SKEMA JAWAPAN

PEPERIKSAAN AKHIR TAHUN 2009

KIMIA

TINGKATAN 6 RENDAH

KERTAS 1

1 D 11 B 21 D 31 D 41 A

2 A 12 B 22 C 32 C 42 B

3 D 13 A 23 A 33 C 43 A

4 D 14 A 24 D 34 D 44 C

5 B 15 C 25 D 35 D 45 D

6 D 16 A 26 A 36 B 46 C

7 D 17 D 27 C 37 B 47 A

8 D 18 A 28 B 38 C 48 B

9 D 19 C 29 C 39 C 49 A

10 A 20 D 30 A 40 A 50 C

SKEMA JAWAPAN

PEPERIKSAAN AKHIR TAHUN 2009

KIMIA

TINGKATAN 6 RENDAH

KERTAS 2

1. (a) X � Graphite , Y � Fullerene , Z � Diamond [3 ]

(b) Z is denser. [1 ]

The carbon atoms in each sheet of X are separated by a large distance. [1 ]

(c) X is able to conduct electricity. [1 ]

Each carbon atom in X has one free delocalized electron which can move when an

electric field is applied. [1 ]

(d) sp2

[1 ]

(e) Carbon dioxide gas [1 ]

(f) C + O2 � CO2 [1 ]

Total : 10 marks 2.

(a) labeled correctly � [2 ]

(b) boiling point shown correctly � [1 ]

(c) O � triple point

B � critical point [2 ]

(d) At point X, only liquid and vapour exist in equiliubrium ;

At point O, all three phases of solid, liquid and vapour exist at equilibrium. [2]

(e) When pressure is increased, the average distance between the substance R is reduced and

substance R solidifies. [2 ]

(f) The melting point of substance R increases with increase in pressure. [1 ]

Total : 10 marks

solid liquid

gas

1.0

Boiling point

3. (a)(i) A gas which obeys ideal gas equation and gas laws. [1 ]

(ii) Volume of gas is negligible & no forces of attraction between the particles. [2 ]

(iii) Hydrogen / Helium [1 ]

(iv) Low pressure and high temperature [2 ]

(b)(i) Graph � [2]

(ii)

[2]

Total : 10 marks

4. (a)(i)

E1 E2 E3 E4 E5 E6 E7

Difference(kJ mol-1

) - 2362 1360 1466 2098 1675 1693

[1]

(ii) Group 1.

The large difference in the first and second ionisation energy shows that there is one

electron in the valence shell. [2]

(iii) The larger ionization energy difference between E4 and E5 indicates the extra stability of

the half filled p3 orbital. [1]

(b)(i) Potassium [1]

(ii) 4s1

[1]

(iii)

(c)(i) Higher [1]

(ii) The electron is removed from an inner quantum shell which is nearer to the

nucleus. [1]

(iii) 3s2 3p

6 [1]

Total : 10 marks

5. (a)(i) Avogadro's law — At a fixed temperature and pressure, the volume of a gas is

directly proportional to the molar amount of gas.

V α n [1]

(ii) V = nRT

P

= 1.0 x 8.31 x 273

1.01 x 105

= 0.02246 m3

= 22460 cm3

[2]

(iii) 22460 cm3. One mole of any gas has the same volume. [2]

(b)(i) PV = nRT

PV= mRT

MR

MR = mRT

PV

= 1.02 g x 8.31NmK-1

mol-1

x 273 K

1.01 x105 Nm

-2 x 196 x 10

-6 m

3

= 116.89 g mol-1

[5]

(ii) SOCl2 = 32 + 16 +2(35.5)

= 119 [2]

[1]

SOCl2 is not an ideal gas.

It has a rather large molecular volume.

Strong attractive forces exist between the molecules. [3]

Total : 15 marks

6. (a) In the ground state, the electron occupies the lowest energy level.

In this state, the atom is stable and does not radiate energy.

When an electrical discharge is applied, the electron absorbs energy and is

excited to higher energy levels.

The excited electron is unstable and thus drops to the ground state, in a single step or a

few steps.

The electron gives up a quantum of energy in the form of light. The light is emitted with

a frequency, v, given by the equation ∆E = hv

The line spectrum formed shows that the energy of the electron is quantised .

(limited to a discrete set of values).

With increase in frequency, the spacing between adjacent lines decrease.

This shows that the energy levels are closer to each other.

The further the energy levels are from the nucleus, the closer the levels are to

each other. [5]

[3]

[4]

At the converging limit, ∆v = 0.

From the graph, when ∆v = 0,

v = 32.6 x 1014

Hz

E = h . v . NA

1000 = (6.63 x 10-

34) x (32.6 x 10

14) x (6.02 x 10

23)

1 000

= 1 300 kJ mol-1

[3]

Total : 15 marks

7. (a) AlF3 has an ionic structure, with ionic bonding between A13+

and F- ions.

PF3 has a simple molecular structure with covalent bonding between P and F atoms

within the molecule, but weak van der Waals forces between the molecules. [3]

(b) The carboxyl group in ethanoic acid can form strong intermolecular hydrogen bonds.

Weak van der Waals forces exist between the ester molecules. [3]

(c) Carbon uses all four of its valence electrons to form two covalent bonds each with an

oxygen atom. The two electron groups are distributed in a linear arrangement.

In SO2, the sulphur atom has three electron groups (one lone pair of electrons and two

covalently bonded oxygen atoms). The electron group geometry is "trigonal planar" but

the molecular geometry is "V-shape". [3]

(d) The melting point of 2-methy1-2- propanol is higher because the molecule is more

compact and can thus be arranged closer and in a more orderly manner. Boiling point

does not depend on the arrangement of molecules. The boiling point of 1-butanol is higher

due to stronger Van der Waals forces because its molecules are more spread out and more

easily polarized. [3]

(e) Copper is a good conductor of electricity because the valence electrons are free to move in

the conduction band. Its conductivity decreases with increasing temperature because

increased vibrations in the crystal lattice impede the free movement of electrons in the

conduction band. The valency and conduction bands in silicon are separated by a small

energy gap. Hence, at room temperature, its electrical conductivity is low. Its conductivity

increases with increase in temperature because more electrons having sufficient energy

can cross the energy gap and move in the conduction band. [3]

Total : 15 marks

8. (a) first ionisation energy: minimum energy required to remove one mole of electron from

one mole of gaseous atom to form one mole of gaseous ion. [2 ]

(b) Ne : 1s2 2s

2 2p

6

F : 1s2 2s

2 2p

5

It is more difficult to remove one electron from Ne which is more stable. [3]

(c) O : 1s2

2s2 2p

4

N : 1s2 2s

2 2p

3

An electron from the p orbital of O can be removed easily compared to N, which has

partially filled orbitals. [3 ]

(d) Average difference between O, F and Ne is 385. So, C should be approximately

1400 – 385 = 1015 [3 ]

(e) (i) Group 14. [1]

� the first 4 ionisation energies show a steady increase, then there is a large

difference. [2]

(ii) ns2 np

2 [1]

Total : 15 marks

9 (a)(i)

tetrahedral pyramidal V-shaped

[3]

(ii) CH4 : London dispersion forces / TD – ID

NH3 , H2O : hydrogen bond, D-D [3]

(iii) -Repulsion from the lone pairs pushes the bond pairs nearer in ammonia and water.

-more repulsion in water which has 2 lone pairs [2]

(iv) Water.

� O is the most electronegative

� O-H bonds are the most polar

� .’. H bond � strongest

Each water molecule can form hydrogen bond to four other molecules since the O

atom has two lone pairs [2]

(b)(i) xx H +

H •x N x• H xx

x• H •x N x• H

H x•

H

ammonia ammonium ion [2]

(ii) Dative covalent bond or coordinate bond is a normal covalent bond in which both the

shared electrons are donated by one of the atoms [2]

(iii) All the N—H bonds are identical & indistinguishable. [1]

Total : 15 marks

10. (a)(i)

[4] (ii) electronegativity : a measure of its ability to attract a shared electron pair [2]

(b) (i) 120 [2 ]

(ii) Hydrogen bond [1]

� bond between H atom and a electronegative atom [3]

(iii) Molecules not dimerised and are hydrated by forming H-bonds with water molecules,

and there is some dissociation into ethanoate and oxonium ions. [3 ]

Prepared by :

……………………………….

(PARAMASIVAM A/L MUNUSAMY)

∂+ ∂-

H3C — Cl ∂+ ∂-

Cl — F

∂- H∂+

CH3 CH2 N

H∂+