Chemistry - Atomic Structure and Bonding

Atomic Structure

Subatomic Particles

Particle	Symbol	Relative Mass	Relative Charge	Location
Proton	$p$	1	+1	Nucleus
Neutron	$n$	1	0	Nucleus
Electron	$e^-$	$\frac{1}{1836}$ (negligible)	-1	Electron shells

The atomic number (proton number) $Z$ equals the number of protons in the nucleus.

The mass number $A$ equals the number of protons plus neutrons:

$A = Z + N$

Where $N$ is the number of neutrons.

Isotopes

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They have identical chemical properties but different physical properties (e.g., different masses).

Examples of isotopes:

• Carbon-12, Carbon-13, Carbon-14
• Hydrogen (protium), Deuterium, Tritium
• Chlorine-35, Chlorine-37

Worked Example 1

Chlorine has two isotopes: Cl-35 (75.8% abundance) and Cl-37 (24.2% abundance). Calculate the relative atomic mass of chlorine.

$A_r = \frac{35 \times 75.8 + 37 \times 24.2}{100} = \frac{2653 + 895.4}{100} = \frac{3548.4}{100} = 35.48$

Mass Spectrometry

A mass spectrometer separates ions based on their mass-to-charge ratio ($m/z$). The stages are:

1. Ionisation: Atoms are ionised by electron bombardment to form positive ions
2. Acceleration: Ions are accelerated by an electric field
3. Deflection: Ions are deflected by a magnetic field (lighter ions are deflected more)
4. Detection: Ions hit a detector, producing a signal proportional to abundance

info

In a mass spectrum, the x-axis is the mass-to-charge ratio ($m/z$) and the y-axis is the relative abundance. For singly charged ions, $m/z$ equals the relative isotopic mass.

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Electron Configuration

Energy Levels and Subshells

Electrons occupy energy levels (shells) numbered $n = 1, 2, 3, \ldots$

Each energy level contains subshells:

• $s$ subshell: holds up to 2 electrons
• $p$ subshell: holds up to 6 electrons
• $d$ subshell: holds up to 10 electrons
• $f$ subshell: holds up to 14 electrons

Order of Filling (Aufbau Principle)

Electrons fill orbitals in order of increasing energy:

$1s \lt 2s \lt 2p \lt 3s \lt 3p \lt 4s \lt 3d \lt 4p \lt 5s \lt 4d \lt 5p \lt 6s \lt 4f \lt 5d \lt 6p$

warning

warning cations. This is because the $4s$ orbital is at a higher energy than $3d$ once electrons are in the $3d$ subshell.

Pauli Exclusion Principle

Each orbital can hold a maximum of 2 electrons with opposite spins.

Hund's Rule

When filling degenerate orbitals (orbitals of the same energy, such as the three $2p$ orbitals), electrons occupy separate orbitals with parallel spins before pairing up.

Worked Example 2

Write the electron configuration of:

• Sodium ($Z = 11$): $1s^2\, 2s^2\, 2p^6\, 3s^1$
• Iron ($Z = 26$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^6$
• Chlorine ($Z = 17$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^5$

Condensed Electron Configuration

For transition metals, use the noble gas core notation:

• Iron: $[\mathrm{Ar}]\, 4s^2\, 3d^6$
• Copper ($Z = 29$): $[\mathrm{Ar}]\, 4s^1\, 3d^{10}$ (exception: full $d$ subshell is more stable)
• Chromium ($Z = 24$): $[\mathrm{Ar}]\, 4s^1\, 3d^5$ (exception: half-full subshells are more stable)

tip

tip (not $4s^2\, 3d^9$). These arise because half-filled and fully-filled $d$ subshells have extra stability.

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The Periodic Table

Periods and Groups

• Periods: Horizontal rows (1 to 7). The period number equals the number of occupied electron shells.
• Groups: Vertical columns. Elements in the same group have the same number of valence electrons and similar chemical properties.

Periodic Trends

Property	Across a Period (Left to Right)	Down a Group (Top to Bottom)
Atomic radius	Decreases	Increases
First ionisation energy	Generally increases	Generally decreases
Electronegativity	Increases	Decreases
Metallic character	Decreases	Increases
Melting point (Groups 1-3)	Increases	Decreases

Atomic Radius

The atomic radius decreases across a period because the increasing nuclear charge pulls electrons closer. It increases down a group because additional electron shells are added.

Ionisation Energy

First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms:

$\mathrm{X}(g) \to \mathrm{X}^+(g) + e^-$

Trends in first ionisation energy:

• Increases across a period: Nuclear charge increases, electrons held more tightly
• Decreases down a group: Outer electrons are further from the nucleus and more shielded

Dips in ionisation energy occur at:

• Group 3 (e.g., Al): electron removed from $p$ subshell (higher energy than $s$)
• Group 6 (e.g., S): electron removed from a paired orbital (electron-electron repulsion)

Worked Example 3

Explain why the first ionisation energy of aluminium ($Z = 13$) is lower than that of magnesium ($Z = 12$).

Magnesium has electron configuration $1s^2\, 2s^2\, 2p^6\, 3s^2$. The electron is removed from the $3s$ subshell.

Aluminium has electron configuration $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^1$. The electron is removed from the $3p$ subshell.

The $3p$ subshell is at a slightly higher energy level than $3s$, so the $3p$ electron is less tightly held and requires less energy to remove.

Electronegativity

Electronegativity is the ability of an atom to attract the bonding pair of electrons in a covalent bond.

• Fluorine is the most electronegative element (Pauling scale: 4.0)
• Electronegativity increases across a period and decreases down a group

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Ionic Bonding

Formation of Ions

Ionic bonds form between metals (which lose electrons to form cations) and non-metals (which gain electrons to form anions).

Lattice Structure

Ionic compounds form giant ionic lattices:

• Ions are arranged in a regular 3D pattern
• Each ion is surrounded by ions of opposite charge
• The electrostatic attraction between oppositely charged ions is the ionic bond
• There are no discrete molecules

Properties of Ionic Compounds

Property	Explanation
High melting and boiling points	Strong electrostatic forces throughout the lattice
Conduct electricity when molten or dissolved	Ions are free to move and carry charge
Do not conduct when solid	Ions are fixed in position
Soluble in polar solvents (e.g., water)	Polar solvent molecules attract and separate ions
Brittle	Shifting layers brings like charges together, causing repulsion

Worked Example 4

Write the formula of magnesium oxide.

Magnesium is in Group 2: $\mathrm{Mg} \to \mathrm{Mg}^{2+} + 2e^-$

Oxygen is in Group 16: $\mathrm{O} + 2e^- \to \mathrm{O}^{2-}$

To balance charges: 1 $\mathrm{Mg}^{2+}$ ion balances 1 $\mathrm{O}^{2-}$ ion.

Formula: $\mathrm{MgO}$

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Covalent Bonding

A covalent bond is formed when two atoms share a pair of electrons. It occurs between non-metal atoms.

Types of Covalent Bonding

Single bond: One shared pair of electrons, e.g., H-H

Double bond: Two shared pairs of electrons, e.g., O=O

Triple bond: Three shared pairs of electrons, e.g., N$\equiv$N

Dative (coordinate) bond: Both electrons in the shared pair come from the same atom, e.g., in the ammonium ion $\mathrm{NH}_4^+$

Bond Polarity

When two atoms with different electronegativities form a covalent bond, the bonding electrons are pulled towards the more electronegative atom, creating a polar bond with a dipole.

• Non-polar covalent: $\Delta\mathrm{EN} \lt 0.5$ (e.g., H-H, Cl-Cl)
• Polar covalent: $0.5 \leqslant \Delta\mathrm{EN} \lt 1.7$ (e.g., H-Cl, H-O)
• Ionic: $\Delta\mathrm{EN} \geqslant 1.7$ (e.g., Na-Cl)

Worked Example: Predicting Bond Polarity

Use electronegativity values to predict the bond polarity of (a) $\mathrm{H_2O}$, (b) $\mathrm{CCl_4}$, and (c) $\mathrm{KBr}$. Given: $\mathrm{H} = 2.1$, $\mathrm{O} = 3.5$, $\mathrm{C} = 2.5$, $\mathrm{Cl} = 3.0$, $\mathrm{K} = 0.8$, $\mathrm{Br} = 2.8$.

Solution

(a) $\mathrm{H_2O}$: $\Delta\mathrm{EN} = 3.5 - 2.1 = 1.4$. This is polar covalent ($0.5 \leqslant 1.4 \lt 1.7$). The oxygen atom carries a partial negative charge ($\delta^-$) and hydrogen carries a partial positive charge ($\delta^+$).

(b) $\mathrm{CCl_4}$: $\Delta\mathrm{EN} = 3.0 - 2.5 = 0.5$. Each C-Cl bond is polar covalent. However, because the molecule is tetrahedral and symmetrical, the individual bond dipoles cancel out. $\mathrm{CCl_4}$ is a non-polar molecule overall.

(c) $\mathrm{KBr}$: $\Delta\mathrm{EN} = 2.8 - 0.8 = 2.0$. This is ionic ($\Delta\mathrm{EN} \geqslant 1.7$). Potassium transfers its electron to bromine, forming $\mathrm{K^+}$ and $\mathrm{Br^-}$.

Shapes of Molecules (VSEPR Theory)

The Valence Shell Electron Pair Repulsion theory predicts molecular shapes based on the idea that electron pairs around a central atom repel each other and arrange themselves as far apart as possible.

Electron Pairs	Shape	Bond Angle	Example
2 bonding pairs	Linear	$180^\circ$	$\mathrm{BeCl}_2$, $\mathrm{CO}_2$
3 bonding pairs	Trigonal planar	$120^\circ$	$\mathrm{BF}_3$
2 bonding, 1 lone	Bent	$\lt 120^\circ$	$\mathrm{SO}_2$
4 bonding pairs	Tetrahedral	$109.5^\circ$	$\mathrm{CH}_4$
3 bonding, 1 lone	Trigonal pyramidal	$\lt 109.5^\circ$	$\mathrm{NH}_3$
2 bonding, 2 lone	Bent	$\lt 109.5^\circ$	$\mathrm{H}_2\mathrm{O}$
5 bonding pairs	Trigonal bipyramidal	$90^\circ, 120^\circ$	$\mathrm{PCl}_5$
6 bonding pairs	Octahedral	$90^\circ$	$\mathrm{SF}_6$

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Lone pairs exert a greater repulsive effect than bonding pairs because they are closer to the nucleus. This reduces bond angles from the ideal values.

Worked Example 5

Predict the shape and bond angle of $\mathrm{NH}_3$.

Nitrogen has 5 valence electrons. Three are used in bonding with hydrogen, leaving one lone pair.

Total electron pairs = 4 (3 bonding + 1 lone pair)

The electron pair geometry is tetrahedral. With one lone pair, the molecular shape is trigonal pyramidal.

The bond angle is approximately $107^\circ$ (less than $109.5^\circ$ due to lone pair repulsion).

Simple Molecular vs Giant Covalent Structures

Simple molecular (e.g., $\mathrm{H}_2\mathrm{O}$, $\mathrm{CO}_2$, $\mathrm{I}_2$):

• Low melting and boiling points (weak intermolecular forces between molecules)
• Do not conduct electricity
• Usually gases or liquids at room temperature

Giant covalent (e.g., diamond, graphite, silicon dioxide):

• Very high melting and boiling points (strong covalent bonds throughout)
• Diamond: hard, insulator (all electrons in bonds)
• Graphite: soft (layers can slide), conducts electricity (delocalised electrons)

Worked Example: Comparing Diamond, Graphite, and $\mathrm{SiO_2}$

Explain why both diamond and $\mathrm{SiO_2}$ have very high melting points, but graphite has a lower (though still high) melting point and conducts electricity.

Solution

Diamond and $\mathrm{SiO_2}$: Both have giant covalent (network) structures with strong covalent bonds in all three dimensions. Melting requires breaking these strong covalent bonds throughout the entire structure, which needs very high temperatures. Neither conducts electricity because all valence electrons are localised in covalent bonds.

Graphite: Has a layered structure. Within each layer, strong covalent bonds hold atoms together (giving a high melting point). Between layers, only weak van der Waals forces act. The melting point is high because the in-plane covalent bonds must be broken, but it is slightly lower than diamond because the layers can slide. Graphite conducts electricity because each carbon atom has one delocalised electron (from the $p_z$ orbital) that is free to move within the layers.

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Intermolecular Forces

Intermolecular forces are weaker than intramolecular (covalent) bonds.

van der Waals Forces (London Dispersion Forces)

• Present between all molecules (including non-polar ones)
• Caused by instantaneous dipoles due to uneven electron distribution
• Strength increases with molecular size (more electrons) and shape (greater surface area contact)

Trend: larger molecules with more electrons have stronger van der Waals forces.

Dipole-Dipole Interactions

• Occur between polar molecules
• The positive end of one molecule attracts the negative end of another
• Stronger than van der Waals forces but weaker than hydrogen bonding

Hydrogen Bonding

A special, strong type of dipole-dipole interaction that occurs when:

1. Hydrogen is covalently bonded to a highly electronegative atom (N, O, or F)
2. The hydrogen atom interacts with a lone pair on another N, O, or F atom

Conditions: H bonded to N, O, or F, and interacting with another N, O, or F.

Examples: $\mathrm{H}_2\mathrm{O}$, $\mathrm{NH}_3$, HF, DNA base pairing.

Effect of Intermolecular Forces on Properties

Property	Strong IMF	Weak IMF
Melting point	High	Low
Boiling point	High	Low
Viscosity	High	Low
Volatility	Low	High

Worked Example 6

Explain why $\mathrm{H}_2\mathrm{O}$ has a higher boiling point than $\mathrm{H}_2\mathrm{S}$, despite $\mathrm{H}_2\mathrm{S}$ having a larger molecular mass.

Both molecules have van der Waals forces, which are stronger for $\mathrm{H}_2\mathrm{S}$ (larger, more electrons).

However, $\mathrm{H}_2\mathrm{O}$ can form hydrogen bonds between molecules (H bonded to O), while $\mathrm{H}_2\mathrm{S}$ cannot (S is not electronegative enough).

Hydrogen bonding in $\mathrm{H}_2\mathrm{O}$ is much stronger than the van der Waals forces in $\mathrm{H}_2\mathrm{S}$, resulting in a higher boiling point for water.

Trends in Boiling Points of Group 17 Halogens

Halogen	Boiling Point	Explanation
$\mathrm{F}_2$	$-188^\circ\mathrm{C}$	Few electrons, weak van der Waals forces
$\mathrm{Cl}_2$	$-34^\circ\mathrm{C}$	More electrons, stronger van der Waals forces
$\mathrm{Br}_2$	$59^\circ\mathrm{C}$	Even more electrons
$\mathrm{I}_2$	$184^\circ\mathrm{C}$	Most electrons, strongest van der Waals forces

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Metallic Bonding

The Sea of Electrons Model

In metallic bonding:

• Metal atoms lose their valence electrons to form positive ions (cations)
• The valence electrons are delocalised and form a "sea" of electrons
• The electrostatic attraction between the cations and the delocalised electrons is the metallic bond

Properties of Metals

Property	Explanation
High melting and boiling points	Strong metallic bonds throughout the lattice
Good electrical conductivity	Delocalised electrons are free to move and carry charge
Good thermal conductivity	Delocalised electrons transfer kinetic energy
Malleable and ductile	Layers of cations can slide without breaking the metallic bonds
Lustrous	Delocalised electrons absorb and re-emit light at all visible wavelengths
Generally high density	Atoms are closely packed

Alloys

An alloy is a mixture of two or more elements, at least one of which is a metal.

• Different-sized atoms disrupt the regular lattice
• Layers cannot slide as easily
• Alloys are harder and stronger than pure metals

Examples: steel (Fe + C), brass (Cu + Zn), bronze (Cu + Sn), solder (Sn + Pb)

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Summary Table

Topic	Key Concept	Example
Atomic structure	Protons, neutrons, electrons	$A = Z + N$
Electron configuration	Aufbau, Pauli, Hund	$1s^2\, 2s^2\, 2p^6\, 3s^1$ for Na
Ionisation energy	Energy to remove outermost electron	Decreases down a group
Ionic bonding	Transfer of electrons, giant lattice	NaCl
Covalent bonding	Sharing of electrons	H-Cl, O=O
VSEPR	Electron pair repulsion	Tetrahedral for $\mathrm{CH}_4$
Intermolecular forces	van der Waals, dipole-dipole, H-bonding	H-bonding in water
Metallic bonding	Delocalised electrons	Malleability of metals

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Exam Tips

• For electron configuration questions, always use the correct order: $4s$ before $3d$ for filling, but $3d$ before $4s$ for ion formation.
• When explaining ionisation energy trends, mention nuclear charge, shielding, and atomic radius.
• In VSEPR questions, count total electron pairs (bonding + lone pairs) first, then determine the shape based on bonding pairs only.
• Remember that hydrogen bonding requires H bonded to N, O, or F specifically.
• When comparing boiling points, consider the type and strength of intermolecular forces, not just molecular mass.
• For ionic compounds, always state that they have high melting points due to strong electrostatic forces throughout the giant lattice.

Exam-Style Practice Questions

Question 1: The first three ionisation energies of an element are 578, 1817, and 2745 kJ/mol. Identify the group of this element.

The large jump between the 3rd and 4th ionisation energies indicates that the fourth electron is removed from a new, inner shell. This means the element has 3 valence electrons, placing it in Group 13 (it is aluminium).

Question 2: Draw the shape of $\mathrm{PCl}_3$ and state its bond angle.

Phosphorus has 5 valence electrons. Three are used for bonding with Cl, leaving one lone pair.

Total electron pairs = 4 (3 bonding + 1 lone pair)

Shape: trigonal pyramidal, bond angle approximately $107^\circ$.

Question 3: Explain why the boiling point of neon is lower than that of argon.

Both are noble gases with only van der Waals forces. Argon has more electrons than neon, so it has stronger van der Waals forces. More energy is required to overcome these forces, giving argon a higher boiling point.

Question 4: Draw the dot-and-cross diagram for $\mathrm{NH}_3$ and predict its shape.

Nitrogen has 5 valence electrons. Three electrons form covalent bonds with three hydrogen atoms (3 shared pairs), and one lone pair remains. The shape is trigonal pyramidal with a bond angle of approximately $107^\circ$.

Question 5: Explain why $\mathrm{SiO}_2$ has a very high melting point.

$\mathrm{SiO}_2$ has a giant covalent structure. Each silicon atom is covalently bonded to four oxygen atoms, and each oxygen atom is bonded to two silicon atoms, forming a continuous 3D network. Breaking this structure requires breaking many strong covalent bonds, which requires a large amount of energy, hence the very high melting point.

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Chemical Bonding Energetics

Bond Enthalpy

Bond enthalpy (bond energy) is the average energy required to break one mole of a particular type of bond in the gaseous state.

Bond	Average Bond Enthalpy (kJ/mol)
C-C	347
C=C	612
C$\equiv$C	838
C-H	413
C-O	358
C=O	745
O-H	463
O=O	498
H-H	436
N$\equiv$N	945
N-H	391

Using Bond Enthalpies to Calculate Enthalpy Changes

$\Delta H = \sum(\mathrm{Bonds broken}) - \sum(\mathrm{Bonds formed})$

• Bonds broken: endothermic (positive value)
• Bonds formed: exothermic (negative value)

Worked Example 7

Calculate the enthalpy change for the reaction: $\mathrm{H}_2(g) + \mathrm{Cl}_2(g) \to 2\mathrm{HCl}(g)$

Bonds broken:

• 1 $\times$ H-H = $436 \mathrm{ kJ/mol}$
• 1 $\times$ Cl-Cl = $243 \mathrm{ kJ/mol}$

Bonds formed:

• 2 $\times$ H-Cl = $2 \times 432 = 864 \mathrm{ kJ/mol}$

$\Delta H = (436 + 243) - 864 = 679 - 864 = -185 \mathrm{ kJ/mol}$

The reaction is exothermic.

Worked Example 8

Calculate the enthalpy of combustion of methane: $\mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \to \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(g)$

Bonds broken:

• 4 $\times$ C-H = $4 \times 413 = 1652 \mathrm{ kJ/mol}$
• 2 $\times$ O=O = $2 \times 498 = 996 \mathrm{ kJ/mol}$

Bonds formed:

• 2 $\times$ C=O = $2 \times 745 = 1490 \mathrm{ kJ/mol}$
• 4 $\times$ O-H = $4 \times 463 = 1852 \mathrm{ kJ/mol}$

$\Delta H = (1652 + 996) - (1490 + 1852) = 2648 - 3342 = -694 \mathrm{ kJ/mol}$

warning

warning formation data instead. Bond enthalpy calculations are less accurate for reactions involving liquids or solids because they only account for gaseous state bonds.

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Giant Ionic Structures in Detail

Properties Explained

The physical properties of ionic compounds can be explained by their giant ionic lattice structure:

High melting and boiling points:

The electrostatic attraction between oppositely charged ions acts throughout the entire lattice. A large amount of energy is required to overcome these strong forces, resulting in high melting and boiling points.

Electrical conductivity:

In the solid state, ions are held in fixed positions and cannot move, so ionic solids do not conduct electricity. When molten or dissolved in water, ions are free to move and carry charge, allowing conductivity.

Solubility in polar solvents:

Water molecules are polar and can attract ions from the lattice surface. The positive end of water molecules (near $\mathrm{H}$) attracts anions, while the negative end (near $\mathrm{O}$) attracts cations. If the hydration energy exceeds the lattice energy, the ionic compound dissolves.

Brittleness:

When a force is applied, layers of ions shift. Ions of the same charge come adjacent to each other, and the repulsive forces cause the crystal to fracture along a cleavage plane.

Lattice Energy

Lattice energy is the energy released when one mole of an ionic compound is formed from its gaseous ions.

Factors affecting lattice energy:

• Ionic charge: Higher charge leads to stronger attraction and larger lattice energy
• Ionic radius: Smaller ions can get closer together, increasing lattice energy

Compound	Ionic Charges	Lattice Energy Trend
$\mathrm{NaCl}$	+1, -1	Lower
$\mathrm{MgO}$	+2, -2	Higher (about 4$\times$ $\mathrm{NaCl}$)

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Electron Configuration and Chemical Behaviour

Valence Electrons

Valence electrons are the electrons in the outermost shell of an atom. They determine the chemical properties of an element.

Group	Valence Electrons	Common Ion	Typical Behaviour
1	1	+1	Loses 1 electron
2	2	+2	Loses 2 electrons
13	3	+3	Loses 3 electrons
15	5	-3	Gains 3 electrons
16	6	-2	Gains 2 electrons
17	7	-1	Gains 1 electron
18	8	None	Noble gas, unreactive

Transition Metals

Transition metals have the following characteristic properties:

• Variable oxidation states (e.g., Fe: +2 and +3; Mn: +2, +4, +7)
• Formation of coloured compounds (due to d-d electron transitions)
• Catalytic activity (e.g., Fe in Haber process, V$_2$O$_5$ in Contact process)
• Formation of complex ions (e.g., $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$)

Worked Example 9

Write the electron configuration of $\mathrm{Fe}^{3+}$.

Fe ($Z = 26$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^6$

$\mathrm{Fe}^{3+}$: Remove 3 electrons. Since $4s$ electrons are lost before $3d$:

$\mathrm{Fe}^{3+}$: $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 3d^5$

Note that $\mathrm{Fe}^{3+}$ has a half-filled $3d$ subshell ($3d^5$), which contributes to its relative stability compared to $\mathrm{Fe}^{2+}$ ($3d^6$).

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Trends Across Period 3

Period 3 elements ($\mathrm{Na}$ to $\mathrm{Ar}$) show clear trends that are frequently examined:

Atomic and Ionic Radii

Element	Na	Mg	Al	Si	P	S	Cl	Ar
Atomic radius (pm)	186	160	143	117	110	104	99	—

Atomic radius decreases across the period because increasing nuclear charge pulls electrons closer.

Melting and Boiling Points

Element	Na	Mg	Al	Si	P	S	Cl	Ar
Melting point ($^\circ\mathrm{C}$)	98	650	660	1410	44	115	-101	-189

• $\mathrm{Na}$, $\mathrm{Mg}$, $\mathrm{Al}$: Metallic bonding, increasing strength (more delocalised electrons)
• $\mathrm{Si}$: Giant covalent structure, very high melting point
• $\mathrm{P}$, $\mathrm{S}$, $\mathrm{Cl}$, $\mathrm{Ar}$: Simple molecular, weak van der Waals forces

Electrical Conductivity

• $\mathrm{Na}$, $\mathrm{Mg}$, $\mathrm{Al}$: Good conductors (metallic bonding with delocalised electrons)
• $\mathrm{Si}$: Semiconductor (conductivity increases with temperature)
• $\mathrm{P}$, $\mathrm{S}$, $\mathrm{Cl}$, $\mathrm{Ar}$: Non-conductors (no mobile charge carriers)

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Additional Worked Examples

Worked Example: Electron Configuration of an Ion

Write the electron configuration of $\mathrm{S^{2-}}$ and explain why it has the same configuration as argon.

Solution

Sulphur ($Z = 16$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^4$

$\mathrm{S^{2-}}$ gains 2 electrons: $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6$

This is the same as argon ($Z = 18$): $1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6$

Sulphur is in Group 16. By gaining 2 electrons to form $\mathrm{S^{2-}}$, it achieves the stable noble gas electron configuration of argon (a full outer shell of 8 electrons).

Worked Example: Predicting Molecular Shape

Predict the shape and bond angle of $\mathrm{SF_4}$.

Solution

Sulphur has 6 valence electrons. In $\mathrm{SF_4}$, 4 are used in bonding with fluorine, leaving 1 lone pair.

Total electron pairs = 5 (4 bonding + 1 lone pair)

Electron pair geometry: trigonal bipyramidal ($sp^3d$ hybridisation)

The lone pair occupies an equatorial position to minimise repulsion. The molecular shape is see-saw (disphenoidal).

Bond angles: approximately $120^\circ$ (equatorial) and $90^\circ$ (axial-equatorial), both slightly reduced from ideal values due to lone pair repulsion.

Worked Example: Intermolecular Forces Comparison

Explain why propanone ($\mathrm{CH_3COCH_3}$, b.p. $56^\circ\mathrm{C}$) has a higher boiling point than propane ($\mathrm{CH_3CH_2CH_3}$, b.p. $-42^\circ\mathrm{C}$), but a lower boiling point than propan-1-ol ($\mathrm{CH_3CH_2CH_2OH}$, b.p. $97^\circ\mathrm{C}$).

Solution

Propanone vs. propane: Propanone has a polar C=O bond, creating permanent dipole-dipole interactions between molecules. Propane is non-polar and has only weak van der Waals forces. Dipole-dipole forces are stronger than van der Waals forces, so propanone has a higher boiling point.

Propanone vs. propan-1-ol: Propan-1-ol has an -OH group and can form hydrogen bonds between molecules. Propanone cannot form hydrogen bonds (it has no H bonded to N, O, or F). Hydrogen bonding is much stronger than dipole-dipole interactions, so propan-1-ol has a higher boiling point.

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Additional Practice Questions

More Exam-Style Problems

Question 6: The first four ionisation energies of boron are 801, 2427, 3660, and 25026 kJ/mol. Explain the large jump between the third and fourth ionisation energies.

The first three electrons are removed from the outer shell (2s and 2p subshells). The fourth electron is removed from the inner 1s shell, which is much closer to the nucleus and experiences much less shielding. This requires significantly more energy, hence the large jump.

Question 7: Explain why the melting point of $\mathrm{MgO}$ is much higher than that of $\mathrm{NaCl}$.

Both have giant ionic lattices, but $\mathrm{Mg}^{2+}$ and $\mathrm{O}^{2-}$ have higher charges than $\mathrm{Na}^+$ and $\mathrm{Cl}^-$. The electrostatic attraction is proportional to the product of the charges: $\mathrm{MgO}$ has $2 \times 2 = 4$ while $\mathrm{NaCl}$ has $1 \times 1 = 1$. Additionally, $\mathrm{Mg}^{2+}$ and $\mathrm{O}^{2-}$ are smaller ions, allowing them to get closer together. Both factors result in stronger ionic bonds and a higher melting point for $\mathrm{MgO}$.

Question 8: Draw the dot-and-cross diagram for $\mathrm{CO}_2$ and explain why it is a linear molecule.

Carbon has 4 valence electrons and forms two double bonds with oxygen atoms (each oxygen has 6 valence electrons). The molecule has no lone pairs on the central carbon atom. With two bonding pairs, the electron pair geometry and molecular shape are both linear with a bond angle of $180^\circ$.

Question 9: Explain why $\mathrm{HF}$ has a higher boiling point than $\mathrm{HCl}$ despite having a lower molecular mass.

$\mathrm{HF}$ can form hydrogen bonds between molecules because hydrogen is bonded to fluorine (highly electronegative). $\mathrm{HCl}$ cannot form hydrogen bonds because chlorine is not electronegative enough. Hydrogen bonding in $\mathrm{HF}$ is much stronger than the van der Waals forces and dipole- dipole interactions in $\mathrm{HCl}$, resulting in a higher boiling point for $\mathrm{HF}$.

Question 10: Explain why diamond is an electrical insulator while graphite is a good conductor.

In diamond, each carbon atom is covalently bonded to four other carbon atoms in a tetrahedral arrangement. All four valence electrons are used in sigma bonds, leaving no delocalised electrons to carry charge. In graphite, each carbon atom is bonded to three others in a planar hexagonal structure. The fourth electron from each carbon is delocalised and free to move throughout the layers, allowing graphite to conduct electricity.

Question 11: Explain the trend in first ionisation energy across Period 3 (Na to Ar).

Ionisation energy generally increases across the period because nuclear charge increases (more protons) while the shielding effect remains similar (same number of inner electron shells). This pulls the outer electrons closer to the nucleus, making them harder to remove. The dip at aluminium is because the electron is removed from the higher-energy $3p$ subshell. The dip at sulfur is because the electron is removed from a paired $3p$ orbital where electron-electron repulsion makes it easier to remove.

Question 12: Write the electron configuration of $\mathrm{Cu}^+$ and explain why it is more stable than $\mathrm{Cu}^{2+}$ in some contexts.

$\mathrm{Cu}$ ($Z = 29$): $[\mathrm{Ar}]\, 4s^1\, 3d^{10}$

$\mathrm{Cu}^+$: $[\mathrm{Ar}]\, 3d^{10}$ (removing the $4s$ electron first)

$\mathrm{Cu}^+$ has a completely filled $3d$ subshell, which is particularly stable due to the symmetrical distribution of electrons. However, $\mathrm{Cu}^{2+}$ ($3d^9$) is more common in aqueous chemistry because of the high hydration energy that compensates for the loss of the stable $3d^{10}$ configuration.

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Advanced Bonding Concepts

Hybridisation

Hybridisation is the concept of mixing atomic orbitals to form new hybrid orbitals that are equivalent in energy and suitable for bonding.

Hybridisation	Geometry	Bond Angle	Example
$sp$	Linear	$180^\circ$	$\mathrm{BeCl}_2$, $\mathrm{C}_2\mathrm{H}_2$
$sp^2$	Trigonal planar	$120^\circ$	$\mathrm{BF}_3$, $\mathrm{C}_2\mathrm{H}_4$
$sp^3$	Tetrahedral	$109.5^\circ$	$\mathrm{CH}_4$, $\mathrm{NH}_3$
$sp^3d$	Trigonal bipyramidal	$90^\circ, 120^\circ$	$\mathrm{PCl}_5$
$sp^3d^2$	Octahedral	$90^\circ$	$\mathrm{SF}_6$

Worked Example 10

Determine the hybridisation of the central atom in $\mathrm{SF}_4$.

Sulphur has 6 valence electrons. In $\mathrm{SF}_4$, four are used for bonding with fluorine, leaving one lone pair. Total electron pairs = 5 (4 bonding + 1 lone pair).

Hybridisation: $sp^3d$ (trigonal bipyramidal electron pair geometry, with the lone pair in an equatorial position, giving a "see-saw" molecular shape).

Molecular Orbital Theory (Brief Overview)

Molecular orbital theory describes bonding in terms of the combination of atomic orbitals to form molecular orbitals that extend over the entire molecule.

Bonding orbitals: Lower in energy than the original atomic orbitals; stabilise the molecule.

Antibonding orbitals: Higher in energy than the original atomic orbitals; destabilise the molecule.

Bond order:

$\mathrm{Bond order} = \frac{1}{2}(\mathrm{bonding electrons} - \mathrm{antibonding electrons})$

• Bond order = 1: single bond
• Bond order = 2: double bond
• Bond order = 1.5: intermediate (e.g., $\mathrm{O}_2^-$)
• Bond order = 0: no bond (molecule does not exist, e.g., $\mathrm{He}_2$)

info

Molecular orbital theory explains why $\mathrm{O}_2$ is paramagnetic (has unpaired electrons in antibonding orbitals), which cannot be explained by simple Lewis structures.

danger

Common Pitfalls

• Confusing atomic number with mass number: Atomic number (proton number) defines the element and equals the number of protons. Mass number equals protons PLUS neutrons. Isotopes of the same element have the same atomic number but different mass numbers. Students frequently mix these up when writing nuclear notation.

• Drawing ionic bonds with shared electrons: Ionic bonding involves the TRANSFER of electrons from metal to non-metal, forming positive and negative ions held by electrostatic attraction. There are NO shared electron pairs in ionic bonds. Shared electrons indicate covalent bonding, which occurs between non-metals.

• Forgetting that metallic bonding is different from both ionic and covalent: Metallic bonding involves a lattice of positive metal ions surrounded by a "sea" of delocalised electrons. It is NOT ionic (no oppositely charged ions) and NOT covalent (no shared electron pairs). This delocalised electron model explains why metals conduct electricity and are malleable.

• Misidentifying bond polarity: A covalent bond is polar if the two atoms have DIFFERENT electronegativities, causing unequal electron sharing. But a polar MOLECULE requires polar bonds AND an asymmetrical shape. CO2 has polar bonds but is a non-polar molecule (linear, symmetrical). H2O is polar (bent shape, dipole moments do not cancel).

Intermolecular Forces in Detail

The strength of intermolecular forces directly affects physical properties:

Boiling points of hydrogen halides:

Compound	Boiling Point ($^\circ\mathrm{C}$)	Dominant IMF
HF	19.5	Hydrogen bonding
HCl	-84.9	Dipole-dipole + van der Waals
HBr	-67.0	Dipole-dipole + van der Waals
HI	-35.4	van der Waals

HF has the highest boiling point due to hydrogen bonding. For HCl, HBr, and HI, boiling points increase with molecular mass (more electrons = stronger van der Waals forces), despite the decreasing dipole moment.

Worked Example 11

Explain why $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$ (ethanol) is soluble in water but $\mathrm{C}_6\mathrm{H}_{12}$ (hexane) is not.

Ethanol has a hydroxyl group (-OH) that can form hydrogen bonds with water molecules, making it miscible. Hexane is a non-polar hydrocarbon with only weak van der Waals forces. The energy released from forming hexane-water interactions is insufficient to overcome the hydrogen bonds between water molecules and the van der Waals forces between hexane molecules. This follows the principle "like dissolves like."

Allotropes of Carbon

Allotrope	Structure	Bonding	Properties
Diamond	Tetrahedral 3D network	Each C bonded to 4 others ($sp^3$)	Hardest natural substance, insulator, high melting point
Graphite	Layered hexagonal sheets	Each C bonded to 3 others ($sp^2$), delocalised electrons	Soft, conductor, lubricant, high melting point
Graphene	Single layer of graphite	Each C bonded to 3 others ($sp^2$)	Strongest material known, conductor, transparent
Fullerene ($\mathrm{C}_{60}$)	Hollow sphere (buckyball)	Each C bonded to 3 others	Molecular solid, semiconductor
Carbon nanotube	Cylindrical graphene	Each C bonded to 3 others	Very strong, conductor

Worked Example 12

Explain why graphite is soft and slippery while diamond is hard.

In graphite, the carbon atoms are arranged in flat hexagonal layers. Within each layer, strong covalent bonds hold atoms together. Between layers, only weak van der Waals forces act, allowing the layers to slide over each other easily. This makes graphite soft and a good lubricant. In diamond, all carbon atoms are covalently bonded in a rigid 3D tetrahedral network with no weak planes, making it extremely hard.

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Problem Set

Problem 1: Boron has two isotopes: B-10 (19.9% abundance) and B-11 (80.1% abundance). Calculate the relative atomic mass of boron.

If you get this wrong, revise: Isotopes and Mass Spectrometry

Solution

$A_r = \frac{10 \times 19.9 + 11 \times 80.1}{100} = \frac{199 + 881.1}{100} = \frac{1080.1}{100} = 10.81$

Problem 2: Write the electron configuration of $\mathrm{Cu^{2+}}$ and explain why it is not $[\mathrm{Ar}]\, 3d^9$.

If you get this wrong, revise: Electron Configuration

Solution

$\mathrm{Cu}$ ($Z = 29$): $[\mathrm{Ar}]\, 4s^1\, 3d^{10}$ (exception: full $d$ subshell is more stable)

When forming $\mathrm{Cu^{2+}}$, both the $4s$ electron and one $3d$ electron are removed:

$\mathrm{Cu^{2+}}$: $[\mathrm{Ar}]\, 3d^9$

Note: this IS $3d^9$. The answer is that $\mathrm{Cu^{2+}}$ does have the configuration $[\mathrm{Ar}]\, 3d^9$. It loses the $4s$ electron first (as $4s$ is at higher energy once $3d$ is occupied), then one $3d$ electron. Despite losing the stable $3d^{10}$ configuration, the high hydration energy of $\mathrm{Cu^{2+}}$ in solution makes this oxidation favourable.

Problem 3: Explain why the first ionisation energy of magnesium ($738 \mathrm{ kJ/mol}$) is higher than that of sodium ($496 \mathrm{ kJ/mol}$), but the second ionisation energy of sodium ($4562 \mathrm{ kJ/mol}$) is much higher than that of magnesium ($1451 \mathrm{ kJ/mol}$).

If you get this wrong, revise: Ionisation Energy

Solution

First IE: $\mathrm{Mg}$ has a higher nuclear charge ($+12$ vs $+11$) and a smaller atomic radius. The outer electron is held more tightly, requiring more energy to remove.

Second IE: $\mathrm{Na}$ has the configuration $[\mathrm{Ne}]\, 3s^1$. After losing the first electron ($\mathrm{Na^+} = [\mathrm{Ne}]$), the second electron must be removed from the stable noble gas core ($1s^2\, 2s^2\, 2p^6$), which requires a huge amount of energy. For $\mathrm{Mg}$, the second electron is still in the $3s$ subshell ($\mathrm{Mg^+} = [\mathrm{Ne}]\, 3s^1$), so it is much easier to remove.

Problem 4: Predict the shape and bond angle of $\mathrm{XeF_4}$.

If you get this wrong, revise: Shapes of Molecules (VSEPR Theory)

Solution

Xenon has 8 valence electrons. Four are used in bonding with fluorine, leaving 2 lone pairs.

Total electron pairs = 6 (4 bonding + 2 lone pairs)

Electron pair geometry: octahedral ($sp^3d^2$)

The two lone pairs occupy opposite positions to minimise repulsion. The molecular shape is square planar with bond angles of $90^\circ$.

Problem 5: Explain why sodium chloride has a high melting point ($801^\circ\mathrm{C}$) while carbon tetrachloride ($\mathrm{CCl_4}$) is a liquid at room temperature (b.p. $77^\circ\mathrm{C}$).

If you get this wrong, revise: Ionic Bonding and Simple Molecular vs Giant Covalent

Solution

$\mathrm{NaCl}$ has a giant ionic lattice. Strong electrostatic forces between $\mathrm{Na^+}$ and \mathrm{Cl^- act throughout the entire lattice, requiring large amounts of energy to overcome.

$\mathrm{CCl_4}$ is a simple molecular substance. The covalent bonds within each molecule are strong, but only weak van der Waals forces act between molecules. These weak intermolecular forces require little energy to overcome, resulting in a low boiling point.

Problem 6: Explain why ammonia ($\mathrm{NH_3}$, b.p. $-33^\circ\mathrm{C}$) has a higher boiling point than phosphine ($\mathrm{PH_3}$, b.p. $-88^\circ\mathrm{C}$), even though phosphine has a larger molecular mass.

If you get this wrong, revise: Hydrogen Bonding

Solution

$\mathrm{NH_3}$ can form hydrogen bonds between molecules because hydrogen is bonded to nitrogen (highly electronegative, EN $= 3.0$). $\mathrm{PH_3}$ cannot form hydrogen bonds because phosphorus (EN $= 2.1$) is not electronegative enough. Hydrogen bonding is much stronger than the van der Waals forces in $\mathrm{PH_3}$, giving $\mathrm{NH_3}$ a higher boiling point despite its lower molecular mass.

Problem 7: Use bond enthalpies to calculate $\Delta H$ for the reaction:

$\mathrm{N_2} + 3\mathrm{H_2} \to 2\mathrm{NH_3}$

Given: $\mathrm{N \equiv N} = 945$, $\mathrm{H - H} = 436$, $\mathrm{N - H} = 391 \mathrm{ kJ/mol}$.

If you get this wrong, revise: Bond Enthalpy

Solution

Bonds broken: $1 \times \mathrm{N \equiv N} + 3 \times \mathrm{H - H} = 945 + 3(436) = 945 + 1308 = 2253 \mathrm{ kJ/mol}$

Bonds formed: $6 \times \mathrm{N - H} = 6(391) = 2346 \mathrm{ kJ/mol}$

$\Delta H = 2253 - 2346 = -93 \mathrm{ kJ/mol}$

The reaction is exothermic.

Problem 8: Explain why metals are good conductors of electricity, whereas ionic solids are not.

If you get this wrong, revise: Metallic Bonding and Properties of Ionic Compounds

Solution

Metals have delocalised electrons that are free to move throughout the metallic lattice. When a potential difference is applied, these mobile electrons carry charge, allowing electrical conduction.

In ionic solids, ions are held in fixed positions in the giant ionic lattice by strong electrostatic forces. Although ions are charged, they cannot move and therefore cannot carry charge. Ionic compounds only conduct when molten or dissolved, when ions are free to move.

Problem 9: Explain the trend in electrical conductivity across Period 3 (Na to Ar).

If you get this wrong, revise: Trends Across Period 3

Solution

$\mathrm{Na}$, $\mathrm{Mg}$, $\mathrm{Al}$: Good conductors due to metallic bonding with delocalised electrons. Conductivity increases from $\mathrm{Na}$ to $\mathrm{Al}$ as more delocalised electrons per atom are available.

$\mathrm{Si}$: Semiconductor; conductivity is much lower than metals but increases with temperature as more electrons are promoted to the conduction band.

$\mathrm{P}$, $\mathrm{S}$, $\mathrm{Cl}$, $\mathrm{Ar}$: Non-conductors. They are simple molecular or monatomic species with no mobile charge carriers.

Problem 10: Explain why graphite can conduct electricity but diamond cannot, even though both are forms of carbon.

If you get this wrong, revise: Allotropes of Carbon

Solution

In diamond, each carbon atom is bonded to four others in a rigid 3D tetrahedral network. All four valence electrons are used in covalent bonds; none are free to move.

In graphite, each carbon atom is bonded to only three others in planar hexagonal layers ($sp^2$ hybridised). The fourth valence electron from each carbon is delocalised and free to move within the layers. These mobile electrons carry charge, allowing graphite to conduct electricity parallel to the layers.

Problem 11: Determine the hybridisation of the central atom in $\mathrm{BF_3}$ and $\mathrm{NF_3}$, and explain why their bond angles differ.

If you get this wrong, revise: Hybridisation and VSEPR Theory

Solution

$\mathrm{BF_3}$: Boron has 3 valence electrons, all used in bonding. 3 bonding pairs, 0 lone pairs. Hybridisation: $sp^2$. Shape: trigonal planar. Bond angle: $120^\circ$.

$\mathrm{NF_3}$: Nitrogen has 5 valence electrons; 3 used in bonding, 1 lone pair. 4 electron pairs (3 bonding + 1 lone). Hybridisation: $sp^3$. Shape: trigonal pyramidal. Bond angle: approximately $107^\circ$ (less than $109.5^\circ$ due to lone pair repulsion).

Despite both having formula $\mathrm{XF_3}$, the lone pair on $\mathrm{N}$ in $\mathrm{NF_3}$ causes a different geometry and smaller bond angle.

Problem 12: Explain why the melting point of $\mathrm{MgO}$ ($2852^\circ\mathrm{C}$) is much higher than that of $\mathrm{NaCl}$ ($801^\circ\mathrm{C}$).

If you get this wrong, revise: Giant Ionic Structures in Detail

Solution

Both have giant ionic lattices, but:

1. Ionic charges: $\mathrm{Mg^{2+}}$ and $\mathrm{O^{2-}}$ have double charges compared to $\mathrm{Na^+}$ and $\mathrm{Cl^-}$. The electrostatic attraction is proportional to the product of charges: $2 \times 2 = 4$ for $\mathrm{MgO}$ vs $1 \times 1 = 1$ for $\mathrm{NaCl}$.

2. Ionic radii: $\mathrm{Mg^{2+}}$ ($72 \mathrm{ pm}$) and $\mathrm{O^{2-}}$ ($140 \mathrm{ pm}$) are smaller than $\mathrm{Na^+}$ ($102 \mathrm{ pm}$) and $\mathrm{Cl^-}$ ($181 \mathrm{ pm}$). Smaller ions can get closer together, increasing the electrostatic attraction.

Both factors give $\mathrm{MgO}$ much stronger ionic bonds and a much higher melting point.

Problem 13: Define the term "first ionisation energy" and explain why the first ionisation energy of sodium is lower than that of neon.

If you get this wrong, revise: Ionisation Energy

Solution

First ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous atoms.

Sodium ($Z = 11$, $[\mathrm{Ne}]\, 3s^1$) has its outermost electron in the 3s subshell, which is further from the nucleus and well-shielded by the neon core. The electron is relatively easy to remove.

Neon ($Z = 10$, $1s^2\, 2s^2\, 2p^6$) has a stable noble gas configuration with a full outer shell. All electrons are in the n=2 shell, closer to the nucleus with less shielding. Removing an electron from a stable, fully-occupied shell requires significantly more energy.

Problem 14: Explain the term "dative covalent bond" and give an example.

If you get this wrong, revise: Covalent Bonding

Solution

A dative covalent bond (coordinate bond) is a covalent bond in which both electrons in the shared pair come from the same atom.

Example: In the ammonium ion ($\mathrm{NH_4^+}$), the fourth N-H bond is a dative bond. The nitrogen atom of ammonia donates its lone pair to form a bond with $\mathrm{H^+}$ (which has no electrons):

$\mathrm{NH_3} + \mathrm{H^+} \to \mathrm{NH_4^+}$

Once formed, the dative bond is indistinguishable from the other three N-H bonds; all four N-H bonds are equivalent.

Problem 15: Explain why the boiling point of $\mathrm{CH_3CH_2OH}$ ($78^\circ\mathrm{C}$) is much higher than that of $\mathrm{CH_3OCH_3}$ ($-24^\circ\mathrm{C}$), even though both have the same molecular formula ($\mathrm{C_2H_6O}$).

If you get this wrong, revise: Hydrogen Bonding and Functional Group Isomerism

Solution

Both compounds are functional group isomers. $\mathrm{CH_3CH_2OH}$ (ethanol) has an -OH group that can form hydrogen bonds between molecules, requiring significant energy to overcome. $\mathrm{CH_3OCH_3}$ (methoxymethane, an ether) has no -OH group; although it has polar C-O bonds, it cannot form hydrogen bonds. Intermolecular forces in methoxymethane are limited to dipole-dipole interactions and van der Waals forces, which are much weaker than hydrogen bonding.

Problem 16: Describe the metallic bonding in sodium and explain why sodium has a lower melting point ($98^\circ\mathrm{C}$) than iron ($1538^\circ\mathrm{C}$).

If you get this wrong, revise: Metallic Bonding

Solution

In sodium, each atom donates one valence electron to form $\mathrm{Na^+}$ ions in a lattice, with delocalised electrons forming a "sea." The metallic bond is the electrostatic attraction between $\mathrm{Na^+}$ ions and the delocalised electrons.

In iron, each atom donates two valence electrons (and some from the 3d subshell), forming $\mathrm{Fe^{2+}}$ ions. Iron has a higher charge on the cations and more delocalised electrons per atom, giving stronger metallic bonds. Also, iron ions are smaller than sodium ions, allowing closer packing and stronger electrostatic attraction. These factors give iron a much higher melting point.

Problem 17: Explain why the first ionisation energy decreases from Be to B and from N to O in Period 2.

If you get this wrong, revise: Ionisation Energy

Solution

Be ($1s^2\, 2s^2$) to B ($1s^2\, 2s^2\, 2p^1$): The electron removed from B is in the $2p$ subshell, which is at a slightly higher energy than the $2s$ subshell (from which Be's electron is removed). The $2p$ electron is less tightly held and requires less energy.

N ($1s^2\, 2s^2\, 2p^3$) to O ($1s^2\, 2s^2\, 2p^4$): In N, each $2p$ orbital has one electron (Hund's rule). In O, one $2p$ orbital has two paired electrons. The paired electrons repel each other, making it easier to remove one of them from O compared to removing an unpaired electron from N.

Problem 18: Use the concept of electronegativity to explain why $\mathrm{HCl}$ is a polar molecule but $\mathrm{Cl_2}$ is not.

If you get this wrong, revise: Bond Polarity

Solution

In $\mathrm{HCl}$, chlorine ($\mathrm{EN} = 3.0$) is more electronegative than hydrogen ($\mathrm{EN} = 2.1$). The bonding electrons are pulled towards chlorine, creating a dipole with $\mathrm{Cl}^{\delta-}$ and $\mathrm{H}^{\delta+}$. Since the molecule is diatomic and the two atoms are different, the bond dipole does not cancel, making $\mathrm{HCl}$ a polar molecule.

In $\mathrm{Cl_2}$, both atoms are identical (same electronegativity). The bonding electrons are shared equally, so there is no dipole. $\mathrm{Cl_2}$ is a non-polar molecule.

Problem 19: Explain the term "allotrope" and give two examples of allotropes of carbon.

If you get this wrong, revise: Allotropes of Carbon

Solution

An allotrope is a different structural form of the same element in the same physical state.

Examples of carbon allotropes:

1. Diamond: 3D tetrahedral network ($sp^3$), extremely hard, insulator
2. Graphite: Layered hexagonal sheets ($sp^2$), soft, conducts electricity
3. Graphene: Single layer of graphite, strongest known material
4. Fullerene ($\mathrm{C_{60}}$): Hollow spherical molecule
5. Carbon nanotube: Cylindrical graphene sheet

Any two of the above.

Problem 20: Calculate the enthalpy change for the reaction $\mathrm{NH_3} + \mathrm{HCl} \to \mathrm{NH_4Cl}$ using bond enthalpies.

Given: $\mathrm{N - H} = 391$, $\mathrm{H - Cl} = 432$, $\mathrm{N - H}$ (in $\mathrm{NH_4^+}$) $\approx 391$, $\mathrm{Cl^-}$ (ionic) -- treat $\mathrm{NH_4Cl}$ as ionic, so bond enthalpies are not directly applicable. Instead, state why bond enthalpies give a poor estimate for this reaction.

If you get this wrong, revise: Bond Enthalpy and Ionic Bonding

Solution

Bond enthalpies give a poor estimate because:

1. $\mathrm{NH_4Cl}$ is an ionic solid, not a molecular gas. Bond enthalpies apply only to gaseous molecules.
2. The reaction involves a change of state (gases forming a solid), and lattice energy is not accounted for.
3. The actual enthalpy change would need to include: bond breaking in $\mathrm{NH_3}$ and $\mathrm{HCl}$ (endothermic), then lattice energy release for $\mathrm{NH_4Cl}$ (exothermic).

The appropriate method is to use standard enthalpies of formation instead:

$\Delta H = \Delta H_f^\circ(\mathrm{NH_4Cl}) - \Delta H_f^\circ(\mathrm{NH_3}) - \Delta H_f^\circ(\mathrm{HCl})$

Problem 21: Explain why helium has the highest first ionisation energy of all elements.

If you get this wrong, revise: Ionisation Energy and Periodic Trends

Solution

Helium has the electron configuration $1s^2$. Its outer electrons are in the first shell ($n = 1$), which is closest to the nucleus. There is minimal shielding (no inner shell electrons). The small atomic radius and high effective nuclear charge mean the electrons are very tightly held. Additionally, the $1s$ subshell is fully occupied, giving helium extra stability as a noble gas. These factors combine to give helium the highest first ionisation energy of all elements ($2372 \mathrm{ kJ/mol}$).