The Periodic Table – Period 3
Chapter 4
The Periodic Table
- Period 3
Preface
Before we begin, this chapter requires quite a lot of memorisation, but it is possible to memorise certain concepts through understanding and frequent application through practice questions.
Graph Trends
These trends are commonly tested in Paper 1 MCQs and should be at your fingertips after much revision.
Test Yourself: [A Level H2 Chemistry 2013 Paper 1 A13]
For the elements in the third period of the Periodic Table, which property decreases consistently from sodium to chlorine?
- electrical conductivity
- ionisation energy
- melting point
- radius of the atom
Solution (with Detailed Explanation for good revision purposes):
Option A:
Electrical conductivity increases from Na to Al due to the increase in number of delocalised electrons from 1 to 3. Si is a semiconductor/metalloid, hence it has low electrical conductivity and there is a sharp decline. As for P, S and Cl, they are simple molecules with no delocalised electrons (charge carriers) and thus the electrical conductivity graph declines to zero. There is an initial increase before the sudden decline in trend, thus Option A is wrong.
Option B: [Recall Atomic Structure - Ionisation Energy]
First ionisation energy generally increases across a period as nuclear charge increases (due to increase in number of protons) and there is a negligible increase in shielding effect (since successive electrons are added to the same valence shell), resulting in increase in effective nuclear charge and valence electrons becoming more strongly attracted to the nucleus. Hence, more energy is needed for removal of valence electrons from every atom.Since there is a general increasing trend, Option B is wrong.
Option D: [Recall Atomic Structure - Atomic Radius]
Atomic radius decreases across the period due to increase in nuclear charge (since there is increase in number of protons) and negligible increase in shielding effect (successive electrons are added to the same electron shell), resulting in increase in effective nuclear charge. Hence, valence electrons are more strongly attracted to the nucleus across the period, showing that atomic radius decreases across the period. Therefore, Option D is the right answer.
Option C:
Melting point generally increases from Na to Si, followed by a sharp decrease from Si to P, and fluctuations between P to Ar. Thus Option C is false too.
Major tip
When attempting MCQs, always try to give an explanation for every option (optimal when studying together and discussing with friends) to maximise and consolidate your learning for the chapter.
Summary Comparison Tables for Easier Revision:
| Oxides | Reactions | pH |
|---|---|---|
| Na₂O (basic) | Na₂O(s) + H₂O(l) → 2NaOH(aq) Reacts vigorously to give alkaline solution |
About 13 |
| MgO (basic) | MgO(s) + H₂O(l) ⇌ Mg(OH)₂(s) Mg(OH)₂(s) + aq ⇌ Mg²⁺(aq) + 2OH⁻(aq) Slightly soluble to give a weakly alkaline solution |
About 9 |
|
Al₂O₃ SiO |
Insoluble in water, no reaction | 7 |
| P₄O₁₀ | P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq) Reacts to give weak tribasic phosphoric acid |
About 2 |
| SO₃ | SO₃(g) + H₂O(l) → H₂SO₄(aq) | About 1 |
Acid-Base Behaviour of Oxides or Hydroxides
There is no need for complete memorisation of all the equations. Some of them can be derived easily, thus leaving space in your memory for other things.
- You need to recognise the acid-base nature of oxides and hydroxides of period 3. Metallic oxides and hydroxides are basic, non-metallic oxides and hydroxides are acidic. As for Al₂O₃ and Al(OH)₃, they are amphoteric in nature.
- When undergoing acid-base reaction, the products are salt and H₂O. Hence, for acid-base reactions involving HCl and basic oxides or hydroxides, those equations can be easily formed and balanced.
The following equations in the table are the ones that you must remember:
| Oxides | Reactions with NaOH |
|---|---|
| Al(OH)₃ | Al₂O₃(s) +2NaOH(aq) +3H₂O(l) → 2Na[Al(OH)₄](aq) Al(OH)₃(s) + NaOH(aq) → Na[Al(OH)₄](aq) |
| SiO₂ | No reaction with dilute NaOH(aq) Reaction with concentrated NaOH at high temperature (extreme conditions): SiO₂(s) + 2NaOH(conc) → Na₂SiO₃(aq) + H₂O(l) |
| P₄O₁₀ | P₄O₁₀(s) +12NaOH(aq) → 4Na₃PO₄(aq) + 6H₂O(l) |
| SO₃ | SO₃(g) +2NaOH(aq) → Na₂SO₄(aq) + H₂O(l) |
Reaction of Chlorides with Water
| Chlorides | Reactions | pH of Solution |
|---|---|---|
| Na₂O | NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq) Dissolves readily, no hydrolysis |
7 |
| MgCl₂ |
Dissolution: weaken The image below may serve as a visual aid for you to understand better: |
6.5 |
| AlCl₃ | Dissolution: AlCl₃(s) + 6H₂O(l) → [Al(H₂O)₆]³⁺(aq) + 3Cl⁻(aq) Hydrolysis: [Al(H₂O)₆]³⁺(aq) ⇌ [Al(H₂O)₅(OH)]²⁺(aq) + H⁺(aq) Dissolves and hydrolyse; Al³⁺ has high charge density, able to polarise, weaken and break O-H bond in H₂O |
3 |
| SiCl₄ |
SiCl₄(l) + 4H₂O(l) → SiO₂.2H₂O(s) + 4HCl(aq) Acidic solution of HCl formed |
2 |
| PCl₅ | Case 1: Limited supply of water (cold) PCl₅(s) + H₂O(l) → POCl₃(s) + 2HCl(aq) Case 2: Large amount of water PCl₅(s) + H₂O(l) → H₃PO₄(aq) + 5HCl(aq) Acidic solution of HCl formed |
1 |
Test Yourself - Application Time:
[2017 CJC H2 Chemistry Prelim Paper 1 Qn 14]
The oxide and chloride of an element E are mixed separately with water. The two resulting solutions have the same effect on litmus.
What is element E?
Solution with Detailed Explanation:
For Option B and C, do note that both SiO₂ and Al₂O₃ are insoluble in water. The resulting solution would therefore be neutral. Both SiCl₄ and AlCl₃ will undergo hydrolysis to give acidic solutions.
PCl₅ undergoes hydrolysis to form H₃PO₄ and HCl, which causes the resulting solution to be acidic and turns blue litmus red. P₄O₁₀ would also undergo hydrolysis to give H₃PO₄, which is an acidic solution.
MgO gives Mg(OH)₂ in water which is weakly basic whereas MgCl₂ is weakly acidic.
Therefore, the answer is Option D.
Tips and tricks:
By observing trends across the period and understanding the reason behind why such trends appear, remembering them becomes an easier feat. Sometimes, blind memorisation can be tough and futile, but understanding the logic behind trends can solidify our understanding and memory.
- Notice that as we go across the period, the oxidation number of oxides increases. Why is that so?
Reason: The valence electrons available for bond formation increases. - Following a similar train of thought, we understand that P and S show several oxidation numbers because they are able to expand their octet due to vacant and energetically accessible 3d orbitals available.
- Across period 3, the nature of period 3 oxides change from basic to acidic, and their structures from giant ionic lattice structure to simple molecular structure. This can be attributed to period 3 elements changing from metals to non-metals across the period.
- Why is Al₂O₃ amphoteric?
Reason: Al₂O₃ is an ionic oxide. Since Al³⁺ has high charge density (∝), it has high polarising power and is able to polarise the small O²⁻ ion, resulting in Al₂O₃ having some covalent character (refer to Chemical Bonding). Mixture of ionic and covalent character results in Al₂O₃ displaying both acidic and basic properties. - Now you may ask, what about AlCl₃?
Even though Al is a metal, AlCl₃ is a simple covalent molecule.Reason: Al3+ has high charge density (∝), it has high polarising power and is able to polarise the large electron cloud of Cl- to a larger extent, compared to the smaller O²⁻ ion that has a smaller electron cloud and is less polarised.