Types of chemical bonds

AP Chemistry· AP Chemistry CED — Molecular and Ionic Compound Structure and Properties· 14 min read

1. Ionic Bonding★★☆☆☆⏱ 5 min

📘 Definition

Ionic Bonding

Ionic bonding forms when complete transfer of one or more valence electrons occurs from a low electronegativity atom (usually a metal) to a high electronegativity atom (usually a nonmetal), producing oppositely charged cations and anions held together by strong electrostatic attraction.

The strength of ionic 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 AP-accepted threshold for predicting ionic bonding is an electronegativity difference $Δ EN > 1.7 - 2.0$ . 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

Given $EN_{Ca} = 1.0$ , $EN_{F} = 4.0$ , $Ca^{2 +}$ radius = 100 pm, $F^{-}$ radius = 133 pm, $EN_{Na} = 0.9$ , $EN_{Cl} = 3.2$ , $Na^{+}$ radius = 102 pm, $Cl^{-}$ 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 $Δ EN$ for Ca-F:
2
$\Delta \text{EN} = 4.0 - 1.0 = 3.0$
3
This is greater than the 2.0 threshold for ionic bonding, so Ca-F is confirmed ionic.
4
Calculate the product of charges for each bond: For Ca-F: $∣ q_{1} q_{2} ∣ = ∣ (+ 2) (- 1) ∣ = 2$ . For Na-Cl: $∣ q_{1} q_{2} ∣ = ∣ (+ 1) (- 1) ∣ = 1$ .
5
Calculate interionic distance $r$ for each: $r_{Ca-F} = 100 + 133 = 233$ pm; $r_{Na-Cl} = 102 + 181 = 283$ pm.
6
Apply the Coulomb's law proportionality:
7
$F \propto \frac{|q_1 q_2|}{r^2}$
8
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 classification. Focus on applying rules, not memorizing numbers.

2. Covalent Bonding (Polar and Nonpolar)★★☆☆☆⏱ 4 min

Covalent bonding forms when two atoms (almost always nonmetals with similar high 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: $Δ EN < 0.5$ — electron density is shared nearly equally between the two atoms, so no permanent partial charge separation
Polar covalent: $0.5 \leq Δ EN \leq 2.0$ — the more electronegative atom pulls the shared electron pair closer, creating a partial negative charge ( $δ^{-}$ ) on the more electronegative atom and a partial positive charge ( $δ^{+}$ ) on the less electronegative atom, forming a bond dipole

A larger $Δ EN$ within the covalent range always means a more polar bond.

📐 Worked Example

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

1
Calculate $Δ EN$ for each bond: (a) $Δ EN = 3.5 - 3.5 = 0$ ; (b) $Δ EN = 3.2 - 2.5 = 0.7$ ; (c) $Δ EN = 2.8 - 1.0 = 1.8$ .
2
Apply classification rules: (a) $0 < 0.5$ → nonpolar covalent; (b) $0.5 \leq 0.7 \leq 2.0$ → 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 $δ^{-}$ partial negative charge.

3. Metallic Bonding★★★☆☆⏱ 5 min

📘 Definition

Metallic Bonding

Metallic bonding is the non-directional 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 bind specific pairs of atoms), metallic bonding is non-directional, which explains why metals are malleable (can be hammered into sheets) and ductile (can be drawn into wires): 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

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 ( $Δ EN \approx 0.1$ ), 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.

4. AP Style Practice Problems★★★☆☆⏱ 6 min

📐 Worked Example

Given the electronegativity values: $EN_{C} = 2.5$ , $EN_{O} = 3.5$ , $EN_{K} = 0.8$ , $EN_{S} = 2.5$ , $EN_{Br} = 2.8$ . 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

1
First calculate $Δ EN$ for each bond: S-S $Δ EN = 2.5 - 2.5 = 0 < 0.5$ → nonpolar covalent. C-O $Δ EN = 3.5 - 2.5 = 1.0$ , which falls between 0.5 and 2.0 → polar covalent. K-Br $Δ EN = 2.8 - 0.8 = 2.0$ , which meets the threshold for ionic bonding between a metal and nonmetal → ionic.
2
The question asks for nonpolar, polar, ionic in order, which matches option B. The other options have the wrong order, so the correct answer is B.

📐 Worked Example

Use the data below to answer the following questions:
(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 ( $Ca^{2 +}$ radius = 100 pm, $O^{2 -}$ 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.

Data: Strontium: EN = 0.95, ionic radius 118 pm, charge +2; Chlorine: EN = 3.2, radius 181 pm, charge -1; Carbon: EN = 2.5

1
(a) Calculate $Δ EN = 3.2 - 0.95 = 2.25$ . This is greater than the 2.0 threshold for ionic bonding, so the bond is ionic. Strontium is an alkaline earth metal that transfers valence electrons to chlorine to form oppositely charged ions.
2
(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.
3
(c) $Δ EN = 3.2 - 2.5 = 0.7$ , 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 $δ^{-}$ partial negative charge.

5. Common Pitfalls

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 $Δ EN$ 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 $Δ EN$ 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 $Δ EN$ 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. Quick Reference Cheatsheet

Bond Type / Property	Classification / Rule	Key AP Exam Notes
Ionic Bond	$Δ EN > 1.7 - 2.0$	Forms between metal + nonmetal; full electron transfer, electrostatic attraction between ions
Nonpolar Covalent Bond	$Δ EN < 0.5$	Forms between two nonmetals; equal electron sharing, no bond dipole
Polar Covalent Bond	$0.5 \leq Δ EN \leq 2.0$	Forms between two nonmetals; unequal sharing, $δ^{-}$ on more electronegative atom
Metallic Bond	No $Δ EN$ threshold	Forms between metal atoms; attraction between metal cations + delocalized valence electron sea
Coulomb's Law (Ionic Strength)	$F \propto \frac{∣ q _{1} q _{2} ∣}{r ^{2}}$	Higher charge product, smaller interionic distance = stronger ionic bond
C-H Bond	$Δ EN \approx 0.4$	Classified as nonpolar covalent on AP exams
Ionic Solid Conductivity	N/A	Do not conduct as solids; conduct only when molten/dissolved
Metallic Solid Conductivity	N/A	Conduct electricity in both solid and liquid states

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Bond classification from EN values
2022 · FRQ
Ionic bond strength comparison

What's Next

This topic is the foundational classification for all structure-property relationships in AP Chemistry, 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 and their bulk properties like melting point, conductivity, and hardness. Without mastering bond type classification, you cannot correctly reason through why different solids have 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.