy white precipitate of AgCl, soluble in \text{NH}_3, shows Cl.

3. Alcohols: butan-1-ol, \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}; butan-2-ol, \text{CH}_3\text{CH}_2\text{CH(OH)CH}_3; 2-methylpropan-1-ol, (\text{CH}_3)_2\text{CHCH}_2\text{OH}; 2-methylpropan-2-ol, (\text{CH}_3)_3\text{COH}. Ethers: ethoxyethane, \text{C}_2\text{H}_5\text{OC}_2\text{H}_5; 1-methoxypropane, \text{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3; 2-methoxypropane, \text{CH}_3\text{OCH(CH}_3)_2. That is seven isomers. Alcohols and ethers are functional isomers; butan-1-ol and butan-2-ol are position isomers; butan-1-ol and 2-methylpropan-1-ol are chain isomers.

4. The inductive effect is the permanent displacement of the \sigma electrons along a chain due to the electronegativity difference between atoms. Alkyl groups (-\text{CH}_3, -\text{C}_2\text{H}_5) push electrons (+I); -\text{Cl}, -\text{NO}_2 and -\text{COOH} pull electrons (-I). In \text{ClCH}_2\text{COOH} the chlorine withdraws electrons and stabilizes the carboxylate ion, so the acid ionizes more easily; in \text{CH}_3\text{COOH} the methyl group pushes electrons and destabilizes the anion. Hence chloroacetic acid is the stronger acid.

5. (i) Reduction of a haloalkane: \text{CH}_3\text{CH}_2\text{Br} + 2[\text{H}] \rightarrow\ [\text{Zn/HCl}] \text{C}_2\text{H}_6 + \text{HBr}. (ii) Wurtz reaction: 2\text{CH}_3\text{Br} + 2\text{Na} \rightarrow\ [\text{dry ether}] \text{C}_2\text{H}_6 + 2\text{NaBr}. (iii) Decarboxylation: \text{CH}_3\text{COONa} + \text{NaOH} \rightarrow\ [\text{CaO},\ \Delta] \text{CH}_4 + \text{Na}_2\text{CO}_3. (iv) Hydrogenation: \text{C}_2\text{H}_4 + \text{H}_2 \rightarrow\ [\text{Ni},\ 573\ \text{K}] \text{C}_2\text{H}_6. (v) Kolbe: 2\text{CH}_3\text{COONa} + 2\text{H}_2\text{O} \rightarrow\ [\text{electrolysis}] \text{C}_2\text{H}_6 + 2\text{CO}_2 + 2\text{NaOH} + \text{H}_2.

6. Markovnikov: when an unsymmetrical reagent HX adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already. \text{CH}_3\text{CH}{=}\text{CH}_2 + \text{HBr} \to \text{CH}_3\text{CHBrCH}_3 (2-bromopropane, via the more stable secondary carbocation). Peroxide effect: in the presence of peroxide only HBr adds the opposite way (free-radical mechanism): \text{CH}_3\text{CH}{=}\text{CH}_2 + \text{HBr} \rightarrow\ [\text{peroxide}] \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} (1-bromopropane).

7. The sp carbon is more electronegative than sp^{2} or sp^{3} carbon (50 % s character), so the terminal \text{C}-\text{H} bond is polarized and its hydrogen is weakly acidic. (a) 2\text{HC}{\equiv}\text{CH} + 2\text{Na} \to 2\text{HC}{\equiv}\text{CNa} + \text{H}_2. (b) \text{HC}{\equiv}\text{CH} + 2[\text{Ag(NH}_3)_2]^{+}\text{OH}^{-} \to \text{AgC}{\equiv}\text{CAg}\downarrow + 4\text{NH}_3 + 2\text{H}_2\text{O} (white precipitate). Ethene gives no precipitate with ammoniacal \text{AgNO}_3, so this distinguishes ethyne from ethene.

8. Kekulé proposed a planar hexagon of six carbon atoms, each bonded to one hydrogen, with alternating single and double bonds, and a rapid shift between two forms. Limitations of the single structure: it predicts two different 1,2-disubstituted isomers (one across a single bond, one across a double bond), but only one is known; and it predicts addition reactions of the double bonds, but benzene undergoes substitution. Evidence for resonance: all six \text{C}-\text{C} bonds are equal (139\ \text{pm}, between single 154 and double 134); the heat of hydrogenation is about 150\ \text{kJ mol}^{-1} against 360 expected for three isolated double bonds, so benzene is stabilized by about 150\ \text{kJ mol}^{-1} (resonance energy).

9. Step 1: generation of the electrophile: \text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \text{NO}_2^{+} + \text{H}_3\text{O}^{+} + 2\text{HSO}_4^{-}. Step 2: attack of the \pi electrons on \text{NO}_2^{+} to form a resonance-stabilized carbocation (arenium ion). Step 3: loss of \text{H}^{+} from the arenium ion restores the aromatic ring: \text{C}_6\text{H}_6 + \text{HNO}_3 \rightarrow\ [\text{conc. H}_2\text{SO}_4,\ 323\ \text{K}] \text{C}_6\text{H}_5\text{NO}_2 + \text{H}_2\text{O}.

10. Alkylation: \text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \rightarrow\ [\text{anhyd. AlCl}_3] \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl} (toluene). Acylation: \text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \rightarrow\ [\text{anhyd. AlCl}_3] \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl} (acetophenone). The \text{AlCl}_3 generates the electrophile \text{R}^{+} or \text{RCO}^{+}.

Group C — Long answer (8 marks each)

  1. Describe three methods for the preparation of ethene and the reactions of ethene with \text{HBr}, water (acid-catalysed), ozone and cold alkaline \text{KMnO}_4. [8]
  2. Describe the preparation of benzene (two methods) and its reactions: nitration, sulphonation, halogenation, hydrogenation and combustion. [8]
  3. How will you convert(a) ethane to ethene,(b) ethene to ethyne,(c) ethyne to acetaldehyde,(d) benzene to toluene? Give equations with conditions. [8]
Answers — Group C, Organic

1. Preparation: (i) dehydration of ethanol: \text{C}_2\text{H}_5\text{OH} \rightarrow\ [\text{conc. H}_2\text{SO}_4,\ 443\ \text{K}] \text{C}_2\text{H}_4 + \text{H}_2\text{O}; (ii) dehydrohalogenation: \text{C}_2\text{H}_5\text{Br} + \text{KOH(alc.)} \to \text{C}_2\text{H}_4 + \text{KBr} + \text{H}_2\text{O}; (iii) partial hydrogenation of ethyne: \text{HC}{\equiv}\text{CH} + \text{H}_2 \rightarrow\ [\text{Pd/BaSO}_4\text{ (Lindlar)}] \text{C}_2\text{H}_4. Reactions: with HBr: \text{C}_2\text{H}_4 + \text{HBr} \to \text{C}_2\text{H}_5\text{Br}; with water: \text{C}_2\text{H}_4 + \text{H}_2\text{O} \rightarrow\ [\text{H}^{+}] \text{C}_2\text{H}_5\text{OH}; with ozone: ethene forms an ozonide, \text{C}_2\text{H}_4 + \text{O}_3 \to \text{ozonide}, which is hydrolysed in the presence of Zn to give two molecules of methanal, 2\text{HCHO}; with cold alkaline \text{KMnO}_4 (Baeyer's reagent): \text{C}_2\text{H}_4 + [\text{O}] + \text{H}_2\text{O} \to \text{HOCH}_2\text{CH}_2\text{OH} (ethane-1,2-diol), the purple colour disappearing.

2. Preparation: (i) sodium benzoate heated with soda lime: \text{C}_6\text{H}_5\text{COONa} + \text{NaOH} \rightarrow\ [\text{CaO},\ \Delta] \text{C}_6\text{H}_6 + \text{Na}_2\text{CO}_3; (ii) ethyne passed through a red-hot iron tube at 873\ \text{K}: 3\text{C}_2\text{H}_2 \to \text{C}_6\text{H}_6; (iii) phenol distilled with zinc dust: \text{C}_6\text{H}_5\text{OH} + \text{Zn} \to \text{C}_6\text{H}_6 + \text{ZnO}. Reactions: nitration — conc. \text{HNO}_3 + conc. \text{H}_2\text{SO}_4, 323\ \text{K}, nitrobenzene; sulphonation — fuming \text{H}_2\text{SO}_4 gives benzenesulphonic acid, \text{C}_6\text{H}_5\text{SO}_3\text{H}; halogenation — \text{C}_6\text{H}_6 + \text{Cl}_2 \rightarrow\ [\text{FeCl}_3] \text{C}_6\text{H}_5\text{Cl} + \text{HCl}; hydrogenation — \text{C}_6\text{H}_6 + 3\text{H}_2 \rightarrow\ [\text{Ni},\ 473\ \text{K}] \text{C}_6\text{H}_{12} (cyclohexane); combustion — 2\text{C}_6\text{H}_6 + 15\text{O}_2 \to 12\text{CO}_2 + 6\text{H}_2\text{O}, with a sooty flame because of the high carbon content.

3. (a) \text{C}_2\text{H}_6 + \text{Cl}_2 \rightarrow\ [\text{UV}] \text{C}_2\text{H}_5\text{Cl} + \text{HCl}; then \text{C}_2\text{H}_5\text{Cl} + \text{KOH(alc.)} \rightarrow\ [\Delta] \text{C}_2\text{H}_4 + \text{KCl} + \text{H}_2\text{O}. (b) \text{C}_2\text{H}_4 + \text{Br}_2 \to \text{BrCH}_2\text{CH}_2\text{Br}; then \text{BrCH}_2\text{CH}_2\text{Br} + 2\text{KOH(alc.)} \rightarrow\ [\Delta] \text{HC}{\equiv}\text{CH} + 2\text{KBr} + 2\text{H}_2\text{O}. (c) \text{HC}{\equiv}\text{CH} + \text{H}_2\text{O} \rightarrow\ [\text{HgSO}_4,\ \text{dil. H}_2\text{SO}_4,\ 333\ \text{K}] [\text{CH}_2{=}\text{CHOH}] \to \text{CH}_3\text{CHO} (the unstable enol rearranges). (d) \text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \rightarrow\ [\text{anhyd. AlCl}_3] \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}.

6.4 Applied Chemistry (units 16–17, 15 hours)

Group B — Short answer (5 marks each)

  1. Describe the stages in producing a new chemical product, from the laboratory to the full-scale plant. [5]
  2. Distinguish between batch and continuous processes, with one example of each. [5]
  3. Distinguish between fixed capital and working capital in a chemical plant. [5]
  4. List five environmental effects of the chemical industry and one control measure for each. [5]
  5. Apply Le Chatelier's principle to choose the conditions for the Haber process. [5]
  6. Write the chemistry of Ostwald's process for nitric acid. [5]
  7. Write the reactions at each electrode in the diaphragm cell for sodium hydroxide, and name the function of the diaphragm. [5]
Answers — Group B, Applied

1. (i) Research and development: the reaction is tested in the laboratory (yield, catalyst, conditions). (ii) Process design and economic evaluation: flow sheet, raw material and energy balance, cost estimate. (iii) Pilot plant: a small-scale unit tests the process, corrosion and safety. (iv) Full-scale plant: design, construction, commissioning and operation. (v) Marketing and monitoring of product quality and environmental impact.

2. Batch: the reactor is charged with a fixed amount of reactants, the reaction is completed, and the product is withdrawn before the next charge; suited to small-volume or varied products (dyes, drugs, soap). Continuous: feed enters and product leaves without interruption, in steady state; suited to large-volume products (ammonia, sulphuric acid). Continuous plants have lower labour cost per tonne and steadier quality; batch plants are more flexible.

3. Fixed capital: the money invested in land, buildings, machinery and installation; it is not consumed in production. Working capital: the money needed to run the plant from day to day — raw materials, wages, fuel, stocks and credit; it is recovered through sales. Total investment = fixed capital + working capital.

4. Air: \text{SO}_2 and \text{NO}_x cause acid rain (use scrubbers, desulphurization). Water: untreated effluent poisons rivers (effluent treatment plant). Soil: solid waste and heavy metals (secure landfill, recycling). Climate: \text{CO}_2 emission (energy efficiency, capture). Ozone: CFCs (replace with safer coolants). Also: eutrophication from fertilizer run-off (controlled use).

5. \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3, \Delta H = -92\ \text{kJ}. Pressure: four moles of gas become two, so high pressure (\approx200\ \text{atm}) favours ammonia. Temperature: the reaction is exothermic, so low temperature favours it, but too low a temperature makes it slow; a compromise of 450–500^{\circ}\text{C} is used. Catalyst: finely divided iron with Mo or \text{Al}_2\text{O}_3 as a promoter speeds the approach to equilibrium. Removal of ammonia by liquefaction pushes the equilibrium forward, and unreacted gases are recycled.

6. (i) 4\text{NH}_3 + 5\text{O}_2 \rightarrow\ [\text{Pt–Rh gauze},\ 800^{\circ}\text{C}] 4\text{NO} + 6\text{H}_2\text{O}, \Delta H = -905\ \text{kJ}. (ii) NO is cooled and oxidized: 2\text{NO} + \text{O}_2 \to 2\text{NO}_2. (iii) \text{NO}_2 is absorbed in water in the presence of excess air: 4\text{NO}_2 + \text{O}_2 + 2\text{H}_2\text{O} \to 4\text{HNO}_3 (50–60\ \% acid, which can be concentrated by distillation with conc. \text{H}_2\text{SO}_4).

7. Brine is fed into the anode compartment. Anode (titanium or graphite): 2\text{Cl}^{-} \to \text{Cl}_2 + 2e^{-}. Cathode (steel): 2\text{H}_2\text{O} + 2e^{-} \to \text{H}_2 + 2\text{OH}^{-}. \text{Na}^{+} passes through the diaphragm to give NaOH solution at the cathode. The porous asbestos diaphragm separates the two compartments and prevents \text{Cl}_2 from reacting with NaOH (which would give hypochlorite and chlorate) and prevents \text{H}_2 and \text{Cl}_2 from mixing explosively.

Group C — Long answer (8 marks each)

  1. Describe the manufacture of ammonia by Haber's process, giving the principle, the conditions with reasons, and a labelled flow sheet. [8]
  2. Describe the manufacture of sodium hydroxide in the diaphragm cell, with a labelled diagram, the electrode reactions and the uses of the products. [8]
Answers — Group C, Applied

1. Principle: \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3, \Delta H = -92\ \text{kJ mol}^{-1}. Raw materials: nitrogen from liquefied air; hydrogen from steam reforming of natural gas (\text{CH}_4 + \text{H}_2\text{O} \to \text{CO} + 3\text{H}_2) or electrolysis of water. Conditions: 200 atm (high pressure favours the side with fewer moles); 450–500^{\circ}\text{C} (a compromise between yield and rate); finely divided iron catalyst with Mo or \text{Al}_2\text{O}_3 promoter; the poisons (e.g. sulphur compounds, CO) are removed from the gases first. Flow sheet: \text{N}_2 + \text{H}_2 (purified, 3 : 1) \to compressor (200 atm) \to heat exchanger \to catalytic converter (450^{\circ}\text{C}) \to cooler and condenser (ammonia liquefies) \to liquid ammonia storage; unreacted \text{N}_2 and \text{H}_2 are recycled to the compressor. The yield per pass is about 15 %, and the overall yield is high because of recycling. Uses: fertilizers (urea, ammonium nitrate and sulphate), nitric acid, explosives, refrigerant.

2. Cell: a rectangular steel tank divided by a porous asbestos diaphragm on a perforated steel cathode; the anode is graphite or titanium. Process: saturated purified brine enters the anode compartment; chlorine is liberated at the anode, 2\text{Cl}^{-} \to \text{Cl}_2 + 2e^{-}; at the cathode, 2\text{H}_2\text{O} + 2e^{-} \to \text{H}_2 + 2\text{OH}^{-}. The brine seeps through the diaphragm into the cathode compartment, where \text{Na}^{+} and \text{OH}^{-} form NaOH; the liquor contains about 10–12 % NaOH and unchanged NaCl. It is evaporated; NaCl crystallizes out and is recycled, and a 50 % NaOH solution results. Overall: 2\text{NaCl} + 2\text{H}_2\text{O} \to 2\text{NaOH} + \text{H}_2 + \text{Cl}_2. The diaphragm prevents the product mixing: otherwise \text{Cl}_2 + 2\text{NaOH} \to \text{NaCl} + \text{NaOCl} + \text{H}_2\text{O}. Uses: NaOH — soap, paper, textiles, aluminium purification; \text{Cl}_2 — water treatment, bleaching, PVC; \text{H}_2 — ammonia, hydrogenation.

6.5 Examiner's note: where Chemistry marks are lost

Area Common loss Remedy
General and Physical Unbalanced equation in a stoichiometry question; forgetting the limiting reactant Balance first; compute moles of each reactant
General and Physical Using ^{\circ}\text{C} in gas laws; R in the wrong units Convert to kelvin; match R to the unit of pressure
Inorganic Missing conditions and colours in equations; wrong drying agent Write the condition over the arrow; learn the drying agents
Organic Wrong IUPAC numbering; Markovnikov applied to the peroxide case Number from the end nearest the functional group; remember HBr only
Applied Flow sheet without labels or conditions Label every vessel; give pressure, temperature and catalyst

Chapter summary

  • Group B and C are 64 of the 75 marks; balanced equations with conditions, and numericals with the formula first, are what the marker ticks.
  • Moles: n = m/M = V/22.4 = N/N_A; percentage yield = actual/theoretical \times100; K_p = K_c(RT)^{\Delta n}; PV = nRT.
  • VSEPR shapes: sp linear, sp^{2} trigonal planar, sp^{3} tetrahedral, pyramidal (107^{\circ}) and bent (104.5^{\circ}), sp^{3}d trigonal bipyramidal, sp^{3}d^{2} octahedral.
  • Inorganic: ring test, Kipp's apparatus, Down's process (fused \text{NaCl} + \text{CaCl}_2), calcination against roasting, plaster of Paris (120^{\circ}\text{C}).
  • Organic: Lassaigne tests; seven isomers of \text{C}_4\text{H}_{10}\text{O}; Wurtz, Kolbe, decarboxylation; Markovnikov and peroxide effect; ethyne is acidic; benzene undergoes electrophilic substitution.
  • Applied: Haber (200 atm, 450–500^{\circ}\text{C}, Fe), Ostwald (Pt–Rh, 800^{\circ}\text{C}), contact (\text{V}_2\text{O}_5), diaphragm cell, Solvay.

Practice questions

Group A — Multiple choice (1 mark each)

  1. The mass of 0.2 mol of \text{H}_2\text{SO}_4 is [1]a9.8\ \text{g}b19.6\ \text{g}c98\ \text{g}d196\ \text{g}
  2. The product of Wurtz reaction of \text{CH}_3\text{CH}_2\text{Br} with sodium is [1]aethanebpropanecbutanedethene
  3. The oxidation number of sulphur in \text{H}_2\text{S} is [1]a-2b0c+2d+4
  4. The catalyst used in the hydration of ethyne to acetaldehyde is [1]a\text{AlCl}_3b\text{HgSO}_4cNidPt
Answers to Group A

1. (b) — 0.2\times98 = 19.6\ \text{g}.

2. (c) — two ethyl groups join: \text{C}_4\text{H}_{10}.

3. (a) — 2(+1) + x = 0.

4. (b) — mercuric sulphate in dilute \text{H}_2\text{SO}_4 (Kucherov reaction).

Group B — Short answer (5 marks each)

  1. How many moles and how many molecules are in 5.6\ \text{L} of \text{CO}_2 at STP? [5]
  2. Write the electronic configuration of \text{Cu} (Z = 29) and of \text{Cu}^{2+}, and state the number of unpaired electrons in each. [5]
  3. Write the products when ammonia reacts with(a) HCl,(b) excess \text{Cl}_2,(c) \text{CuSO}_4 solution. [5]
  4. Write the structure and IUPAC name of the product when propene reacts with HBr in the presence of peroxide. [5]
Answers to Group B

1. n = 5.6/22.4 = 0.25 mol; molecules = 0.25\times6.022\times10^{23} = 1.5\times10^{23}.

2. Cu: [\text{Ar}]3d^{10}4s^{1}, one unpaired electron (the 4s). \text{Cu}^{2+}: [\text{Ar}]3d^{9}, one unpaired electron.

3. (a) \text{NH}_3 + \text{HCl} \to \text{NH}_4\text{Cl} (white fumes). (b) \text{NH}_3 + 3\text{Cl}_2 \to \text{NCl}_3 + 3\text{HCl} (excess chlorine); with excess ammonia, 8\text{NH}_3 + 3\text{Cl}_2 \to \text{N}_2 + 6\text{NH}_4\text{Cl}. (c) \text{CuSO}_4 + 4\text{NH}_3 \to [\text{Cu(NH}_3)_4]\text{SO}_4 (deep blue; a pale blue precipitate of \text{Cu(OH)}_2 first).

4. Anti-Markovnikov addition: \text{CH}_3\text{CH}_2\text{CH}_2\text{Br}, 1-bromopropane.

Group C — Long answer (8 marks each)

  1. Describe the preparation and chemical properties of ammonia. [8]
  2. Explain Rutherford's atomic model and its limitations, and show how Bohr's model removed them. [8]
Answers to Group C

1. Preparation: laboratory — 2\text{NH}_4\text{Cl} + \text{Ca(OH)}_2 \to \text{CaCl}_2 + 2\text{NH}_3 + 2\text{H}_2\text{O}, dried with CaO; industrial — Haber process. Properties: colourless gas with a pungent smell, very soluble in water (fountain experiment), lighter than air; basic: \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^{+} + \text{OH}^{-}; \text{NH}_3 + \text{HCl} \to \text{NH}_4\text{Cl}; reducing: 2\text{NH}_3 + 3\text{CuO} \to 3\text{Cu} + \text{N}_2 + 3\text{H}_2\text{O}; burns in oxygen with a yellowish flame, 4\text{NH}_3 + 3\text{O}_2 \to 2\text{N}_2 + 6\text{H}_2\text{O} (with Pt catalyst, NO is formed); complex formation with \text{Cu}^{2+} and \text{Ag}^{+}. Uses: fertilizers, nitric acid, refrigerant, cleaning agent.

2. Rutherford: from \alpha-particle scattering by gold foil — most pass straight through, a few deflect, very few bounce back — he proposed a small, dense, positive nucleus with electrons revolving around it. Limitations: (i) a revolving electron should radiate energy continuously and spiral into the nucleus, making the atom unstable; (ii) it should give a continuous spectrum, but atoms give line spectra. Bohr's remedy: electrons move only in certain stationary orbits of fixed energy without radiating; angular momentum mvr = nh/2\pi; radiation is emitted or absorbed only when an electron jumps between orbits, \Delta E = h\nu, which explains the line spectrum of hydrogen. Bohr's model fails for atoms with more than one electron and does not explain the splitting of lines (Zeeman and Stark effects).