Chemistry Form 2: C. PERIODICITY OF BONDING AND STRUCTURE

C. Periodicity of Bonding and Structure

The periodic table does not classify elements simply as metals and non-metals. Instead, it arranges them in order of increasing atomic numbers.

However, based on the structure and bonding of the elements in the periodic table:

• The top right-hand corner of about twenty elements are non-metals.
• To the left of each non-metal is an element that shows characteristics of both metals and non-metals. These elements are called semi-metals/metalloids. They include Boron, Silicon, Germanium, Arsenic, and Tellurium.
• All other elements in the periodic table are metals.
• Hydrogen is a non-metal with metallic characteristics, such as donating or losing its outer electron to form a cation/H⁺ ion.
• Bromine is the only known natural liquid non-metal element at room temperature and pressure.
• Mercury is the only known natural liquid metal element at room temperature and pressure.
• Carbon-graphite is a semi-metal/metalloid, while carbon-diamond is a pure non-metal; both are allotropes of carbon (the same element).

a) Sketch of the periodic table showing metals, metalloids, and non-metals

Metals Metalloids Non-metals

b) Periodicity in the physical properties of elements across periods 2 and 3

Study Table I and Table II below:

Table I (Period 2)

Table II (Period 3)

From Table I and Table II above:

1. Explain the trend in atomic radius along/across a period in the periodic table

Observation: The atomic radius of elements in the same period decreases successively across the period from left to right.

Explanation: Across the period from left to right, there is an increase in nuclear charge due to the addition of protons in the nucleus and electrons entering the same energy level.

The increase in nuclear charge increases the effective nuclear attraction on the outer energy level, pulling it closer to the nucleus successively across the period. For example:

• From the tables above, the atomic radius of Sodium (0.157 nm) is larger than that of Magnesium (0.137 nm). This is because Magnesium has a stronger effective nuclear attraction on the outer energy level than Sodium, hence pulls the outer energy level closer to its nucleus.
• The rate of decrease in atomic radius becomes smaller as the atom becomes heavier. For example, the atomic radius of Magnesium from Sodium falls by (0.157 nm – 0.137 nm) = 0.02, while the atomic radius of Chlorine from Sulfur falls by (0.104 nm – 0.099 nm) = 0.005. This is because adding one more proton to 11 already present causes a greater proportional change in nuclear attraction power for Magnesium than adding one more proton to 16 already present in Sulfur to Chlorine.
• Period 3 elements have more energy levels than Period 2 elements. Therefore, they have larger atomic radii than corresponding Period 2 elements in the same group.

2. Explain the trend in ionic radius along/across a period in the periodic table

Observation: The ionic radius of elements in the same period decreases successively across the period from left to right for the first three elements, then increases drastically, then slowly decreases again.

Explanation: Across the period from left to right, elements change from electron donors/losers (reducing agents) to electron acceptors (oxidizing agents).

• An atom forms a stable ion by either gaining extra electrons or donating outer electrons.
• Metals form stable ions by donating all the outer energy level electrons, thus losing the outer energy level. For example, the sodium ion has one less energy level than the sodium atom. The ion is formed by the sodium atom donating all the outer energy level electrons, making the ion smaller than the atom.
• Ionic radius therefore decreases across the period from Lithium to Boron in period 2 and from Sodium to Aluminium in period 3. This is because the loss of electrons increases the effective nuclear attraction on the remaining electrons.
• Non-metals form stable ions by gaining extra electrons in the outer energy level. The extra electrons increase repulsion among electrons and reduce the effective nuclear attraction on the outer energy level. The outer energy level therefore expands to accommodate the extra repelled electrons. The more electrons gained, the greater the repulsion and expansion, resulting in a larger ionic radius. For example, the nitrogen ion has three more electrons than the nitrogen atom, causing the outer energy level to expand.
• Ionic radius decreases from group IV onwards from left to right because the number of electrons gained to form ions decreases across the period. For example, the nitrogen ion has a larger ionic radius than oxygen.

3. Explain the trend in melting and boiling points of elements in a period in the periodic table

Observation: The melting and boiling points of elements rise up to the elements in Group IV (Carbon/Silicon) along the period, then continuously fall.

Explanation: Melting and boiling points depend on the packing of the structure making the element and the strength of the bonds holding the atoms or molecules together.

Across the period (2 and 3), the structure changes from giant metallic, giant atomic/covalent to simple molecular.

• For metals, the number of delocalized electrons increases across the period, resulting in stronger metallic bonds and structures, thus requiring more heat or energy to weaken.
• The strength of a metallic bond also depends on atomic radius. The melting and boiling points decrease as the atomic radius of metals increases due to decreased packing of larger atoms. For example, the melting and boiling points of Lithium are lower than those of Beryllium because Beryllium has more delocalized electrons and hence a stronger metallic bond. Lithium’s melting and boiling points are higher than Sodium’s because Lithium atoms are smaller and better packed.
• Carbon-graphite, carbon-diamond in period 2, and Silicon in period 3 form very well-packed giant atomic/covalent structures held together by strong covalent bonds, resulting in very high melting and boiling points. Both carbon allotropes have smaller atomic radii than Silicon, leading to higher melting and boiling points due to better packing.
• Non-metals from group V onwards form simple molecules joined by weak intermolecular van der Waals forces, requiring little energy to weaken, leading to low melting and boiling points. The strength of these forces decreases with decreasing atomic radius, lowering melting and boiling points across the period. For example, Nitrogen has a higher melting and boiling point than Oxygen because of stronger intermolecular forces due to its larger atomic radius.
• Rhombic sulphur exists as a puckered ring of S₈ atoms which are well packed. Before melting, the ring breaks and forms long chains that entangle, causing the unusually high melting and boiling points of Rhombic sulphur.
• Sulphur and phosphorus exist as allotropes. Sulphur exists as Rhombic and monoclinic forms, with Rhombic being stable at room temperature. Phosphorus exists as white and red forms, with white phosphorus stable at room temperature.

4. State and explain the trend in density of elements in a period in the periodic table

Observation: Density increases up to the elements in group IV then falls successively across the period.

Explanation: Density is the mass per unit volume occupied by matter.

• For metals, stronger metallic bonds and more delocalized electrons ensure a well-packed giant metallic structure that occupies less volume, resulting in higher density. The more delocalized electrons across the period, the higher the density. For example, Aluminium has a higher density than Sodium because it has more delocalized electrons, forming a more compact structure.
• Carbon-graphite, carbon-diamond, and silicon in group IV form well-packed giant atomic/covalent structures joined by strong covalent bonds, occupying less volume per given mass. Carbon-graphite forms a less well-packed hexagonal planar structure joined by van der Waals forces, so its density (2.25 g/cm³) is less than that of carbon-diamond (3.53 g/cm³) and silicon (2.33 g/cm³). Carbon-diamond has a smaller atomic radius than silicon, resulting in better packing and higher density. Carbon-diamond is the hardest known natural substance due to its high density.
• For non-metals, the strength of van der Waals forces decreases with decreasing atomic radius across the period, reducing the mass occupied by a given volume of atoms. For example, phosphorus has a higher atomic radius than chlorine and argon, resulting in stronger intermolecular forces and higher density.

5. State and explain the trend in thermal and electrical conductivity of elements in a period in the periodic table

Observation: Conductivity increases across groups I, II, and III, then decreases in group IV and drastically decreases from group V to VIII.

Explanation:

• Metals have free delocalized electrons responsible for thermal and electrical conductivity. Conductivity increases with the number of delocalized electrons but decreases with increasing temperature. For example, Aluminium, with three delocalized electrons per atom, has the highest conductivity in period 3.
• Carbon-graphite also has free fourth valence electrons delocalized within its layers, responsible for its electrical conductivity.
• Silicon and carbon-diamond do not conduct electricity but conduct heat. Their closely packed giant tetrahedral structures allow heat transfer between atoms. Thermal conductivity increases with temperature.
• All other non-metals are poor conductors of heat and electricity because their molecules lack free or mobile electrons.

Periodicity of the oxides of elements along/across period 3

The table below summarizes some properties of the oxides of elements in period 3 of the periodic table.

1. All the oxides of elements in period 3 except those of sulphur and chlorine are solids at room temperature and pressure.

2. Across the period, bonding of the oxides changes from ionic in sodium oxide, magnesium oxide, and aluminium oxide (which show both ionic and covalent properties) to covalent in the rest of the oxides.

3. Across the period, the structure of the oxides changes from giant ionic structures in sodium oxide, magnesium oxide, and aluminium oxide to giant atomic/covalent structure in silicon (IV) oxide. The rest of the oxides form simple molecular structures.

4. Sodium oxide and magnesium oxide are basic/alkaline in nature. Aluminium oxide is amphoteric (shows both acidic and basic characteristics). The rest of the oxides are acidic.

5. Ionic oxides have very high melting and boiling points because of the strong electrostatic attraction in the giant ionic crystal lattice. The melting and boiling points increase from sodium oxide to aluminium oxide as the number of electrons involved in bonding increases, strengthening the ionic bond.

6. Silicon (IV) oxide has a well-packed giant atomic/covalent structure joined by strong covalent bonds, resulting in very high melting and boiling points.

7. Phosphorus (V) oxide, sulphur (IV) oxide/sulphur (VI) oxide, and dichlorine heptoxide exist as simple molecules joined by weak van der Waals forces, resulting in low melting and boiling points.

8. Ionic oxides conduct electricity in molten and aqueous states but not in solid state because ions are fixed in solids but free in molten or aqueous states. Sodium oxide, magnesium oxide, and aluminium oxide are good conductors when molten or dissolved.

9. Covalent oxides do not conduct electricity in any state because they lack free ions. Phosphorus (V) oxide, sulphur oxides, and dichlorine heptoxide are insulators.

10. Silicon (IV) oxide is a poor conductor of heat in solid state due to its closely packed structure.

11. Electropositivity decreases and electronegativity increases across the period, making oxides less ionic and more covalent.

12. The change from giant ionic to giant atomic/covalent to simple molecular structures leads to differences in reactions with water, acids, and alkalis:

• Reaction with water
• a) Ionic oxides react with water to form alkaline solutions, e.g.:

• I. Sodium oxide reacts with water forming sodium hydroxide:
  Na₂O(s) + H₂O(l) → 2NaOH(aq)
• II. Magnesium oxide reacts slowly with water forming magnesium hydroxide:
  MgO(s) + 2H₂O(l) → Mg(OH)₂(aq)
• III. Aluminium oxide does not react with water.

• b) Non-metallic oxides are acidic and react with water to form weakly acidic solutions:
• I. Phosphorus (V) oxide reacts with water forming phosphoric (V) acid:
  P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq)
  P₂O₅(s) + 3H₂O(l) → 2H₃PO₄(aq)
• II. Sulphur (IV) oxide reacts with water forming sulphurous (IV) acid:
  SO₂(g) + H₂O(l) → H₂SO₃(aq)
  Sulphur (VI) oxide reacts with water forming sulphuric (VI) acid:
  SO₃(g) + H₂O(l) → H₂SO₄(aq)
• III. Dichlorine oxide reacts with water forming hypochlorous acid:
  Cl₂O(g) + H₂O(l) → 2HClO(aq)
• IV. Dichlorine heptoxide reacts with water forming perchloric acid:
  Cl₂O₇(l) + H₂O(l) → 2HClO₄(aq)
• c) Silicon (IV) oxide does not react with water but reacts with hot concentrated alkalis to form silicate salts:
  SiO₂(s) + 2NaOH(aq) → Na₂SiO₃(aq) + H₂O(l)

• Reaction with dilute acids
• a) Ionic oxides react with dilute acids to form salt and water (neutralization), e.g.:
• Na₂O(s) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l)
• MgO(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂O(l)
• Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)
• Aluminium oxide is amphoteric and reacts with hot concentrated alkalis to form complex salts:
  Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2NaAl(OH)₄(aq)
  Al₂O₃(s) + 2KOH(aq) + 3H₂O(l) → 2KAl(OH)₄(aq)
• b) Acidic oxides do not react with dilute acids.

Periodicity of the Chlorides of elements along/across period 3

The table below summarizes some properties of the chlorides of elements in period 3 of the periodic table.