Author: Andrew Massawe

1.	(a)	     H	   C
  11.1 / 1	88.9 / 12  (1)
= 11.1	= 7.4
    1.5	    1  (1)
Empirical formula  C_2H_3   (1)	3
 
(b)	HI has more electrons  (1)
has greater induced–dipole–induced dipole / vdW forces  (1)	2
 
(c)	(i)	pyramidal  
	Need to show evidence of three dimensional or state it is pyramidal with two dimensional diagram  (1)
3 bond pairs and 1 lone pair to get as far apart as possible  (1)	2
 
(ii)	tetrahedral  
	Need to show evidence of three dimensional or state it is tetrahedral with two dimensional diagram  (1)
4 bond pairs around aluminium as far apart as possible  (1)	2
 
(d)	Amount of phosphine	= 8.0/24000  (1)
 = 3.33 × 10^–4 mol
Number of molecules of phosphine   = 6.0 × 10^23 × 3.33 × 10^–4  (1)	2
 = 2.0 × 10^20
[11]

2.	(a)	(i)	number of protons (in the nucleus)/ proton number (1)
not ‘number of electrons’ or ‘number of protons in an element’ (1)
(ii)	Electronic configuration differs from previous element by an electron in a d (sub) shell or orbital / d-shell is filling / d electron is last electron (1)
Allow outer electron is d / highest energy electron is d	1
(iii)	Forms at least one ion/compound with partially full / incomplete d sub shell (1)
1

(b)	1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^5  or  3d^54s^2 (1)	1
(c)	(i)	The heat/energy/enthalpy change needed to remove one mole of electrons (1)
from (1 mole) of gaseous (chlorine) atoms (1)
Correct equation i.e. Cl(g) → Cl^+(g) + e^– can score second mark.	2
(ii)	Increasing slopes (1)
Jump after 7 (1)
Jump after 15 (1)
Ignore small jumps in the correct places. The points do not need to be joined.
3
 
(d)	There are several way of doing this calculation; the following is one way. All other valid ways score full marks
43.7/55 = 0.795       56.3/35.5  =  1.59 (1)
0.795/0.795  =  1       1.59/0.795  =  2 (1)
MnCl_2 (1)  This is a stand alone mark
or
MnCl_2 and some correct working (3)
Note:
If a candidate gets a formula MnCl_x where × is between 2 and 7 because they have made a chemical error, they can score a maximum of 1 mark. If the error is mathematical they can score a maximum of 2 marks
3
[12]

3.	(a)	(i)	fizzing/ effervescence
metal disappears /gets smaller
floats/ moves around on surface
melts/ turns into ball
any 2
do not allow ‘dissolves’	2
 
(ii)	2Na + 2H_2O → 2NaOH + H_2
species (1) balance (1)	2
 
(b)	amount Na = 3.0123 = 0.13 mol (1)
amount H_2 = 0.065 mol (1)
vol H_2 = 0.065 × 24 dm^3 = 1. 6 (dm^3) (allow 1.56, 1.57 or 1.565) (1)
answers consequential on equation in (a)(i)
If units quoted and are wrong final mark lost	3
[7]

4.	(a)		Na	Cl 33.3 / 35.5	O
	21.6/23		45.1/16	(1)
	= 0.939	= 0.938	= 2.82
		÷ by smallest		(1)
	1	1	3
	NaClO_3
Could argue from formula and calculate back to shown percentages 
for full marks.	2
 
(b)	(i)	3OCl^− → 2Cl^− + ClO_3
species (1) balance (1)
Fully balanced molecular equation score / mark only	2
ignore spectator sodium ions in ionic equation if on both sides
 
(ii)	Identification of oxidation states (1) +1 +5 −1
Identification of an oxidation reaction (1) +1 to +5
Identification of a reduction reaction (1) +1 to −1
Disproportionation because (Cl in) OCI^− both oxidised or reduced (1)
Final mark can be awarded for a simple definition of
disproportionation related to chlorine.
n. b. reference to a single atom of chlorine not acceptable.	4
 
(c)	(i)	Cl_2 + 2e^− → 2Cl^− (1) (*)	1
(ii)	2Br^− → Br_2 + 2e^− (1) (*)	1
	(*) or multiples / negative charge on e not required
[10]

5.	(a)	(i)	 = 0.00258 / 2.58 × 10^–3 / 0.0026 / 0.002577 (1)	1
 
(ii)	0.00258 / same number of moles as calculated in (i) (1)
0.00258 ×  (1) = 0.110 (mol dm^–3) (1) units not required
consequential on (i) possible answers 0. 11, 0.110, 0.1111	3
 
(b)	= 8% (1)
	but allow  × 100 = 4%	1

(c)	W	Weighing	must be evidence of two weighings at some point in 
		the process (1)
P	Preparation	Rinsing out one piece of relevant apparatus
	correctly (1) (*)
D	Dissolve	Dissolve in water in beaker / volumetric flask (1) (*)
R	Rinse	Rinse beaker and add washing to volumetric flask /
	rinse funnel (if solid straight to volumetric flask) (1) (*)
V	Volumetric	Volumetric / standard / graduated flask (1)
flask	DO NOT AWARD IF CANDIDATE USES VOLUMETRIC
	FLASK TO MEASURE OUT 250 cm^3 (*)
 
M	250 cm^3	Making up to mark / exactly 250 cm^3 of solution (1) (*)
S	Shake	Shake / invert / mix final solution (1) n.b. this is at end (*)
C	concentration =  (1)
H	Safety
(solution of) acid is corrosive and appropriate safety precaution
e.g. wear eye protection and/or  gloves (1)
(*) Max 5 marks
[13]

6.	(a)	(i)	2Ca + O_2 → 2CaO (1)	1
(ii)	Na_2O + H_2O → 2NaOH (1)	1
(iii)	Na_2O + 2HCl → 2NaCl + H_2O
Species (1) Balancing (1)	2
 
(b)	(Thermal stability) increases (1) (with some
attempt at a reason)
	(cat)ion size increases / (cat)ion charge density 
decreases (1)
polarises or distorts carbonate ion / anion /
electron cloud less. (1)	3
[7]

7.	(a)	(i)	moles of KNO_3 = 10.1/101 = 0.100 (1)	1
Allow 0.1/0.10

(ii)	moles of KOH = 0.100 (1)
or answer from (i)– could be shown in calculation below.
volume = 0.1 × 1000/2 = 50.0(cm^3) (1)	2
Consequential on (i); allow 50
 
(iii)	moles of O_2 = 0.1/2 = 0.0500 (1) i.e. divide by 2
vol O_2 = 0.05 × 24 = 1.2 (dm^3) (1) i.e. × by 24	2
	consequential on (ii) or (i)
 if use wrong unit eg mol dm^–3 max (1)
 
(b)	(i)	Percentage of oxygen = 29.1 % (1) stand alone
	K	O
	70.9/39	29.1/16 (1)
		i.e. divide by A_r
	1.82	1.82
	1	1
KO (1)	3
	If assume KO and prove it (Max 2)
 
(ii)	M_r (= 22/0.2) = 110 (1)
(M_r of KO = 55 so) molecular formula = K_2O_2) (1)	2
[10]

8.	(a)	(i)	•	Energy / enthalpy change per mole (1)
•	required to remove an electron (1)
from / mole of gaseous atoms (1)	3
(ii)	•	The nuclear charge on K is greater than on Na (1)
•	the outer electron is further from the nucleus (1)
but there is more shielding around K than Na (1)	3
 
(b)	(i)	4.56 / 71 (1) = 0.0642 (1)mol	2
(ii)	Answer from (i) − 2 (1) = 0.0321 mol	1
(iii)	Answer from (ii) × 24 (1) 0.771 dm^3	1
(iv)	Answer from (iii) × 3/2 (1) 1.16 dm^3	1
[11]

9.	(a)	(i)	NO
(ii)	CrO	2
(b)	Ag^+, CrO or names	1
(c)	2Ag^+(aq)+ CrO(aq) → Ag_2CrO_4(s)
Formulae and balancing (1)
State symbols (1)	2
[5]

10.	(a)	% oxygen 100 – (31.84 + 28.98) = 39.18 (1)
	K	Cl	O
	31.84/39	28.98/35.5	39.18/16	Divide by A_r (1)
	0.8164	0.8163	2.448	Divide by smallest
	
	Alternative multiply by A_r + SUM (1)    Calc % (1)    39.18 for 0 (1)	3
 
(b)	Ratio of moles or mass (1)
Moles of A or relative mass of A (1)
× 24 or volume of O_2 (1) (not stand alone)
E.g.
2 moles of A gives 3 mols of oxygen (1) 
1.0g of A 1.00/122.5 moles of A
therefore 1.00 × 3/122.5 × 2 moles of oxygen
volume of oxygen = 1.00 × 3 × 24 / 122.5 × 2
                             = 0.294 dm^3
 
OR
1.00g of A gives 0.3918 g of oxygen (1)
0.3918 g of oxygen = 0.03918/32 moles of oxygen = 0.0122 moles (1)
0.0122 moles of oxygen = 0.0122 × 24 dm^3 of O² = 0.293 dm^3 (1)
	2-4 significant figure in answer allowed	3
[6]

11.	(a)	(i)	4gS = 1/8 mol / 0.125 mol	1
(ii)	1/8mol S → 1/8 mol SO_2 (stated or implied) (1)
Volume = 24/8 = 3 dm^3 / 3.0 dm^3/ 3.00 dm^3 / 3000 cm^3 (1)
–1 for incorrect/missing units	2
 
(b)	 → 	2
[5]

12.	(a)	Moles = 2.20 / 101 (1)
Concentration = 0.0218/ 0.05 = 0.436 (mol dm^–3) but must be in mol dm^–3 (1)	2
 
(b)	(i)	Mass = (0.0218 ×) 85 (1)
=1.85 (g) (1)
Answer could be in kg but units must be quoted	2
 
(ii)	Vol = (24 ×) 0.01089 (1)
= 0.261 (dm³) (1)
Can quote volume in any feasible unit
In (a) and ( b) intermediate values (if stated) and final answers to
calculations must be given to 2 – 4 s.f.
Penalise SF once	2
 
(c)	Stability increases followed by an attempt at an explanation (1)
	because charge density of cation/ metal ion/ group 1 ion decreases 
OR size increases but charge stays the same (1)
	so polarises/distorts/ weakens bonds within nitrate/ negative ion/ anion less (1)	3
 
(d)	       (1)
       (1)
(2:1:3)   ie K_2CO_3 (1)
If ÷ 19 in first step ie K_4CO_3   2 (out of 3)	3
(working must be shown)
[12]

13.	Number of molecules
= 12/24 (1) × 6×10^23
= 3×10^23 (1)	2

14.	(a)	H_2SO_4(aq) + CuCO_3(s) → CuSO_4(aq) + CO_2(g) + H_2O(l)
Reactants (1)
Products (1)
No/wrong state symbols 1max
If H_2CO_3 product 1 max	2
H_2SO_4(I) 1 max
 
(b)	(i)	Measuring cylinder / pipette / burette
Volumetric flask (0)	1
(ii)	2.5×10^–2 / 0.025	1
 
(c)	(i)	2.5 × 10^–2 × 123.5 (1) × 1.1 = 3.4 / 3.40 / 3.396 g (1)
ALLOW TE from (b)(ii)	2
(ii)	To prevent the reaction mixture from frothing out of the beaker	1
(iii)	Filter (to remove unreacted copper(II) carbonate)	1
 
(d)	(i)	249.5	1
(ii)	actual mass/ mol × 100 / = 3.98 / 249.5 × 100
method
theoretical mass / mol	2.5 × 10^–2
	OR method using masses =    3.98       =  3.98   etc
(1)
	0.025×249.5     6.2375	2
	answer 63.8 / 64 % (1)
ALLOW TE from (a) / (c) (I) / (d)(i)
 
(e)	Toxic/irritant/enzyme inhibitors	1
[12]

15.	(a)	(i)	ALLOW 3 or 4 sig figs – penalise once only
MUST be some working
	moles P = 93/31 = 3.0 (1)
moles PCl_3 also = 3.0 (1)
mass PCl_3 = 137.5 × 3.0 = 412.5 / 413 (g) (1)
OR alternative route
Max 2 if wrong units	3
 
(ii)	moles Cl_2 = 3/2 × 3 = 4.5 (1)
volume of C1_2 = 4.5 × 24 = 108 (dm^3) (1) - consequential on 1^st mark	2
(iii)	Cl_2 with attempt at reason (1)
because gains electrons / ox. no. becomes more negative / oxidation
number decreases / 0 → –1
OR
P loses electrons / oxidation number increases / 0 → +3 (1)	2
 
(b)	(i)	Outer shell of P in a molecule (1)
Cl lone pairs / six more electrons around each Cl (1)
Lone pair must be in the same space.	2
(ii)	Trigonal pyramidal diag. (1)
Must be some attempt to show 3-D. A poor diagram can be rescued by a correct name.
	100 – 108° (1) NOT consequential	2
(c)	Tetrahedral	1
[12]

16.	ALLOW multiples in (i), (ii) and (iii) and (c)
	(a)	(i)	Ca → Ca ^2+ + 2e^– / Ca – 2e^– → Ca ^2+	1
(ii)	2H_2O + 2e^– → H_2 + 2OH^–	1
(iii)	Ca + 2H_2O → Ca^2+ + H_2 + 2OH^–
Or        → Ca(OH)_2 + H_2
NOT consequential on (i) and (ii)	1
 
(iv)	(Vigorous) bubbles/fizzes/effervesces (1)
(white) ppt./suspension / cloudy / milky (1)
IGNORE “gas evolved”/ “hydrogen produced”/ names
IGNORE calcium dissolves/gets smaller
If flame described 1 max	2

(b)	Increases	1
(c)	2Na + Cl_2 → 2NaCl
species (1)
balancing (1) – dependent on 1^st mark	2
[8]

17.	2Mg(NO_3)_2(s) → 2MgO(s) + 4NO_2(g) + O_2(g)
ALLOW whole equation divided by 2 (i. e.½ O_2(g)) or other correct multiples	1
[10]

18.	(a)	Reduction is electron gain (1) IGNORE any reference to oxygen and hydrogen
ALLOW decrease in oxidation number/state	1
(b)	In (i) and (ii) allow multiples
(i)	Fe^3+ + e^– → Fe^2+	^1
(ii)	2I^– → I_2 + 2e^–	Do NOT allow I for ½I_2 	_1
(iii)	2Fe^3+ + 2I^– → 2Fe^2+ + I_2
	Stand alone
This equation must be correct for the mark and not just a combination
of (i) and (ii)	1
(c)	(i)	Cl is (+) 1 in NaClO (1),
(+) 5 in NaClO_3 (1),
and –1 in NaCl (1)
ACCEPT as Roman numerals	3
(ii)	Chlorine is both oxidised and reduced
OR
The chlorine’s oxidation number goes from +1 to +5 and –1.
consequential on oxidation numbers in (i) provided that chlorine has
gone both up and down	1
 
(d)	(i)	Amount of N_2 = 54 dm^3/20 dm^3 mol^–1 = 2.7 (mol)	1
(ii)	Amount of NaN_3 = 2.7 × 2/3 = 1.8 mol	1
(iii)	117 g (2)
Salvage marks: 
M_r of NaN_3 = 65 g mol^–1 (1)
Mark consequentially (i–iii)
2–4 SF
117 with no unit (1)
117 with wrong unit (1)	2
[12]

19.	Only penalise wrong or missing units once in parts (a) & (b).
(a)	24 dm^3 OR 24 000 cm^3	1
(b)	48 dm^3 OR 48 000 cm^3	1
[2]

20.	(a)	(ionic) precipitation	1
(b)	(i)	(2)NH_4^+ and Cr_2O_7^2-	2
(ii)	Cr_2O_7^2-((aq)) + 2NH_4^+((aq)) → (NH_4)_2Cr_2O_7((s))
State symbols not required	1
(iii)	The orange colour would move towards the anode / + / left	1
 
(c)	(i)	18 × 2 + 52 × 2 + 16 × 7 = 252 (g / g mol^-1)
Penalise incorrect units eg 252 g^-1 in (i) and (ii) only once.	1
(ii)	0.1 mol has a mass of 25.2 (g)
ALLOW TE	1
(iii)	100 cm^3 / 0.1 dm^3 must have units	1
(iv)	Filter (1)
Wash with (small quantity) / (cold) water (1)
Dry between filter papers / in a warm oven (< 40° C) / in a dessicator (1)	3
(v)	Some remains in solution	)
Some lost on washing	)  Any two
Transfer loss eg on glassware, filter paper	)	2
[13]

21.	(a)	(i)	% of oxygen = 45.1% (1)
54.9/39 = 1.41  and  45.1/16 = 2.82 (1)
1.41/1.41 = 1   and   2.82/1.41 = 2   (hence KO_2) / 1.41 : 2.82 ≡ 1 : (1)
2
MUST have some working	3
Correct inductive reasoning (3)
(ii)	-0.5 / – ½ / -.5	1

(b)	KNO_3 because K^+ / potassium ion has larger radius / is larger - ion essential (1)
but same charge - stated or K^+ and Li^+ given
[lower charge density scores 1 out of the first two marks]
“Charge density” on its own (1) UNLESS term is explained (2)
polarises/distorts nitrate/negative ion/anion less (1)
OR weakens bonds in nitrate less
NOT weakens ionic bond
If LiNO_3 more stable (0)	3
[4]

22.	(a)	(i)	moles silicon = 10/28 = 0.357 (1)
moles SiCl_4 = 0.357 (1)
mass = 0.357 × 170 = 60.7 / 60.69 (g) ALLOW 2- 4 SF (1)
OR by mass ratio
Units not required but if given must be correct.
Correct answer with some recognisable working (3)
Correct answer with no working (1)	3
 
(ii)	moles chlorine = 2 × 0.357	moles Si × 2 (1)	(1)
∴ vol = 0.714 × 24.0 = 17.1 (dm^3)	moles Cl_2 × 24 (1)	(1)
ALLOW TE from (i)
ALLOW 2 – 4 SF
Units not required, but if given must be correct
Correct answer with some recognisable working (2)
Correct answer with no working (1)
Penalise SF once only across (i) and (ii)		2
Penalise units once across (i) and (ii)

ratio

10/28

10/28
= 0.357

10/28
= 0.36

10/28
= 0.4
loses SF mark

mass | 60.69 | 60.71 | 61.2 | 68
vol | 17.14 | 17.14 | 17.3 | 19.2

(b)	
 (1)
	Wedges not required    e.g.
Atoms can be represented by circles etc provided there are 4 of one type and
1 of another
	tetrahedral (1)
	Any angle in range 109 – 109.5 ° (1)
degree symbol can be shown on diagram (1)
	4 (bond) pairs of electrons / 4 bonding pairs (1)
NOT bonds
NOT atoms
NOT groups of electrons
	Repel to position of minimum repulsion / potential energy NOT “Equal
repulsion” (1)
OR Repel to position of maximum separation	5
4^th mark cannot be awarded if atoms referred to
 
(c)	(i)	Si and Cl have different electronegativities / Cl attracts the bonding	1
electrons very / more strongly / Si less electronegative than Cl / Cl
very electronegative
(ii)	symmetrical molecule / chlorines equally spaced (1)
bond polarities / dipoles / vectors cancel
OR
Centres of positive and negative charge coincide / vectors cancel. (1)	2
[13]

23.	(a)	(i)	Filter (1)
	Evaporate some of the filtrate by boiling / heating (1)
	Leave to crystallise / cool (collect crystals) (1)
	Dry between sheets of filter paper / blotting / dessicator / warm
oven (1)
NOT “dabbing” / “patting” on its own
NOT “hot oven”
NOT “oven”
If temperature quoted, must be < 70 °C
	Stages must be in correct order.
Mark until procedure fails
Can score remaining 3 marks even if initial filtration has not been
carried out	4
 
(ii)	BaCO_3(s) + 2HCl(aq) → BaCl_2(aq) + H_2O(l) + CO_2(g)
ALLOW H_2CO_3(aq)
	balanced equation (1)
state symbols (1)
	BaCl_2.2H_2O(aq/s) acceptable, providing extra H_2O(l) on left
ALLOW 2^nd mark provided a sensible but unbalanced equation is
given.	2
 
(iii)	moles of HCl used =
((25/1000) × 1.0)
= 0.025 / 2.5 × 10^–2
IGNORE units	1
 
(iv)	M_r [BaCl_2.2H_2O(s)] = 137 + 71 + 36
= 244 (g mol^–1 )	1
 
(v)	Moles of BaCl_2.2H_2O = 0.5 × 0.025	= 0.0125
Mass of crystals = 0.0125 × 244	= 3.05 / 3.1 (g)
IGNORE units
ALLOW transferred error from (ii), (iii) and (iv)	1
 
(vi)	Any one
	BaCl_2 lost in the (saturated) filtrate when crystals collected /
OWTTE (1)
	Transfer loss/ OWTTE (1)
	Loss when washing (1)
	NOT incomplete reaction/ inaccurate measurement of materials /
spillage on its own BUT neutral otherwise	1
 
(b)	(i)	(Apple) green / yellow-green NOT yellow	1
(ii)	Pt/nichrome (wire)/ceramic rod / spatula
NOT nickel / chromium wire
NOT wire of indeterminate material	1
[12]

24.	(a)	(i)	moles Na = 92 / 23 = 4 = moles NaCl
mass NaCl = 4 × 58.5 = 234 (g)
Penalise use of atomic numbers once
Incorrect answer scores  (1) only if moles (NaCl) mentioned
	OR
23g Na ⇒ 58.5g NaCl  (1)
Mass NaCl  (1)
	2
 
(ii)	 = 0.40 mol dm^–3	OR	 g dm^–3
	consequential on (a)(i)
units required	1
 
(iii)	moles chlorine = 2
vol = 2 × 24 = 48 (dm^3)
Consequential on (a)(i)
	Correct answer (some working)  (2)
Correct answer (no working)  (1)
Incorrect answer scores  (1) only if moles of Cl_2 mentioned	2
 
(b)	Regular pattern or lattice of (sodium) ions
	in a sea of electrons / delocalised electrons
ALLOW “cloud of electrons”
	electrons are mobile / free to move (under an applied
potential and so conduct electricity)
NOT ‘free’ on its own or carry the charge	3
 
(c)	(i)	Energy (allow enthalpy) required per mole
to remove 1 electron (per atom)
from gaseous atoms
OR
X(g) → X^+(g) + e
Species  (1)
State symbols  (1) – only on correct equation
Electron affinity defined  (0)	3
 
(ii)	chlorine has more protons / nucleus more positive
	Same shielding / same number of inner electrons/atomic
radius less     ALLOW outer electron(s) in same shell
(so more energy required)
	OR
effective nuclear charge increases  (1)	2
[13]

25.	(a)	Number of moles /   (1)
If candidate does first part only, working must be shown
	Number of atoms = 3.01 × 10^23  (1)
ACCEPT 3.0 OR  3 OR 3.010(×10^23)
NOT 3.01^23
	If all working shown, allow TE for 2^nd mark    Ignore units
Correct answer with no working (2)	2
 
(b)	(i)	2Li((s)) + 2H^+((aq))  → 2Li^+((aq)) + H_2((g))
ALLOW multiples     Ignore state symbols	1
 
(ii)	
	(1)	(1)
Allow all dots or all crosses on Cl^–
Max 1 if no/wrong charges
If covalent (0)
Do NOT penalise if electrons not shown in pairs
Maximum 1 if Li and Cl not labelled
Li and Cl symbols can go below diagram
Square brackets not essential
Allow number of protons/positive charges in nucleus as
alternative to symbols for Li and Cl	2

(iii)	Any two from:
	Temp 298 K/ 25 °C OR “at a specified temperature”
Unit of temperature needed
NOT “room temperature”
(Acid/solution)  concentration 1 mol dm^–3 / 1 molar 
	Pressure 1 atm / 10^5 Pa / 1.01 × 10^5 Pa/10^2 k Pa /
101 k Pa/10^5 N m^–2 / 76 cm Hg
NOT “pressure of hydrogen” OR “pressure of reactants”
NOT atmospheric pressure
	Must be the most stable/usual/normal physical states
NOT “standard states”
If more than 2 conditions given, deduct 1 mark for each incorrect answer	2
[7]

26.	(a)	(i)	–1/–l, 0           –1/–l, 0
minus can be either side, sub or superscript
	iodine no’s correct (1)
chlorine no’s correct (1)	2
 
(ii)	chlorine oxidation number goes down/goes from 0 to –1, so reduced (1)
	iodine oxidation number goes up/goes from –1 to 0, so oxidised (1)	2
Mark consequentially on (a)(i)
 
(iii)	moles NaI =  (1)
	moles I_2 = 0.1 (1)
	mass of I_2 = 0.1 × 254 = 25.4 (g) (1)
	OR
300g NaI (1) → 254g I_2 (1)
	30.0 × = 25.4(g) (1)
	Correct answer with some working (3)
Use of atomic numbers 2 max
Penalise wrong units	3
 
(iv)	vol = 0.1 × 24 = 2.4 (dm^3)	1
If not 2.4, check for consequential on (a)(iii)

(b)	(i)	black/grey/grey-black (1)
NOT blue-black
NOT purple
IGNORE shiny/silvery
Solid (1)	2
 
(ii)	I(g) → I^+(g) + e^(–)      OR          I(g) – e^(−) → I^+(g)
	species (1)
state symbols (1) - award state symbols mark only if species correct
and in correct place, or if wrong halogen used
If I_2 OR ½I_2 (0)	2
[12]

27.	(a)	(i)	4 pairs of electrons /2 lone pairs and 2 bond pairs (1)
	so electron pairs arranged tetrahedrally
OR
Arranged to give maximum separation/minimum repulsion (1)	2
 
(ii)	103 – 105 ^(°) (1)
	lone pair repulsion> bond pair repulsion (1)	2
 
(b)	(i)	trigonal planar diagram (1)
e.g	two opposite wedges gets (1)
	three wedges of two types gets (1)
	one wedge only gets (0)
IGNORE name
	120 ^(°) marked on diagram (1) - stand alone	2
 
(ii)	B and Cl have different electronegativities / Cl more
electronegative than B	1
OR different electronegativities explained
 
(iii)	Dipoles (or vectors) cancel/symmetrical molecule/centres
of positive and negative charges coincide
IGNORE polarity cancels	1
 
(iv)	Induced-dipole(-induced dipole)/dispersion/London/v der Waals/vdw
Temporary or instantaneous can be used instead of induced
	NOT “dipole” forces
NOT permanent dipole
NOT dipole-dipole	1
 
(c)	 = (0.481)	 = (2.40) (1)
 = 1	 = 5   ,   so PCl_5  (1)
	Use of atomic number max 1	2
[11]

28.	(a)	To make sure the decomposition/ reaction is complete / all the carbon
dioxide has been given off.
Reference to burning (0)
NOT “maximum CO_2”	1
	IGNORE significant figures in (b) and (c)
 
(b)	(i)	2.2(0) (g)	1
 
(ii)	= 0.05(00) mark is for ÷ 44	1
 
(iii)	0.05(00)	1
 
(iv)	(1) = 115 (g mol^–1)	1
 
(v)	115 – (12 + 48) = 55	1
Consequential BUT answer must be sensible
 
(c)	(i)	Molar mass error = = (±) 1(.05) (1)	1
Consequential on (b)(iv)
ALLOW a range of 2 × error
 
(ii)	114 to 116	1
Consequential on (i)
 
(iii)	54 to 56	1
Consequential on (ii)

(iv)	“Could be Mn or Fe”	1
Consequential on (iii)
MUST be metals and must give all possible in range
[10]

29.	(a)	(1)
	= 6.03 × 10^23 (1)
	–1 mark for SF error
Final answer must be 6.03 × 10^23 for 2nd mark
Correct answer with no working (2)
6 × 10^23/ 6.02 × 10^23 quoted with no working (0)
Error in method, max (1)	2
 
(b)	80 is the average mass of Br atoms / isotopes
OR
There must be another/at least one Br isotope of mass greater than
80/with more than 45 neutrons
NOT naturally occurring isotope has mass 80	1
[3]

30.	(a)
| Neutrons | Electrons
24 12Mg | | 12

26
12Mg

14

24 12Mg^2+ | | 10
	1 mark each number	3
Accept words or numbers

(b)
		1s	2s	2p	3s	3p
Mg | | | | | ↑↓ | ↑↓ | ↑↓ | | ↑↓ | | | |
		1s	2s	2p	3s	3p
Cl | | | | | ↑↓ | ↑↓ | ↑↓ | | ↑↓ | | ↑↓ | ↑↓ | ↑
	Arrows can be
 for ↑
 for ↓	2
Accept both arrows up or both down
Reject numbers
 
(c)	Mg(s) + Cl_2(g) → MgCl_2(s)
	Formulae (1)
State symbols (1) – only if formulae correct or near miss for MgCl_2
(e.g. MgCl/Mg_2Cl)	2
Accept multiples
Accept Mg^2+(Cl^–)_2(s)
Reject “Mg^2++ 2Cl^–” for MgCl_2
(0 mark)
 
(d)	 (1)
= 71 (1)
Any unit max 1
2^nd mark consequential on fraction provided 70, 72 and 74 used	2
Accept answer ≥ 2 SF
Reject use of Ar (0 mark)
Reject just “71” with no working (0 mark)
 
(e)	 moles (1)
	X 30.6 = 2.04 dm^3 (1)
Answer with no working 1 max	2
Accept consequential if wrong answer to (d) used.
Accept 71 used when (d) incorrect
Accept answer ≥ 2 SF
Reject no or incorrect unit of volume (loses 1 mark)

(f)	Type – Metallic(1)
	Attraction between Mg^2+ (1)
	And (surrounding) sea of electrons/delocalised electrons (1)
	Stand alone	3
Accept cations/positive ions /magnesium ions
Reject atoms/nuclei/ions
“force between” if used instead of “attraction”
 
(g)	Ionic (1)

	Correct charges and number of ions (1)
Correct electronic structures (1)
Stand alone	3
Accept diagram without brackets
Accept Mg with no electrons shown
ie [Mg]^2+
Reject any suggestion of electrons being shared
Reject [Mg^•]^+
[17]

31.	(a)	(i)	23 + 3 × 14 = 65(g)
Ignore units e.g. g mol^–1, g/mol	1

(ii)	48 dm^3 = 2 moles (1)
allow TE from (a)(i)
allow 87 g/86.67 g
Reject 86 g
86.6 g
86.6666666 g
	number of moles of NaN_3 = 2/3 × 2 = 4/3
	mass = 4/3 × 65 = 86.7 g (1)
ALLOW 2,3 or 4 SF
Accept correct answer with no working (2)
	If 2 moles of N2 seen anywhere award 1^st mark	2
 
(b)	Formation of sodium which is reactive
with water/air / oxygen (1)
Reject –1 if discuss poisonous flammability of N_2 as well as correct problems with sodium
	to produce hydrogen which is flammable / NaOH which is
corrosive (1)
Reject sodium is poisonous
	Max 1 if only discuss sodium and air	2
[5]

32.	(a)	(i)	(18 × 1.35) = 24.3/ 24.30 (kJ) (1)	1
 
(ii)	kJ from 1 mole =  /  / 24.3 × 88 (1)
	∆H = – 2140 (3SF) (kJ mol^–1) (1)
Second mark must have negative sign and 3SF
Allow TE from incorrect value in (i)	2
Accept ∆H = – 2138.4/ –2138/+2138 (kJ mol^–1)for 1 mark
 
(iii)	Incomplete combustion / combustion to C or CO. Not
complete combustion (1)	1
Reject not all of the propane burns.
Reject comments on accuracy of equipment.

(b)	(i)

+6490 = ∆H_c + (6 × 805 + 8 × 464)
∆H_c	=+6490 – 4830 – 3712
	= –2052 kJ mol^–1
Accept –2050 (kJ mol^–1)
	Balancing cycle with 5O2 and 10 O(g) (1)
∆H_1 = (6 × 805 + 8 × 464) = (+) 8542 (kJ mol^–1) (1)
Final value –2052 (kJ mol^–1) (1)
IGNORE SF
Allow TE from an incorrectly calculated ∆H_1 if method clear.	3
 
(ii)	H_2O is gas in equation/ not standard state
OR mean bond energies differ from bond energies in these
compounds / Environment in these compounds changes bond
energies from the mean.	1
Accept H_2O is liquid in ∆H combustion calculation but ges in bond energy calculation.
Reject “mean bond energies are used” without qualification
Reject all the substances are in the gasous state
 
(c)	(i)	Free radical (1) substitution (1)	2
Accept reverse order
 
(ii)	2C_3H_7• → C_6H_14 (1)
	Two (propyl) radicals may combine /a radical and a
molecule may produce C_6H_14 (in a propagation step) (1)	2
Accept C_3H_7• + C_3H_7• → C_6H_14 (1)
Accept multiples
Accept C_3H_7• + C_3H_8 → C_6H_14 + H• (1)
Full mechanisms may be shown

(d)	(i)	2-chloropropane: white precipitate/ solid /cloudiness (1)
Reject white colour
Reject creamy
	2-iodopropane: yellow precipitate / solid (1)	2
Accept ppt appears slowly with chloropropane and quickly with iodopropane (1)
Accept pale yellow precipitate
Accept white colour and yellow colour – 1 out of 2
 
(ii)	Ag^+(aq) + I^–(aq) → AgI (s)
	Formulae (1)
State symbols (1)
Minor error in formula with correct state symbols max 1	2
Accept max 1 out of 2 for wrong halide
 
(iii)	Propan-2-ol/ CH_3CH(OH)CH_3 (1)	1
Accept displayed formula

Reject molecular formula C_3H_7OH
[17]

33.	(a)	The number of atoms in 12g of ^12C (2)
The number of atoms in 1 mole of ^12C (2)	2
Accept number of atoms in 1 mole (1) of atoms / stated monatomic substance (1)
OR Number of molecules in 1 mole (1) of molecules / stated molecular substance (1)
OR Number of electrons in 1 mole (1) of electrons (1)
OR Number of particles in 1 mole (1) max
If answer just quotes the number it does not score it is in the question.
Reject number of particles in 1 mole of a substance

(b)	(i)	1.907 g of Z contains 2.87 × 10^22
Accept moles of Z = 2.87 × 10^22 / 6.02 × 10^23 = (0.04767) (1)
	1.907 × 6.02 × 10^23/ 2.87 × 10^22 is 1 mol (1)
= 40.(0) (1) No units but allow 40 g/mol
IGNORE s.f. in answer
Accept atomic mass = 1.907/moles = 40.(0) (1)
	Allow 39.7 for 2 marks this is rounding 0.04767 to 2 sig figs in calc
Allow 38.14 for 1 mark as this is rounding to 1 sig fig.
Correct answer with some working (2)	2
 
(ii)	Ar / Argon
Consequential on (i) but must be nearest group 0	1
 
(c)	(i)	Amount hydrogen peroxide produced =  = 0.09088 (moles) (1)
Amount of potassium superoxide = 0.09088 × 2 (moles) (1)
34g of H_2O_2 requires 2 × 71g of KO_2 (1)
3.09g requires  (1) or (2) if this is start line
= 12.9 g (1)
If round 0.09088 to 0.09 can score 2 for 12.78
	mass of potassium superoxide = 0.09088 × 2 × 71
= 12.9g (1) / 13g
incl unit but ignore S.F. (note = 6.45 g scores 2 marks)
	Correct answer with some working scores 3 marks	3
 
(ii)	Volume of oxygen =  = 2.18 dm^3 (1) IGNORE s.f. &
do not penalise lack of units twice
Allow error carried forward. i.e if omit to ×2 in part (i)
only penalise it in part (i) not here	1
Accept 2.2
[9]

34.	[CH_3COOH] =  × 0.020 = 0.8(0) (mol dm^–3)
Accept correct answer with no working.
[1]

35.	(a)	(1s^2) 2s^2 2p^6 3s^2 3p^6 ……….... (1)
..…… 3d^8 4s^2 / 4s^2 3d^8 (1)	2
 
(b)	Mr[Ni] = (58 × 0.6902) + (60 × 0.2732) + (62 × 0.0366)
Or correct fraction using percentages (1)
	= 58.6928 (calculator value)
= 58.7 (3 s.f.) (1)
No 2^nd mark if units given, e.g. g, % etc	2
Accept 58.6928 / 58.693 / 58.69
Mr[Ni] = 59, if working shown
Reject 60 /Incorrectly rounded answer, e.g. 58.692, 58.70, 58.6
 
(c)	5.9 g of nickel =  = 0.10 (mol) (1)
From equation, 0.40 mol of CO required
Volume of CO = 0.40 × 24 = 9.6 dm^3 / 9600 cm^3 (1)
Correct units needed for 2nd mark, eg dm^3 mol^–1/dm^–3 (0)
Allow TE for 2^nd mark, from wrong number of moles
(i.e. 4 × number of moles × 24 dm^3)	2
Allow full marks for answer based on A_r[Ni] calculated in (b)
 
(d)	(i)	Ni + 2H^+ → Ni^2+ + H_2
Ignore state symbols	1
Accept Ni + 2H_3O^+ → Ni^2+ + 2H_2O + H_2 (1)
 
(ii)	Stop marking when operation no longer “works”, e.g.
distil/add CaSO_4/boil solution to dryness
	Boil/heat (NOT warm) to drive off some of the water/to
concentrate (not to dryness) (1)
	leave/set aside for some time/overnight (to crystallise) /
allow to cool (must be evident that some solution remains
afterwards) (1)
	Collect crystals by decantation/filtration/use of tweezers (1)
	Dry crystals between (sheets of) filter paper (must imply
an “active process” – leaving on filter paper isn’t enough) /
use of warm oven, not just “oven” (1)
	4 key points → 3
3 key points → 2
2 key points → 1
1 or 0 key point → 0	3

(iv)	Mr[NiSO_4.7H_2O]= 59 + 32 + 64 + (7 × 18)
= 281 (g mol^–1) (1)
2.95 g of Ni =  = 0.050 mol
Mass of crystals formed = 0.050 × 281 = 14.1 (g) (1)
Ignore units, e.g. g mol^–1
Allow TE from incorrect M_r[NiSO_4.7H_2O], i.e. 0.05 × M_r
Allow full credit for answer based on accurate A_r[Ni], e.g. 58.7	2
Accept 14.05 g answer to between 2 and 4 sig. fig
[12]

36.	(a)	(i)	The ability of an atom/element/ species to attract the electrons (1)
Accept “Power/extent” instead of “ability”
Accept “pulls toward/draws” instead of “attract”
Reject molecule
	in a covalent bond/bond pair/shared electrons (1)	2
 
(ii)	The molecule is symmetrical / tetrahedral (1)
Reject too small a difference in electronegativity
	So bond polarity/dipoles cancels
OR
centres of positive and negative charge coincide (1) – stand alone	2
Accept diagrams showing vectors
Reject charge cancels
 
(iii)	Dispersion/Induced dipole /London
OR
temporary/instantaneous dipole	1
Accept van der Waals/vdw
Reject dipole-dipole
Reject hydrogen bond

(b)	(i)	Ignore sig. figs UNLESS rounded to 1SF
	700 g TMP =  (1) = 6.14 mol
Reject moles 2C_8H_18 =  = 3.07
	Moles of oxygen = 12.5 × 6.14 (1) = 76.75
	Volume of oxygen = 12.5 × 6.14 × 24 = 1842 dm^3 (1)
Units essential
Working must be checked i.e.
3.07 × 25 × 24 = 1842 dm^3 (2)
3.07 × 12.5 × 24 = 921 dm^3 (1)
Accept 1840/1800 dm^3
Accept 1830 if 6.14 rounded to 6.1
	OR 228 g of TMP need 25 × 24 dm^3 of oxygen (1)
	∴ 700 g of TMP need  of oxygen (1)
	= 1842 dm^3 (1)
	Units essential
[Working must be checked]	3
 
(ii)	Ignore sig. figs UNLESS rounded to 1SF
	Moles of CO_2 = 8 × 6.14 (1) = 49.12
	Mass of CO_2 = 8 × 6.14 × 44 = 2161 g (1)
Units essential but don’t penalise if already penalised in (i)
Accept 2160/2200 or 2147 / 2150 / 2100 if 6.14 rounded to 6.1
	OR
228 g of TMP give 44 × 16 g CO_2 (1)
∴ 700g of TMP give  g of CO_2
= 2161 g (1)
	Could be consequential on (i)	2
[10]

37.	(i)	Ignore sig figs unless they round to 1 sig.fig during calculation
Incorrect /absent units in final answer penalise only once in part (i)/(ii)
	7.19 g of PCl_5 =  mol (1)
2 × 31 g of P produce 2 × 208.5 g of PCl_5 (1)
	(= 0.03448)
(1 mol of PCl_5 from 1 mol of P)
	Mass of P = 0.03448 × 31 = 1.07 g (1)
7.19 g of PCl_5 from 
= 1.07g (1)
	Penalise use of Atomic Number only once
Answer with no working scores 2	2
Allow 0.034 but NOT 0.035
 
(ii)	Mark consequentially on part (i)
	Moles of chlorine needed = 0.03448 × 2.5 (1)
Accept 2 × 208.5 g of PCl_5 produced from 5 × 24 dm^3 of Cl_2 (1)
	Volume = 24 × 0.03448 × 2.5 = 2.07 dm^3 (1)
Value and unit necessary
	Value consequential on their calculated/stated moles of chorine × 24
Answer with no working scores 2	2
7.19 g PCl_5 produced from  = 2.07 dm^3 (1)
Just 24 × 2.5 = 60 dm^3 scores zero
[4]

Reject –1 if discuss poisonous flammability of N_2 as well as correct problems with sodium
	to produce hydrogen which is flammable / NaOH which is
corrosive (1)
Reject sodium is poisonous
	Max 1 if only discuss sodium and air	2
[5]

38.	(a)	A	1
 
(b)	D	1
[2]

39.	(a)	C	1
 
(b)	D	1
 
(c)	A	1
 
(d)	D	1
[4]

40.	(a)	QWC (i) & (iii)
	(Lattice of) positively charged ions/ ions with 2+charge (1)
	held together by (electrostatic) attraction to delocalised electrons (1)
	Delocalised electrons /free electrons/ electrons in sea of electrons
are free to move and carry charge / current (1)	3
Reject incorrect use of the terms atom or molecule for ion.
Reject incorrect descriptions of delocalised electrons.
 
(b)	(i)	Mg^2+ shown as 2,8 (1)	1
 
(ii)	F^– shown as 2,8 (1)	1
 
(c)	When molten/ when dissolved in water so that ions can move/
lattice breaks down (1)	1
Reject dissolved in other solvents.
Reject reference to atoms or molecules rather than ions.
 
(d)	(i)	77% ^24Mg, 10% ^25Mg, 13% ^26Mg (1)
	Average atomic mass
 = 24.36 = 24.4 g (1)	2
 
(ii)	Have same electron configuration	1
Reject same number of electrons in outer orbit
 
(e)	(i)	1.20 × 10^–9 mol of Mg per dm^3 (1)
(1.20 × 10^–9 × 24.3 × 10^–3) =
2.92 × 10^–11 / 29.2 × 10^–12 (g) (1)
max 1 for more/less than 3 significant figures eg 2.916	2
 
(ii)	Hydrogen because it has the least number of electrons per atom	1
[12]

41.	(a)	(6.02 × 10^23 × )=	1
1.25 × 10^24 / 1.254 × 10^24 / 1.26 × 10^24
Allow TE from a	1
 
(b)	M_r = (23 + 42) = 65 (1)
Mass = (2 × 65 × ) (1)
= 90/ 90.3g (1) Allow TE from (c)	3
Reject wrong unit eg kg
 
(c)	decrease	1
 
(d)	QWC (i) & (iii)
	Sodium is hazardous (1)
	May go on fire with water/ produces flammable gas with water/
produces explosive gas with water/ produces strong alkali with
water/ reacts with moisture on skin and becomes hot /corrosive (1)
	2^nd mark depends on reference to sodium	2
Reject unspecific comments about sodium being poisonous / toxic / flammable without reference to water.
[8]

42.	C
[1]

St John's RC Comprehensive School