Chemical Bonding and Molecular Structure

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Chemical Bonding and Molecular Structure — Key Facts for JAMB

Ionic Bonding: Formed by electron transfer from metal to non-metal. Cations: $Na^+, Mg^{2+}, Al^{3+}$; Anions: $Cl^-$. Lattice energy increases with charge and decreasing ionic radius. $NaCl$ (rock salt structure), $CsCl$ (body-centred cubic), $ZnS$ (zinc blende).

Covalent Bonding: Formed by sharing of electron pairs. Single bond: 1 shared pair. Double bond: 2 shared pairs. Triple bond: 3 shared pairs. Bond length decreases as bond order increases: $C-C$ (154 pm) > $C=C$ (134 pm) > $C\equiv C$ (120 pm).

Electronegativity (Pauling Scale): Measures tendency to attract bonding electrons. Fluorine = 4.0 (most electronegative), caesium = 0.79 (least among common elements). Non-polar covalent: $\Delta EN < 0.4$. Polar covalent: $0.4 < \Delta EN < 1.7$. Ionic: $\Delta EN > 1.7$.

VSEPR Shapes:

• Linear (180°): $CO_2$, $BeCl_2$, $C_2H_2$
• Trigonal planar (120°): $BF_3$, $C_2H_4$
• Bent (~$104.5°$): $H_2O$
• Trigonal pyramidal (~$107°$): $NH_3$
• Tetrahedral (109.5°): $CH_4$
• Trigonal bipyramidal: $PCl_5$
• Octahedral: $SF_6$

⚡ Exam tip: Lone pair electrons repel more than bonding pairs. This is why bond angles in $H_2O$ ($104.5°$) and $NH_3$ ($107°$) are smaller than the tetrahedral angle of $109.5°$.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Chemical Bonding and Molecular Structure — JAMB UTME Study Guide

Lewis Structures: Draw dots for valence electrons around each atom. Each bond = 2 electrons. Try to satisfy octet (2 for H).

Examples:

• $O_2$: $O::O$ (double bond, each O has 4 nonbonding + 2 bonding = 6 electrons — actually double bond means each O has 6 bonding+nonbonding = 8 total). However $O_2$ has 2 unpaired electrons (triplet oxygen, paramagnetic) — explained by MOT, not Lewis.
• $N_2$: $N\equiv N$ (triple bond, each N has 2 nonbonding electrons)
• $CO_2$: $O=C=O$ (linear, double bonds)

Formal Charge: $FC = V - N - B/2$ where $V$ = valence electrons, $N$ = nonbonding electrons, $B$ = bonding electrons. For $CO$: formal charges: C (+1), O (-1) in one resonance structure. Actual structure has a coordinate bond from C→O with formal charges of 0 on both when C has one lone pair and forms a triple bond.

Hybridisation:

• $sp^3$: 4 electron domains — methane $CH_4$ (tetrahedral)
• $sp^2$: 3 electron domains — ethene $C_2H_4$ (trigonal planar, each C)
• $sp$: 2 electron domains — ethyne $C_2H_2$ (linear, each C)

Intermolecular Forces:

1. Van der Waals / London dispersion forces: Present in ALL molecules. Strength ∝ molecular mass. Temporary dipole-induced dipole interactions. For isomeric alkanes, straight chain has slightly higher BP due to more surface contact.
2. Dipole-dipole: Between polar molecules. $HCl$ (μ = 1.08 D) is polar.
3. Hydrogen bonding: F, O, or N bonded to H. $H_2O$ (b.p. 100°C), $NH_3$ (b.p. -33°C), $HF$ (b.p. 19.5°C). Ethanol ($C_2H_5OH$) has BP 78°C vs. dimethyl ether ($CH_3OCH_3$) BP -24°C — huge difference due to H-bonding.

Metallic Bonding: Electron sea model: positive metal ions in a delocalised sea of valence electrons. Explains: electrical conductivity, malleability, ductility, metallic lustre.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Chemical Bonding and Molecular Structure — Comprehensive Chemistry Notes

Bond Energy and Enthalpy: Bond enthalpy (bond dissociation energy): energy required to break one mole of a specific bond in a gaseous molecule. Example: $D(H-H) = 436$ kJ/mol. For reaction: $\Delta H = \sum D_{broken} - \sum D_{formed}$.

Example: $CH_4 + Cl_2 → CH_3Cl + HCl$:

• Bonds broken: 4 C-H ($4×414$) + Cl-Cl ($243$) = 1899 kJ
• Bonds formed: 3 C-H ($3×414$) + H-Cl ($431$) = 1673 kJ
• $\Delta H = 1899 - 1673 = +226$ kJ/mol (endothermic)

But actually the reaction proceeds due to the favourable entropy change and the stability of the products. The initiation step ($Cl_2 \rightarrow 2Cl^\cdot$) requires UV light.

Molecular Orbital Theory:

Energy order for diatomic molecules (from Li₂ onward): $\sigma_{1s} < \sigma^{1s} < \sigma{2s} < \sigma^{2s} < \pi{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi^_{2p_x} = \pi^{2p_y} < \sigma^*{2p_z}$

Electron configurations:

• $Li_2$: $(σ2s)^2$, bond order = 1 (stable)
• $Be_2$: $(σ2s)^2(σ^*2s)^2$, bond order = 0 (does not exist)
• $B_2$: $(σ2s)^2(σ^*2s)^2(π2p_x)^1(π2p_y)^1$, bond order = 1, paramagnetic (2 unpaired)
• $C_2$: $(σ2s)^2(σ^*2s)^2(π2p_x)^2(π2p_y)^2$, bond order = 2, diamagnetic
• $N_2$: $(σ2s)^2(σ^*2s)^2(π2p_x)^2(π2p_y)^2(σ2p_z)^2$, bond order = 3, diamagnetic. This explains $N_2$‘s inertness — triple bond is very strong.
• $O_2$: $(σ2s)^2(σ^*2s)^2(σ2p_z)^2(π2p_x)^2(π2p_y)^2(π^*2p_x)^1(π^*2p_y)^1$, bond order = 2, paramagnetic (2 unpaired electrons — explains why liquid $O_2$ is paramagnetic)
• $F_2$: $(σ2s)^2(σ^*2s)^2(σ2p_z)^2(π2p_x)^2(π2p_y)^2(π^*2p_x)^2(π^*2p_y)^2$, bond order = 1, diamagnetic

Lattice Energy — Born-Haber Cycle: For $MgCl_2$: $Mg(s) → Mg(g)$ (atomisation) $ΔH = 150$ kJ/mol; $Mg(g) → Mg^{2+}(g) + 2e^-$ (IE₁ + IE₂ = 2184$ kJ/mol); $Cl_2(g) → 2Cl(g)$ (bond dissociation = 244 kJ/mol); $Cl(g) + e^- → Cl^-(g)$ (EA = -349 kJ/mol); $Mg^{2+}(g) + 2Cl^-(g) → MgCl_2(s)$ (lattice energy = -2527 kJ/mol). Net: $\Delta_f H° = 150 + 2184 + 244 + 2(-349) + (-2527) = -641$ kJ/mol.

Hybridisation — Percentage Character: $sp^3$: 25% s, 75% p. Bond angle: 109.5°. $sp^2$: 33.3% s, 66.7% p. Bond angle: 120°. $sp$: 50% s, 50% p. Bond angle: 180°.

Resonance:

• Ozone $O_3$: two equivalent resonance structures with one single and one double bond. Actual O-O bond length = 128 pm (between single 148 and double 121).
• Carbonate ion $CO_3^{2-}$: three equivalent resonance structures, all C-O bonds equal (141 pm).
• Benzene $C_6H_6$: two Kekulé structures, all C-C bonds = 140 pm.

Dipole Moment Calculation: $\mu = Q \times r$. For HCl: bond length = 127 pm = $1.27 \times 10^{-10}$ m. Partial charges: $\delta = \mu/r = (1.08 \times 3.336 \times 10^{-30})/(1.27 \times 10^{-10}) = 2.84 \times 10^{-20}$ C. Fraction of electron charge: $2.84 \times 10^{-20}/(1.6 \times 10^{-19}) = 0.178$ e. So about 18% charge separation.

JAMB Pattern Analysis: Common questions: (1) Identify type of bonding from electronegativity difference, (2) VSEPR shape prediction, (3) MO theory bond order for $O_2$, $N_2$, $F_2$, (4) Hydrogen bonding in water, alcohols, (5) Formal charge on Lewis structures. JAMB 2022: “Which of the following has the highest boiling point: (a) $CH_4$ (b) $C_2H_6$ (c) $C_3H_8$ (d) $n-C_4H_{10}$?” Answer: $n-C_4H_{10}$ (d) — highest molecular mass, strongest London dispersion forces.

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📊 JAMB Exam Essentials

Detail	Value
Questions	180 MCQs (UTME)
Subjects	4 subjects (language + 3 for course)
Time	2 hours
Marking	+1 per correct answer
Score	400 max (used for university admission)
Registration	January – February each year

🎯 High-Yield Topics for JAMB

• Use of English (Grammar + Comprehension) — 60 marks
• Biology for Science students — 40 marks
• Chemistry (Organic + Physical) — 40 marks
• Physics (Mechanics + Optics) — 35 marks
• Mathematics (Algebra + Geometry) — 40 marks

📝 Previous Year Question Patterns

• Q: “The process of photosynthesis requires…” [2024 Biology]
• Q: “The electronic configuration of Fe is…” [2024 Chemistry]
• Q: “Find the value of x if 2x + 5 = 15…” [2024 Mathematics]

💡 Pro Tips

• Use of English carries the most weight — master grammar rules and comprehension strategies
• JAMB syllabus is your Bible — questions come directly from it. Download and use it.
• Past questions are highly predictive — repeat patterns appear every year
• For Science students, Biology and Chemistry are high-scoring if you study NCERT-level content

🔗 Official Resources

• JAMB Official
• JAMB Syllabus

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