2021 2022 - Tutorial MeKa KumBe - Chapter 3.0 Periodic Table

SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015
TUTORIAL 3
3.1: Classification Of Elements

1. a) Period, group, and block

ELEMENT ELECTRONIC CONFIGURATION GROUP PERIOD BLOCK
A 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 17 4 p
B 5 4 d
C 1s2 2s2 2p6 3s2 3p6 3d3 4s2 2 4 s
D 1s2 2s2 2p6 3s2 3p6 4s2 2 3 s
E 1s2 2s2 2p6 3s2 1 2 s
1s2 2s1

b) Arrangement in periodic table
The elements are arranged in the order of increasing proton number.

c) same group element
Both elements C and D have the same number of valence electrons.

d) same period element
Both elements A and C have the same highest principal quantum number, n.

2. a) Ground state valence electronic configuration and stable oxidation number

U : 3s2 3p5, Stable oxidation number: -1

W : 4s1, Stable oxidation number: +1

b)

i. noble gas = V iv. element that forms acidic oxide = R/S/T/U/V

ii. alkaline earth metal = X v. element that forms basic oxides = W/X

iii. group 14 element = R vi. element that forms amphoteric oxides = Q

3.2: Periodicity

1. Trend in atomic radius and ionic radius down group 1
As we move down group 1:
• principle quantum number, n increases.
• number of shells increases.
• more inner electrons are present to shield the valence electrons from the nucleus
(shielding effect/screening effect increases).
• attraction between nucleus and outermost electrons becomes weaker.
• atomic radius and ionic radius increases.

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015

2. Size of ions and their respective neutral atom
Size of Na+ < Na.
• When electron is removed, the electron- electron repulsion decrease but nuclear
charge remain same.
• The attraction between nucleus towards remaining electron increase.
• Size of Na+ ion is smaller than Na atom.

Size of Cl− > Cl.
• When electron is added, the mutual electron repulsion increase.
• Domain of electron cloud enlarge.
• Size or Cl- ion is larger than Cl atom.

3. Ascending order of ionic radius
Size of Si4+ < Al3+ < Mg2+ < Na+
• Proton number of Si4+ > Al3+ > Mg2+ > Na+
• Nuclear charge of Si4+ > Al3+ > Mg2+ > Na+
• Attraction between nucleus towards remaining electron of Si4+ > Al3+ > Mg2+ > Na+
• Size of Si4+ < Al3+ < Mg2+ < Na+

4. a) Define

The first ionisation energy of an atom is the minimum energy required to remove one

mole of electron from one mole of gaseous atom to form one mole of positive ion.

Energy + X(g) X +(g) + e-

b) Graph

Successive Ionization Energy Of Carbon Atom

IE6
IE5

IE4
IE3
IE2
IE1

From the1graph,2 3 4 5 6 Number of electrons removed

• Successive ionization energy of carbon increase gradually from IE1 to IE4.

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015

• Because as one electron is removed, the remaining electrons felt stronger
attraction towards nucleus, therefore the removal of the next electron requires
more energy.

• The fifth ionization energy, IE5 shows drastic increase, reflects much greater
energy needed to remove the fifth electron which is in the inner shell (closer to
the nucleus).

• So carbon has 4 valence electrons.

c) compare IE1 Na and Al
Na: 1s2 2s2 2p6 3s1
Al: 1s2 2s2 2p6 3s23p1
• Na has bigger atomic size than Al,
• The nucleus attraction towards the valence electron of Na weaker than Al,
• Therefore, less energy needed to remove the valence electron of Na than in Al.

d) IE Be > B
Beryllium (Be): 1s2 2s2 has higher IE1 than Boron (B): 1s2 2s2 2p1
• The 2p subshell of B is at a higher energy than the 2s subshell in Be.
• In B, the electron in 2p subshell is well shielded by 1s and 2s electrons. The
attraction between nucleus and electron in 2p subshell is weaker.
• Less energy required to remove the electron in 2p subshell in B. Therefore, B
has lower first ionisation energy than Be.

e) Trend IE down G1
Down group 1:
• Principal quantum number/number or shell increases, shielding effect increases
• Attraction of nucleus towards the valence electron becomes weaker.
• Less energy needed to remove the valence electron.
• Ionization energy decreases.

5. a) Define
The first ionisation energy of an atom is the minimum energy required to remove one
mole of electron from one mole of gaseous atom to form one mole of positive ion.
Energy + X(g) → X +(g) + e-

The second ionisation energy of an atom is the minimum energy required to remove
one mole of electron from one mole of positive ion in the gaseous state.
Energy + X +(g) → X 2+(g) + e-

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015

b) element that form stable +1 ion
M, because there is sharp increase in IE2. It indicates that the second electron is
removed from the inner shell, so it has one valence electron.

c) i. same group element

Element K and L, group 14.

ii. compare IE

The first ionisation energy for element L is lower than K because L is below K

in the periodic table. The atomic size of L is bigger and the shielding effect is

greater than K. Thus, the attraction of nucleus on the valence electron in L is

weaker than K.

d) determine group J
Group 13.
Sharp increase in IE4, shows that the fourth electron is removed from the inner shell.
So there are 3 valence electrons of J and valence electronic configuration of J is
ns2np1.

e) compare IE1 of N and O
7N : 1s2 2s2 2p3 (half-filled 2p orbital )
8O : 1s2 2s2 2p4 (partially-filled 2p orbital)

• When 7N loses an electron, it must come from the half-filled 2p orbital which
is more stable than the partially-filled 2p orbital in 8O.

• More energy required to remove electron in the stable half-filled 2p orbital of
7N. Therefore, 7N has higher first ionisation energy than 8O.

6. Explain electronegativity
Across a period:
• Proton number increases
• Effective nuclear charge increases
• The attraction between the nucleus and the valence electrons become stronger
• It is easier for the atom to attract electron towards itself.
• Electronegativity increases
Down a group:
• Number of shell increases
• The shielding effect becomes stronger as more inner electrons are present to
shield the valence electrons from the nucleus
• The attraction between the nucleus and valence electrons becomes weaker
• It is harder for the atom to attract electron towards itself.
• Electronegativity decreases.

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015
7. a) classify
Basic oxide: MgO, Na2O
Acidic oxide: SiO2, P4O10, SO3, Cl2O
Amphoteric oxide: Al2O3

b) equation

i. Na2O(s) + H2O(l) → 2NaOH(aq)

ii. SO3(g) + H2O(l) → H2SO4(aq)

iii. As a base:

Al2O3(s) + 6HCl(aq) 2AlCl3(aq) + 3H2O(l)

As an acid:

Al2O3(s) + 2NaOH(aq) + 3H2O(l) 2NaAl(OH)4 (aq)

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
MEKA 3 CHEMISTRY SK015

3.1: Classification of elements

1. fill in the blank

ELEMENT ELECTRONIC GROUP PERIOD BLOCK
CONFIGURATION
F 18 2 p
G 1s2 2s2 2p6 1 3 s
H 1s2 2s2 2p6 3s1 7 4 d
I 1s2 2s2 2p6 3s2 3p6 4s2 3d5 13 4 p
J 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 3 4 d
1s2 2s2 2p6 3s2 3p6 4s2 3d1

2. a) State group and period
K : group 16, period 2
L : group 18, period 2
M : group 1, period 3
N : group 13, period 3

b) i. Ascending order of atomic radius

L<K<N<M

ii. Ascending order of IE1

M<N<K<L

c) Inert

L

d) formula compound
M2K

3.2 Periodicity

1. group and block
Group 13, block p.
There is a drastic increase in ionization energy when the 4th electron was removed. It shows
that the 4th electron is from inner shell. Therefore, element Z has 3 valence electron, its
valence electronic configuration is ns2 np1.

2. a) Define
Electronegativity is the relative tendency of an atom to attract electrons to itself when
covalently bonded with another atom.

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015

b) Ascending order of electronegativity
L<K<M<N
• N is in period 2 of the periodic table with smallest atomic radius since N has
least number of shell and has less shielding effect. The attraction between
nucleus and its valence electrons is the strongest. Thus it has the greatest
electronegativity.
• Proton number of L<K<M.
• Effective nuclear charge in L<K<M.
• Atomic size of L>K>M.
• The nucleus attraction towards valence electron in L<K<M.
• Thus, the electronegativity of L<K<M.

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015
KUMBE 3
3.1: Classification of elements

1. a) period and group
O : group 13, period 3
P : group 17, period 3

b) molecular formula
OCl3

c) molecular formula and type of bond
HP, ionic bond

d) molecular formula
OP3

3.2: Periodicity

1. Compare size
Na+: 1s22s22p6
S2-: 1s22s22p63s23p6
• Number of shell in S2- > Na+
• Sheilding effect in S2- > Na+
• Ionic size S2- > Na+

2. a) Trend first ionization energy
Based on the graph, the first ionization energy across period 3 increases from Na to

Cl
because across the period:
• Proton number increases
• Nuclear charge increases
• Atomic radius decreases
• Attraction between nucleus towards valence electron increases
• High energy needed to remove the first electron from Na to Cl;

However, first ionization energy of Al < Mg and S<P
Al: 1s2 2s2 2p6 3s23p1
Mg: 1s2 2s2 2p6 3s2

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015

• The 3p subshell of 13Al is at a higher energy than the 3s subshell in 12Mg.
• In 13Al, the electron in 3p subshell is well shielded by 1s, 2s, 2p and 3s

electrons. The attraction between nucleus and electron in 3p subshell is weaker.
• Less energy required to remove an electron in 3p subshell in 13Al. Therefore,

13Al has lower first ionisation energy than 12Mg.

S: 1s2 2s2 2p6 3s23p4
P: 1s2 2s2 2p6 3s23p3
Half-filled 3p orbital in P is more stable than partially-filled 3p orbital in S. Less energy
needed to remove an electron from partially-filled 3p orbital in S.

b) Variation of atomic radii
The atomic radii across period 3 decreases from Na to Cl because across the period:
• Proton number increases
• Nuclear charge increases
• Attraction between nucleus towards valence electron increases
 Atomic radii decreases

3. a) state
X

b) determine group
Group 13.
There is a drastic increase in ionization energy when the 4th electron was removed. It
shows that the 4th electron is from inner shell. Therefore, element Y has 3 valence
electron, its valence electronic configuration is ns2 np1.

c) formula oxide
Y2O3

d) compare IE1 and IE2
When 1st valence electron is removed;
• the mutual repulsion decreases but nuclear charge remain the same.
• The remaining valence electrons experience a greater nuclear charge.
• It is difficult to remove the 2nd electron because the 2nd electron is from the
inner shell and it is held more tightly to the nucleus. Therefore, more energy
is needed to remove another electron from the positively charge ion.
 The second IE is higher than first IE

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
CHEMISTRY SK015

4. a) State group and period
Group 13, period 3.
There is a drastic increase in IE4. The 4th electron is removed from the inner shell. So,
Z has 3 valence electron. Its valence electronic configuration is ns2np1.
There are 2 drastic increase (IE4 and IE12). Therefore Z has 3 shell.

b) electronic configuration
1s22s22p63s23p1

c) oxidation number
+3

5. Identify
D: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1
• D is in Period 4, p block element
• Has 3 valence electrons, one of it is unpaired.

E: 1s2 2s2 2p6 3s2 3p6 3d3 4s2
E2+: 1s2 2s2 2p6 3s2 3p6 3d3

• E is in Period 4, d block element.
• E removed 2 electrons from 4s orbital to form E2+ ion.
• E2+ ion has 3 unpaired valence electrons.

6. a) i. Same group

M and T. They have the same number of valence electrons.

ii. name

Group 1

iii. highest melting point

M.

• M & T are metals; S is a transition metal; P is metalloid, Q & R are non-

metals.

• Metals has strong metallic bond.

• Metallic bond (attraction between nucleus and electron sea) in M is

stronger than in T because the size (no. of shell) of M is smaller than T.

• More energy is needed to overcome the attraction.

• Thus melting point M the highest.

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SESSION 2020/2021 TOPIC 3: PERIODIC TABLE
b) i. CHEMISTRY SK015
ii.
Character S
Transition metal.
Stable electronic conf.
1s2 2s2 2p6 3s2 3p6 3d10 4s1
The completely-filled 3d orbitals give more stability to element S, than in its
expected electronic configuration, which has partially-filled (3d9) orbitals.

c) i. Product

Oxide compound with molecular formula P2Q3.

ii. property

Amphoteric

iii. equations
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3 H2O (l)
Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4 (aq)

d) not react with other element
R. It is group 18 element (noble gas). All of its orbitals are completely filled thus
generally R is an inert element.

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