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WORKBOOK ANSWERSOCR AS/A-level ChemistryFoundations in chemistry Periodic table

This Answers document provides suggestions for some of the possible answers that might be given for the questions asked in the workbook. They are not exhaustive and other answers may be acceptable, but they are intended as a guide to give teachers and students feedback.

Module 2 Foundations in chemistryAtoms and reactions

Atomic structure and isotopes1

a

Particle Relative charge Relative mass

Proton (p) +1 1

Neutron (n) 0 1

Electron (e) –1 Approximately 1/2000

1 mark for each correct column.

© John Older and Mike Smith 2015 Philip Allan for Hodder Education 1

b

1 mark for each correct column.

2

a The relative atomic mass is the weighted mean mass of an atom of the element compared to 1/12th of the mass of an atom of carbon-12 which is taken as exactly 12. (2)

The word ‘weighted’ is advisable but not required.

b The weighted mean mass is (100/140) × 85 + (40/140) × 87 (1)

= 60.7 + 24.9 = 85.6 (1)

3 Let x be the percentage contributed by the 20Ne then 22Ne will contribute (100 – x) (1)

Then:

(1)

2200 – 21x = 2018

21x = 182

x = 91% and contribution of 22Ne is 9% (1)

This is not the only way of doing this problem. Some could argue that the difference in mass of the two isotopes is 2 yet the mixture has an average of just 0.18 more than the lightest component (i.e. the 20Ne). Therefore the 22Ne is contributing 0.18/2 percent of the average which is 9% while the 20Ne must contribute 91%. Exam mark schemes will allow for any correct procedure that is used.

Compounds, formulae and equations4 RbI

Rb2CO3

SrSe The periodic table is needed for this one

Rb2Se

1 mark for each correct response

© John Older and Mike Smith 2015 Philip Allan for Hodder Education

Atom or ion

Number of protons

7 20 19 35 35

Number of electrons

10 18 19 36 36

Number of neutrons

7 20 20 44 46

2

5

a 27 protons, 32 neutrons, 27 electrons (1)

b Co(NO3)2 (1)

Co2(SO4)3 (1)

c CoCO3(s) + 2HCl(aq) → CoCl2(aq) + H2O(l) + CO2(g) (1)

d i Co(s) + 2AgNO3(aq) → Co(NO3)2(aq) + 2Ag(s) (1)

d ii Co(s) + 2Ag+(aq) → Co2+(aq) + 2Ag(s) (1)

For d ii the state symbols are essential for the mark.

e i 60Ni has 2 more neutrons than 58Ni. (1)

e ii 60Co has 27 protons, 33 neutrons, 27 electrons while 60Ni has 28 protons, 32 neutrons, 28 electrons. (1)

Therefore as 60Co breaks down into 60Ni it gains 1 proton and loses 1 neutron and 1 electron. (1)

Exam-style question1

a The relative isotopic mass is the mass of an atom of an isotope of the element compared to the 1/12th of the mass an atom of carbon-12 which is taken as exactly 12. (2)

b Ga has 31 protons, 38 neutrons and 31 electrons (1)

Ga3+ has 31 protons, 38 neutrons and 28 electrons (1)

c The weighted mean mass is (60.2/100) × 69 + (39.8/100) × 71 (1)

= 41.54 + 28.26 = 69.8 (1)

Amount of substance1

a The number of carbon-12 atoms that are needed to obtain a mass of exactly 12 g (1)

The molar mass of a substance is the mass of one mole of a substance. (1) It has units of g mol–1 (1)

b i moles of Al = 10/27 = 0.37 mol (1)

number of Al atoms = 0.37 × 6.02 × 1023 = 2.2 × 1023 (1)

b ii 5/(23 + 35.5) = 0.085(1)

number of chloride ions = 0.085 × 6.02 × 1023 = 5.1 × 1022 (1)


2

a The mixture is separated by filtering and collecting the solid in the filter funnel (1).

The solid is washed by passing some of the organic solvent through the solid in the funnel. (1)

The solid is then left to dry (1)

b The mass of iodine used was 26.38 – 5.00 = 21.38g (1)

The amounts in mol of the tin and iodine used were:

Sn: 5.00/118 .7 = 0.0421 mol and I: 21.38/126.9 = 0.168 mol (1)

Remember that even though iodine occurs naturally as a molecule I2 for these calculations it is the amount in mol of iodine atoms that is required. So you must divide by its atomic mass.

For every 1 mol of Sn, 4 mol of I is required.

Therefore the empirical formula of the tin iodide is SnI4. (1)

3

a Mass of FeSO4 = 24.98 – 20.23 = 4.75g

and mass of H2O in the crystals = 28.91 – 24.98 = 3.93g (1)

The amounts in mol are FeSO4:

4.75/151.9 = 0.0313 mol

and H2O:

3.93/18 = 0.218 mol (1)

Therefore for every 1 mol of FeSO4 there are 0.218/0.0313 = 6.98 or 7 mol of H2O

So x = 7 (1)

And the formula of hydrated iron(II) sulfate is FeSO4.7H2O.

b Heat the crystals to constant mass (1)

4

a Bubbles of gas (effervescence) would be seen (1) and the white solid would dissolve/react to form a colourless solution (1).

Be careful to make sure it is a description that you give when the question asks ‘what would you see’ and not a theoretical comment such as the magnesium carbonate reacts to form carbon dioxide and magnesium chloride solution.

b MgCO3(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) + CO2(g).

1 mark for the correctly balanced equation and 1 mark for correct state symbols.

c At RTP, 1 mol of gas has a volume of 24.0 dm3. This is given on the data sheet.

142 cm3 will be 142/24000 mol = 0.00592 mol (1)


d 1 mol of CO2(g) is formed from 1 mol of magnesium carbonate.

So 0.00592 mol of magnesium carbonate has been used.

0.50g of magnesium carbonate is therefore 0.00592 mol (1)

And the mass of 1 mol of magnesium carbonate = mass of solid/amount in mol = 85 (to two sf) (1)

In practice you would get the marks if you simply gave the arithmetic with little or no explanation. It is however sensible to provide some structure to your answers and this will help you if questions are more complicated.

e The calculation assumes that all the magnesium carbonate has been reacted and it is therefore essential to make sure enough hydrochloric acid is present for this to be true. (1)

The examiners would allow variations of this wording that gave an equivalent explanation.

5

a C8H18(l) + 12½ O2(g) → 8CO2(g) + 9H2O(g)

1 mark for the correctly balanced equation and 1 mark for correct state symbols.

The mark for the equation would also be given for 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g). Don’t forget that C8H18 is a liquid so the state symbol is (l) but water vapour has been specified so for H2O it is (g).

b 2 dm3 of C8H18 would require 25 dm3 of oxygen. (1)

This is directly from the equation using the balancing coefficients.

c The total volume would be 34 dm3. (1)

At 500oC water is a vapour and therefore the volume can be deduced from the coefficients of the equation. 1dm3 C8H18(l) would produce 17 dm3 of products.

d i C8H18(l) + 11½ O2(g) → 6CO2(g) + 2CO(g) + 9H2O(g) (1)

d ii The volume of the products would be the same (1)

6 1 mol of NaOH is 40 g

5.0 g of sodium hydroxide is 5.0/40 = 0.125 mol (1)

concentration = amount in mol/volume = 0.125 ÷ (500/1000) = 0.25 mol dm-3 (1)

7 Amount in moles = concentration × volume in dm3

= 0.1 × (25/1000) = 0.0025 (or 2.5 × 10-3) mol (1)

8 The dilution required is 2/0.1 = 20 (1)

The original solution must be diluted 20 times (20 × 50 = 1000 cm3) hence 950 cm3 of water should be added to the 50 cm3 to get 1000 cm3 of the diluted solution. (1)

9

a CaCO3(s) → CaO(s) + CO2(g) (1)


The question does not specify that state symbols are required for this equation and these would therefore probably not be required for the mark. It is however good practice to include them in all equations.

b 1 mole of a gas has a volume of 24.0 dm3 at 273 K and standard pressure.

The amount in mol of calcium carbonate used is:

mass/mass of 1 mol = 1.40/100.1 = 0.0140 mol (1)

The calculator value is 0.013986013 but it can reasonably be rounded to 3 sf.

0.0140 mol of calcium carbonate produces 0.0140 mol of carbon dioxide which would have a volume of 0.0140 × 24000 cm3 = 336 cm3 at RTP (1)

To complete this calculation you should maintain the calculator volume of 0.013986013 and multiply this by 24000 to give the answer 335.664312 which would be 336 cm3 if rounded to 3 sf.

Using V1/T1 = V2/T2 gives the volume at 60oC (333 K) as (333/298) × 336 = 375 cm3 (1)

c 1 mol of barium carbonate has a higher mass than 1 mol of calcium carbonate. 1.40g of barium carbonate is a smaller amount in mol than calcium carbonate. (1)

This would therefore give a smaller volume of carbon dioxide when decomposed (1)

Just writing smaller on its own would probably not get you any marks. The explanation is essential.

10

a Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) (1)

b At 298C 1 mol of carbon monoxide has a volume of 24000 cm3

Using V1/T1 = V2/T2 would give the volume at 1500oC (1773K) of (1773/298) × 24.0 = 143 dm3 (1 for the calculation and 1 for giving the answer to 3 significant figures)

c 40g of iron is mass/mass of 1 mol = 40/55.8 = 0.717 mol (1)

Calculator value is 0.716845878.

2 mol of iron is formed from 3 mol of carbon monoxide so 0.717 mol of iron is formed from 1.075 mol of carbon dioxide (1)

The volume of 1.075 mol of carbon monoxide at 1500oC is 1.075 × 143 = 154 dm3 (to 3 sf)

This uses the answer to part b. The examiner would allow you an ‘error carried forward’ for this part so that if your answer to part b was incorrect you could still get the mark here.

The volume of carbon monoxide required is therefore 154 × 106 dm3 (1)

As with the previous question the marks would be awarded for the completing the calculations and the explanation is not really required)

d The volume of the carbon dioxide produced will be the same as the volume of carbon monoxide required.


154 × 106 dm3 (1)

Exam-style questions1

a 1s2 2s2 2p6 3s2 3p6 3d8 4s2 or 1s2 2s2 2p6 3s2 3p6 4s2 3d8 (1)

b The mass of the atom depends on the sum of its neutrons and protons (1),

The relative atomic mass includes the contributions of all of the isotopes present. (1)

Co has a heavy isotope which causes its relative atomic mass to exceed that of nickel (1)

In questions asking for an explanation examiners will allow for different wording of an answer as long as it provides a correct explanation.

c The mass of bromine in nickel bromide is 7.44 – 2.00 = 5.44g

The following are the mol of Ni and Br in nickel bromide

Ni 2.00/58.7 = 0.0341 mol and Br 5.44/79.9 = 0.0681 mol (1)

The formula of nickel bromide is therefore NiBr2 (1)

d i 2NaOH(aq) + Ni(NO3)2(aq) → Ni(OH)2(s) + 2NaNO3(aq) (1)

State symbols are not requested in the question and therefore do not have to be included in the equation given.

d ii Ni2+(aq) + 2OH–(aq) → Ni(OH)2(s) (1)

State symbols are required and are always essential in ionic equations.

2

a MCO3(s) → MO(s) + CO2(g) (1)

b V1/T1 = V2/T2

483/353 = V2/298 (V2 is the volume at the room temperature of 25oC) (1)

V2 = 407.7 cm3 (1)

Remember the temperature must be in K.

c 2.51g has a volume of 407.7 cm3

1 mol has a volume of 24 000

The molar mass of MCO3 is (2.51 × 24 000)/407.7

= 147.8 g mol–1 (1)

The relative atomic mass of M is therefore 147.8 – (12 +48) = 87.8 (1)

This corresponds to the element strontium (RAM 87.6)


d The mass of water in the hydrated strontium chloride is 3.70 – 2.20 = 1.50g (1)

The molar mass of SrCl2 = 87.6 + 2(35.5) = 158.6 g mol–1 and H2O = 18 g mol–1

Amounts in mol are therefore SrCl2 = 2.20/158.6 = 0.01387 mol and H2O = 1.50/18 = 0.08333 mol (1)

Mol of H2O for each mol of SrCl2 = 0.08333/0.01387 = 6

Therefore the formula of hydrated strontium chloride is SrCl2.6H2O (1)

Acids1

a A solution of an acid with a large mass of acid dissolved into a small volume of water is concentrated (1)

but which is only partially dissociated into ions is weak (1)

b i Any organic acid such as ethanoic acid, CH3COOH (1)

CH3COOH(aq) ⇌ CH3COO–(aq) + H+(aq) (1)

The equilibrium sign is essential.

b ii CH3COOH(aq) + NaOH(aq) CH3COO–Na+(aq) + H2O (1)

It is best to indicate that sodium ethanoate is ionic by placing the charges on the ions.

2

a An alkali is a soluble base (1)

b NH3(g) + H2O(l) ⇌ NH4+(aq) + OH–(aq) (1)

It is good to include the equilibrium sign but in this context it is not required.

c i There would be bubbling/effervescence (1)

c ii Na2CO3(aq) + 2HNO3(aq) 2NaNO3(aq) + CO2(g) + H2O(l) (1)

c iii H=(aq) + OH– H2O(l) (1)

State symbols are essential.

3

a The marks would be for:

using a safety bulb

filling the pipette so that the bottom of the meniscus sits on the mark on the neck of the pipette


letting the solution run out naturally into a flask and then either touching the surface of the solution with the bottom of the pipette OR touching the side of the flask (3)

b Each reading of the burette will have a possible error of 0.05 cm3. Therefore the total error is 0.1 cm3 (1)

The percentage error is therefore (0.1/23.80) × 100 = 0.42% (1)

c The solution of sodium carbonate will be more concentrated than it should be. (1)

This means more sulfuric acid will be required to reach the end-point of the titration. (1)

Therefore it will appear to be more dilute than it really is. (1)

Other wording of the explanation if correct would also be given credit.

4

a Na2CO3 + 2HCl 2NaCl + CO2 + H2O (1)

b i The amount in mol of sodium carbonate = 0.25 × (25/1000) = 0.00625 mol (n = cV, where V is in dm3) (1)

b ii From the equation, 1 mol of Na2CO3 reacts with 2 mol of HCl

Therefore 0.00625 mol of Na2CO3 reacts with 2 × 0.00625 = 0.0125 mol of HCl (1)

b iii The 0.0125 mol of HCl must be contained in the 18.5 cm3 added.

So the concentration of the hydrochloric acid = 0.0125/0.0185 = 0.676 mol dm-3 (c = n/V) (1)


a

Titration number Rough 1 2 3

Final volume/cm3 23.00 22.30 22.60 44.95

Initial volume/cm3 0.00 0.00 0.00 22.45

Volume used/cm3 23.00 22.30 22.60 22.50

Marks are for correcting:

22.32 to 23.30 (22.32 is to a greater accuracy than is possible with the burette)

0 to 0.00 (0 also has a 2 decimal place accuracy)

22.5 to 22.50 (22.5 must also be given a 2 decimal place accuracy)

b 22.55 cm3 (1)

c Amount of hydrochloric acid used is 0.100 × (22.55/1000) = 0.002255 mol (1)


Equation is NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) so the amount of NaOH must also be 0.002248 (1)

This is in the 25.00 cm3 use so the concentration of NaOH is (1000/25) × 0.002255

= 0.0902 mol dm–3 (1) which is 0.0902 to three significant figures (1)

2

a CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(l) + H2O(g) (1) + state symbols (1)

b The marble would have disappeared and no more bubbling would be observed. (1)

c i

1 mark for labelled axes, 1 mark for appropriate scale, 1 mark for correctly plotted points.

c ii 1 mark for lines correctly drawn

Volume of acid = 5.0 cm3 (allow answers within 0.2 cm3)

d The 5.0 cm3 of hydrochloric acid will contain (5/1000) × 2.00 mol = 0.01 mol (1)

From the equation this will react with half the amount in mol of marble = 0.005 mol (1)

1 mol of CaCO3 has a mass of 100.1 g so 0.005 mol has a mass of 0.50 g (1)

e After 5.0 cm3 of acid has been added the acid is neutralised so the 8 cm3 in the boiling tube contains 3 cm3 of acid (1)

Concentration of the acid is therefore (3/8) × 2.0 = 0.75 mol dm–3 (1)

f The error is (2/27) × 100 = 7.4% (1)

g The carbon dioxide will be (slightly) soluble in the water (1)


Redox1

a There would be bubbling/effervescence (1)

The grey zinc would dissolve/react to form a colourless solution (1)

b Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g) (1)

Zn(s) has an oxidation number of zero, but in ZnSO4 its oxidation number is +2. It has therefore been oxidised. (1)

Hydrogen has an oxidation number of 1 in H2SO4, but as H2 it is zero. It has therefore been reduced. (1)

This question could also answered in terms of zinc losing two electrons to become Zn2+ (1) and the hydrogen ions in sulfuric acid receiving electrons to become H2 (1).

2

a 2FeCl2(aq) + Cl2(g) 2FeCl3(aq) (1)

b The oxidation number of Fe in FeCl2 is +2. The oxidation number of Fe in FeCl3 is +3. The Fe2+ has therefore been oxidised. (1)

Alternatively it is sufficient just to state that in the reaction Fe2+ becomes Fe3+ and has therefore been oxidised.

The chlorine molecule, (Cl2) had an oxidation number of zero but in FeCl3 its oxidation number is –1. Therefore it has been reduced. (1)

3

a A salt is a compound formed when an acid has one or more of its hydrogen ions replaced by either a metal ion or an ammonium ion. (1)

b Cu(s) + 4HNO3(aq) Cu(NO3)2(aq) + 4NO2(g) + 2H2O(l) (1)

c Cu has an oxidation number of 0 but in Cu(NO3)2 its oxidation number is +2 . Hence it has been oxidised. (1)

Nitrogen is HNO3 has an oxidation number of +5 but in NO2 its oxidation number is +4. Hence it has been reduced (1)

Therefore the reaction is a redox reaction and the copper has been oxidised and the NO3

– ion has been reduced. (1)

You might be awarded the mark if you did not put the ‘+’ sign before the positive oxidation numbers but it is certainly not good practice to omit them.

Exam-style questions

1

a 35p, 44n, 35e (1)


b The other isotope has a mass number of 81. (1) Since the relative atomic mass is 79.9 it must be present as approximately 50% (or slightly less) of the naturally occurring sample) (1)

c i Ag+(aq) + Br–(aq) AgBr(s) (1)

c ii There is excess silver nitrate (1)

Therefore the mass of the precipitate depends on the amount in mol of the potassium bromide, This is (25/1000) × 0.1 = 0.0025 mol (1)

So 0.0025 mol of silver bromide is formed.

The molar mass of silver bromide = 107.9 + 79.9 = 187.8 g mol–1 (1)

So the mass of silver bromide formed = 0.47g (1)

d The bromide has been oxidised, oxidation number change: from –1 to 0 (1)

element that has been reduced sulfur oxidation number change: from +6 to +4 (1)

Electrons, bonding and structure

Electron structure1 Li 1s2 2s1

O 1s2 2s2 2px2 2py

1 2pz1

Si 1s2 2s2 2p6 3s2 3px1 3py

1

S2– 1s2 2s2 2p6 3s2 3p6

Ca2+ 1s2 2s2 2p6 3s2 3p6

1 mark for each correct answer.

2

a A 2s orbital is the same shape but larger (1)

A better answer would say that the distance from the nucleus to the area of greatest electron density is greater for the 2s orbital.

b

(1)


a A region around the nucleus of an atom that can hold up to a maximum of two electrons. (1)


b Neon 1s2 2s2 2p6 (1)

Phosphorus 1s2 2s2 2p6 3s2 3px1 3py

1 3pz1 or 1s2 2s2 2p6 3s2 3p3 (1)

c i Neon has two shells of orbitals completely filled but sodium has one electron in a 3s orbital.

The 3s electron is further from the nucleus and is shielded from the nucleus by the two complete shells of electrons. (1)

Because of this the attraction of the nucleus for this electron is less and therefore its first ionisation energy is less (1)

c ii Mg has an electron structure of 1s2 2s2 2p6 3s2 but aluminium is 1s2 2s2 2p6 3s2 3px1.

Because of its shape the 3p electron is slightly further from the nucleus than the 3s electrons (1)

The 3s electrons also shield the 3p electron from the nucleus and it is therefore lost more readily (1)

c iii Silicon has one more 3p electron than aluminium but the shielding from the nucleus is similar. (1)

However, it has one more proton too therefore the attraction between the 3p electrons and the nucleus is stronger so that its first ionisation energy is greater. (1)

It could also be answered as: The atomic radii decrease across a period but the shielding is the same (1) but the number of protons in nucleus increases so there is an increase in the effect of the nucleus for Si (1).

c iv Phosphorus has a complete set of half-filled 3p orbitals, but sulfur has one p-orbital containing two electrons (1). There is a small repulsion between this electron pair which makes sulfur slightly easier to ionise. (1)

Bonding and structure1

a Ionic bonding occurs when electrons are transferred from metal atoms to non-metal atoms. (1)

b Electrostatic attraction between positive and negative (oppositely charged) ions (1)

c The (strong) electrostatic attraction between a shared pair of electrons and the nuclei of both bonded atoms (1)

d Same as a covalent bond “The strong electrostatic attraction between a shared pair of electrons and the nuclei of both bonded atoms” but one atom provides both of the shared pair of electrons. (1)


2

Ionic Covalent Metallic

Melting point (high or low)

High / Low* High / Low* High / Low*

High because strong ionic bonds throughout the giant ionic lattice – all of which have to be broken hence needs high energy

Low because only weak intermolecular bonds between the separate molecules have to be broken

High because strong metallic bonds throughout the giant metallic lattice – all of which have to be broken hence needs high energy

Conductivity

(high or low)

High / Low* High / Low* High / Low*

Poor conductor when solid as the ions(charge carriers) are not mobile

Good conductor when molten or when dissolved in water because the ions are mobile

Poor conductor as there are no mobile charge carriers (either electrons or ions)

Good conductor because has mobile charge carriers (delocalised electrons) even when solid

*delete as appropriate

3

Ammonia (NH3) Calcium oxide (CaO)

Phosphonium ion (PH4+) Carbonyl chloride (COCl2)

4

A has a giant ionic lattice because it dissolves in water but not in a non-polar solvent such as hexane and conducts electricity only when molten as the ions are not mobile when solid

B is a simple covalent molecule because it is insoluble in water but soluble in a non-polar solvent and doesn’t have any mobile charge carriers (electrons or ions)


C has a giant covalent lattice because it is insoluble in all solvents and doesn’t have any mobile charge carriers (electrons or ions)

D has a giant metallic lattice because it is insoluble in all solvents and it has mobile charge carriers (electrons) when solid and when liquid

5 Electron pairs repel ; the shape depends on the number and the type of electron pairs around the central atom ; lone pairs repel more than bonded pairs (3)

6

a ammonia, NH3

diagram to show that there are 4 pairs (3 bonded and 1 lone) around the central N

Shape is pyramidal and bond angle is approx. 107o

b ammonium ion, NH4+

diagram to show that there are 4 pairs (all 4 bonded) around the central N

Shape is tetrahedral and bond angle is approx. 109.5o

c amide ion, NH2–

diagram to show that there are 4 pairs (2 bonded and 2 lone) around the central N

Shape is angular and bond angle is approx. 104o

7 It is the ability of an atom to attract the bonding electrons in a covalent bond. (1)

8 BCl3, diagram to show that there are 3 pairs (all bonded) around the central B

Shape is trigonal planar , which is symmetrical and the dipoles cancel so the molecule is non-polar .

PCl3, diagram to show that there are 4 pairs (3 bonded and 1 lone) around the central P.

Shape is pyramidal , which is not symmetrical and the dipoles do not cancel so the molecule is polar .

9

a

b polar molecules B, D and E non-polar molecules A and C

10


a hydrogen chloride gas: permanent dipole–dipole intermolecular forces because of the difference in electronegativity and induced dipole–induced dipole because of the movement of the electrons (2)

b hydrogen gas: induced dipole–induced dipole because of the movement of the electrons only because the bond is non-polar (2)

c ammonia gas: permanent dipole–dipole intermolecular forces because of the difference in electronegativity which are sufficiently strong to be classified as hydrogen bonds and induced dipole–induced dipole because of the movement of the electrons (2)

11 Water has a higher than expected boiling point (or melting point) because the H-bonds have to be broken when boiling (or melting) H2O .

Ice is less dense than water because the H2O molecules in ice are held in a lattice and the molecules are further apart in ice than they are in water . (4)

Exam-style questions

1

a

(2)

The metallic bond is the electrostatic attraction between the positive ions in the lattice and the mobile electrons (1)

b i 2Ca(s) + O2(g) 2CaO(s) state symbols = (2)

b ii

Oxygen Calcium oxide

b iii Covalent bonding which is the electrostatic attraction between the shared pair of electron and the nuclei of the two oxygen atoms (1)


c Calcium is a good conductor in both the solid and the liquid state because in both states the outer electrons are delocalised throughout the lattice and act as mobile charge carriers

Calcium oxide only conducts when liquid as in the solid state the ions are held in fixed positions and cannot function as mobile charge carriers. When liquid the ions are free to move.

2

a

Table 1 Hydrides of group 14

Compound Boiling point/K Number of electrons

CH4 112 6 + 4 =10

SiH4 161 14 + 4 =18

GeH4 178 32 + 4 =36

SnH4 221 50 + 4 = 54

Table 2 Hydrides of group 16

Compound Boiling point /K Number of electrons

H2O 373 8 + 2 =10

H2S 213 16+ 2 =18

H2Se 231 34 + 2 =36

H2Te 270 52 + 2= 54

b i More electrons therefore more induced dipole–dipole interactions hence more energy required to break the intermolecular force (2)

b ii

Two marks for diagram


Electronegativity decreases down the group so the dipoles decrease down the group

Water has the largest dipole and can form hydrogen bonds which require more energy to break

c i

c ii Both have same number of electrons.

H2S is not symmetrical and SiH4 is symmetrical.

H2S is polar and SiH4 is (almost) non-polar.

H2S has both permanent dipole–permanent dipole and induced dipole–induced dipole intermolecular forces but SiH4 only has induced dipole–induced dipole intermolecular forces.


Module 3 Periodic tableThe periodic table

Periodicity1

a In rows and columns in order of their increasing atomic number (1)

b Oxygen has a structure 1s2 2s2 2px2 2py

1 2pz1

Sulfur has a structure 1s2 2s2 2p6 3s2 3px2 3py

1 3pz1 (1)

Because the outer structure has the same pattern they are placed in the same column (1)

c i The energy required to remove one electron from the ground state of each atom in a mole (1) of gaseous oxygen atoms to form a mole of gaseous O+ ions. (1)

c ii The shielding from the nucleus is similar for both elements and the atomic radii decreases (1)

Fluorine has one more electron than oxygen but also one more proton and therefore the attraction between the electrons and the nucleus is greater in fluorine (1)

2

a i Si2+(g) Si3+(g) + e– (1)

State symbols must be included.

a ii 1s2 2s2 2p6 3s1 (1)

a iii Once one electron has been removed the size of the ion will decrease (1). Also Si+ has one more proton than electron and as a result of both effects the attraction between the protons in the nucleus of the atom and the 3s electron will be therefore be greater (1)

a iv There is a large increase in the ionisation energy when the ionisation Si4+(g) Si5+(g) + e– occurs. (1) This suggests the 5th electron is being removed from a shell closer to the nucleus. (1)

b Solid silicon consists of a giant lattice of atoms (1) covalently bonded together (1)


3

(1)

OR Magnesium is able to conduct electricity as the electrons are mobile between the magnesium ions (1)

a As shown below a diagram should show carbon in six-membered rings (1) creating a network (1)

Only a few interlinking rings need be shown.

Graphene is able to conduct electricity because mobile electrons from each carbon atom are present on the surface the sheet of carbon atoms (1)

b Each carbon atom in graphene and in graphite is bonded to three other carbon atoms.

Both form hexagonal planes.

Both have delocalised electrons/

Any two answers.


a The answer can be expressed in your own words but six key points must be clearly identified.

Lithium has a giant metallic lattice/metallic bonding (1) but with only 1 delocalised electron for each atom its melting point is not very high. (1) (for 2nd marking point also allow strong metallic bonds throughout the lattice)

Diamond has a giant covalent lattice (1) consisting of carbon atoms covalently bonded to four other carbon atoms (1). To melt diamond these bonds must be broken this takes a lot of energy and hence its melting point is high. (1)


Neon consists of atoms with very little attraction (id–id only) for other neon atoms and therefore its melting point is very low (1)

b

Element Mg Al Si P

Type of structure Giant metallic Giant metallic Giant covalent

Simple molecular

2 marks completely correct, 1 mark if only one error.

c

Marks for Al lower than Mg (1), Si and P higher than Al (1), S lower than P (1).

The marks are for the trend not the precise placing of the crosses.

Group 21

a 20Ca 1s2 2s2 2p6 3s2 3p6 4s2 (1)

b 38Sr2+ 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 (1)

2

a The energy required to remove 1 electron from each atom in 1 mole of gaseous atoms resulting in the formation of 1 mole of gaseous 1+ ions. (2)

b i Ba(g) Ba+(g) + e– and state symbols (2)

b ii Be+(g) Be2+(g) + e– and state symbols (2)


3

a Ionisation energy decreases down the group because down the group atomic radius increase

Shielding increases and the effect of the nuclear charge decreases (4)

b i Second ionisation energy decreases down the group (1)

b ii 1+ ion is smaller than the atom nuclear charge is the same so the effect of the nuclear charge increases hence it is more difficult to remove the second electron (3)

b iii The 3rd electron is removed from an inner shell much closer to the nucleus with less shielding and the nuclear charge is the same so the effect of the nuclear charge increases greatly (4)

4 React by losing electrons as go down the group 1st & 2nd ionisation energies decrease down because the radii increases, shielding increase, effect of nuclear charge decreases (4)

5

a 2Mg(s) + O2(g) 2MgO(s) (1)

b Sr(s) + 2H2O(l) Sr(OH)2(aq) + H2(g) (1)

c Ba(s) + 2HCl(aq) BaCl2(aq) + H2(g) (1)

d Ba(s) + 2CH3COOH(aq) Ba(OOCCH3)2(aq) + H2(g) (1)

e) Ba(s) + 2CH3CH(OH)COOH(aq) Ba(OOCCH(OH)CH3)2(aq) + H2(g) (1)

6

a Sr(s) + 2H2O(l) Sr2+(aq) + 2OH–(aq) + H2(g) (1)

b Ba(s) + 2H+(aq) Ba2+(aq) + H2(g) (1)

c Ba(s) + 2H+(aq) Ba2+(aq) + H2(g) (1)

d Ba(s) + 2H+(aq) Ba2+(aq) + H2(g) (1)

7

a Equation BaO(s) + H2O(l) Ba(OH)2(aq) (1)

Ionic equation BaO(s) + H2O(l) Ba2+(aq) + 2OH–(aq) (2)

b Equation Mg(OH)2(s) + H2SO4(l) MgSO4(aq) + 2 H2O(l) (1)

Ionic equation Mg(OH)2(s) + 2H+(aq) Mg2+(aq) + 2H2O(l) (2)

c Equation 3CaCO3(s) + 2H3PO4(aq) Ca3(PO4)2 + 3H2O(l) + 3CO2(g) (2)

Ionic equation 3CO32-(aq) + 6H+(aq) 3H2O(l) + 3CO2(g) (1)


The halogens (group 17)1

a

The molecule is made of two atoms (2)

b Bonding is covalent

Non-polar because each Cl has the same electronegativity (2)

2

a

Halogen Fluorine Chlorine Bromine Iodine

Boiling point/oC –188 –35 59 184

Physical state at 25oC

Gas Gas Liquid Solid

b Halogens are diatomic non-polar moleculesintermolecular forces are induced dipole-induced dipole only number of electrons increases down the group hence so does number of induced dipole-induced dipoles therefore boiling point increases down group (3)

3

a 35Br 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 (1)

b 17Cl– 1s2 2s2 2p6 3s2 3p6 (1)

4 Reactivity decreases down the group because they react by gaining electrons Radii increases down the group shielding increases therefore effect of nuclear charge decreases (5)


5

a

Fluoride (F−) Chloride (Cl−) Bromide (Br−) Iodide (I−)

Fluoride (F2

Chlorine (Cl2) x

Bromine (Br2) x x

Iodine (I2) x x x

1 mark for all 4 ticks and 1 mark for the crosses (2)

b Cl2 + 2Br– 2Cl– + Br2 (1)

c i The colourless solution turns brown due to a precipitate of iodine. (1)

c ii The iodine dissolves in the organic upper layer which is purple/violet. (1)

6

a Simultaneous oxidation and reduction of the same element (2)

b i Cl2(g) + H2O(l) ⇌ HCl(aq) + HClO(aq)

1 mark for the equation and one for the state symbols (2)

b ii Cl2(g) + H2O(l) ⇌ 2H+(aq) + Cl–(aq) + ClO–(aq) (2)

b iii Benefits — kills bacteria and makes water safe to drink/potable

Risks — chlorine could react with any dissolved organic material and might form toxic chlorinated hydrocarbons (2)

7 Equation Cl2 + 2NaOH NaCl + NaClO + H2O

Conditions NaOH must be dilute and must be cold (2)

8

a H2S(g) + Cl2(aq) S(s) + 2HCl(aq)

1 mark for the equation and 1 for the state symbol (2)

b S changes from oxidation state -2 to 0 so it has been oxidised therefore the Cl2 molecules must be the oxidising agent (or Cl changes from oxidation state 0 to –1 and therefore is the oxidising agent) (2)


9

Reaction Ionic equation Observation

Initial Dil. NH3 Conc. NH3

KCl(aq) + AgNO3(aq)

Cl–(aq) + Ag+(aq) AgCl(s) White solid Colourless solution

Colourless solution

CaCl2(aq) + AgNO3(aq)

Cl–(aq) + Ag+(aq) AgCl(s) White solid Colourless solution

Colourless solution

NaBr(aq) + AgNO3(aq)

Br–(aq) + Ag+(aq) AgBr(s) Cream solid Cream solid

Colourless solution

BaI2(aq) + AgNO3(aq)

I–(aq) + Ag+(aq) AgI(s) Yellow solid Yellow solid Yellow solid

4 marks for the equations 2 marks for each correct set of observations.

Qualitative analysis10

a Reagent Dilute acid such as HCl(aq)/H+(aq)

Equation 2H+(aq) + CO32-(aq) H2O(l) + CO2(g)

Observation(s) bubbles/ effervescence (3)

b Reagent aqueous silver nitrate, AgNO3(aq)/Ag+(aq)

Equation Br–(aq) + Ag+(aq) AgBr(s)

Observation(s) a cream coloured precipitate/solid (3)

c Reagent aqueous barium nitrate, Ba(NO3)2(aq) or aqueous barium chloride, BaCl2(aq)/Ba2+(aq)

Equation Ba2+(aq) + SO42-(aq) BaSO4(s)

Observation(s) a white coloured precipitate/solid (3)

d The sequence should be test for carbonate then sulfate and finally halide because BaCO3 and Ag2SO4 are both insoluble. (2)



a i Sr(s) + 2H2O(l) Sr(OH)2(aq) + H2(g) (2)

a ii Ca(s) + 2H2O(l) Ca2+(aq) + 2OH–(aq) + H2(g) (1)

a iii Reactivity increases down the group because it is easier to lose electrons as they are further from the nucleus with more shielding which decrease the effect of the nuclear charge (3)

b i 2Sr + O2 2SrO (1)

b ii Sr3N2 (1)

c i SrCO3(aq) + 2HCl(aq) SrCl2(aq) + H2O(l) + CO2(g)

A B (2)

c ii CO32–(aq) + 2H+(aq) H2O(l) + CO2(g)

B (1)

c iii SrSO4(s) (1)

c iv Silver chloride or AgCl(s)

White solid/precipitate (2)

2

a i Cream precipitate/solid (1)

a ii Br–(aq) + Ag+(aq) AgBr(s) (1)

b i Concentrated ammonia (1)

b ii The amount/density of precipitate would decrease because the cream AgBr(s) would dissolve leaving behind a yellow precipitate of AgI(s) (2)

b iii The bromine reacted with the aqueous iodide ions to produce a brown precipitate of iodine solid. When shaken with hexane the iodine dissolved in the upper organic layer to produce a purple/violet solution. (3)

b iv Br2(aq) + 2I–(aq) 2Br–(aq) + I2(s) (1)

3

a i 195/24 000 = 8.125 × 10–3 (1)

a ii 8.125 × 10–3 (1)

a iii Molar mass = (0.71/195)/8.125 × 10–3 = 87.4 g mol–1 (2)

a iv CO3 has mass 12 + 16 + 16 + 16 = 60 therefore mass of group 2 metal = 87.4 – 60 = 27.4 which is closest to Mg. (2)

b i/ii


Trial 1 2 3

Final burette reading/cm3 0.00 0.00 0.00 24.20

Initial burette reading/cm3 25 24.90 24.20 48.50

Titre/cm3 25 24.90 24.20 24.30

Titres used to calculate mean ()

Mean titre/cm3 24.25

(2)

c i MCO3 + 2HCl MCl2 + H2O + CO2 (1)

c ii 0.012125 (1)

c iii 0.012125/2 = 6.06250 × 10–3 (1)

c iv 6.06250 × 10–3 × 10 = 6.06250 × 10–2 (1)

c v 5.10/6.06250 × 10–2 = 84.12 (1)

c vi 84.12 – 60 = 24.12 therefore it is Mg (1)

d The first method appears to be the less accurate — possible sources of error:

Escape of gas while replacing the bung

CO2 is soluble in water (2)

4

a i Cl2(g) + H2O(l) ⇌ HClO + HCl (1)

a ii Chlorine is toxic or risk of formation of chlorinated hydrocarbons (1)

b i 2Li(s) + Cl2(g) 2LiCl(s) (1)

b ii Ag+(aq) + Cl–(aq) AgCl(s) (1)

b iii White precipitate which would dissolve in the dilute ammonia (2)

c Ag Cl = 107.9 + 35.5 = 143.4

Moles of AgCl = 7.17/143.4 = 0.05

Molar mass of group 1 chloride = 6.05/0.05 = 121

Mass of group 1 metal = 121 – 35.5 = 85.5 = Rb (4)


d

1 mark each for the three labels and the diagram.

e 2KClO3(s) 2KCl(s) + 3O2(g)

Molar mass of KClO3 is 122.6 g mol–1 (1)

Moles of KClO3 = 0.005 mol (1)

Moles of oxygen 0.0075 (1)

Volume of oxygen = 180 cm3 (1)


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