DK014

CN- (aq) + MnO4- (aq) → CNO- (aq) + MnO2 (s)

Step 1 Oxidation: CN- (aq) → CNO- (aq)
Separate the equation into Reduction: MnO4- (aq) → MnO2 (s)
two half-reactions
Step 2 CN- (aq) → CNO- (aq)
Balance each half- H2O (l) + CN- (aq) → CNO- (aq)
reaction. H2O (l) + CN- (aq) → CNO- (aq) + 2H+ (aq) + 2e-
o Balance the elements 2OH-(aq) + H2O(l) + CN-(aq) → CNO-(aq) + 2H+ (aq)+2OH- (aq) + 2e-
2OH-(aq) + CN-(aq) → CNO-(aq) + H2O(l) + 2e-
other than O and H
o Add H2O to balance O MnO4- (aq) → MnO2 (s)
MnO4- (aq) → MnO2 (s) + 2H2O (l)
atom 4H+ (aq) + MnO4- (aq) → MnO2 (s) + 2H2O (l)
o Add H+ to balance H 3e- + 4H+ (aq) + MnO4- (aq) → MnO2 (s) + 2H2O (l)
3e- +4H+(aq) +4OH-(aq)+MnO4-(aq) → MnO2(s) + 2H2O(l) + 4OH-(aq)
atoms 3e- + 2H2O (l) + MnO4- (aq) → MnO2(s) + 4OH- (aq)
o Add e- to balance the

charges
o Add OH- to both side to

neutralise all the H+
o Cancel water that

appears at both side of

reactants and products

Step 3 (2OH-(aq) + CN-(aq) → CNO-(aq) + H2O(l) + 2e-)  3
Multiply each half-reaction 6OH-(aq) + 3CN-(aq) → 3CNO-(aq) + 3H2O(l) + 6e-
by an integer, so that
number of electron lost in (3e- + 2H2O (l) + MnO4- (aq) → MnO2(s) + 4OH- (aq))  2
one half-reaction equals 6e- + 4H2O (l) + 2MnO4- (aq) → 2MnO2(s) + 8OH- (aq)
the number gained in the
other. 6OH-(aq) + 3CN-(aq) → 3CNO-(aq) + 3H2O(l) + 6e-
Step 4 6e- + 4H2O (l) + 2MnO4- (aq) → 2MnO2(s) + 8OH- (aq)
Combine the two half- 3CN-(aq) + H2O(l) +2MnO4-(aq) → 3CNO-(aq) +2MnO2(s) +2OH-(aq)
reactions and cancel out
the species that appear on Element/charge Left Right
both sides of the equation. Mn 2 2
Step 5 H 2 2
Check to ensure the C 3 3
number of atoms and the O 9 9
charges are balanced. N 3 3
-5 -5
Charge

EXERCISE:
1. Determine the oxidation number of sulfur in (a) H2S, (b) S8, (c) SCl2, (d) Na2SO3, (e) SO42-

2. Balance the following chemical equation:
a. Fe2O3 + HCl → FeCl3 + H2O
b. NH3 + CuO → Cu + N2 + H2O
c. C2H5OH + O2 → CO2 + H2O

47 | Chapter 4 Chemistry 1 DK014

d. AgNO3 + Na2CrO4 → Ag2CrO4 + NaNO3
e. Al2(SO4)3 + KOH → Al(OH)3 + K2SO4

3. Balance the following redox equation:
a. Zn(s) + H+(aq) → Zn2+(aq) + H2(g)
b. S2O32- + I2 → S4O62- + I-
c. C2O42- + MnO4- → CO2 + Mn2+ (acidic solution)
d. Cr2O72- + Fe2+ + H+ → Cr3++ Fe3+ + H2O
e. CrO2- + ClO- → CrO42- + Cl- (basic solution)

Answer:
1. -2, 0, +2, +4, +6
2. Fe2O3 + 6HCl → 2FeCl3 + 3H2O

2NH3 + 3CuO → 3Cu + N2 + 3H2O
C2H5OH + 3O2 → 2CO2 + 3H2O
2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3
Al2(SO4)3 + 6KOH → 2Al(OH)3 + 3K2SO4

3. Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
2S2O32- + I2 →S4O62- + 2I-
5C2O42- + 2MnO4- + 16H+(aq) → 10CO2 + 2Mn2+ + 8H2O
Cr2O72- + 6Fe2+ + 14H+ → 2Cr3++ 6Fe3+ + 7H2O
2CrO2- + 3ClO- + 2OH- → 2CrO42- + 3Cl- + H2O

4.2 STOICHIOMETRY
▪ Stoichiometry is the relationships among the quantities of reactants and products involved in

chemical reactions.
▪ How to “read” chemical equations:

o We read the + sign as “react with” and the arrow as “produces”
o The chemical formula at left-hand side of the arrow is reactant and right-hand side is
product.

Let consider to the folowing chemical reaction;
2 Mg(s) + O2 (g) → 2 MgO (s)

We can read as;
o 2 Mg atoms reacts with 1 O2 molecule to produce 2 formula units of MgO
o 2 mol Mg react with 1 mol O2 to produce 2 mol MgO
o 48.6 g Mg reacts with 32.0 g O2 to produce 80.6 g MgO

EXAMPLE 1:

2 Mg(s) + O2 (g) → 2 MgO (s)

By referring to the chemical reaction above answer the following;
i. how many moles of MgO produce from the complete reaction of 3.4 mol of O2? Assume
there is more than enough Mg.
ii. How many grams of MgO form when 20.0 g of Mg reacts completely with O2?

48 | Chapter 4 Chemistry 1 DK014

iii. How many moles of O2 gas are used when 40.0 g of MgO formed? Determine the gas
volume at STP

Answer: find:
moles of MgO
i. Plan:
given:

ii. mole of O2

1 mol O2 produces 2 mol MgO

So, 3.4 mol O2 produce 2 × 3.4 2
1 2

= 6.8

iii. Plan: mole of Mg mole of MgO Find:
mass of MgO
given: (use molar (use
mass of Mg mass) coefficient) (use molar mass)
iv.
Find:
mole of O2 (use
=
molar volume at
= 20.0 STP)

24.3 /

= 0.823

2 mol Mg produce 2 mol MgO

So, 0.823 mol Mg produce 2 × 0.823
2

= 0.823

= ×
= 0.823 × 40.3 /
= 33.2

v. Plan:

given: mole of MgO mole of O2
mass of MgO
(use molar (use coefficient)
mass)

vi.
=



40.0
= 40.3 /

= 0.993

2 mol MgO produced by 1 mol O2
So, 0.993 mol MgO produced by 1 2 × 0.993

2

= 0.496 2

1 mol of O2 gas occupies 22.4 dm3 at STP

49 | Chapter 4 Chemistry 1 DK014

So, 0.496 mol O2 occupies 22.4 3 × 0.496 2
1

= 11.11 3

EXAMPLE 2:
If 0.05M sulphuric acid was neutralised by 25.00 mL of 0.0976 M sodium carbonate, what would
be the volume of the sulphuric acid needed?

Answer:

given: mole mole H2SO4 Find:
Na2CO3
Volume of Na2CO3 = 25.0 (use coefficent Volume H2SO4 needed
mL (use from balance
MV) chemical equation) (use mole &
Molarity of Na2CO3 = concentration)
0.0976 M
Plan:
Molarity of H2SO4 = 0.05 M

2 3 = × ( )
= 0.0976 × 0.025

= 2.44 × 10−3

Na2CO3 (aq) + H2SO4 (aq) → Na2SO4 (aq) + CO2 (g) + H2O (l)

From the equation;

1 mol Na2CO3 react with 1 mol H2SO4

So, 2.44  10-3 mol Na2CO3 react with 1 2 4 × 2.44 × 10−3 2 3
1 2 3
= 2.44 × 10−3 2 4

2 4 = 2 4
2 4
2.44 × 10−3
= 0.05 /

= 0.0488

Or alternative way; Using mole ratio of Na2CO3 and H2SO4

2 4 2 4 = 2 4
2 3 2 3 2 3

0.05 × 2 4 = 1
0.0976 × 25.00 1

2 4 = 48.8
Limiting reactant and excess reactant
▪ Limiting reactant is reactant that is completely consumed or used up first in a reaction and
▪ limit the amount of products formed.
▪ Excess reactant is reactant that still remain after the reaction complete.

The Cheese Sandwich Analogy
Supposed you wish to make several sandwiches using one slice of cheese and two slices
50 | Chapter 4 Chemistry 1 DK014

of bread for each. Using Bd=bread, Ch=cheese, we can repesent the recipe for making a
sandwich like chemical equation:
2 Bd + Ch  Bd2Ch

+
If you have ten slices of bread and seven slices of cheese, you can only make five
sandwiches and will have two slices of cheese left over. The amount of bread available
limits the number of sandwiches.
Limiting reactant = bread (reactant that completely consumed)
Excess reactant = cheese (reactant that left over)

▪ Steps to determine limiting reactant:
1. Write complete equation; let consider xA + yB  zC
2. Calculate moles of reactants, A and B.
3. Calculate amount of reactant B required to react completely with reactant A
4. Compare the amount of B required (needed) with that given in the system.
5. If B (given) < B(needed), B = limiting reactant, A= excess reactant
If B (given) > B(needed), B = excess reactant  A = limiting reactant

EXAMPLE 3:
Determine limiting and excess reactant in the reaction between 3 mol of hydrogen and 2 mol
of flourine in the formation of hydrogen flouride.

Answer H2 (g) + F2 (g)  2 HF (g)
Step 1
Write the balanced equation Mol of H2 = 3 mol
Step 2 Mol of F2 = 2 mol
Calculate moles for both reactant 1 mol H2 needs 1 mol F2
Step 3 So, 3 mol H2 need 3 mol F2
Calculate mole of F2 required to
react completely with H2 (or vise Mol of F2 needed = 3 mol
versa) Mol of F2 given = 2 mol
Step 4
Compare amount of F2 required Mol F2 needed > Mol F2 given
(needed) with that is given So, F2 is limiting reactant
Step 5 H2 is excess reactant
Determine limiting and excess
reactant

EXAMPLE 4:
For the following reaction:

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

18.0 g NH3 and 90.0 g CuO are allowed to react.
i. Determine the limiting reactant.
ii. Calculate the mass of N2 gas produced.
iii. Determine the mass of the excess reactant remain after the completion of the reaction.

51 | Chapter 4 Chemistry 1 DK014

Answer:
i. Plan:

Given: mole NH3 and mole reactant compare Find:
CuO given required if one of it mole of
mass of NH3 is limiting reactant reactant Limiting
= 18.0 g (use required reactant
(use coeffient from with that is
mass of CuO ) balance chemical given
= 90.0 g
equation)

18.0
3 = 17.0 /

= 1.059

90.0
= 79.6 /

= 1.131

From the equation:

2 mol NH3 react with 3 mol CuO

So, 1.059 mol NH3 react with 3 × 1.059 3
2 3

= 1.589 (needed)

Mole of CuO needed (1.589) > mole of CuO given (1.131)
So, Cuo is limiting reactant.

ii. Plan: mole of N2 Find:
produced mass of N2 produced
Given:
CuO is limiting (use coeffient from (use )
balance chemical
reactant
equation)

3 mol CuO produce 1 mol N2
So, 1.131 mol CuO produce 1 2 × 1.131

3

= 0.377 2

2 = 0.377 × 28.0 /
= 10.56

iii. Plan: mole of NH3 needed mole of NH3 remain Find:

Given: (use coeffient from (substract NH3 mass of
CuO is limiting balance chemical needed from mole NH3 remain

reactant equation) NH3 given)

3 mol CuO react with 2 mol NH3

So, 1.131 mol CuO produce 2 3 × 1.131
3

52 | Chapter 4 Chemistry 1 DK014

= 0.754 3 (needed)

3 = 3 − 3
= 1.059 − 0.754
= 0.305

3 = 0.305 × 17.0 /
= 5.18

Theoretical yield and actual yield

▪ Theoretical yield is the quantity of product calculated to form using balanced chemical
equation when all limiting reactant is consumed

▪ Actual yield is the amount product actually obtained in a reaction.
▪ Actual yield almost always less than theoretical yield due to some reactants may not react,

or they may react in different from the desired (side reaction).

Percentange yield
▪ The percent of the actual yield of a product to its theoretical yield.


= ℎ × 100%

EXAMPLE 5:
If 3.7 g sodium metal (Na) and 4.3 g chlorine gas (Cl2) react to form NaCl, what is the theoretical
yield? If 5.5 g NaCl was formed, what is the percentage yield?

Answer: determine Find: Find:
Plan: limiting percentage yield
reactant theoretical
Given: yield of (use
mass Na = 3.7 g NaCl
Mass Cl2 = 4.3 g
Actual yield of NaCl = (use
limiting
5.5 g reactant)

3.7
= 23.0 /

= 0.16

4.3
2 = 71.0 /

= 0.061

2 Na (s) + Cl2 (g) → 2 NaCl (s)

From the equation:
2 mol Na react with 1 mol Cl2
So, 0.16 mol Na react with 1 2 × 0.16

2

53 | Chapter 4 Chemistry 1 DK014

= 0.08 2 (needed)

Mole of Cl2 needed (0.08) > mole of Cl2 given (0.016)
So, Cl2 is limiting reactant.

1 mol Cl2 produces 2 mol NaCl

So, 0.061 mol Cl2 produce 2 × 0.061 2
1 2

= 0.122

= 0.122 × 58.5 /
= 7.137 (theoretical yield)


= ℎ × 100%

5.5
= 7.137 × 100%

= 77.0 %

EXERCISE:
1. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al + 3Cl2 → 2AlCl3

a. Determine the limiting reactant and excess reactant 5 mol of aluminum react with 5 mol
of chlorine gas?

b. Determine mass of excess reactant remain after reaction was completed.
c. Calculate mass of AlCl3 produced.
d. Calculate % yield of AlCl3 if AlCl3 obtained was 400 g

2. For the neutralization reaction between:
2NaOH + H2SO4 → Na2SO4 + 2H2O

a. If 25 mL of 0.2 M of sulphuric acid react with 10.0 mL of 0.1 M of sodium hydroxide,
determine the limiting and excess reactant?

b. Determine mass of excess reactant remain after reaction was completed.
c. Calculate mass of Na2SO4 produced.
d. Calculate % yield of Na2SO4 if Na2SO4 obtained was 0.025 g

3. In the production of ammonia in Haber process, 100 mL of both hydrogen and nitrogen
gas are mixed at STP.
3H2 + N2 → 2NH3
a. Determine the limiting and excess reactant?
b. Determine mass of excess reactant remain after reaction was completed.
c. Calculate mass of ammonia produced.
d. Calculate % yield of ammonia if ammonia obtained was 0.04 g

4. For the combustion of sucrose:

54 | Chapter 4 Chemistry 1 DK014

C12H22O11 + 12O2 → 12CO2 + 11H2O
there are 10.0 g of sucrose and 10.0 g of oxygen reacting.
a. Determine the limiting and excess reactant?
b. Determine mass of excess reactant remain after reaction was completed.
c. Calculate mass of CO2 produced.
d. Calculate % yield of CO2 if CO2 obtained was 10.0 g
Answer:
1.Cl2 is limiting reactant, 45.1g, 444.56 g, 89.98 %
2. NaOH is limiting reactant, 0.44 g, 0.0711 g, 35.16 %
3.H2 is limiting reactant, 0.0832 g, 0.0505 g, 79.2 %
4. O2 is limiting reactant, 1.094 g, 13.75 g, 72.73 %

TUTORIAL 4

1. The oxidation number of manganese in KMnO4 is
A. +4
B. +1
C. +3
D. +7

55 | Chapter 4 Chemistry 1 DK014

2. Oxygen has an oxidation number of -2 in
A. O2
B. Na2O2
C. OF2
D. NO2

3. An Oxidation number can be
A. positive
B. negative
C. zero
D. All of Above

4. The elements whose oxidation number is increased are
A. reduced
B. hydrogenated
C. oxidized
D. oxygenated

5. The oxidation state of any free element is always
A. 1
B. 2
C. 3
D. 0

6. When the following equation is balanced with the smallest possible integer coefficients, what
is the coefficient of H+?
__ Cr + __ H+ → __ Cr3+ + __ H2
A. 2
B. 3
C. 5
D. 6

7. In the following reaction, the oxidation number of Fe changes from ____ to ____.
Fe2O3 + 3CO → 2Fe + 3CO2

A. 0, +2
B. +3, 0
C. doesn't change
D. +2, 0

8. MnO4- + HSO3- → MnO42- + SO42-
For the above reaction , the oxidizing agent is _____ and the reducing agent is _____.

A. MnO4- , HSO3-
B. OH-, HSO3-
C. MnO4-, OH-
D. HSO3-, MnO4-

9. In the oxidation-reduction reaction below
ClO3- + Cl- → Cl2 + ClO2

A. Cl is oxidized and O is reduced.
B. Cl is both oxidized and reduced.
C. O is oxidized and Cl is reduced.
D. O is both oxidized and reduced.

10. The following reaction occurs in acidic aqueous solution:
H3AsO4 + Zn → AsH3 + Zn2+

56 | Chapter 4 Chemistry 1 DK014

H3AsO4 is a(n) __________ agent in which the oxidation number of As is ____.
A. reducing, +3
B. oxidizing, +5
C. reducing, +5
D. oxidizing, +3

11. Which one of the following reactions is an oxidation reduction reaction?
A. 3CuO + 2NH3 → 3Cu + 3H2O + N2
B. 3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 3H2O
C. CaSO3 + 2HCl → CaCl2 + H2O + SO2
D. NaCl + AgNO3 → AgCl + NaNO3

12. Balance the following equation using minimum integral coefficients:
NH3 + O2 → NO2 + H2O

The stoichiometric coefficient for oxygen gas O2 is:
A. 1
B. 3
C. 7
D. 5

13. What is a limiting reactant?
A. the reactant that is in excess
B. the product that you can make the most of
C. the reactant that determines how much product can be made
D. the amount of reactants that react with each other

14. Which of the following statements is FALSE for the chemical equation given below in which
nitrogen gas reacts with hydrogen gas to form ammonia gas assuming the reaction goes to
completion?
N2 + 3H2 → 2NH3
A. The reaction of one mole of H2 will produce 2/3 moles of NH3.
B. One mole of N2 will produce two moles of NH3.
C. One molecule of nitrogen requires three molecules of hydrogen for complete reaction.
D. The reaction of three moles of hydrogen gas will produce 17 g of ammonia.

SECTION B

1. Find the oxidation number of (Ans: +7)
a) manganese in a permanganate ion, MnO4- ? (Ans: +6)
b) sulfur in sodium sulphate, Na2SO4 ?

2. Balance the equations
a) H2SO4 + Fe → Fe2(SO4)3 + H2
b) Na2CO3 + HCl → NaCl + CO2 + H2O
c) Mg3N2 + H2O → MgO + NH3

3. Balance the redox equations
a) SnO2 + H2 → Sn + H2O

57 | Chapter 4 Chemistry 1 DK014

b) MnO4-(aq) + Cl-(aq) → Mn2+(aq) + Cl2(aq)
c) Zn(s) + KNO3(aq) → NH3(aq) + K2Zn(OH)4

4. Consider the following equation:

Cu + AgNO3 → Cu(NO3)2 + Ag

a) Balance the above equation.

b) Calculate the number of moles of Cu needed to react with 3.50 moles of AgNO3.

(Ans:1.75 mol)

c) If 90.5 g of Ag were produced, how many grams of Cu had reacted

(Ans: 26.67 g)

5. Nitrogen monoxide reacts with oxygen to form nitrogen dioxide

2NO(g) + O2(g) → 2NO2(g)

In an experiment, 0.886 mol of nitrogen monoxide reacts with 0.503 mol of oxygen to

produce nitrogen dioxide.

a) Determine the limiting reactant. (Ans: NO)

b) Calculate the mass of nitrogen dioxide produced. (Ans: 40.8 g)

6. An aqueous solution of MgSO4 is added to an aqueous solution of BaCl2 to form BaSO4

and MgCl2. If 1.75 g of MgSO4 and 2.75 g of BaCl2 are used,

a) Determine the limiting reactant (Ans:BaCl2)

b) Calculate the mass of BaSO4 produced after the reaction has completed.

(Ans:3.08 g)

7. 0.0869 mol of sodium is reacted with 0.0345 mol of chlorine gas to produce sodium

chloride.
2Na + Cl2 → 2NaCl

a) Determine the limiting reactant. (Ans:Cl2)

b) Calculate the mass of sodium chloride produced in this reaction. (Ans:4.037 g)

c) If 3.7500 g of sodium chloride is produced after this experiment.

What is the percentage yield? (Ans:92.89%)

8. Disulfur dichloride, S2Cl2 is used to vulcanize rubber. It can be made by treating

molten sulphur with gaseous chlorine:
S8(l) + Cl2(g) → S2Cl2(l)

a) Balance the chemical equation.

b) Starting with a mixture of 1.32 mol of sulfur and 4.44 mol of Cl2, determine the limiting

reactant.

(Ans:Cl2)

58 | Chapter 4 Chemistry 1 DK014

c) If the percentage yield of the reaction is 89.3 %, what is the actual yield can be
obtained from the experiment?
(Ans: 536.06 g)

9. In an experiment, 1.46 g of magnesium is added into 160.00 mL of 0.50 mol L-1
hydrochloric acid. The reaction involved is:
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
a) Determine the limiting reactant for the above reaction. (Ans: HCl)
b) Calculate the percentage yield if 672.00 mL of hydrogen gas is obtained at STP.
(Ans:75 %)

10. Adipic acid, H2C6H8O4 is produced by a reaction between cyclohexane and excess
oxygen. The equation for the reaction is :
2C6H12(l) + 5O2(g) → 2 H2C6H8O4(l) + 2H2O(l)
a) If 45.0 g of cyclohexane is used, calculate the theoretical yield of adipic acid.
(Ans:78.3 g)
b) Given the actual yield of adipic acid is 63.5 g, what is the percentage yield of adipic
acid.
(Ans:81.1 %)

Revised by:
Dr Elmi Sharlina binti Md Suhaimi
Zanarina Binti Sapiai

59 | Chapter 4 Chemistry 1 DK014

CHAPTER 5: ELECTRONIC CONFIGURATION

GENERAL OVERVIEW

i. principle
quantum
number (n)

5.1 Quantum a) Define the ii. angular
numbers of term orbital momentum
electrons
b) Sate and quantum
5.0 ELECTRONIC describe all number (l);other
CONFIGURATION
the four term for l are
5.2 Electronic quantum azimuthal or
configuration of an numbers of subsidiary
an electron in
atom an orbital quantum
number
c. Sketch the
shapes of iii. magnetic
s,p,and d quantum
orbitals with
the correct number (m);and
orientation
iv. electron spin
quantum number

(s)

a. State Aufbau principle, Hund's Rule and
Pauli exclusion principle.

b. Apply the rules in (a) to fill electrons into
atomic orbitals

c. Write the electronic configuration of atoms
and monoatomic ions using spdf notation and

orbital diagram.

*limit the proton number (Z) 1 to 36

** exclude chromium and cooper.

60 | C h a p t e r 5 C h e m i s t r y 1 D K 0 1 4

5.1 QUANTUM NUMBERS OF ELECTRON
▪ An orbital is a three-dimensional region in space
around the nucleus where there is a high
probability of finding an electron.
▪ Orbital differ from one another in their energy and
in the shape and spatial orientation of their electron
cloud.
▪ For H atom, an atomic orbital is specified by 4
quantum numbers:

1. Principal quantum number (n), indicates the energy level of the electron.
2. Angular momentum quantum number (ℓ), indicates the shape of the

orbital.
3. Magnetic quantum number (m), describes the orientation of the orbital in

space.
4. Electron spin quantum number (s), represents the spin direction of

electron on its own axis.
▪ These quantum numbers describe the energy level of an orbital, define the

shape of orbital and orientation of the region in space where the electron is most
likely to be found.

PRINCIPAL QUANTUM NUMBER (n)
▪ The principle quantum number, n, designates the principal energy level.
▪ The value of n determines the size and energy of an atomic orbital.
▪ Indicates relative size of orbital  relative distance of electron from nucleus.
▪ As n increases, the energy of the electron increases and the electron is farthest

from the nucleus.
▪ The principal quantum number may have on positive integers: n= 1,2,3……, ∞
▪ For H atom: n ↑ energy level ↑

61 | C h a p t e r 5 C h e m i s t r y 1 D K 0 1 4

Orbital size:
3s > 2s > 1s

ANGULAR MOMENTUM QUANTUM NUMBER (ℓ)

▪ Each principal shell (principal quantum number, n) includes one or more sublevels

(subshells).
▪ Each subshell has an angular momentum quantum number, ℓ. It also called

azimuthal / subsidiary / orbital quantum number.
▪ Indicates shape and type of orbital.
▪ ℓ = an integer from 0 to (n – 1)
▪ ℓ depends on the value of n. For example, n=4, ℓ = 0, 1, 2, and 3.

▪ There are four types of orbitals, known as s,p,d, and f orbitals.
(a) s orbital are spherical in shape.
(b) p orbital are shaped like dumb-bells
(c) d orbitals are shaped like cloverleaf

n ℓ Subshell

10 1s

0 2s
2 2p

1

0 3s
3 3p

1

2 3d

z z z
x x x
y y
y
p –orbital dumbbell d - orbital cloverleaf
s – orbital sphere

62 | C h a p t e r 5 C h e m i s t r y 1 D K 0 1 4

MAGNETIC QUANTUM NUMBER (m)

▪ The magnetic quantum number m determines the orientation in space of the

electrons cloud surrounding the nucleus.
▪ m depends on the value of ℓ
▪ m = an integral value from -ℓ through 0 to +ℓ. For example:

If ℓ =0 , then m = 0
If ℓ = 1 , then m = -1,0,+1
If ℓ = 2 , then m = -2,-1,0,+1,+2
▪ For a given type of orbital with the angular momentum quantum number ℓ, there are
(2ℓ + 1) different spatial orientation for the orbitals. For example:

i. There is only one spatial arrangement (m=0) for the s orbital (ℓ = 0) since

2(0) + 1 = 1.

ℓ=0 m=0 1 possible orbitals (s subshell)

ii. There are three different spatial arrangements (m = -1, 0, +1) for the p
orbitals (ℓ =1) since 2 (1) + 1 = 3.

ℓ = 1 m = –1, 0, +1

iii. There are five spatial arrangements ( m = -2,-1,0,+1,+2) for the d orbitals
(ℓ = 2) since 2 (2) + 1 = 5.

ℓ=2

m = –2, –1, 0, +1, +2 5 possible orbitals (d subshell)

ELECTRON–SPIN QUANTUM NUMBER (s)

▪ When a charged particle spins on its axis, a magnetic field is produced. An electron
can spin on its own axis. Thus, an electron has magnetic properties that correspond
to those of a charged particle spinning on its axis. An electron can only spin in either
one of two directions, clockwise or anticlockwise.

▪ The spin of an electron is described by the electrons spin quantum number ,s,
which can only have one of two possible values, +½ or –½.

▪ Electrons that have the same value of s (that is both +½ or both –½) are said to have
parallel spins.

▪ Electrons that have different values of s (one is +½ and the other is –½) are said to
have opposite spin.

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POINT TO REMEMBER

▪ Each allowed combination of n, ℓ and m values, specifies one atomic orbital in term of
size (energy), shape and spatial orientation.

▪ Hierarchical relationship:

llimit ℓ llimit m

n

Example:

n=2 possible ℓ values = 0, 1

ℓ=1 possible m values = – 1, 0, +1

ℓ=0 possible m values = 0

▪ Maximum number of electron in the shell can be determining by 2n2.

n ℓ m Orbitals Number of - Total
orbitals numbers of e-
(name) Number of e
1 (2n2)
10 0 1s 2 2
1
20 0 2s 3 2 8
1 -1,0,+1 2p 6
1
30 0 3s 3 2
3p 5
1 -1,0, +1 3d 6 18
10
2 -2,-1,0,+1,+2

SHAPE OF ATOMIC ORBITAL

s Orbital
▪ The s orbital is represented by ℓ = 0
▪ It has spherical shape with the nucleus at the center.
▪ The ℓ = 0, there is only one value of m = 0, which refer to one s orbital

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▪ The size of s orbital becomes larger as the value of n increases

SHAPE OF s ORBITAL

2s
or

p Orbital

▪ The p orbitals are represented by ℓ = 1
▪ Each p orbitals has dumbbell shaped and separated by a node at the nucleus
▪ The ℓ =1 , there are three possible values of

m = -1,0,+1, which refer to three p orbitals,( px , py , pz )
▪ As n increases, the p orbitals get larger

SHAPE OF p ORBITAL

Px Py Pz
Or

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d Orbital

▪ The d orbitals are represented by ℓ = 2.
▪ There are five possible m values: -2, -1, 0, +1, +2, which correspond to five d

orbitals with five different orientations
▪ All the d orbitals do not look alike

SHAPE OF d ORBITAL

or

dxz

dxy

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EXAMPLE 1:
Specify the ℓ and m values for n = 4.

Answer:

When, n = 4
ℓ = 0, 1, 2, 3

ℓ m
0 0
1 -1,0,+1
2 -2,-1,0,+1,+2
3 -3,-2,-1,0,+1,+2,+3

EXAMPLE 2:
Give all possible m values for orbitals that have each of the following:

a) ℓ = 2
b) n = 1
c) n = 4, ℓ = 3

Answer:

a) When ℓ = 2,
Possible values of m = -2, -1, 0, +1, +2

b) When n = 1 ℓ = 0
ℓ = 0 Possible values of m = 0

c) When n = 4 ℓ = 3
ℓ=3
Possible values of m = -3, -2, -1, 0, +1, +2, +3

EXAMPLE 3:

Give the name, magnetic quantum numbers, and number of orbitals for each subshell with
the following quantum numbers:

a) n = 3 , ℓ = 2
b) n = 2 , ℓ = 0
c) n = 5 , ℓ = 1
d) n = 4 , ℓ = 3
e)
Answer:

n ℓ Name of orbital Possible m values Number
a) 3 5
b) 2 2 3d -2,-1,0,+1,+2 1
c) 5 3
d) 4 0 2s 0 7

1 5p -1,0,+1

3 4f -3,-2,-1,0,+1,+2,+3

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EXAMPLE 4:

Are the following quantum number combinations allowed?
If not, show two ways to correct them.

a) n = 1; ℓ = 0; m = –1
b) n = 4; ℓ = 3; m = –1

Answer:

a) n = 1; ℓ = 0; m = –1
When n = 1; ℓ = 0
ℓ = 0; m = 0 m ≠ –1
So, the combination is not allowed.
Correction:
n = 1; ℓ = 0; m = 0
or
n = 2; ℓ = 1; m = –1

b) n = 4; ℓ = 3; m = –1
When n = 4; ℓ = 0, 1, 2, 3
ℓ = 3; m = –3, –2, –1, 0, +1, +2, +3

So, the combination is allowed

EXAMPLE 5:

For the following subshells give the values of the quantum numbers (n, ℓ, m) and the
number of orbitals in each subshell:

a) 3p
b) 6s
c) 5d
d) 4f

Answer:

a) 3p (3 orbital)
n = 3, ℓ = 1, m= -1, 0 +1 (1 orbital)

b) 6s
n = 6, ℓ = 0 , m = 0

c) 5d (5 orbitals)
n = 5, ℓ = 2 , m = +2,+1,0, -1,-2

d) 4f (7 orbitals)
n = 4, ℓ = 3 , m = +3,+2,+1,0, -1,-2,-3

EXERCISE 1:

Write an acceptable value for each of the missing quantum numbers
a) n = 3, ℓ = ?, m = 2
b) n = ?, ℓ = 2, m = –1
c) n = 4, ℓ = 2, m = ?
d) n =?, ℓ = 0, m = ?

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EXERCISE 2:
Write the set of four quantum numbers for an electron in a 3p orbital.

EXERCISE 3:
Sketch any two orbitals of an electron characterized by the principal quantum number of 3
and the azimuthal quantum number of 2.

5.2 ELECTRONIC CONFIGURATION OF ATOM

▪ The electronic configuration of an atom or ion describes how electrons are distributed
among the various orbitals in the principal shells and subshells of the atom or ion.

▪ We can denote the electronic configuration by using an orbital diagram or spdf
notation.

▪ It shows how the electrons are distributed among the various atomic orbitals
▪ Example:
H atom (ground state)

n value 1 Number of electrons in

1s the orbital or subshell

Subshell or orbital

Representing Electronic Configuration

Method 1: Orbital diagram
▪ The oxygen atom has eight electrons. The orbital diagram of the oxygen atom is as
follows.
▪ Each orbital is represented by a box and each electron by an arrow.
▪ The arrow head indicates the direction of spin. An arrow pointing upward ( )
represent an electron with a positive spin quantum number (s = +½) and an arrow
pointing downward ( ) represent an electron with s = -½.
▪ Box form

O:
8
▪ Platform form

O:
8

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Method 2: spdf notation Number of electrons in the subshells
Azimuthal quantum number, 
8O : 1s2 2s2 2p4

Principal quantum number, n

▪ In spdf notation, the electronic configuration is denoted by writing the symbol for the

occupied subshell and adding a superscript to indicate the number of electrons in
that subshell.

Rules Used to Fill Electrons Into Atomic Orbitals

1. Aufbau Principle

▪ Electrons fill the lowest energy orbitals first and other orbitals in order of increasing
energy.

n=1 1s 2p
n=2 2s

n=3 3s 3p 3d 1s 2s 2p
n=4 4s 4p 4d
n=5 5s 5p 5d 4f
5f

The order of filling energy orbitals with electrons:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s

KEEP IN MIND

▪ Degenerate orbitals are orbitals with the same energy.
▪ Example:

2px = 2py = 2pz

3dxy = 3dyz = 3dxz = 3dx2- 2 = 3dz2
y

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Relative Energy Level of Atomic Orbitalls

5s 4d
n=5 4p

3d

Energy 4s
n=4 3p

3s 2p
n=3

2s
n=2

1s
n=1

Orbital energy levels in a many-electron atom

EXAMPLE 1:
Which of the following pairs is lower in energy for the case of many–electron atoms:

a) 2s , 2p
b) 3p , 3d
c) 3s , 4s
d) 4d , 5f
e) 3d , 4s

Answer:

a) 2s < 2p
b) 3p < 3d
c) 3s < 4s
d) 4d < 5f
e) 4s < 3d

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2. Pauli Exclusion Principle

▪ This principle states that only two electrons may occupy the same orbital and
these two electrons must have opposite spins.

▪ Chemists use the symbol or to denote the electron with clockwise spin and
or to denote the electron with anticlockwise spin.

▪ For example, for helium, proton number = 2 and number of electrons = 2.
▪ These two electrons must occupy the 1s orbital, because a 1s orbital has the lowest

energy (Aufbau principle).

No two electrons in the same atom have the same four quantum numbers
@

Each electron must have a different set of quantum numbers

EXAMPLE:

He atom 1s 1s 1s
1st electron (1, 0, 0, +½) (1, 0, 0, -½) (1, 0, 0, +½)
(1, 0, 0, -½) (1, 0, 0, -½)
2nd electron (1,0,0, +½ )
correct

3. Hund’s Rule

▪ Hund’s rule states that when electrons are placed in a set of orbitals with equal, or
degenerate, energies (for example, 2px, and 2py, and 2pz orbitals), the electrons must
occupy them singly with parallel spins before they occupy the orbitals in pairs.
In other words, an atom tends to have as many unpaired electrons as possible.

▪ Degenerate orbitals are orbitals with the same energy level.
▪ Most stable arrangement of electrons in orbital of a subshell is the one with the

greatest number of parallel spin
▪ Thus electrons fill each and all degenerate orbital singly before they pair up.

EXAMPLE:
C (Z = 6)

1s 2s 2p

number of parallel spin = 2

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▪ There are number of ways to show the electronic configuration of the atom or ion of
an element.

▪ For example, the proton number of magnesium is 12 so an atom of magnesium has
12 electrons.

▪ Two electron will fill up the 1s orbital, two other electrons will fill up the 2s orbital , six
electrons will fill up the 2p orbital and the rest of the electrons will be in the 3s
orbitals.

▪ The electronic configuration of magnesium can be written in two ways.

Mg: 1s2 2s2 2p6 3s2

Mg:

1s 2s 2p 3s
Or

Mg : 10[Ne] 3s2

where 10 [Ne] represent 1s2 2s2 2p6

Electronic configurations of d block elements
▪ There are 10 elements in the first row of d block elements in the periodic table:

scandium(Sc),titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron(Fe),
cobalt (Co),nickel (Ni),cooper(Cu), and zinc (Zn).
▪ Consider the d-block element, titanium, with proton number 22. The electronic
configuration of titanium is

Titanium, Ti (Z = 22)
Electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d2

1s 2s 2p 3s 3p 4s

3d
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▪ Electrons are added to the d orbitals in accordance with Hund’s rule.
▪ 4s is more stable than 3d orbital in terms of energy. However, as soon as the electron

occupy the 3d orbitals, the 4s electrons are repelled to higher energy.
▪ Electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2 3d2 (by applying Aubau principle)
▪ Electron configuration: 1s2 2s2 2p6 3s2 3p6 3d2 4s2 (due to energy level after repulsion)

Note: Both configurations can be used!
▪ When a d- block element forms simple ions, the electrons from the 4s orbitals are

removed first before the 3d electrons.

▪ For example :

26 Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2

Fe 2+ : 1s2 2s2 2p6 3s2 3p6 3d6

Fe 3+: 1s2 2s2 2p6 3s2 3p6 3d5

The Anomalous Electronic Configurations of Chromium and Copper
▪ For d subshells that are half - filled or fully filled are particularly stable.
▪ Because of this, the electronic configuration of chromium (Cr) (proton number 24) is

1s2 2s2 2p6 3s2 3p6 3d5 4s1 and NOT 1s2 2s2 2p6 3s2 3p6 3d4 4s2
▪ Instead of adding a second electron into the 4s orbitals, the electron is placed in a 3d

orbital so that the 3d orbitals are half-filled.
▪ The electronic configuration of copper (Cu) is 1s2 2s2 2p6 3s2 3p6 3d10 4s1
▪ and NOT 1s2 2s2 2p6 3s2 3p6 3d9 4s2 .
▪ A second electron is not added to the 4s orbital but is added to a 3d orbital so that

the 3d orbitals are fully filled.

EXAMPLE 2:

Write the electron configuration of K (Z = 19) and Mg (Z = 12).

Answer:
K : 1s2 2s2 2p6 3s2 3p6 4s1
Mg: 1s2 2s2 2p6 3s2

EXAMPLE 3:

Write the electron configuration of the following transition metal:
a) V5+ (V; Z = 23)
b) Ti4+ (Ti; Z = 22)
c) Sc3+ (Sc; Z = 21)

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Answer:

a) Electron configuration:
V : 1s2 2s2 2p6 3s2 3p6 4s2 3d3
or
V : 1s2 2s2 2p6 3s2 3p6 3d3 4s2

V 5+ (18 electrons)
V 5+ : 1s2 2s2 2p6 3s2 3p6

b) Ti4+ (Ti; Z = 22)

Ti (22 electrons)
Electron configuration:
1s2 2s2 2p6 3s2 3p6 4s2 3d2

Or
1s2 2s2 2p6 3s2 3p6 3d2 4s2
Ti4+ (18 electrons)
Ti4+: 1s2 2s2 2p6 3s2 3p6

c) Sc3+ (Sc; Z = 21)
Sc (21 electrons)
Electron configuration:
1s2 2s2 2p6 3s2 3p6 4s2 3d1
Or
1s2 2s2 2p6 3s2 3p6 3d1 4s2
Sc3+ (18 electrons)
Sc3+: 1s2 2s2 2p6 3s2 3p6

EXAMPLE 4:

Draw the orbital diagram and write the electron configuration for Be atom ( Z = 4 )
and O atom ( Z = 8 ).

Answer:

Be (Z=4):

1s 2s

Electron configuration: 1s2 2s2

O (Z = 8) 2p
1s 2s

Electron configuration: 1s2 2s2 2p4

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EXERCISE 1:

State the quantum numbers n, ℓ and m of the 3dx2 - 2 orbital and draw its shape.
y

EXERCISE 2:

Write electronic configuration and orbital diagram of the following atoms.

Element Proton Electronic Configuration Orbital Diagram
Number
Boron
Sodium (Z)
Flourine 5
Phosphorous 11
Aluminium 9
Calcium 15
13
20

EXERCISE 3:
Write the electronic configuration for the following atoms/ions using spdf notation of :

i. Na ii. Na+ iii. Cl iv. Cl-

v. Al vi. Al3+

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TUTORIAL 5

SECTION A

1. “No two electrons in an atom can have the same four quantum numbers” is a statement
called
A. Pauli exclusion principle.
B. Bohr’s atomic postulate.
C. Hund’s rule.
D. Aufbau principle.

2. The electronic configuration of a ground state for Ti (Z=22) atom is
A. [Ar]4s23d2
B. 1s22s22p63s23d5
C. [Ne]3s23d7
D. [Ar]4s13d3

3. Which of the following electronic configuration belongs to an element that forms a simple
ion with the charge of -2?
A. 1s22s22p4
B. 1s22s22p63s23p64s23d1
C. 1s22s22p5
D. 1s22s22p63s23p64s23d7

4. Which of the following represents the electronic configuration of a d-block element?
A. …3s23p4
B. …3d54s2
C. …3s23p64s2
D. …3d104s24p1

5. The figure shows an orbital:

z
x

y

The most probable set of quantum numbers for an electron to occupy the above orbital
is
A. n=2, l=1, m=2, s = +1/2
B. n=2, l=2, m=2, s = +1/2
C. n=3, l=0, m=1, s = +1/2
D. n=3, l=2, m=2, s = +1/2
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6. Which of the following statements is correct?
A. Each electron in an atom has a set of three quantum numbers
B. The magnetic quantum number defines the spatial orientation of the orbital.
C. The angular momentum quantum number may be a negative or positive number.
D. The total number of orbitals in a shell is 2n2.

7. What is the possible quantum number(s) for the valence electrons of aluminium (Z=13)
at ground state?
I. n=3, l=0, m=0, s=+1/2
II. n=3, l=1, m=-1, s=+1/2
III. n=3, l=2, m=-1, s=+1/2

A. I
B. I and II
C. II and III
D. I, II and III

8. Which of the following does not have paired electrons in the p orbitals of its atom?
A. Carbon (Z=6)
B. Oxygen (Z=8)
C. Aluminium (Z=13)
D. Silicon (Z=14)

9. Which of the following has the electronic configuration: 1s22s22p6 ?
I. Ne
II. O2-
III. Na+

A. I
B. I and II
C. II and III
D. I, II and III

10. Which of the following has one unpaired electron in its ground state?
I. Na
II. Al
III. Cl

A. I
B. I and II
C. II and III
D. I, II and III

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SECTION B

5.1 : QUANTUM NUMBERS OF ELECTRONS
1. Define orbitals.
2. State the four quantum numbers of an electron in an orbital and describe the physical

meaning of the four quantum numbers.
3. Which of the following orbitals are allowed?

(a) 2d (b) 7s (c) 3f (d) 4p
4. Which of the following quantum numbers are not allowed? Explain your answer.

(a) (1, 1, 0, +1/2) (b) (3, 1,-2, +1/2)

(c) (2, 1, 0, +1/2) (d) (2, 0,0, +1)

5. Draw the shape of the following orbitals.

s, px , py, dxz , dx2-y2 and dz2

5.2 : ELECTRONIC CONFIGURATION OF ATOM
6. (a) State Aufbau’s principle.

(b) Arrange the following orbitals in order of increasing energy.

4dxy, 3dxy, 3dyz, 4pz, 3pz, 3py, 2py, 3s, 2s, 1s, 4s
7. (a) State Hund’s rule and Pauli’s exclusion principle.

(b) Based on both the principle and rule, show how the electronic configuration of
element 23V is built.

(c) Draw the dxy and dx2 – 2 orbital in 23V and give the possible azimuthal and magnetic
y

quantum numbers for each of the orbitals.

8. State the quantum numbers n, ℓ and m of the 3dx2- 2 orbital and draw its shape.
y

9. The sets of quantum numbers below represent the four outermost electrons of element
Y at ground state.
n=3, ℓ =1, m=-1, s=+1/2
n=3, ℓ =1, m=-1, s=-1/2
n=3, ℓ =1, m=0, s=+1/2
n=3, ℓ =1, m=+1, s=+1/2

(a) Draw the orbital diagram of element Y.

(b) Draw the shapes of the orbitals for the electrons above.

(c) State the stable oxidation number of the ion formed by element Y.

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10 Element X has a proton number of 21.
. (a) Write the set(s) of quantum numbers for each of its valence electron(s).

(b) What is the electronic configuration of X+ ion?
11 The orbitals of the first two principal energy levels of atoms are shown below.
.

In an atom of element P, orbitals A, B, C and E are full of electrons while orbitals D are
half full of electrons.

(a) Write the electronic configuration of P.
(b) State the four quantum number for an electron that occupies orbital B and E.
(c) State the quantum number n, ℓ, and m for D orbital.
12 Phosphorus is an element in period 3.
(a) Write the electronic configuration of phosphorus.
(b) Draw its orbital diagram
(c) How many unpaired electrons does it have?
13 Element M has 15 protons.
(a) Write the electronic configuration of M 3-.
(b) Give a set of quantum numbers for an electron in the 3s orbital
(c) State two differences between 2s and 3s orbital

14 E is an element with proton number of 21
(a) Write the electronic configuration of E.
(b) Draw the shapes of orbitals occupied by the valence electrons.
(c) Give the sets of quantum numbers for the electrons that occupy the fourth shell.

Revised by:
Suhaina Binti Mohd Yatim
Zanarina Binti Sapiai

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CHAPTER 6: PERIODIC TABLE

GENERAL OVERVIEW

6.1 a) Specify periods, groups and
Classification blocks
(s, p, d, f )
of element
b) Deduce the position of elements
in the periodic table from the
electronic configuration

6.0 PERIODIC a) Describe the i. across period 2 & 3
TABLE variation in atomic
radii ii. down the group 1, 2, 13
to 18.

*include effective nuclear
charge and shielding effect

b) Compare the atomic radius of an element and
it's corresponding ionic radius.

* exclude the comparison of radii for
isoelectronic ions.

6.2 Periodicity c) Define electronegativity

d) Explain the variation in electronegativity:
i. across period 2 & 3
ii. down the group 1,2 13 & 18

e) Define first ionisation energy

f) Explain the variation in first ionisation energy:
i. across period 2 & 3
ii. down the group 1,2 13 & 18
*include the irregulation of the first ionisation
energy between Be & B and between N & O

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6.1 Classification of element
▪ The element in periodic table are arranged in order of increasing proton number (or

atomic number).
▪ The position of elements can be determined by using electronic configuration.
▪ Diagram 1 show that the the position of group and period in the periodic table.

block s group 13-18
block p
group
1-2

block d
group 3-12

G
R
O
U block f
P

Diagram 1

Group:
▪ A group is a vertical column of the periodic table.
▪ There are a total of 18 groups numbered as 1, 2, and 13 to 18. The other ten groups (3

to 12) are transition metals.
▪ Elements in the same group have the same number of valence electrons, thus same

chemical properties.

Block:
▪ Elements in the periodic table can be classified into four blocks according to their

valence electron configuration. (refer Diagram 1)
▪ These blocks are block s & p (main block), d and f.

s-block:
▪ The s-block is on the left side of the conventional periodic table and is composed of

elements from the first two columns.
▪ the nonmetals hydrogen and helium and the alkali metals (in group 1) and alkaline

earth metals (group 2).
▪ Valence electrons configuration: ns1 to ns2

p-block:
▪ The p-block is on the right side of the periodic table and includes elements from the

six columns beginning with column 13 to 18 (or group 13 to 18).
▪ Helium, which is in the top of column 18, is not included in the p-block.
▪ Valence electrons configuration: ns2 np1 to ns2 np6

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d-block:
▪ The d-block elements are found in the middle of the period table.
▪ Include element from column 3 to 12 (or group 3 to 12).
▪ Also known as transition metals.
▪ These elements called transition elements because its form a bridge between the

chemically active metals of s-block elements and non-metals elements of p-block.
▪ Valence electrons configuration: ns2 (n-1)d1 to ns2 (n-1)d10

f-block:
▪ also known by the name of inner transition elements.
▪ f-block elements are divided into two series, namely lanthanides and actinides.

EXAMPLE 1:
The elements in Group 1:

element electronic configuration valence valence electronic
electrons
Li (Z=3) 1s2 2s1 configuration
Na (Z=11) 1s2 2s2 2p6 3s1 1 (ns1)
K (Z=23) 1s2 2s2 2p6 3s2 3p6 4s1 1 2s1
1 3s1
4s1

EXAMPLE 2:
The elements in Group 2:

element electronic configuration valence valence electronic
electrons
4Be 1s2 2s2 configuration
12Mg 1s2 2s2 2p6 3s2 2 (ns2)
20Ca 1s2 2s2 2p6 3s2 3p6 4s2 2 2s2
2 3s2
4s2

EXAMPLE 3:
The elements in Group 13:

element electronic configuration valence valence electronic
electrons
5B 1s2 2s2 2p1 configuration
13Al 1s2 2s2 2p6 3s2 3p1 3 (ns2np1)
31Ga 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 3 2s2 2p1
3 3s2 3p1
4s2 4p1

EXAMPLE 4:
The elements in Group 17:

element electronic configuration valence valence electronic
electrons
9F 1s2 2s2 2p5 configuration
17Cl 1s2 2s2 2p6 3s2 3p5 7 (ns2 np5)
35Br 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 7 2s2 2p5
7 3s2 3p5
4s2 4p5

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EXAMPLE 5:
The elements in d-block::

element electronic configuration valence valence electronic
electrons
21Sc 1s2 2s2 2p6 3s2 3p6 4s2 3d1 configuration
23V 1s2 2s2 2p6 3s2 3p6 4s2 3d3 3 ns2 (n-1)d1 to10
30Zn 1s2 2s2 2p6 3s2 3p6 4s2 3d10 5
12 4s2 3d1
4s2 3d3
4s2 3d10

EXAMPLE 6:
Element 39Y have electronic configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d1
Determine valence electrons, valence electronic configuration and group for 39Y.

(Answer 3, 5s2 4d1, 3)

EXERCISE 1:
Classify the following elements into its appropriate group and block.

A: 1s2 2s2 2p6 3s2 3p6
B: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
C: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
D: 1s2 2s2 2p6 3s2 3p6 4s2 3d5

EXERCISE 2:
Given that element M and N has proton number 17 and 19 respectively.

a) Write electronic configuration of element M and N.
b) Determine the

i) Period
ii) Group
iii) Block

Period:
▪ A period is the name given to a horizontal row of the periodic table.
▪ The periodic table has seven periods and numbered as 1 to 7.
▪ All of the elements in a period have the same number of highest principle quantum

number, n.
▪ Elements of the same period have the same number of electron shells.

EXAMPLE 7:

Classify the following elements into its appropriate period.

P: 1s2 2s2 2p6 3s2 3p6
Q: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
R: 1s2 2s2
S: 1s2 2s2 2p6 3s2 3p6 3d34s2

(Answer: P=3, Q=4, R=2, S=4)

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EXAMPLE 8:

The electronic configurations for elements M and N are given as follows:

M 1s 2 2s 2 2p 6 3s 2 3p 1
N 1s 2 2s 2 2p 6 3s 2 3p 5

a) State the group and the period for elements M and N.
b) What is the molecular formula for the chloride compound formed by element M?
c) Write the molecular formula for the compound formed when elements M and N react.

Answer: (a) M = group 13, period 3; N = group 17, period 3
(b) MCl3
(c) MN3

6.2: Periodicity

▪ The atomic radius of an element is determined by two factors which is effective

nuclear charge, Zeff and shielding effect.

• Effective nuclear charge, Zeff is the net positive charge experienced by valence

electrons.
▪ Zeff = Z – S
▪ Z = number of proton and S = number of electrons filled at the inner shell.
▪ When Zeff increases, the nucleus attraction towards electrons become stronger.
▪ The nucleus pulls the outer electrons closer.
▪ The atomic radius becomes smaller.

EXAMPLE 1:
Explain why size of atom 17Cl is smaller than 12Mg.

Answer:

17Cl : 1s2 2s2 2p6 3s2 3p5 Zeff : 17-10 = +7
12Mg:1s2 2s2 2p6 3s2 Zeff : 12-10 = +2

• The atomic radius of Cl is smaller because the nucleus pulls the outer electrons

closer with a charge of +7

• The atomic radius of Mg is bigger than Cl because the nucleus can only pull the outer

electrons closer with a charge of +2.

Shielding effect

▪ The shielding effect is caused by the mutual repulsion between electrons of inner
shell and the electrons occupying valence shell.

▪ Also known as the screening effect.
▪ It also occurs between electrons in the same shell but is less effective compared to

that of electrons in the different shells.

▪ Generally: Shielding effect ∝ Atomic radius

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▪ So, more the shielding⇒more the outer electrons are shielded ⇒The attractive
pull by the nucleus on the outer electrons decreases⇒ Thus, the size ( radius) of

the atom increases.

EXAMPLE 2:
Why is potassium bigger than sodium?

Answer:
• The element sodium has the electron configuration 1s22s22p63s1. The outer energy
level is n = 3 and there is one valence electron. The attraction between this lone
valence electron and the nucleus with 11 protons is shielded by the other 10 core
electrons.
• The electron configuration for potassium is 1s22s22p63s23p64s1. While there are more
protons in a potassium atom, there are also many more electrons shielding the outer
electron from the nucleus. The outermost electron, 4s1, therefore, is held very
loosely. Because of shielding, the nucleus has less control over this 4s1 electron than
it does over a 3s1 electron.

▪ Variation in Atomic Radii:

Across a period Down a group
- proton number increases. - no of shell, n increases.
- effective nuclear charge, Zeff - shielding effect increases.
- attraction between valence electrons
increases.
- attraction between valence and nucleus weaker.
- atomic size increases.
electrons and nucleus stronger.
- atomic size smaller.

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EXAMPLE 3:
Explain the change in atomic radius from

(i) Li to Rb
(ii) Na to Ar

(PSPM SK016 2006/2007)

Answer: Atomic radius increase from Li to Rb because of increasing no of shell and
(i) increasing screening effect in Rb. Attraction between nucleus and outer electron
weaker. So the size is increased.
(ii) Atomic radius decreased from Na to Ar because proton number increase, Zeff
increased. Attraction between nucleus and outer electron stronger. So the size is
decreased.

EXERCISE 1:

Elements 11A, 12B and 16C are found in periodic table. Which element has the largest atomic
radius? Explain your answer.

(PSPM SK016 2010/2011)

(a) Compare the atomic radius of an element and it's corresponding ionic radius.
* exclude the comparison of radii for isoelectronic ions.

• Introduction of cation and anion.

Cation Anion
- Anions are atoms or radicals (groups
- Cations are formed when a neutral of atoms), that have gained electrons
atom loses an electron. - consequently giving it a net negative
charge.
- The alkali metals (group I) lose a single
electron to form a cation with a 1+ - The number of electrons gained, and
charge. so the charge of the ion.

- The alkaline earth metals (group II - Chlorine (Cl) gains one electron to
elements) lose two electrons to form a
2+ charge. become Cl-, while oxygen (O) gains two

- Aluminium, a member of the group III, electrons to become O2-.
loses three electrons to form a 3+
charge. - Example:

- Examples: Cl + e- Cl-
Na Na+ + e-
Al Al3+ + 3e- O + 2e- O2-

N + 3e- N3-

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Radius (pm)EXAMPLE 4:
• Observe the graph Diagram 1 below:

Diagram 1

• Generally, the size of neutral atomic
radius is bigger than its cation.

• Meanwhile, The size of neutral atomic
radius is smaller than its anion.

Proton number

▪ Why the size of atom Na is bigger than its cation?

Answer:

▪ 11Na : 1s2 2s2 2p6 3s1 & 11Na+ : 1s2 2s2 2p6
▪ Cation of Na+ is formed when an atom loses its valence electron.
▪ Neutral atom (Na) & cation (Na+) have same no. of proton, but Na+ contains less

electrons/ less no of shell than its neutral atom.
▪ Therefore, for Na+ the attractive forces between nucleus and remaining electrons is

greater than in the neutral atom.
▪ Thus, the size of cation is smaller than its neutral atom.

EXAMPLE 5:
▪ Why the size of atom Cl is smaller than the size of its anion?

Answer:

▪ 17Cl : 1s2 2s2 2p6 3s2 3p5 & 17Cl- : 1s2 2s2 2p6 3s2 3p6
▪ Anion is formed when an atom accept electron.
▪ Neutral atom (Cl) & cation (Cl-) have same no. of proton, but Cl- contains more

electrons than its neutral atom.
▪ Therefore, for Cl- the mutual repulsion between the electrons increases.
▪ The repulsion caused the electron cloud spread out and the outer orbital expand.
▪ Thus, the size of anion is larger than its neutral atom.

EXERCISE 2:
• Choose the larger particle in each pair:
(a) Al or Al3+
(b) S or S2–

Answer:
(a) Size of Al atom is larger than Al3+.
(b) Size of S2– atom is larger than S atom.

EXERCISE 3:
Define
(i) atomic radius.
(ii) Describe the periodic trend for atomic radius.

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(b) Electronegativity
▪ Electronegativity is a relative ability of a bonded atom to attract the shared electrons.
▪ Imagine a game of tug-of-war. If the two teams are of equal strength, the rope stays

centered.
▪ If one team is stronger, the rope is pulled in that team's direction.
▪ If one team is overwhelmingly stronger, the weaker team is no longer able to hold

onto the rope and the entire rope ends up on the side of the stronger team.

▪ This is analogous to chemical bonds. If the two atoms of the bond are of equal
electronegativity, the electrons are equally shared.

▪ Meanwhile, if one atom is more electronegative, the electrons of the bond are more
attracted to that atom.

(c) Variation in electronegativity
▪ Generally, the arrangement of electronegativity of elements in periodic table as

shown in Diagram 2 below, with the most electronegative element is Fluorine.

Diagram 2
(i) Variation of electronegativity accross period 2 and 3.

• Generally, the electronegativity of elements increase across the period 2 and 3.
• When across the period, proton number increases.
• Effective nuclear charge, Zeff increases.
• Attraction between valence electrons and nucleus stronger.
• Atomic size decreases.
• Ability of an atom to attract the shared electrons increases.
• So, electronegativity increases.

(ii) Variation of electronegativity going down the group 1, 2 and 13 to 18.
▪ When going down the group, no of shell, n increases.
▪ Shielding effect increases.
▪ Attraction between valence electrons and nucleus become weaker.
▪ Atomic size increases.
▪ Ability of an atom to attract the shared electrons decreases.
▪ So, the electronegativity decreases.

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EXAMPLE 6:
Electronegativity of Lithium (Li) is smaller than Oxygen (O). Why? Explain your answer.

Answer:
• Proton number of Li (Z=3) is smaller than O (Z=8).
• Zeff of O > Zeff of Li.
• Attraction between valence electrons and nucleus stronger in Oxygen.
• Size of O ˂ Li.
• Ability of an atom O to attract the shared electrons increases.
• So, electronegativity of O is bigger than Li.

EXAMPLE 7:
Electronegativity of Beryllium (Be) is larger than Calcium (Ca). Why? Explain your answer.

Answer:
• When going down the group, no of shell, n increases. That means, Ca have more
shells than Be.
• The increase of shells, means the size become bigger in Ca.
▪ Shielding effect increases in Ca. So, attraction between valence electrons and
nucleus become weaker.
▪ Ability of an atom to attract the shared electrons decreases.
▪ So, the electronegativity decreases from Be to Ca.

EXERCISE 4:
On the Pauling scale the electronegativities of nitrogen and oxygen are respectively 3.0 and
3.5. Why is oxygen more electronegative than nitrogen?

( Question refference: https://www.chemguide.co.uk/atoms/questions/q-electroneg.pdf )
( Answer refference: https://www.chemguide.co.uk/atoms/questions/a-electroneg.pdf )

EXERCISE 5:
By thinking about where the following atoms are in the Periodic Table, sort them into order
of increasing electronegativity:

Aluminium, Barium, Boron, Caesium, Calcium, Carbon, Fluorine.

( Question refference: https://www.chemguide.co.uk/atoms/questions/q-electroneg.pdf )
( Answer refference: https://www.chemguide.co.uk/atoms/questions/a-electroneg.pdf )

d) Ionisation energy
• Ionisation energy is the minimum energy required to remove an electron from a

gaseous atom or ion in its ground state.

Minimum energy required
to remove electron

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The First Ionisation Energy (IE1):
• Is the minimum energy (in kJ/mol) required to remove an electron from a gaseous
atom in its ground state.

Example : Mg(g) → Mg+(g) + e- H= 736 kJ/mol

Additional note:
• The Second Ionisation Energy (IE2):
• Is the minimum energy required to remove an electron from positive gaseous ion in
its ground state.

Example : Mg+(g) →Mg2+(g) + e- H= 1450 kJ/mol

(d) Variation in first ionisation energy.
• Generally, across the period ionisation energy increases.
• Down the group, ionisation energy decreases.

(i) Variation of IE1 across period 2 and 3
• Generally proton number increases when across the period.
• Effective nuclear charge, Zeff increases.
• Attraction between valence electrons and nucleus become stronger.
• ionisation energy increases

(ii) Variation of IE1 down the group 1, 2 and 13 to 18.
• When down the group, no of shell, n increases.
• Shielding effect increases.
• Attraction between valence electrons and nucleus become weaker.
• So, ionisation energy decreases.

Irregulation of IE1 in Beryllium (Be) and Boron (B).
• Generally, across the period ionisation energy increases.
• However, there are special cases against this general rule. For example, the
ionisation energy of of Boron (B) in group 13 is in fact lower than Be (Beryllium) in
group 2. Why?

• Following are the electronic configurations of two elements:
5B : 1s2 2s2 2p1
4Be : 1s2 2s2

• The orbital diagram for these two elements are:

5B ____ ____ ____
1s 2s 2p

4Be ____ ____
1s
2s
(half-filled) – have extra stability – required more energy

• In Boron, valence electron do not have half-filled configuration. So, it is required less
energy to remove electron from outermost shell.

• However, Beryllium have half-filled configuration in outermost shell. This
phenomenon make the 2s orbital have extra stability. So, it is required more energy
in order to remove the electron from this orbital.

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Irregulation of IE1 in Nitrogen (N) and Oxygen (O).
• Another case of application in half-filled phenomenon is the first ionization energy
between Nitrogen and Oxygen.
• Why IE1 of N (group 15) is higher than O (group 16)?
• Following are the electronic configurations of two elements:

N: 1s2, 2s2, 2p3 and O: 1s2, 2s2, 2p4

• Orbital diagram of N and O as shown below:

7N ____ ____ ____ ____ ____
1s 2s 2p

(half-filled configuration)

8O ____ ____ ____ ____ ____
1s 2s 2p

• In Nitrogen, valence electron have half-filled configuration. This phenomenon make
the 2s orbital have extra stability So, it is required more energy to remove electron
from outermost shell.

• However, oxygen do not have half-filled configuration in outermost shell. So, it is
required less energy in order to remove the electon from this orbital.

EXAMPLE 8:
Based on their positions in the periodic table, predict which has the largest first ionization
energy: 12Mg, 3Li, 5B, 8O, 15P. Explain your answer.

Answer:
•O
• Because O in terms of position in periodic table, the element located extremely right
(group 16) together with smallest no of period (period 2).

EXAMPLE 9:
Explain, using shielding and effective nuclear charge, why F has a larger first ionization than
O.

Answer:

• F: Zeff = 9 – 2 = 7
• O: Zeff = 8 – 2 = 6

• The greater the effective nuclear charge, the more difficult it is to remove a valence
electron.

• It requires more energy to remove an electron from F than from O.

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EXERCISE 6:
The table below shows elements and their proton number.

Element Proton number
P 8
Q 10
R 11
S 13

a) Arrange the element in increasing order of first ionisation energy.
Explain your answer.

b) Which element has the highest electronegativity?

EXERCISE 7:
Generally, across the period ionisation energy increases. However, Sulphur (16S) has a
smaller first ionization energy than expected. Explain why?.

Answer refference:

file:///C:/Users/KMKtDesk/Downloads/WorkedSolutions-3-04-PeriodicTrends-
IonizationEnergy.pdf

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TUTORIAL 6

SECTION A

1. What name is given to the vertical columns in the periodic table?

A. Groups
B. Periods
C. Blocks
D. Noble gas

2. What group is aluminium in?

A. 1
B. 2
C. 13
D. 14

3. Which ion below has the largest radius?

A. Cl–
B. K+
C. Br–
D. F–

4. Why are the elements in Group 18 unreactive?

A. They have very large atoms.
B. They have even numbers of electrons.
C. They have full outer shells.
D. They have half-filled configuration.

5. Atomic radius generally increases as we move __________.

A. down a group and from right to left across a period
B. up a group and from left to right across a period
C. down a group and from left to right across a period
D. up a group and from right to left across a period

6. When group 2 ions form, the outer electrons are found in

A. s orbitals
B. p orbitals
C. d orbitals
D. f orbitals

7. The atomic radius of main-group elements generally increases down a group because

A. effective nuclear charge increases down a group
B. effective nuclear charge decreases down a group
C. effective nuclear charge zigzags down a group
D. the principal quantum number of the valence orbitals increases

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8. Of the following atoms, which has the largest first ionization energy?

A. Br
B. O
C. C
D. P

9. Which of the elements below has the largest electronegativity?

A. Si
B. Mg
C. P
D. S

10. Of the choices below, which gives the order for first ionization energies?

A. Cl > S > Al > Ar > Si
B. Ar > Cl > S > Si > Al
C. Al > Si > S > Cl > Ar
D. Cl > S > Al > Si > Ar

11. Ability of atom to attract electrons towards itself is called

A. Electron affinity
B. Electronegativity
C. Ionization energy
D. Shielding effect

12. Which of the following statement (s) about the Modern Periodic Table are incorrect
i. The elements in the Modern Periodic Table are arranged on the basis of their
decreasing atomic number.
ii. The elements in the Modern Periodic Table are arranged on the basis of their
increasing atomic masses.
iii. Isotopes are placed in adjoining group (s) in the Periodic Table.
iv. The elements in the Modern Periodic Table are arranged on the basis of their
increasing atomic number.

A. (i) only
B. (i), (ii) and (iii)
C. (i), (ii) and (iv)
D. (iv) only

13. Physical properties of the elements in periodic table depends on the

A. size of atom
B. size of proton
C. size of neutron
D. no. of electrons

14. The elements in group I are not usually found free in nature because

A. they are very reactive.
B. they are very unreactive.
C. they are all man made.
D. they are rare earth metals.

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15. Name the family of elements highlighted in the Periodic Table below

A. Alkali metals
B. Halogens
C. Alkaline earth metals
D. Metals

16. If an element has one to three valence or outer level electrons, then it is a(n):

A. Nonmetal
B. Halogen
C. Noble Gas
D. Metal

17. Each element in the periodic table is assigned a number called the atomic number.
This is equal to

A. the number of neutrons in each atom of that element.
B. the number of protons plus the number of electrons in each atom of that element.
C. the number of protons in each atom of that element.
D. the number of protons and neutrons in each atom of that element.

18. Why are the elements fluorine, chlorine and iodine placed in the same group of the
Periodic Table?

A. They all readily react with oxygen.
B. They are all metals.
C. They have the same number of electrons in their outer shell.
D. They have the same density.

19. What is meant by a group in the Periodic Table?

A. A vertical row of elements
B. The family of metals in the Periodic Table
C. The family of non-metals in the Periodic Table
D. A horizontal row of elements

20. Which of the following statements is/are correct about the group 17 elements?
(i) They are known as the halogens.
(ii) They all have the same melting point.
(iii) They have similar chemical properties.

A. Statements (i) and (ii)
B. Statement (i) only
C. Statements (i) and (iii)
D. Statements (ii) and (iii)

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