DEPARTMENT OF CHEMISTRY, CFS, IIUM
SEMESTER III, 2013/2014
QUIZ 4
CHE 0315
NAME:……………………………………………MATRIC
NO………………………..GROUP:…………
Answer all questions.
[30 mins]
PART I [5 marks]
1
Which quantum number describes the energy and size of the orbitals?
A
B
C
D
1
Principle
Angular momentum
Magnetic
Electron spin
Which quantum number describes the orientation of an orbital? C/L1/4a
A
B
C
D
2
A/L1/4a
Which quantum number describes the shapes of the orbitals? B/L1/4a
A
B
C
D
1
Principle
Angular momentum
Magnetic
Electron spin
Principle
Angular momentum
Magnetic
Electron spin
Which of the following reactions represents the first electron affinity of
fluorine?
C/L1/4l
A
B
C
D
F2 (g) + 2e-  2F- (g)
F (g)  F+ (g) + eF (g) + e-  F- (g)
F- (g) + e-  F (g)
2
Which of the following reactions represents the first electron affinity of
chlorine?
C/L1/4l
A
B
C
D
2
Cl2 (g) + 2 e-  2Cl- (g)
Cl (g)  Cl+ (g) + eCl (g) + e-  Cl- (g)
Cl- (g) + e-  Cl (g)
Which of the following reactions represents the first electron affinity of
bromine?
C/L1/4l
A
B
C
D
Br2 (g) + 2 e-  2Br- (g)
Br (g)  Br+ (g) + eBr (g) + e-  Br- (g)
Br- (g) + e- Br (g)
3
Which of the following have their valence electrons in the same shell?
C/L2/4e
A
K, As, Br
B
B, Si, As
C
N,As, Bi
D
He, Ne, F
3
Which of the following have their valence electrons in the same shell?
D/L2/4e
A
Ca, Ge, Kr
B
Si, As,Te
C
Ne, Xe, I
D
P, Sb, Bi
3
Which of the following have their valence electrons in the same shell?
B/L2/4e
A
Mg, P, Cl
B
O, Se, Te
C
C, P, Se
D
He, Ne, F
4
Which of the following is not isoelectronic with Kr?
A
B
C
D
As3+
Se2Rb+
Sr2+
A/L2/4g
4
Which of the following is isoelectronic with Ar?
A
B
C
D
4
5
5
5
Fe2+
F
BrCa2+
Which of the following pairs consists of isoelectronic species ?
A
B
C
D
D/L2/4g
A/L2/4g
Cl- and K+
Mn2+ and Ar
Na+ and K+
Cl- and S
Which of the following represent electron configurations does not obey the
Pauli exclusion principle?
B / L3 /4d
I
II
III
IV
[Ne] 3s1 3p5
[Ne] 3s3 3p2
[Kr] 4d12 5s2 5p3
[Ar] 3d10 4s2 4p2
A
C
I, II
III, IV
B
D
II, III
I, IV
Which of the following represent electron configurations does not obey the
Pauli exclusion principle?
A/ L3 /4d
I
II
III
IV
[Ne] 3s2 3p7
[Kr] 4d12 5s2 5p3
[Ar] 3d10 4s2 4p5
[Ar] 3d10 4s2 4p2
A
C
I, II
III, IV
B
D
II, III
I, IV
Which of the following represent electron configurations does not obey the
Pauli exclusion principle?
B / L3 /4d
I
II
III
IV
[Ne] 3s1 3p5
[Xe] 6s2 5d12
[Kr] 4d12 5s2 5p3
[Ar] 3d10 4s1 4p2
A
C
I, II
III, IV
B
D
II, III
I, IV
PART II [10 MARKS]
[SET 1]
1. (a)
Give the period number and group number for element X with the
following electronic configuration;
[2]/L1/4f
1s2 2s2 2p2
Period 2 & Group 14
(a)
Draw all the ‘p’ orbitals occupy in the electronic configuration above
[2]/L1/4b
2px
2py
2. The graph shows the five successive ionization energies in kJ/mol for element
Q
40000
30000
20000
10000
0
1
2
3
4
5
(a)
Identify the group number of element Q
G 13 or 3A
[1]/L2/4k
(b)
Explain your answer in (a)
[2]/L2/4k
Because there is a big jump between IE3 and IE4,[1]
means three electron is removed from the valence shell
and indicating that the element has three valence
electrons.[1]
(c)
Write the condensed electronic configuration of Q if it is in
Period 3
[1]/L2/4j
[Ne] 3s2 3p1
3. Across period 2, the IE for nitrogen is higher than oxygen. Explain.
[2]/L3/4i
This is due to the stability of half filled 2p orbitals in nitrogen. Thus,
a higher energy is required to remove the first electron[1].
Meanwhile, for the oxygen, it has an additional electron-electron
repulsions in its 2p orbitals . Thus, its easier to remove the first
electron. [1]
[SET 2]
1. (a)
Give the period number and group number for element X with the
following electronic configuration.
[2]/L1/4f
1s2 2s2 2p3
Period 2 & Group 15
(b)
Draw all the ‘s’ orbitals occupy in the electronic configuration above
[2]/L1/4b
1s
2s
2. The graph shows the five successive ionization energies in kJ/mol for element
Q
25000
20000
15000
10000
5000
0
1
2
3
4
5
(d)
Identify the group number of element Q
G 14 or 4A
[1]/L2/4k
(e)
Explain your answer in (a)
[2]/L2/4k
Because there are a big jump between IE4 and IE5,[1]
means four electron is removed from the valence shell
and indicating that the element has four valence
electon.[1]
(f)
Write the condensed electronic configuration of Q if it is in
Period 3
[1]/L2/4j
[Ne] 3s2 3p2
4. Across period 3, the IE1 for phosphorus is higher than sulphur. Explain.
[2]/L3/4i
This is due to the stability of half filled 2p orbitals in phosphorus. Thus,
a higher energy is required to remove the first electron[1]. Meanwhile,
for the sulphur, it has an additional electron-electron repulsions in its
2p orbitals . Thus, its easier to remove the first electron.[1]
[SET 3]
1. (a)
Give the period number and group number for element X with
the following electronic configuration.
[2/L1/4f]
1s2 2s2 2p4
Period 2 & Group 16
(b)
Draw all the ‘s’ orbitals occupy in the electronic configuration above
[2/L1/4b]
1s
2s
1. The graph shows the five successive ionization energies in kJ/mol for element Q
16000
14000
12000
10000
8000
6000
4000
2000
0
1
2
3
4
5
(g)
Identify the group number of element Q
G 2 or 2A
[1/L2/4k]
(h)
Explain your answer in (a)
[2/L2/4k]
Because there are a big jump between IE2 and IE3,[1]
means two electron is removed from the valence shell
and indicating that the element has two valence
electon.[1]
(i)
Write the condensed electronic configuration of Q if it is in
Period 4
[1/L2/4j]
[Ar] 4s2
2. Across period 3, the IE1 for Magnesium is higher than Aluminium. Explain.
[2/L3/4i]
Mg : 1s22s22p63s2
Al: 1s22s22p63s2 3p1 [1]
Removing of electron from 3p orbital is easier than 3s since it get
shielded by the inner orbitals and further from nucleus, thus make
the electrostatic forces become weaker.[1]