🔗 Unit 2: Molecular and Ionic Compound Structure and Properties

Understand how atoms bond together and how molecular shape determines chemical behavior.

Exam Weight: 7–9% | Topics 2.1–2.7

📑 In This Unit

2.1 — Types of Chemical Bonds

2.2 — Intramolecular Force and Potential Energy

2.3 — Structure of Ionic Solids

2.4 — Structure of Metals and Alloys

2.5 — Lewis Diagrams

2.6 — Resonance and Formal Charge

2.7 — VSEPR and Bond Hybridization

📝 Practice Multiple-Choice Questions (10)

2.1 — Types of Chemical Bonds

Chemical bonds form when atoms transfer or share electrons to achieve a more stable electron configuration (usually a noble-gas configuration). The type of bond that forms depends on the electronegativity difference between the atoms involved. There are three main types of chemical bonds, and understanding how they differ is critical for predicting the properties of substances.

Ionic Bonds

Formed by the transfer of electrons from a metal to a nonmetal.
Driven by the large difference in electronegativity between the two atoms (ΔEN > 1.7 is a rough guideline).
Produce cations (+) and anions (−) held together by electrostatic attraction.
Result in crystal lattice structures with high melting points.
Example: NaCl — Na loses 1 e⁻ to become Na⁺; Cl gains 1 e⁻ to become Cl⁻.

Covalent Bonds

Formed by the sharing of electron pairs between two nonmetals.
Nonpolar covalent: Equal sharing when ΔEN = 0 or very small (e.g., H₂, O₂, N₂).
Polar covalent: Unequal sharing when there is a moderate ΔEN (e.g., H₂O, HCl, NH₃). The more electronegative atom pulls the shared electrons closer, gaining a partial negative charge (δ−), while the other gets a partial positive charge (δ+).
Bond order: Single bonds share 1 pair (2 e⁻), double bonds share 2 pairs (4 e⁻), triple bonds share 3 pairs (6 e⁻).
Higher bond order = shorter bond length = stronger bond.

Metallic Bonds

Formed between metal atoms that share a “sea” of delocalized valence electrons.
This electron sea explains conductivity, malleability, ductility, and luster.
Metallic bond strength increases with more valence electrons and smaller atomic radius.

Predicting bond type from electronegativity:
• ΔEN ≈ 0: Nonpolar covalent (e.g., N₂, O₂, Cl₂)
• 0 < ΔEN < 1.7: Polar covalent (e.g., H₂O, HCl, CO₂)
• ΔEN > 1.7: Ionic (e.g., NaCl, MgO, CaF₂)
Note: These are guidelines, not strict cutoffs. Bond character is a continuous spectrum from pure covalent to pure ionic.

Comparison Table:

Property	Ionic	Covalent	Metallic
Particles	Cations & anions	Molecules	Cations in e⁻ sea
Melting Point	High	Low to moderate	Variable (often high)
Conductivity (solid)	No	No (usually)	Yes
Conductivity (liquid/dissolved)	Yes	No (usually)	Yes
Solubility in water	Often soluble	Varies (polar in polar)	Insoluble

2.2 — Intramolecular Force and Potential Energy

When two atoms approach each other to form a bond, their potential energy changes. Understanding this energy relationship is key to understanding bond strength, stability, and why bonds form in the first place. The potential energy curve (also called a Morse curve for covalent bonds) is one of the most important diagrams in AP Chemistry.

The Potential Energy Curve

As two atoms approach each other from a large distance:

At large distances: No significant interaction; potential energy is approximately zero (baseline).
As they get closer: Attractive forces (nucleus of one atom attracting electrons of the other) lower the potential energy. The system becomes more stable.
At the bond length (r₀): The minimum potential energy is reached. This is the equilibrium bond length — the most stable configuration.
Too close: Nucleus-nucleus repulsion and electron-electron repulsion dominate. Potential energy rises steeply — the atoms repel each other.

Key definitions from the PE curve:
• Bond energy (D): The depth of the potential energy well (measured from the baseline to the minimum). It is the energy required to completely separate the bonded atoms. Always a positive value.
• Bond length (r₀): The internuclear distance at the energy minimum. This is the distance at which the system is most stable.

Relationship: Shorter bonds are stronger (deeper potential energy well, higher bond energy).

Bond order and strength:
Triple bonds > Double bonds > Single bonds (in terms of both strength and shortness)

Bond	Bond Energy (kJ/mol)	Bond Length (pm)
C—C (single)	347	154
C=C (double)	614	134
C≡C (triple)	839	120
N—N (single)	160	145
N=N (double)	418	125
N≡N (triple)	945	110

Coulomb’s Law and Ionic Bonds

For ionic bonds, the potential energy is governed by Coulomb’s Law. The electrostatic potential energy between two ions is:

E = k · q₁q₂ / r

• For opposite charges (cation + anion), E is negative (stable, attractive interaction).
• The potential energy is more negative (more stable) when charges are larger and when ions are closer together.
• This explains why MgO (2+/2−, small ions) has a much stronger ionic bond than NaCl (1+/1−, larger ions).

AP Exam Application

You may be asked to compare PE curves for different bonds. The bond with the deeper well (more negative minimum) is stronger. The bond with the shorter equilibrium distance (minimum shifted to the left) is shorter. These two features generally go together.

2.3 — Structure of Ionic Solids

Ionic compounds do not exist as discrete molecules. Instead, they form crystal lattices — repeating three-dimensional arrangements of alternating cations and anions, held together by electrostatic forces in all directions. The formula of an ionic compound (e.g., NaCl) represents the simplest ratio of ions, not a single molecule.

The NaCl crystal lattice — each Na⁺ (small) is surrounded by 6 Cl⁻ (large) ions and vice versa. This is the rock salt structure. (Wikimedia Commons)

Properties of Ionic Solids

Property	Explanation
High melting/boiling points	Strong electrostatic forces between many ions require significant energy to overcome.
Brittle	A mechanical force shifts layers, bringing like charges next to each other → repulsion → the crystal shatters.
Conduct electricity when dissolved or molten	Ions become free to move and carry electric charge. Solid ionic compounds do NOT conduct (ions are locked in place).
Soluble in polar solvents	Water molecules surround and stabilize ions through ion-dipole interactions (hydration). Energy of hydration must overcome lattice energy.

Lattice energy is the energy released when gaseous ions form a solid ionic crystal (or the energy needed to separate a crystal into gaseous ions). It depends on:
• Charge: Higher charges → much higher lattice energy (the effect of charge is squared in Coulomb’s law)
• Size: Smaller ions → higher lattice energy (ions are closer together)

Ranking lattice energies: MgO (3850 kJ/mol) > LiF (1037 kJ/mol) > NaCl (786 kJ/mol) > KBr (682 kJ/mol) > CsI (600 kJ/mol)
Notice: MgO is dramatically larger because it has 2+/2− charges. The others are all 1+/1− and differ only in size.

Coordination Number

The coordination number is the number of nearest neighbors surrounding each ion in the crystal lattice. In NaCl, each Na⁺ is surrounded by 6 Cl⁻ ions, and each Cl⁻ is surrounded by 6 Na⁺ ions (coordination number = 6). Different crystal structures have different coordination numbers, which affects the packing efficiency and density.

2.4 — Structure of Metals and Alloys

Metals have a unique bonding model: positive metal cations arranged in a regular lattice, surrounded by a “sea” of delocalized electrons that are free to move throughout the structure. This is called the metallic bonding model or the electron sea model.

The free-electron (electron sea) model of metallic bonding. Delocalized electrons move freely around fixed metal cations. (Wikimedia Commons)

Properties Explained by the Electron Sea Model

Property	Explanation
Electrical conductivity	Delocalized electrons move freely through the lattice, carrying charge when a voltage is applied.
Thermal conductivity	Free electrons transfer kinetic energy (heat) rapidly throughout the metal.
Malleability & ductility	When force is applied, layers of cations slide past each other without breaking bonds — the electron sea simply readjusts.
Luster	Free electrons absorb incoming photons and re-emit them at many wavelengths, producing a shiny appearance.
High melting points (many metals)	Strong metallic bonds; strength increases with more valence electrons and smaller cation radius.

Metallic Bond Strength

Metallic bond strength depends on:

Number of valence electrons: More delocalized electrons = stronger metallic bonding. Transition metals (with d electrons) often have very strong metallic bonds.
Cation size: Smaller cations pack more tightly, increasing the attraction to the electron sea.
Example: Tungsten (W) has an extremely high melting point (3422°C) due to many valence electrons and strong metallic bonds.

Alloys

Alloys are mixtures of metals (and sometimes nonmetals). There are two main types:

Substitutional alloys: Atoms of similar size replace metal atoms in the lattice.
• Example: Brass (Cu + Zn) — Cu and Zn atoms are similar in size
• Example: Sterling silver (Ag + Cu)
• The foreign atoms disrupt the regular arrangement, making it harder for layers to slide → harder and stronger than the pure metal

Interstitial alloys: Smaller atoms fit into the gaps (interstices) between larger metal atoms.
• Example: Steel (Fe + C) — small carbon atoms fit between iron atoms
• The small atoms “lock” the layers in place, dramatically increasing hardness but decreasing ductility

Key point: Both types of alloys are generally harder and stronger than pure metals because the disrupted lattice prevents easy sliding of atomic layers. However, they are typically less malleable and ductile.

2.5 — Lewis Diagrams

Lewis dot diagrams (Lewis structures) show how valence electrons are distributed in a molecule or polyatomic ion. They are essential for predicting molecular shape, polarity, bond order, and reactivity. Mastering Lewis structures is one of the most important skills in AP Chemistry.

Steps to Draw a Lewis Structure

Count total valence electrons. Add valence electrons from each atom. For anions, add extra e⁻ equal to the charge; for cations, subtract e⁻.
Identify the central atom. Usually the least electronegative atom (never H, rarely F). It goes in the center with other atoms arranged around it.
Draw single bonds from the central atom to each surrounding atom. Each bond uses 2 e⁻.
Distribute remaining electrons as lone pairs, starting with the outer atoms. Give each atom an octet (H only needs 2).
If the central atom lacks an octet, convert lone pairs on outer atoms into double or triple bonds until the octet is satisfied.
Check: Total electrons in structure must equal the total valence electrons counted in Step 1.

Example — CO₂:
Total VE = 4 (C) + 6 + 6 (two O) = 16 e⁻
C is central. Single bonds to each O: O—C—O (uses 4 e⁻, 12 remaining)
Place remaining 12 e⁻ as lone pairs on O: each O gets 3 pairs (6 e⁻)
Check: each O has octet, but C only has 4 e⁻. Not enough!
Move one lone pair from each O to form double bonds: O=C=O
Now C has 8 e⁻ (4 bonding pairs), each O has 8 e⁻ (2 bonding + 2 lone pairs). Total: 16 e⁻.

Example — NH₄⁺ (ammonium ion):
Total VE = 5 (N) + 4(1) (four H) − 1 (positive charge) = 8 e⁻
N is central with 4 H atoms bonded (4 single bonds = 8 e⁻)
All 8 electrons are used. N has an octet, each H has 2 e⁻. Done!
Put brackets around the structure with a + charge: [structure]⁺

Exceptions to the Octet Rule

Exception	Description	Examples
Incomplete octet	Central atom has <8 e⁻. Occurs with B and Be (they are electron-poor).	BF₃ (B has 6 e⁻), BeCl₂ (Be has 4 e⁻)
Expanded octet	Central atom has >8 e⁻. Only possible for elements in Period 3+ (they have empty d orbitals available).	PCl₅ (P has 10 e⁻), SF₆ (S has 12 e⁻), XeF₄ (Xe has 12 e⁻)
Odd-electron species	Molecules with an odd total number of valence electrons. At least one atom cannot have a full octet.	NO (11 e⁻), NO₂ (17 e⁻), ClO₂

Coordinate (Dative) Covalent Bonds

A coordinate covalent bond forms when both shared electrons come from the same atom. Once formed, it is identical to any other covalent bond. Example: NH₃ donating its lone pair to H⁺ to form NH₄⁺. The N—H bond formed is a coordinate covalent bond, but all four N—H bonds in NH₄⁺ are equivalent.

2.6 — Resonance and Formal Charge

Resonance Structures

When a molecule can be represented by two or more valid Lewis structures that differ only in the placement of electrons (not atoms), these are called resonance structures. The actual molecule is a resonance hybrid — a blend of all contributing structures. No single Lewis structure accurately represents the molecule.

Example — Ozone (O₃):
Structure 1: O=O—O (double bond on left, single on right)
Structure 2: O—O=O (single bond on left, double on right)

The real O₃ has two equivalent bonds, each with a bond order of 1.5 (between a single and double bond). The actual bond length is between that of a single and double O—O bond.

Example — Carbonate ion (CO₃²⁻):
CO₃²⁻ has 3 equivalent resonance structures. In each, one C=O double bond and two C—O single bonds, just in different positions.
Bond order = 4 total bonds ÷ 3 positions = 4/3 ≈ 1.33
All three C—O bonds are identical in the real ion (same length, same strength).

Important examples of resonance:

NO₃⁻ (nitrate): 3 equivalent resonance structures; bond order = 4/3
SO₃ (sulfur trioxide): 3 equivalent resonance structures
Benzene (C₆H₆): 2 resonance structures with alternating single/double bonds; actual structure is a delocalized ring with bond order 1.5
NO₂⁻ (nitrite): 2 equivalent resonance structures; bond order = 3/2 = 1.5

Formal Charge

Formal charge is a bookkeeping tool that helps determine the best Lewis structure when multiple non-equivalent structures are possible. It assumes electrons in a bond are shared equally.

Formal Charge = Valence electrons − Lone pair electrons − (1/2)(Bonding electrons)

Rules for choosing the best Lewis structure:
1. The structure with formal charges closest to zero on all atoms is preferred.
2. Negative formal charges should be on the more electronegative atom.
3. Adjacent atoms with the same sign of formal charge are unfavorable (destabilizing).
4. The sum of all formal charges must equal the overall charge of the molecule or ion.

Example — SCN⁻ (thiocyanate): Which structure is best?

Structure A: S=C=N (FC: S = 0, C = 0, N = −1) ← Best! Negative FC on more electronegative N
Structure B: S≡C—N (FC: S = −1, C = 0, N = 0)
Structure C: S—C≡N (FC: S = 0, C = 0, N = −1)

Structures A and C both place the −1 on N (more electronegative). Structure A is preferred because it has double bonds rather than a single + triple combination, making the formal charges closest to zero.

2.7 — VSEPR and Bond Hybridization

The Valence Shell Electron Pair Repulsion (VSEPR) model predicts molecular geometry based on the principle that electron groups around a central atom repel each other and arrange themselves to minimize this repulsion. An electron group can be a single bond, double bond, triple bond, or lone pair (multiple bonds count as ONE group).

VSEPR molecular geometries based on steric number and number of lone pairs. (Wikimedia Commons)

Step-by-Step: Predicting Molecular Geometry

Draw the Lewis structure for the molecule.
Count the steric number = total number of electron groups around the central atom (bonds + lone pairs). Remember: double and triple bonds each count as ONE group.
Determine electron geometry from the steric number (using the table below).
Determine molecular geometry from the number of bonding groups and lone pairs.
Predict bond angles (lone pairs compress bond angles slightly below the ideal value).

Complete VSEPR Geometry Table

Steric #	Bonding	Lone Pairs	Electron Geometry	Molecular Geometry	Bond Angle	Example
2	2	0	Linear	Linear	180°	CO₂, BeCl₂
3	3	0	Trigonal planar	Trigonal planar	120°	BF₃, SO₃
3	2	1	Trigonal planar	Bent	~118°	SO₂, O₃, NO₂⁻
4	4	0	Tetrahedral	Tetrahedral	109.5°	CH₄, SiH₄, NH₄⁺
4	3	1	Tetrahedral	Trigonal pyramidal	~107°	NH₃, PCl₃, H₃O⁺
4	2	2	Tetrahedral	Bent	~104.5°	H₂O, H₂S, OF₂
5	5	0	Trigonal bipyramidal	Trigonal bipyramidal	90°, 120°	PCl₅, PF₅
5	4	1	Trigonal bipyramidal	Seesaw	<90°, <120°	SF₄
5	3	2	Trigonal bipyramidal	T-shaped	<90°	ClF₃, ICl₃
5	2	3	Trigonal bipyramidal	Linear	180°	XeF₂, I₃⁻
6	6	0	Octahedral	Octahedral	90°	SF₆
6	5	1	Octahedral	Square pyramidal	<90°	BrF₅, IF₅
6	4	2	Octahedral	Square planar	90°	XeF₄, ICl₄⁻

Why lone pairs matter: Lone pairs occupy more space than bonding pairs (they are held closer to the nucleus and spread out more). This means lone pairs compress bond angles below ideal values. Example: CH₄ = 109.5°, NH₃ = 107° (1 lone pair), H₂O = 104.5° (2 lone pairs). More lone pairs = smaller bond angles.

Hybridization

Hybridization describes the mixing of atomic orbitals on a single atom to form new hybrid orbitals that match the observed molecular geometry. The number of hybrid orbitals formed always equals the number of atomic orbitals mixed (which equals the steric number).

Comparison of sp, sp², and sp³ hybrid orbitals formed by mixing s and p atomic orbitals. (Wikimedia Commons)

Steric Number	Hybridization	Geometry	Orbitals Mixed	Example
2	sp	Linear	1s + 1p	CO₂, C₂H₂, BeCl₂
3	sp²	Trigonal planar	1s + 2p	BF₃, C₂H₄, SO₃
4	sp³	Tetrahedral	1s + 3p	CH₄, NH₃, H₂O
5	sp³d	Trigonal bipyramidal	1s + 3p + 1d	PCl₅, SF₄
6	sp³d²	Octahedral	1s + 3p + 2d	SF₆, XeF₄

Sigma (σ) and Pi (π) Bonds:
• A single bond = 1 sigma bond
• A double bond = 1 sigma + 1 pi bond
• A triple bond = 1 sigma + 2 pi bonds
• Sigma bonds form from head-on overlap of hybrid orbitals
• Pi bonds form from side-by-side overlap of unhybridized p orbitals
• Pi bonds prevent rotation around the bond axis (important for cis/trans isomerism)

Example — C₂H₄ (ethylene): Each C is sp² hybridized. The C=C contains 1 σ + 1 π bond. The molecule is planar because the pi bond locks the geometry.

Molecular Polarity

A molecule is polar if it has polar bonds AND those bond dipoles do NOT cancel out. Molecular geometry determines whether dipoles cancel:

Polar molecules: H₂O (bent — dipoles don’t cancel), NH₃ (pyramidal), HCl, SO₂
Nonpolar molecules: CO₂ (linear — dipoles cancel), BF₃ (trigonal planar — symmetric), CH₄ (tetrahedral — symmetric), CCl₄
Any molecule with lone pairs on the central atom and polar bonds will usually be polar (asymmetric charge distribution).

📝 Practice Multiple-Choice Questions

Test your knowledge of Unit 2. Click "Show Answer" to reveal the correct choice and explanation.

1. Which of the following compounds contains only ionic bonds?

(A) H₂O
(B) NaCl
(C) CO₂
(D) CH₄

Answer: (B)
NaCl is formed by the transfer of an electron from Na (metal) to Cl (nonmetal), creating Na⁺ and Cl⁻ held by electrostatic attraction. The others are all covalent compounds (nonmetal-nonmetal bonds).

2. On a potential energy curve for the formation of a covalent bond, the bond length corresponds to:

(A) The point where potential energy is zero
(B) The internuclear distance at the potential energy minimum
(C) The point where the curve begins to rise steeply
(D) The maximum of the curve

Answer: (B)
The bond length (r₀) is the internuclear distance at the potential energy minimum — the most stable configuration. The bond energy is the depth of this well.

3. Which of the following ionic compounds has the highest melting point?

(A) NaCl
(B) KBr
(C) MgO
(D) CsI

Answer: (C)
MgO has the highest lattice energy (and therefore highest melting point) because it has 2+ and 2− charges and small ionic radii. The others are all 1+/1− with larger ions.

4. Steel is harder than pure iron because:

(A) Carbon atoms increase the number of delocalized electrons
(B) Small carbon atoms in the interstices prevent layers of iron atoms from sliding past each other
(C) Carbon atoms replace iron atoms in the lattice
(D) The carbon atoms form covalent bonds with iron

Answer: (B)
Steel is an interstitial alloy. Small carbon atoms fit into the gaps between larger iron atoms, disrupting the regular lattice and preventing layers from sliding — making the alloy harder but less ductile.

5. How many total lone pairs of electrons are in the Lewis structure of XeF₂?

(A) 6
(B) 9
(C) 3
(D) 12

Answer: (B)
Total VE = 8 (Xe) + 2(7) (two F) = 22 e⁻. Two Xe—F bonds use 4 e⁻. Remaining 18 e⁻ = 9 lone pairs: 3 on each F (6 pairs) + 3 on Xe (3 pairs) = 9 lone pairs total. XeF₂ has an expanded octet with 3 lone pairs on the central Xe.

6. Which of the following best describes the bond order of each C—O bond in the carbonate ion (CO₃²⁻)?

(A) 1
(B) 4/3
(C) 3/2
(D) 2

Answer: (B)
CO₃²⁻ has 3 resonance structures, each with one C=O double bond and two C—O single bonds. Total bond order = (1 + 1 + 2)/3 = 4/3 ≈ 1.33 for each C—O bond. All three bonds are equivalent in the resonance hybrid.

7. What is the molecular geometry of SF₄?

(A) Tetrahedral
(B) Square planar
(C) Seesaw
(D) Trigonal pyramidal

Answer: (C)
SF₄ has 5 electron groups (4 bonds + 1 lone pair) around S. Electron geometry = trigonal bipyramidal. With 4 bonding pairs and 1 lone pair (in the equatorial position), the molecular geometry is seesaw.

8. What is the hybridization of the central atom in XeF₄?

(A) sp³
(B) sp³d
(C) sp³d²
(D) sp²

Answer: (C)
XeF₄ has 6 electron groups around Xe (4 bonds + 2 lone pairs). Steric number = 6, so hybridization = sp³d². The molecular geometry is square planar.

9. How many sigma (σ) and pi (π) bonds are in the molecule C₂H₂ (acetylene)?

(A) 3 sigma, 2 pi
(B) 5 sigma, 0 pi
(C) 2 sigma, 3 pi
(D) 4 sigma, 1 pi

Answer: (A)
H—C≡C—H: Each C—H single bond = 1 σ. The C≡C triple bond = 1 σ + 2 π. Total: 3 sigma + 2 pi = 5 bonds. Each carbon is sp hybridized (linear, 180°).

10. Which of the following molecules is polar?

(A) CO₂
(B) CCl₄
(C) BF₃
(D) NH₃

Answer: (D)
NH₃ is trigonal pyramidal (3 bonds + 1 lone pair on N). The lone pair creates an asymmetric charge distribution, so the bond dipoles do NOT cancel → NH₃ is polar. CO₂ (linear), CCl₄ (tetrahedral), and BF₃ (trigonal planar) are all symmetric, so their dipoles cancel → nonpolar.

💡 Study Tip for Unit 2

Unit 2 is heavily tested on the AP exam. Make sure you can draw Lewis structures quickly and correctly (including exceptions to the octet rule), assign formal charges to determine the best structure, and predict molecular geometry and polarity from VSEPR. Know the hybridization table cold — steric number = hybridization. Always remember that lone pairs compress bond angles below ideal values. Practice identifying sigma and pi bonds in double and triple bonds. Understanding these concepts is essential for Unit 3 (intermolecular forces).

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