Atom The Smallest Part Of An Element That Can Exist On Its Own

Sub-atomic Relative Relative Mass No E(element) E = hf Energy = Planck’s constant x frequency particle charge mass Atomic No Atomic number Proton number Proton +1 1 Mass number Proton number plus the number of neutrons in the Electron -1 negligible nucleus of its atom Neutron 0 1 Hydration Chemical combination of water and another substance Solvent Substance in which other substances are dissolved Solute Substance dissolved in another substance(solvent)to form a solution Empirical formula Ratio of atoms of different elements present in a molecule of a compound, in their lowest terms as whole numbers (C4H4 has CH empirical formula) Molecular formula Actual number of atoms of each element in a molecule of a compound Dibasic acid One which has 2 replaceable H atoms per molecule Isotopes Atoms having the same atomic number but different mass numbers - As the number of protons increases, the number of neutrons increases relatively faster, so small atoms have proton and neutron numbers which are comparable whereas large atoms have more neutrons than protons - Neutrons reduce repulsive forces between positive protons - Departure from optimum range of ratio of protons to neutrons will lead to nuclear instability(radioactivity) - No of electrons in outer(valence)shell(and sometimes the shell next to the outer shell)and the IE’s & EA’s for an atom will determine the chemistry of the element - Outer shell electrons determine the chemistry of an element as they can get close to the outer shell electrons of other atoms, so can be transferred or shared. The inner shell electrons are tightly held and shielded from the electrons in other atoms/molecules - Isotopes don’t affect the electron number or structure so don’t affect the chemistry of the element, but have varying rates of reaction - 12C is the only atom with relative atomic mass which is an exact whole number because 12C is chosen as a standard and given a relative atomic mass of exactly 12 Nuclide A nuclear species of given mass number and atomic number Radionuclide Radioactive nuclide Relative atomic mass Average mass of an atom of an element compared with 1/12 of the mass of an atom of carbon-12 isotope Relative isotopic mass Mass of an atom of an isotope of an element compared with 1/12 of the mass of an atom of carbon-12 isotope Relative molecular mass Sum of relative atomic masses 1 Electrons are emitted from the filament, accelerated and used to bombard the gaseous sample which is at very low pressure 2 Sample molecules have electrons knocked off them by bombarding electrons forming positive ions. The molecules can also fragment or rearrange giving different positive ions 3 Positive ions accelerated by an electric field 4 Positive ions deflected by a magnetic field in a circular path whose radius depends on mass/charge ratio and strength of the field. The machine sweeps over the chosen mass range by altering the magnetic field and hence ions reaching the detector are separated according to their mass 5 Positive ions(Species: radical cation/molecular ion/fragment ion) detected and relative amounts calculated by the machine - Chamber vacuumed to prevent air molecules obstructing the passage of particles through the mass spectrometer - Likelihood of 2+ ions being produced in the mass spectrometer is small, because the chance of a molecule being ionised by electron impact is already quite small. When it has been ionised it is pulled away by the potential gradient leading to the magnet which also helps to reduce the possibility of a second impact - Important to use minimum possible energy to ionise a sample in a mass spectrometer to prevent formation of double positive ions rather than single positive ions(or to prevent fragmentation of molecular substances) (%Abundance x Mass) + (%Abundance x Mass) +….. Total % composition = Relative atomic mass RAM Molecular ion peak (The molecular ion peak is not always the most intense and maybe absent) Highest significant peak where the molecule has lost 1 electron but has not broken up Base peak Highest peak in the mass spectrum - If the sample is an element each line represents an isotope of the element - Molecules broken up into fragments make fragmentation patterns on the mass spectrum(used to identify molecules and structure) - Relative masses are just numbers, no units - No element in the periodic table has a relative atomic mass that is a whole number because relative atomic mass is an average so it’s not usually a whole number - Mass spectrometer with gas-liquid chromatography used in forensic work for analysis of complex mixtures, samples introduced to gas liquid chromatograph and various constituents separated, output gases then led to mass spectrometer, fragmentation patterns compared with large database of patterns from known substances - Once the electron has passed the final anode, it doesn’t decelerate and go back to the anode because it has either collided with a gaseous molecule or passed on to earth(ground)at the other end of the chamber - It is a good approx to consider that relative atomic mass of 6Li+ determined in a mass spectrometer is the same as that of 6Li because mass of the electron lost from 6Li to form 6Li+ is negligible (1) Calculate RAM from these isotopes and their % compositions 54Fe 5.8%, 56Fe 91.6%, 57Fe 2.2%, 58Fe 0.33% (1) (54x5.8) + (56x91.6) + (57x2.2) + (58x0.33) 5.8 + 91.6 + 2.2 + 0.33 = 55.91(2dp) 1st Ionisation energy M(g) à M+(g) + e– Energy/enthalpy change per mole to remove an electron from each atom in the gas phase to form a singly positive ion 2nd Ionisation energy Energy/enthalpy change per mole for the process, M+(g) à M2+(g) + e– 1st Electron affinity X(g) + e– à X–(g) Energy/enthalpy change per mole for each atom in the gas phase to gain an electron to form a singly negative ion - Negative(exothermic), since the electron is attracted by the positive charge on the atoms nucleus 2nd Electron affinity Energy/enthalpy change per mole for the process, X–(g) + e– à X2–(g) - Positive(endothermic), since energy needed to overcome repulsion between the electron and negative ions

Energy level Electrons in atoms can only have certain amounts of energy, groups of electrons can exist with roughly the same amount of energy, these positions of roughly similar amounts of energy are called energy levels Size of IE depends on • Nuclear charge • Atomic radius • Electron shielding, energy level Successive IE’s increase because electrons are being removed from increasingly positive ions and so the attractive forces are greater - Large jumps in IE’s arise from a large increase in attraction, corresponding to an electron being removed from a new energy level significantly closer to the nucleus(proving that electrons are arranged in shells) IE’s increase across periods(left to right) • Number of protons increasing, meaning stronger nuclear attraction • Extra electrons are at roughly the same energy level, even if the outer electrons are in different orbital types • Little extra shielding effect, little extra distance to lessen the attraction from the nucleus IE’s decrease down groups • Each element down a group has an extra electron shell • Extra inner shells means extra distance of outer electrons from the nucleus, and greater shielding from the attraction of the nucleus, overall reducing nuclear attraction Atomic radius decreases across period 3(left to right) Nuclear charge increases, electrons pulled closer to the nucleus, electrons are all added to the same outer shell n Subshell Number of electrons Subshell 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f Max number of electrons 2 2 6 2 6 10 2 6 10 14

Principal quantum number(n) shell numbers Subshell 1st shell has no subshell Orbital s subshell has 1 orbital, p has 3, d has 5 ‘Aufbau’ ‘build up’ principle • Electrons are added to the lowest energy orbital available • One at a time • With no more than 2 electrons occupying one orbital • If there are several orbitals of the same energy available then electrons enter these orbitals singly so as to be as far apart as possible • Halogens have high IE’s so they don’t form positive ions but negative ions because they have one electron less than a full shell • Anomalously low EA for F due to repulsion of the incoming electron from a concentrated electric field of a small atom

Quantum mechanics The electron in an atom behaves as a wave which is a mathematical construction, not a particle Atomic orbital IS the electron/pair of electrons, the volume in which the electron has a 95% probability of being found(no such thing as an empty orbital) Spin is a property of an electron. The 2 electrons in an orbital have opposite spins, helping to counteract the repulsion between their negative charges(spin pairing) 2 2 Z Symbol 1s 2s 2p 3s 3p 3d 4s 4p Group II Be 1s 2s 1st IE = 900 kJmol–1 1 H 1 2 2 1 st Group III B 1s 2s 2px 1 IE = 799 kJmol–1 2 He 2 2 2 1 1 1 Group V N 1s 2s 2px 2py 2pz 1st IE = 1400 kJmol–1 3 Li 2 1 Group VI O 1s22s22p 22p 12p 1 st –1 4 Be 2 2 x y z 1 IE = 1310 kJmol 5 B 2 2 1 N 2s 2p O 2s 2p 6 C 2 2 2 7 N 2 2 3 • In B the 2p electron easier to remove than the 2s electron from 8 O 2 2 4 Be because subshells that are full are more stable 9 F 2 2 5 • In B the 2p orbital is further from the nucleus 10 Ne 2 2 6 • The 2p orbital is screened not only by the 1s2 electrons but also 11 Na 2 2 6 1 partially by the 2s2 electrons 12 Mg 2 2 6 2 - These factors are strong enough to override the effect of the 13 Al 2 2 6 2 1 increased nuclear charge resulting in the IE to drop slightly 14 Si 2 2 6 2 2 • Screening identical and electron being removed is from an 15 P 2 2 6 2 3 identical orbital 16 S 2 2 6 2 4 • N structure is symmetrical and more stable than that in O 17 Cl 2 2 6 2 5 • Repulsion between 2 electrons in the same orbital means an 2 18 Ar 2 2 6 2 6 electron in the 2px pair is easier to remove 19 K 2 2 6 2 6 1 20 Ca 2 2 6 2 6 2 (q) An atom contains 5 protons and 5 neutrons, give the symbol for

21 Sc 2 2 6 2 6 1 2 this atom including the mass number (a) B 22 Ti 2 2 6 2 6 2 2 (q) Formula of the compound formed between this element and 23 V 2 2 6 2 6 3 2 chlorine (a) BCl3 24 Cr 2 2 6 2 6 5 1 25 Mn 2 2 6 2 6 5 2 26 Fe 2 2 6 2 6 6 2 27 Co 2 2 6 2 6 7 2 28 Ni 2 2 6 2 6 8 2 29 Cu 2 2 6 2 6 10 1 30 Zn 2 2 6 2 6 10 2 31 Ga 2 2 6 2 6 10 2 1 32 Ge 2 2 6 2 6 10 2 2 33 As 2 2 6 2 6 10 2 3 34 E 2 2 6 2 6 10 2 4 35 Br 2 2 6 2 6 10 2 5 36 Kr 2 2 6 2 6 10 2 6 (1)(i)Mass spectrum of HCl, peak at mass 36 is molecular ion (H 35Cl)+ Chlorine has only 2 isotopes 35Cl, 37Cl What particle is responsible for peak mass 38? (1)(i) (H 37Cl)+ (ii)How do you explain the fact that the height of the peaks at mass 36 is 3 times as high than the peak at mass 38 (ii)There is 3 times as much 35Cl to 37Cl and therefore 3 times as much H35Cl to H37Cl (2)(a)Mass spectrum of methane, peak at mass 16 is molecular ion(CH)+ Explain peaks of relative mass 1, 2, 12, 13, 14, 15, 17 Relative mass 1 2 12 13 14 15 17 1 + 2 + 12 + 12 1 + 12 1 + 12 1 + 12 1 2 + ( H) ( H) ( C) ( C H) ( C H2) ( C H3) ( C H3 H) (2)(a)Bromine consists of 2 isotopes, mass numbers 79 and 81. A sample of Br2(g) was examined in a mass spectrometer. Identify the species responsible for the peak at m/e = 160 (2)( 79Br81Br)+ Peak at m/e Relative abundance (2)(b)For a particular sample of copper two peaks were obtained in the mass spectrum 65 + 63 69.1 (i)Give the formula of the species responsible for the peak at m/e = 65 (i) Cu 65 30.9 (ii)State why two peaks, at m/e values of 63 and 65, were obtained in the mass spectrum (ii)2 different isotopes (3)How would accelerating field and magnetic field differ in its affect on X+ and X2+ ? (3)The accelerating force and deflecting field on X+ will be twice that on X+ (4)2 reasons why particles must be ionised before being analysed in a mass spectrometer? (4)Have to be accelerated, then deflected (5)Boron, relative atomic mass 10.8 gives 2 peak mass spectrum, m/z=10 and m/z=11 Calculate the ratio of the heights of the 2 peaks (5)10x + 11(1 – x) = 10.8, x = 0.2, y =1 – x = 0.8 ratio of heights = 1:4 (7)Explain why K has lower 1st IE than Na (7)• Electron being removed is further from the nucleus • More shielding • Reduces attraction of the nucleus 1 –

(8)What force causes the scattering of α particles by nuclei? l 6 (8)α particles and nuclei both positively charged thus electrostatic forces of repulsion o m

(9)Explain why all isotopes of Mg have the same chemical properties J 5 k /

(9)• Same number of electrons in all Mg isotopes • Outer electron structure y 4 g determines chemical properties r e

(10)Using subshell notation, give electronic configuration of K atom and K ion e 3 2 2 6 2 6 1 + 2 2 6 2 6 (10)K 1s 2s 2p 3s 3p 4s K 1s 2s 2p 3s 3p n o i t 2

(11)State why 2 peaks at m/e values of 63 and 65 were obtained in a mass spectrum a s i

of an element (11)2 different isotopes n

o 1 th 4+ 5+ i (12)Write equation for 5 ionisation of Na (12)Na (g) à Na (g) + e– g (14)The logarithm of successive IE’s for Mg across the page o 0 L Explain what this graph tells you about the electron arrangement in the Mg atom 0 1 2 3 4 5 6 7 8 9 10 11 12 (14)Two/big jumps show 3 different shells present, shows 2.8.2 Number of electron removed Find the empirical formula of the compound containing C 22.02% H 4.59% Br 73.39% by mass Atomic ratio Simplified atomic ratio C 22.02/12= 1.835 1.835/0.917 = 2 H 4.59/1= 4.59 1.835:4.59:0.917 4.59/0.917 = 5 Br 73.39/80= 0.917 0.917/0.917 = 1 Empirical formula is thus C2H5Br (1)Compound X contains only B and H, % by mass of B in X is 81.2% In mass spectrum of X the largest value peak of m/z is at 54 Calculate empirical and molecular formula

(1)B:H = 81.2/10.8 : 18.8/1, 7.51:18.8, 1:2.5, 2:5 Empirical formula is B2H5 Mr(B2H5) = 26.6 Molecular formula = B4H10 (2)Hydrazine(empirical formula NH2)mass spectrum of this compound shows a molecular ion peak at m/e 32, show the molecular formula of hydrazine is N2H4 (2)Relative Molar Mass = 32 n(N + 2H)= 32 n(14 + 2)= 32 n= 2 Molecular formula = 2 × NH2 = N2H4 Equations are: • Internationally understood • Quantitative • Shorter than the same information given in words Equations balance for mass and total charge LHS & RHS have the same number of each type of atom, if there are 2 positives on LHS there must be 2 on RHS Ionic equations: 1 Write soluble ionic compounds with the ions separated 2 Write insoluble ionic and covalent compounds as usual 3 Cross out spectator ions(ions which appear on both sides of the equation)

NaCl(aq) + AgNO3(aq) à NaNO3(aq) + AgCl(s) Silver nitrate silver chloride(white ppt) + – + – + – Ions: Na (aq) + Cl (aq) + Ag (aq) + NO3 (aq) à Na (aq) + NO3 (aq) + AgCl(s) Deleting spectator ions: Ag+(aq) + Cl–(aq) à AgCl(s)

2KMnO4(aq) + 8H2SO4(aq) + 10FeSO4(aq) à 2MnSO4(aq) + 5Fe2(SO4)3(aq) + K2SO4(aq) + 8H2O(l) potassium manganate(VII) iron(II)sulphate + – + 2– 2+ 2+ 3+ 2– + Ions: 2K (aq) + 2MnO4 (aq) + 16H (aq) + 18SO4 (aq) + 10Fe (aq) à 2Mn (aq) + 10Fe (aq) + 18SO4 (aq) + 2K (aq) + 8H2O(l) – + 2+ 2+ 3+ Deleting spectator ions: MnO4 (aq) + 8H (aq) + 5Fe (aq) à Mn (aq) + 5Fe (aq) + 4H2O(l)

23 –1 Avogadro constant NA is 6.02 x 10 mol (L, Loschmidt’s number) - At RTP, 298K, 100kPa, 1 mol of gas occupies 24dm3 - Volumes of all gases are equal under the same conditions and contain the same number of particles - 1 mol of any substance • is 6.02 x1023 particles of it • is its relative molecular/atomic mass in g(molar mass, gmol–1) - Fe(s) + S(s) à FeS(s) Contain the same number of particles

Indicator pKind pH range Half way colour Litmus 6.5 5-8 Dark purple Methyl orange 3.7 3.1-4.4 Orange Phenolphthalein 9.3 8.3-10 Pale pink Titration/quantitative/volumetric analysis(analysing amount of substance present) • Volumes of both solutions and concentration of one of them known • Complete reaction between 2 substances, means concentration of the other solution can be found Titration procedure: • Rinse out the burette with distilled water followed by a little of the solution to be used in it Ensure that any water in the burette does not dilute the solution, if not then the titration would be too large NOT “wrong”  first titration must only be considered ‘rough’ • Fill the burette(so there’s no air bubbles and reading at eye-level whereby the bottom of the meniscus is level with the zero mark) • Wash out the conical titration flask with distilled water • Rinse out the pipette with distilled water followed by a little of the solution to be used in it • Use the pipette to measure out the required volume of solution into the conical flask – Avoiding air bubbles – Reading pipette at eye-level whereby the bottom of the meniscus is level with the mark – Touching the side of the flask with the tip of the pipette, leaving a drop in the tip of the pipette • Add 1 or 2 drops of indicator to the solution in the conical flask. Place the flask on the white tile under the burette • Add acid to alkali, by swirling the conical flask, then drop-wise towards the end point whereby the indicator colour changes by the addition of one drop • Carry out a rough titration, then 2 accurate titrations which agree to within 0.1cm3 of each other • Record results as a statement and table Molar solution of a substance(moldm–3, gdm–3)One mole of a substance has water added until the volume of the solution is 1dm3 Acid/base titrations(Acid & base reacted with a suitable indicator)finds the purity of a substance or produces a standard solution for use in another titration Standard solution One which can be made of known concentration by weighing out the primary standard(solute) Primary standard(solute)must: 1 Be available commercially in a high state of purity 2 Be stable over long periods of time 3 Not be volatile(so losses due to evaporation during weighing don’t occur) 4 Not decompose when dissolved in water

5 Not absorb water or CO2 from the atmosphere

Making a s tandard solution of Na2CO3(aq) of known concentration: 1 1.25g of pure anhydrous Na2CO3 is dissolved in a beaker with distilled water 2 Using a funnel and a glass rod, the solution is transferred to a 250cm3 graduated flask 3 The beaker is washed a couple of times with distilled water and the washings are added to the graduated flask 4 Make the mixture up to ‘the mark’ with distilled water 5 Stopper the flask and shake to ensure that the solution is homogenous

(1)Na2CO3(s) + 2HCl(aq) à 2NaCl(aq) + H2O(l) + CO2(g) 3 3 3 12.5g of Na2CO3 in 1dm solution, 25cm of this titrated with HCl, 23.45cm of HCl was required, concentration of the acid? –1 –3 –1 –3 (1)Na2CO3= 106gmol 12.5gdm /106gmol = 0.118moldm (concentration) 3 –3 0.025dm x 0.118moldm = 0.00295mol(Amount of Na2CO3) 1 mol of Na2CO3 requires 2mol HCl Amount of HCl= 0.00295mol x2= 0.0059mol 0.0059mol/0.0235dm3= 0.251moldm–3 3 (2)1g Na2CO3 dissolved in water and volume made to 200cm Portions of 25cm3 of this solution were titrated with 0.12moldm–3HCl solution What volume was required?

(2)Volume required depends on indicator used, assuming methyl orange used Na2CO3 + 2HCl à 2NaCl + H2O + CO2 –1 –3 –3 –2 Amount of Na2CO3 = 1g/106gmol = 9.434 x10 mol Amount of HCl = 2 x 9.434 x10 mol = 1.887 x10 mol Volume of HCl required = 1.887x10–2mol/0.12moldm–3 = 0.157dm3 3 3 (3)1cm of (conc)H2SO4 in a 500cm graduated flask made up to the mark with pure water 25cm3 portions titrated with 0.1moldm–3 NaOH(aq), 19.8cm3 of NaOH needed. Concentration of original acid? 3 –3 –3 (3)H2SO4(aq) + 2NaOH(aq) à Na2SO4(aq) + 2H2O(l) Amount of NaOH = 0.0198dm x 0.1moldm = 1.98 x10 mol –3 –4 Since 1 mol H2SO4 requires 2 mol NaOH Amount of H2SO4 = 1.98 x10 mol/2 = 9.9 x10 mol –4 3 –4 3 3 –3 9.9 x10 mol in 25cm , 9.9x10 x 500/25, 0.0198mols in 1cm of (conc)H2SO4 19.8mols in 1dm conc = 19.8moldm (4)Back-titration(concerning substances which are insoluble)used to find purity of sample of chalk which is insoluble in water A solution can’t be used so it’s reacted with a known amount of excess acid, acid remaining then titrated with standard alkali 1.5g of chalk reacted with (excess)50cm3 of 1moldm–3 HCl When reaction ceased, solution transferred to a 250cm3 graduated flask and made to the mark with pure water 25cm3 of the solution titrated with 0.1moldm–3 NaOH(aq) 24.5cm3 required, % purity of the chalk? –1 (4)CaCO3(s) + 2HCl(aq) à CaCl2(aq) + H2O(l) + CO2(g) HCl(aq) + NaOH(aq) à NaCl(aq) + H2O(l) CaCO3 = 100gmol Amount of NaOH = amount of HCl unreacted = 0.0245dm3 x 0.1moldm–3 = 2.45 x10–3mol in 25cm3 Total amount of HCl unreacted = 2.45 x10–3mol x 250/25 = 0.0245mol Original amount of HCl taken = 0.05dm3 x 1moldm–3 = 0.05mol Amount of HCl used to react with the CaCO3 = (0.05 – 0.0245)mol = 0.0255mol Amount of CaCO3 = 0.0255mol/2 = 0.0128molMass of –1 CaCO3 = 0.0128mol x 100gmol = 1.280g % purity of CaCO3 =1.28g/1.5g x 100 = 85.3% (1)Marble reacted with HCl acid, mass loss was 2.33g, what volume of CO2 was evolved? –1 (1)CaCO3(s) + 2HCl(aq) à CaCl2(aq) + CO2(g) + H2O(l) CaCO3 = 2.33g/100gmol = 0.0233mol, 0.0233mol CO2 produced 3 –1 3 Volume of CO2 = 0.0233mol x 24dm mol = 0.56dm 3 3 (2)10cm of a hydrocarbon C4Hx reacts with an excess of oxygen at 150°C, 1 atm. The products occupy a volume 10cm greater than the reactants at this temperature and pressure. Find x (2)C4Hx(g) + (4 + x/4)O2(g) à 4CO2(g) + x/2H2O(g) Change in volume = (volume of products – volume of reactants) Avogadro’s rule: 1 volume + (4 + x/4)volumes à 4 volumes + x/2 volumes 10cm3 + (4 + x/4)10cm3 à 40cm3 + 5xcm3 (40cm3 + 5xcm3) – (50cm3 + 2.5x cm3) = 10cm3 2.5x = 20 hence x = 8

Na2CO3(s) was dissolved in distilled water and this solution was put into a conical flask and three drops of methyl orange indicator added, titrated against HCl acid until the end point was reached Na2CO3 + 2HCl  2NaCl + H2O + CO2 (3)Describe the colour change that tells when the end point has been reached (3)Solution will go from yellow à orange Iron metal reacts with copper(II)sulphate solution to form copper metal and iron ions An experiment was performed to find which of the two equations is correct Cu2+(aq) + Fe(s)  Cu(s) + Fe2+(aq) 3Cu2+(aq) + 2Fe(s)  3Cu(s) + 2Fe3+(aq) • Powdered iron of mass 1.4g was placed in a beaker and excess copper(II)sulphate solution was added • Mixture stirred for 5 minutes • The contents of the beaker were then poured into a funnel containing a weighed piece of filter paper • The beaker and the residue were washed with cold water, and the copper and the filter paper were left overnight to dry • Next day they were weighed, and the copper was found to have a mass of 1.65g (a)Calculate the mass of copper that should be produced from 1.4g of iron if: (i)Equation I is correct (ii)Equation II is correct (a)(i)Amount of iron = 1.4/56= 0.025mol, 0.025×63.5= 1.59g (ii)Amount of copper = 0.025×3/2=0.0375mol, 0.0375×63.5= 2.38g (b)Which iron ion was produced in the reaction? (b)Fe2+ (c)(i)Suggest why the experimental value of the mass of copper was slightly different from the value you calculated in (a) (c)(i)copper/filter paper was still wet (ii)Suggest one way in which the accuracy of this experiment could have been improved (ii)Improved drying, wash with suitable solvent(propanone or ethanol), suction filtration (d)Why is it essential to use excess copper(II) sulphate solution? (d)So that all the iron reacts NOT “the reaction is complete” (a)0.25g of sulphamic acid NH2SO3H(s) was dissolved in distilled water in a conical flask 23.45cm3 NaOH(aq) required to react with sulphamic acid solution NH2SO3H(aq) + NaOH(aq)  NH2SO3Na(aq) + H2O(l) (i)Calculate the amount of sulphamic acid in 0.25g Mr(NH2SO3H) = 97 (i)0.25/97 = 0.00258mols (ii)Calculate concentration of NaOH (ii)0.00258mols × 1000/23.45 = 0.110moldm–3 (b)Balance used to weigh the sulphamic acid is accurate to ±0.01g Calculate % error in mass of sulphamic acid weighed 0.01 (b) 2  0.01  100 = 8% allow  100 = 4% 0.25 0.25 - Properties depend on the nature of bonds and how these bonds are distributed throughout the material - Bonds are formed to attain greater stability, atoms/molecules rearrange their electrons to give lower energy arrangements (by electron loss, gain or sharing) - If a solid has a regular structure, it’s a crystal, the structure is a crystal lattice Chemical bond A force of attraction between atoms, ions or molecules Ionic bond The electrostatic attraction between + and – ions, formed by complete transfer of electron Positive ions(cations)are attracted to the (-)cathode during electrolysis(metal atoms which have lost one or more electrons(K+, Ca+) Negative ions(anions)are attracted to the positive anode during electrolysis Covalent bond Sharing a pair of electrons, one pair to a bond (rather than complete transfer) - Covalent bonds are non-dative bonds, 2 one-electron orbitals overlapping giving a 2 electron bonding orbital(electron density increases between the bonded atoms) Double covalent bond Sharing of 2 pairs of electrons in a bond Dative covalent bond Both electrons in a covalent bond are donated from the same atom to an accepting atom(a 2 electron orbital donating electron density into an empty orbital on the accepting atom) - No difference in length, strength, between a normal covalent bond and a dative covalent bond Molecular orbitals Overlapping orbitals where electron density extends over at least 2 atoms ‘Octet rule’ Hydrogen obtains 2 electrons in its outer shell and other atoms obtain 8 Intramolecular bonds • Covalent Intermolecular bonds • Van der Waals forces charge • Ionic • Dipole dipole attraction charge density = volume • Metallic • Hydrogen bonds strongest Electronegativity(EN) Power(of an atom)to attract(the pair of)electrons in a covalent bond - EN affects bond length with larger differences giving shorter bonds - More electronegative atoms attract shared electrons more towards, itself and acquire a partial negative charge • Electronegativity decreases going down a group(most electronegative element is fluorine) • Electronegativity increases across period 3, elements on the LHS lose electrons and elements on the RHS gain electrons to achieve a stable structure * Ionic bonds are partially covalent when EN is small * Covalent bonds are partially ionic(creation of dipoles)when EN is large Polarisability The ease with which the electron cloud of an anion is distorted by a cation so there’s electron sharing • Smaller and higher the charge(higher the charge density)on the cation, the more polarising it is • Larger and higher the charge on the anion, the more easily it is polarised Features favouring ionic bonding: • Large cation metal, of low charge, having a low IE • Large anion non metal, of low charge, having a high EA

- Large anions most stable with large cations Small cations most stable with small anions - A covalent bond is polar if electrons in the bond are unequally shared

+ – Lattice Enthalpy Hlatt Energy change per mole for exothermic process M (g) + X (g) à MX(s) • Enthalpy/heat energy released when gaseous ions come together to form 1 mole of solid - Lattice enthalpy is a measure of strength of bonding in an ionic substance(relevant when considering solubility of an ionic compound) Calculated value assumes no polarisation, presence of covalence increases lattice enthalpy Complex ion A metal ion associated with a No of anions or neutral - - molecules(ligands) (H2O NH3 Cl CN ) Ligand Anions/molecules firmly bonded to the central cation. Each ligand contains at least one atom with a lone pair of electrons. These can be donated to the central cation forming a co-ordinate (dative)bond. The ligand is said to be co-ordinated to the central ion.

x+ Hydration of metal ions In solution metal ions attract water molecules forming complex ions([M(H2O)6] ) The metal ion is joined to 6 water ligands by dative covalent bonds. The water molecules donate a lone pair of electrons to empty orbitals on the metal ion and an octahedral complex forms 2+ ( [Cu(H2O)6] (pale blue) Ions formed by transition metals are usually coloured) – 2– 4– Simple anions Non metals (Cl , O ) Complex anions Where groups around the central metal ion are negative([Fe(CN) 6] ) + Polyatomic cations Several atoms bonded covalently, the whole structure having a positive charge (NH4 ) 2– – – – Polyatomic anions Several atoms bonded covalently, the whole structure having a negative charge(SO4 , NO3 , CHCOO , MnO4 ) derived from acids by the loss of one or more hydrogen ions –1 • Cation increases in size Group (II) Cation Hlatt kJmol • Charge density decreases Chlorides Radius Experimental Calculated Difference

• Less polarising MgCl2 72 – 2526 – 2326 200

• Covalence decreases CaCl2 100 – 2258 – 2223 35

SrCl2 113 – 2156 – 2127 29

BaCl2 136 – 2056 – 2033 23 –1 • Anion increases in size Magnesium Anion Hlatt kJmol • More polarisable Halides Radius Experimental Calculated Difference

(outer electrons further from the nucleus, less tightly held, MgF2 133 – 2957 – 2913 44 more prone to distortion) MgCl2 180 – 2526 – 2326 200

• Covalence increases MgBr2 195 – 2440 – 2097 343

MgI2 215 – 2329 – 1944 385 • Bonds in solids vibrate about a mean position in the crystal lattice which arise because the crystal isn’t at absolute zero, amplitude of vibrations increase with temperature • Forces between molecules in liquids are no different in type from those in solids, difference is that the particles in liquids are more energetic than those in solids so bonds aren’t particularly directional and liquids have no particular shape

• At Tm vibrations in solid become sufficient to overcome forces holding the crystal together • At Tb vapour pressure of the liquid is the same as the external pressure and bubbles of vapour produced throughout the liquid, heat being put into the liquid is being used by particles to overcome the interparticle forces to separate from each other and escape the liquid(evaporation) so temp remains constant Bt of liquids depends on • Magnitude of interparticle forces • Masses of particles 2bp linear shape 3bp trigonal planar shape (molecule is flat) 4bp tetrahedral shape

5bp Trigonal 6bp octahedron shape 2bp 2lp bent shape bipyramid shape

3bp 1lp pyramidal shape 3bp 2lp T shaped

P H H H Electron lone pairs Electron pairs not involved in bonding 1 Electron pairs in the valence shell(bp or lp)are as far apart as possible to minimise repulsion/reach lowest energy state 2 Order of repulsion between electron pairs is: bp–bp less than bp–lp less than lp–lp 3 Multiple bonds behave as single bonds(no difference between dative and single bonds) - A molecule will not be polar if it’s symmetrical and dipoles of molecule cancel Hybrid orbital A type of atomic orbital that results when 2 or more atomic orbitals of an isolated atom mix, describes orbitals in covalently bonded atoms(hybrid orbitals don’t exist in isolated atoms), in a set are equivalent, and form identical bonds Hybridisation A model that describes the changes in the atomic orbitals of an atom when it forms a covalent compound (Excited C atoms, 4 orbitals rearranged into 4 identical hybrid orbitals(sp3 hybrid orbitals)4 bonds aren’t identical unless you start from 4 identical orbitals) sp hybrid orbital One of the 2 hybrid orbitals formed by hybridisation of an s orbital and p orbital sp 2 hybrid orbital One of the 3 hybrid orbitals formed by hybridisation of an s orbital and 2p orbitals, same shape as sp3 but lies in one plane at 120 ° sp 3 hybrid orbital One of the 4 hybrid orbitals formed by hybridisation of an s orbital and 3p orbital

Sigma bond σ Direct overlap of orbitals Double bond Contains 1σ and 1p bond Triple bond Contains 1σ and 2p bonds

- Ethene, only 3 orbitals hybridised rather than all 4, one 2s electron, two 2p electrons, other 2p electron unchanged(sp2 hybrid orbitals) - All double bonds will consist of a pi bond and a σ bond - In 3rd diagram σ bonds shown using lines, each line representing one pair of shared electrons - 2 C atoms linked by overlap of sp2 hybrid orbitals, making a σ bond and a pi bond formed by sideways overlap of the non-hybridised p atomic orbitals. - Pi bond • sideways overlap of two p-orbitals • to give a two part orbit above and below a  bond (likely to be attacked by electrophiles, they are weaker than σ bonds) Pi bond formed by sideways overlap of the non hybridised p atomic orbitals

2 s atomic orbitals overlap σ bond

Hydrogen bond Electrostatic attraction between a strongly  + H atom attached covalently to a highly electronegative element F/N/O and a strongly  – F/N/O atom on another molecule, H bonds are longer than covalent bonds

- As water cools, (long)H bonds form in greater quantity, the open & ordered structure of ice gives it a lower density than liquid water - Compounds which can H bond with water are very soluble (glucose, has OH groups that H bond with water) - Extensive H bonds with F/N/O cause higher Bt’s

Dipole-Dipole forces(permanently polar molecules)  + and  – parts of the molecules attract electrostatically giving Bt’s higher than those of non polar molecules of similar size. For large molecules dispersion forces can exceed dipole-dipole attraction Dispersion/Van der waals forces(non polar molecules)Temporary dipoles form between molecules because of mobile electron density within the molecule.  + on one molecule will induce a  – on a nearby one and so on, tho there’s a net attraction between molecules Higher forces: • More electrons • Larger area of contact(larger molecules/atoms) • Linear instead of branched chains - Descending noble gases or hydrides, bigger van der waals forces as • Atomic/molecular size increases • More shells of electrons Giant molecular substances Diamond, • High Mt, hard, stiff as have to break strong covalent bonding throughout whole structure • Good thermal conductor as it readily transmits vibration • Poor electrical conductor as no ions or free electrons as they are held tightly between atoms • Insoluble in water, organic solvents as attractions between solvent molecules and carbon atoms will never be strong enough to overcome the strong covalent bonds Silica Lower Mt as has longer(therefore weaker)bonds across whole structure Graphite • Lower density than diamond because of space between the sheets • High Mt as have to break strong covalent bonding throughout whole layered structures • Electrical conductivity along layers, but not at right angles Graphite Electrons aren’t localised between carbon atoms and are free to move along but not between layers, van der waals forces attract the layers together • Insoluble in water, organic solvents as attractions between solvent molecules and carbon atoms will never be strong enough to overcome the strong covalent bonds Diamond Mt4000°C SiO2 Mt1700°C • Soft, slippery as layers of giant molecules can slide over one another - Used in pencils Ionic substances • High Mt as have to break strong electrostatic bonds throughout lattice of oppositely charged ions • Hard, brittle as layers of crystal may slide so ions of same charge come next to one another and repel • Solid doesn’t conduct electricity as no charge carriers available to move • Molten does conduct electricity as mobile ions can move • Soluble in water Ice Molecular covalent substances • Bonded strongly within the molecule but weakly between molecules • Properties depend on intermolecular forces • Non-conductors of electricity (no free electrons, all used in bonding) Metallic bonding • Malleability from non-directional nature of bonds • Electrical conductivity(outer electrons in metals aren’t localised in bonds but free to move around whole lattice formed from + ions) Polymers All long chains which maybe cross-linked by covalent bonds or held by H bonds or dispersion forces(or in silicates forces of ionic attraction between the negatively charged chains of silicate and positive metal ions) - Mechanical properties of the polymer depend on • extent of cross-linking • whether crystallites can form - Pure substances have sharp Mt’s but polymers don’t(usually impure substances), melting over a range of temperatures instead as varying chain lengths of molecules means that they will have a melting range Synthetic organic polymers • Made from alkenes by addition reactions or from reactions between organic molecules which have 2 functional groups which can undergo condensation reactions • They are mixtures since chain lengths vary(no sharp Mt) Polymers formed by radical polymerisation • LDPE is branched, cross-linked • Few crystallites, flexible, translucent Polymers formed by highly controlled type of polymerisation • HDPE, ziegler natta catalyst • More crystallites, stiffer, opaque Natural polymers(organic)polysaccharides(cellulose, glycogen), nucleic acids, proteins, all produced by condensation reactions Inorganic polymers Silicates, phosphoric acid on heating (1)Why & which elements lose electrons? (1)Metals because • Usually they have 1, 2, 3 electrons • By losing these electrons they achieve full outer shells and become more stable • In energy terms, it’s easier to lose these electrons than gain more electrons (2)Why & which elements gain electrons? (2)Non metals because • Usually they have 5, 6, 7 electrons • By gaining electrons they achieve full outer shells and become more stable • In energy terms, it’s easier to gain these electrons than lose the outer electrons (3)Suggest reasons why NaCl vapour is regarded as a collection of ion pairs rather than as NaCl molecules (3)If ion pairs collide there’s nothing to stop them “changing partners” whereas if covalent molecules collide, it’s improbable (4)What happens to an electrostatically charged rod next to a polar liquid like water? (4)It’ll move towards the rod, because polar liquids contain molecules with permanent dipoles. Doesn’t matter if the rod is positively or negatively charged. Polar molecules in the liquid can turn around so the oppositely charged end is attracted towards the rod (5)Why does water have a partial charge? (5)Oxygen has a higher electronegativity than hydrogen (6)(a)State the difference in density between solid ice and liquid water and describe how the presence of H bonds accounts for this (a)• Water is more dense than solid ice • The H bonds in solid ice which holds the molecules together are in fixed positions and lead to an open structure • In water the H bonds are constantly being broken and made (b)Explain the structure of ice, include a diagram (b)• Covalent in water molecules • H bonds between molecules • each water with four waters around – tetrahedral (7)Describe in terms of the position and motion of particles, what happens when some MgCl2(s) is heated from RT to just above Mt (7)• At RT the ions are in a fixed positions in a lattice • As heat is applied the ions vibrate more • Eventually ions have enough energy to overcome electrostatic attraction • Ions break free and are able to move as solid melts

(9)Sketch a graph of temp vs time as a (1)(a)What part of the NH3 molecule enables it to form a dative covalent bond? substance is heated from (a)Lone pair on the nitrogen just below Mt to just above Bt (b)List intermolecular forces between molecules of ammonia (b)H bonds and dispersion forces 2+ (2)Name 2 elements in [Mg(H2O)6] which are joined by a covalent bond (2)H and O 2+ (3)Name 2 elements in [Mg(H2O)6] which are joined by a dative covalent bond (3)O and Mg (1)Why is MgI2 more covalent than MgCl2? (1)Because I– ion is larger than Cl– ion so more easily polarised leading to covalency (2)Metal/Non-metal compounds usually ionic yet solid aluminium chloride has many covalent characteristics because? (2)Small radius, large + charge of Al3+, means high polarising ability giving covalent character pulling electron density away from Cl– creating a mostly covalent bond

(1)Explain why the Mt of Mg is higher than that of Na (1)• Mg ions have larger charge(density) than Na, Mg contributes 2 electrons per atom to the ‘sea’ of electrons, Na has fewer delocalised electrons, Na is a larger atom • Hence Mg ions have greater attraction for the ‘sea’ of electrons than Na • Melting requires energy to overcome this attraction, meaning a higher Mt (3)Explain why Bt of phosphine is lower than that for ammonia (3)• Phosphine doesn’t have H bonds • Lack of H bonds not compensated by increased induced dipole-dipole forces

(4)State the strongest type of intermolecular force present in samples of CH4 (4)Van der waals forces (1)Explain how covalent structure of iodine leads to it having a low Mt (3)Weak intermolecular forces require little energy to break (2)Glucose, poly(ethene), Forces? Properties as a consequence? (2)Glucose, covalent, H bonding, dipole-dipole • Unusually high Mt for its size • H bonds between polar OH groups and water makes it water soluble Poly(ethene), covalent, dispersion • Molecules of different sizes lead to a Mt range • Non-conductor of electricity • Large covalent molecules insoluble in any solvent because, solvent–solute interactions aren’t strong enough to overcome the large dispersion forces between such large molecules + (1)State and explain the shape of the ammonia ion NH4 (1)• Tetrahedral • Has 4 pairs of bonding electrons • Repel as far away from each other as possible • Bt increases Noble Relative atomic Tb/K Group 4 Relative Tb/K • Relative atomic gases mass Hydrides atomic mass mass and size He 4.0 4 CH4 16.0 81.6 increases Ne 20.2 27 SiH4 32.1 161

Ar 39.9 87 GeH4 76.6 185

Kr 83.8 121 SnH4 122.7 221

Xe 131.3 166 PbH4 probably doesn’t exist Periodicity Regular periodic variations of properties of elements with atomic number Periodic Law Elements are arranged in order of increasing atomic number in the periodic table - All elements within a period(row) have the same number of electron shells - All elements within a group(column)have the same number of electrons in their outer shell indicated by the group number Period 3 Na Mg Al Si P S Cl Ar

Tm/K 371 922 933 1683 317 392 172 84

Tb/K 1156 1380 2740 2328 553 718 231 87 Conductivity 0.35 0.36 0.61 10–18 10–17 10–23 relative to Ag =1 –1 Ha kJmol 107 148 326

Structure Metallic Giant covalent P4 S8 Cl2 Monatomic (gives monatomic vapours) (gives monatomic Molecular covalent vapour) Atoms get smaller & higher Mt across period • Number of protons increasing meaning stronger nuclear attraction • Extra electrons are at roughly the same energy level even if the outer electrons are in different orbital types • Little extra shielding effect, little extra distance to lessen the attraction from the nucleus • Smaller atoms can pack more closely which means better orbital overlap • More electrons in the valence shell to delocalise around the metal lattice Sharp rise in Mt between Mg & Na and only small rise between Mg & Al • Alkali metals are body centred cubic packing that doesn’t bring atoms as close as possible • Mg & Al hexagonal close packing Mt’s depend on dispersion forces Phosphorus Weak van der waals forces so Mt is low Sulphur Highest Mt, largest van der waals forces, largest molecule Chlorine Lowest Mt, weakest van der waals forces, smallest molecule, simple diatomic molecules with no permanent dipoles Argon Argon, monatomic, stable electron arrangement, small intermolecular dispersion van der waals forces so Mt is low Grp 1(alkali metals) Grp 2(alkaline earth metals)S block metal compounds ionic, Ox No’s(+1 & +2 respectively)only because going above these entails an IE input which couldn’t be recovered from LE of resulting solid - Very reactive, low density, low Mt, low EN, soft because of weak metallic bonding, compounds usually colourless unless transition metal is present in the anion - Ionic radius smaller than atomic radius for s block elements because loss of outer electrons results in loss of outer shell Grp 1 Atomic radius Ionic Density Mt °C Bt °C Abundance 1st IE 2nd IE Flame pm radius gcm3 ppm kJmol –1 kJmol –1 colours pm Li 133 60 0.53 181 1330 65 520 7298 Carmine red Na 157 95 0.97 98 890 28300 496 4563 Yellow K 203 133 0.86 Increase 63 774 25900 419 3051 Lilac Rb 216 148 1.53 39 688 310 403 2632 Decrease Colourless Cs 235 169 1.88 29 690 7 376 2420 Colourless Grp 2 3rd IE kJmol –1 Be 89 31 1.85 1278 2477 6 900 1757 14800 Colourless Mg 136 65 1.74 649 1110 20900 738 1451 7740 White Ca 174 97 1.54 839 1487 36300 590 1145 4940 Brick red Sr 191 113 2.6 Increase 769 1380 150 550 1064 4120 Decrease Crimson red Ba 198 135 3.5 725 1640 430 503 965 3390 Apple green Group 1 Group 2 Densitiy, hardness, Mt’s higher than Group 1 because • Atoms the largest of the period • Atomic radius smaller in Group 2 • Body centred cubic packing • Hexagonal close packing(except Ba) • 2nd IE’s a lot larger than 1st because removing 2nd electron • 2 electrons per atom for metallic bonding requires breaking into inert gas structure, which has less • Extra proton shielding, shell is closer to the nucleus, so nuclear attraction is much larger • Grp 1 more reactive than Grp 2 because, 1 less proton & only needs to lose 1 electon to lose to achieve a full outer shell Body centred cubic Hexagonal close packed arrangement arrangement /face centred cubic lattice Descending Group 1 & 2 • Density rises because mass of the atom increases more rapidly than its size - Mt & hardness decreases, reactivity increases • Atomic radius increases, smaller charge density, delocalised electrons more spread out • Reduced attraction of + ions to ‘sea’ of delocalised electrons, less energy to break bonds - IE & EN decreases • Extra electron shells shielding outer electrons from nuclear charge • Outer electrons further away from the nucleus and less attracted • Overall, outweighs increase in protons, reducing nuclear attraction Acids proton donors + + Acids mixed with water release H (never alone in water)which combine with H2O(hydrated)to form hydroxonium ions H3O + – HCl(g) + H2O(l) à H (aq) + Cl (aq) HCl(g) doesn’t release hydrogen ions until it meets water so HCl(g) isn’t an acid Strong acids HCl + water à H+ + Cl– Ionise(dissociate)almost completely, nearly every H atom released to become a hydrated proton, lots of H+(aq) ions + – Weak acids H2CO3 + water H + HCO3 weak acid equilibrium lies to the left so most of the acid won’t be ionised Ionise(dissociate)slightly, little H atoms released, little H+(aq)ions Strong acids and concentrated acids OR weak acids and dilute acids aren’t the same • ‘Strong’ ‘weak’ refers to how much the acid has ionised • ‘Concentrated’ ‘dilute’ refers to molcm–3of the acid Bases proton acceptors, make OH– hydroxyl ions when dissolved in water Salt A compound formed by replacing hydrogen in an acid by a metal

Metal + Acid à Salt + H2 Metal carbonate + Acid à Salt + CO2 + H2O Mg(s) + H2SO4(aq) à MgSO4(aq) + H2(g) Na2CO3(s) + 2HCl(aq) à 2NaCl(aq) + CO2(g) + H2O(l) + + Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) Na2CO3(s) + 2H (aq) à 2Na (aq) + CO2(g) + H2O(l) + 2+ Mg(s) + 2H (aq) à Mg (aq) + H2(g) CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l) + – Acid/Base neutralisation reaction Ionic equation for neutralisation H (aq) + OH (aq) à H2O(l)

Metal Oxide(alkali) + Acid à Salt + H2O Metal hydroxide(alkali) + Acid à Salt + H2O

Grp 1 M2O(s) + 2HCl(aq) à 2MCl(aq) + H2O(l) Grp 1 MOH(aq) + HCl(aq) à MCl(aq) + H2O(l) Grp 2 MO(s) + 2HCl(aq) à MCl2(aq) + H2O(l) Grp 2 M(OH)2(aq) + 2HCl(aq) à MCl2(aq) + 2H2O(l) S block metals and oxygen All oxides are basic except Be is amphoteric Radicals Atoms/ions with a lone unpaired electron, very reactive Group 1 reacts with oxygen in air at RT, silvery when cut they rapidly tarnish and become dull due to oxide coating Li, Na, K stored in paraffin oil, while Rb, Cs which are more reactive are stored in sealed containers Li reacts with nitrogen in the air 6Li(l) + N2(g) à 2Li3N(s) Heat Remaining alkali metals form superoxides 4Li(s) + O2(g) à 2Li2O(s)white 2Na(s) + O2(g) à Na2O2(s)pale yellow 2K(s) + O2(g) à KO2(s)yellow 2– 2– – O lithium oxide O2 sodium peroxide O2 potassium superoxide(radical) Group 2 heat heat 2Mg(s) + O2(g) à 2MgO(s) (except for Ba) 4Ba(s) + 2O2(g) à 2BaO(s) + BaO2(s) In reaction with oxygen, Li behaves more Grp 2 than Grp 1 behaving like Mg, diagonal relationship(BeO & Al2O3 amphoteric) S block metals and water Group 1 Li Bubbles, floats, melts, gets smaller Metal hydroxides dissolve to form colourless solutions that Na Bubbles, floats, melts, gets smaller are strongly alkaline(hence alkali metals) K Bubbles, floats, melts, gets smaller 2Na(s) + 2H2O(l) à 2NaOH(aq) + H2(g) Rb Explodes violently with water

Group 2 Metal hydroxides less soluble, ppt white suspension, metals don’t float Ca(s) + 2H2O(l) à Ca(OH)2(s) + H2(g) - Be doesn’t react at all with water or steam - Mg reacts slightly with cold water, quicker with hot water/steam , reaction stops quickly because Mg(OH)2 barrier formed on Mg preventing further reaction Mg(s) + 2H2O(l) à Mg(OH)2(s) + H2(g) Mg(s) + H2O(g) à MgO(s) + H2(g) - MgO rather than Mg(OH)2 formed because Mg(OH)2 is thermodynamically unstable with regard to MgO + H2(g) If formed it would decompose to MgO + H2(g) Transition Metal Aluminium reacts with water, appears not to due to protective surface oxide layer

2Al(s) + 6H2O(l) à 2Al(OH)3 + 3H2(g) S block metals and chlorine All metal chlorides ionic(except Be because of high electronegativity, forms a polymer with dative covalent bonds) Group 1 2Na(s) + Cl2(g) à 2NaCl(s) Group 2 Mg(s) + Cl2(g) à MgCl2(s)

MgCl2(s) + H2O(l) Mg(OH)Cl(s) + HCl(aq) hydrolysis

All ionic chlorides soluble in water. MgCl2 has some covalent character shown by difference between Born Haber cycle value for the LE and the calculated value S block oxides with water All solutions alkaline owing to OH– Reactions of the anion independent of the cation therefore reactions same for Grps 1 & 2 2– – + – Oxides O (s) + H2O(l) à 2OH (aq) 2Li (aq) + 2OH (aq) Li2O(s) + H2O(l) à 2LiOH(aq) 2– – Peroxides O2 (s) + 2H2O(l) à 2OH (aq) + H2O2(aq)Na2O2(s) + 2H2O(l) à 2NaOH(aq) + H2O2(aq) – – Superoxides 2O2 (s) + 2H2O(l) à 2OH (aq) + H2O2(aq) + O2(g) 2KO2(s) + 2H2O(l) à 2KOH(aq) + H2O2(aq) + O2(g)

Group 2 Cation Hydroxide solubility Sulphate solubility radius/pm mol per 100g water mol per 100g water Mg2+ 65 2.00 x10–5 1.83 x10–1 Ca2+ 99 1.53 x10–3 4.66 x10–3 Sr2+ 113 3.37 x10–3 7.11 x10–5 Ba2+ 135 1.50 x10–2 Increase 9.43 x10–7 Decrease Thermal stability Ability of a material to decompose under heat stress. More ionic, more thermally stable, more heat before decomposition - Thermal stability of a carbonate will depend on the stability of the carbonate lattice compared with the oxide lattice at the same temperature, as cation size changes LE’s(strength of bonding in an ionic substance)of carbonates & oxides change by different factor - Thermal stability increases down the group as cations have a larger ionic radius so charge density decreases, so lower polarising power distorting carbonate/nitrate anion less Group 1 - Group 1 cations larger, less charge, compounds more thermally stable than Group 2 so differences in LE between the carbonate & oxide aren’t sufficient to allow decomposition of the carbonates at Bunsen temperatures(except Li which has the smallest cation) Li2CO3(s) Li2O(s) + CO2(g) - Nitrates all decompose 2NaNO3(s) à 2NaNO2(s) + O2(g) (except Li) 4LiNO3(s) à 2Li2O(s) + 4NO2(g) + O2(g) (white) Nitrite (pale yellow) Larger cations give nitrites(smaller anions than nitrates, with higher LE, but not too small like the oxide) Group 2 Limestone Heat - Carbonates all decompose CaCO3(s) à CaO(s) + CO2(g) Reaction used in cement manufacture and extraction of iron - Nitrates all decompose 2Ca(NO3)2(s) à 2CaO(s) + 4NO2(g) + O2(g) (white) (brown) Small cations form stronger lattices with oxide than with the larger nitrate

2AgNO3  2Ag + 2NO2 + O2

Test of ease of decomposition of nitrates How long it takes until O2 or NO2 (toxic brown gas(fume cupboard))is produced

Test of ease of decomposition of carbonates How long it takes until CO2 is produced S block metals are far too reactive to occur native(as the uncombined metal) all are extracted by electrolysis of their molten chlorides so conversion to the chlorides necessary for extraction(except Na, K) - NaCl obtained by solution mining(water pumped into underlying salt strata and brine obtained, can lead to land subsidence) NaCl in seawater is at a low concentration K obtained from soluble mineral carnallite Li, Rb, Cs in insoluble aluminosilicate minerals - Be in beryl(pale green insoluble aluminosilicate)used as a semi-precious stone Mg in seawater & in carnallite & dolomite Ca in limestone, marble, chalk, Sr, Ba in their insoluble sulphates Flame test 1 Clean end of platinum/nichrome wire by dipping it into (conc)HCl and burning off impurities in a roaring bunsen flame until there’s no persistent flame colouration 2 Moisten the end of the clean wire with (conc)HCl and then dip into the sample to be tested 3 Hold the sample at the edge of a roaring bunsen flame Lithium Carmine red Calcium Brick red Sodium Yellow Strontium Crimson Potassium Lilac Barium Apple green - Heat energy absorbed from the flame causes electrons to be excited within the metal(move to higher energy levels) When these return to lower energy levels they emit light of characteristic frequencies and colour can be observed, or analysed in a spectrometer(shows a spectrum)which gives a line emission spectrum, light appears as a series of coloured lines, each line of light is a particular level and wavelength, which is evidence that electrons can only be found at particular energy levels(shells and subshells) within an atom, Na light is virtually monochromatic - • In astronomy spectra can be used to analyse the atmosphere of other planets • Sodium vapour used in yellow street lamps • NaCl used as food flavouring and preservative • Sodium carbonate used in manufacture of glass • NaOH used in manufacture of soaps, detergents, bleaches • Potassium nitrate used in fertilisers • Magnesium hydroxide used in antacid indigestion powders to neutralize excess stomach acid • Calcium carbonate and calcium oxide used in agriculture to reduce acidity of soil and improve fertility • Barium sulphate used in X-ray scans (1)Ar insulator because? (1)• Monatomic • Full outer shells of electrons tightly held in place so completely uninterested in bonding

(2)Why does calcium stop reacting with (dil)H2SO4 after a few seconds even though it didn’t react initially (2)• Calcium sulphate • Forms an insoluble/protective layer (1)State relative thermal stability of potassium nitrate and calcium nitrate and explain how it’s related to the sizes and charges of the ions involved (1) • Calcium nitrate less thermally stable/decomposes more easily than potassium nitrate • Ca cation smaller • with double charge • polarises nitrate more • bonds in nitrate more easily broken/oxygen atom more attracted (1)Describe a test to distinguish between LiCl(s) and NaCl(s) (1)Flame test (2)State how a flame test would distinguish between Ca(NO3)2 and Ba(NO3)2 (2)Ca brick red, Ba (apple)green (1)(a)(i)Identify one of the elements that is composed of simple molecules at RT (i)P or S or Cl 1800

(2)Name the type of bonding in (i)CaCl2 (ii)HCl(g) (2)(i)Ionic (ii)Covalent K 1600

(3)(i)2 factors which affect the polarising power of cations? (i)ionic radius, charge e

r 1400 u t

(1)(a)2 bottles are clearly labeled ‘sulphate’. The solid in bottle A dissolves easily in a 1200 r e water but none of the solid in bottle B appears to dissolve when added to water p 1000 m

Which of these two bottles contains barium sulphate? (a)B e 800 t

(b)Bottle C, labeled ‘magnesium carbonate’ When a sample is heated a colourless gas g 600 n i t is produced that turns limewater cloudy. State whether this label is correct and explain l

(b)LimewatermilkyCO2 MgCO3 decomposes on heating to CO2 label correct m 200 0 Na Mg Al Si P S Cl Ar element

Oxidation Is Loss Reduction Is Gain Oxidation and reduction occur together in redox reactions, each element in a compound is treated as an ion

Synthesis reactions 2 or more simple substances combine to form a more complex substance 2H2 + O2 à 2H2O Decomposition reaction Complex substance breaks down into its more simple parts 2H2O à 2H2 + O2 Oxidation Chemical process in which electrons are removed from an atom, ion or compound Oxygen is added/hydgrogen is lost from a substance Oxidising agent Acceptor of electrons in a redox reaction, the best being fluorine Reducing agent The reactant that gives up electrons in a redox reaction Oxidation number(state) Charge on an atom if the element/compound were ionic Disproportionation A reaction in which an element/compound simultaneously undergoes both oxidation and reduction

1 Elements have Ox No zero(H2, Br2, Na, Be, K) 2 Monatomic ions Ox No same as charge 3 Ox No of F is always (-1), H is (+1) in most compounds, O(–2) • Sum of Ox No’s in a neutral molecule must add up to zero or the charge on the ion – • Ox No’s usually integers (O2 (Ox No = –½)) • The more electronegative atom has the (-)Ox No, C in CO2 is (+4), in CH4 is (–4), because carbon more electronegative than hydrogen 2– SO4 overall oxidation = –2 Ox state of O = –2 total = –8 Ox state of S = +6 Fe4O4 overall oxidation = 0 Ox state of O = –2 total = –6 Ox state of Fe = +3 When half reactions combine number of electrons in each must be the same(one or both half reactions may need to be multiplied) – – NO2 becomes NO3 with oxidising agent potassium manganate(VII) with H2SO4 (electrons and H ions don’t combine to form hydrogen since electrons aren’t free but given to an oxidising agent) – + – 2+ – – – + 2 half reactions: MnO4 + 8H + 5e à Mn + 4H2O NO2 + H2O à NO3 + 2e + 2H – + – 2+ – – – + Balancing equations: (x2) 2MnO4 + 16H + 10e à 2Mn + 8H2O (x5) 5NO2 + 5H2O à 5NO3 + 10e + 10H – – + 2+ – Overall: 2MnO4 + 5NO2 + 6H à 2Mn + 5NO3 + 3H2O – – (1)(i) Write ionic half equation for reduction of bromine to bromide ions (i)Br2 + 2e à 2Br (ii) Write ionic half equation for oxidation of Fe2+ ions to Fe3+ ions (ii)Fe2+ à Fe3+ + e– 2+ 2+ – 3+ (iii) Hence write overall ionic equation for reaction of Fe ions with bromine (iii)Br2 + 2Fe à 2Br + 2Fe

‘Halogen’ Means salt former, because of large number of ionic, salt compounds Group 7 forms Enthalpy of dissociation Enthalpy change when 1 mole of a gaseous substance is broken up into free gaseous atoms(measure of strength of covalent bonds, bond enthalpy(+ve)) State at RT Mt °C Bt °C Atomic Ionic 1st IE EA Bond Colour in Colour in radius radius kJmol –1 kJmol –1 dissociation water hexane ppm ppm enthalpy F Pale yellow gas -220 -188 72 136 1680 -340 58 Cl Green gas -101 -34.7 99 181 1260 -364 242 Colourless Colourless Br Brown liquid -7.2 58.8 114 195 1140 -342 Increase 193 Decrease Orange Brown I Dark grey solid 114 184 133 216 1010 -314 151 Brown Purple

Highly reactive non-metals, strong oxidising agents, natural state is covalent diatomic molecules(Cl2), Vdw forces between molecules - Covalency means low solubility in water, but dissolves easily in organic compounds(hexane) – – - Halogens in (+)Ox states don’t form (+)ions but are bonded covalently to more electronegative atoms as oxyanions OCl & ClO3 - EA & bond dissociation enthalpies peak at Cl, normally the shorter a bond the stronger it is, but with F the non bonding electrons are brought so close that they repel, weakening the bond(partly why F is very reactive) EN & oxidising power descending group K/NaCl(aq) K/NaBr(aq) K/NaI(aq)(colourless)

Oxidising strengths seen by displacement reactions Cl2(aq)Colourless X Orange Br2(aq)formed Brown I2(aq) formed with halide ions Br2(aq)Orange X X Brown I2(aq) formed Cl2(g) + 2NaI(aq) à 2NaCl(aq) + I2(aq) I2(aq)Brown X X X Halogens undergo disproportionation with alkalis COLD HOT

X2 + 2NaOH(aq) à NaXO + NaX + H2O 3X2 + 6NaOH à NaXO3 + 5NaX + 3H2O – – – – – – X2 + 2OH à XO + X + H2O 3X2 + 6OH à XO3 + 5NX + 3H2O Ox State of X 0 +1 –1 0 +5 –1 Chlorine test • Swimming pool smell, moist litmus paper blue à red à bleached • Moist starch iodide paper, iodide ions oxidised to iodine, so turns blue-black – – • Cl2 will oxidise Br to Br2 and I to I2, solutions turn orange/brown, excess Cl2 used then I2(s)ppts – – Cl2(aq) + 2Br (aq) à Br2(aq) + 2Cl (aq) obtains bromine from sea water, used for leaded petrol, antifreeze – – – – Cl2 + 2e à 2Cl 2Br à Br2 + 2e Bromine test • Slower than chlorine, moist litmus paper blue à red à bleached • Moist starch iodide paper, iodide ions oxidised to iodine, so turns blue-black – • Br2 will oxidise I to I2, solution darkens, immiscible organic solvent(hexane) can show presence of iodine, turning solution purple • Bromine will react with paper moistened with fluorescein dye turning it scarlet, product is eosin(ingredient of red ink) Iodine test • No effect on litmus • Moist starch iodide paper and starch solution turns blue-black which neither chlorine or bromine will do • Iodine is purple in immiscible organic solvents(hexane, methylbenzene) which have no oxygen. Presence and test of oxygen in a solvent gives a browner cast - Halides are compounds with a halogen ion(KI, HCl, NaBr) - Hydrogen halides(covalent hydrides H–X)acidic colourless gases, white misty fumes in moist air, giving white fumes with NH3(g) + Polar molecules, high solubility, strong acids due to high Hhyd of the H ions and relatively small halide ions which compensate for the bond dissociation energy of the molecule, extent to which dissociation occurs dependent on H–X + – + – HX(aq) + H2O(l) à H3O (aq) + X (aq) HCl(g) + H2O(l) à H (aq) + Cl (aq) HCl, HBr, HI all strong acids, stronger down the group • Reducing power of halide increases descending • Larger atomic radius, electrons further away from the nucleus • Extra shielding Bond enthalpy Strength of the bond, energy needed to make the bond(Average of bond dissociation enthalpies of a bond) –1 Bond enthalpy for C–H bond is average value from breaking all 4 C–H bonds CH4(–416kJmol ) Value taken from Ha(CH4) H–F H–Cl H–Br H–I Bond enthalpy (kJmol –1) 562 431 366 299 • HF acid oddly behaves as a weak acid, H–F bond is the strongest, but not enough to explain the difference in acid strength This reaction prevents H–F molecules from dissociating, F– ions form H bonds with undissociated HF molecules H–F(aq) + F–(aq) à [F----- H–F] – (aq) 2– • HF not normally corrosive but dissolves glass and gives burns, from formation of SiF6 ion which is water soluble, other halide ions are too large to fit around the Si atom, HF solutions kept in poly(ethane)bottles Halide ion test(Not F) (Since AgF is water soluble) • Acidify with (dil)HNO3(aq) to ensure removal of ions which would give a spurious ppt – + • Add AgNO3(aq), silver halide ppt formed X (aq) + Ag (aq) AgX(s) • Add (dil)/(conc)NH3 to ppt + + + Reaction reduces concentration of Ag ions Ag (aq) + 2NH3(aq) [Ag(NH3)2] (aq) – Cl AgNO3(aq) (dil)NH3 to ppt OR • Oxidise to respective halogen White ppt AgCl forms Ppt dissolves leaving colourless solution • Acidify with (dil)HNO3(aq) to ensure removal of ions – Br AgNO3(aq) (conc)NH3 to ppt which would give a spurious ppt Cream ppt AgBr forms Ppt dissolves leaving colourless solution • Add immiscible organic solvent(hexane) – I AgNO3(aq) (conc)NH3 to ppt • Add sodium chlorate(I) solution and shake Yellow ppt AgBr forms Ppt insoluble • Br– = brown organic layer, I– = purple organic layer Halide salt with (conc)H2SO4 at RT • Li/NaCl(s) + H2SO4(l) à HCl(g) + Li/NaHSO4(s)sodium hydrogen sulphate – – Cl2 not produced, Fl and Cl ions too weak reducing agents to reduce (conc)H2SO4 • Bromide ions: NaBr(s) + H2SO4(l) à HBr(g) + NaHSO4(s) 2HBr(g) + H2SO4(l) à Br2(g) + SO2(g) + 2H2O(l) – – + – – + Ionic equations 2Br à Br2 + 2e H2SO4 + 2H + 2e à SO2 + 2H2O 2Br + 2H + H2SO4 à Br2 + SO2 + 2H2O • Iodide ions: Dark grey solid iodine produced, mixture is brown, with purple iodine vapour – – – Brown colour due to production of I3 I2 + I I3 NaI(s) + H2SO4(l) à HI(g) + NaHSO4(s) Firstly 2HI(g) + H2SO4(l) à I2(s) + SO2(g) + 2H2O(l) Then 6HI(g) + H2SO4(l) à 3I2(s) + S(s) + 4H2O(l) Finally 8HI(g) + H2SO4(l) à 4I2(s) + H2S(g) + 4H2O(l) H2S(g) has ‘bad egg smell’ + – – Chlorine with water Cl2(g) + H2O(l) HCl(aq) + HClO (aq) Cl2(g) + H2O(l) 2H (aq) + ClO (aq) + Cl (aq) – + HClO(aq) + H2O(l) ClO (aq) + H3O (aq) Heat – – – 3NaClO à NaClO3 + 2NaCl(aq) 3ClO (aq) à 2Cl (aq) + ClO3 (aq) – NaCl in bleach doesn’t matter, however bleach shouldn’t be mixed with other materials as sodium chlorate(I) will oxidise Cl to Cl2 ClO– + 2H+ + Cl  Cl2 + H2O Sodium chlorate(I) used as disinfectant/water treatment, bleach • Oxyacids(HClO and HClO3) oxyanions are oxidising agents in acidic solution – + – – – + – – ClO (aq) + 2H (aq) + 2e à Cl (aq) + H2O(l) ClO3 (aq) + 6H (aq) + 6e à Cl (aq) + 3H2O(l) Electrolysis of (conc)NaCl(aq)(brine) (sea water can’t be used because concentrations of NaCl too low) Cell with 2 compartments separated by a membrane. Solution contains Na+ ions Cl– ions H+ ions OH– ions – – + – Titanium anode(+) oxidation 2Cl (aq) à Cl2(g) + 2e Steel cathode(–) reduction 2H (aq) + 2e à H2(g) Na+ and OH– ions remain so 3rd product is NaOH(aq) contaminated by NaCl which must be separated Overall 2NaCl(aq) + 2H2O(l) à 2NaOH(aq) + H2(g) + Cl2(g) Uses of chlorine Water/sewage treatment/sterilisation, PVC/HCl manufacture Organochlorine manufacture, chlorine reacts with many organic compounds, can give pesticides, though organochlorine causes problems such as toxicity and ozone layer depletion from CFC’s + (1)Explain why HCl(g) is soluble in H2O(l) (1)HCl reacts with H2O(l) and H ions formed (2)Which of the elements Cl or Br is the stronger oxidising agent? Explain the importance of this in the extraction of bromine from seawater(2)• Chlorine • Can except electrons from Br–

(3)Br2(aq) will oxidise Fe2+ ions to Fe3+ ions (i)Write the ionic half-equation for the reduction of bromine to bromide ions (i)Br2 + 2e  2Br– (ii)Write the ionic half-equation for the oxidation of Fe2+ ions to Fe3+ ions (ii)Fe2+  Fe3+ + e (iii)Hence write the overall ionic equation (iii)Br2 + 2Fe2+  2Br + 2Fe3+ (d)Chlorine and bromine react with NaOH(aq) in a similar way at RT (i)Write equation for bromine reaction with NaOH(aq) (i)Br2 + 2NaOH  NaBr + NaBrO + H2O (ii)What type of reaction is this? (ii)Disproportionation H2S is produced when (conc)H2SO4 is added to NaI(s), but SO2 is produced when (conc)H2SO4 is added to NaBr(s) (1)(i)Write an ionic half-equation for oxidation of Cl– to Cl2 (i)2Cl  Cl2 + 2e– (ii)Write an ionic half-equation for reduction of OCl– to Cl– in acidic conditions (ii)OCl– + 2H+ + 2e  Cl– + H2O (iii)Combine the two ionic half-equations to show the effect of adding acid to bleach (iii)OCl– + 2H+ + Cl  Cl2 + H2O (1)(a)What would happen when (i)KCl +(conc)H2SO4 (i) • Reaction occurs • Misty fumes • HCl doesn’t reduce sulphur in H2SO4 acid (ii)KI + (conc)H2SO4 (ii) • Reaction takes place • Iodine produced black solid/purple vapour • smell of sulphur or hydrogen sulphide • No HI produced and • hydrogen iodide reduces sulphur in sulphuric acid OR sulphuric acid can oxidise hydrogen sulphide (b)Test to show that HCl(g) not HI(g) was given off? (b) • Dissolve in water • Add AgNO3(aq) • White ppt soluble in (dil)NH3 if chloride • Yellow ppt insoluble in (conc)NH3 if iodide (2)(i)Observations when (conc)H2SO4 is added to • Lithium chloride(s) • Sodium bromide(s) • Potassium iodide(s) (2)(i) White/steamy fumes (ii) Brown/orange liquid (iii) Purple vapour/dark solid

(3)(a)Observations when Br2(aq)is added to a solution of KI (3)(a)Solution from colourless to dark brown/black solid produced – – (4) Observations when Cl2 + 2I à I2 + 2Cl (4)Colourless solution turns brown (5)Equation for reaction between LiCl(s) and (conc)H2SO4 (5)LiCl(s) + H2SOl  LiHSO4(s) + HCl(g) OR 2LiCl(s) + H2SO4(l)  Li2SO4(s) + 2HCl(g) (1)Hydrazine is manufactured by reacting ammonia with sodium chlorate(I) 2NH3 + NaOCl  N2H4 + NaCl + H2O The reaction is a two stage process The 1st stage is NH3 + NaOCl  NaOH +NH2CL Write the equation for the second stage (1)NH3 + NH2Cl+ NaOH  N2H4 + NaCl + H2O