C
L
3
6
Cambridge Chemistry Challenge Lower 6th
June 2011
Marking scheme for teachers
(please also read the additional instructions)
Page 1
1(a)
leave
blank
name of isomer
cyclopropane
class of compound
cycloalkane
name of isomer
class of compound
4
alkene
propene
[ answers can be either way round ]
(b)
Structure of A
Structure of F
Br
(c)
Structure of B
2
Br
Structure of G
Br
Br
2
(d) (i) Equation for combusion:
C3H6(g) + 9/2 O2(g)
3CO2(g) + 3H2O(g)
[ quantities may be doubled ]
1
(ii) Standard enthalpy of formation of A:
[9/2 O2(g)] + 3C(s) + 3H2(g)
3(–393.5) 3(–241.8)
ΔfH (A)
+ [9/2 O2(g)]
–2058
3CO2(g) + 3H2O(g)
ΔfH (A) = 2058 – 3(393.5) – 3(241.8) = +152.1 kJ mol–1
[ one mark if correct value but wrong sign;
also one mark if equation is doubled and value is doubled ]
2
Page 2
leave
blank
(d) (iii) Standard enthalpy of formation of B:
ΔfH (A) = 2091 – 3(393.5) – 3(241.8) = +185.1 kJ mol–1
[ one mark if correct value but wrong sign;
also one mark if equation is doubled and value is doubled ]
(e)
2
Standard enthalpy change for reaction B → A:
ΔrH = 152.1 –185.1 = –33 kJ mol–1
1
[or ΔrH = –2081 – (–2058) = –33 kJ mol–1 ]
(f) (i)
Structure(s) of A with one D.
D
D
4
D
D
[ +1 for each correct structure
–1 for any additional repetition]
(ii)
Structure(s) of B with one D.
D
1
[ +1 for correct structure,
0 if more than one structure is drawn]
Page 3
(g)
leave
blank
Structure of A1:
D
1
D
Structures of X1 and X2:
(h)
D
D
D
D
D
D
D
D
2
[ answers can be either way round ]
Structures of A2 and A3:
(i)
D
D
D
2
D
[ answers can be either way round ]
(j) (i)
Structure of Y:
D
D
D
1
D
(ii)
Structures of A4 and A5:
D
D
D
2
D
[ answers can be either way round ]
Page 4
(k)
Structure of A6:
D
D
D
D
D
2
D
(l)
leave
blank
Structure of Z:
Structure of A7:
D
1
D
(m)
NO plane of
symmetry
Plane of
symmetry
Structure(s)
Structure(s)
NO
rotational
symmetry
D
D
2 marks for each correct structure in the correct box
1 mark for a correct structure but in the wrong box (up to 4 marks)
−1 mark for any duplicate structure (down to zero)
Structure(s)
Structure(s)
8
D
Rotational
symmetry
D
(enantiomers)
D
D
D
D
Total 38
Page 5
2(a)
leave
blank
(i) oxidation state
(ii) shape:
octahedral
+4
(b)(i)
Formula for W
NaF
(ii)
1
Formula for X
SiF4
(iii)
shape:
2
tetrahedral
Equation for formation of W and X
Na2SiF6(s)
(c)(i)
2
Formula for Z
Formula for Y
CaF2
1
2NaF(s) + SiF4(g)
SiH4
3
Equation for formation of Y and Z
2CaH2(s) + SiF4(g)
(ii)
2CaF2(s) + SiH4(g)
Equation for combustion of Z:
SiH4(g) + 2O2(g)
(d)
SiO2(s) + 2H2O(l)
1
Equation for decomposition of Z
1
SiH4(g)
Si(s) + 2H2(g)
Page 6
(e)(i)
leave
blank
calculate n
n = (8 × 1/8) + (6 × 1/2) + 4 = 8
(ii)
2
number of atoms in sphere
n atoms in a3 pm3 ≡ a3 × 10−36 m3
volume of sphere = V cm3 ≡ V × 10−6 m3
atoms in volume V =
V n × 1030
a3
3
2 marks for expression
3rd mark if with factor of × 1030
(iii)
Expression for Avogadro constant
atoms in m g =
V n × 1030
a3
atoms in 1 g =
V n × 1030
ma3
atoms in Ar g =
3
Ar V n × 1030
m a3
2 marks for expression
3rd mark if with factor of × 1030
Page 7
(f)
leave
blank
Si–Si bond length
z
a/2
xy =
a2 + a2
4
4
=
yz =
a2 + a2
2
4
=
y
x
a/2
a
2
3a
2
4
yz = twice bond length
Si−Si bond length = 3 a pm ( =0.433a pm)
4
[ 3 marks if correct but no unit; 2 marks for twice the answer;
1 mark for some Pythagorean working but wrong answer ]
(g)
Ar for silicon
Ar = (1 − 41.2×10−6 − 1.3×10−6 ) × 27.97692653
+ ( 41.2×10−6 × 28.97649470 )
+ ( 1.3×10−6 × 29.97377017 )
= 27.97697031
2
[ 1 mark for some correct working but wrong answer ]
(h)
Calculated value for Avogadro constant
putting values into expression gives
6.02214096 × 1023
1
1 mark if this answer. No carry forward.
Total 26
Page 8