Chemistry: Unit-wise Objective Question Bank

Group A of the NEB Grade 11 Chemistry paper is eleven one-mark multiple-choice questions, drawn from across the seventeen units of the CDC syllabus. This chapter drills every unit in order. Each section opens with a high-yield facts table (the formulae, reagents and trends the objective questions keep returning to), then gives six to eight MCQs, then an answer key in which every answer carries a one-line reason and, where a wrong option encodes a real misconception, names it.

5.1 Foundation and Fundamentals

High-yield fact	Value / form
1 amu	\tfrac{1}{12} of the mass of one carbon-12 atom (1.66\times10^{-27}\ \text{kg})
Relative molecular mass	sum of the relative atomic masses of all atoms in the formula
Molecular formula	n\times empirical formula; n = \dfrac{\text{molar mass}}{\text{empirical formula mass}}
Mass percentage of an element	\dfrac{\text{mass of the element in one mole}}{\text{molar mass}}\times100
Radical	a group of atoms behaving as one unit and carrying a charge, e.g. \text{NH}_4^{+}, \text{SO}_4^{2-}

1. One atomic mass unit is equal to [1]athe mass of one hydrogen atomb\tfrac{1}{12} the mass of a carbon-12 atomcthe mass of one proton exactlydthe mass of one electron
2. The empirical formula of glucose, \text{C}_6\text{H}_{12}\text{O}_6, is [1]a\text{CH}b\text{C}_2\text{H}_4\text{O}_2c\text{CH}_2\text{O}d\text{C}_3\text{H}_6\text{O}_3
3. The percentage of oxygen in water (\text{H}=1, \text{O}=16) is [1]a11.1 %b50 %c88.9 %d94.4 %
4. Which of the following is a radical? [1]a\text{NaCl}b\text{NH}_4^{+}c\text{CO}_2d\text{H}_2\text{O}
5. The relative molecular mass of \text{CaCO}_3 (\text{Ca}=40, \text{C}=12, \text{O}=16) is [1]a56b84c100d116
6. A compound has empirical formula \text{CH} and relative molecular mass 78. Its molecular formula is [1]a\text{C}_2\text{H}_2b\text{C}_4\text{H}_4c\text{C}_6\text{H}_6d\text{C}_8\text{H}_8

Answers — 5.1

1. (b) — the definition of the amu.

2. (c) — divide the subscripts 6:12:6 by 6.

3. (c) — 16/18 = 0.889; (a) is the hydrogen percentage, a classic distractor.

4. (b) — \text{NH}_4^{+} is a charged group that moves as a unit.

5. (c) — 40 + 12 + 48 = 100; (b) is \text{MgCO}_3.

6. (c) — 78/13 = 6, so \text{C}_6\text{H}_6.

5.2 Stoichiometry

High-yield fact	Value / form
Avogadro constant	6.022\times10^{23}\ \text{mol}^{-1}
Molar volume at STP (273 K, 1 atm)	22.4\ \text{L mol}^{-1}
Vapour density	\text{V.D.} = \dfrac{M}{2}
Laws	conservation of mass (Lavoisier); definite proportions (Proust); multiple proportions (Dalton); reciprocal proportions (Richter); Gay-Lussac's law of gaseous volumes
Percentage yield	\dfrac{\text{experimental yield}}{\text{theoretical yield}}\times100
Limiting reactant	the reactant used up first; it fixes the amount of product

1. The number of molecules in 0.5 mol of a gas is [1]a6.022\times10^{23}b3.011\times10^{23}c1.2\times10^{24}d6.022\times10^{22}
2. The volume at STP of 8\ \text{g} of oxygen (\text{O}_2) is [1]a2.8\ \text{L}b5.6\ \text{L}c11.2\ \text{L}d22.4\ \text{L}
3. The vapour density of a gas of molar mass 44\ \text{g mol}^{-1} is [1]a22b44c88d11
4. \text{H}_2 + \text{Cl}_2 \to 2\text{HCl}. The mass of HCl formed from 2\ \text{g} of hydrogen with excess chlorine is [1]a36.5\ \text{g}b71\ \text{g}c73\ \text{g}d146\ \text{g}
5. In CO and \text{CO}_2, 12\ \text{g} of carbon combines with 16\ \text{g} and 32\ \text{g} of oxygen. This illustrates the law of [1]aconservation of massbdefinite proportionscmultiple proportionsdreciprocal proportions
6. If the theoretical yield is 20\ \text{g} and 15\ \text{g} is obtained, the percentage yield is [1]a25 %b75 %c80 %d133 %
7. Which postulate of Dalton's atomic theory is no longer accepted? [1]aatoms combine in simple whole-number ratiosbatoms of different elements differ in masscatoms are indivisibledcompounds form by combination of atoms

Answers — 5.2

1. (b) — 0.5\times6.022\times10^{23}.

2. (b) — 8/32 = 0.25 mol; 0.25\times22.4 = 5.6\ \text{L}.

3. (a) — M/2 = 22.

4. (c) — 2\ \text{g} is 1 mol of \text{H}_2, giving 2 mol HCl = 73\ \text{g}; (a) forgets the factor of two.

5. (c) — the oxygen masses 16 : 32 = 1 : 2 for a fixed mass of carbon.

6. (b) — 15/20\times100.

7. (c) — atoms are divisible into electrons, protons and neutrons.

5.3 Atomic Structure

High-yield fact	Value / form
Quantum numbers	n (size, energy), l (shape: 0 to n-1), m_l (orientation: -l to +l), m_s (\pm\tfrac{1}{2})
Electrons in a shell	2n^{2}
Orbitals in s, p, d, f	1, 3, 5, 7
de Broglie wavelength	\lambda = \dfrac{h}{mv}
Heisenberg	\Delta x\cdot\Delta p \ge \dfrac{h}{4\pi}
Exceptions	Cr: [\text{Ar}]3d^{5}4s^{1}; Cu: [\text{Ar}]3d^{10}4s^{1}
Rules	Aufbau (lowest energy first); Pauli (no two electrons with the same four quantum numbers); Hund (single occupancy of degenerate orbitals first)

1. The number of orbitals in a d subshell is [1]a1b3c5d7
2. The maximum number of electrons in the shell with n = 3 is [1]a8b10c18d32
3. The electronic configuration of chromium (Z = 24) is [1]a[\text{Ar}]3d^{4}4s^{2}b[\text{Ar}]3d^{5}4s^{1}c[\text{Ar}]3d^{6}d[\text{Ar}]4s^{2}4p^{4}
4. The quantum number that determines the shape of an orbital is [1]aprincipalbazimuthalcmagneticdspin
5. Electrons occupy degenerate orbitals singly before pairing, according to [1]aHund's rulebPauli's exclusion principlecAufbau principledHeisenberg's principle
6. The number of unpaired electrons in \text{Fe}^{3+} (Z = 26) is [1]a1b3c4d5
7. Heisenberg's uncertainty principle states that [1]aelectrons move in fixed orbitsbposition and momentum cannot both be known exactlycno two electrons have the same four quantum numbersdmatter has no wave nature

Answers — 5.3

1. (c) — l = 2 gives 2l + 1 = 5 orbitals.

2. (c) — 2n^{2} = 18.

3. (b) — a half-filled d subshell is more stable; (a) is the unadjusted Aufbau answer.

4. (b) — l gives s (spherical), p (dumb-bell).

5. (a) — Hund's rule of maximum multiplicity.

6. (d) — \text{Fe}\ 3d^{6}4s^{2} loses the 4s pair and one 3d electron, leaving 3d^{5}: five unpaired.

7. (b) — (a) is Bohr's model and (c) is Pauli's principle.

5.4 Classification of Elements and the Periodic Table

High-yield fact	Value / form
Modern periodic law	properties of elements are periodic functions of their atomic numbers
Effective nuclear charge	Z_{eff} = Z - S (S = shielding constant)
Across a period (left to right)	atomic radius falls; ionization energy and electronegativity rise; metallic character falls
Down a group	atomic radius rises; ionization energy and electronegativity fall
Electronegativity (Pauling)	F = 4.0 (highest); O = 3.5; N = 3.0; Cl = 3.0
Electron affinity	Cl has the largest magnitude, not F (small size, repulsion)

1. The most electronegative element is [1]aoxygenbchlorinecfluorinednitrogen
2. Which has the largest atomic radius? [1]aLibNacKdRb
3. Across a period from left to right the first ionization energy generally [1]aincreasesbdecreasescstays constantdfirst falls then rises sharply
4. The halogen with the highest electron affinity (magnitude) is [1]aFbClcBrdI
5. The correct order of ionic radii of the isoelectronic ions \text{Na}^{+}, \text{Mg}^{2+}, \text{Al}^{3+} is [1]a\text{Na}^{+} > \text{Mg}^{2+} > \text{Al}^{3+}b\text{Al}^{3+} > \text{Mg}^{2+} > \text{Na}^{+}call equald\text{Mg}^{2+} > \text{Na}^{+} > \text{Al}^{3+}
6. An element of atomic number 17 belongs to [1]as-block, group 1bd-block, group 7cp-block, group 17dp-block, group 7

Answers — 5.4

1. (c) — F has the highest Pauling value, 4.0.

2. (d) — radius grows down the group; Rb is the lowest in the list.

3. (a) — the nuclear charge rises with no extra shell (small dips at Group 13 and 16 are exceptions).

4. (b) — fluorine's small 2p subshell repels the incoming electron, so Cl exceeds F.

5. (a) — same electrons, more protons pull harder: \text{Na}^{+} is the largest.

6. (c) — Cl, [\text{Ne}]3s^{2}3p^{5}, is group 17 of the p-block (IUPAC numbering); (d) uses the old numbering.

5.5 Chemical Bonding and Shapes of Molecules

Species	Shape	Hybridization	Bond angle
\text{BeF}_2, \text{CO}_2	linear	sp	180^{\circ}
\text{BF}_3	trigonal planar	sp^{2}	120^{\circ}
\text{CH}_4	tetrahedral	sp^{3}	109.5^{\circ}
\text{NH}_3	trigonal pyramidal	sp^{3}	107^{\circ}
\text{H}_2\text{O}	bent (V-shaped)	sp^{3}	104.5^{\circ}
\text{PCl}_5	trigonal bipyramidal	sp^{3}d	90^{\circ}, 120^{\circ}
\text{SF}_6	octahedral	sp^{3}d^{2}	90^{\circ}

1. The shape of the ammonia molecule is [1]atetrahedralbtrigonal pyramidalcplanardlinear
2. The hybridization of boron in \text{BF}_3 is [1]aspbsp^{2}csp^{3}dsp^{3}d
3. The bond angle in water is about [1]a90^{\circ}b104.5^{\circ}c109.5^{\circ}d120^{\circ}
4. Which molecule has zero dipole moment? [1]a\text{H}_2\text{O}b\text{NH}_3c\text{CO}_2d\text{HCl}
5. The strongest hydrogen bonding occurs in [1]aHClb\text{H}_2\text{S}cHBrdHF
6. A coordinate covalent bond is present in [1]a\text{CH}_4b\text{NH}_4^{+}cNaCld\text{H}_2
7. The shape of \text{SF}_6 is [1]atetrahedralbtrigonal bipyramidalcoctahedraldsquare planar

Answers — 5.5

1. (b) — three bond pairs and one lone pair; (a) ignores the lone pair.

2. (b) — three bond pairs and no lone pair give sp^{2}, trigonal planar.

3. (b) — two lone pairs compress the angle below 109.5^{\circ}.

4. (c) — the two C=O dipoles of the linear molecule cancel.

5. (d) — F is the most electronegative and smallest; HF forms the strongest H-bonds.

6. (b) — the lone pair of N is donated to \text{H}^{+}.

7. (c) — six bond pairs, no lone pair.

5.6 Oxidation and Reduction

High-yield fact	Value / form
Oxidation	loss of electrons; increase in oxidation number
Reduction	gain of electrons; decrease in oxidation number
Rules	free element = 0; H usually +1; O usually -2 (-1 in peroxides); sum = charge on the ion
Common values	S in \text{H}_2\text{SO}_4: +6; Mn in \text{KMnO}_4: +7; Cr in \text{K}_2\text{Cr}_2\text{O}_7: +6; Cl in \text{HClO}_4: +7
Faraday's laws	m = ZIt; 1\ \text{F} = 96\,500\ \text{C} deposits one equivalent
Electrolysis	oxidation at the anode, reduction at the cathode

1. The oxidation number of chromium in \text{K}_2\text{Cr}_2\text{O}_7 is [1]a+3b+4c+6d+7
2. In 2\text{Mg} + \text{O}_2 \to 2\text{MgO} the oxidizing agent is [1]aMgb\text{O}_2cMgOdnone
3. The average oxidation number of sulphur in \text{Na}_2\text{S}_2\text{O}_3 is [1]a0b+2c+4d+6
4. The charge needed to deposit one mole of copper from \text{Cu}^{2+} is [1]a96\,500\ \text{C}b144\,750\ \text{C}c193\,000\ \text{C}d289\,500\ \text{C}
5. Reduction takes place at the [1]acathodebanodecsalt bridgedelectrolyte
6. The oxidation number of manganese is highest in [1]a\text{MnO}_2b\text{MnSO}_4c\text{K}_2\text{MnO}_4d\text{KMnO}_4

Answers — 5.6

1. (c) — 2x + 2(+1) + 7(-2) = 0 gives x = +6.

2. (b) — \text{O}_2 gains electrons and so oxidizes Mg.

3. (b) — 2(+1) + 2x + 3(-2) = 0 gives x = +2 on average.

4. (c) — \text{Cu}^{2+} + 2e^{-} \to \text{Cu} needs 2\times96\,500.

5. (a) — reduction is gain of electrons at the cathode (RED CAT).

6. (d) — \text{MnO}_2 +4, \text{MnSO}_4 +2, \text{K}_2\text{MnO}_4 +6, \text{KMnO}_4 +7.

5.7 States of Matter

High-yield fact	Value / form
Boyle's law	P_1V_1 = P_2V_2 at constant T
Charles' law	\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} at constant P (T in kelvin)
Ideal gas equation	PV = nRT; R = 8.314\ \text{J mol}^{-1}\text{K}^{-1} = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}
Dalton's law	P_{total} = P_1 + P_2 + \cdots
Graham's law	\dfrac{r_1}{r_2} = \sqrt{\dfrac{M_2}{M_1}}
Real gases deviate	at high pressure and low temperature
Solids	efflorescent loses water of crystallization to air (\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}); deliquescent absorbs water and dissolves (\text{CaCl}_2, NaOH)

1. A gas occupies 2\ \text{L} at 1\ \text{atm}. At constant temperature and 4\ \text{atm} it occupies [1]a0.25\ \text{L}b0.5\ \text{L}c1\ \text{L}d8\ \text{L}
2. The ratio of the rates of diffusion of \text{H}_2 and \text{O}_2 under the same conditions is [1]a1:4b1:16c4:1d16:1
3. The value of R in \text{L atm mol}^{-1}\text{K}^{-1} is [1]a8.314b0.0821c1.987d62.4
4. A gas at 27^{\circ}\text{C} is heated at constant pressure until its volume doubles. The new temperature is [1]a54^{\circ}\text{C}b300\ \text{K}c327^{\circ}\text{C}d600^{\circ}\text{C}
5. An efflorescent substance is [1]a\text{Na}_2\text{CO}_3\cdot10\text{H}_2\text{O}banhydrous \text{CaCl}_2cNaCldconc. \text{H}_2\text{SO}_4
6. A real gas behaves most ideally at [1]ahigh pressure, low temperatureblow pressure, low temperatureclow pressure, high temperaturedhigh pressure, high temperature
7. The boiling point of a liquid is the temperature at which [1]avapour pressure is 1\ \text{atm} alwaysbvapour pressure equals the external pressurecthe liquid freezesdthe surface tension is zero

Answers — 5.7

1. (b) — P_1V_1 = P_2V_2 gives V_2 = 2\times1/4 = 0.5\ \text{L}.

2. (c) — \sqrt{32/2} = 4; the lighter gas diffuses faster.

3. (b) — (a) is in joules; (d) is in \text{L mmHg}-type units.

4. (c) — T doubles in kelvin: 300 \to 600\ \text{K} = 327^{\circ}\text{C}; (a) doubles the Celsius value, the common trap.

5. (a) — washing soda loses water on standing.

6. (c) — molecules are far apart and fast, so forces and volume matter least.

7. (b) — (a) is the normal boiling point only.

5.8 Chemical Equilibrium

High-yield fact	Value / form
Law of mass action	the rate of a reaction is proportional to the product of the active masses (concentrations) of the reactants
For aA + bB \rightleftharpoons cC + dD	K_c = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}
K_p and K_c	K_p = K_c(RT)^{\Delta n}; \Delta n = n_{gas,products} - n_{gas,reactants}
Le Chatelier	a system at equilibrium opposes any change imposed on it
Catalyst	speeds both rates equally; does not change K or the position

1. For which reaction is K_p = K_c? [1]a\text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI}b\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3c\text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2d2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3
2. For \text{N}_2\text{O}_4 \rightleftharpoons 2\text{NO}_2, \Delta n is [1]a-1b0c+1d+2
3. Increasing the pressure on \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 [1]ashifts it to the leftbshifts it to the rightchas no effectdchanges K_c
4. A catalyst in a reversible reaction [1]ashifts the equilibrium to the rightbraises K_cchelps equilibrium to be reached faster without changing itdstops the reverse reaction
5. The K_c expression for \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 is [1]a\dfrac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]}b\dfrac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}c[\text{PCl}_3][\text{Cl}_2]d[\text{PCl}_5]
6. At chemical equilibrium [1]athe forward rate equals the reverse ratebthe reaction stopscconcentrations of reactants and products are equaldthe forward rate is zero

Answers — 5.8

1. (a) — \Delta n = 2 - 2 = 0.

2. (c) — 2 - 1 = +1.

3. (b) — the right side has fewer gas moles (2 against 4).

4. (c) — a catalyst changes the rate, not the position.

5. (b) — products over reactants, each to the power of its coefficient.

6. (a) — equilibrium is dynamic; (c) confuses equal rates with equal concentrations.

5.9 Chemistry of Non-metals

High-yield fact	Value / form
Heavy water	\text{D}_2\text{O} (deuterium oxide); used as a moderator in reactors
Ozone	\text{O}_3; bent; depleted by CFCs; turns moist starch–iodide paper blue
Ring test	brown ring of [\text{Fe}(\text{H}_2\text{O})_5\text{NO}]^{2+} at the junction of acid and \text{FeSO}_4 solution
Kipp's apparatus	\text{FeS} + \text{H}_2\text{SO}_4 \to \text{FeSO}_4 + \text{H}_2\text{S}
Oxidation number of N	+5 in \text{HNO}_3; +3 in \text{HNO}_2; -3 in \text{NH}_3
Halogens	\text{F}_2 and \text{Cl}_2 gases; \text{Br}_2 liquid; \text{I}_2 solid; oxidizing power \text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2
Fullerene	\text{C}_{60}, buckminsterfullerene
Thiosulphate	\text{Na}_2\text{S}_2\text{O}_3\cdot5\text{H}_2\text{O} ("hypo")

1. The brown ring in the nitrate ring test is due to [1]a\text{FeSO}_4b[\text{Fe}(\text{H}_2\text{O})_5\text{NO}]^{2+}c\text{Fe}_2\text{O}_3d\text{NO}_2
2. The gas prepared in a Kipp's apparatus from \text{FeS} is [1]a\text{H}_2b\text{CO}_2c\text{H}_2\text{S}d\text{SO}_2
3. Depletion of the ozone layer is mainly caused by [1]a\text{CO}_2bchlorofluorocarbonsc\text{CH}_4d\text{N}_2
4. Heavy water has the formula [1]a\text{D}_2\text{O}b\text{H}_2\text{O}_2c\text{H}_2^{18}\text{O} onlyd\text{HDO} only
5. The allotrope of carbon containing 60 carbon atoms is [1]adiamondbgraphitecbuckminsterfullerenedcharcoal
6. The oxidation number of nitrogen in \text{HNO}_3 is [1]a+1b+3c+4d+5
7. The only halogen that is a liquid at room temperature is [1]achlorinebbromineciodinedfluorine
8. The formula of sodium thiosulphate is [1]a\text{Na}_2\text{SO}_4b\text{Na}_2\text{SO}_3c\text{Na}_2\text{S}_2\text{O}_3d\text{NaHSO}_4

Answers — 5.9

1. (b) — nitric oxide coordinates to \text{Fe}^{2+}.

2. (c) — \text{FeS} with dilute acid gives hydrogen sulphide; (b) comes from marble.

3. (b) — chlorine radicals from CFCs catalyse \text{O}_3 \to \text{O}_2.

4. (a) — deuterium oxide.

5. (c) — the 60-atom cage.

6. (d) — 1 + x - 6 = 0 gives +5.

7. (b) — bromine; mercury is the only liquid metal.

8. (c) — (a) is sulphate and (b) is sulphite, close distractors.

5.10 Chemistry of Metals

High-yield fact	Value / form
Calcination	heating the ore in limited or no air (carbonates, hydroxides)
Roasting	heating the ore in excess air (sulphides)
Flux and slag	gangue + flux \to slag (\text{CaO} + \text{SiO}_2 \to \text{CaSiO}_3)
Thermite	\text{Fe}_2\text{O}_3 + 2\text{Al} \to 2\text{Fe} + \text{Al}_2\text{O}_3
Down's process	electrolysis of fused NaCl (with \text{CaCl}_2) for sodium
Plaster of Paris	\text{CaSO}_4\cdot\tfrac{1}{2}\text{H}_2\text{O}; Epsom salt \text{MgSO}_4\cdot7\text{H}_2\text{O}; quick lime CaO; bleaching powder \text{CaOCl}_2
Refining	poling (copper); electro-refining (Cu, Zn)

1. The formula of plaster of Paris is [1]a\text{CaSO}_4\cdot2\text{H}_2\text{O}b\text{CaSO}_4\cdot\tfrac{1}{2}\text{H}_2\text{O}c\text{CaSO}_4d\text{CaCO}_3
2. Calcination is heating an ore [1]ain limited or no airbin excess aircwith carbondin an electric current
3. In Down's process the electrolyte is [1]aaqueous NaClbmolten NaOHcfused NaCl with \text{CaCl}_2daqueous NaOH
4. Epsom salt is [1]a\text{CaSO}_4\cdot2\text{H}_2\text{O}b\text{MgSO}_4\cdot7\text{H}_2\text{O}c\text{ZnSO}_4\cdot7\text{H}_2\text{O}d\text{FeSO}_4\cdot7\text{H}_2\text{O}
5. In the thermite process the reducing agent is [1]acarbonbhydrogenccarbon monoxidedaluminium
6. The substance formed when a flux combines with gangue is [1]amattebslagcalloydamalgam
7. Poling is used in the refining of [1]aironbcoppercaluminiumdzinc

Answers — 5.10

1. (b) — gypsum (a) loses three-quarters of its water on heating to 120^{\circ}\text{C}.

2. (a) — (b) is roasting.

3. (c) — water would give NaOH and \text{H}_2 instead; \text{CaCl}_2 lowers the melting point.

4. (b) — magnesium sulphate heptahydrate.

5. (d) — Al is a stronger reducing agent than Fe.

6. (b) — slag is the fusible product removed from the furnace.

7. (b) — blister copper is stirred with green wood poles to remove \text{Cu}_2\text{O}.

5.11 Bio-inorganic Chemistry

Metal	Role
Mg	central ion of chlorophyll
Fe	haemoglobin (oxygen carrier), cytochromes
Co	vitamin \text{B}_{12}
Zn	carbonic anhydrase and many enzymes
Na, K	nerve impulse; sodium–potassium pump moves 3\ \text{Na}^{+} out and 2\ \text{K}^{+} in
Ca	bones, teeth, muscle contraction, blood clotting
Toxic metals	Hg (Minamata disease), Pb, As, Cd

1. The metal at the centre of chlorophyll is [1]aironbmagnesiumccalciumdzinc
2. The metal in haemoglobin is [1]aironbcopperccobaltdmagnesium
3. The \text{Na}^{+}/\text{K}^{+} pump moves [1]a2 \text{Na}^{+} out and 3 \text{K}^{+} inb3 \text{K}^{+} out and 2 \text{Na}^{+} inc3 \text{Na}^{+} out and 2 \text{K}^{+} indequal numbers of each
4. Vitamin \text{B}_{12} contains [1]aironbzincccobaltdnickel
5. Minamata disease is caused by poisoning with [1]aleadbmercuryccadmiumdarsenic
6. The enzyme carbonic anhydrase contains [1]azincbironccopperdmanganese

Answers — 5.11

1. (b) — \text{Mg}^{2+} sits in the porphyrin ring (iron is in haem).

2. (a) — \text{Fe}^{2+} binds \text{O}_2.

3. (c) — three sodium out, two potassium in, per ATP.

4. (c) — cobalamin.

5. (b) — methylmercury in fish.

6. (a) — \text{Zn}^{2+} in the active site.

5.12 Basic Concept of Organic Chemistry

High-yield fact	Value / form
Tetracovalency	carbon forms four covalent bonds (sp^{3}, sp^{2}, sp)
Catenation	self-linking of carbon atoms into chains and rings
Homologous series	consecutive members differ by -\text{CH}_2-
Functional groups	-\text{OH} alcohol; -\text{CHO} aldehyde; >\text{C=O} ketone; -\text{COOH} acid; -\text{NH}_2 amine
Octane number	iso-octane = 100; n-heptane = 0
Cetane number	rates diesel fuel; cetane (hexadecane) = 100
Cracking	breaking large hydrocarbons into smaller, more useful ones

1. The property of carbon to form long chains is called [1]acatenationbisomerismcallotropydpolymerization
2. The functional group of alcohols is [1]a-\text{CHO}b-\text{OH}c-\text{COOH}d-\text{NH}_2
3. Successive members of a homologous series differ by [1]a\text{CH}_4b\text{CH}_2c\text{C}_2\text{H}_4d\text{CH}
4. The octane number of iso-octane is [1]a0b50c100d120
5. The cetane number is a measure of the quality of [1]apetrolbdieselckerosenedLPG
6. Cracking is the process of [1]abreaking large hydrocarbon molecules into smaller onesbjoining small moleculescremoving sulphurddistilling crude oil
7. The valency of carbon is [1]a2b3c4d6

Answers — 5.12

1. (a) — catenation; isomerism is the same formula with different structures.

2. (b) — (a) is the aldehyde group.

3. (b) — one methylene unit.

4. (c) — the standard for 100; n-heptane is 0.

5. (b) — the ignition quality of diesel; (a) uses octane number.

6. (a) — (d) describes fractional distillation.

7. (c) — tetravalent.

5.13 Fundamental Principles of Organic Chemistry

High-yield fact	Value / form
Lassaigne's test	N: Prussian blue; S: black PbS with lead acetate; halogen: curdy white (Cl), pale yellow (Br), yellow (I) precipitate with \text{AgNO}_3
Isomerism	chain, position, functional, metamerism, tautomerism (structural); geometrical and optical (stereo)
Optical isomerism	needs a chiral carbon (four different groups)
Fission	homolytic gives free radicals; heterolytic gives ions
Inductive effect	-\text{CH}_3 is +I; -\text{Cl}, -\text{NO}_2 are -I
Carbocation stability	3^{\circ} > 2^{\circ} > 1^{\circ} > \text{CH}_3^{+}

1. The IUPAC name of \text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_3 is [1]apentaneb2-methylbutanec3-methylbutanedisopentene
2. In Lassaigne's test, a Prussian blue colour shows [1]anitrogenbsulphurcchlorinedphosphorus
3. Ethanol and dimethyl ether are [1]achain isomersbposition isomerscfunctional isomersdmetamers
4. Which compound shows geometrical isomerism? [1]apropenebbut-2-enecethenedbut-1-ene
5. Which of the following is an electrophile? [1]a\text{NH}_3b\text{OH}^{-}c\text{H}_2\text{O}d\text{BF}_3
6. Which group shows a -I effect? [1]a-\text{Cl}b-\text{CH}_3c-\text{C}_2\text{H}_5d-\text{CH}(\text{CH}_3)_2
7. Which molecule contains a chiral carbon? [1]a\text{CH}_3\text{CH}_2\text{OH}b\text{CH}_3\text{COOH}c\text{CH}_3\text{CH(OH)COOH}d\text{CH}_3\text{CH}_2\text{CH}_3
8. Homolytic fission of a covalent bond produces [1]acations and anionsbfree radicalsccarbanions onlydcarbocations only

Answers — 5.13

1. (b) — the longest chain is butane with a methyl on C-2; (c) numbers from the wrong end.

2. (a) — \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 forms from cyanide.

3. (c) — \text{C}_2\text{H}_6\text{O} with -\text{OH} against -\text{O}-.

4. (b) — each alkene carbon of but-2-ene has two different groups.

5. (d) — \text{BF}_3 is electron-deficient; the others have lone pairs.

6. (a) — Cl withdraws electrons; alkyl groups push them (+I).

7. (c) — the middle carbon carries H, \text{CH}_3, OH and COOH.

8. (b) — each fragment keeps one electron.

5.14 Hydrocarbons

High-yield fact	Value / form
Wurtz	2\text{CH}_3\text{Br} + 2\text{Na} \to \text{C}_2\text{H}_6 + 2\text{NaBr}
Kolbe	electrolysis of sodium acetate gives ethane at the anode
Markovnikov	in \text{HX} addition, H goes to the carbon already holding more H
Peroxide effect	anti-Markovnikov, and only for HBr
Ethyne	\text{CaC}_2 + 2\text{H}_2\text{O} \to \text{C}_2\text{H}_2 + \text{Ca(OH)}_2
Terminal alkyne	acidic: gives white silver acetylide with ammoniacal \text{AgNO}_3, red copper acetylide with ammoniacal \text{Cu}_2\text{Cl}_2
Tests for unsaturation	bromine water: orange to colourless; Baeyer's reagent (cold alkaline dilute \text{KMnO}_4): purple to colourless

1. Propene reacts with HBr (no peroxide) to give mainly [1]a1-bromopropaneb2-bromopropanec1,2-dibromopropanedpropane
2. The Wurtz reaction of \text{CH}_3\text{Cl} with sodium gives [1]aethanebmethanecpropanedethene
3. Ethyne gives with ammoniacal \text{AgNO}_3 [1]ano reactionba red precipitateca white precipitateda yellow precipitate
4. Baeyer's reagent is [1]aacidified \text{K}_2\text{Cr}_2\text{O}_7bcold alkaline dilute \text{KMnO}_4cbromine waterdTollens' reagent
5. Which does not decolourise bromine water? [1]aethenebethynecpropenedethane
6. Heating ethanol with conc. \text{H}_2\text{SO}_4 at 170^{\circ}\text{C} gives [1]aethenebethanecdiethyl etherdethyne
7. Kolbe's electrolysis of sodium acetate gives [1]amethanebethenecethanedpropane
8. The peroxide effect is observed with [1]aHClbHBrcHIdall halogen acids

Answers — 5.14

1. (b) — H adds to the \text{CH}_2 end, Br to the middle carbon (Markovnikov).

2. (a) — 2\text{CH}_3\text{Cl} + 2\text{Na} \to \text{C}_2\text{H}_6; the product has twice the carbons.

3. (c) — silver acetylide, white.

4. (b) — alkaline cold dilute permanganate.

5. (d) — ethane is saturated.

6. (a) — dehydration at 170^{\circ}\text{C}; at 140^{\circ}\text{C} the product is the ether.

7. (c) — 2\text{CH}_3\text{COO}^{-} \to \text{C}_2\text{H}_6 + 2\text{CO}_2 + 2e^{-}.

8. (b) — only the H–Br bond splits homolytically under peroxide.

5.15 Aromatic Hydrocarbons

High-yield fact	Value / form
Hückel's rule	planar, cyclic, conjugated with (4n + 2)\ \pi electrons
Benzene	six \pi electrons; all C–C bonds equal (139\ \text{pm}); resonance hybrid
Nitration	conc. \text{HNO}_3 + conc. \text{H}_2\text{SO}_4 at about 323\ \text{K} gives nitrobenzene
Friedel–Crafts	anhydrous \text{AlCl}_3 with an alkyl or acyl halide
Directing groups	-\text{OH}, -\text{CH}_3, -\text{NH}_2: ortho/para; -\text{NO}_2, -\text{COOH}, -\text{CHO}: meta
Preparation	sodium benzoate heated with soda lime; ethyne through a red-hot tube

1. The number of \pi electrons in benzene is [1]a2b4c6d8
2. Hückel's rule for aromaticity is [1]a4n \pi electronsb(4n+2) \pi electronsc2n \pi electronsdn^{2} \pi electrons
3. The nitrating mixture is [1]aconc. \text{HNO}_3 and conc. \text{H}_2\text{SO}_4bdilute \text{HNO}_3c\text{HNO}_3 and HCld\text{NaNO}_3 and water
4. The catalyst in the Friedel–Crafts reaction is [1]aNibanhydrous \text{AlCl}_3c\text{V}_2\text{O}_5dPt
5. In nitrobenzene the -\text{NO}_2 group directs an incoming electrophile to the [1]aortho positionbpara positioncmeta positiondortho and para
6. Benzene is obtained by heating sodium benzoate with [1]aNaOHbsoda limeczinc dustdconc. \text{H}_2\text{SO}_4

Answers — 5.15

1. (c) — one \pi electron from each of the six carbons.

2. (b) — benzene has n = 1.

3. (a) — the sulphuric acid generates \text{NO}_2^{+}.

4. (b) — a Lewis acid; it must be anhydrous.

5. (c) — -\text{NO}_2 deactivates the ring and directs meta.

6. (b) — decarboxylation with soda lime (\text{NaOH} + \text{CaO}); plain NaOH is not enough.

5.16 Fundamentals of Applied Chemistry

High-yield fact	Value / form
Stages of a new product	research in the laboratory, then a pilot plant, then a full-scale plant
Batch process	the reactor is charged, the reaction runs, and the product is removed before the next charge
Continuous process	feed enters and product leaves without a break (ammonia, sulphuric acid)
Fixed capital	land, buildings, machinery
Working capital	raw materials, wages, stocks
Environmental impact	\text{SO}_2 gives acid rain; fertilizer run-off gives eutrophication

1. A batch process is one in which [1]afeed enters and product leaves continuouslybthe reactor is charged, the reaction is completed, and the product is removedconly gases reactdno catalyst is used
2. A pilot plant is [1]athe full production plantba plant for waste onlyca small-scale plant used to test a process before scale-upda laboratory flask
3. Which is an example of fixed capital? [1]araw materialsbwagescland and machinerydfuel in stock
4. Large-volume products such as ammonia are usually made by [1]abatch processingbcontinuous processingchand mixingdfermentation only
5. Acid rain is mainly caused by [1]a\text{CO}_2b\text{SO}_2 and \text{NO}_xc\text{O}_3d\text{N}_2
6. Eutrophication is caused by excess [1]anitrates and phosphatesboxygencchlorinedcarbon monoxide

Answers — 5.16

1. (b) — (a) describes a continuous process.

2. (c) — it tests a process on a small scale first.

3. (c) — raw materials and wages are working capital.

4. (b) — steady, high-volume production.

5. (b) — \text{SO}_2 and \text{NO}_x form sulphuric and nitric acids in the air.

6. (a) — nutrient run-off causes algal blooms.

5.17 Modern Chemical Manufactures

Process	Key facts
Haber (ammonia)	\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3, \Delta H = -92\ \text{kJ}; Fe catalyst, Mo or \text{Al}_2\text{O}_3 promoter; about 200\ \text{atm} and 450–500^{\circ}\text{C}
Ostwald (nitric acid)	4\text{NH}_3 + 5\text{O}_2 \to 4\text{NO} + 6\text{H}_2\text{O} over Pt–Rh gauze at about 800^{\circ}\text{C}; then 2\text{NO} + \text{O}_2 \to 2\text{NO}_2; 3\text{NO}_2 + \text{H}_2\text{O} \to 2\text{HNO}_3 + \text{NO}
Contact (sulphuric acid)	2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 over \text{V}_2\text{O}_5 at about 450^{\circ}\text{C}; \text{SO}_3 absorbed in conc. \text{H}_2\text{SO}_4 to give oleum
Diaphragm cell (NaOH)	brine electrolysed; \text{Cl}_2 at the anode, \text{H}_2 and NaOH at the cathode; asbestos diaphragm
Solvay (soda ash)	\text{NaCl} + \text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O} \to \text{NaHCO}_3 + \text{NH}_4\text{Cl}; 2\text{NaHCO}_3 \to \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}; by-product \text{CaCl}_2

1. The catalyst used in the Haber process is [1]a\text{V}_2\text{O}_5bfinely divided ironcplatinumdnickel
2. The catalyst used in Ostwald's process is [1]aFeb\text{V}_2\text{O}_5cPt–Rh gauzed\text{MnO}_2
3. In the contact process \text{SO}_3 is absorbed in [1]awaterbconc. \text{H}_2\text{SO}_4cdilute \text{H}_2\text{SO}_4dNaOH
4. The by-product of the Solvay process is [1]a\text{NH}_3b\text{CaCl}_2cNaOHd\text{CaCO}_3
5. In the diaphragm cell, chlorine is liberated at the [1]aanodebcathodecdiaphragmdoutlet
6. The promoter in the Haber process is [1]a\text{V}_2\text{O}_5bPtcModNi
7. The usual temperature of the Haber process is about [1]a25^{\circ}\text{C}b200^{\circ}\text{C}c450^{\circ}\text{C}d1000^{\circ}\text{C}
8. The temperature of the catalytic oxidation of \text{NH}_3 in Ostwald's process is about [1]a100^{\circ}\text{C}b450^{\circ}\text{C}c800^{\circ}\text{C}d1500^{\circ}\text{C}

Answers — 5.17

1. (b) — \text{V}_2\text{O}_5 belongs to the contact process.

2. (c) — platinum–rhodium, a trap for those who choose Pt alone.

3. (b) — water gives an acid mist that is hard to condense.

4. (b) — formed when \text{NH}_4\text{Cl} is treated with \text{Ca(OH)}_2 to recover ammonia.

5. (a) — 2\text{Cl}^{-} \to \text{Cl}_2 + 2e^{-} at the anode.

6. (c) — molybdenum (or \text{Al}_2\text{O}_3) improves the iron catalyst.

7. (c) — a compromise between rate and yield for an exothermic reaction.

8. (c) — the gauze glows red-hot, so the reaction is self-sustaining.

Chapter summary

• Group A is eleven one-mark MCQs drawn from all seventeen units; no unit is safe to skip, and the question types are: a name or definition, a formula or reagent, a one-step calculation, or a misconception.
• Stoichiometry: N_A = 6.022\times10^{23}; molar volume 22.4\ \text{L} at STP; \text{V.D.} = M/2; percentage yield = actual/theoretical \times100.
• Atomic structure: 2n^{2} electrons per shell; 2l+1 orbitals per subshell; Cr and Cu are the 3d exceptions; Hund, Pauli and Aufbau.
• Periodic trends: radius falls and IE and EN rise across a period; the reverse down a group; F is the most electronegative, Cl has the highest electron affinity.
• Shapes: linear sp, planar sp^{2}, tetrahedral sp^{3}, pyramidal and bent (lone pairs), sp^{3}d and sp^{3}d^{2} for \text{PCl}_5 and \text{SF}_6.
• Redox: oxidation is loss of electrons; \text{Cr} = +6 in dichromate, \text{Mn} = +7 in permanganate; 1\ \text{F} = 96\,500\ \text{C}.
• Gases: P_1V_1 = P_2V_2, PV = nRT, r_1/r_2 = \sqrt{M_2/M_1}; K_p = K_c(RT)^{\Delta n}.
• Inorganic: ring test, Kipp's apparatus, Down's process, plaster of Paris \text{CaSO}_4\cdot\tfrac{1}{2}\text{H}_2\text{O}; Haber (Fe), Ostwald (Pt–Rh), contact (\text{V}_2\text{O}_5), Solvay.
• Organic: Markovnikov and the peroxide effect (HBr only); Wurtz and Kolbe give alkanes; Baeyer and bromine water test unsaturation; Hückel (4n+2); -\text{NO}_2 directs meta.

Practice questions

Group A — Multiple choice (1 mark each)

1. The number of moles in 11\ \text{g} of \text{CO}_2 is [1]a0.125b0.25c0.5d4
2. The shape of the methane molecule is [1]aplanarbtetrahedralcpyramidaldlinear
3. The gas liberated at the cathode during electrolysis of brine in a diaphragm cell is [1]achlorineboxygenchydrogendnitrogen
4. The hybridization of carbon in ethyne is [1]asp^{3}bsp^{2}cspdsp^{3}d

Answers to Group A

1. (b) — 11/44 = 0.25 mol.

2. (b) — four bond pairs, 109.5^{\circ}.

3. (c) — 2\text{H}_2\text{O} + 2e^{-} \to \text{H}_2 + 2\text{OH}^{-} at the cathode.

4. (c) — each carbon forms one \sigma bond to H and one to C, with two \pi bonds.

Group B — Short answer (5 marks each)

1. Calculate the mass of \text{CaO} obtained by heating 50\ \text{g} of \text{CaCO}_3. [5]
2. State Le Chatelier's principle and apply it to the Haber process for pressure and temperature. [5]
3. Write the Lewis structure idea and shape of \text{H}_2\text{O} and explain why its bond angle is below 109.5^{\circ}. [5]
4. Distinguish between ethane, ethene and ethyne by two chemical tests. [5]

Answers to Group B

1. \text{CaCO}_3 \to \text{CaO} + \text{CO}_2; 50\ \text{g} = 0.5 mol; one mole of CaCO₃ gives one mole of CaO; mass of CaO = 0.5\times56 = 28\ \text{g}.

2. If a system at equilibrium is disturbed, it shifts to oppose the disturbance. \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 (\Delta H < 0, fewer moles on the right): high pressure favours ammonia; low temperature favours ammonia but slows the rate, so about 450^{\circ}\text{C} with an Fe catalyst is the compromise.

3. Oxygen has six valence electrons, two used in bonds, so there are two bond pairs and two lone pairs (sp^{3}). Lone pair–lone pair and lone pair–bond pair repulsion exceeds bond pair–bond pair repulsion, squeezing the angle to 104.5^{\circ}; the shape is bent.

4. Bromine water: ethene and ethyne decolourise it, ethane does not. Ammoniacal \text{AgNO}_3: only ethyne gives a white precipitate. Baeyer's reagent is decolourised by ethene and ethyne but not ethane.

Group C — Long answer (8 marks each)

1. Describe the manufacture of sulphuric acid by the contact process with the reactions, conditions and a flow sheet. [8]
2. Describe the Solvay process for sodium carbonate, giving the reactions and the role of ammonia recovery. [8]

Answers to Group C

1. Steps: (i) Sulphur dioxide: \text{S} + \text{O}_2 \to \text{SO}_2 (or roasting pyrites, 4\text{FeS}_2 + 11\text{O}_2 \to 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2); the gas is purified and dried. (ii) Oxidation: 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3, \Delta H = -197\ \text{kJ}, over \text{V}_2\text{O}_5 at 400–450^{\circ}\text{C} and 1–2 atm; excess oxygen and removal of heat favour the yield. (iii) Absorption: \text{SO}_3 is dissolved in conc. \text{H}_2\text{SO}_4 to give oleum: \text{SO}_3 + \text{H}_2\text{SO}_4 \to \text{H}_2\text{S}_2\text{O}_7. (iv) Dilution: \text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \to 2\text{H}_2\text{SO}_4. Water is not used for direct absorption because it forms a mist. Flow sheet: burner \to purifier \to dryer \to converter (\text{V}_2\text{O}_5) \to absorption tower \to oleum tank \to dilution.

2. Brine is saturated with \text{NH}_3 in an ammonia tower. In the carbonating (Solvay) tower \text{CO}_2 is passed in: \text{NaCl} + \text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O} \to \text{NaHCO}_3 + \text{NH}_4\text{Cl}; sparingly soluble \text{NaHCO}_3 separates and is filtered. Heating gives soda ash: 2\text{NaHCO}_3 \to \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} (the \text{CO}_2 is recycled). \text{CO}_2 comes from heating limestone, \text{CaCO}_3 \to \text{CaO} + \text{CO}_2, and the lime slakes to \text{Ca(OH)}_2. Ammonia is recovered: 2\text{NH}_4\text{Cl} + \text{Ca(OH)}_2 \to 2\text{NH}_3 + \text{CaCl}_2 + 2\text{H}_2\text{O}. Overall: 2\text{NaCl} + \text{CaCO}_3 \to \text{Na}_2\text{CO}_3 + \text{CaCl}_2; ammonia is recycled, so only \text{CaCl}_2 is lost.