DSE Chemistry Diagnostic: Bonding

Unit Test 1: VSEPR for Complex Molecules

Question

(a) Draw the Lewis structure of chlorine trifluoride ($ClF_{3}$) and determine its molecular geometry. [3 marks]

(b) A student predicts that $ClF_{3}$ is trigonal planar because it has three bonding pairs. Explain why this prediction is incorrect. [2 marks]

(c) Predict the bond angles in $ClF_{3}$ and explain any deviation from the ideal angle. [3 marks]

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Worked Solution

(a) Chlorine is the central atom. $Cl$ has 7 valence electrons; each $F$ contributes 1 for bonding. Total valence electrons = $7 + 3 \times 7 = 28$.

Lewis structure: $Cl$ forms three single bonds with three $F$ atoms (using 6 electrons). Remaining electrons = $28 - 6 = 22$, i.e. 11 lone pairs. Each $F$ gets 3 lone pairs (using $3 \times 6 = 18$ electrons). Remaining on $Cl$: $22 - 18 = 4$ electrons = 2 lone pairs.

Electron domain geometry: trigonal bipyramidal (5 domains: 3 bonding pairs + 2 lone pairs).

Molecular geometry: The two lone pairs occupy equatorial positions (to minimise repulsion at 90 degrees). This leaves one equatorial and two axial positions occupied by $F$ atoms.

Molecular shape: T-shaped.

(b) The student only counted bonding pairs and ignored the lone pairs on the central atom. VSEPR theory requires counting all electron domains (bonding pairs + lone pairs) around the central atom. $ClF_{3}$ has 5 electron domains, not 3.

(c) In an ideal trigonal bipyramid, the equatorial-equatorial angle is 120 degrees and the axial-equatorial angle is 90 degrees.

In $ClF_{3}$:

• The two lone pairs in equatorial positions repel the bonding pairs more strongly than bonding pairs repel each other.
• The $F_{eq}-Cl-F_{ax}$ angles are compressed below 90 degrees (approximately 87.5 degrees).
• The $F_{ax}-Cl-F_{ax}$ angle is approximately 175 degrees (compressed from 180 degrees).

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Unit Test 2: Molecular Polarity

Question

Consider the following molecules: $CF_{4}$, $CH_{2}F_{2}$, $CHF_{3}$.

(a) For each molecule, determine whether the $C-F$ bond is polar. [1 mark]

(b) Determine which of the three molecules is/are polar overall. Explain your reasoning using molecular geometry. [4 marks]

(c) Explain why $CF_{4}$ has a boiling point of $-128\degree C$ while $CH_{2}F_{2}$ has a boiling point of $-52\degree C$, even though $CF_{4}$ has a higher molar mass. [3 marks]

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Worked Solution

(a) All $C-F$ bonds are polar because fluorine ($EN = 4.0$) is significantly more electronegative than carbon ($EN = 2.5$). The dipole moment of each $C-F$ bond points from $C$ towards $F$.

(b) $CF_{4}$: Tetrahedral geometry. The four $C-F$ bond dipoles are arranged symmetrically and cancel out. Non-polar.

$CHF_{3}$: Tetrahedral geometry. Three $C-F$ dipoles point towards $F$ atoms and one $C-H$ dipole points towards $H$. The three $C-F$ dipoles have a resultant that is partially cancelled by the $C-H$ dipole, but the cancellation is not complete because the magnitudes differ ($C-F$ dipole $\neq C-H$ dipole). Polar.

$CH_{2}F_{2}$: Tetrahedral geometry. Two $C-F$ dipoles and two $C-H$ dipoles. The resultant dipole depends on the vector sum. In $CH_{2}F_{2}$, the $C-F$ and $C-H$ bonds can be arranged such that the dipole moments do not fully cancel (the molecule has $C_{2v}$ symmetry, not $T_{d}$). Polar.

(c) $CF_{4}$ is non-polar and only has London dispersion forces between molecules. $CH_{2}F_{2}$ is polar and additionally has permanent dipole-dipole interactions between molecules. Although $CF_{4}$ has stronger London forces (higher molar mass), the dipole-dipole interactions in $CH_{2}F_{2}$ more than compensate, giving it a significantly higher boiling point.

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Unit Test 3: Intermolecular Forces and Hydrogen Bonding

Question

(a) Explain why ammonia ($NH_{3}$) has a higher boiling point ($-33\degree C$) than phosphine ($PH_{3}$, $-88\degree C$), despite phosphine having a higher molar mass. [3 marks]

(b) A student claims that hydrogen bonding exists between molecules of hydrogen chloride ($HCl$). Evaluate this claim. [2 marks]

(c) Explain why water ($H_{2}O$) has a significantly higher boiling point ($100\degree C$) than hydrogen sulphide ($H_{2}S$, $-60\degree C$), even though both are Group 16 hydrides. [3 marks]

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Worked Solution

(a) $NH_{3}$ can form hydrogen bonds between molecules because:

1. It contains hydrogen covalently bonded to nitrogen (a highly electronegative atom with lone pairs).
2. The $N-H\cdots N$ hydrogen bonds are strong intermolecular forces.

$PH_{3}$ cannot form hydrogen bonds because phosphorus is not sufficiently electronegative ($EN = 2.1$). $PH_{3}$ only has London dispersion forces and weaker dipole-dipole interactions. The hydrogen bonding in $NH_{3}$ far outweighs the stronger London forces in $PH_{3}$.

(b) The claim is incorrect. Although $HCl$ has a polar $H-Cl$ bond, hydrogen bonding only occurs when hydrogen is bonded to fluorine, oxygen, or nitrogen ($F$, $O$, $N$). Chlorine is not electronegative enough ($EN = 3.0$ vs $F = 4.0$, $O = 3.5$, $N = 3.0$ -- actually $N = 3.0$ is borderline, but nitrogen's small size allows orbital overlap; chlorine is too large). $HCl$ has dipole-dipole interactions and London dispersion forces only.

(c) $H_{2}O$ can form extensive hydrogen bonding because:

1. Each water molecule has two $O-H$ bonds (two hydrogen bond donors) and two lone pairs on oxygen (two hydrogen bond acceptors).
2. This creates a three-dimensional network of hydrogen bonds in liquid water and ice.

$H_{2}S$ cannot form hydrogen bonds (sulphur is not electronegative enough and too large). $H_{2}S$ only has dipole-dipole forces and London dispersion forces. The hydrogen bonding network in water is exceptionally strong, resulting in a dramatically higher boiling point.

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Integration Test 1: Bonding Type from Physical Properties

Question

Four substances have the following properties:

Substance	Melting Point ($\degree C$)	Electrical Conductivity (solid)	Electrical Conductivity (molten)	Solubility in Water
$W$	801	No	Yes	High
$X$	170	No	No	Insoluble
$Y$	$-183$	No	No	Slightly soluble
$Z$	1085	Yes	Yes	Insoluble

(a) Identify the type of bonding in each substance. [4 marks]

(b) For substance $W$, explain why it conducts electricity when molten but not when solid. [2 marks]

(c) Substance $X$ has a higher melting point than substance $Y$. Explain this difference in terms of the types of intermolecular forces present. [3 marks]

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Worked Solution

(a) $W$: Ionic bonding. High melting point, conducts when molten (mobile ions), soluble in water (hydration of ions).

$X$: Simple covalent (molecular). Moderate melting point (but well above room temperature, suggesting stronger IMF -- likely hydrogen bonding or strong dipole-dipole), does not conduct in any state, insoluble in water.

$Y$: Simple covalent (molecular) with weak IMF. Very low melting point, non-conducting, only slightly soluble.

$Z$: Metallic bonding. High melting point, conducts in both solid and molten states (delocalised electrons), insoluble in water.

(b) In solid $W$ (ionic lattice), the ions are fixed in position and cannot move, so electricity cannot be conducted. When molten, the ionic lattice breaks down and the ions become mobile, allowing them to carry charge.

(c) $X$ has a higher melting point than $Y$, indicating stronger intermolecular forces. Possible explanations:

• $X$ has hydrogen bonding (e.g., a molecule with $O-H$ or $N-H$ groups) while $Y$ only has London dispersion forces.
• Alternatively, $X$ may have a larger electron cloud or greater molecular mass giving stronger London forces.
• $X$ may have strong permanent dipole-dipole interactions that $Y$ lacks.

Without knowing the identity of the substances, the general explanation is that $X$ has stronger IMF (either hydrogen bonding, stronger dipole-dipole interactions, or significantly stronger London forces due to larger molecular size) requiring more energy to overcome.

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Integration Test 2: VSEPR + Polarity + IMF

Question

Consider the molecule $XeF_{4}$.

(a) Draw the Lewis structure of $XeF_{4}$ and determine its molecular geometry. [3 marks]

(b) Is $XeF_{4}$ polar or non-polar? Explain your reasoning. [2 marks]

(c) $XeF_{4}$ is a solid at room temperature (m.p. $117\degree C$) while $XeF_{2}$ is also a solid (m.p. $129\degree C$). A student argues that $XeF_{4}$ should have a lower melting point because it has no net dipole moment. Evaluate this argument. [3 marks]

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Worked Solution

(a) $Xe$ has 8 valence electrons; each $F$ contributes 1 for bonding. Total = $8 + 4 \times 7 = 36$.

$Xe$ forms 4 bonds (8 electrons), leaving $36 - 8 = 28$ electrons = 14 lone pairs. Each $F$ gets 3 lone pairs (12 lone pairs, 24 electrons). Remaining on $Xe$: $28 - 24 = 4$ electrons = 2 lone pairs.

Electron domains: 6 (4 bonding + 2 lone pairs).

Electron domain geometry: Octahedral.

The two lone pairs occupy axial positions (opposite each other, 180 degrees apart, minimising repulsion).

Molecular geometry: Square planar.

(b) $XeF_{4}$ is non-polar. Although each $Xe-F$ bond is polar (F is more electronegative), the four bond dipoles are arranged symmetrically in a square planar geometry and cancel out completely. The resultant dipole moment is zero.

(c) The student's argument is flawed. Melting point depends on the overall strength of intermolecular forces, not just the presence of a permanent dipole. $XeF_{4}$ and $XeF_{2}$ both have significant London dispersion forces due to the large electron cloud of xenon (high molar mass, many electrons). These London forces can be very strong and are the dominant factor determining melting points for these heavy molecules. Additionally, $XeF_{4}$ molecules are more polarizable (more electrons, larger electron cloud), which can lead to stronger London forces. The absence of a permanent dipole does not necessarily mean weaker overall IMF.

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Integration Test 3: Bond Energy and Properties

Question

The bond dissociation energies for selected bonds are:

Bond	Bond Energy (kJ/mol)
$C-C$	346
$C=C$	614
$C\equiv C$	839
$C-H$	413
$C-F$	485
$C-Cl$	327
$F-F$	158
$Cl-Cl$	242

(a) The $C=C$ double bond is not exactly twice the $C-C$ single bond energy. Explain why. [2 marks]

(b) Calculate the enthalpy change for the reaction:

$CH_{4}(g) + 4Cl_{2}(g) \rightarrow CCl_{4}(g) + 4HCl(g)$

[3 marks]

(c) The $F-F$ bond is significantly weaker than the $Cl-Cl$ bond despite fluorine being more electronegative. Explain this anomaly. [2 marks]

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Worked Solution

(a) A $C=C$ double bond consists of one sigma ($\sigma$) bond and one pi ($\pi$) bond. The $\pi$ bond is formed by sideways overlap of $p$ orbitals, which is less effective than the head-on overlap forming the $\sigma$ bond. Therefore, the $\pi$ bond is weaker than the $\sigma$ bond, and $C=C$ energy ($614$) $\lt 2 \times 346 = 692$.

(b) Bonds broken (endothermic):

• $4 \times C-H = 4 \times 413 = 1652$ kJ/mol
• $4 \times Cl-Cl = 4 \times 242 = 968$ kJ/mol
• Total bonds broken = $1652 + 968 = 2620$ kJ/mol

Bonds formed (exothermic):

• $4 \times C-Cl = 4 \times 327 = 1308$ kJ/mol
• $4 \times H-Cl = 4 \times 431 = 1724$ kJ/mol (using $H-Cl = 431$)
• Total bonds formed = $1308 + 1724 = 3032$ kJ/mol

$\Delta H = \text{Bonds broken} - \text{Bonds formed} = 2620 - 3032 = -412 \text{ kJ/mol}$

The reaction is exothermic.

(c) In $F_{2}$, the fluorine atoms are very small, so the two bonding electrons are very close together and experience strong inter-electronic repulsion. In $Cl_{2}$, the chlorine atoms are larger, so the bonding electrons are further apart and the repulsion is weaker. The small size of fluorine also means the lone pairs on each $F$ atom are close to the bonding region, creating additional lone pair-bond pair repulsion that weakens the $F-F$ bond.