SUGGESTED ANSWER PSPM I SK015

STRIVE FOR EXCELLENCE PRE PSPM I ANSWER SCHEME

QUESTION 1

a)

Average Atomic mass of chlorine

= ( × ) + ( × ) @



= 35.5 amu

Relative atomic mass of chlorine

= 35.5

1 × 12
12

= 35.5

b) i) Element C H
Mole ratio = 7.85 mol = 6.29 mol

Simplest ratio 6.29 mol 6.29 mol
= 1.25 x 4 =1x4
n = 128 g mol-1 = 2
64 g mol-1 5 4
Empirical formula C5H4

Molecular formula: C10H8

ii) % w/w of H2O2 = 30.0 % = m of solute x 100 %
m of solution

assume m of solution = 100 g

m of solute H2O2 = 30.0 g

n of solute H2O2 = 30.0 g = 0.8824 mol
34 g mol-1

m of solvent = 100 - 30 = 70.0 g

molality = n of solute (1M)
m of solvent (kg)

= 0.8824 mol
0.070 kg

= 12.61 m

c) Reduction: 5 H2O2 + 10 H+ + 10 e- → 10H2O @ H2O2 + 2 H+ + 2 e- → 2 H2O

Oxidation: 6 H2O + Br2 → 2BrO3- + 12 H+ + 10 e-

Overall: 5H2O2 + Br2 → 4 H2O + 2 BrO3- + 2 H+

d) i) Mole of Na = a 38.0 g
23.0 gmol-1

= 1.652 mol

6 mole of Na produced 2 mole of Na3N

1.652 mole of Na produced 1.652 mol Na x 2 mol Na3N
6 mol Na

= 0.5507 mol Na3N

Theoretical mass of Na3N = 0.5507 x [ 3(23.0) + 14]
= 45.71 g

% yield = Actual yield cx 100%
Theoretical yield

75.1= Actual yield x 100%
45.71 g

Actual yield of Na3N = 34.33 g

ii) 6 mole of Na reacted with 1 mole of N2

1.652 mole of Na reacted with 1.652 mol Na x 1 mol N2
6 mol Na

= 0.2753 mol N2

Mole of N2 remained = 0.30 mol – 0.2753 mol
= 0.0247 mol

QUESTION 2

a) i) The line V of electron transition from n = 3 to n = 2

ii) The wavelength of line W.

1 = RH  1 − 1  , n1  n2
λ n12 n22

1 = 1.097 x 107 m -1  1 - 1 
λ  22 42 

= 2.057 x 106 m-1

λ = 4.86 x 10-7 m

iii) ∆E = hc

= (6.6256 x 10-34) (3.0 x 108)
(4.34 x 10-7)

= 4.58 x 10-19 J

Line X released 4.58 x 10-19 J of energy.

iv)
✓ 1 M axis and show gap between energy level n to n

✓ 1 M Show transition arrow with Label X, Y and Z

b) i) 29Cu: 1s22s22p63s23p64s13d10
The reason for anomaly:
Completely filled 3d orbitals (3d10) are stable compared with partially filled
orbitals.

ii) A set of quantum numbers for an electron in 3d orbital.
(n,Ɩ,m,s) = (3,2, -2,+1/2)
Or (3,2, -2,-1/2) Or (3,2, -1,+1/2) Or (3,2, -1,-1/2) Or (3,2, 0,+1/2) Or
(3,2, 0,-1/2) Or (3,2, +1,+1/2) Or (3,2, +1,-1/2) Or (3,2, +2,+1/2) Or
(3,2, +2,-1/2)

iii) The shape of orbital of the fifth electron in the orbitals.

✓ 1 M label & shape

QUESTION 3

a) ∙∙ ∙∙ :S≡C−∙S∙ : // :S∙∙−C≡S:
:S=C=S: ∙∙(+1) (0) (-1) ∙∙

(0) (0) (0) (-1) (0) (+1)

Structure I (√1M) Structure II (√1M)

(all correct formal charge:√1M)

Most plausible: structure I
All atoms have zero formal charge.

b) i) :F∙∙−Xe−F∙∙:
∙∙ ∙∙

Valence electron configuration of Xe: 5s25p6
Valence electron configuration of F: 2s22p5

Orbital diagram for central atom Xe:

Ground state : ↑↓ ↑↓ ↑↓ ↑↓

5s 5p ↑ 5d
Promotion of electron: ↑↓ ↑↓ ↑↓ ↑ 5d
empty 5d orbitals
Hybridisation state: 5s 5p ↑
↑↓ ↑↓ ↑↓ ↑

sp3d hybrid orbital

Hybrid orbitals shape (sp3d and 2p)
Label all orbitals
Position of lon(ii)

ii) Non-polar molecule

The Xe-F bond is a polar bond. Bond dipoles can cancel each other and
resultant dipole moment, = 0

c) Aluminium
More valence electron and smaller size than Mg and Na
Attraction between positive ions and delocalized electron stronger, stronger
metallic bond

QUESTION 4

a) (i) = 2 + 2

2 = 748 − 17.5
=730.5 torr @ 0.9611 atm

(ii) PV = nRT

Mole of O2, n = 2 2


= 0.9611 0.307

0.08206 293.15

= 0.01227 mol

Mass of O2 = 0.01227 mol x 32 g mol-1
= 0.3925 g

b) ✓ Acetone has van der Waals forces between molecules
✓ Ethanol has hydrogen bond between molecules

✓ Hydrogen bond is stronger than van der Waals forces

c) i) The density of water is higher in liquid phase.

ii) OB line is negative slope because when high pressure the volume is smaller.
Water favour to the smaller volume which is liquids state.

QUESTION 5

i) 0.50
[SO2Cl2]i = 5.0 = 0.1 M

SO2Cl2(g) ⇌ SO2(g) + Cl2(g)

[ ]i (M) 0.1 0 0
[ ]Δ (M)
-x +x +x
[ ]eq (M)
0.1 - x x x

1
M

x = 0.016, so [SO2]eq = 0.016M
[Cl2]eq = 0.016M
= 0.084M
[SO2Cl2]eq

Kc = [SO2][Cl2]

[SO2Cl2]
(0.016)(0.016)
= (0.084)
= 3.0476 × 10-3

ii) Concentration

SO2Cl2 1M
SO2 & Cl2 1M

Time

iii) Based on Le Chatelier’s Principle,
▪ System must increase the temperature by releasing more heat to

surrounding, exothermic reaction is favorable.
▪ Equilibrium position must shift to the left.
▪ SO2Cl2 produce more.
▪ Kc value is decreased.

QUESTION 6

a) HNO2 (aq) + H2O (aq) ⇌ NO2- (aq) + H3O+ (aq)

[ ]i / M 0.01 - 00
∆[ ] / M -x - +x +x
0.01 - x - xx
[ ]eq / M

pH = - log [H3O+]
2.67 = - log [H3O+]
x = [H3O+] = 2.138 x 10-3 M

Ka = [H3O+] [NO2-]
[HNO2]

= (2.138 x 10-3) (2.138 x 10-3)
(0.01 – 2.138 x 10-3)

= 5.814 x 10-4

b) HCOOH (aq) + OH- (aq) → HCOO- (aq) + H2O (l)

ni (mol) 0.70M x 0.50 L 0.05 0.50M x 0.50 L -
∆n (mol) = 0.35 - 0.05 = 0.25 -
- 0.05 + 0.05 -

nf (mol) 0.30 0 0.30

pH = pKa + log [HCOO− ]
[ ]

= - log 1.77 x 10-4 + log [0.301 ]

[0.301 ]

= 3.75

c) HCN(aq) + OH-(aq) → CN-(aq) + H2O(l)

ni (mol) (0.25M)(0.025L) (0.5M)(0.02L) 0 -
=6.25x10-3 +6.25x10-3 -
∆n (mol) -6.25x10-3 =0.01 6.25x10-3 -
nf (mol) -6.25x10-3
0 3.75x10-3

total volume = 0.02 +0.025 = 0.045 L

[OH-] = 3.75 10−3 = 0.083 M

0.045

pOH = -log [OH-]

= - log 0.083
= 1.08

pH = pKw – pOH
= 14.00 – 1.79

= 12.92

d) Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

[ ]i / M - 0 1x10-3
∆[ ] / M - +s +2s
- s 1x10-3 + 2s
[ ]eq / M

s very small, 1x10-3 + 2s ≈ 1x10-3
Ksp = [Mg2+][OH-]2

= (s) x (1x10-3)2
= 1x10-3 s = 1.6x10-13

s = 1.6 x 10-7 M

so, the molar solubility of Mg(OH)2 is 1.6 x 10-7 M
Mg(OH)2 is more soluble in water compared to NaOH.

QUESTION SUGGESTED ANSWER PSPM I 2016/2017
NO.
Molarity = Mol of solute(mol)__ = 2.5 M @ mol L-1
1 (a) Volume of solution(L)

Assume that Volume of solution = 1L = 1000 mL

Mol of solute (Na2CO3) = 2.5 mol

Mass of solute (Na2CO3) = 2.5 mol x 106

= 265 g

Density = Mass of solution (g)___ =1.430 g mL-1

Volume of solution(mL)

Mass of solution(Na2CO3)= 1430g

Mass of solution = mass of solute (Na2CO3) + mass of solvent(Ethanol)

mass of solvent(Ethanol) = 1430 g – 265 g

= 1165 g

= 1.165 g

Molality = Mol of solute (Na2CO3) (mol)__
Mass of solvent (Ethanol)(Kg)

= 2.5 mol

1.165Kg

= 2.146 m @ molal

(b) i. 4Zn 4 Zn2+ + 8e
ii.
10H+ + NO3- + 8e NH4+ + 3H2O

______________________________________________

4Zn +10H+ + NO3- 4 Zn2+ + NH4+ + 3H2O

Mol of NO3- = 25.0mL x 0.5M
1000

= 0.0125 mol

1 mol of NO3- ≡ 4 mol of Zn
0.0125 mol of NO3- ≡ 0.0125 x 4

1
= 0.05 mol of Zn

Mass of Zn = 0.05 mol x 65.4 gmol-1
= 3.27 g

% purity of the Zinc sample = 3.27 g x 100%
4.0 g

= 81.75%

2 (a) i. XO2 XO3

X O
OO
X
OO

ii. X-O bond is polar.
O and X both are from group 16 but O is located at higher position than O.
Since electronegativity decreases going down the group, then O is more electronegative than X.

iii. XO2 XO3
i.
(b) X O
OO
X
• Molecular geometry is bent. OO
• Vectors cannot cancel out.
• The combined vectors is not zero, • Molecular geometry is symmetrical.
• The polarity of bond is identical and
µ≠0
in opposite direction, therefore can
PCl3 cancel one another out.
• The sum vectors is zero, µ = 0.
Cl
Cl P Cl POCl3

Cl

Cl P Cl

O

ii. sp3

iii. P ground state Cl 3p
3s 3p 3s

P excited state

3s 3p

P hybrid state

sp3

sp3

P sp3 Cl
sp3 sp3 p

Cl Cl
p

p

iv. PCl3 POCl3

• Has 4 pairs of e- around the central • Has 4 pairs of e- around the central
atom, e- arrangement is tetrahedral. atom, e- arrangement is tetrahedral.

• Has 3 bonding pair and 1 lone pair e-. • All are bonding pair e-.

• According to VSEPR, repulsion • According to VSEPR, the repulsion is

between lone pair-bonding pair is equal.
• Bond angle is 109.5o
greater than bonding pair-bonding

pair.
• The bonding pair e- is pushed closer

towards each other.
• Bond angle is <109.5o

3 (a) i. PT = PO2 + PH2O
PH2O = 765 – 25.2 mmHg
= 739.8 mmHg

ii. PV = nRT

n= (738/760 atm) (0.250 L) .

(0.08206 atm L mol-1 K-1) (26+273.15 K)

= 9.91 x 10-3 mol

iii. 2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
3 mol O2 ≡ 2 mol KClO3
nKClO3 = ⅔ x 9.91 x 10-3
= 6.61 x 10-3
mass KClO3 = 6.61 x 10-3 x [ 39.1 + 35.5 + (3x16)]
= 0.810 g

4 (a) i.

[ ]i/ M 0.16 - -
/M -x +x +x
x x
[ ]f/ M (0.16 – x)

Therefore, [ ] = [ ]=

[ ] [H3O+] = 1 x 10-14 M2

 [H3O+] = 1.33 x 10-12 M

ii. pH = -log [H3O+]
= -log (1.33 x 10-12)
= 11.90

iii.

(b) Pb(NO3)2 (aq) + 2NaI (aq) → PbI2 (aq) + 2NaNO3 (aq)

[Pb(NO3)2] = 0.2 M
[NaI] = 1.6 x 10-3M

PbI2(s) Pb2+(aq) + 2I-(aq)

0.2 M 1.6 x 10-3M

Qsp= [Pb2+][I-]2
= 5.12 x 10-7

Qsp > Ksp

- Solution is supersaturated.

- Precipitate will form.

5 (a) A ground state electron of an element is an outer orbital electron that is in the lowest energy
(b) state possible for that electron. An excited state electron has absorbed additional energy and
exists at a higher energy level than a ground state electron.

Energy of electron at the excited level

E = -RH/n2
= -2.18 x 10-18/62
=-6.055 x 10-20 J/1000
= -6.055 x 10-23kJ x (6.02x 1023)
=-36.45 kJ/mol

Energy emitted

E = RH(1/ni2-1/nf2)

= 2.18 x 10-18(1/62 -1/32) x (6.02 x 1023)/1000

= 109.6 kJ/mol

Shortest wavelength

1/ = RH (1/n12 -1/n22) n1 < n2

1/ = 1.097 x 107 (1/32-1/2)

 = 8.2 x 10-7 x 109
 = 820.45nm

Ionization energy is minimum energy required to remove 1 mol electron from 1 mol of gaseous
atom to form 1 mol of positively charge gaseous ion.

Energy

1 2 3 4 No. of electron

Figure shown the sharp increase between lE 3 to lE4. 3 electrons in outermost shell
and fourth electron in inner shell.

6 (a) For CH3F molecule:

Element No. of valence electrons, e-
C 4e- x 1 = 4e-
H 1e- x 3 = 3e-
F 7e- x 1 = 7e-

TOTAL = 14e-

Lewis structure:

Number of electron groups = 4
Electron pair arrangement = tetrahedral
Molecular shape = tetrahedral

Each C-H bond and C-F bond are polar. However, C-F bond has greater dipole moment
than C-H bond due to F atom is more electronegative than H atom.

The four dipole moments do not cancel each other.
The resultant dipole moment, µ is not equal to zero.
Therefore, the molecule of CH3F is a polar molecule.

For CF4 molecule:

Element No. of valence electrons, e-
C 4e- x 1 = 4e-
F 7e- x 4 = 28e-

TOTAL = 32e-

Lewis structure:

Number of electron groups = 4
Electron pair arrangement = tetrahedral
Molecular shape = tetrahedral

Each C-F bond is polar and has the same dipole moment.
Since the molecule is symmetrical, each C-F bond dipole moment cancels each other.

The resultant dipole moment, µ is equal to zero.
Therefore, the molecule of CF4 is a non-polar molecule.

(b)
Element No. of valence electrons, e-
I 7e- x 1 = 7e-
F 7e- x 5 = 35e-
TOTAL = 42e-
Lewis structure:

Based on the Lewis structure, the electron pair arrangement is octahedral since there are six
electron pairs around the central atom I which is consist of five bonding pairs and one lone
pair.
According to the VSEPR theory, the repulsion between lone pair electrons and bonding pair
electrons is greater than the repulsion between the bonding pair electrons and bonding
pair electrons.
In order to minimise the repulsions, the six electron pairs are arranged as follows:

Therefore, the molecular geometry of IF5 molecule is square pyramidal with all of the bond
angles are similar which is less than 90º.

Valence orbital diagram in iodine, I:

Ground ↑↓ ↑↓ ↑↓ ↑ 5d
state: 5p
5s

Excited ↑↓ ↑↑↑ ↑↑
state: 5p 5d
5s

Hybrid ↑↓ ↑↑↑↑ ↑
state: sp3d2orbitals

5d

The six sp3d2 hybrid orbitals are produced by the hybridisation of one 5s, three 5p and two
5d orbitals of the valence shell of the iodine atom.

Valence orbital diagram in fluorine, F:

Ground ↑↓ ↑↓ ↑↓ ↑
state: 2p
2s

Each I-F bond is formed by the overlapping of one sp3d2 orbital of the iodine atom with one
2p orbital of the fluorine atom.

The results of these overlapping formed five σ bonds.
One of the sp3d2 orbitals of the iodine atom is occupied by lone pair electrons.

< 90°

Electron pair arrangement: Octahedral
Molecular geometry: Square pyramidal

7 (a) PV = nRT

n of C2H6 =

=

= 0.012 mol (available)
PV = nRT

n of O2 =

=

= 0.0497 mol (available)
From balanced equation ;

2 mol C2H6 7 mol O2
0.012 mol C2H6 0.012 x 7/2

= 0.042 mol O2 (needed)
moles of O2 needed < moles of O2 available

Limiting reactant is C2H6

nO2 remained = 0.0497 mol – 0.042 mol

= 0.0077 mol
From balanced equation;

2 mol C2H6 4 mol CO2
0.012 mol C2H6 = 0.024 mol CO2
2 mol C2H6 6 mol H2O
0.012 mol C2H6 = 0.036 mol H2O

nT = nO2 + nCO2 n+ H2O

= 0.0077 + 0.024 + 0.036
= 0.0677 mol

PT = nTRT

V

= 0.0677 mol x 0.08206 Latmmol-1K-1 x 300.15K
7.5 L

= 0.2223 atm x 760mmHg
1 atm

= 168.948 mmHg

Kp = Kc (RT)n
(b)

Kc =

= 6.4 x 10-6______
(0.08206 x 500.15)5-4

= 1.5594 x 10-7

[ Cl2 ] = 0.50 mol
2.0 L

= 0.25 M

[H2O] = 0.40 mol
2.0 L

= 0.20 M

[ HCl] = 0.50 mol
2.0 L

= 0.25 M

[O2] = 0.015 mol
2.0 L

= 0.0075M

Qc = [HCl]4[O2]
[Cl2]2[H2O]2

= (0.25)4(0.0075)
(0.25)2(0.2)2

= 0.0117

- Qc > Kc
- The reaction is not at equilibrium.
- The equilibrium position will shift to left to reestablish the equilibrium.

8 i) HNO2(aq) + H2O (l) NO2- (aq) + H3O + (aq)

Dissociation of weak acid, HNO2

ii) HNO2 (aq) + NaOH (aq) NaNO2 (aq) + H2O (l)

Neutralisation reaction

iii) HNO2 (aq) + NaNO2 (aq)

Acidic buffer solution

20.0ml of 0.08M HNO2 was added to 40.0ml of water.
Mol of HNO2 = 0.08 x 0.02 = 1.6 x 10-3 mol
[HNO2] = 1.6 x 10-3/ 0.06 = 0.0266 M

20.0ml of 0.08M HNO2 + 40.0 ml of 0.03M NaOH.
20.0ml of 0.08M HNO2 + to 40.0ml 0.03 M NaNO2.

SK 016/2

SUGGESTED ANSWER PSPM I 2017/2018
SECTION A [60 marks]

1 (a) A 25.0 mL of 0.050 M silver nitrate solution is mixed with 25.0 mL of
0.050 M of calcium bromide solution to give 0.105 g of solid silver bromide.
(i) Write the balanced chemical equation for this reaction.

2AgNO3 + CaBr2 2AgBr + Ca(NO3)2

(ii) Determine the limiting reactant.

n AgNO3 = 0.05 x 25
1000

 1.25 x103 mol

From equation,
n AgNO3 ≡ n CaBr2

2 mol AgNO3 ≡ 1 mol CaBr2
1.25 x 10-3 mol AgNO3 ≡ 1.25 x103 mol AgNO3 x 1mol CaBr2

2 mol AgNO3
= 6.25 x 10-4 mol CaBr2
number of mol CaBr2 needed < number of mol CaBr2 available
limiting reactant is AgNO3
(iii) Calculate the percentage yield of silver bromide.

From equation,
2 mol AgNO3 ≡ 2 mol AgBr

1.25 x 10-3 mol AgNO3 ≡ 1.25 x 10-3 mol AgBr
mass AgBr = 1.25 x 10-3 mol x 187.8 gmol-1

= 0.2348 g

% yield  actualyield x 100 [9 marks]
theoritical yield

 0.105 x 100
0.2348

 44.72%

1

SK 016/2

(b) An organic compound contains only carbon, hydrogen and oxygen with a
mass composition of 41.40% C, 3.47% H and 55.13% O. If 0.05 mol of
this compound weighs 5.80 g, determine its
(i) empirical formula

% mass C H O
Moles
ratio 41.40 3.47 55.13
41.40  3.45 3.47  3.47 55.13  3.45

12 1 16
1 1 1

Empirical formula is CHO

(ii) molecular formula
(CHO)n = molar mass

(12n  n  16n)  5.80 g
0.05 mol

29n 116
n4

Molecular formula is C4H4O4

[6 marks]

2 (a) The azide ion, N3 - , exists in several resonance forms.
(i) Draw possible resonance structures for the azide ion.

(-1) (-1) (0) (-2) (-2) (0)
N- N+ N- N N+ N2- N2-N+ N
(-1)
Structure 2 Structure 3
Structure 1

(ii) Use formal charges to select the most stable structure.
Structure 1.

2

SK 016/2

(iii) Using Lewis structure, explain why N3 – ion exists, whereas
trifluoride ion, F3 - , does not exist.

F F F0

F is in period 2, central atom cannot form expanded octet, so
F3- ion does not exist.
N is in period 2, central atom obey octet rule, so N3- ion exists.

[9 marks]

(b) Chloroform, CHCl3, is a common organic solvent. If H in CHCl3 is replaced
by Cl, it becomes CCl4, a toxic solvent. For each of CHCl3 and CCl4
compound.

(i) draw the molecular shape.

CHCl3
No. of valence electron :

C= 4

H= 1

3 Cl = 3  7 = 21

Total = 26

Lewis structure :

H H
Cl C Cl C

Cl Cl Cl Cl

molecular geometry :

tetrahedral

3

SK 016/2

CCl4
No. of valence electron :

C= 4
4 Cl = 4  7 = 28
Total = 32
Lewis structure :

Cl  Cl
Cl C Cl Cl
C
Cl Cl Cl

molecular geometry : tetrahedral

(ii) show the bond polarity, H
CHCl3 C
Cl Cl Cl
CCl4
Cl
C
Cl Cl Cl

(iii) predict the polarity.

The magnitude of C-H dipole moment is not equal to C-Cl dipole
moment. The four dipole moments do not cancel out each other,
  0. Therefore CHCl3 is an asymmetry and a polar molecule.

4

SK 016/2

All C-Cl dipole moments are equal in magnitude. The four dipole
moments are arranged in tetrahedral shape causes them to cancel
out each other,  = 0. Therefore CCl4 molecule is a symmetry and a
non-polar molecule.

[6 marks]

3. A 3.0 L flask at 298 K contains a mixture of 2.0 mol helium gas and 3.0 mol
xenon gas. The van der Waals constants for helium and xenon are given in
TABLE 3.

TABLE 3

Gas a (L2 atm mol-2) b (L mol-1)

Helium 0.03421 0.02370

Xenon 4.194 0.05105

Given the van der Waals equation: ( + 22 ) ( − ) =

(a) State two (2) postulates of the kinetic molecular theory of an ideal gas in
relationship to parameters a and b as in TABLE 3.
[2 marks]

Kinetic molecular theory of an ideal gas in relationship to:

parameter a -The attractive and repulsive forces between gas molecules
are negligible.
@
The intermolecular forces are negligible.

parameter b -The combined volume of all molecules of the gas is
negligible relative to volume of the gas container.

@
The total volume of all gas molecules is negligible compared to the volume
in which the gas is contained.

(b) Which gas is expected to behave ideally and which gas exhibits a marked
deviation from ideal behaviour?

[2 marks]

5

SK 016/2

Helium is expected to behave ideally while xenon exhibits a marked
deviation from ideal behaviour.

(c) Explain your answer in 3(b) by referring to the significance of a and b
values in TABLE 3.
[5 marks]

Helium has smaller value of a which means the attractive and repulsive
forces between helium atoms are very weak and almost negligible. Helium
also has a smaller value of b which means the combined volume of all
helium atoms is negligible compared to the volume of the gas container.
Therefore, helium behaves almost like an ideal gas.

Whereas, xenon has a bigger value of a which means the attractive and
repulsive forces between xenon atoms are stronger and significant. A
bigger value of b means that the combined volume of all xenon atoms is
significant compared to the volume of the gas container. Therefore, xenon
deviates from ideal gas behaviour.

(d) Calculate total pressure of the gas in the mixture using an ideal gas
equation.
[3 marks]

PT = nTRT

V

= (2.0  3.0)mol  0.08206L atmmol1 K1  298K
3.0 L

= 41 atm
@
PHe = nHeRT

V

= 2.0 mol  0.08206L atmmol1 K1  298K
3.0 L

= 16.3 atm
PXe = nXeRT

V

= 3.0mol  0.08206L atmmol1 K1  298K
3.0 L

6

SK 016/2

= 24.5 atm
PT = 16.3 atm + 24.5 atm

= 41 atm

(e) Using the van der Waals equation, the total pressure of the system is
calculated to be 38.1 atm. Explain the reasons behind the differences
between this value and the pressure calculated in 3(d).
[3 marks]

Pressure of a gas is the result of the collisions between the gas molecules
with the wall of their container. The higher the frequency and impact of the
collisions, the higher the gas pressure.

Helium and xenon are real gases with weak attractive forces between
atoms. These forces lessen the impact of collision of a given atom with the
wall of the container. Therefore, the actual pressure calculated using the
van der Waals equation is lower than the pressure calculated using the
ideal gas equation.

The combined volume of gas molecules is NOT the reason in this case
because bigger gas molecules reduces the unoccupied volume and
therefore increases the gas pressure.

4 (a) In a series of acids of the halogen group, hydroiodic acid, HI, is the
strongest acid while hydrofluoric acid, HF, is the weakest.
(i) Explain what is meant by strong and weak acids.

Strong acid is an acid that dissociates completely, while weak acid
is an acid that dissociates partially.

(ii) Briefly explain the main factor in determining the strength of the
acid.

The polarity of the bond formed between atom F and H. The weaker
the bond, the easily it can dissociates to produce high concentration
of H3O+.

(iii) Calculate the degree of dissociation,  of 0.20 M HF solution.
[Ka = 6.8 x 10-4 at 25C]
[8 marks]

7

SK 016/2

[ ] / HF (aq) + H2O (l) ⇌ F- (aq) + H3O+ (aq)
M

I 0.20

C -x +x +x
E 0.20 – x xx

= [ 3 +][ −]
[ ]

6.8 × 10−4 = ( )2
(0.20− )

x1 = 0.01133 , x2 = -0.01200 (rejected)

degree of dissociation,  =

0.20

= 0.01133

0.20

= 0.057

(b) Acetic acid, CH3COOH, is a weak acid with Ka = 1.8 x 10-5. It dissociates
in water according to the following equation:

CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)

(i) Rewrite the above equation and label the conjugate acid and
conjugate base.

CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)

acid base conjugate base conjugate acid

(ii) Determine the pH of 0.10 M.

[ ]/M CH3COOH (aq) + H2O (l) ⇌ CH3COO- (aq) + H3O+ (aq)
I
C 0.10 0 0
E
-x +x +x

0.10 – x xX

8

SK 016/2

= [ 3 −][ 3 +]
[ 3 ]

1.8 × 10−5 = ( )2
(0.10− )

Since Ka is very small, assume 0.10 – x  0.10

1.8 × 10−5 = ( )2
0.10

x = 1.34 x 10-3 M = [H3O+]

pH = - log [H3O+]
= - log 1.34 x 10-3
= 2.87

(iii) Predict the percentage of dissociation when the acid concentration
is increased.
Percent of dissociation decrease when the acid concentration is
increased.

[7 marks]

SECTION B [40 marks]

5 (a) Sketch the energy level diagram to show the electronic transitions which
give rise to the first five lines in the Lyman series of the hydrogen atom.
Explain.

[10 marks]

9

Energy level diagram SK 016/2

energ n=6
y n=5
n=4
n=3
n=2

n=1

Line spectrum of Lyman series.

When energy is supplied to an atom, electron at the ground state will
absorb the energy. This electron will excite to higher energy level and the
electron is now at its excited state. The electron is unstable and will fall
back to lower energy level, n = 1 by releasing specific amount of energy
with specific wavelength in the form of photon. The photon released is
passed through a prism and fall on the photograph plate to be recorded as
line spectrum.

10

SK 016/2

5 (b) Element Y is in Period 3 of the Periodic Table. The first six successive
ionization energies of element Y are given in Table 5.

Ionization First TABLE 5 Fourth Fifth Sixth
4963 6274 21267
energy 1011 Second Third
(kJ mol-1) 1907 2914

Determine the group for Y and explain your answer. Write the valence
electronic configuration and the valence orbital diagram of Y. Explain how
the Pauli exclusion principle, Aufbau principle and Hund’s rule are applied
in drawing the valence orbital diagram of element Y.

[10 marks]

• Sharp increase occur at IE6 (IE6>>IE5).
• The sixth electron is removed from inner shell.
• Element Y has 5 valence electron.
• Element Y in group 15
• Valence electronic configuration of Y : 3s2 3p3
• Valence orbital diagram :

• Pauli exclusion principle state that no two electrons can have same set
of four quantum number. Hence, only 2 electrons are located at same
orientation with different spin.

• Aufbau principle state that in the ground state of an atom or ion,
electrons fill atomic orbitals of the lowest energy level before higher
energy level. Energy level 3s orbital is lower than 3p orbital.

• Hund’s rule state that when electron are filled on degenerate orbital
(such as 3p orbital), electron are filled singly with same spin before its
paired.

11

SK 016/2

6 (a) Sodium bicarbonate, NaHCO3 is used in baking as raising agent. When
dissolved in water, the salt dissociates into ions. Write the Lewis structure
for the bicarbonate ion, HOCO2- , and draw its resonance structures.

Lewis structure:

-

O
H OCO

Resonance Structure:

- -

O O
H OCO H OCO

Determine the shape of the HOCO2- ion and the hybridization of central
atoms. Draw and label the overlapping of the orbitals to show the
formation of the covalent bonds.

[12 marks]

Molecular Geometry for the HOCO2- ion = Trigonal planar

Type of hybridisation:

2 -

O

H OCO

1

i. O1 = sp 3
ii. O2 = sp2
iii. C = sp2

12

SK 016/2

Orbital overlapping:

H=
1s

O3 = 2p
2s

O2

GS= 2p
2s

ES = 2p
2s

ES =

SP2 2p

O1

GS= 2p
2s

ES = 2p
2s

ES =

SP3

C

GS= 2p
2s

ES = 2p
2s

ES =

SP2 2p

13

SK 016/2

Orbital Overlaping:

SP2 SP2
2p
O

SP2

 
SP2
O
SP3 SP3 2p C SP2
SP3 SP2 2p
O 

 SP3

H
1s

(b) Metallic compounds have some physical properties which are different
from covalent compounds. State four (4) properties and relate these
properties to metallic bonding.

[8 marks]

Four properties of metallic compound:

i) High melting and boiling point

Metallic compounds have strong electrostatic attraction between positively
charged metal ion and negatively charge electron sea. More energy
needed to break this attraction.

ii) Able to conduct electricity

Metallic compounds have free moving electrons.

iii) Malleable

Delocalized electrons enable the metal atoms to roll over each other.
Metal atoms only slide to each other.

iv) Luster

Photons of light do not penetrate very far into the surface of a metal and
are typically reflected on the metallic surface.

14

SK 016/2

7 (a) Carbon dioxide can be used to extract caffeine under supercritical condition.
Some of the physical properties of carbon dioxide are shown below.

Triple point temperature -51°C
Triple point pressure 5 atm
Critical temperature 31°C
Critical pressure 73 atm

What is meant by triple point and critical point?

Triple point: Temperature and pressure at which solid, liquid and gas
simultaneously exist in equilibrium.
Critical Temperature and pressure at which gaseous phase and
point: liquid phase are indistinguishable.

Using the data given above, sketch a labeled phase diagram for carbon dioxide.
Show the supercritical point in the diagram.

Phase diagram of carbon dioxide

Pressure (atm) Supercritical
point

73

5
1

-78.5 -51 31
15 Temperature (°C)

SK 016/2

Using the phase diagram, under what conditions can dry ice be formed to act as
coolants?
Dry ice can be formed to act as coolants under atmospheric pressure (1 atm)
and the temperature of -78.5°C. Under these conditions, dry ice breaks down
and turns directly into carbon dioxide gas rather than a liquid. The super-cold
temperature and the sublimation feature make dry ice great for refrigeration or
as coolants. For example, if we want to send something frozen across the
country, we can pack it in dry ice. It will be frozen when it reaches its
destination, and there will be no messy liquid left over like we would have with
normal ice.

[10 marks]
(b) In an experiment, 0.10 mol of N2O4 and 0.55 mol of NO2 are mixed in a closed

vessel at 350 K with a total pressure of 2.0 atm. At this temperature, the
equilibrium constant, Kp is 3.89. The reaction is given as follows:

N2O4 (g) ⇌ 2NO2 (g)
In which direction will the reaction proceed to reach equilibrium?

16

SK 016/2

P  X  PN2O4
N 2O4 total

 n N 2O4  Ptotal
ntotal

 0.1 mol   2.0 atm
0.10  0.55 mol

 0.3077 atm

PNO2  X NO2  Ptotal

  1  X N2O4  Ptotal

 1  0.1538 2.0 atm

 1.6923 atm

 QP P2  1.69232  9.31
NO2 0.3077 

PN 2O4

Since QP > KP:
 The system is not at equilibrium
 Initially, there are more NO2 in the reaction mixture
 To reach equilibrium QP = KP, the reaction will shift backward

If the above experiment is carried out at 500 K, the new KP is 1700. Is the
reaction exothermic or endothermic? Explain.

 At higher temperature, the value of KP increases

 The system consumes more N2O4 and produces more NO2

 This indicates that at higher temperature, the equilibrium system is

disturbed

 According to Le Chatelier’s principle, equilibrium position will be shifted to

the right as to lower the temperature of the system by absorbing the

added heat

 Therefore, the forward reaction is endothermic [10 marks]

17

SK 016/2

8 (a) Hypochlorous acid, HOCl, is a monoprotic acid. An aqueous solution of
0.028 M HOCl has a pH of 4.5. Calculate the Ka value for this acid.
A buffer solution is prepared by mixing 100 mL of 0.04 M HOCl with 100
mL of 0.02 M NaOCl. Calculate the pH of the solution. Explain how a
mixture of HOCl and sodium hypochlorite, NaOCl, solution behaves as a
buffer solution when a small amount of strong acid is added.
[12 marks]

HOCl(aq) + H2O(l) OCl- (aq) + H3O+(aq)

[ ]i 0.028 - 0 0
[ ]c -x - +x +x
[ ]f 0.028 – x - x
x

pH = - log [H3O+]
4.5 = - log x

x = 3.16 x 10-5 M

[HOCl] = 0.02797M
[OCl-] = [H3O+] = 3.16 x 10-5 M

 Ka  OCl - H3O
[HOCl]
   3.16 x105 3.16 x105
0.02797
 3.57 x 108

n HOCl = 0.040.1 = 4 x 10-3 mol

[HOCl] = 4 x 103  0.02M
0.2

n NaOCl = 0.020.1 = 2 x 10-3 mol

[NaOCl] = 2 x 103  0.01M
0.2

18

SK 016/2

OCl  

pH = - log Ka + log HOCl

= - log 3.57x 10-8 + log 0.01
0.02

= 7.15

Buffer solution contain HOCl act as an acid and OCl- that acts as a base.
When small amount of strong acid is added to it, OCl- will neutralized it.

Thus, the pH is not much affected.

OCl- (aq) + H3O+(aq) HOCl(aq)

(b) The solubility product, Ksp for calcium sulphate, CuSO4, is 2.0 x 10-5 at 250C.
Calculate the molar solubility of CuSO4 in water and in 0.10 M sodium sulphate,
Na2SO4. State common ion effect on the solubility of CaSO4.
[8 marks]

CaSO4(s) Ca2+(aq) + SO42-(aq)

xx

Ksp = [Ca2+][SO42-]

2  105 = x2

x = 4.47 x 10-3 M

Molar solubility in water is 4.47 x 10-3 M

CaSO4(s) Ca2+(aq) + SO42-(aq)

y y + 0.10 M

Ksp = [Ca2+][SO42-]

2  105 = (y)( y + 0.10) ; assume x<< 0.10, y+ 0.10 = 0.10

2  105 = (y)( 0.10)

y = 2 x 10-4 M

Molar solubility in Na2SO4 is 2 x 10-4 M

19

SK 016/2
 Solubility of CaSO4 in pure water higher than in 0.10 M Na2SO4 solution,

SO42- (common ion) is present.
 The equilibrium position shifts backward.
 Solubility of CaSO4 decreases.
 The present of common ion, SO42- reduces the solubility of CaSO4.
.

20

SUGGESTED ANSWER PSPM I
20182019

1. a) Bromine has proton number of 35. FIGURE below shows a mass spectrum of
bromine.

i. Write the notations for all isotopes of bromine.

79 Br and 81 Br
35 35

ii. Calculate the relative atomic mass of bromine.

(isotopic mass  abundance)

Average atomic mass =

 abundance

7951  81 49
= 49  51

= 79.98
∴ relative atomic mass = 79.98

[4 marks]

b) A reagent bottle contains a stock solution of 0.90% by mass of sodium chloride,
NaCl. The density of the solution is 1.00 g cm-3. Calculate
i. The mole fraction of NaCl.
Assume mass of solution = 100 g
Mass of NaCl = 0.90 g

Mole of NaCl = 0.90 g = 0.0154 mol
58.5 g mol

Mass of H2O = 100 g – 0.90 g = 99.1 g

Mole of H2O = 99.1 g = 5.506 mol

18 g mol

∴Mole fraction of NaCl = 0.0154 mol mol

0.0154  5.506

= 2.79×10-3

1

SUGGESTED ANSWER PSPM I
20182019

ii. The molality of NaCl solution.

Molality = Mole of NaCl (mol)
mass of solvent (kg)
0.0154 mol

=

0.0991 kg

= 0.155 m

iii. The volume of the stock solution required to prepare 100 mL of 0.01 M NaCl
solution.

Mass of solution (g)
Density of solution = volume of solution (cm3)

1.00 g cm-3 = 100 g
volume of solution (cm3)

Volume of solution = 100 cm3

Molarity = Mole of NaCl (mol)
Volume of solution (L)

= 0.0154 mol
0.1 L

= 0.154 M

Mole of diluted NaCl solution = (0.1 L)(0.01 M)
= 1×10-3 mol

1×10-3 mol

∴Volume NaCl solution needed =

0.154 M

= 6.49×10-3 L

[9 marks]

c) Silicon tetrachloride, SiCl4 can be prepared by heating silicon in excess chlorine gas.

Si(s)  2Cl2(g)  SiCl4(l)

i. Calculate the mass of silicon needed to produce 400 g SiCl4 if the percentage
yield is 42.5%.

42.5% = 400 g 100

theoretical yield

Theoretical yield (mass of SiCl4) = 941.18 g

Mole of SiCl4 = 941.18 g = 5.533 mol

170.1 g mol

1 mol SiCl4 ≡ 1 mol Si
5.533 mol SiCl4 ≡ 5.533 mol Si

∴Mass of Si = (5.533 mol)(28.1 g/mol) = 155.5 g
2

SUGGESTED ANSWER PSPM I
20182019

ii. If 15 mol of chlorine is used, determine the amount (mole) of unreacted
chlorine.
1 mol SiCl4 ≡ 2 mol Cl2
5.533 mol SiCl4 ≡ 11.066 mol Cl2
∴ unreacted Cl2 = (15 - 11.066) mol = 3.934 mol
[8 marks]

3

SUGGESTED ANSWER PSPM I
20182019

2. a) The wavelength that produces a line, B in Brackett series is 2165.6 nm.
i. Determine the transition that forms the B line.

1  RH  1  1   n2
  n12 n22  , n1
 

1  1.097 107 m1  1  1 
2.1656106 m  42 n22 
 

n1 = 7

∴n =7 to n=4

ii. Calculate the energy emitted for the transition.

E  RH  1  1 
 ni2 n2f 

= 2.18 1018 J  1  1 
 72 42 

= - 9.18×10-20 J

iii. Another line, C was form with wavelength of 1817.5 nm. Explain
qualitatively, whether line B or line C, has higher energy emitted.
 Line C has higher energy emitted than line B.
 Energy is inversely proportional to wavelength.
 Line C has smaller wavelength than line B.
[8 marks]

b) The electronic configuration of element D is 1s2 2s2 2p6 3s2 3p3.
i. Give a set of quantum number for the 9th electron.
(n=2, l=1, m=0, s=+1/2)

ii. Draw the orbital diagram of the valence electrons.

3s 3p

iii. Draw and label the 3D shape of orbitals occupied by the valence electrons.
zz

xy xy
3s 3px

4

SUGGESTED ANSWER PSPM I
20182019

zz

x yx y
3py 3pz

iv. Explain the filling of the valence electrons in 2(b)(iii) according to the
appropriate rule(s)/principle(s).
 Aufbau principle.
 Electrons added to the lower energy orbital first. The valence electrons
are fill the 3s orbital before 3p orbital. 3s orbital has lower energy than
3p orbital.
 Hund’s rule.
 When electrons are, add into degenerate orbitals, each orbitals are,
filled singly with electron of the same spin before it is pair. 3p orbitals
are degenerate orbitals. Therefore, the remaining 3 valence electrons is
filled singly in each orbital with the same spin.
 Pauli’s Exclusion Principle.
 No two electrons can have same set of quantum number. Only 2
electrons can occupy the same atomic orbitals, thus 2 electrons must be
filled in 3s orbital with opposite spin.
[13 marks]

5

SUGGESTED ANSWER PSPM I
20182019

3. a) Ammonia, NH3 and boron triflouride, BF3 are covalent compounds. NH3 and BF3
react to form H3NBF3 molecule.
i. Explain why NH3 obeys octet rule but BF3 does not.

HNH HBH

HH
 N in NH3 is surrounded by 8 valence electrons while B in BH3 is

surrounded by only 6 valence electrons.
 BH3 is incomplete octet.

ii. Show the formation of H3NBF3 molecule using Lewis dot symbol and label

the bond formed.

HH HH

+H N BH HN BH

HH HH
Dative bond

[5 marks]

b) Oxygen difluoride, OF2 is a strongly oxidizing colourless gas.
i. Determine the molecular geometry of this molecule.

FO

F

 Number of electron pair around oxygen atom: 4
 Electron pair arrangement: tetrahedral
 2 bonding pair and 2 lone pair.
 Based on VSEPR theory, the electrons pair will be located as far as

possible to minimize the repulsion among them. The lone pair-lone pair
repulsion > lone pair-bonding pair repulsion > bonding pair-bonding pair
repulsion.
 Molecular geometry: V-shape

FO

F

ii. Explain whether OF2 is a polar or non-polar molecule.
 Polar molecule.
 F is more electronegative than O. Therefore, O-F bond is polar.

FO

F

 Since, molecule OF2 is unsymmetrical, thus, dipole moment can’t cancel
each other, μ≠0.
[8 marks]

6

SUGGESTED ANSWER PSPM I
20182019

c) Aluminium and sodium are metals.
i. Explain the formation of metallic bond in sodium using the electron sea
model.

Na+ Na+ Na+ Na+ Na+
e e e e ee

Na+ Na+ Na+ Na+ Na+
ee e eee

Na+ Na+ Na+ Na+ Na+
e ee

 Metallic bond form in Na metal is the electrostatic forces between
positively charge ion, Na+ and negatively charge free moving electrons.

ii. Why aluminium has higher boiling point than sodium?
 Aluminium ion, Al3+ is smaller than sodium ion, Na+. Thus, Al3+ has
stronger nucleus attraction towards the free moving electrons.
 Al has 3 valence electrons and sodium has only 1 valence electron.
 Strength of metallic bond is directly proportional to the number of
valence electron and inversely proportional to size of ion. Thus, Al has
stronger metallic bond than Na.
[4 marks]

4. a) A 10-L cylinder contains 4 g of hydrogen gas and 28 g of nitrogen gas. If the
temperature is 31oC,
i. Determine the total pressure of the gas mixture.
Mole of hydrogen gas = 2 mol
Mole of nitrogen gas = 1 mol
Total mol = 3 mol

PT = nT RT
V

 3 mol 0.08206 atm L mol-1 K-1 304.15 K

= 10 L

= 7.49 atm

ii. Calculate the partial pressure of hydrogen gas.

PH2  X H2 PT

  2   7.49 atm
 3 

= 4.99 atm

iii. What will happen if the gaseous mixture is heated to 550oC?

 At high temperature, the average kinetic energy of gas particles will

increase and cause the reaction between hydrogen and nitrogen to

occur producing gas ammonia.

7