Chemical Bonding (Advanced)

1. VSEPR Theory

Principle

The Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular geometry based on the repulsion between electron pairs (both bonding and lone pairs) around a central atom.

Postulates:

1. Electron pairs arrange themselves to minimise repulsion.
2. Repulsion order: lone pair--lone pair $\gt$ lone pair--bonding pair $\gt$ bonding pair--bonding pair.
3. Lone pairs occupy more space than bonding pairs because they are held by one nucleus and spread closer to the central atom.
4. Multiple bonds are treated as a single region of electron density (but repel more strongly than single bonds).

Electron Pair Geometries

Electron pairs	Geometry	Bond angle
$2$	Linear	$180\degree$
$3$	Trigonal planar	$120\degree$
$4$	Tetrahedral	$109.5\degree$
$5$	Trigonal bipyramidal	$120\degree, 90\degree$
$6$	Octahedral	$90\degree$

Molecular Shapes from VSEPR

Steric number	Lone pairs	Shape	Bond angle	Example
$2$	$0$	Linear	$180\degree$	$\mathrm{BeCl}_2$, $\mathrm{CO}_2$
$3$	$0$	Trigonal planar	$120\degree$	$\mathrm{BF}_3$, $\mathrm{SO}_3$
$3$	$1$	Bent (V-shaped)	$\lt 120\degree$	$\mathrm{SO}_2$, $\mathrm{O}_3$
$4$	$0$	Tetrahedral	$109.5\degree$	$\mathrm{CH}_4$, $\mathrm{NH}_4^+$
$4$	$1$	Trigonal pyramidal	$\lt 109.5\degree$	$\mathrm{NH}_3$, $\mathrm{PH}_3$
$4$	$2$	Bent	$\lt 109.5\degree$	$\mathrm{H}_2\mathrm{O}$, $\mathrm{H}_2\mathrm{S}$
$5$	$0$	Trigonal bipyramidal	$120\degree, 90\degree$	$\mathrm{PF}_5$, $\mathrm{PCl}_5$
$5$	$1$	Seesaw	$\lt 90\degree, \lt 120\degree$	$\mathrm{SF}_4$
$5$	$2$	T-shaped	$\lt 90\degree$	$\mathrm{ClF}_3$
$5$	$3$	Linear	$180\degree$	$\mathrm{XeF}_2$, $\mathrm{I}_3^-$
$6$	$0$	Octahedral	$90\degree$	$\mathrm{SF}_6$, $\mathrm{PF}_6^-$
$6$	$1$	Square pyramidal	$\lt 90\degree$	$\mathrm{BrF}_5$, $\mathrm{IF}_5$
$6$	$2$	Square planar	$90\degree$	$\mathrm{XeF}_4$, $\mathrm{ICl}_4^-$

Bond Angle Distortions

Lone pair repulsion compresses bond angles. The more lone pairs, the greater the compression:

Molecule	Expected	Actual	Reason
$\mathrm{NH}_3$	$109.5\degree$	$107\degree$	One lone pair compresses angles
$\mathrm{H}_2\mathrm{O}$	$109.5\degree$	$104.5\degree$	Two lone pairs compress angles more

Electronegativity effects also distort angles: more electronegative terminal atoms pull bonding pairs away from the central atom, reducing repulsion and decreasing bond angles.

Common Pitfalls

• Confusing steric number (total regions of electron density) with the number of atoms bonded.
• Placing lone pairs in equatorial positions of trigonal bipyramidal arrangements (equatorial positions minimise $90\degree$ repulsions).
• Forgetting that double and triple bonds count as one region of electron density.

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2. Hybridization

Definition

Hybridization is a mathematical model that combines atomic orbitals on the central atom to form equivalent hybrid orbitals that explain observed geometries.

Types of Hybridization

Hybridization	Geometry	Orbitals combined	Angle
$sp$	Linear	$1s + 1p$	$180\degree$
$sp^2$	Trigonal planar	$1s + 2p$	$120\degree$
$sp^3$	Tetrahedral	$1s + 3p$	$109.5\degree$
$sp^3d$	Trigonal bipyramidal	$1s + 3p + 1d$	$120\degree, 90\degree$
$sp^3d^2$	Octahedral	$1s + 3p + 2d$	$90\degree$

Determining Hybridization

Count the regions of electron density (steric number) around the central atom:

$$\mathrm{Steric number} = \mathrm{bonded atoms} + \mathrm{lone pairs}$$

Steric number	Hybridization
$2$	$sp$
$3$	$sp^2$
$4$	$sp^3$
$5$	$sp^3d$
$6$	$sp^3d^2$

Examples

Molecule	Central atom	Steric number	Hybridization	Shape
$\mathrm{CH}_4$	C	$4$	$sp^3$	Tetrahedral
$\mathrm{NH}_3$	N	$4$	$sp^3$	Trigonal pyramidal
$\mathrm{H}_2\mathrm{O}$	O	$4$	$sp^3$	Bent
$\mathrm{BF}_3$	B	$3$	$sp^2$	Trigonal planar
$\mathrm{CO}_2$	C	$2$	$sp$	Linear
$\mathrm{SF}_6$	S	$6$	$sp^3d^2$	Octahedral
$\mathrm{PCl}_5$	P	$5$	$sp^3d$	Trigonal bipyramidal

Sigma and Pi Bonds

• Sigma ($\sigma$) bond: formed by head-on overlap of orbitals along the internuclear axis. Always the first bond between two atoms. Can be formed by $s$-$s$, $s$-$p$, $p$-$p$, or hybrid-hybrid overlap.
• Pi ($\pi$) bond: formed by sideways overlap of parallel $p$-orbitals. Always the second (or third) bond in a multiple bond.

A double bond consists of one $\sigma$ and one $\pi$ bond. A triple bond consists of one $\sigma$ and two $\pi$ bonds.

$$\mathrm{C=C}: 1\sigma + 1\pi, \qquad \mathrm{C\equiv C}: 1\sigma + 2\pi$$

Common Pitfalls

• Hybridization describes the central atom's orbitals, not the entire molecule.
• Lone pairs are included in the steric number when determining hybridization.
• Pi bonds do not participate in hybridization — they involve unhybridized $p$-orbitals.

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3. Molecular Orbital Theory

Principle

Molecular orbital (MO) theory treats the molecule as a whole. Atomic orbitals combine to form molecular orbitals that are delocalised over the entire molecule.

Rules for Forming MOs

1. Atomic orbitals must have similar energy.
2. Atomic orbitals must overlap effectively.
3. The number of MOs equals the number of atomic orbitals combined.
4. MOs that are lower in energy than the parent AOs are bonding; those higher are antibonding.
5. Each MO can hold two electrons with opposite spins.

MO Diagrams for Diatomic Molecules

For homonuclear diatomics ($\mathrm{O}_2$, $\mathrm{F}_2$, and beyond):

$$\sigma_{1s} \lt \sigma_{1s}^* \lt \sigma_{2s} \lt \sigma_{2s}^* \lt \sigma_{2p_z} \lt \pi_{2p_x} = \pi_{2p_y} \lt \pi_{2p_x}^* = \pi_{2p_y}^* \lt \sigma_{2p_z}^*$$

For $\mathrm{B}_2$, $\mathrm{C}_2$, $\mathrm{N}_2$ (and $\mathrm{Li}_2$, $\mathrm{Be}_2$), the $\sigma_{2p_z}$ and $\pi_{2p}$ levels are reversed:

$$\cdots \lt \pi_{2p_x} = \pi_{2p_y} \lt \sigma_{2p_z} \lt \pi_{2p_x}^* = \pi_{2p_y}^* \lt \sigma_{2p_z}^*$$

Bond Order

$$\mathrm{Bond order} = \frac{N_{\mathrm{bonding}} - N_{\mathrm{antibonding}}}{2}$$

Molecule	Bond order	Bond type	Magnetic
$\mathrm{H}_2$	$1$	Single	Diamagnetic
$\mathrm{He}_2$	$0$	Does not exist	—
$\mathrm{Li}_2$	$1$	Single	Diamagnetic
$\mathrm{B}_2$	$1$	Single	Paramagnetic
$\mathrm{C}_2$	$2$	Double	Diamagnetic
$\mathrm{N}_2$	$3$	Triple	Diamagnetic
$\mathrm{O}_2$	$2$	Double	Paramagnetic
$\mathrm{F}_2$	$1$	Single	Diamagnetic
$\mathrm{Ne}_2$	$0$	Does not exist	—

Key Successes of MO Theory

• Explains why $\mathrm{O}_2$ is paramagnetic (two unpaired electrons in $\pi^*$ orbitals) — VSEPR and valence bond theory cannot explain this.
• Explains why $\mathrm{He}_2$ and $\mathrm{Ne}_2$ do not exist (bond order = 0).
• Predicts correct bond orders and bond lengths.

Common Pitfalls

• The ordering of $\sigma_{2p}$ and $\pi_{2p}$ reverses between $\mathrm{N}_2$ and $\mathrm{O}_2$.
• Bond order of zero means the molecule is unstable.
• Antibonding electrons weaken the bond more than bonding electrons strengthen it.

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4. Metallic Bonding

The Sea of Electrons Model

In a metal, the valence electrons are delocalised over the entire lattice of positive metal ions. These delocalised electrons form an "electron sea" that holds the metal cations together through electrostatic attraction.

Properties Explained by Metallic Bonding

Property	Explanation
Electrical conductivity	Delocalised electrons can move freely under an applied potential
Thermal conductivity	Delocalised electrons transfer kinetic energy
Malleability	Cations can slide past each other without breaking the metallic bond
Ductility	Same as malleability — layers of cations can shift
High melting points	Strong electrostatic attraction between cations and electrons
Luster	Delocalised electrons absorb and re-emit light at all visible wavelengths

Factors Affecting Metallic Bond Strength

Factor	Effect
More delocalised electrons	Stronger bonding
Smaller ionic radius	Stronger bonding (cations closer to electrons)
Higher ionic charge	Stronger bonding

Comparison	Stronger bonding	Reason
Na vs Al	Al	Al contributes 3 electrons, Na contributes 1
Mg vs Ca	Mg	$\mathrm{Mg}^{2+}$ is smaller than $\mathrm{Ca}^{2+}$

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5. Band Theory

Formation of Energy Bands

When a large number of atoms are brought together (as in a metal), the discrete atomic energy levels broaden into energy bands due to orbital overlap.

• Valence band: the highest occupied band at $0\mathrm{ K}$.
• Conduction band: the lowest unoccupied band.
• Band gap ($E_g$): the energy difference between the top of the valence band and the bottom of the conduction band.

Classification of Solids

Type	Band gap	Conductivity	Examples
Metal	$E_g = 0$	High; electrons easily promoted	$\mathrm{Cu}$, $\mathrm{Fe}$, $\mathrm{Na}$
Semiconductor	Small $E_g$	Moderate; temperature-dependent	$\mathrm{Si}$ ($1.1\mathrm{ eV}$), $\mathrm{Ge}$ ($0.7\mathrm{ eV}$)
Insulator	Large $E_g$	Very low	Diamond ($5.5\mathrm{ eV}$), $\mathrm{SiO}_2$

Semiconductor Behaviour

In an intrinsic semiconductor at $0\mathrm{ K}$, the valence band is full and the conduction band is empty. As temperature increases:

1. Some electrons gain enough energy to cross the band gap.
2. These electrons enter the conduction band and carry charge.
3. The holes left in the valence band also contribute to conduction.

$$n_i \propto \exp\!\left(-\frac{E_g}{2k_BT}\right)$$

Doping

• n-type semiconductor: doped with a Group 15 element (e.g., P in Si). Extra electrons are donated to the conduction band.
• p-type semiconductor: doped with a Group 13 element (e.g., B in Si). Holes are created in the valence band.

Common Pitfalls

• The band gap is a property of the bulk material, not individual atoms.
• Metals have overlapping valence and conduction bands, not a zero band gap with separate bands.
• In p-type semiconductors, it is holes (positive charge carriers) that conduct, not the dopant atoms themselves.

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Practice Problems

Problem 1

Predict the shape and bond angles of $\mathrm{ClF}_3$ and explain why the lone pairs occupy equatorial positions.

Solution:

Chlorine has $7$ valence electrons. In $\mathrm{ClF}_3$: $3$ bonding pairs + $2$ lone pairs = $5$ regions of electron density (trigonal bipyramidal electron pair geometry).

The lone pairs occupy equatorial positions because this minimises the number of $90\degree$ lone pair--bonding pair repulsions. If the lone pairs were axial, they would experience three $90\degree$ repulsions each. In equatorial positions, they experience only two $90\degree$ repulsions each.

Molecular shape: T-shaped. Bond angles: approximately $87.5\degree$ and $175\degree$ (slightly compressed from $90\degree$ and $180\degree$ due to lone pair repulsion).

Problem 2

Construct the MO diagram for $\mathrm{O}_2$ and use it to explain the paramagnetism and bond order.

Solution:

Electron configuration of $\mathrm{O}_2$ ($12$ valence electrons):

$$\sigma_{2s}^2\, (\sigma_{2s}^*)^2\, \sigma_{2p_z}^2\, \pi_{2p_x}^2\, \pi_{2p_y}^2\, (\pi_{2p_x}^*)^1\, (\pi_{2p_y}^*)^1$$

Bond order $= \frac{10 - 6}{2} = 2$

The two unpaired electrons in the degenerate $\pi_{2p}^*$ orbitals make $\mathrm{O}_2$ paramagnetic. This is confirmed experimentally: liquid oxygen is attracted to a magnet.

Problem 3

Explain why the melting point of aluminium ($660\degree\mathrm{C}$) is much higher than that of sodium ($98\degree\mathrm{C}$).

Solution:

Aluminium has the electron configuration $[\mathrm{Ne}]\, 3s^2\, 3p^1$ and contributes three delocalised electrons to the metallic bond. Sodium has $[\mathrm{Ne}]\, 3s^1$ and contributes only one. More delocalised electrons produce stronger electrostatic attraction between the cations and the electron sea. Additionally, $\mathrm{Al}^{3+}$ is smaller than $\mathrm{Na}^+$, so the charge density is higher and the ions are closer together. Both factors increase the strength of the metallic bond and therefore the melting point.

Problem 4

Silicon has a band gap of $1.1\mathrm{ eV}$. Calculate the minimum wavelength of light required to excite an electron from the valence band to the conduction band.

Solution:

E = \frac`\{hc}`{\lambda} \implies \lambda = \frac`\{hc}`{E}$$\lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{1.1 \times 1.602 \times 10^{-19}} = \frac{1.988 \times 10^{-25}}{1.762 \times 10^{-19}} = 1.13 \times 10^{-6}\mathrm{ m} = 1130\mathrm{ nm}$$

This is in the infrared region.

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Worked Examples

Worked Example: VSEPR prediction for an oxoanion

Predict the molecular geometry, bond angles, hybridization, and polarity of the $\mathrm{SO}_4^{2-}$ ion.

Solution

Step 1: Draw the Lewis structure.

Sulfur has 6 valence electrons. Each oxygen contributes 6. The $-2$ charge adds 2 electrons. Total: $6 + 4(6) + 2 = 32$ valence electrons.

Sulfur forms double bonds with all four oxygens (expanding the octet since S is in Period 3): no lone pairs on sulfur. Each oxygen has 2 lone pairs.

$\mathrm{O}_2\mathrm{S}(=\mathrm{O})_2^{2-}$

Step 2: Determine the steric number.

Bonding pairs on S: 4. Lone pairs on S: 0. Steric number = 4.

Step 3: Predict geometry and hybridization.

Electron pair geometry: tetrahedral. Molecular shape: tetrahedral. Hybridization: $sp^3$.

Step 4: Bond angles.

All $\mathrm{O\!-\!S\!-\!O}$ bond angles are approximately $109.5\degree$ (ideal tetrahedral). The double bonds repel slightly more than single bonds, so angles may deviate by a few degrees, but the overall geometry remains tetrahedral.

Step 5: Polarity.

The $\mathrm{S\!-\!O}$ bond dipoles are symmetrically arranged and cancel. $\mathrm{SO}_4^{2-}$ is nonpolar despite having polar bonds.

Worked Example: MO theory for a heteronuclear diatomic

Use MO theory to determine the bond order and magnetic properties of NO. Compare the bond order and bond length of NO with those of $\mathrm{NO}^+$.

Solution

NO has 11 valence electrons (5 from N, 6 from O). Since oxygen is to the right of nitrogen in the periodic table, the $\mathrm{O}_2$/$\mathrm{F}_2$ energy ordering applies:

$\sigma_{2s}^2\, (\sigma_{2s}^*)^2\, \sigma_{2p_z}^2\, \pi_{2p_x}^2\, \pi_{2p_y}^2\, (\pi_{2p_x}^*)^1$

Bonding electrons: $2 + 2 + 2 + 2 + 2 = 10$. Antibonding electrons: $2 + 1 = 3$.

$\mathrm{Bond\ order} = \frac{10 - 3}{2} = 3.5$

One unpaired electron in $\pi_{2p_x}^*$: NO is paramagnetic.

For $\mathrm{NO}^+$: 10 valence electrons (one fewer than NO).

$\sigma_{2s}^2\, (\sigma_{2s}^*)^2\, \sigma_{2p_z}^2\, \pi_{2p_x}^2\, \pi_{2p_y}^2$

$\mathrm{Bond\ order} = \frac{10 - 2}{2} = 4.0$

$\mathrm{NO}^+$ has no unpaired electrons: diamagnetic.

Comparison: $\mathrm{NO}^+$ has a higher bond order (4.0 vs 3.5), so it has a shorter bond length and higher bond dissociation energy than NO. Removing an electron from an antibonding orbital strengthens the bond.

Worked Example: Sigma and pi bond counting

How many sigma ($\sigma$) and pi ($\pi$) bonds are in a molecule of $\mathrm{CH}_3\mathrm{CH}=\mathrm{CHC}\equiv\mathrm{N}$? Identify the hybridization of each carbon atom.

Solution

Step 1: Draw the structure.

$\mathrm{CH_3 - CH = CH - C \equiv N}$

Step 2: Count bonds by type.

Bond	Type	$\sigma$	$\pi$
$\mathrm{C_1\!-\!C_2}$	Single	1	0
$\mathrm{C_2=C_3}$	Double	1	1
$\mathrm{C_3\!-\!C_4}$	Single	1	0
$\mathrm{C_4\equiv N}$	Triple	1	2
$\mathrm{C_1\!-\!H}$ (x3)	Single	3	0
$\mathrm{C_3\!-\!H}$	Single	1	0

Total: $8\sigma$ bonds, $3\pi$ bonds.

Step 3: Hybridization of each carbon.

Carbon	Regions of e$^-$ density	Hybridization
$\mathrm{C_1}$ ($\mathrm{CH_3\!-\!}$)	4 (3 C--H + 1 C--C)	$sp^3$
$\mathrm{C_2}$ ($\mathrm{=\!CH\!-\!}$)	3 (1 C--C + 1 C=C + 1 C--H)	$sp^2$
$\mathrm{C_3}$ ($\mathrm{-\!CH\!=\!}$)	3 (1 C=C + 1 C--C + 1 C--H)	$sp^2$
$\mathrm{C_4}$ ($\mathrm{-\!C\!\equiv}$)	2 (1 C--C + 1 C$\equiv$N)	$sp$

Worked Example: Band gap and semiconductor classification

Diamond has a band gap of $5.5\mathrm{ eV}$ and silicon has a band gap of $1.1\mathrm{ eV}$. (a) Calculate the minimum wavelength of light required to excite an electron across the band gap in each material. (b) Classify each material. (c) Explain why diamond is transparent to visible light but silicon is not.

Solution

Step 1: Calculate the threshold wavelengths.

$\lambda = \frac{hc}{E}$

For diamond ($E = 5.5\mathrm{ eV} = 5.5 \times 1.602 \times 10^{-19}\mathrm{ J}$):

$\lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{5.5 \times 1.602 \times 10^{-19}} = \frac{1.988 \times 10^{-25}}{8.81 \times 10^{-19}} = 2.26 \times 10^{-7}\mathrm{ m} = 226\mathrm{ nm}$

For silicon ($E = 1.1\mathrm{ eV}$):

$\lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{1.1 \times 1.602 \times 10^{-19}} = \frac{1.988 \times 10^{-25}}{1.76 \times 10^{-19}} = 1.13 \times 10^{-6}\mathrm{ m} = 1130\mathrm{ nm}$

Step 2: Classification.

Diamond ($5.5\mathrm{ eV}$): insulator (band gap far exceeds thermal energy at room temperature). Silicon ($1.1\mathrm{ eV}$): semiconductor (band gap comparable to visible photon energies).

Step 3: Transparency explanation.

Visible light has wavelengths $400$--$700\mathrm{ nm}$. Diamond's threshold ($226\mathrm{ nm}$) is in the ultraviolet, so all visible photons have insufficient energy to excite electrons across the band gap. Visible light passes through unabsorbed: diamond is transparent.

Silicon's threshold ($1130\mathrm{ nm}$) is in the infrared. All visible photons have more than enough energy to promote electrons to the conduction band, so visible light is absorbed. Silicon appears metallic-grey and opaque.

Worked Example: Metallic bonding and melting point comparison

Explain why tungsten ($\mathrm{W}$) has a melting point of $3422\degree\mathrm{C}$ while sodium ($\mathrm{Na}$) melts at $98\degree\mathrm{C}$, despite both being metals.

Solution

The strength of metallic bonding depends on three factors:

1. Number of delocalized electrons per atom:

   • $\mathrm{W}$: electron configuration $[\mathrm{Xe}]\, 4f^{14}\, 5d^4\, 6s^2$, contributing up to 6 valence electrons to the delocalized sea.
   • $\mathrm{Na}$: configuration $[\mathrm{Ne}]\, 3s^1$, contributing only 1 valence electron.

2. Ionic radius (charge density):

   • $\mathrm{W}^{6+}$ has a much smaller ionic radius than $\mathrm{Na}^+$ because tungsten is in Period 6 but has a much higher nuclear charge pulling electrons inward.
   • Smaller ionic radius means the cations are closer to the electron sea, increasing electrostatic attraction.

3. Ionic charge:

   • $\mathrm{W}$ can achieve high oxidation states ($+6$), creating cations with much greater charge density than $\mathrm{Na}^+$.

The combined effect of more delocalized electrons, smaller ionic radius, and higher ionic charge produces a much stronger metallic bond in tungsten. This is reflected in its extremely high melting point, hardness, and tensile strength compared to sodium.

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Common Pitfalls

• Counting lone pairs incorrectly for VSEPR: The steric number includes both bonding pairs and lone pairs. $\mathrm{NH}_3$ has 3 bonds + 1 lone pair = 4 regions, giving tetrahedral electron pair geometry and trigonal pyramidal molecular shape.

• Misidentifying the central atom: In molecules like $\mathrm{OF}_2$, oxygen is the central atom (less electronegative than fluorine). In $\mathrm{SCl}_4$, sulfur is central. Always place the least electronegative atom (excluding hydrogen) at the centre.

• Assigning hybridization based on bonded atoms only: Hybridization depends on the total steric number (bonded atoms + lone pairs). $\mathrm{NH}_3$ has steric number 4, so its hybridization is $sp^3$, not $sp^2$.

• Reversing the MO energy ordering between $\mathrm{N}_2$ and $\mathrm{O}_2$: For $\mathrm{B}_2$, $\mathrm{C}_2$, $\mathrm{N}_2$ (and lighter), $\pi_{2p}$ is lower than $\sigma_{2p_z}$. For $\mathrm{O}_2$, $\mathrm{F}_2$, $\mathrm{Ne}_2$, the ordering reverses. The boundary is between $\mathrm{N}_2$ and $\mathrm{O}_2$.

• Treating antibonding electrons as equivalent to bonding electrons: Each antibonding electron weakens the bond by the same amount a bonding electron strengthens it. An electron in a $\pi^*$ orbital reduces the bond order by 0.5.

• Assuming metals always have high melting points: Mercury ($\mathrm{Hg}$, $Z = 80$) is a liquid at room temperature (m.p. $-39\degree\mathrm{C}$) because it has a large ionic radius and only 2 valence electrons, producing weak metallic bonding despite being in Period 6.

• Placing lone pairs in axial positions of trigonal bipyramidal: Lone pairs always occupy equatorial positions to minimise $90\degree$ repulsions. Axial lone pairs would experience three $90\degree$ interactions; equatorial lone pairs experience only two.

• Confusing $p$-type and $n$-type doping: $n$-type means extra negative charge carriers (electrons from Group 15 dopant). $p$-type means positive charge carriers (holes from Group 13 dopant). The dopant itself does not conduct; it provides carriers.

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Exam-Style Problems

1. Predict the molecular geometry and bond angles of $\mathrm{BrF}_5$. Draw the structure, clearly indicating the position of the lone pair. State the hybridization of bromine and explain why the lone pair occupies an axial position. [Medium]

2. Construct the MO diagram for $\mathrm{C}_2$ and determine: (a) the bond order, (b) whether it is paramagnetic or diamagnetic, (c) the number of unpaired electrons. Explain why $\mathrm{C}_2$ has a higher bond order than $\mathrm{B}_2$. [Hard]

3. For the molecule $\mathrm{SF}_4$: (a) Predict its molecular geometry. (b) Would you expect the axial or equatorial S--F bonds to be longer? Justify using VSEPR theory. (c) Is the molecule polar? Explain. [Medium]

4. Gallium arsenide (GaAs) is doped with zinc (Zn, Group 12). (a) Explain whether this produces $n$-type or $p$-type semiconductors. (b) Describe the mechanism of charge conduction at room temperature. (c) How does increasing temperature affect the conductivity? [Medium]

5. Explain why the bond angle in $\mathrm{H}_2\mathrm{O}$ ($104.5\degree$) is smaller than in $\mathrm{NH}_3$ ($107\degree$), which is smaller than the tetrahedral angle ($109.5\degree$). Reference specific VSEPR repulsion orders ($\mathrm{LP\!-\!LP} \gt \mathrm{LP\!-\!BP} \gt \mathrm{BP\!-\!BP}$) in your explanation. [Medium]

6. For the molecule HCN: (a) Draw the Lewis structure. (b) Determine the hybridization of each atom. (c) State the number of sigma and pi bonds. (d) Predict the H--C--N bond angle. (e) Is the molecule polar? [Easy]

7. Use MO theory to explain why $\mathrm{He}_2$ does not exist as a stable molecule, while $\mathrm{He}_2^+$ has been detected experimentally. Include bond order calculations for both species and a sketch of the MO energy level diagram. [Hard]

8. Explain why the melting point of magnesium ($650\degree\mathrm{C}$) is higher than that of sodium ($98\degree\mathrm{C}$) but lower than that of aluminium ($660\degree\mathrm{C}$). Reference the number of delocalized electrons, ionic charge, and ionic radius in your answer. [Medium]

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Worked Examples (Expanded)

Worked Example: Dipole Moment Calculation

The $\mathrm{H\!-\!Cl}$ bond has a bond length of $127\;\mathrm{pm}$ and a bond dipole moment of $1.08\;\mathrm{D}$. Calculate the percent ionic character of the bond. ($1\;\mathrm{D} = 3.336 \times 10^{-30}\;\mathrm{C \cdot m}$; charge of electron $= 1.602 \times 10^{-19}\;\mathrm{C}$)

Solution

If the bond were $100\%$ ionic, the dipole moment would be:

$\mu_{\mathrm{ionic}} = q \times d = (1.602 \times 10^{-19}\;\mathrm{C}) \times (127 \times 10^{-12}\;\mathrm{m}) = 2.035 \times 10^{-29}\;\mathrm{C \cdot m}$

Converting to debye: $\mu_{\mathrm{ionic}} = \frac{2.035 \times 10^{-29}}{3.336 \times 10^{-30}} = 6.10\;\mathrm{D}$

The actual dipole moment is $1.08\;\mathrm{D}$, so the percent ionic character is:

$\%\;\text{ionic} = \frac{1.08}{6.10} \times 100 = 17.7\%$

This result demonstrates that even highly polar bonds like $\mathrm{H\!-\!Cl}$ have significant covalent character. Purely ionic bonding is an idealisation; real bonds exist on a continuum.

Worked Example: Lewis Structure and Formal Charge for an Unusual Species

Draw the Lewis structure of $\mathrm{XeO}_3$, determine the formal charge on each atom, predict the molecular geometry, and state whether the molecule is polar.

Solution

Step 1: Count valence electrons. Xe: $8$ valence electrons. Each O: $6$ valence electrons. Total: $8 + 3(6) = 26$ valence electrons.

Step 2: Build the skeleton. Xe is the central atom. Form single bonds to all three oxygens (using 6 electrons). Distribute remaining electrons to satisfy octets: $26 - 6 = 20$ remaining. Each oxygen needs 6 more electrons (3 lone pairs). That accounts for $3 \times 6 = 18$ electrons. Remaining: $20 - 18 = 2$ electrons, which go on Xe as a lone pair.

Step 3: Check formal charges.

Formal charge $= V - N - B/2$ where $V$ = valence electrons, $N$ = non-bonding electrons, $B$ = bonding electrons.

• Xe: $8 - 2 - 6/2 = 8 - 2 - 3 = +3$
• Each O: $6 - 6 - 2/2 = 6 - 6 - 1 = -1$

Net formal charge: $+3 + 3(-1) = 0$. This is valid but has high formal charges.

Step 4: Minimise formal charges by forming double bonds.

Convert one lone pair on each oxygen into a bonding pair, forming three Xe=O double bonds:

$\mathrm{O}=\mathrm{Xe}(=\mathrm{O})_2 \text{ with one lone pair on Xe}$

• Xe: $8 - 2 - 12/2 = 8 - 2 - 6 = 0$
• Each O: $6 - 4 - 4/2 = 6 - 4 - 2 = 0$

All formal charges are zero — this is the preferred structure.

Step 5: Geometry and polarity.

Steric number = 3 bonding regions + 1 lone pair = 4. Electron pair geometry: tetrahedral. Molecular shape: trigonal pyramidal.

The Xe=O bonds are polar (Xe and O have different electronegativities), and the molecule lacks a plane of symmetry (the lone pair breaks the symmetry). Therefore $\mathrm{XeO}_3$ is polar.

Worked Example: Resonance and Bond Order in Polyatomic Ions

Draw all significant resonance structures for the nitrate ion ($\mathrm{NO}_3^-$). Calculate the average N--O bond order. Explain why the measured N--O bond length ($124\;\mathrm{pm}$) is intermediate between a typical N--O single bond ($136\;\mathrm{pm}$) and N=O double bond ($115\;\mathrm{pm}$).

Solution

$\mathrm{NO}_3^-$ has $5 + 3(6) + 1 = 24$ valence electrons.

Three equivalent resonance structures can be drawn, each with one N=O double bond and two N--O single bonds, with the double bond rotating among the three oxygen atoms. The negative charge is delocalised over the three oxygens.

The resonance hybrid has partial double-bond character between N and each O.

$\text{Bond order} = \frac{1\;\text{(double bond)} + 2\;\text{(single bonds)}}{3\;\text{(N--O bonds)}} = \frac{1 + 2}{3} = \frac{3}{3} = 1.33$

The average bond order of $1.33$ means each N--O bond has approximately $1/3$ double-bond character and $2/3$ single-bond character. The observed bond length of $124\;\mathrm{pm}$ is between the single-bond ($136\;\mathrm{pm}$) and double-bond ($115\;\mathrm{pm}$) values, consistent with a bond order of $1.33$.

This demonstrates a key principle of resonance: the real structure is an average of all resonance forms, and the electron density is delocalised, producing bonds that are stronger and shorter than single bonds.

Worked Example: Lattice Energy and Born-Lande Equation

Use the Kapustinskii equation to estimate the lattice energy of $\mathrm{KBr}$:

$U = \frac{1200.2 \cdot z_+ z_- \cdot v}{r_+ + r_-}\left(1 - \frac{34.5}{r_+ + r_-}\right)\;\mathrm{kJ/mol}$

where $v$ is the number of ions per formula unit ($v = 2$ for $\mathrm{KBr}$), $r_+$ and $r_-$ are ionic radii in pm, and $z_+, z_-$ are ion charges. Given: $r(\mathrm{K}^+) = 138\;\mathrm{pm}$, $r(\mathrm{Br}^-) = 196\;\mathrm{pm}$.

Solution

$r_+ + r_- = 138 + 196 = 334\;\mathrm{pm}$

$U = \frac{1200.2 \times 1 \times 1 \times 2}{334}\left(1 - \frac{34.5}{334}\right)$

$U = \frac{2400.4}{334}\left(1 - 0.1033\right)$

$U = 7.187 \times 0.8967 = 6.445\;\mathrm{kJ/mol} \times 10^2 = 644.5\;\mathrm{kJ/mol}$

Wait — the Kapustinskii equation as written gives the answer in $\mathrm{kJ/mol}$ directly when radii are in pm:

$U = \frac{1200.2 \times 2}{334}\left(1 - \frac{34.5}{334}\right) = \frac{2400.4}{334}(0.8967) = 7.187 \times 0.8967 = 6.44$

$U \approx -644\;\mathrm{kJ/mol}$

The experimental lattice energy of $\mathrm{KBr}$ is approximately $-679\;\mathrm{kJ/mol}$. The Kapustinskii estimate is within about $5\%$, demonstrating its utility for predicting lattice energies when Born-Haber data are unavailable.

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Exam-Style Problems (Expanded)

Problem 9: Quantitative -- Formal Charge and Stability of Lewis Structures

Sulfur dioxide ($\mathrm{SO}_2$) has two resonance structures: one with a double bond and a coordinate covalent bond, and one with two double bonds. (a) Draw both resonance structures and calculate the formal charge on each atom in both. (b) Identify the more stable structure and justify using formal charge analysis. (c) The measured S--O bond length in $\mathrm{SO}_2$ is $143\;\mathrm{pm}$. A typical S--O single bond is $155\;\mathrm{pm}$ and S=O is $143\;\mathrm{pm}$. What does this suggest about the contribution of each resonance form?

Problem 10: Extended Response -- Molecular Polarity and Physical Properties

$\mathrm{PF}_5$ has trigonal bipyramidal geometry and is non-polar, while $\mathrm{IF}_5$ has square pyramidal geometry and is polar. (a) Draw the 3D structures of both molecules, clearly showing the position of lone pairs. (b) Explain why $\mathrm{PF}_5$ is non-polar but $\mathrm{IF}_5$ is polar, using symmetry arguments. (c) Predict which molecule has the higher boiling point and explain your reasoning. (d) How do the VSEPR geometries of $\mathrm{PF}_5$ and $\mathrm{IF}_5$ differ in terms of the position of lone pairs in the trigonal bipyramidal vs octahedral electron pair geometries?

Problem 11: Quantitative -- MO Theory Bond Length Correlation

Use MO theory to rank the following diatomic species in order of increasing bond length: $\mathrm{O}_2$, $\mathrm{O}_2^+$, $\mathrm{O}_2^-$, $\mathrm{O}_2^{2-}$. For each species, state the bond order and number of unpaired electrons. Explain the trend in bond length using the concept of bond order.

Problem 12: Extended Response -- Hybridization in Carbon Compounds

For each of the following molecules, identify the hybridization of every atom (except H): (a) $\mathrm{H}_2\mathrm{C}=\mathrm{C}=\mathrm{CH}_2$ (allene), (b) $\mathrm{HC}\equiv\mathrm{C}\!-\!\mathrm{C}\equiv\mathrm{N}$ (cyanoacetylene), (c) $\mathrm{CH}_3\mathrm{COOH}$ (acetic acid). In each case, count the total number of sigma bonds and pi bonds. Explain why allene is not planar, while acetic acid has two different C--O bond lengths.

Problem 13: Extended Response -- Semiconductor Devices

A p-n junction diode is fabricated by doping silicon with boron (p-type) on one side and phosphorus (n-type) on the other. (a) Explain the formation of the depletion region at the junction. (b) Explain why the diode conducts current in forward bias but not in reverse bias, with reference to band bending at the junction. (c) Calculate the minimum photon energy required to generate an electron-hole pair in silicon ($E_g = 1.1\;\mathrm{eV}$) and state the corresponding maximum wavelength. (d) Explain why silicon solar cells have a theoretical maximum efficiency of approximately $30\%$ and discuss one approach to exceeding this limit.

Problem 14: Quantitative -- Enthalpy of Atomisation and Bond Energy

The standard enthalpy of atomisation of methane is $+1660\;\mathrm{kJ/mol}$ (i.e., $\mathrm{CH}_4(g) \to \mathrm{C}(g) + 4\mathrm{H}(g)$). Given that the standard enthalpy of formation of $\mathrm{CH}_4(g)$ is $-75\;\mathrm{kJ/mol}$, the bond dissociation energy of $\mathrm{H}_2$ is $+436\;\mathrm{kJ/mol}$, and the enthalpy of sublimation of graphite is $+717\;\mathrm{kJ/mol}$: (a) Calculate the mean C--H bond enthalpy in methane. (b) Explain why the first C--H bond dissociation energy ($+439\;\mathrm{kJ/mol}$) differs from the mean value. (c) The C--H bond enthalpy in $\mathrm{CHCl}_3$ is $413\;\mathrm{kJ/mol}$. Explain why this is different from the value in methane.

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Common Pitfalls (Expanded)

• Counting lone pairs incorrectly for expanded octet species: Elements in Period 3 and beyond can have more than 8 valence electrons (expanded octet). For $\mathrm{SF}_6$, sulfur has $12$ valence electrons (6 bonds, 0 lone pairs). Do not force an octet on central atoms that can accommodate more.

• Assuming that polar bonds always make a molecule polar: $\mathrm{CCl}_4$, $\mathrm{CO}_2$, and $\mathrm{BF}_3$ all have polar bonds but are non-polar molecules because the bond dipoles cancel due to molecular symmetry. Always check the geometry before assigning polarity.

• Misidentifying coordinate (dative) covalent bonds: A coordinate bond is formed when both electrons in the bond come from the same atom. Once formed, a coordinate bond is indistinguishable from a normal covalent bond. In $\mathrm{NH}_4^+$, the fourth N--H bond is a coordinate bond, but all four N--H bonds are equivalent in the ammonium ion.

• Using the wrong MO energy ordering for heteronuclear diatomics: For heteronuclear diatomics (e.g., $\mathrm{CO}$, $\mathrm{NO}$), the energy levels are not exactly degenerate as in homonuclear diatomics. The more electronegative atom contributes lower-energy AOs, and the bonding orbitals are closer in energy to the AO of the more electronegative atom. However, for a rough analysis, the homonuclear ordering is often used as an approximation.

• Confusing band gap with bond gap: The band gap is a bulk solid-state property arising from the interaction of many atoms. It is not the same as the energy gap between bonding and antibonding molecular orbitals in a single molecule (which is related to bond dissociation energy).

• Assuming all p-type dopants produce the same effect: The effectiveness of doping depends on the concentration of dopant atoms and their energy levels relative to the valence band. Deep-level impurities (e.g., gold in silicon) create energy levels in the middle of the band gap and act as recombination centres, reducing carrier lifetime — the opposite of what is desired for semiconductor devices.

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If You Get These Wrong, Revise:

• Atomic structure and electron configurations → Review ./atomic-theory
• States of matter and physical properties → Review ./states-of-matter
• Periodic trends in electronegativity → Review ./periodicity
• Organic molecular geometry → Review ./organic-chemistry-advanced