Bonding and Intermolecular Forces

Chemical Bonds (CED Units 2-3)

Types of Bonds

Bond Type	Description	Electronegativity Difference
Ionic	Transfer of electrons; metal + nonmetal	$\gt 1.7$
Polar Covalent	Unequal sharing of electrons	$0.4$ — $1.7$
Nonpolar Covalent	Equal sharing of electrons	$\lt 0.4$

The distinction between ionic and covalent bonding is not always sharp. Bonds with electronegativity Differences near the boundary ($\approx 1.7$) have significant ionic and covalent character. Bonding Is better described as a continuum rather than a discrete set of categories.

Ionic Bonding

Ionic compounds consist of cations and anions held together by electrostatic attraction in a Three-dimensional lattice.

Lattice energy ($U$): the energy released when gaseous ions form one mole of solid ionic Compound:

$$U \propto \frac{|z_+ \cdot z_-|}{r_+ + r_-}$$

Higher charges and smaller ionic radii lead to larger (more negative) lattice energy. This is a Direct consequence of Coulomb’s law: the electrostatic attraction is proportional to the product of The charges and inversely proportional to the distance between them.

Example: MgO has a much higher lattice energy than NaCl because Mg$^{2+}$ and O$^{2-}$ carry Double charges compared to Na$^+$ and Cl$^-$.

Derivation: Born-Haber Cycle and Lattice Energy

The Born-Haber cycle applies Hess’s law to calculate lattice energy from measurable quantities. For NaCl:

\Delta H_f^\circ = \Delta H_{\mathrm{sub} + IE_1 + \frac{1}{2}D_{\mathrm{Cl_2} + EA_{\mathrm{Cl} + U

Solving for $U$:

U = \Delta H_f^\circ - \Delta H_{\mathrm{sub} - IE_1 - \frac{1}{2}D_{\mathrm{Cl_2} - EA_{\mathrm{Cl}

Each term represents a step in forming the ionic solid from its elements. The lattice energy $U$ is the largest (most negative) term, reflecting the strong electrostatic attraction in the Ionic lattice.

Covalent Bonding

A covalent bond forms when atoms share one or more pairs of electrons. The shared electrons are Attracted to both nuclei simultaneously, which lowers the potential energy and holds the atoms Together.

• Single bond: 1 shared pair ($\sigma$ bond)
• Double bond: 2 shared pairs (1 $\sigma$ + 1 $\pi$)
• Triple bond: 3 shared pairs (1 $\sigma$ + 2 $\pi$)

Bond Energy and Bond Length

• Bond length decreases as bond order increases (more shared electrons pull nuclei closer).
• Bond energy increases as bond order increases (more shared electrons = stronger bond).

Bond Type	Typical Length (pm)	Typical Energy (kJ/mol)
C—C	154	347
C=C	134	614
C$\equiv$C	120	839

The trend is consistent: as bond order increases, the nuclei are pulled closer together (shorter Bond) and the bond becomes stronger (higher energy). This is because each additional shared pair Adds more electron density between the nuclei, increasing the net attractive force.

Lewis Structures (CED Unit 2)

Rules for Drawing Lewis Structures

1. Count total valence electrons.
2. Draw the skeletal structure (least electronegative atom is central, except H which is never central).
3. Connect atoms with single bonds (each uses 2 electrons).
4. Complete octets of terminal atoms first.
5. Place remaining electrons on the central atom.
6. If the central atom lacks an octet, form double or triple bonds.
7. Check: total electrons = valence electrons.

Formal Charge

\mathrm{Formal Charge = V - N - \frac{B}{2}

Where $V$ = valence electrons of the free atom, $N$ = nonbonding electrons on the atom, $B$ = Bonding electrons shared by the atom.

The best Lewis structure minimises formal charges and places negative formal charges on more Electronegative atoms. If formal charges must be non-zero, adjacent atoms should not carry the same Sign charge.

Worked Example. Draw the Lewis structure for \mathrm{NO_3^-.

Total valence electrons: $5 + 3(6) + 1 = 24$.

Connect N to each O with single bonds: $6$ electrons used, $18$ remaining.

Complete octets: $3 \times 6 = 18$ electrons. All 24 used.

Formal charges: N has $5 - 0 - 6/2 = +2$. Each O has $6 - 6 - 2/2 = -1$.

Better structure: form one N=O double bond. N has $5 - 0 - 8/2 = +1$. Double-bonded O has $6 - 4 - 4/2 = 0$. Each single-bonded O has $6 - 6 - 2/2 = -1$.

Total: $+1 + 0 + (-1) + (-1) = -1$. Three resonance structures are possible, each with the double Bond on a different oxygen.

Resonance

When multiple Lewis structures (resonance structures) are possible, the actual molecule is a hybrid Of all resonance forms. The actual bond order is the average.

Resonance stabilises a molecule: the more resonance structures, the lower the energy. Benzene (\mathrm{C_6\mathrm{H_6) is a classic example, with six equivalent resonance structures giving It exceptional stability.

Exceptions to the Octet Rule

1. Odd-electron molecules: \mathrm{NO (11 valence electrons), \mathrm{ClO_2 (19 valence electrons). These have an unpaired electron and are called radicals.
2. Electron-deficient: \mathrm{BeCl_2 (4 valence electrons), \mathrm{BF_3 (6 valence electrons). These stable molecules have fewer than 8 electrons around the central atom.
3. Expanded octets: Elements in period 3 and beyond can have more than 8 electrons because they have accessible d orbitals (e.g., \mathrm{SF_6$$\mathrm{PCl_5$$\mathrm{XeF_4).

VSEPR Theory (CED Unit 2)

Valence Shell Electron Pair Repulsion (VSEPR) predicts molecular geometry based on the repulsion Between electron pairs (both bonding and lone pairs). The key idea is that electron pairs arrange Themselves to be as far apart as possible.

Electron Domain Geometry

Electron Domains	Geometry	Bond Angles
2	Linear	$180^\circ$
3	Trigonal planar	$120^\circ$
4	Tetrahedral	$109.5^\circ$
5	Trigonal bipyramidal	90^\circ$$120^\circ
6	Octahedral	$90^\circ$

Molecular Geometry (Accounting for Lone Pairs)

Domains	Bonding	Lone	Shape	Example
2	2	0	Linear	\mathrm{CO_2
3	3	0	Trigonal planar	\mathrm{BF_3
3	2	1	Bent	\mathrm{SO_2
4	4	0	Tetrahedral	\mathrm{CH_4
4	3	1	Trigonal pyramidal	\mathrm{NH_3
4	2	2	Bent	\mathrm{H_2\mathrm{O
5	5	0	Trigonal bipyramidal	\mathrm{PCl_5
5	4	1	Seesaw	\mathrm{SF_4
5	3	2	T-shaped	\mathrm{ClF_3
5	2	3	Linear	\mathrm{XeF_2
6	6	0	Octahedral	\mathrm{SF_6
6	5	1	Square pyramidal	\mathrm{BrF_5
6	4	2	Square planar	\mathrm{XeF_4

Lone Pair Repulsion Order

Lone pair—lone pair $\gt$ lone pair—bond pair $\gt$ bond pair—bond pair

This is why bond angles decrease when lone pairs are present (e.g., \mathrm{H_2\mathrm{O at $104.5^\circ$ vs ideal $109.5^\circ$). A lone pair occupies more space than a bonding pair because It is attracted to only one nucleus, whereas a bonding pair is attracted to two nuclei and is Therefore more compressed.

Axial vs. Equatorial Positions in Trigonal Bipyramidal

In a trigonal bipyramidal geometry, there are two types of positions: axial (90$^\circ$ from three Equatorial atoms) and equatorial (120$^\circ$ from two equatorial atoms, 90$^\circ$ from two axial Atoms). Lone pairs preferentially occupy equatorial positions because this minimises repulsion (only Two 90$^\circ$ interactions instead of three).

Worked Example. Determine the molecular geometry of \mathrm{SF_4.

S has 6 valence electrons + 4 from bonds = 10 electrons = 5 electron domains. Four are bonding Pairs, one is a lone pair. This gives a seesaw geometry. The lone pair occupies an equatorial Position, and the two axial fluorine atoms are bent slightly away from the lone pair.

Molecular Polarity (CED Unit 3)

A molecule is polar if:

1. It contains polar bonds, AND
2. The molecular geometry does not cancel out the bond dipoles.

Dipole Moments

$$\vec{\mu} = q \cdot d$$

Where $q$ is the partial charge and $d$ is the distance between charges. The dipole moment is a Vector quantity; the overall molecular dipole is the vector sum of all bond dipoles.

Examples

Molecule	Polar Bonds?	Symmetric?	Polar?
\mathrm{CO_2	Yes (C=O)	Yes (linear)	No
\mathrm{H_2\mathrm{O	Yes (O—H)	No (bent)	Yes
\mathrm{CCl_4	Yes (C—Cl)	Yes (tetrahedral)	No
\mathrm{NH_3	Yes (N—H)	No (trigonal pyramidal)	Yes
\mathrm{BF_3	Yes (B—F)	Yes (trigonal planar)	No
\mathrm{CH_2\mathrm{Cl_2	Yes	No (tetrahedral, not symmetrical)	Yes

Intermolecular Forces (CED Unit 3)

Intermolecular forces (IMFs) are weaker than intramolecular bonds but determine physical properties Like boiling point, melting point, and solubility.

Types of IMFs (Weakest to Strongest)

1. London Dispersion Forces (LDF): Present in all molecules. Caused by temporary, instantaneous dipoles arising from fluctuations in electron distribution. Strength increases with molecular size (number of electrons) and molecular surface area.

2. Dipole-Dipole Forces: Between polar molecules. The positive end of one dipole attracts the negative end of another. These are permanent (unlike LDFs, which are transient).

3. Hydrogen Bonding: A special, strong type of dipole-dipole interaction between H bonded to N, O, or F and a lone pair on N, O, or F. Hydrogen bonding is about 5-10% as strong as a covalent bond.

4. Ion-Dipole Forces: Between an ion and a polar molecule. Important in solutions (dissolving ionic compounds in water).

Trends in IMF Strength

• LDF increases with molar mass (more electrons = larger polarizability) and molecular surface area (longer chain = stronger LDF; branched = weaker LDF for same formula mass).
• Boiling point generally increases with IMF strength.
• Hydrogen bonding is responsible for anomalously high boiling points of water, ammonia, and hydrogen fluoride compared to other hydrides in their groups.

Worked Example: Comparing IMF Strengths

Arrange in order of increasing boiling point: \mathrm{C_2\mathrm{H_6 (ethane), \mathrm{CH_3\mathrm{OH (methanol), \mathrm{CH_3\mathrm{OCH_3 (dimethyl ether).

\mathrm{C_2\mathrm{H_6 (-89^{\circ}\mathrm{C) $\lt$ \mathrm{CH_3\mathrm{OCH_3 (-24^{\circ}\mathrm{C) $\lt$ \mathrm{CH_3\mathrm{OH (65^{\circ}\mathrm{C).

Ethane has only LDFs. Dimethyl ether has LDFs and dipole-dipole forces. Methanol has LDFs, Dipole-dipole, and hydrogen bonding (O-H group). Hydrogen bonding makes methanol the strongest.

Solubility: “Like Dissolves Like”

Polar solvents (e.g., water) dissolve ionic and polar solutes (via ion-dipole and dipole-dipole Interactions). Nonpolar solvents (e.g., hexane) dissolve nonpolar solutes (via LDFs). Ionic Compounds do not dissolve in nonpolar solvents because the energy gained from ion-induced dipole Interactions is insufficient to overcome the lattice energy.

Summary: IMF Types and Properties

IMF Type	Present In	Strength Order	Effect on Boiling Point
London (LDF)	All molecules	Weakest	Increases with size
Dipole-dipole	Polar molecules	Moderate	Moderate increase
Hydrogen bonding	H bonded to N, O, or F	Strongest	Large increase
Ion-dipole	Ionic + polar solvent	Strongest IMF	Dissolution

Hybridization (CED Unit 2)

sp Hybridization

• 2 electron domains
• Geometry: linear
• Example: \mathrm{BeCl_2$$\mathrm{CO_2$$\mathrm{C_2\mathrm{H_2
• One s orbital + one p orbital = two sp hybrid orbitals; two p orbitals remain unhybridized

sp2 Hybridization

• 3 electron domains
• Geometry: trigonal planar
• Example: \mathrm{BF_3$$\mathrm{C_2\mathrm{H_4
• One s orbital + two p orbitals = three sp$^2$ hybrid orbitals; one p orbital remains unhybridized (forms the pi bond)

sp3 Hybridization

• 4 electron domains
• Geometry: tetrahedral
• Example: \mathrm{CH_4$$\mathrm{NH_3$$\mathrm{H_2\mathrm{O
• One s orbital + three p orbitals = four sp$^3$ hybrid orbitals

sp3d and sp3d2 Hybridization

• 5 and 6 electron domains respectively
• Examples: \mathrm{PCl_5 (sp3d), \mathrm{SF_6 (sp3d2)
• Require d orbitals, so only available for period 3 and beyond

Sigma and Pi Bonds

• Sigma ($\sigma$) bonds: Head-on overlap of orbitals along the internuclear axis. Single bonds are always sigma bonds. Sigma bonds allow free rotation.
• Pi ($\pi$) bonds: Side-to-side overlap of parallel p orbitals, above and below the internuclear axis. Found in double and triple bonds. Pi bonds restrict rotation.
• A double bond = 1 $\sigma$ + 1 $\pi$.
• A triple bond = 1 $\sigma$ + 2 $\pi$.

Hybridization and Bonding in Ethene and Ethyne

Ethene (\mathrm{C_2\mathrm{H_4): Each carbon is sp$^2$ hybridised. The three sp$^2$ orbitals Form three sigma bonds (two C—H and one C—C). The remaining unhybridized p orbital on each carbon Overlaps to form a pi bond. The C=C double bond is planar, and rotation is restricted.

Ethyne (\mathrm{C_2\mathrm{H_2): Each carbon is sp hybridised. The two sp orbitals form two Sigma bonds (one C—H and one C—C). The remaining two unhybridized p orbitals on each carbon form Two pi bonds. The molecule is linear.

Worked Example: Bonding in Benzene

Benzene (\mathrm{C_6\mathrm{H_6) has six carbon atoms in a ring. Each carbon is sp$^2$ Hybridised. Three sp$^2$ orbitals form two C—H sigma bonds and one C—C sigma bond. The remaining p Orbital on each carbon overlaps with its neighbours to form a delocalised pi system above and below The ring. The six C—C bonds are equivalent, each with bond order 1.5.

Bonding Continuum and Partial Ionic Character

The distinction between ionic and covalent bonding is not absolute. Bonds exist on a continuum from Pure covalent (zero electronegativity difference) to ionic (large electronegativity difference).

Percent ionic character can be estimated from the electronegativity difference:

\%\mathrm{ ionic character \approx \left(1 - e^{-0.25(\Delta\chi)^2}\right) \times 100

$\Delta\chi$	% Ionic Character	Bond Example
0.0	0%	H-H
0.4	4%	C-H
0.9	19%	H-Cl
1.7	51%	Na-Cl
2.1	67%	Mg-O
3.0	89%	Cs-F

Even a bond like Na-Cl has some covalent character because the chloride ion is polarizable and the Sodium cation distorts the electron cloud. Conversely, even H-Cl has some ionic character.

Worked Example: Formal Charge Calculations

Example. Draw the Lewis structure for \mathrm{N_2\mathrm{O and determine the formal charges.

Total valence electrons: $5 + 5 + 6 = 16$.

Three possible resonance structures:

Structure 1: \mathrm{N\equiv\mathrm{N-\mathrm{O: N(left) = $5 - 2 - 6/2 = 0$; N(right) = $5 - 0 - 8/2 = +1$; O = $6 - 6 - 2/2 = -1$.

Structure 2: \mathrm{N=\mathrm{N=\mathrm{O: N(left) = $5 - 4 - 4/2 = -1$; N(right) = $5 - 0 - 8/2 = +1$; O = $6 - 4 - 4/2 = 0$.

Structure 3: \mathrm{N-\mathrm{N\equiv\mathrm{O: N(left) = $5 - 6 - 2/2 = -2$; N(right) = $5 - 0 - 6/2 = +2$; O = $6 - 2 - 6/2 = +1$.

Structure 1 is the best because the formal charges are closest to zero and the negative charge is on The more electronegative atom (oxygen).

Detailed Worked Example: VSEPR with Multiple Lone Pairs

Example. Determine the molecular geometry of \mathrm{XeF_4.

Xe has 8 valence electrons + 4 from bonds = 12 electrons = 6 electron domains. Four are bonding Pairs and two are lone pairs.

Electron domain geometry: octahedral. The two lone pairs occupy axial positions (to minimise 90 Degree repulsions). This gives a square planar molecular geometry with 90 degree bond angles.

Worked Example: Formal Charge to Determine Best Structure

Example. Draw the best Lewis structure for \mathrm{N_2\mathrm{O (nitrous oxide) and Determine the formal charges.

Total valence electrons: $5 + 5 + 6 = 16$.

Three possible resonance structures:

Structure 1: \mathrm{N\equiv\mathrm{N-\mathrm{O: N(left) FC = 0; N(right) FC = +1; O FC = -1.

Structure 2: \mathrm{N=\mathrm{N=\mathrm{O: N(left) FC = -1; N(right) FC = +1; O FC = 0.

Structure 3: \mathrm{N-\mathrm{N\equiv\mathrm{O: N(left) FC = -2; N(right) FC = +2; O FC = +1.

Structure 1 is the best because the formal charges are closest to zero and the negative charge is on The more electronegative atom (oxygen). Structure 3 can be eliminated because it has the largest Formal charges.

Worked Example: VSEPR for Complex Molecules

Example. Determine the geometry around each central atom in \mathrm{CH_3\mathrm{COOH (acetic Acid).

Carbon 1 (\mathrm{CH_3): 4 bonding domains, 0 lone pairs. Tetrahedral, 109.5 degrees.

Carbon 2 (\mathrm{COOH): 3 bonding domains (one C-C, one C=O, one C-O), 0 lone pairs. Trigonal Planar, 120 degrees.

Oxygen (in C=O): 2 bonding domains, 2 lone pairs. Bent, approximately 120 degrees (sp2).

Oxygen (in C-OH): 2 bonding domains, 2 lone pairs. Bent, approximately 109.5 degrees (sp3).

Worked Example: Polarity Analysis

Example. Is \mathrm{SF_4 polar?

S has 5 electron domains (4 bonding, 1 lone pair). Seesaw geometry. The bond dipoles do not cancel Because the geometry is not symmetric (the lone pair distorts the structure). Therefore, \mathrm{SF_4 is polar.

Compare with \mathrm{XeF_4: 6 electron domains (4 bonding, 2 lone pairs). Square planar. The bond Dipoles of the four Xe-F bonds cancel in pairs because the molecule is symmetric. \mathrm{XeF_4 Is nonpolar.

Derivation: Why Hydrogen Bonds Are Directional

Hydrogen bonds are directional because they require a specific geometry: the hydrogen must be Colinear with the two electronegative atoms (donor-H…acceptor angle close to 180 degrees). This Maximises the electrostatic attraction between the partial positive hydrogen and the lone pair on The acceptor. Deviation from linearity weakens the hydrogen bond significantly.

This directionality explains many of water’s unique properties. In ice, each water molecule forms Four hydrogen bonds in a tetrahedral arrangement, creating an open lattice structure with lower Density than liquid water.

Bonding Continuum and Partial Ionic Character

The distinction between ionic and covalent bonding is not absolute. Bonds exist on a continuum from Pure covalent (zero electronegativity difference) to ionic (large electronegativity difference).

Percent ionic character can be estimated from the electronegativity difference:

\%\mathrm{ ionic character \approx \left(1 - e^{-0.25(\Delta\chi)^2}\right) \times 100

$\Delta\chi$	% Ionic Character	Bond Example
0.0	0%	H-H
0.4	4%	C-H
0.9	19%	H-Cl
1.7	51%	Na-Cl
2.1	67%	Mg-O
3.0	89%	Cs-F

Even a bond like Na-Cl has some covalent character because the chloride ion is polarizable and the Sodium cation distorts the electron cloud. Conversely, even H-Cl has some ionic character.

Worked Example: Bond Type Prediction

Predict the bond type and percent ionic character for the C-O bond in methanol.

$\Delta\chi = 3.44 - 2.55 = 0.89$. This falls in the polar covalent range (0.4 to 1.7).

\%\mathrm{ ionic \approx (1 - e^{-0.25(0.89)^2}) \times 100 = (1 - e^{-0.198}) \times 100 = (1 - 0.820) \times 100 = 18\%.

The C-O bond in methanol is polar covalent with approximately 18% ionic character.

Common Pitfalls

1. Drawing incorrect Lewis structures. Always count valence electrons and verify formal charges.
2. Confusing electron domain geometry with molecular geometry. Electron domain geometry includes lone pairs; molecular geometry only considers atom positions.
3. Forgetting that hydrogen bonding requires H bonded to N, O, or F. H bonded to C or S does not participate in hydrogen bonding.
4. Confusing polarity of bonds with polarity of molecules. A molecule with polar bonds can be nonpolar if the geometry is symmetric (e.g., \mathrm{CCl_4).
5. Misidentifying the central atom. The central atom is the least electronegative (except H, which is never central).
6. Incorrect hybridization. The hybridization matches the number of electron domains, not the number of atoms bonded.
7. Forgetting expanded octets. Only elements in period 3 and beyond can exceed an octet.
8. Counting sigma and pi bonds incorrectly. Every bond has at least one sigma bond. Only additional bonds (second and third) are pi bonds.
9. Confusing LDF strength with dipole-dipole strength. For large molecules, LDFs can be stronger than dipole-dipole forces.
10. Assuming all molecules with hydrogen form hydrogen bonds. H must be bonded to N, O, or F.
11. Placing lone pairs in axial positions of a trigonal bipyramid. Lone pairs always go equatorial to minimise repulsion.
12. Forgetting that resonance structures are not real. The actual molecule is a hybrid; no single resonance structure exists independently.

Comparison Table: IMF Types and Boiling Points

Substance	Molar Mass	IMF Types	Boiling Point (^{\circ}\mathrm{C)
\mathrm{CH_4	16	LDF only	-161
\mathrm{NH_3	17	H-bonding, LDF	-33
\mathrm{H_2\mathrm{O	18	H-bonding, LDF	100
\mathrm{Ne	20	LDF only	-246
\mathrm{HF	20	H-bonding, LDF	20
\mathrm{Ar	40	LDF only	-186
\mathrm{HCl	36.5	Dipole-dipole, LDF	-85
\mathrm{H_2\mathrm{S	34	Dipole-dipole, LDF	-60

This table shows that hydrogen bonding produces dramatically higher boiling points than Other IMF types for similar molar masses.

Summary Table: Types of Chemical Bonds Compared

Property	Ionic	Nonpolar Covalent	Polar Covalent	Metallic
Constituents	Cations and anions	Shared electron pairs	Unequal sharing	Cations + delocalised e
Electron transfer	Complete	None	Partial	Complete (delocalised)
Electronegativity difference	$\gt 1.7$	$\lt 0.4$	0.4 — 1.7	N/A
Melting point	High	Low	Low	High
Electrical cond.	Molten/dissolved	None	None	Yes (always)
Solubility	Polar solvents	Nonpolar solvents	Both	Insoluble
Example	NaCl	\mathrm{O_2	HCl	Cu, Fe

Summary Table: Hybridization Quick Reference

Electron Domains	Hybridization	Geometry	Bond Angles	Examples
2	sp	Linear	180$^\circ$	\mathrm{BeCl_2, \mathrm{CO_2
3	sp$^2$	Trigonal planar	120$^\circ$	\mathrm{BF_3, \mathrm{C_2\mathrm{H_4
4	sp$^3$	Tetrahedral	109.5$^\circ$	\mathrm{CH_4$$\mathrm{NH_3$$\mathrm{H_2\mathrm{O
5	sp$^3$D	Trigonal bipyramidal	90$^\circ$120$^\circ$	\mathrm{PCl_5
6	sp$^3$D$^2$	Octahedral	90$^\circ$	\mathrm{SF_6$$\mathrm{XeF_4

Practice Questions

1. Draw the Lewis structure for \mathrm{XeO_3 and determine its molecular geometry and polarity.

2. Arrange in order of increasing boiling point: \mathrm{F_2$$\mathrm{Cl_2$$\mathrm{Br_2 \mathrm{I_2. Explain your reasoning.

3. Explain why \mathrm{NH_3 has a higher boiling point than \mathrm{PH_3.

4. Determine the hybridization, electron domain geometry, and molecular geometry of \mathrm{SF_4.

5. Draw all resonance structures for \mathrm{NO_3^- and determine the average N—O bond order.

6. Which has a higher boiling point and why: $n$-pentane or neopentane (2,2-dimethylpropane)?

7. For each molecule, predict whether it is polar or nonpolar: \mathrm{BrF_5$$\mathrm{XeF_4 \mathrm{IF_3$$\mathrm{PF_5.

8. Draw the Lewis structure for \mathrm{ClO_4^-Determine the formal charge on each atom, and describe the molecular geometry.

9. Explain why $n$-butanol has a much higher boiling point than diethyl ether, despite having the same molecular formula (\mathrm{C_4\mathrm{H_{10}\mathrm{O).

10. Describe the bonding in \mathrm{O_3 including hybridization, sigma and pi bonds, and the concept of resonance.

11. Explain, using the concept of hybridization, why the H—C—H bond angle in methane is $109.5^\circ$ but the H—N—H bond angle in ammonia is $107^\circ$.

12. Predict the molecular geometry of \mathrm{I_3^- and explain why the central iodine atom can have more than 8 electrons.

13. Compare and contrast the types of intermolecular forces present in liquid \mathrm{CH_3\mathrm{F and liquid \mathrm{CH_3\mathrm{OH.

14. Draw the Lewis structure for \mathrm{SF_6Determine the formal charges, and explain why sulfur can accommodate 12 electrons around it.

15. Which of the following can form hydrogen bonds with water: \mathrm{CH_3\mathrm{OH \mathrm{CH_3\mathrm{OCH_3$$\mathrm{CH_3\mathrm{CH_3? Explain.

16. Describe the bonding in the nitrate ion (\mathrm{NO_3^-), including hybridization, resonance, and bond order.

17. Explain why \mathrm{CCl_4 is nonpolar despite having four polar C—Cl bonds, while \mathrm{CHCl_3 is polar.

18. Determine the hybridization of the central atom and the molecular geometry of \mathrm{BrF_3.

19. Arrange in order of increasing boiling point and explain: \mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{CH_3 \mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{OH$$\mathrm{HOCH_2\mathrm{CH_2\mathrm{OH.

20. Draw the Lewis structure for \mathrm{ClF_3 and explain why the molecule has a T-shaped geometry rather than a trigonal planar geometry.

21. For the molecule \mathrm{SO_3Draw the Lewis structure, determine the hybridization of sulfur, and explain why all three S—O bonds have the same length despite one being a double bond in the Lewis structure.

22. Explain why \mathrm{HF has a higher boiling point than \mathrm{HCl even though \mathrm{HCl has a larger molar mass.

23. Draw the Lewis structure for \mathrm{XeF_4Determine the formal charge on each atom, and explain the square planar geometry.

24. Calculate the number of sigma and pi bonds in \mathrm{H_2\mathrm{C=\mathrm{CH-\mathrm{C\equiv\mathrm{N.

25. Explain why the bond angle in \mathrm{H_2\mathrm{S ($92^\circ$) is smaller than the bond angle in \mathrm{H_2\mathrm{O ($104.5^\circ$), even though both have the same number of electron domains and lone pairs.

26. Draw all resonance structures for the carbonate ion (\mathrm{CO_3^{2-}) and determine the average C—O bond order.

27. Predict the hybridization and molecular geometry of \mathrm{ICl_4^-.

28. Which compound in each pair has the higher boiling point? Explain your reasoning in each case: (a) \mathrm{CH_3\mathrm{OH or \mathrm{CH_3\mathrm{SH (b) \mathrm{C_2\mathrm{H_6 or \mathrm{C_4\mathrm{H_{10} (c) \mathrm{NH_3 or \mathrm{PH_3.

29. Draw the Lewis structure for \mathrm{PO_4^{3-} and determine the formal charge on each atom. What is the hybridization of phosphorus?

30. Explain, using VSEPR theory, why the \mathrm{F-\mathrm{Xe-\mathrm{F bond angles in \mathrm{XeF_4 are all $90^\circ$.

31. Calculate the percent ionic character of the H-F bond. Is it more accurate to describe this bond as covalent or ionic?

32. Draw the Lewis structure for \mathrm{ClO_2^-Determine the molecular geometry, and predict whether the ion is polar.

33. Explain why the boiling point of \mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{CH_2\mathrm{OH (117^{\circ}\mathrm{C) is higher than that of \mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{CH_3 (0^{\circ}\mathrm{C) by more than can be explained by the difference in molar mass alone.

34. For the molecule \mathrm{BF_3\mathrm{NH_3Determine the hybridization of both boron and nitrogen, and identify the type of bond formed between them.

35. Explain why carbon tetrachloride (\mathrm{CCl_4) does not conduct electricity in any state, whereas molten sodium chloride does.

Practice Problems

Question 1: Lewis structures and formal charge

Draw the best Lewis structure for the \mathrm{SO_4^{2-} ion, showing all formal charges. Explain Why this structure is preferred over alternative arrangements. Calculate the average S-O bond order.

Answer

The best Lewis structure has sulfur as the central atom with four equivalent resonance structures. Each S-O bond is shown as a single bond in the skeleton, with sulfur making double bonds to two Oxygens and single bonds to the other two (with formal charges). The actual structure is a resonance Hybrid.

Formal charge calculation: \mathrm{FC = V - N - B/2.

In the resonance hybrid with two double bonds and two single bonds:

• S: V = 6$$N = 0$$B = 10$$\mathrm{FC = 6 - 0 - 5 = +1.
• Double-bonded O: V = 6$$N = 4$$B = 4$$\mathrm{FC = 6 - 4 - 2 = 0.
• Single-bonded O: V = 6$$N = 6$$B = 2$$\mathrm{FC = 6 - 6 - 1 = -1.

Net charge: $+1 + 0 + 0 + (-1) + (-1) = -2$. Correct for \mathrm{SO_4^{2-}.

The four resonance structures delocalise the double bonds, making all S-O bonds equivalent. Average Bond order = $(2 + 2 + 1 + 1) / 4 = 6/4 = 1.5$. This minimises formal charge and maximises the Number of bonds, which is energetically favourable.

Question 2: Molecular geometry and polarity

Determine the molecular geometry and polarity of \mathrm{XeF_4. Explain whether it has a net Dipole moment.

Answer

\mathrm{XeF_4 has 8 valence electrons from Xe plus $4 \times 7 = 28$ from F, minus 2 for the Charge (if any — this is neutral, so 36 total, 18 pairs).

Xe has 4 bonding pairs and 2 lone pairs. The electron domain geometry is octahedral. The molecular Geometry is square planar (the lone pairs occupy axial positions, 180 degrees apart).

\mathrm{XeF_4 is nonpolar. Although each Xe-F bond is polar (F is more electronegative), the four Bonds are arranged symmetrically in a square plane. The bond dipoles cancel out because they point In opposite directions. The lone pairs are opposite each other (axial) and do not create a net Dipole.

Question 3: Intermolecular forces and boiling points

Arrange the following compounds in order of increasing boiling point and explain your reasoning: \mathrm{CH_4$$\mathrm{CH_3\mathrm{OH$$\mathrm{CH_3\mathrm{Cl \mathrm{CH_3\mathrm{NH_2.

Answer

Increasing boiling point: \mathrm{CH_4 \lt \mathrm{CH_3\mathrm{Cl \lt \mathrm{CH_3\mathrm{NH_2 \lt \mathrm{CH_3\mathrm{OH.

\mathrm{CH_4: Only London dispersion forces (nonpolar, smallest molar mass). Lowest boiling Point.

\mathrm{CH_3\mathrm{Cl: Has dipole-dipole interactions (polar molecule) in addition to London Forces. The C-Cl bond is polar.

\mathrm{CH_3\mathrm{NH_2: Has hydrogen bonding (N-H bonds) plus London forces and dipole-dipole Interactions. Hydrogen bonding with N is weaker than with O because N is less electronegative.

\mathrm{CH_3\mathrm{OH: Has the strongest hydrogen bonding (O-H bonds are more polar than N-H Bonds) plus London forces and dipole-dipole interactions. Highest boiling point.

The dominant factor is hydrogen bonding: compounds with O-H hydrogen bonding have higher boiling Points than those with N-H hydrogen bonding, which in turn have higher boiling points than compounds With only dipole-dipole or London forces.

Question 4: Hybridization and bond angles

The molecule \mathrm{SF_4 has a see-saw molecular geometry. Identify the hybridization of the Central sulfur atom, draw its shape, and predict the bond angles. Explain why the axial and Equatorial bond lengths differ.

Answer

Sulfur in \mathrm{SF_4 has 5 electron domains (4 bonding pairs + 1 lone pair). The hybridization Is $sp^3d$ (one s, three p, and one d orbital combine).

The electron domain geometry is trigonal bipyramidal. The lone pair occupies an equatorial position To minimise repulsion (equatorial has two 90 degree interactions; axial has three). The molecular Geometry is see-saw.

Bond angles: The equatorial F-S-F angle is less than $120^\circ$ (compressed by the lone pair, Closer to $101^\circ$). The axial F-S-F angle is $180^\circ$. Axial-equatorial angles are less than $90^\circ$.

The axial bonds are longer than the equatorial bonds because axial bonds experience greater Repulsion from the three equatorial bonding pairs at 90 degrees. The equatorial bonds experience Repulsion from only two axial bonds at 90 degrees, making them shorter and stronger.

Question 5: Lattice energy and ionic radii

Arrange the following ionic compounds in order of increasing lattice energy and explain the trend: \mathrm{NaCl$$\mathrm{MgO$$\mathrm{NaBr$$\mathrm{MgS.

Answer

Increasing lattice energy: \mathrm{NaBr \lt \mathrm{NaCl \lt \mathrm{MgS \lt \mathrm{MgO.

Lattice energy depends on: (1) the charges on the ions (higher charge = higher lattice energy) and (2) the ionic radii (smaller ions = higher lattice energy, by Coulomb’s law).

\mathrm{NaBr and \mathrm{NaCl have $+1/-1$ charges. \mathrm{Br^- is larger than \mathrm{Cl^-So \mathrm{NaBr has the lowest lattice energy.

\mathrm{MgO and \mathrm{MgS have $+2/-2$ charges. The $+2/-2$ compounds have much higher Lattice energy than the $+1/-1$ compounds (lattice energy is proportional to the product of the Charges).

Between \mathrm{MgO and \mathrm{MgS: \mathrm{O^{2-} is smaller than \mathrm{S^{2-}So \mathrm{MgO has the highest lattice energy.

Worked Examples

Example 1: Newton’s second law

A $2.0\,\text{kg}$ object is pulled across a rough horizontal surface by a horizontal force of $15\,\text{N}$. The frictional force is $5.0\,\text{N}$. Calculate the acceleration.

Solution:

$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 15 - 5.0 = 10\,\text{N}$

$a = \frac{F_{\text{net}}}{m} = \frac{10}{2.0} = 5.0\,\text{m\,s}^{-2}$