JAWAPAN PSPM 20122013

Suggested Answer

Peperiksaan Semester
Program Matrikulasi

PSPM

SUGGESTED ANSWER PSPM 1
SESI 2012/2013

1. a) TABLE 4 shows the atomic structure of four particles, represented by the

letter A to D. The particles are atoms and ions.

TABLE 4

Particle Electrons Protons Neutrons

A6 66

B 12 12 12

C 10 12 12

D6 68

Use the letter A to D to answer the following questions.
i. Which two particles are atom and ion of the same element? Explain.

B and C
Same number of protons and neutrons but different number of
electrons.

ii. Which two particles are isotopes of the same element? Explain.

A and D
Same number of electrons and protons but different number of neutrons
or nucleon.

b) Compound E is a hydrocarbon, CxHy. When 6.84 g of this compound is burnt

completely in pure oxygen, 21.5 g of CO2 and 8.87 g of H2O are obtained. Determine

the empirical formula of compound E. If the molar mass of the compound is 128.2

gmol-1, what is its molecular formula? [6 m]

Elements C H
Mass(g) 5.86 0.98
Number of mole 0.4887 mol 0.98 mol
Mole ratio
1 2

Empirical formula = CH2 17

[CH2]n = 128
n=9

Molecular formula = C9H18

b) The reaction between the solutions of permanganate ion, MnO4-, and oxalate ion,
C2O42-, in acidic medium is shown below:

MnO4-(aq) + C2O42-(aq) MnO2(s) + CO2(g)

Write:
i. the half-reactions for the reduction and oxidation reactions respectively.

Reduction:

MnO4- + 4H+ + 3e MnO2 + 2H2O

Oxidation:

C2O42- 2CO2 + 2e

ii. the balanced redox equation for the reaction.

2 x [MnO4- + 4H+ + 3e MnO2 + 2H2O]

3 x [C2O42- 2CO2 + 2e]

2MnO4- + 3C2O42- + 8H+ 2MnO2 + 6CO2 + 4H2O

2. a) For each of the molecules given; PCl3, PCl5 and POCl3,
i. draw a Lewis structure and name the shape based on molecular geometry.

Lew is Structure: Lew is Structure: Lew is Structure:
Cl
Cl Cl
Cl P Cl Cl Cl P Cl
Molecular Geometry: O
P Cl
Trigonal pyramidal Cl Cl Molecular Geometry:

Molecular Geometry: tetrahedral

trigonal bipyramidal

ii. deduce the polarity and predict the Cl-P-Cl bond angle of each molecules.
PCl3: less than 109.5o, polar
PCl5: 120o, non-polar
POCl3: 109.5o, polar

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b) Show the formation of a dative bond in the reaction of;

AlCl3 + Cl- AlCl4-

Cl - Cl -
+Al Cl Cl Al Cl dative bond
Cl
Cl
Cl

3. a) In the offshore base oil drilling operation, methane gas, CH4, was found at the
bottom of the sea at 480oC and a pressure of 12.8 atm.

i. calculate the volume of the gas needed to be transferred to a container in the
refinery mill with a volume of 1.80 x 103 L at a pressure of 1.0 atm and
temperature of 22oC.

( )( )

= 359 L

ii. determine the mass of the gas being transferred.

( )( )( )
( )( )

b) i. define vapour pressure and boiling point.

Vapour pressure:
Pressure exerted by vapour molecules to it liquid at equilibrium in a
sealed container.

Boiling Point:
Temperature at which the vapour pressure of the liquid is equal to the
external atmospheric pressure.

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ii. based on TABLE 5, arrange the compounds in order of increasing boiling
point. Explain your answer.
TABLE 5
Compound
Propane, CH3CH2CH3
Ethanol, CH3CH2OH
Dimethyl ether, CH3OCH3
Ethanoic acid, CH3COOH

CH3CH2CH3 < CH3OCH3 < CH3CH2OH < CH3COOH
 Strength of intermolecular forces directly proportional to boiling
point.
 Propane is non-polar molecule so it can form weak London
Dispersion forces between its molecules.
 Dimethyl ether is polar molecule, so it can form dipole-dipole
forces between its molecule.
 Dipole-dipole forces stronger than London dispersion forces.
 Both ethanol and ethanoic acid can form hydrogen bond.
 Hydrogen bond stronger than Van der Waals forces.
 Ethanoic acid can form more hydrogen bond compare to ethanol.

4. a) What is meant by end point and equivalence point?
End point:
pH in titration at which the indicator change colour.

Equivalence point:
The pH at which the no. of mole of H+ is stoichiometrically equivalence to the no.
of mole of OH-.

b) A titration of 25.0 mL solution of 0.15 M HNO3 requires 37.5 mL of NaOH to reach the
equivalence point.
i. calculate the pH of the solution before the addition of NaOH.

HNO3(aq) H+(aq) + NO3-(aq)
[ ]i 0.15 --
[ ]f -
0.15 0.15

pH = - log [H+]
= - log (0.15)
= 0.8239

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iii. calculate the concentration of NaOH solution needed to neutralize the HNO3
solution.

( )( ) ()

c) A solution contains 0.175 M manganese(II) ions and 0.182 M lead(II) ions. The Ksp
for manganese(II) sulphide is 1.4 x 10-15 and for lead(II) sulphide is 8.0 x 10-28.

i. Write the solubility product expression for both sulphides.

[ ][ ] [ ][ ]

ii. Calculate the maximum concentration of sulphide ions present in the solution
without manganese ions being precipitated.

[ ][ ]

PbS(s) Pb2+(aq) + S2-(aq)
ss

[ ][ ]

[ ][ ]
()
M

5. a) Describe the formation of emission spectrum for hydrogen atom. Show and
label the first three series of electron transitions between energy levels.
An electron in a hydrogen atom is transferred from n=5 to n=3. Calculate the energy
of the photon emitted and the wavelength of the spectral line produced.

When energy supplied, the electron at the ground state (n=1) absorbed energy. It
will excite from lower energy level to a higher energy level. Electron at the
excited state is unstable. It will fall back to lower energy level, n=1 and release
specific amount of energy with specific wavelength. Hence the line spectrum
formed.

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energy

Paschen n=7
n=6
Balmer n=5
n=4
n=3

n=2

Lyman n=1

()
()

Energy emitted is

( )( )
()

b) An element Q has proton number of 8. Draw the orbital diagram of the element.
Explain the two rules applied in arranging the electrons in the orbitals.

8Q: 2p
1s 2s

Aufbau principle:
Electrons are filled into the orbitals according to increase energy level.

Hund’s rule:
The 4 electrons in 2p orbital are spread out with parallel spin before pairing up.

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d) Given four elements W, X, Y and Z with their proton numbers of 11, 12, 13 and 18
respectively. Identify the group and period of the elements in the Periodic Table.
Arrange the elements according to increasing atomic radii and explain the trend.

Elements Electronic configuration Group Period
W 1s2 2s2 2p6 3s1 1 3
X 1s2 2s2 2p6 3s2 2 3
Y 13 3
Z 1s2 2s2 2p6 3s2 3p1 18 3
1s2 2s2 2p6 3s2 3p6

Increasing order of atomic radii: Z < Y < X < W

W, X, Y, Z are elements located in the same period.
Across period, the proton no. increase from W to Z causes the effective nuclear
charge increases from W to Z, the nucleus attraction towards the valence
electrons increase from W to Z. Therefore, the atomic radii increase from left to
the right across the same period.

6. a) By using Lewis dot symbol, show the formation of the compounds, BCl3 and
LiCl, from their respective atoms. Identify the types of bond in each compound.
Explain the hybridization state of the central atom in the BCl3 molecule. State its
molecular geometry.

B+ Cl Cl
Cl B Cl

Cl

Cl

+Li Cl Li+ -
Cl

Type of bond:
BCl3 : Covalent Bond
LiCl : Ionic bond

BCl3 : There are 3 bonding pairs / 3 σ bonds. Thus, type of hybridization is
sp2 23

B (ground state) : 2p
2s

B (excited state) : 2p
2s

B (hybrid state) :
sp2 2p

Molecular geometry : Trigonal Planar

3p

Cl 

sp2 

sp2 3p
sp2 B Cl



Cl
3p

b) Formic acid is widely used in the preservation of biological specimens. Its chemical
structure is shown below:

O

H C OH

Draw and describe the formation of bonds by involving the overlapping of orbitals.
What is the possible intermolecular forces that exist between these acid molecules?

Type of hybrid: O1 O2: sp3
C: sp2 H COH

2

O1: sp2

C(ground state): 2p O1(ground state): 2p O2(ground state): 2p
2s 2s 2s

C(excited state): 2p O1(excited state): 2p O2(excited state): 2p
2s 2s 2s

C(hybrid state): p O1(hybrid state): O2(hybrid state):
sp2
sp2 p sp3

sp2 

O1 

sp2
sp2

sp2  sp3

sp2 sp3 sp3 sH

C O2 sp3 

sp2



s
H

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Intermolecular Forces: Hydrogen bond.
7. a) Carbon dioxide, CO2, has a triple triple point at 5.2 atm and -57oC and a critical

point at 72.9 atm and 31oC. Sketch a phase diagram of CO2.

Pressure (atm)

72.9 B C
Solid
liquid
5.2 T

A gas
Temperature (OC)

-57 31

Based on the diagram;
i. explain liquid CO2 can be obtained

Liquid CO2 can be obtained only at pressure higher than 5.2 atm.

ii. the formation of solid CO2 (dried ice) at room condition.
At room condition (1 atm, 28oC), CO2 exist in gas form. Impossible to
form solid at room conditions. Triple point below room temperature and
above 1 atm. To form solid, temperature must be reduced to below 57oC
and pressure must be increased to above 5.2 atm.

b) decomposition of ammonium hydrogen sulphide, NH4HS, on heating in a sealed flask
is an endothermic process.

NH4HS(s) H2S(g) + NH3(g)

If Kp for this reaction is 0.11 at 25oC, determine the partial pressure of NH3
and total pressure in the flask (atm) at equilibrium.
Based on the Kp value, calculate Kc for this reaction at the same temperature.

NH4HS(s) H2S(g) + NH3(g)
n(eq) xx

( )( )

25

Using Le Chatelier’s principle, explain how increasing the temperature would affect
the equilibrium.

When temperature increase, the equilibrium position will shift forward to absorb
more heat in order to overcome the high temperature.

What will happen to the pressure of NH3 if some H2S is removed from the flask?

When some of H2S is removed, equilibrium position will shift forward to produce
more H2S in order to increase the concentration of H2S and pressure of H2S will
increase.

8. a) Define a buffer solution.
Buffer solution is a mixture of salt and weak acid/weak base which can maintain
the pH of the solution when small amount of strong acid or strong base is added
to the solution.

b) Explain how the pH of a buffer solution made from a weak acid and its conjugate
base changes when;
i. the acid dissociation constant, Ka, of the weak acid increases.
When Ka increase, pH will decrease.

ii. the acid concentration is decreased relative to the concentration of its
conjugate base.
The ratio [Conj. Base]/[acid] will increase. The log term increase and pH
should increase.

c) calculate the pH of a buffer solution prepared by adding 5.0 g of sodium acetate,
CH3COONa, to 250 mL of 0.20 M acetic acid, CH3COOH. What is the pH of the
buffer solution after addition of 0.01 M sodium hydroxide, NaOH, solution? [Ka acetic
acid = 1.8 x 10-5]

[ ]

[CH3COOH] = 0.20 M ]
]
[
[

()
()

26

OH- + CH3COOH CH3COO- + H2O
0.2439
[ ]i 0.01 0.20 0.2539

[ ]f 0 0.19

[]
[]

()
()

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