Chemistry Answers Chapter 3

Chapter 3 Exercises Element Period Group 4 C (a) helium 1 18 7 (b) (c) chlorine barium 3 6 17 2 (a) Half the distance between the nuclei of neighbouring atoms of the same element. (d) francium 7 1 1 2 (a) Periods are rows and groups are columns. (b) 1s22s22p63s23p3 The valence energy level is the third principal energy level, so the element is in period 3. It has the 3p3 configuration, so it is in the third group of the p block, which is Group 15. 3 Element 51 is antimony (Sb), which is in Group 15. Its valence electrons are 5s25p3, and so it has five valence electrons. 5 B 6 C (b) (i) The noble gases do not form stable ions and engage in ionic bonding so the distance between neighbouring ions cannot be defined. (ii) The atomic radii decrease from Na to Cl. This is because the number of inner, shielding, electron is is constant (10) but the nuclear charge increases from +11 to +17. As we go from Na to Cl, the increasing effective nuclear charge pulls the outer electrons closer. 9 10:01 Answers Si4+ has an electronic configuration of 1s22s22p6 where Si4– has an electronic configuration of 1s22s22p63s23p6. Si4+ has two occupied energy levels and Si4− has three and so Si4− is larger. 9 A 13 10 B 11 C 12 32 D (a) The electron in the outer electron energy level (level 4) is removed to form K+. The net attractive force increases as the electrons in the third energy level experience a greater effective nuclear charge. (b) P3− has electronic configuration of 1s22s22p63s23p6 whereas Si4+ has an electronic configuration of 1s22s22p2. P3– has one more principal energy level than Si4+ so its valence electrons will be further from the nucleus and it will have a larger ionic radius. (c) The ions have the same electron configuration, 1s22s22p63s23p6: both have two complete shells; the extra protons in Na+ attract the electrons more strongly. 14 Phosphorus exists as molecules with four atoms: P4. Sulfur exists as molecules with eight atoms: S8. There are stronger London dispersion forces between the larger S8 molecules as there are more electrons. 15 D 16 C 17 Cl− > Cl > Cl+ 18 B 19 C 20 D 22 Sodium floats on the surface; it melts into a sphere; there is fizzing/effervescence/bubbles; sound is produced; solution gets hot; white smoke is produced. 21 B 23 D 24 The reactivities of the alkali metals increase but those of the halogens decrease. 25 C 26 D 27 D 29 A 30 B 31 D A (b) Structure and bonding MgO (s) SiO2 (quartz) (s) P4O10 (s) molecular, covalent bonding; weak van der Waals’ forces between molecules; P4O10 is larger molecule and so has stronger London dispersion forces and a higher melting point than SO2 SO2 (g) molecular, covalent bonding; weak van der Waals’ forces between molecules; SO2 is smaller molecule and so has weaker London dispersion forces and a higher melting point than P4O10 (c) Oxide MgO giant structure ionic bonding; strong attraction between oppositely charged ions giant structure covalent bonding; strong covalent bonds throughout structure pH of solution Equations alkaline MgO(s) +H2O(l) → Mg(OH)2(aq) SiO2 neutral – oxide (quartz) is insoluble 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) 28 (a) State under standard conditions 8 P4O10 acidic P4O10(s) + 6H2O(l) → 4H3PO4(aq) SO2 acidic SO2(l) + H2O(l) → H2SO3(aq) (d) (i) Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l) (ii) Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq) 33 The oxides of Na and Mg are basic; the oxide of Al is amphoteric; the oxides of Si to Cl are acidic. Ar forms no oxide. Na2O + H2O → 2NaOH SO3 + H2O → H2SO4 10 10:01 34 3d 4s 46 47 Ti3+ Ni2+ Zn2+ 35 D 38 (a) 1s22s22p63s23p63d104s2 B 37 (c) The element does not form ions with partially filled d orbitals. 40 44 Calcium has one oxidation state: +2 (typical of Group 2). Chromium has common oxidation states of +2, +3 and +6. Calcium(II) and chromium(VI) have noble gas electron configurations, which are typically stable. However, the extremely high charge density of chromium(VI) makes it unstable and other oxidation states are more common. The chromium(II) oxidation state has lost its outer 4s electron and one 3d electron. Chromium(III) forms when the atom loses its 4s electron and two 3d electrons. C 41 C 42 D (c) They can be easily removed by filtration from the reaction mixture and re-used. 43 48 D 50 Chromium has the electron configuration [Ar]3d54s1; it has six unpaired electrons, which is the maximum number for the series. Zn has the [Ar]3d10 configuration with no unpaired electrons. 51 In a complex the d sub-level splits into two energy levels due to the presence of the ligand’s lone pair of electrons. The energy difference between the two sets of d orbitals depends on the oxidation state of the central metal, the number of ligands and the identity of the ligand. Electron transitions between d orbitals result from the absorption of energy from the visible region of the electromagnetic spectrum. The wavelength (colour) of light absorbed depends on the size of the splitting between the two sets of d orbitals. C (a) Zn (b) (i) Fe3O4: +2.67 (ii) MnO4−: +7 (iii) CrO42−: +6 (a) +2 (b) The N atoms adopt a square planar arrangement. (c) The planar structure allows oxygen molecules easy access to the iron ion, which can accept a lone pair of electrons from an oxygen moledule and form a coordinate bond. This bond is not strong, so it the process is easily reversible. This allows the complex to absorb oxygen where oxygen is in high concentrations (i.e. in the lungs) but release oxygen in tissues with low oxygen concentrations. 49 D As the two complexes both contain a cobat ion in the +2 oxidation state the difference in colour is due to the identity of the ligands (H2O vs Cl–1) and the coordination number (6 in [Co(H2O)6]2+ and 4 in [CoCl4]2–), which changes from H2O to Cl–. (iv) [FeCN6]4−: +2 45 (a) Homogeneous catalysts are in the same state of matter as the reactants; heterogeneous catalysts are in a different state from the reactants. (b) They provide a surface for the reactant molecules to come together with the correct orientation. D (b) 1s22s22p63s23p63d10 39 (b) V2O5 (c) Pt or Pd Sc3+ 36 (a) Ni 52 (a) difference in nuclear charge of metal (ion) (b) difference in oxidation number (c) difference in ligand 53 Fe2+ has configuration [Ar]3d6 and Zn2+ is [Ar]3d10. Colour in transition metal complexes is due to the splitting of the d subshell into two sets of d orbitals with different energy levels; 11 10:01 Answers the absorption of visible light results in electrons being excited from the lower energy set to the higher energy set and the colour observed is complementary to the colour (wavelength) of light absorbed. Light can only be absorbed if the d orbitals are partially filled and the higher energy set has an empty or partially filled orbital that can accept an electron from the lower energy set. Fe2+ has partially filled d orbitals and so electronic transitions can occur from the lower energy set to the higher energy set with the absorption of visible light and it appears colour in solution. In Zn2+ all of the d orbitals are fully occupied so an electronic transition cannot occur from the lower energy set to the higher energy set so it is unable to absorb visible light and Zn2+ is not coloured in solution. Challenge yourself 1 Ytterbium, yttrium, terbium, erbium 2 Two liquids, 11 gases 3 ‘Metalloid’ refers to the properties of certain elements in relation to the periodic table. ‘Semiconductor’ refers to the physical properties of materials (including alloys, compounds). There is a partial overlap between the two sets. 4 1s22s22p63s23p64s23d104p65s24d105p64f76s2 or [Xe]4f76s2 5 The entropy change is positive as there are more particles on the right-hand side. 6 The broad absorption spectrum of the complex ions should be contrasted with the sharp lines of atomic spectra (discussed in Chapter 2). Both phenomena are due to electronic transitions, but the spectrum of a complex ion is affected by the surrounding ligands as the complex ion also has vibrational and rotational energy levels. This allows the complex ion to absorb a wider range of frequencies due to the large number of vibrational and rotational excited states also available. Because the absorption of complex ions is measured in solution, interactions with the solvent further increase the number of energy levels present in the complex ion and the number of associated frequencies it can absorb, resulting in the broad absorption bands observed. Fe2+ not in its highest oxidation state and so can be oxidized by removal of d electron; Zn2+ in its highest oxidation state and so can’t be oxidized (and so can’t act as reducing agent). 54 λmax= 525 nm. The colour absorbed is green; the colour transmitted is red. 55 (a) [Ar] 3d6 (b) The splitting would be greater for [Fe(CN)6]4−. 56 (a) [Fe(H2O)6]3+ is yellow and [Cr(H2O)6]3+ is green; the colours they show are complementary to the colours of light they absorb; colour is caused by transitions between the two sets of d orbitals in the complex; the different metals in the two complexes cause the d orbitals to split differently as they have different nuclear charges and this results in different wavelengths (colours) of light being absorbed. (b) The oxidation state of the central ion is different in the two complexes and this affects the size of the d orbital splitting due to the different number of electrons present in d orbitals. Fe2+ has the electron configuration [Ar]3d6 and Fe3+ has the electron configuration [Ar]3d5. The isolated gaseous ions do not have vibrational or rotational energy levels available to them and will only absorb energy of the exact wavelength required to move an electron from a lower energy to a higher energy atomic orbital, generating discrete line spectra. Practice questions 1 C 2 A 3 B 4 B 5 A 6 B 7 D 8 C 12 10:01 9 13 D 10 D 11 C 12 C 17 (a) the amount of energy required to remove one (mole of) electron(s) [1] from (one mole of) an atom(s) in the gaseous state Na+ has greater net positive charge/same number of protons pulling smaller number ofelectrons [1] [1] (b) greater positive charge on nucleus / greater number of protons / greater core charge [1] greater attraction by Mg nucleus for electrons (in the same shell) / smaller atomic radius 14 (b) Si4+: 10 e− in 2 (filled) energy levels / electron arrangement 2.8 / OWTTE [1] P3−: 18 e− in 3 (filled) energy levels / electron arrangement 2.8.8, thus larger / OWTTE [1] [1] Na2O(s) + H2O(l) → 2NaOH(aq) [1] SO3(l) + H2O(l) → H2SO4(aq) [1] OR Si4+ has 2 energy levels whereas P3− has3 / P3− has one more (filled) energy level Na2O is basic and SO3 is acidic [1] (a) solution becomes yellow/orange/brown/ darker [1] 18 chlorine is more reactive than iodine (and displaces it from solution) / OWTTE [1] (b) 2O2− → O2 + 4e− / O2− → 21O2 + 2e− [1] Accept e instead of e−. (b) no colour change / nothing happens as fluorine is more reactive than chlorine / OWTTE [1] (a) atomic number / Z [1] (c) (i) basic [1] Allow alkaline. (ii) Na2O + H2O → 2NaOH / Na2O + H2O → 2Na+ + 2OH− [1] Do not accept Accept nuclear charge / number of protons. (b) Across period 3: increasing number of protons / atomic number / Z / nuclear charge (a) in the solid state ions are in fixed positions / there are no moveable ions / OWTTE [1] Do not accept answer that refers to atoms or molecules. Allow correct equation Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(s) for second mark or stating that iodine/I2 is formed. 16 [1] Si4+ has 10 e− whereas P3− has 18 e− / Si4+ has fewer electrons / P3+ has more electrons [1] State symbols are not needed. 15 (a) Na: 11 p, 11/2.8.1 e− and Na+: 11 p, 10/2.8 e− OR Na+ has 2 shells/energy levels, Na has 3 / OWTTE [1] (atomic) radius/size decreases / same shell/ energy level / similar shielding/screening (from inner electrons) [1] (a) first ionization energy: M(g) → M+(g) + e−/e OR the (minimum) energy (in kJ mol−1) to remove one electron from a gaseous atom OR the energy required to remove one mole of electrons from one mole of gaseous atoms [1] No mark for shielding/screening or shielding/ screening increases. periodicity: repeating pattern of (physical andchemical) properties [1] 19 [1] Noble gases: do not form bonds (easily) / have a full/stable octet/shell/energy level / cannot attract more electrons [1] Do not accept ‘inert’ or ‘unreactive’ without reference to limited ability/inability to form bonds or attract electrons. (b) 2.8.8 [1] Two of: the outer energy level/shell is full; theincreased charge on the nucleus; great(est) attraction for electrons [2] (c) 17 p in Cl nucleus attract the outer level more than 11 p in Na nucleus / greater nuclear charge attracts outer level more [1] 13 10:01 Answers (c) +2 Allow converse for Na. Do not accept ‘has larger nucleus’. (d) S2− has one proton less/smaller nuclear charge so outer level held less strongly / OWTTE [1] 23 Allow converse for chloride. Do not accept ‘has larger nucleus’. (e) the radii of the metal atoms increase (from Li → Cs) (so the forces of attraction are less between them) / OWTTE [1] 20 [1] Only penalize once if roman numerals are used or if written as 2+ or 3+. (a) [Fe(CN)6]4− +2 [1] (b) [FeCl4]− +3 [1] Award [1 max] if answers given as 2+ and 3+, 2and 3 or II and III. electron transitions between (split, partially filled)d orbitals [1] the forces of attraction between halogen molecules are van der Waals forces [1] absorption depends on energy difference between the split d orbitals [1] these forces increase with increasing mass/ number of electrons [1] water molecules replaced by ammonia molecules [1] ammonia (ligands) increase the splitting betweenthe d orbitals/larger energy difference [1] (a) complex (ion) / the charge is delocalized over all that is contained in the brackets 24 [1] (b) colour is due to energy being absorbed when electrons are promoted within the split d orbitals OR the colour observed is the complementary colour to the energy absorbed / OWTTE [1] absorption moves to shorter wavelength/higher frequency/towards blue end of spectrum [1] [max 3] 25 (a) Accept either answer for the first mark. changing the ligand / coordination number / geometry [1] changes the amount the d orbitals are split/ energy difference between the d orbitals / OWTTE [1] 21 (b) splitting increases/greater each NH3/ligand donates an electron pair (toCu2+) [1] NH3 replace H2O ligands around Cu2+ ion/ around central ion [1] forming coordinate (covalent)/dative covalent bond [1] [max 3] 22 [1] NH3 is a Lewis base and Cu2+ is a Lewis acid [1] (a) +2 [1] (b) +3 [1] [1] NH3 has greater electron/charge density / NH3 higher in spectrochemical series / NH3 stronger base [1] Allow converse argument for H2O. Do not award second mark for stating that NH3 is a stronger ligand or has a smaller size. If ‘decreases’ is given for the first part of the answer then no mark can be scored in the second part. 14 10:01