Bonding and Structure — Diagnostic Tests

Unit Tests

UT-1: VSEPR Theory with Multiple Lone Pairs

Question:

(a) Draw the Lewis structures and predict the shapes and bond angles of $\text{XeF}_4$, $\text{BrF}_3$, and $\text{ICl}_4^-$.

(b) Explain why the bond angle in $\text{NH}_3$ ($107^\circ$) is greater than in $\text{H}_2\text{O}$ ($104.5^\circ$), even though both molecules have four electron domains around the central atom.

Solution:

(a)

$\text{XeF}_4$: Xenon has 8 valence electrons. Four bonds to F use 4 electrons, leaving 4 electrons (2 lone pairs).

• Electron domains: 6 (4 bonding pairs + 2 lone pairs)
• Electron domain geometry: octahedral
• Molecular shape: square planar (lone pairs occupy axial positions to minimise repulsion)
• Bond angles: $90^\circ$ and $180^\circ$ (F--Xe--F)

$\text{BrF}_3$: Bromine has 7 valence electrons. Three bonds to F use 3 electrons, leaving 4 electrons (2 lone pairs).

• Electron domains: 5 (3 bonding pairs + 2 lone pairs)
• Electron domain geometry: trigonal bipyramidal
• Molecular shape: T-shaped (lone pairs occupy equatorial positions to minimise repulsion at $90^\circ$)
• Bond angles: $90^\circ$ and $180^\circ$ (F--Br--F)

$\text{ICl}_4^-$: Iodine has 7 valence electrons + 1 from the negative charge = 8 total. Four bonds use 4 electrons, leaving 4 electrons (2 lone pairs).

• Electron domains: 6 (4 bonding pairs + 2 lone pairs)
• Electron domain geometry: octahedral
• Molecular shape: square planar (lone pairs occupy axial positions)
• Bond angles: $90^\circ$ and $180^\circ$ (Cl--I--Cl)

(b) Both $\text{NH}_3$ and $\text{H}_2\text{O}$ have a tetrahedral electron domain geometry. In $\text{NH}_3$ there is one lone pair, while in $\text{H}_2\text{O}$ there are two lone pairs. Lone pair--bonding pair (LP--BP) repulsion is greater than bonding pair--bonding pair (BP--BP) repulsion because lone pairs occupy more space (they are attracted to only one nucleus). In $\text{H}_2\text{O}$, the two lone pairs repel the bonding pairs more strongly, compressing the H--O--H angle to $104.5^\circ$, compared to $\text{NH}_3$ where only one lone pair compresses the H--N--H angle to $107^\circ$.

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UT-2: Intermolecular Forces and Physical Properties

Question:

Explain the following observations:

(a) Ethanol ($\text{C}_2\text{H}_5\text{OH}$, $M_r = 46$) boils at $78\,^\circ\text{C}$, while dimethyl ether ($\text{CH}_3\text{OCH}_3$, $M_r = 46$) boils at $-24\,^\circ\text{C}$, despite having the same molecular formula and molar mass.

(b) $\text{HF}$ ($M_r = 20$) boils at $20\,^\circ\text{C}$, while $\text{HCl}$ ($M_r = 36.5$) boils at $-85\,^\circ\text{C}$, despite $\text{HCl}$ having greater van der Waals forces.

(c) Water ($\text{H}_2\text{O}$) has a higher boiling point ($100\,^\circ\text{C}$) than hydrogen sulphide ($\text{H}_2\text{S}$, $-60\,^\circ\text{C}$), but the boiling points of $\text{HCl}$ ($-85\,^\circ\text{C}$) and $\text{HBr}$ ($-67\,^\circ\text{C}$) increase down Group 17 as expected.

Solution:

(a) Both molecules have the same molecular formula $\text{C}_2\text{H}_6\text{O}$ (they are functional group isomers). Ethanol has an $-$OH group capable of hydrogen bonding (H bonded to the highly electronegative O), while dimethyl ether has no $-$OH group and can only form dipole-dipole interactions and van der Waals forces. Hydrogen bonds are significantly stronger than dipole-dipole interactions and van der Waals forces, so more energy is required to separate ethanol molecules, giving it a much higher boiling point.

(b) Although $\text{HCl}$ has greater van der Waals forces (due to more electrons and larger electron cloud), $\text{HF}$ can form hydrogen bonds between molecules (H bonded to F, the most electronegative element). The hydrogen bonds in HF are strong enough to overcome the van der Waals advantage of $\text{HCl}$, resulting in a much higher boiling point for HF. Each HF molecule can form an average of approximately two hydrogen bonds (unlike water which forms four).

(c) Water can form hydrogen bonding (each molecule can form up to four H-bonds: two as donor and two as acceptor), while $\text{H}_2\text{S}$ cannot form hydrogen bonds (S is not electronegative enough). The hydrogen bonding in water dominates, making its boiling point anomalously high.

For $\text{HCl}$ and $\text{HBr}$, neither can form significant hydrogen bonding (Cl is insufficiently electronegative, and the H--Cl bond is not polar enough). Therefore, the normal trend of increasing boiling point with increasing $M_r$ (due to stronger van der Waals forces from more electrons and larger electron clouds) applies. The hydrogen bonding anomaly only applies to $\text{HF}$, $\text{H}_2\text{O}$, and $\text{NH}_3$ in Period 2.

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UT-3: Ionic vs Covalent Character Continuum

Question:

(a) Aluminium chloride ($\text{AlCl}_3$) sublimes at $178\,^\circ\text{C}$ and the vapour consists of $\text{Al}_2\text{Cl}_6$ molecules. In the solid state, it has a relatively low melting point and does not conduct electricity when molten. Explain these properties in terms of bonding.

(b) Explain why $\text{AlCl}_3$ behaves as a Lewis acid and write an equation for its reaction with $\text{Cl}^-$.

(c) Compare and contrast the bonding in $\text{AlCl}_3$ with that in $\text{NaCl}$ and $\text{CCl}_4$.

Solution:

(a) $\text{Al}^{3+}$ is a small, highly charged cation with high charge density. It polarises the electron cloud of the $\text{Cl}^-$ anion significantly (Fajans' rules: small, highly charged cations distort large anions). This causes electron density to be drawn towards the $\text{Al}$, giving the bond significant covalent character rather than being purely ionic.

In the solid state, $\text{AlCl}_3$ forms a layer lattice with some covalent character, explaining the low melting point and lack of conductivity when molten. On sublimation, the $\text{Al}_2\text{Cl}_6$ dimer forms, held together by dative covalent bonds where chlorine atoms donate lone pairs to the electron-deficient aluminium atoms. This molecular nature of the vapour confirms the predominantly covalent bonding.

(b) $\text{Al}$ in $\text{AlCl}_3$ has only six electrons in its valence shell (three bonds, no lone pair), making it electron-deficient and therefore a Lewis acid (electron pair acceptor):

$\text{AlCl}_3 + \text{Cl}^- \to \text{AlCl}_4^-$

The chloride ion donates a lone pair to form a dative covalent bond, giving Al a complete octet.

(c) $\text{NaCl}$: Predominantly ionic bonding. $\text{Na}^+$ is a large, singly-charged cation with low charge density, so it does not significantly polarise $\text{Cl}^-$. Forms a giant ionic lattice with high melting point, conducts when molten.

$\text{CCl}_4$: Purely covalent bonding. Carbon and chlorine have similar electronegativity and both are non-metals. Forms simple molecules with weak van der Waals forces between them. Low melting point, does not conduct.

$\text{AlCl}_3$: Intermediate position on the ionic-covalent continuum. The high charge density of $\text{Al}^{3+}$ introduces significant covalent character into what might naively be expected to be an ionic compound. The result is molecular behaviour (low melting point, no conductivity) despite the large electronegativity difference.

Integration Tests

IT-1: Bonding and Structure in Determining Solubility (with Acids and Bases)

Question:

(a) Explain why $\text{NaCl}$ is soluble in water but insoluble in hexane ($\text{C}_6\text{H}_{14}$).

(b) Explain why $\text{CCl}_4$ is insoluble in water but miscible with hexane.

(c) Glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) is highly soluble in water despite being a large molecule. Explain this observation.

Solution:

(a) For $\text{NaCl}$ to dissolve, the strong ionic bonds in the lattice must be broken and the ions must be hydrated. Water molecules are polar and can surround the $\text{Na}^+$ and $\text{Cl}^-$ ions with the appropriate partial charges (oxygen atom of water orientated towards $\text{Na}^+$, hydrogen atoms towards $\text{Cl}^-$). The energy released from ion-dipole interactions (hydration enthalpy) compensates for the lattice energy. Hexane is non-polar and cannot solvate ions, so $\text{NaCl}$ is insoluble.

The general rule is: "like dissolves like" -- ionic and polar solutes dissolve in polar solvents.

(b) $\text{CCl}_4$ is a non-polar molecule (tetrahedral symmetry means the bond dipoles cancel). Water is polar and forms hydrogen bonds between its molecules. To dissolve $\text{CCl}_4$ in water, hydrogen bonds between water molecules would need to be broken and replaced by weaker dipole-induced dipole interactions, which is energetically unfavourable. Hexane is non-polar, so only weak van der Waals forces need to be overcome and similar weak London dispersion forces are formed in their place, making the process energetically favourable.

(c) Glucose has five $-$OH groups and one $=$O group. Each $-$OH group can form hydrogen bonds with water molecules, and the ring oxygen also contributes to polarity. Although glucose is a large molecule, the numerous hydrogen bonding sites allow extensive solute-solvent interactions that compensate for the energy needed to separate water molecules. The large number of hydroxyl groups makes glucose highly hydrophilic despite its size.

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IT-2: Bond Polarity, Molecular Polarity, and Intermolecular Forces (with Organic Chemistry)

Question:

Consider the following molecules: $\text{CH}_3\text{Cl}$, $\text{CH}_2\text{Cl}_2$, $\text{CHCl}_3$, $\text{CCl}_4$.

(a) For each molecule, state whether it is polar or non-polar, explaining your reasoning with reference to molecular shape and bond polarity.

(b) Arrange the four molecules in order of increasing boiling point and explain the trend.

(c) Predict which molecule would have the greatest solubility in water and justify your answer.

Solution:

(a)

• $\text{CH}_3\text{Cl}$: Polar. Tetrahedral geometry with one polar C--Cl bond. The bond dipole is not cancelled by the three less polar C--H bonds (the C--Cl bond moment is much larger than C--H). Net dipole moment towards Cl.

• $\text{CH}_2\text{Cl}_2$: Polar. Tetrahedral geometry. Although it might appear symmetrical, the two C--Cl bond dipoles do not fully cancel the two C--H bond dipoles because the bond moments have different magnitudes. The resultant dipole bisects the Cl--C--Cl angle.

• $\text{CHCl}_3$: Polar. Tetrahedral geometry. The three C--Cl bond dipoles are partially cancelled by the single C--H bond, but the three Cl atoms create a net dipole towards the Cl face of the molecule.

• $\text{CCl}_4$: Non-polar. Perfect tetrahedral symmetry. The four identical C--Cl bond dipoles cancel completely (vector sum is zero).

(b) Increasing boiling point: $\text{CH}_3\text{Cl} \lt \text{CH}_2\text{Cl}_2 \lt \text{CHCl}_3 \lt \text{CCl}_4$

Despite $\text{CCl}_4$ being non-polar, it has the highest boiling point because it has the most electrons ($4 \times 17 + 6 = 74$ electrons) and therefore the strongest van der Waals forces. Van der Waals forces increase with the number of electrons (and hence $M_r$). The additional dipole-dipole interactions in the other molecules are small compared to the effect of increasing electron count.

(c) $\text{CHCl}_3$ would have the greatest solubility in water among these four. Although none are highly water-soluble, $\text{CHCl}_3$ can engage in weak dipole-dipole interactions with water through its polar C--H bond (the hydrogen is slightly positive and can interact with the lone pairs on water's oxygen). This is analogous to very weak hydrogen bonding. $\text{CCl}_4$ is non-polar and hydrophobic, while $\text{CH}_3\text{Cl}$ has some polarity but fewer interaction sites.

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IT-3: Giant Covalent Structures and Bond Energy Calculations (with Thermodynamics)

Question:

(a) Diamond and graphite are both giant covalent allotropes of carbon. Explain why diamond is an electrical insulator while graphite is a conductor, referring to the bonding in each.

(b) Diamond has a standard enthalpy of atomisation of $+717\,\text{kJ mol}^{-1}$ and graphite has $+716\,\text{kJ mol}^{-1}$. Calculate the enthalpy change for the conversion of diamond to graphite.

(c) Given that the bond enthalpy of a C--C single bond is $+347\,\text{kJ mol}^{-1}$, estimate the average bond enthalpy of a C--C bond in diamond. Each carbon atom in diamond forms four bonds, and each bond is shared between two atoms.

Solution:

(a) In diamond, each carbon atom forms four $\sigma$ bonds in a tetrahedral arrangement. All valence electrons are localised in these strong covalent bonds, leaving no delocalised electrons available to carry current. Diamond is therefore an insulator.

In graphite, each carbon atom forms three $\sigma$ bonds to three other carbon atoms in a planar hexagonal arrangement (trigonal planar). The fourth valence electron of each carbon is delocalised in a system of $\pi$-orbitals above and below the planes. These delocalised electrons are free to move within the layers, allowing graphite to conduct electricity (parallel to the layers, but not perpendicular to them).

(b) Using Hess's law:

$\text{C(diamond)} \to \text{C(g)} \quad \Delta H = +717\,\text{kJ mol}^{-1}$ $\text{C(graphite)} \to \text{C(g)} \quad \Delta H = +716\,\text{kJ mol}^{-1}$

Reversing the second equation:

$\text{C(g)} \to \text{C(graphite)} \quad \Delta H = -716\,\text{kJ mol}^{-1}$

Adding:

$\text{C(diamond)} \to \text{C(graphite)} \quad \Delta H = +717 + (-716) = +1\,\text{kJ mol}^{-1}$

The conversion is very slightly endothermic, meaning graphite is thermodynamically more stable than diamond at standard conditions.

(c) In diamond, each carbon atom forms 4 bonds, but each bond is shared between 2 atoms. So each carbon atom effectively has $4/2 = 2$ bonds.

The enthalpy of atomisation ($+717\,\text{kJ mol}^{-1}$) represents the energy required to break all bonds per mole of carbon atoms. Since each carbon atom is involved in 2 effective bonds:

$\text{Average C--C bond enthalpy in diamond} = \frac{717}{2} = 358.5\,\text{kJ mol}^{-1}$

This is slightly higher than the standard C--C bond enthalpy of $347\,\text{kJ mol}^{-1}$, reflecting the fact that diamond's rigid tetrahedral network creates slightly stronger bonds due to the constrained geometry and lack of bond rotation.

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Additional Practice Problems

UT-4: Predicting Molecular Polarity

Question: For each of the following molecules, predict whether it is polar or non-polar, and explain your reasoning:

(a) $\mathrm{BF}_3$ (trigonal planar) (b) $\mathrm{NF}_3$ (trigonal pyramidal) (c) $\mathrm{SF}_4$ (see-saw) (d) $\mathrm{XeF}_4$ (square planar)

Solution:

(a) Non-polar. $\mathrm{BF}_3$ has trigonal planar geometry with three identical B--F bonds. The bond dipoles are equal in magnitude and arranged at $120^\circ$ to each other. The vector sum of the bond dipoles is zero, so the molecule has no net dipole moment (1 mark).

(b) Polar. $\mathrm{NF}_3$ has trigonal pyramidal geometry (one lone pair on N). The N--F bond dipoles do not cancel because the molecule is not symmetrical. The lone pair also contributes to the asymmetry. The net dipole moment points from the N towards the base of the pyramid, but the lone pair creates a dipole in the opposite direction. The net dipole is actually small and points from F to N (1 mark).

(c) Polar. $\mathrm{SF}_4$ has a see-saw geometry with one lone pair. The axial and equatorial bonds have different bond angles ($180^\circ$ vs $120^\circ$), and the lone pair distorts the geometry. The bond dipoles do not cancel, giving a net dipole moment (1 mark).

(d) Non-polar. $\mathrm{XeF}_4$ has square planar geometry with two lone pairs. The four Xe--F bonds are arranged symmetrically at $90^\circ$ in a plane. The bond dipoles cancel in pairs (opposite bonds have equal and opposite dipoles), giving a net dipole moment of zero (1 mark).

UT-5: Ionic Character and Electronegativity

Question: Use electronegativity values to estimate the percentage ionic character of the bonds in $\mathrm{BeCl}_2$ ($\Delta\chi = 1.57$), $\mathrm{MgCl}_2$ ($\Delta\chi = 1.85$), and $\mathrm{AlCl}_3$ ($\Delta\chi = 1.55$). Comment on the bonding in $\mathrm{AlCl}_3$.

Solution:

The empirical relationship between electronegativity difference and percent ionic character is approximately: $\%\,\text{ionic} \approx 16|\Delta\chi| + 3.5|\Delta\chi|^2$.

$\mathrm{BeCl}_2$: $\%\,\text{ionic} \approx 16(1.57) + 3.5(1.57)^2 = 25.1 + 8.6 = 33.7\%$

$\mathrm{MgCl}_2$: $\%\,\text{ionic} \approx 16(1.85) + 3.5(1.85)^2 = 29.6 + 12.0 = 41.6\%$

$\mathrm{AlCl}_3$: $\%\,\text{ionic} \approx 16(1.55) + 3.5(1.55)^2 = 24.8 + 8.4 = 33.2\%$

$\mathrm{AlCl}_3$ has relatively low ionic character despite $\mathrm{Al}$ being a metal. In the solid state, $\mathrm{AlCl}_3$ forms a layer lattice with significant covalent character (it sublimes at $178^\circ\mathrm{C}$ rather than melting, suggesting molecular $\mathrm{Al}_2\mathrm{Cl}_6$ units in the gas phase). The high charge density of the small $\mathrm{Al}^{3+}$ ion polarises the large $\mathrm{Cl}^-$ ion, drawing electron density towards the cation and increasing covalent character (Fajans' rules) (1 mark).

IT-4: Bonding and Physical Properties

Question: Explain why $\mathrm{SiO}_2$ has a melting point of $1713^\circ\mathrm{C}$ while $\mathrm{CO}_2$ sublimes at $-78^\circ\mathrm{C}$, even though both C and Si are in Group 4 and both compounds contain oxygen.

Solution:

$\mathrm{SiO}_2$ has a giant covalent (macromolecular) structure in which each silicon atom is bonded to four oxygen atoms in a tetrahedral arrangement, and each oxygen atom bridges two silicon atoms ($\text{--Si--O--Si--}$). This creates a continuous three-dimensional network of strong covalent bonds ($\text{Si--O}: 452\,\mathrm{kJ/mol}$) throughout the entire crystal. To melt $\mathrm{SiO}_2$, a large number of these strong covalent bonds must be broken, requiring a large amount of energy (2 marks).

$\mathrm{CO}_2$ consists of simple discrete molecules with double bonds ($\text{C}= \text{O}: 799\,\mathrm{kJ/mol}$). The covalent bonds within each molecule are strong, but the molecules are held together only by weak London dispersion forces between molecules. Melting (or sublimation) only requires overcoming these weak intermolecular forces, not breaking the covalent bonds within the molecules (1 mark).

The key difference is that silicon (Period 3) can expand its octet and form four single bonds to oxygen, creating the extended network. Carbon (Period 2) cannot expand its octet beyond eight electrons and forms stable double bonds instead, producing discrete $\mathrm{O}=\text{C}=\text{O}$ molecules (1 mark).