JAWAPAN UPS SEM 1

Suggested Answer
Ujian Pertengahan Semester

UPS

1. a) i. SUGGESTED ANSWER UPS 1
ii. SESI 2011/2012

No. of proton = 27
No. of neutron = nucleon number – proton number

= 52 – 27
= 25

∑
∑

u

= 52.11
b)

c) i. (required)
ii Mole O2 available (0.503 mol) > mole O2 required (0.443 mol)
therefore, O2 is excess reactant ad NO is limiting reactant

Mass of NO2 produced = 0.886 x [14+2(16)]
= 40.8 g

2

2. a) i. From n=4 to n=2

ii. -Electron at ground state absorbs energy and excites to higher energy level,
excited state, n=4.

- Electron at excited state is unstable.
- Electrons falls back to lower energy level (n=2) by releasing energy in the form

of light/photon with specific wavelength.

iii.

b) i. 9 orbitals
ii. 1s2 2s2 2p6 3s2 3p1

iii. x

y

x z x
y 3s y

or x z
y 3py
z
or
3pz
z

3px

3. a) Proton number of Q = 11
b) T
c) P<S<R<Q<T

d)  P is smallest since it has only 2 shells.
e)  S, R and Q are in the same shell but S has higher proton number/higher
f)
effective nuclear charge than R and Q.
R3-, S2-, T+
R has half-filled orbital which is more stable than partially-filled orbital in S.
Oxides of P, R and S are acidic.

3

4. a) i.

[Al ]3+ 3[ F ]- Cl
Al Cl

Cl

ii Cl Cl Cl

Al Al

Cl
Cl Cl

b) BeH2 is an incomplete octet molecule due to hydrogen that forms duplet/formal
charge zero on Be.
SF6 is an expanded octet molecule due to the existence of empty d orbitals.

c) O-- 2- 0 2-
S2+ O- -
- O- O

- O S O-

O- - O

0

Structure A Structure B

The most plausible structure is B because it bears lower formal charge on each
atom.

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SUGGESTED ANSWER UPS 1
SESI 2012/2013

1 a) i. Proton number is the number of protons in the nucleus of an atom.

ii.

b) Assume that,mass of solution = 100 g
mass of solute = 85 g

mass of solvent = 15 g

n of H3PO4 = 85 g
= 98 g mol-1

0.867 mol

Molality = n of solute (mol)

mass of solvent (kg)

= 0.867 mol

0.015 kg

= 57.8 m

c) i. n of C6H12 = 45.0 g
= 84 g mol-1

0.536 mol

From equation,  2 mol H2C6H8O4
2 mol C6H12  0.536 mol H2C6H8O4
0.536 mol C6H12

Mass of H2C6H8O4 = 0.536 mol x 146.0 g mol-1
= 78.3 g

ii. Percentage yield = actual yield x 100
= percentage yield
= 63.5 x 100
78.3
81.1%

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2 a) i. Infrared region

ii. Line A

() ()

() ()
= 1.097 x 107 (0.0469)
= 1.94 x 10-6 m

b) i. E : 1s2 2s2 2p6 3s2 3p6 4s2 3d1

ii.
z

y or any one of 3d orbitals

x 3dz2
4s

iii. (n, l, m, s) = (4, 0, 0, +½) and (4, 0, 0, -½)

3 a) i. Group 1 Period 3 Block s
ii. P and R
iii. Q
iv. R < S < P < Q

The first ionization energy of P & Q> R&S
Because The number of shell of P & Q< R&S
The shielding effect of P & Q< R&S
The attraction between nucleus and valence electron of P & Q> R&S
The atomic radius of P & Q< R&S

First ionization energy P<Q & First ionization energy R<S
Because effective nuclear charge of P>Q and effective nuclear charge of R>S
The attraction between nucleus and valence electron of P > Q and R>S
The atomic radius of P < Q and R <S

v. Q

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b) i. valence electron = 4

Drastic increase in ionization energy to remove the fifth electron /
The fifth electron located at inner shell / The ratio IE5 to IE4 is the highest /
The fourth electron is removed from outermost shell

ii. Group 14

iii. 1s2 2s2 2p2

4 a) i. BeCl2 : covalent bond
MgCl2 : ionic bond

ii.

xx xx 2+ xx -

x Cl Be Cl x Mg 2 x Cl x
x xx x x x

xx xx

b) i. Structure I Structure II

xx xx

x N N O x x N N O x
x x x x

xx xx

ii. (-2) (+1) (+1) (0) (+1) (-1)

iii. Structure II.
because the -1 formal charge is on the more electronegative atom, oxygen.

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1. a) SUGGESTED ANSWER UPS 1
b) SESI 2013/2014

i. Isotope is two or more atoms of the same element that have the same
proton number but different nucleon number.

ii. )
( )(

u

Relative atomic mass  192 u
1
2 x 12 u

 192

()

c) i. 3V2O5(s) + 10Al(s) 6V(s) + 5Al2O3(s)
ii 3V2O5(s) + 10Al(s) 6V(s) + 5Al2O3(s)

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2. a) i. Paschen series, nf = 3 )
()
ii.
b) i. ( ) ( )(

ii. ∴ The excited state is n = 5
iii. Infrared region.
X: 1s22s22p63s23p64s23d104p1
X3+
1s22s22p63s23p63d10

z

xy z z

4s
z

x yx y xy
4pz
4px 4py

* Choose any one from 4p orbital.

3. a) i. P: Period 3, Group 14
ii. S
iii. R
iv. RS

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b) i. Y: Group 14, Block P
ii. It is because:
 Drastic increase occur between ionization energy 4 and 5.
 5th electron come from the inner shell.
 Y has 4 valence electron.
Valence electronic configuration: ns2np2

Y: 1s22s22p63s23p2

4. a) i. - - -

O IO O I+ O O IO
O O O

ii -

iii O IO
b) O

This structure has the lowest formal charge on each atom.

IO3- is an expanded octet ion.
Due to existence of empty d orbitals.

ionic bond

+
H

HN H -
Cl

H

dative bond covalent bond

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SUGGESTED ANSWER UPS 1
SESI 2014/2015

1. a) i. Isotope No. of protons No. of neutrons
107Ag 47 60
109Ag 47 62

ii. Average atomic mass = isotopic mass  abundance
 abundance

= 106.91 51.50  108.90 48.50  107.88 amu
51.50  48.50

b) i. n HCl = 0.08 mol n Mg = 0.06 mol

2 mol HCl ≡ 1 mol Mg
0.08 mol HCl ≡ 0.04 mol Mg

Moles of Mg available (0.06 mol) > moles of Mg needed (0.04 mol)
∴ Mg is excess reactant and HCl is limiting reactant.

ii. 2 mol HCl ≡ 1 mol H2
0.08 mol HCl ≡ 0.04 mol H2

V of H2 at STP = 0.04 mol × 22.4 L mol−1 = 0.896 L

% yield = Actual yield 100% = 672 103 100% = 75%
Theoretical yield 0.896

E   1  1   2.18 1018  1  1  = 2.18 × 10-18 J atom−1
RH    12 2 
2. a) n 2 n 2 
 i f

IE = (6.02 × 1023)( 2.18 × 10-18) = 1312 kJ mol−1

b) i. Heisenberg’s uncertainty principle states that it is impossible to know
simultaneously both the position and momentum of an electron.

ii. Electron moves in circular orbit around the nucleus.

c) i. It does not obey Aufbau Principle;
Electron must be filled into orbital with lower energy (2s) before start filling into
orbital with higher energy (2p).

ii. It does not obey Pauli exclusion principle;
No two electrons can have same set of quantum numbers

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3. a) i. Group = 16, Period = 3, Block = p

ii. RO3
iii. Acidic
iv. RO3  H2O  H2RO4

b) - Q has higher ionization energy than R.
- because the first electron of Q is removed from more stable half-filled 3p
orbital whereas R is removed from less stable partially filled 3p orbital due to
electrostatic repulsion.

c) - Effective nuclear charge of R is greater than Q.
- Attraction of nucleus towards valence electrons of R is stronger than Q.

4. a) i.

H H+
HNH
H N + +
H
H

H

ii. - - -
b) i. NNN NNN
NNN
AB C

The most stable structure is A.

F F F
FN F FB F F Cl F

ii. BF3 : incomplete octet
ClF3 : expanded octet

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SUGGESTED ANSWER UPS 1
SESI 2015/2016

1. a) i. Given:

Electrons = 18
Neutrons = 24

Proton number = 18+3 = 21
Nucleon number = 24+21 = 45

ii. Isotope notation,
Y45

21

b) i. Given:

V of Pb(NO3)2 solution needed = 250 mL
[Pb(NO3)2] = 0.25 mol L−1

Mol of Pb(NO3)2 = 0.25 mol L−1×0.25 L
= 0.0625 mol

∴ Mass of Pb(NO3)2 = 0.0625 mol x 331.2 g mol−1
= 20.7 g

ii. Two main steps to prepare solution
 20.7 g of Pb(NO3)2 is dissolved with a small amount of distilled water in a
beaker and transferred into 250 mL volumetric flask.
 Distilled water is added into the flask to the calibrated mark.

c) species that undergo reduction,
MnO4−, MnO2

Explanation: The oxidation number of Mn decrease from +7 to +4.

2. a) i. Given: Paschen series, nf = 3
 = 1094 nm

Region of electromagnetic spectrum: Infrared region

13

ii Energy level of excited state, ni

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

1  1 1 
 32 ni2 
   1094109 m  
 1.097 107 m1 

ni = 6

Transition of electron : ni = 6 to nf = 3

b) i. Definition orbital,
Orbital is three-dimensional region around nucleus with high probability of
finding electrons.

ii. Valence electronic configuration for arsenic,
4s24p3

iii. Set of quantum numbers for one valence electron in s orbital
n=4 l=0 m=0 s=+1/2

c) shape of d-orbital that lie on the axes,

z

3. a) i. xy

3dx2  y2

Definition second ionization energy.
The minimum amount of energy required to remove one mole of electrons
from one mole of unipositive ion in the gaseous state.

ii. Electronic configuration of element Z & Group 14
1s22s22p63s23p2

iii. Size of Ca2+ < K+ < Cl- < S2-

 They are isoelectronic species.
 Proton number of Ca2+ > K+ > Cl- > S2-.
 Nuclear charge of Ca2+ > K+ > Cl- > S2-.
 Nucleus attraction towards outer electrons in Ca2+ > K+ > Cl- > S2-

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4. a) i. Definition of electrovalent bond (ionic bond),
Electrostatic attraction between positively and negatively charge ions.

ii. formation of AlF3, F 3+ -
F Al 3 F
Al +

F

Iii. Type of stability : Noble gas configuration.

b) i. Resonance structure of NCO−

- - -
OCN
OCN OCN

ii. most plausible structure,

-
OCN
-1 0 0

Reason:
Negative formal charge resides on O atom which is more electronegative than
N atom.

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