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Transcript
Mass No E(element)
Sub-atomic Relative Relative
E = hf Energy = Planck’s constant x frequency
particle
charge
mass
Atomic number Proton number
Atomic No
Mass number Proton number plus the number of neutrons in the
Proton
+1
1
nucleus of its atom
Electron
-1
negligible
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)  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)  M2+(g) + e–
1st Electron affinity X(g) + e–  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–  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
1s 2s 2p 3s 3p 3d 4s 4p 4d 4f
n Subshell Number of electrons Subshell
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)
Z
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
Symbol
H
He
Li
Be
B
C
N
O
F
Ne
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
E
Br
Kr
1s
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2s
2p
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
2
3
4
5
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
3s
3p
3d
4s
4p
Group II
Group III
Group V
Group VI
N 2s
2p
Be
B
N
O
1s22s2
1s22s22px1
1s22s22px12py12pz1
1s22s22px22py12pz1
O
2s
1st IE = 900 kJmol–1
1st IE = 799 kJmol–1
1st IE = 1400 kJmol–1
1st IE = 1310 kJmol–1
2p
• In B the 2p electron easier to remove than the 2s electron from
Be because subshells that are full are more stable
• In B the 2p orbital is further from the nucleus
• The 2p orbital is screened not only by the 1s2 electrons but also
partially by the 2s2 electrons
- These factors are strong enough to override the effect of the
increased nuclear charge resulting in the IE to drop slightly
• Screening identical and electron being removed is from an
identical orbital
• N structure is symmetrical and more stable than that in O
• Repulsion between 2 electrons in the same orbital means an
electron in the 2px2 pair is easier to remove
1
2
2
1
2
2
2
3
2
4
2
5
2
6
2
6
1
(q) An atom contains 5 protons and 5 neutrons, give the symbol for
2
6
2
this atom including the mass number (a) B
2
6
1
2
(q) Formula of the compound formed between this element and
2
6
2
2
chlorine (a) BCl3
2
6
3
2
2
6
5
1
2
6
5
2
2
6
6
2
2
6
7
2
2
6
8
2
2
6
10
1
2
6
10
2
2
6
10
2
1
2
6
10
2
2
2
6
10
2
3
2
6
10
2
4
2
6
10
2
5
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
(1H)+ (2H)+ (12C)+ (12C1H)+ (12C1H2)+ (12C1H3)+ (12C1H32H)+
(2)(a)Bromine consists of 2 isotopes, mass numbers 79 and 81. A sample of Br 2(g) was examined in a mass spectrometer. Identify
the species responsible for the peak at m/e = 160
(2)( 79Br81Br)+
(2)(b)For a particular sample of copper two peaks were obtained in the mass spectrum
Peak at m/e
Relative abundance
(i)Give the formula of the species responsible for the peak at m/e = 65 (i)65Cu+
63
69.1
(ii)State why two peaks, at m/e values of 63 and 65, were obtained in the mass spectrum
65
30.9
(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
Log ionisation energy /kJ mol –1
6
(8)What force causes the scattering of α particles by nuclei?
(8)α particles and nuclei both positively charged thus electrostatic forces of repulsion
5
(9)Explain why all isotopes of Mg have the same chemical properties
(9)• Same number of electrons in all Mg isotopes • Outer electron structure
4
determines chemical properties
(10)Using subshell notation, give electronic configuration of K atom and K + ion
3
(10)K 1s22s22p63s23p64s1 K+ 1s22s22p63s23p6
2
(11)State why 2 peaks at m/e values of 63 and 65 were obtained in a mass spectrum
of an element (11)2 different isotopes
1
(12)Write equation for 5th ionisation of Na (12)Na4+(g)  Na5+ (g) + e–
(14)The logarithm of successive IE’s for Mg across the page
0
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 B 2H5
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 × NH 2 = 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)  NaNO3(aq) + AgCl(s)
Silver nitrate
silver chloride(white ppt)
Ions: Na+(aq) + Cl–(aq) + Ag+(aq) + NO3–(aq)  Na+(aq) + NO3–(aq) + AgCl(s)
Deleting spectator ions: Ag+(aq) + Cl–(aq)  AgCl(s)
2KMnO4(aq) + 8H2SO4(aq) + 10FeSO4(aq)  2MnSO4(aq) + 5Fe2(SO4)3(aq) + K2SO4(aq) + 8H2O(l)
potassium manganate(VII)
iron(II)sulphate
Ions: 2K+(aq) + 2MnO4–(aq) + 16H+(aq) + 18SO42–(aq) + 10Fe2+(aq)  2Mn2+(aq) + 10Fe3+(aq) + 18SO42–(aq) + 2K+(aq) + 8H2O(l)
Deleting spectator ions: MnO4–(aq) + 8H+(aq) + 5Fe2+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Avogadro constant NA is 6.02 x 1023mol–1(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 x10 23 particles of it
• is its relative molecular/atomic mass in g(molar mass, gmol–1)
- Fe(s) + S(s)  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 standard solution of Na2CO3(aq) of known concentration:
1 1.25g of pure anhydrous Na 2CO3 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)  2NaCl(aq) + H2O(l) + CO2(g)
12.5g of Na2CO3 in 1dm3 solution, 25cm3 of this titrated with HCl, 23.45cm3 of HCl was required, concentration of the acid?
(1)Na2CO3= 106gmol–1 12.5gdm–3/106gmol–1 = 0.118moldm–3(concentration)
0.025dm3x 0.118moldm–3 = 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
(2)1g Na2CO3 dissolved in water and volume made to 200cm3
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  2NaCl + H2O + CO2
Amount of Na2CO3 = 1g/106gmol–1 = 9.434 x10–3mol
Amount of HCl = 2 x 9.434 x10–3mol = 1.887 x10–2mol
–2
–3
Volume of HCl required = 1.887x10 mol/0.12moldm = 0.157dm3
(3)1cm3 of (conc)H2SO4 in a 500cm3 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)H2SO4(aq) + 2NaOH(aq)  Na2SO4(aq) + 2H2O(l) Amount of NaOH = 0.0198dm3 x 0.1moldm–3 = 1.98 x10–3mol
Since 1 mol H2SO4 requires 2 mol NaOH Amount of H2SO4 = 1.98 x10–3mol/2 = 9.9 x10–4mol
9.9 x10–4mol in 25cm3, 9.9x10–4 x 500/25, 0.0198mols in 1cm3 of (conc)H2SO4 19.8mols in 1dm3 conc = 19.8moldm–3
(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?
(4)CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) CaCO3 = 100gmol–1
3
Amount of NaOH = amount of HCl unreacted = 0.0245dm 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 CaCO 3 = 0.0255mol/2 = 0.0128molMass of
CaCO3 = 0.0128mol x 100gmol–1 = 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 CO 2 was evolved?
(1)CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l) CaCO3 = 2.33g/100gmol–1 = 0.0233mol, 0.0233mol CO2 produced
Volume of CO2 = 0.0233mol x 24dm3mol–1 = 0.56dm3
(2)10cm3 of a hydrocarbon C4Hx reacts with an excess of oxygen at 150°C, 1 atm. The products occupy a volume 10cm3 greater than
the reactants at this temperature and pressure. Find x
(2)C4Hx(g) + (4 + x/4)O2(g)  4CO2(g) + x/2H2O(g) Change in volume = (volume of products – volume of reactants)
Avogadro’s rule: 1 volume + (4 + x/4)volumes  4 volumes + x/2 volumes 10cm3 + (4 + x/4)10cm3  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  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 NH 2SO3H(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
(b) 2  0.01  100 = 8%
allow 0.01  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)  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.
Hydration of metal ions In solution metal ions attract water molecules forming complex ions([M(H 2O)6]x+)
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
( [Cu(H2O)6]2+(pale blue) Ions formed by transition metals are usually coloured)
Simple anions Non metals (Cl–, O2–)
Complex anions Where groups around the central metal ion are negative([Fe(CN) 6]4–)
Polyatomic cations Several atoms bonded covalently, the whole structure having a positive charge (NH 4+)
Polyatomic anions Several atoms bonded covalently, the whole structure having a negative charge(SO 42–, NO3–, CHCOO–, MnO4–)
derived from acids by the loss of one or more hydrogen ions
• Cation increases in size
Group (II) Cation
Hlatt kJmol –1
• 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
• Anion increases in size
Magnesium Anion Hlatt kJmol –1
• 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
5bp Trigonal
bipyramid shape
3bp trigonal planar shape (molecule is flat)
4bp tetrahedral shape
6bp octahedron shape
3bp 1lp pyramidal shape
2bp 2lp bent 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
sp2 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 °
sp3 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 1 bond
Triple bond Contains 1σ and 2 bonds
- Ethene, only 3 orbitals hybridised rather than all 4, one 2s electron, two 2p electrons, other 2p electron unchanged(sp 2 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 nonhybridised 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 MgCl 2(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)Name 2 elements in [Mg(H2O)6]2+ which are joined by a covalent bond (2)H and O
(3)Name 2 elements in [Mg(H2O)6]2+ 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 Al 3+, 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 CH 4
(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
• Relative atomic
mass and size
increases
Noble
Relative atomic
Tb/K
Group 4
Relative
Tb/K
gases
mass
Hydrides atomic mass
He
4.0
4
CH4
16.0
81.6
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
107
148
326
Ha kJmol –1
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
3
–1
pm
radius gcm
ppm
kJmol 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
63
774 25900
419
3051
Lilac
Increase
Decrease Colourless
Rb
216
148 1.53
39
688 310
403
2632
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
Decrease Crimson red
Sr
191
113 2.6
550
1064
4120
Increase 769 1380 150
Ba
198
135 3.5
725 1640 430
503
965
3390
Apple green
Group 2 Densitiy, hardness, Mt’s higher than Group 1 because
Group 1
• 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)  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  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  Salt + H2
Metal carbonate + Acid  Salt + CO2 + H2O
Mg(s) + H2SO4(aq)  MgSO4(aq) + H2(g)
Na2CO3(s) + 2HCl(aq)  2NaCl(aq) + CO2(g) + H2O(l)
Na2CO3(s) + 2H+(aq)  2Na+(aq) + CO2(g) + H2O(l)
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g)
Acid/Base neutralisation reaction Ionic equation for neutralisation H+(aq) + OH–(aq)  H2O(l)
Metal Oxide(alkali) + Acid  Salt + H2O
Metal hydroxide(alkali) + Acid  Salt + H2O
Grp 1 M2O(s) + 2HCl(aq)  2MCl(aq) + H2O(l) Grp 1 MOH(aq) + HCl(aq)  MCl(aq) + H2O(l)
Grp 2 MO(s) + 2HCl(aq)  MCl2(aq) + H2O(l)
Grp 2 M(OH)2(aq) + 2HCl(aq)  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)  2Li3N(s)
Heat
Remaining alkali metals form superoxides
4Li(s) + O2(g)  2Li2O(s)white 2Na(s) + O2(g)  Na2O2(s)pale yellow
2K(s) + O2(g)  KO2(s)yellow
O2– lithium oxide
O22– sodium peroxide
O2– potassium superoxide(radical)
Group 2
heat
heat
2Mg(s) + O2(g)  2MgO(s)
(except for Ba) 4Ba(s) + 2O2(g)  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
Rb
Explodes violently with water
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
Group 2 Metal hydroxides less soluble, ppt white suspension, metals don’t float Ca(s) + 2H2O(l)  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)  Mg(OH)2(s) + H2(g) Mg(s) + H2O(g)  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)  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)  2NaCl(s)
Group 2 Mg(s) + Cl2(g)  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
Oxides O2–(s) + H2O(l)  2OH–(aq)
2Li+(aq) + 2OH–(aq) Li2O(s) + H2O(l)  2LiOH(aq)
Peroxides O22–(s) + 2H2O(l)  2OH–(aq) + H2O2(aq)
Na2O2(s) + 2H2O(l)  2NaOH(aq) + H2O2(aq)
Superoxides 2O2–(s) + 2H2O(l)  2OH–(aq) + H2O2(aq) + O2(g)
2KO2(s) + 2H2O(l)  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
2+
–3
Ca
99
1.53 x10
4.66 x10–3
2+
–3
Sr
113
3.37 x10
7.11 x10–5
2+
–2
–7
Decrease
Ba
135
1.50 x10
Increase 9.43 x10
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)  2NaNO2(s) + O2(g) (except Li) 4LiNO3(s)  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)  CaO(s) + CO2(g)
Reaction used in cement manufacture and extraction of iron
- Nitrates all decompose 2Ca(NO3)2(s)  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)(a)2 bottles are clearly labeled ‘sulphate’. The solid in bottle A dissolves easily in
water but none of the solid in bottle B appears to dissolve when added to water
Which of these two bottles contains barium sulphate? (a)B
(b)Bottle C, labeled ‘magnesium carbonate’ When a sample is heated a colourless gas
is produced that turns limewater cloudy. State whether this label is correct and explain
(b)Limewatermilky CO2 MgCO3 decomposes on heating to CO2  label correct
melting temperature / K
(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)H 2SO4 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)CaCl 2 (ii)HCl(g) (2)(i)Ionic (ii)Covalent
1600
(3)(i)2 factors which affect the polarising power of cations? (i)ionic radius, charge
1400
1200
1000
800
600
400
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  2H2O
Decomposition reaction Complex substance breaks down into its more simple parts
2H2O  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
SO42– 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 half reactions:
MnO4– + 8H+ + 5e–  Mn2+ + 4H2O
NO2– + H2O  NO3– + 2e– + 2H+
Balancing equations:
(x2)
2MnO4– + 16H+ + 10e–  2Mn2+ + 8H2O (x5)
5NO2– + 5H2O  5NO3– + 10e– + 10H+
Overall:
2MnO4– + 5NO2– + 6H+  2Mn2+ + 5NO3– + 3H2O
(1)(i) Write ionic half equation for reduction of bromine to bromide ions
(i)Br2 + 2e–  2Br–
2+
3+
(ii) Write ionic half equation for oxidation of Fe ions to Fe ions
(ii)Fe2+  Fe3+ + e–
2+
(iii) Hence write overall ionic equation for reaction of Fe ions with bromine (iii)Br2 + 2Fe2+  2Br– + 2Fe3+
‘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
Brown
Decrease Orange
Increase 193
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)
K/NaCl(aq) K/NaBr(aq)
K/NaI(aq)(colourless)
EN & oxidising power descending group
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)  2NaCl(aq) + I2(aq)
I2(aq)Brown
X
X
X
Halogens undergo disproportionation with alkalis
COLD
HOT
X2 + 2NaOH(aq)  NaXO + NaX + H2O
3X2 + 6NaOH  NaXO3 + 5NaX + 3H2O
X2 + 2OH–
 XO– + X– + H2O
3X2 + 6OH–  XO3– + 5NX– + 3H2O
Ox State of X
0
+1
–1
0
+5
–1
Chlorine test • Swimming pool smell, moist litmus paper blue  red  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)  Br2(aq) + 2Cl–(aq) obtains bromine from sea water, used for leaded petrol, antifreeze
Cl2 + 2e–  2Cl–
2Br–  Br2 + 2e–
Bromine test • Slower than chlorine, moist litmus paper blue  red  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)  H3O+(aq) + X–(aq)
HCl(g) + H2O(l)  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)
Bond enthalpy for C–H bond is average value from breaking all 4 C–H bonds CH4(–416kJmol –1) 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)  [F----- H–F] – (aq)
• HF not normally corrosive but dissolves glass and gives burns, from formation of SiF 62– 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
which would give a spurious ppt
Br – AgNO3(aq)
(conc)NH3 to ppt
Cream ppt AgBr forms
Ppt dissolves leaving colourless solution • Add immiscible organic solvent(hexane)
• Add sodium chlorate(I) solution and shake
I – AgNO3(aq)
(conc)NH3 to ppt
• Br– = brown organic layer, I– = purple organic layer
Yellow ppt AgBr forms Ppt insoluble
Halide salt with (conc)H2SO4 at RT
• Li/NaCl(s) + H2SO4(l)  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)  HBr(g) + NaHSO4(s)
2HBr(g) + H2SO4(l)  Br2(g) + SO2(g) + 2H2O(l)
Ionic equations 2Br–  Br2 + 2e–
H2SO4 + 2H+ + 2e–  SO2 + 2H2O
2Br–+ 2H+ + H2SO4  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)  HI(g) + NaHSO4(s) Firstly 2HI(g) + H2SO4(l)  I2(s) + SO2(g) + 2H2O(l)
Then
6HI(g) + H2SO4(l)  3I2(s) + S(s) + 4H2O(l)
Finally 8HI(g) + H2SO4(l)  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  NaClO3 + 2NaCl(aq) 3ClO–(aq)  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–  Cl–(aq) + H2O(l)
ClO3–(aq) + 6H+(aq) + 6e–  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)  Cl2(g) + 2e–
Steel cathode(–) reduction 2H+(aq) + 2e–  H2(g)
+
–
rd
Na and OH ions remain so 3 product is NaOH(aq) contaminated by NaCl which must be separated
Overall 2NaCl(aq) + 2H2O(l)  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)OCl– + 2H+ + Cl  Cl2 + H2O
(iii)Combine the two ionic half-equations to show the effect of adding acid to bleach
(1)(a)What would happen when (i)KCl +(conc)H2SO4 (i) • Reaction occurs • Misty fumes • HCl doesn’t reduce sulphur in H 2SO4 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–  I2 + 2Cl– (4)Colourless solution turns brown
(5)Equation for reaction between LiCl(s) and (conc)H 2SO4
(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