Chemistry Answers Chapter 4

Chapter 4 Exercises 1 δ lead nitrate, Pb(NO3)2 12 δ barium hydroxide, Ba(OH)2 (c) Cl F potassium hydrogencarbonate, KHCO3 δ δ (e) H N (d) O (a) KBr (b) ZnO (c) Na2SO4 (d) CuBr2 (e) Cr2(SO4)3 (f) AlH3 H 2.2 difference = 0.4 C 2.6 Cl 3.2 difference = 0.6, more polar (b) Si 1.9 Li 1.0 difference = 0.9 Si 1.9 Cl 3.2 difference = 1.3, more polar (c) N 3.0 Cl 3.2 difference = 0.2 N 3.0 (a) tin(II) phosphate Mg 1.3 difference = 1.7, more polar F F C F Cl (b) titanium(IV) sulfate (c) manganese(II) hydrogencarbonate 14 (a) H (b) F (d) barium sulfate (e) mercury sulfide (a) Sn H H (c) H C C H H H (b) Ti 4+ (c) Mn2+ (d) Ba2+ H (e) H C (e) Hg+ 5 A 3B 2 6 (d) H C H Cl Cl P Cl (f) H C (a) 16 (b) 24 Mg 12: electron configuration [Ne]3s2 (c) 32 (d) 8 Br 35: electron configuration [Ar]3d104s24p5 (e) 20 (f) 26 15 The magnesium atom loses its two electrons from the 3s orbital to form Mg2+. Two bromine atoms each gain one electron into their 4p subshell to form Br−. The ions attract each other by electrostatic forces and form a lattice with the empirical formula MgBr2. 7 B 9 Test the melting point: ionic solids have high melting points. 8 Test the conductivity: ionic compounds in aqueous solution are good conductors. D 11 16 H+ 17 (a) H O H  O O N O (c) O N (e) O S H H O H + O N (b) C H +  (d) O O D Test the solubility: ionic compounds usually dissolve in water but not in hexane. 10 O δ H (a) C 2.6 ammonium chloride, NH4Cl 4 δ O H 13 calcium phosphate, Ca3(PO4)2 2+ C δ copper sulfate, CuSO4 3 δ (b) O Br δ magnesium carbonate, MgCO3 2 δ δ (a) H O O 2 O O 18 O (f) H H N N H H (a) 105° bond angle, shape is bent (b) 109.5° bond angle, shape is tetrahedral (c) 180° bond angle, shape is linear C 15 10:01 Answers (d) 107° bond angle, shape is trigonal pyramidal (c) London (dispersion) forces (e) 120° bond angle, triangular planar (d) dipole–dipole, London (dispersion) forces (f) 107° bond angle, trigonal pyramidal 30 (g) 105° bond angle, shape is bent 19 (a) 120° bond angle, shape is trigonal planar (b) 120° bond angle, shape is trigonal planar (c) 180° bond angle, shape is linear B (a) malleability, thermal conductivity, thermal stability (d) 120° bond angle, shape is bent (b) light, strong, forms alloys (e) 105° bond angle, shape is bent (c) thermal conductivity, thermal stability, noncorrosive (a) 4 (b) 3 or 4 (c) 2 (d) 4 (d) light, strong, non-corrosive 33 (a) polar (b) non-polar (c) polar (d) non-polar (e) non-polar (f) polar (g) non-polar (h) non-polar 34 cis isomer has a net dipole moment 23 CH3OH < CO32− < CO2 < CO 24 NO3− has three resonance structures, HNO3 has two resonance structures; N–O bonds shorter in HNO3 Similarities: strong, high melting points, insoluble in water, non-conductors of electricity, good thermal conductors. Differences: diamond is stronger and more lustrous; silicon can be doped to be an electrical conductor. 27 (a) linear, 180° (b) triangular pyramidal, 107° (c) bent, 105° 22 26 (i) anodizing: increasing the thickness of the surface oxide layer helps resist corrosion (ii) alloying: mixing Al with other metals such as Mg and Cu increases hardness and strength (i) non-polar 25 (d) HCl 32 (e) 3 21 (b) H2S (c) Cl2 31 (f) 107° bond angle, shape is trigonal pyramidal 20 (a) C2H6 Graphite and graphene have delocalized electrons that are mobile and so conduct electrical charge. In diamond all electrons are held in covalent bonds and are not mobile. A metal B giant molecular C polar molecular D non-polar molecular E ionic compound 28 A 29 (a) London (dispersion) forces (b) H bonds, dipole–dipole, London (dispersion) forces (d) tetrahedral, 109.5° (e) octahedral, 90° (f) seesaw, 117° 35 (a) 6 (b) 6 (c) 6 (d) 5 (e) 2 or 5 36 (a) 90° (b) 107° (c) 90° 37 38 39 (a) polar (b) polar (c) non-polar (d) polar (e) non-polar (f) non-polar (a) polar (b) non-polar (c) non-polar (d) non-polar (e) non-polar (f) polar In BF3 all the atoms have formal charge of 0. B: FC = 3 – 3 = 0 each F: FC = 7 – (1 + 6) = 0 Zero formal charge represents the most stable, preferred structure, so this is favoured despite violating the octet rule by having fewer than 8 electrons around Be. 16 10:01 40 (i) S: FC = 6 – 4 = +2 each O: FC = 6 – (1 + 6) = –1 2 O O S O angles of 109.5° give the puckered shape. In C6H6 (benzene) the carbon atoms are all sp2 hybridized, forming a planar triangular arrangement with bond angles of 120°. O (ii) S: FC = 6 – 6 = 0 each O: FC = 6 – (1 + 6) = –1 each O: FC = 6 – (2 + 4) = 0 2 O O S O 42 1 Aluminium oxide is less ionic than MgO due to a smaller difference in electronegativity. It has some partially covalent character, which means the comparison with more ionic oxides is not fully valid. 2 F2 has lower bond enthalpy than expected from its atomic radius due to repulsion. The bond length is so short that the lone pairs in the two atoms repel each other, weakening the bond. 3 when bonded to F, e.g. in OF2 4 Run each solution out from separate burettes, and see whether the stream of liquid is deflected in the presence of a charged rod. Only the polar solution will show deflection. O The structure with 12 electrons round the S atom has FC = 0 and so is the preferred structure. 41 Challenge yourself O–O bonds in O3 are weaker than in O2, due to lower bond order, therefore dissociation occurs with lower energy light (longer wavelength). O3 breakdown is catalyzed by NOx and CFCs in the atmosphere, e.g. Test solubility with ionic and covalent solutes. The polar solution will be a better solvent for polar/ionic solutes; the non-polar solution for covalent/non-polar solutes. radical is produced. 5 The high thermal conductivity of diamond is because of its strong covalent bonds. When heated the bonds becoming vibrationally excited, and as they are all connected heat energy could be readily transferred through the network from one bond to the next. Silicon is similarly a good thermal conductor – which is why computer chips need to be cooled. 6 Diamonds are kinetically stable with respect to graphite, as the conversion has a very high activation energy (see Chapter 6). So the reaction generally occurs too slowly to be observed. 7 It is difficult to know the number of valence electrons a transition metal has. Treating bonds from ligands as pure covalent molecules results in transition metals in complex ions with large negative formal charges. The formal charge model may not be useful for complex ions, as the values obtained do not make much sense. catalyst for ozone destruction. 43 Electrons in a sigma bond are most concentrated in the bond axis, the region between the nuclei. Electrons in a pi bond are concentrated in two regions, above and below the plane of the bond axis. 44 (b) H– F in HF (c) Cl– Cl in Cl2 (d) C– H in CH4 (e) C–H in C2H4 (f) C– H in C2H2 (g) C–Cl in C2H3Cl (a) sp (b) sp3 45 2 (c) sp2 (d) sp (e) sp2 46 In C6H12 (cyclohexane) the carbon atoms are sp3 hybridized, each forming a tetrahedral arrangement with two neighbouring carbon atoms and two hydrogen atoms. The bond 17 10:01 Answers 8 In the polar winter small amounts of water vapour freeze into ice crystals in the atmosphere, which provide a reactive surface. Reactions on the crystals’ 1 C 17 (a) A 2 A 3 4 5 B 6 A 7 B 8 D 9 C 10 A 11 B 12 C 13 D 14 B 15 A 16 Hydrogen bonding in butan-1-ol; stronger than dipole–dipole attractions in butanal. [2] Accept converse argument. Do not penalize ‘dipole–dipole bonding’ instead of ‘dipole– dipole attractions’. A PBr3 (i) Lewis structure: Br P SF6 (i) Lewis structure: Br F F F S Br F Allow x’s, dots or lines to represent electrons. Allow x’s, dots or lines to represent electrons. Penalize missing lone pairs on terminal atoms once only for the two Lewis structures. (ii) Shape: trigonal/triangular pyramidal F F Penalize missing lone pairs on terminal atoms once only for the two Lewis structures. (ii) Shape: octahedral Bond angle: less than 109.5° Bond angle: 90° Allow any angle less than 109.5° but greater than or equal to 100° (experimental value is 101°). Ignore extra correct bond angles (e.g. 90° and 180° scores but not 90° and 120°). (iii) Polarity: polar (iii) Polarity: non-polar and and Explanation: net dipole (moment) / polar PBr bonds and molecule not symmetrical/ bond dipoles do not cancel / asymmetric distribution of electron cloud / Explanation: no net dipole (moment) / polar SF bonds but molecule symmetrical/ bond dipoles cancel / symmetric distribution of electron cloud / OWTTE P Br Br Br / OWTTE [8] 18 10:01 (iv) in cis two carboxylic acid groups close together so on heating cyclic anhydride forms (with elimination of water) / OWTTE [1] Do not allow ECF in this question from incorrect Lewis structure. Allow [1] max for stating that PBr is polar and SF is non-polar without giving a reason or if explanations are incorrect. Allow converse argument for trans. Allow polar bonds do not cancel for PBr3 and polar bonds cancel for SF6. (c) O of OH sp3 and O of C=O sp2 Do not allow asymmetric molecule as reason for PBr3 or symmetric molecule for SF6 as reason alone. (b) (i) σ bond: end-on/axial overlap with electron density between the two carbon atoms/nuclei / end-on/axial overlap of orbitals so shared electrons are between atoms / OWTTE π bond: sideways/parallel overlap of p orbitals with electron density above and below internuclear axis/σ bond / sideways/ parallel overlap of p orbitals so shared electrons are above and below internuclear axis/σ bond / OWTTE [2] Marks can be scored from a suitable diagram. Award [1 max] for stating end-on/axial overlap for σ and sideways/parallel overlap for π only i.e. without mentioning electron density OR stating electron density between the two atoms/nuclei for σ and above and below internuclear axis for π. C C H s p p (ii) 11 σ and 3 π 18 (a) Award [2 max] for three of the following features: Bonding Graphite and C60 fullerene: covalent bonds and van der Waals’/London/dispersion forces Diamond: covalent bonds (and van der Waals’/London/dispersion forces) Delocalized electrons Graphite and C60 fullerene: delocalized electrons Diamond: no delocalized electrons Structure Diamond: network/giant structure / macromolecular / three-dimensional structure and Graphite: layered structure / two-dimensional structure / planar C60 fullerene: consists of molecules / spheres made of atoms arranged in hexagons/ pentagons Bond angles Graphite: 120° and Diamond: 109° C60 fullerene: bond angles between 109–120° Allow Graphite: sp2 and Diamond: sp3. Allow C60 fullerene: sp2 and sp3. p p H [1] Oxygens must be identified. [1] (iii) (strong) intermolecular hydrogen bonding in trans but (strong) intramolecular hydrogen bonding in cis so attraction between different molecules is less (hence lower melting point) [1] Allow between molecules for intermolecular and within molecules for intramolecular. Number of atoms each carbon is bonded to Graphite and C60 fullerene: each C atom attached to 3 others Diamond: each C atom attached to 4 atoms/ tetrahedral arrangement of C (atoms) [6 max] (b) (i) network/giant structure / macromolecular each Si bonded covalently to 4 oxygen atoms and each O atom bonded covalently to 2 Si atoms / single covalentbonds [2] 19 10:01 Answers Ethanol contains a hydrogen atom bonded directly to an electronegative oxygen atom / hydrogen bonding can occur between two ethanol molecules / intermolecular hydrogen bonding in ethanol; the forces of attraction between molecules are stronger in ethanol than in methoxymethane / hydrogen bonding strongerthan van der Waals’/dipole-dipole attractions. max [3] Award [1 max] for answers such as network-covalent, giant-covalent or macromolecular-covalent. Both M1 and M2 can be scored by a suitable diagram. (ii) Silicon dioxide: strong/covalent bonds innetwork/giant structure/macromolecule Carbon dioxide: weak/van der Waals’/ dispersion/London forces between molecules [2] Award [2] max if covalent bonds breaking during boiling is mentioned in the answer. (c) triple (covalent) bond one electron pair donated by oxygen to carbon atom / dative (covalent)/coordinate (covalent) bond [2] Award [1 max] for representation of C ≡O. Penalize only once if no reference given to intermolecular nature of hydrogen bonding or van der Waals. 20 Award [2] if CO shown with dative covalent bond. (d) < Note: need the answer to part 18(d) of this question> (a) (bond formed by) sideways overlap of p orbitals. [2] (b) C1 is sp3 and C2 is sp2. [1] 21 Lewis structure Name of shape S F F S SF2 F F bent/angular/V shaped 2 lone pairs on S required for the mark F (e) < Note: need the answer to part 18(e) of this question> F S Seesaw/ distorted tetrahedral F SF4 F 1 lone pair on S required for the mark F F SF6 19 Methoxymethane is very weakly polar/weak van der Waals’/dipole–dipole forces exist between methoxymethane molecules. Accept alternatives to van der Waals’ such as London and dispersion forces F octahedral F Accept square bipyrimidal S F F [6] Penalise missing lone pairs on fluorine atoms once in correct structures only. 20 0 10:01 For Lewis structures candidates are not expected to draw exact shapes of molecules. 2 23 (a) (i) F Do not allow ECF for wrong Lewis structures. F F Si 22 (a) (i) Cl P Cl / Cl P Cl Cl Cl F trigonal pyramid in the range of 100–108°   N H / (ii) N H H H [3] octahedral/octahedron/square bipyramidal 90° / 90° and 180° + O N O (ii) F F / F Xe F F Penalize missing charge, missing bracket once only in (i) and (ii). F F Lone pairs required for BOTH (i) and (ii). square planar 90° [3] Penalize once only if electron pairs are missed off outer atoms (b) (i) sigma bonds are formed by end on/axial overlap of orbitals with electron density between the two atoms/nuclei pi bonds are formed by sideways overlap of parallel p orbitals with electron density above and below internuclear axis/σ bond [2] Accept suitably annotated diagrams (ii) 8 sigma/σ 1 pi/π [2] (iii) 109°/109.5° 120° [3] Allow dots, crosses or lines in Lewis structures. Xe F [3] linear 180° Must include minus sign for the mark. bent/V-shaped in the range of 100–106° [3] (iii) F F [2] (iv) sp hybridization 1 sigma and 2 pi sigma bond formed by overlap between the two sp hybrid orbitals (on each of the two carbon atoms) / pi bonds formed by overlap between remaining p orbitals (on each of the two carbon atoms) / diagram showing 2 sp hybrid orbitals and 2 p orbitals [3] (b) NO2: N O O Award [1] for correct representation of the bent shape and [1] for showing the net dipole moment, or explaining it in words (unsymmetrical distribution of charge). CO2: O C O Award [1] for correct representation of the linear shape and for showing the two equal but opposite dipoles or explaining it in words (symmetrical distribution of charge). [3] For both species, allow either arrow or arrow with bar for representation of dipole moment. Allow correct partial charges instead of the representation of the vector dipole moment. Ignore incorrect bonds. Lone pairs not needed. (c) Structure: network/giant lattice / macromolecular / repeating tetrahedral units 21 1 10:01 Answers Bonding: (single) covalent (bonds) [2] It is not necessary to identify which part refers to structure and bonding specifically. (iv) < Note need answers to Qu23 part (d) part (iv)] (d) (i) methanamide O H H C [2] N H (ii) mixing/joining together/combining/ merging of atomic orbitals to form molecular/new orbitals / orbitals of equal energy [1] (iii) < Note need answers to Qu23 part (d) part (iii) 24 (a) (i) from left N = +1, N = –1, all others = 0 [1] (ii) from left C = +1, 2nd N = –1 formal charges show same differences but structure (i) will be more important because C has stable octet [1] (b) (i) from left O = –1, Cl = +1, O = –1 (ii) O = –1, Cl = 0, O = 0 structure (ii) is preferred due to less difference in formal charge 2 [1] [1] [1] [1] 10:01