Types of chemical bonds — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Ionic bonds, nonpolar covalent bonds, polar covalent bonds, metallic bonds, bond polarity classification, electronegativity difference rules, and Coulomb’s law for ionic bond strength, aligned to the AP Chemistry CED.

You should already know: Periodic trends for electronegativity, Coulomb's law for electrostatic interactions, basic valence electron configuration of atoms.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Types of chemical bonds?

Types of chemical bonds refers to the classification of distinct intramolecular attractive interactions that hold atoms and ions together in chemical compounds and solid materials. All chemical bonds arise from electrostatic attractions between negatively charged electrons and positively charged atomic nuclei, but differ in the degree of electron sharing or transfer between the bonded particles. This topic is part of Unit 2: Molecular and Ionic Compound Structure and Properties, which accounts for 7-9% of the total AP Chemistry exam score. Bond classification appears in both multiple-choice (MCQ) and free-response (FRQ) sections, and is almost always linked to questions asking you to connect microscopic bonding to macroscopic properties. Unlike many topics, this is not a calculation-heavy topic, but it requires precise application of classification rules to novel compounds, which is a common point of difficulty for students.

2. Ionic Bonding

Ionic bonding forms when there is a complete transfer of one or more valence electrons from an atom with low electronegativity and low ionization energy (almost always a metal) to an atom with high electronegativity and high electron affinity (almost always a nonmetal). This transfer produces positively charged cations and negatively charged anions, which are held together by strong electrostatic attraction. The strength of this attraction follows Coulomb’s law: $$F=k\frac{|q_1{q}_2|}{r^2}$$ where $q_1$ and $q_2$ are the charges of the two ions, $r$ is the distance between the centers of the ions, and $k$ is Coulomb’s constant. The general rule for predicting ionic bonding is an electronegativity difference ($\Delta \text{EN}={\text{EN}}_{\text{more EN}}-{\text{EN}}_{\text{less EN}}$) greater than 1.7-2.0, which is the standard threshold accepted on the AP exam. Ionic compounds form crystalline lattice structures, are hard, brittle, and conduct electricity only when molten or dissolved (not as solids, since ions are fixed in place).

Worked Example

Problem: Given ${\text{EN}}_{\text{Ca}}=1.0$, ${\text{EN}}_{\text{F}}=4.0$, ${\text{Ca}}^{2+}$ radius = 100 pm, ${\text{F}}^{-}$ radius = 133 pm, ${\text{EN}}_{\text{Na}}=0.9$, ${\text{EN}}_{\text{Cl}}=3.2$, ${\text{Na}}^{+}$ radius = 102 pm, ${\text{Cl}}^{-}$ radius = 181 pm. (1) Confirm that the Ca-F bond is ionic, (2) compare the strength of the Ca-F ionic bond to the Na-Cl ionic bond using Coulomb’s law.

1. Calculate $\Delta \text{EN}$ for Ca-F: $\Delta \text{EN}=4.0-1.0=3.0$. This is greater than the 2.0 threshold for ionic bonding, so Ca-F is confirmed to be ionic.
2. Calculate the product of charges for each compound: For CaF₂, $|q_1{q}_2|=|(+2)(-1)|=2$. For NaCl, $|q_1{q}_2|=|(+1)(-1)|=1$.
3. Calculate interionic distance $r$ for each: $r_{\text{Ca-F}}=100+133=233$ pm; $r_{\text{Na-Cl}}=102+181=283$ pm.
4. Apply Coulomb’s proportionality $F\propto \frac{|q_1{q}_2|}{r^2}$: Ca-F has a larger charge product and smaller interionic distance than Na-Cl, so Ca-F has a much stronger ionic attraction than Na-Cl.

Exam tip: AP will never ask you to memorize electronegativity values — they will always provide any EN values you need for calculation. Focus on applying the rules, not memorizing numbers.

3. Covalent Bonding (Polar and Nonpolar)

Covalent bonding forms when two atoms (almost always nonmetals, with high and similar electronegativities) share pairs of valence electrons to achieve a stable full valence shell. The attraction arises from the shared electron pair being pulled toward the positively charged nuclei of both bonded atoms. Covalent bonds are split into two subclasses based on electronegativity difference:

• Nonpolar covalent: $\Delta \text{EN}<0.5$ — electron density is shared nearly equally between the two atoms, so no permanent partial charge separation.
• Polar covalent: $0.5\le \Delta \text{EN}\le 2.0$ — the more electronegative atom pulls the shared electron pair closer, creating a partial negative charge (${\delta }^{-}$) on the more electronegative atom and a partial positive charge (${\delta }^{+}$) on the less electronegative atom, forming a bond dipole. A larger $\Delta \text{EN}$ within the covalent range always means a more polar bond.

Worked Example

Problem: Classify the following bonds as nonpolar covalent, polar covalent, or ionic: (a) O=O, (b) C-Cl, (c) Li-Br. Given: ${\text{EN}}_{\text{O}}=3.5$, ${\text{EN}}_{\text{C}}=2.5$, ${\text{EN}}_{\text{Cl}}=3.2$, ${\text{EN}}_{\text{Li}}=1.0$, ${\text{EN}}_{\text{Br}}=2.8$. For any polar covalent bonds, identify which atom is ${\delta }^{-}$.

1. Calculate $\Delta \text{EN}$ for each bond: (a) $\Delta \text{EN}=3.5-3.5=0$; (b) $\Delta \text{EN}=3.2-2.5=0.7$; (c) $\Delta \text{EN}=2.8-1.0=1.8$.
2. Apply classification rules: (a) $0<0.5$ → nonpolar covalent; (b) $0.5\le 0.7\le 2.0$ → polar covalent; (c) $1.8$ is above the 1.7 threshold for ionic bonding (between metal and nonmetal) → ionic.
3. For the polar covalent C-Cl bond, chlorine is more electronegative than carbon, so chlorine carries the ${\delta }^{-}$ partial negative charge.

Exam tip: The C-H bond has a $\Delta \text{EN}$ of ~0.4, so it is always classified as nonpolar covalent on the AP exam — don’t overthink the small difference and misclassify it as polar.

4. Metallic Bonding

Metallic bonding is the electrostatic attractive interaction between positively charged metal cations and a delocalized "sea" of mobile valence electrons shared collectively across the entire solid metal lattice. Metals have low ionization energy, so their valence electrons are not tightly held to individual atoms, leading to delocalization. Unlike ionic and covalent bonds, which are directional (bind specific pairs of atoms), metallic bonding is non-directional, which explains the key properties of metals: malleability (can be hammered into sheets) and ductility (can be drawn into wires) because layers of cations can slide past each other without breaking the bonding network. Delocalized mobile electrons also make metallic solids good conductors of electricity and heat in both solid and liquid states. Metallic bonding occurs between atoms of the same pure metal or between different metal atoms in alloys.

Worked Example

Problem: A student observes that bronze, an alloy of tin and copper, conducts electricity in solid form. The student claims that this conductivity is due to ionic bonding between the tin and copper atoms. Refute this claim, identify the correct bond type, and explain how the bonding explains conductivity.

1. Ionic bonding requires a large electronegativity difference and full electron transfer to form oppositely charged ions, which almost always occurs between a metal and a nonmetal. Both tin and copper are metals, with very similar electronegativities ($\Delta \text{EN}\approx 0.1$), so ionic bonding cannot form.
2. Solid ionic compounds cannot conduct electricity, because all ions are fixed in place in the crystalline lattice and cannot move to carry charge. Bronze conducts electricity in solid form, so it cannot be ionic.
3. The correct bond type is metallic bonding: valence electrons from both tin and copper are delocalized into a shared sea of electrons surrounding the metal cations.
4. The mobile delocalized electrons are free to move through the solid lattice when a voltage is applied, which allows bronze to conduct electricity, matching the observation.

Exam tip: AP FRQs always require you to link the microscopic bonding structure to the macroscopic property. Don’t just write "metals conduct electricity" — explicitly mention delocalized mobile electrons to earn full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Automatically classifying any bond between a metal and nonmetal as ionic, regardless of electronegativity difference. Why: Students memorize the "metal + nonmetal = ionic" rule and forget that high oxidation state metals have significant covalent character. Correct move: Always use given electronegativity values to calculate $\Delta \text{EN}$ and confirm bond type, don’t rely solely on element classification.
• Wrong move: Confusing bond polarity with molecular polarity, concluding that any molecule with polar bonds is a polar molecule. Why: The topics are taught back-to-back, so students mix up bond-level vs molecular-level properties. Correct move: When asked to classify a bond, only consider the $\Delta \text{EN}$ of that individual bond — molecular polarity is a separate question that requires accounting for molecular symmetry.
• Wrong move: Stating that a larger interionic distance leads to a stronger ionic bond when using Coulomb’s law. Why: Students forget that interionic distance is in the denominator of the force equation, so they flip the relationship. Correct move: Always write the proportionality $F\propto \frac{|q_1{q}_2|}{r^2}$ before comparing bond strengths to avoid flipping the relationship.
• Wrong move: Classifying the C-H bond as polar covalent because carbon is slightly more electronegative than hydrogen. Why: Students overthink the small electronegativity difference and forget the AP classification threshold. Correct move: C-H has a $\Delta \text{EN}$ of ~0.4, so it is always classified as nonpolar covalent for the AP exam.
• Wrong move: Describing metallic bonding as attraction between free protons and electrons. Why: Students oversimplify the structure of metal atoms. Correct move: Always describe metallic bonding as the electrostatic attraction between positively charged metal cations (not protons) and a delocalized sea of valence electrons to earn full credit on FRQs.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Given the electronegativity values: ${\text{EN}}_{\text{C}}=2.5$, ${\text{EN}}_{\text{O}}=3.5$, ${\text{EN}}_{\text{K}}=0.8$, ${\text{EN}}_{\text{S}}=2.5$, ${\text{EN}}_{\text{Br}}=2.8$. Which ordered list correctly matches: nonpolar covalent bond, polar covalent bond, ionic bond? A) C-O, K-Br, S-S B) S-S, C-O, K-Br C) K-Br, S-S, C-O D) S-S, K-Br, C-O

Worked Solution: First calculate $\Delta \text{EN}$ for each bond: S-S $\Delta \text{EN}=2.5-2.5=0<0.5$ → nonpolar covalent. C-O $\Delta \text{EN}=3.5-2.5=1.0$, which falls between 0.5 and 2.0 → polar covalent. K-Br $\Delta \text{EN}=2.8-0.8=2.0$, which meets the threshold for ionic bonding between a metal and nonmetal → ionic. The question asks for nonpolar, polar, ionic in order, which matches option B. The other options have the wrong order: A is polar, ionic, nonpolar; C is ionic, nonpolar, polar; D is nonpolar, ionic, polar. The correct answer is B.

Question 2 (Free Response)

Use the data in the table below to answer the questions that follow:

(a) Classify the bond between Sr and Cl, justify your answer with a calculation. (b) Predict whether the Sr-Cl ionic bond is stronger or weaker than the ionic bond in calcium oxide (${\text{Ca}}^{2+}$ radius = 100 pm, ${\text{O}}^{2-}$ radius = 140 pm, charges +2 and -2). Justify your answer with Coulomb’s law. (c) Classify the bond between C and Cl, identify which atom carries the partial negative charge, justify your answer.

Worked Solution: (a) Calculate $\Delta \text{EN}=3.2-0.95=2.25$. This is greater than the 2.0 threshold for ionic bonding. Strontium is an alkaline earth metal with low ionization energy, so it transfers valence electrons to chlorine to form ${\text{Sr}}^{2+}$ and ${\text{Cl}}^{-}$ ions held by electrostatic attraction. The bond is ionic. (b) From Coulomb’s law, ionic bond strength $F\propto \frac{|q_1{q}_2|}{r^2}$. For Sr-Cl: $|q_1{q}_2|=|(+2)(-1)|=2$, $r=118+181=299$ pm. For Ca-O: $|q_1{q}_2|=|(+2)(-2)|=4$, $r=100+140=240$ pm. Sr-Cl has a smaller charge product and larger interionic distance, so it has weaker electrostatic attraction. The Sr-Cl ionic bond is weaker than the Ca-O ionic bond. (c) $\Delta \text{EN}=3.2-2.5=0.7$, which falls between 0.5 and 2.0, so the bond is polar covalent. Chlorine is more electronegative than carbon, so it pulls the shared electron pair closer. Chlorine carries the ${\delta }^{-}$ partial negative charge.

Question 3 (Application / Real-World Style)

Lithium-ion batteries use graphite (a pure carbon solid) as the anode material. Graphite is made of carbon atoms covalently bonded into 2D sheets, with weak interactions between sheets. A battery manufacturer tests a new anode made of pure lithium metal, which is held together by metallic bonding. Explain why the pure lithium anode conducts electricity better than the graphite anode, using your knowledge of bond types.

Worked Solution: Pure lithium metal is held together by metallic bonding, which consists of a lattice of lithium cations surrounded by a delocalized sea of mobile valence electrons. These free electrons can easily move through the material to carry electric charge when a voltage is applied, giving high conductivity. In graphite, carbon atoms are held together by covalent bonds, with only one delocalized electron per carbon atom between the covalent sheets. This results in lower mobility of charge carriers compared to the fully delocalized electron sea in metallic lithium. As a result, metallic lithium has higher electrical conductivity than graphite, making it a better anode material for high-performance batteries.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational classification for all structure-property relationships in AP Chemistry, and is required for nearly every subsequent topic in the course. Next in Unit 2, you will connect bond type to the structure of different solid types (ionic, covalent network, molecular, metallic) and their bulk properties like melting point, conductivity, and hardness. Without mastering bond type classification, you cannot correctly reason through why different solids have such different physical properties, a common weighted FRQ topic. Beyond Unit 2, bond polarity from this topic is required to understand intermolecular forces in Unit 3, and bond strength (derived from Coulomb’s law for ionic bonds) is used to calculate reaction enthalpies in Unit 5.

intramolecular and intermolecular forces structure of crystalline solids bond enthalpy and reaction enthalpy intermolecular forces and properties