Atoms form chemical bonds to achieve lower potential energy and greater stability.
Key principle: Atoms bond to achieve a stable electron configuration, typically a filled valence shell (octet rule for most elements, duet for hydrogen).
Three Main Types of Bonds
1. Ionic Bonds
Definition: Electrostatic attraction between oppositely charged ions
Formation:
Complete transfer of electrons from metal to nonmetal
Metal loses electrons → becomes cation (+)
Nonmetal gains electrons → becomes anion (-)
Opposite charges attract
Characteristics:
Form between metals and nonmetals
Large difference in electronegativity (ΔEN > 1.7)
Form crystalline lattice structures
High melting and boiling points
📚 Practice Problems
1Problem 1easy
❓ Question:
Classify the following bonds as ionic, polar covalent, or nonpolar covalent: (a) K-Cl, (b) C-O, (c) Cl-Cl. Use electronegativity values: K = 0.8, Cl = 3.0, C = 2.5, O = 3.5
💡 Show Solution
Solution:
Given: Electronegativity values and three bonds to classify
Find: Bond type for each
Step 1: Calculate ΔEN for each bond
(a) K-Cl:
Explain using:
📋 AP Chemistry — Exam Format Guide
⏱ 3 hours 15 minutes📝 67 questions📊 3 sections
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💡 Key Test-Day Tips
✓Memorize common polyatomic ions
✓Practice dimensional analysis
✓Know your gas laws
⚠️ Common Mistakes: Types of Chemical Bonds
Avoid these 3 frequent errors
🌍 Real-World Applications: Types of Chemical Bonds
See how this math is used in the real world
📝 Worked Example: Stoichiometry — Limiting Reagent
Problem:
2 mol of H2 reacts with 1 mol of O2. How many grams of water are produced? Which is the limiting reagent? (2H2+O2→2H2O)
Understand ionic, covalent, and metallic bonds, including how electronegativity differences determine bond type and properties.
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Are there practice problems for Types of Chemical Bonds?
Conduct electricity when molten or dissolved (ions are mobile)
Metallic bonds: Delocalized electrons, metal + metal
Electronegativity difference predicts bond type
Bond strength: Depends on bond order, atomic size, and electronegativity
ΔEN=∣3.0−0.8∣=2.2
(b) C-O:ΔEN=∣3.5−2.5∣=1.0
(c) Cl-Cl:ΔEN=∣3.0−3.0∣=0.0
Step 2: Apply classification rules
ΔEN Range
Bond Type
0 - 0.4
Nonpolar covalent
0.4 - 1.7
Polar covalent
> 1.7
Ionic
Step 3: Classify each bond
(a) K-Cl: ΔEN = 2.2
ΔEN > 1.7 → Ionic bond
K is metal, Cl is nonmetal ✓
K loses electron → K⁺
Cl gains electron → Cl⁻
(b) C-O: ΔEN = 1.0
0.4 < ΔEN < 1.7 → Polar covalent bond
Both C and O are nonmetals ✓
O is more electronegative → partial negative charge (δ-)
C has partial positive charge (δ+)
(c) Cl-Cl: ΔEN = 0.0
ΔEN = 0 → Nonpolar covalent bond
Identical atoms share electrons equally
No charge separation
Example: Cl₂ molecule
Answer:
(a) K-Cl: Ionic
(b) C-O: Polar covalent
(c) Cl-Cl: Nonpolar covalent
Summary:
Large ΔEN (>1.7) indicates electron transfer (ionic), moderate ΔEN (0.4-1.7) indicates unequal sharing (polar covalent), and small ΔEN (<0.4) indicates equal sharing (nonpolar covalent).
2Problem 2medium
❓ Question:
Explain why sodium chloride (NaCl) has a much higher melting point (801°C) than water (H₂O, melting point 0°C), even though both compounds involve bonding between atoms.
💡 Show Solution
Solution:
Given: NaCl melting point = 801°C, H₂O melting point = 0°C
Find: Explain the large difference
Step 1: Identify bond types
NaCl:
Na: metal (electronegativity = 0.9)
Cl: nonmetal (electronegativity = 3.0)
ΔEN = 2.1 → Ionic bond
H₂O:
H: nonmetal (electronegativity = 2.1)
O: nonmetal (electronegativity = 3.5)
ΔEN = 1.4 → Polar covalent bond
Step 2: Describe structure differences
NaCl:
Forms ionic crystal lattice
Each Na⁺ surrounded by 6 Cl⁻ ions
Each Cl⁻ surrounded by 6 Na⁺ ions
3D network of strong electrostatic attractions
Must break many strong ionic bonds to melt
H₂O:
Forms discrete molecules (H-O-H)
Covalent bonds within molecules (strong)
Hydrogen bonds between molecules (weaker)
Must break intermolecular forces to melt, NOT covalent bonds
Step 3: Compare forces that must be overcome
To melt NaCl:
Must overcome ionic bonds between Na⁺ and Cl⁻
Very strong electrostatic attractions (∝ 1/r²)
Lattice energy is very high (~787 kJ/mol)
To melt H₂O:
Must overcome hydrogen bonds between molecules
Much weaker than ionic or covalent bonds (~20 kJ/mol per H-bond)
Covalent O-H bonds remain intact when ice melts
Step 4: Energy comparison
Breaking ionic bonds in NaCl:
Requires ~787 kJ/mol (lattice energy)
Very high temperature needed (801°C)
Breaking H-bonds in ice:
Requires ~6 kJ/mol (enthalpy of fusion)
Low temperature needed (0°C)
Answer:
NaCl has a much higher melting point because melting requires breaking strong ionic bonds in the crystal lattice, while melting ice only requires breaking weak hydrogen bonds between water molecules. The covalent bonds within water molecules remain intact during melting.
Key distinction:
Intramolecular forces (ionic/covalent bonds): very strong
Melting ionic compounds requires breaking intramolecular forces, while melting molecular compounds typically only requires breaking intermolecular forces.
Additional note: If you had to break the O-H covalent bonds in water (like in decomposition), that would require much more energy (~460 kJ/mol per bond), comparable to ionic bonds.
3Problem 3hard
❓ Question:
Diamond (C) and graphite (C) are both composed entirely of carbon atoms but have very different properties. Diamond is extremely hard with a very high melting point (~3550°C), while graphite is soft and has a lower melting point (~3600°C but sublimes). Both conduct heat well, but only graphite conducts electricity. Explain these differences in terms of bonding and structure.
💡 Show Solution
Solution:
Given: Two allotropes of carbon with different properties
Find: Explain differences using bonding and structure
Step 1: Describe diamond structure
Diamond:
3D network covalent structure
Each C atom bonded to 4 other C atoms
Tetrahedral geometry (sp³ hybridization)
All electrons in localized C-C single bonds
No free electrons
Bonding diagram:C−C−C−C (in 3D tetrahedral network)
3D network means cutting requires breaking many bonds
All atoms rigidly locked in position
No planes of weakness
Graphite - soft:
Layers can slide past each other
Only weak London forces between layers (~10 kJ/mol)
C-C bonds within layers are strong
But layers separate easily (like a deck of cards)
This is why graphite is used as lubricant
Step 4: Explain electrical conductivity
Diamond - does NOT conduct:
All 4 valence electrons per C in localized bonds
No free electrons to carry charge
All electrons held tightly in σ bonds
Insulator
Graphite - conducts electricity:
Delocalized π electrons above/below planes
One electron per C is mobile within layers
Electrons can move parallel to layers
Similar to metallic bonding (electron sea within layers)
Conductor
Step 5: Explain melting point
Both have very high melting points:
Diamond: ~3550°C (actually sublimes at very high T)
Graphite: ~3600°C (sublimes above ~3700°C at low pressure)
Why both are high:
Both must break strong C-C covalent bonds
Network covalent structures throughout
Diamond: 3D network
Graphite: 2D networks in each layer
Why similar values:
Both involve breaking C-C bonds (~348 kJ/mol)
Graphite layers are strongly bonded internally
Weak interlayer forces don't affect melting (that's not what melts)
Step 6: Explain thermal conductivity
Both conduct heat well:
Diamond: Atoms in rigid 3D network vibrate and transfer energy
Graphite: Delocalized electrons + rigid layers transfer energy
Answer Summary:
Property
Diamond
Graphite
Explanation
Structure
3D network, sp³
Layers, sp²
Hybridization difference
Hardness
Extremely hard
Soft
3D network vs. sliding layers
Electrical conductivity
No
Yes (in plane)
No free e⁻ vs. delocalized π electrons
Melting point
Very high (~3550°C)
Key insight:
Same element (carbon), but different bonding arrangements (sp³ vs. sp²) and structures (3D network vs. layered) lead to dramatically different macroscopic properties. This demonstrates that properties depend not just on atomic composition, but on how atoms are connected.
▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.