AS-LEVEL PAPER 1 PP17 MS

1.   (a)    Carbon / C
If M1 incorrect, CE = 0 / 3
1
Fewest protons / smallest nuclear charge / least attraction between protons (in the nucleus) and electrons / weakest nuclear attraction to electrons
Allow comparative answers.
Allow converse answers for M2
1
Similar shielding
Allow same shielding.
1
(b)     Increase
1
Oxygen / O
If not oxygen, then cannot score M2, M3 and M4
1
Paired electrons in a (2)p orbital
If paired electrons in incorrect p orbital, lose M3 but can award M4
1
(Paired electrons in a p orbital) repel
1
(c)     C(g) → C+(g) + e(−)
OR
C(g) + e (−) → C+(g) + 2e (−)
OR
C(g) − e (−) → C+(g)
Ignore state symbols for electron.
1
(d)     (More energy to) remove an electron from a (more) positive ion / cation
Allow electron closer to the nucleus in the positive ion.
1
(e)     Lithium / lithium / Li
If formula given, upper and lower case letters must be as shown.
1
[10]

2.        (a)     P = 100 000 (Pa) and V = 5.00 x 10–3 (m3)
M1 is for correctly converting P and V in any expression or list Allow 100 (kPa) and 5 (dm3) for M1.
1
		
M2 is correct rearrangement of PV = nRT
1
= 0.202 moles (of gas produced)
This would score M1 and M2.
Therefore  = 0.0404 moles B2O3
M3 is for their answer divided by 5
1
Mass of B2O3 = 0.0404 x 69.6
M4 is for their answer to M3 x 69.6
1
= 2.81 (g)
M5 is for their answer to 3 sig figures.
2.81 (g) gets 5 marks.
1
(b)     B + 1.5 Cl2 → BCl3
Accept multiples.
1
3 bonds
1
Pairs repel equally/ by the same amount
Do not allow any lone pairs if a diagram is shown.
1
(c)     (i)      43.2/117.3 (= 0.368 moles BCl3)
1
0.368 x 3 (= 1.105 moles HCl)
Allow their BCl3 moles x 3
1
Conc HCl = 
Allow moles of HCl × 1000 / 500
1
= 2.20 to 2.22 mol dm–3
Allow 2.2
Allow 2 significant figures or more
1
(ii)     H3BO3 + 3NaOH → Na3BO3 + 3H2O
Allow alternative balanced equations to form acid salts.
Allow H3BO3 + NaOH → NaBO2 + 2H2O
1
(d)     
Mark is for both Mr values correctly as numerator and denominator.
1
8.98(%)
Allow 9(%).
1
Sell the HCl
1

(e)     Alternative method
Cl = 86.8%
Cl = 142 g
1
  B                 Cl
           
  B                 Cl
           
1
1.22           2.45 or ratio 1:2 or BCl2
2:4 ratio
1
BCl2 has Mr of 81.8 so
81.8 x 2 = 163.6
Formula = B2Cl4
B2Cl4
Allow 4 marks for correct answer with working shown.
Do not allow (BCl2)2
1
[20]
 
3.      (a)     Moles HCl =  =  (1) (= 0.537)
Concentration =  (1)
= 2.15 (mol dm–3) (1)
Conseq on  correct
min 2 d.p. 2.14 to 2.15
Ignore wrong units
A.E. lose one mark
3
(b)     (i)       = 5.7(1) × 10–3 (mol) (1)
5.7 to 5.71 × 10–3
(ii)      = 2.85 × 10–3 (mol) (1)
Conseq
(iii)      = 138 (1)
Conseq
(iv)    Relative atomic mass of M: 138 ‑ 60 = 78 (1)
 = 39 (1)
Identify of M: Potassium or K or K+ (1)
Conseq
If 78 = Mr then M = selenium
6
[9]

4.	(a)     Macromolecular / giant covalent / giant molecule
Not giant atomic
1
(b)     No delocalised electrons / no free ions / no free charged particles
1
(c)     SiO2 + 6HF  H2SiF6 + 2H2O
Accept multiples
1
[3]

5.	(a)      (i)     reduction OR reduced OR redox OR reduction–oxidation
Not “oxidation” alone
1
(ii)	Fe3+	+	3e–		Fe	

Ignore state symbols
Do not penalise absence of charge on electron
Credit Fe3+      Fe   –   3e–
Credit multiples
1
(b)     (i)      Because (one of the following)
CO is not the only product OR
Reference to “incomplete combustion to form CO” does not answer the question
(Some) complete combustion (also)occurs OR
CO2 is (also) formed
Further oxidation occurs
1
(ii)     The enthalpy change / heat (energy) change at constant pressure in a reaction
is independent of the route / path taken (and depends only on the initial and
final states)
1
(iii)    M1   The enthalpy change / heat change at constant pressure when 1 molof a compound / substance / element
For M1, credit correct reference to molecule/s or atom/s
M2   is burned completely / undergoes complete combustion in (excess)
oxygen
M3   with all reactants and products / all substances in standard states
For M3
Ignore reference to 1 atmosphere
OR   all reactants and products / all substances in normal / specified states
under standard conditions / 100 kPa / 1 bar and specified T / 298 K
3

(c)     M1 (could be scored by a correct mathematical expression which must 
have all ∆H symbols and the ∑)
Correct answer gains full marks
Credit 1 mark ONLY for –1 (kJ mol–1)
M1   ∆Hr = ∑∆Hf (products) – ∑∆Hf (reactants)
Credit 1 mark ONLY for – 27 (kJ mol–1) i.e. assuming value for Fe(l) = 0
       OR correct cycle of balanced equations with 2Fe, 3C and 3O2
M2   ∆Hr   = 2(+14) + 3(– 394) – (– 822) – 3(–111)
                 = 28 –1182 + 822 + 333
                 (This also scores M1)
M3        = (+) 1 (kJ mol–1)
(Award 1 mark ONLY for – 1)
(Award 1 mark ONLY for – 27)
For other incorrect or incomplete answers, proceed as follows
•        check for an arithmetic error (AE), which is either a transposition error or an incorrect multiplication; this would score 2 marks (M1 and M2)
•        If no AE, check for a correct method; this requires either a correct cycle with 2Fe, 3C and 3O2 OR a clear statement of M1 which could be in words and scores only M1
3
(d)     (i)      C(s)     +     O2(g)     CO2(g)
State symbols essential
Possible to include C(s, graphite)
1
(ii)     These two enthalpy changes are for the same reaction / same equation /
same reactants and products
Penalise reference to CO2 being produced by a different route
OR
They both make one mole of carbon dioxide only from carbon and oxygen 
(or this idea clearly implied)
“both form CO2” is not sufficient (since other products might occur e.g.CO)
OR
The same number and same type of bonds are broken and formed
1
[12]

6.              (a)     (i)     SiCl4     +     2H2          Si     +     4HCl
Ignore state symbols
Credit multiples
Penalise ionic HCl
1
(ii)     Reducing agent / reductant / reduces SiCl4 / reduces (silicon) / electron
donor
1
(b)     
 
2MgO	+	Si		2Mg	+	SiO2	

Ignore state symbols
Credit multiples
1
[3]
 
7.             (i)      2MoS2 + 7O2 → 2MoO3 + 4SO2
OR
MoS2 + 3O2 → MoO3 + 2SO2
Allow multiples
Ignore state symbols
1
(ii)     MoO3 + 3H2 → Mo + 3H2O
Allow multiples
Ignore state symbols
1
 [2]
 
8.     (a)     (i)      MnO2 (+) 4
1
(ii)     MnO2 + 4H+ + 2e–  Mn2+ + 2H2O
Or multiples
Ignore state symbols
Credit electrons subtracted from RHS
Ignore absence of charge on e
1
(iii)     Iodide ion(s) is/are oxidised because they have lost electron(s)
Do not penalise reference to iodine; the mark is for electron loss
1
(b)     (i)      M1       Cl2         0
M2    HClO   (+) 1
2
(ii)     M1    Equilibrium will shift/move to the right
OR    L to R
OR    to favour the forward reaction
OR    to produce more HClO
M2    Consequential on correct M1
          To oppose the loss of HClO
OR    replaces the HClO (that has reacted)
for M2
NOT just “to oppose the change”
2

(c)     (i)      The answers can be in either order
M1    2Br–  Br2 + 2e–
M2    4H+ + SO42– + 2e–  SO2 + 2H2O
OR
          2H+ + H2SO4 + 2e–  SO2 + 2H2O
NOT multiples
Ignore state symbols
Credit electrons subtracted from incorrect side
Ignore absence of charge on e
2
(ii)     KCl + H2SO4  KHSO4 + HCl
OR
          2KCl + H2SO4  K2SO4 + 2HCl
Credit ionic equations
1
(iii)     For M1 and M2, chloride ions are weaker reducing agents
than bromide ions, because
M1    Relative size of ions
Chloride ions are smaller than bromide ions OR
chloride ion electron(s) are closer to the nucleus
OR chloride ion has fewer (electron) shells/levels
OR chloride ion has less shielding (or converse for
bromide ion)
M2    Strength of attraction for electron being lost
Outer shell/level electron(s) OR electron(s) lost
from a chloride ion is more strongly held by the
nucleus compared with that lost from a bromide
ion (or converse for bromide ion)
If the forces are described as intermolecular or Van der Waals then CE = 0
Ignore general reference to Group 7 trend
For M1 accept reference to chlorine/bromine or reference to atoms of these but NOT “chloride/bromide atoms” or “chlorine/bromine molecules”
For M2 insist on reference to the correct ions
This is the expected answer, but award credit for a candidate who gives a correct explanation in terms of hydration enthalpy, electron affinity and atomisation enthalpy.
2
[12]
 
9.    (a)      sulfuric acid / H2SO4
1
(b)      hydriodic acid / HI    OR    hydrobromic acid / HBr
1
(c)      add dilute ammonia solution
Notes 
* do not allow ‘concentrated ammonia’ or ‘ammonia’
1
precipitate / ppt  disappears / dissolves    OR    colourless solution forms
1
(d)      would react with the acid / no gas evolved in tests
1
[5]

10.	
	[4]