Chemical Bonding — Diagnostic Tests

Unit Tests

UT-1: VSEPR and Molecular Geometry

Question: Draw the Lewis structures and predict the shapes and bond angles of $\text{XeF}_4$, $\text{SF}_4$, and $\text{ClF}_3$. Identify the deviation from ideal bond angles where applicable.

Solution:

$\text{XeF}_4$: Xe has 8 valence electrons. Four bonding pairs to F atoms, two lone pairs. Electron geometry: octahedral. Molecular shape: square planar. Bond angles: $90^\circ$ and $180^\circ$. The lone pairs are trans to each other (minimising repulsion), so bond angles are exactly $90^\circ$.

$\text{SF}_4$: S has 6 valence electrons. Four bonding pairs, one lone pair. Electron geometry: trigonal bipyramidal. The lone pair occupies an equatorial position (minimises repulsion at $90^\circ$ from only two bonds instead of three). Molecular shape: seesaw. Axial-equatorial bond angles are compressed below $90^\circ$ (lone pair repulsion) and the equatorial-equatorial angle is less than $120^\circ$. Actual axial-equatorial: $\approx 85^\circ$ (from experimental data).

$\text{ClF}_3$: Cl has 7 valence electrons. Three bonding pairs, two lone pairs. Electron geometry: trigonal bipyramidal. Both lone pairs occupy equatorial positions. Molecular shape: T-shaped. Bond angles: $\approx 87.5^\circ$ (slightly less than $90^\circ$ due to lone pair repulsion).

---

UT-2: Hybridisation Determination

Question: Determine the hybridisation of the central atom in $\text{BF}_3$, $\text{NH}_3$, $\text{H}_2\text{O}$, $\text{XeF}_2$, and $\text{BrF}_5$. For each, state the number of electron domains and the predicted geometry.

Solution:

Molecule	Valence e$^-$ on central	Electron domains	Hybridisation	Geometry
$\text{BF}_3$	3 (B) + 3 (from bonds) = 6, 3 pairs	3	$sp^2$	Trigonal planar
$\text{NH}_3$	5 (N) + 3 (from bonds) = 8, 4 pairs	4	$sp^3$	Trigonal pyramidal
$\text{H}_2\text{O}$	6 (O) + 2 (from bonds) = 8, 4 pairs	4	$sp^3$	Bent
$\text{XeF}_2$	8 (Xe) + 2 (from bonds) = 10, 5 pairs	5	$sp^3d$	Linear
$\text{BrF}_5$	7 (Br) + 5 (from bonds) = 12, 6 pairs	6	$sp^3d^2$	Square pyramidal

Hybridisation rule: number of electron domains = number of hybrid orbitals: 2 ($sp$), 3 ($sp^2$), 4 ($sp^3$), 5 ($sp^3d$), 6 ($sp^3d^2$).

---

UT-3: Intermolecular Force Hierarchy

Question: Explain why the boiling points of the hydrogen halides are: HF (19.5 $^\circ$C) $\gt$ HCl ($-85\ ^\circ$C) $\lt$ HBr ($-67\ ^\circ$C) $\lt$ HI ($-35\ ^\circ$C). Why does HF deviate from the trend? Rank all the intermolecular forces present in each substance.

Solution: For HCl, HBr, and HI, the only significant intermolecular forces are London dispersion forces (LDF). LDF increase with molecular mass (more electrons = larger, more polarisable electron cloud). So boiling point increases HCl $\lt$ HBr $\lt$ HI.

HF is anomalous because it forms strong hydrogen bonds. The small size and high electronegativity of F create a highly polar H--F bond with a large partial positive charge on H and partial negative charge on F. The H atom in one HF molecule is strongly attracted to the lone pairs on F in a neighbouring molecule: $\text{F-H} \cdots \text{F-H}$. Hydrogen bonding (typically $5$--$40\ \text{kJ mol}^{-1}$) is much stronger than LDF ($0.05$--$40\ \text{kJ mol}^{-1}$) and dipole-dipole forces ($5$--$20\ \text{kJ mol}^{-1}$).

Force ranking:

• HF: hydrogen bonding $\gg$ dipole-dipole $\gt$ LDF
• HCl: dipole-dipole $\gt$ LDF
• HBr: dipole-dipole $\gt$ LDF
• HI: LDF only (dipole-dipole is negligible as electronegativity difference is small)

Integration Tests

IT-1: Bonding and States of Matter (with States of Matter)

Question: Explain the following observations using concepts of bonding and intermolecular forces: (a) $\text{C}_6\text{H}_{14}$ (hexane) and water are immiscible. (b) Ethanol ($\text{C}_2\text{H}_5\text{OH}$) is fully miscible with water. (c) $\text{C}_6\text{H}_{12}\text{O}_6$ (glucose) is highly soluble in water despite being a large molecule.

Solution:

(a) Hexane is non-polar (only LDF between molecules). Water is highly polar with strong hydrogen bonding. When mixed, water molecules would need to break hydrogen bonds to accommodate hexane, and hexane molecules would need to break LDF to accommodate water. The energy released from new hexane-water interactions (weak dipole-induced dipole) is far less than the energy required to break the existing strong hydrogen bonds. Hence, the two liquids remain separated.

(b) Ethanol has a polar $-\text{OH}$ group that can form hydrogen bonds with water. The $-\text{C}_2\text{H}_5$ portion is non-polar but small enough that the energy gained from $-\text{OH} \cdots \text{H}_2\text{O}$ hydrogen bonding compensates for breaking water-water hydrogen bonds. Ethanol and water can form a homogeneous solution.

(c) Glucose has five $-\text{OH}$ groups and one $-\text{C}=\text{O}$ group, all capable of forming hydrogen bonds with water. Despite its size, the extensive hydrogen bonding between glucose and water provides enough enthalpy of solvation to overcome the lattice energy holding solid glucose together and the hydrogen bonds broken in water. The large number of hydroxyl groups makes the overall interaction highly favourable.

---

IT-2: Hybridisation and Bond Strength (with Energetics)

Question: The bond enthalpy of a C--C single bond is $347\ \text{kJ mol}^{-1}$, a C=C double bond is $614\ \text{kJ mol}^{-1}$, and a C$\equiv$C triple bond is $839\ \text{kJ mol}^{-1}$. The second bond (first $\pi$ bond) adds $267\ \text{kJ mol}^{-1}$ but the third bond (second $\pi$ bond) only adds $225\ \text{kJ mol}^{-1}$. Explain this trend using hybridisation concepts.

Solution:

A C--C single bond uses $sp^3$ hybrid orbitals on each carbon. These have 25% $s$ character and point directly at each other, maximising overlap and creating a strong $\sigma$ bond.

A C=C double bond uses $sp^2$ hybrid orbitals (33% $s$ character) for the $\sigma$ bond and unhybridised $p$ orbitals for the $\pi$ bond. The higher $s$ character of $sp^2$ orbitals makes them slightly shorter and the $\sigma$ bond slightly stronger than $sp^3$. The additional $\pi$ bond is formed by side-on overlap of $p$ orbitals, which is less effective than head-on overlap -- hence the $\pi$ bond is weaker than the $\sigma$ bond.

A C$\equiv$C triple bond uses $sp$ hybrid orbitals (50% $s$ character) for the $\sigma$ bond and two perpendicular sets of $p$ orbitals for two $\pi$ bonds. The $sp$ orbitals are even shorter and stronger. However, having two $\pi$ bonds means four electrons are in the region between the nuclei, creating increased electron-electron repulsion. This mutual repulsion between the two $\pi$ bonds means the second $\pi$ bond contributes less additional strength than the first. The diminishing returns explain why the second $\pi$ bond adds only $225\ \text{kJ mol}^{-1}$.

---

IT-3: Polarity and Solubility (with Organic Chemistry)

Question: Predict whether each compound is soluble in water, explaining your reasoning: (a) $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$ (butan-1-ol), (b) $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$ (pentane), (c) $\text{CH}_3\text{COOH}$ (ethanoic acid). Discuss the role of "like dissolves like" in terms of intermolecular forces.

Solution:

(a) Butan-1-ol: Moderately soluble (approximately $80\ \text{g/L}$ at $25\ ^\circ$C). The $-\text{OH}$ group can hydrogen bond with water, but the four-carbon non-polar chain is relatively large. The hydrophobic effect of the alkyl chain partially counteracts the hydrogen bonding, limiting solubility compared to shorter alcohols.

(b) Pentane: Essentially insoluble. Pentane is non-polar with only LDF. No favourable interactions with polar water molecules exist; dissolving would require breaking water-water hydrogen bonds with no compensating strong interactions.

(c) Ethanoic acid: Fully miscible with water. The $-\text{COOH}$ group has both a highly polar $\text{C}=\text{O}$ and an $-\text{OH}$ capable of hydrogen bonding. The small methyl group ($-\text{CH}_3$) is the only non-polar portion and is small enough not to significantly hinder solubility. In solution, ethanoic acid can even dimerise less than in the pure liquid, further favouring dissolution.

"Like dissolves like" means that solubility is maximised when the intermolecular forces in the solute are similar to those in the solvent. Water has strong hydrogen bonding, so solutes that can also participate in hydrogen bonding (or strong dipole-dipole interactions) dissolve readily.