Refresh KidLearning
LESSON 02 / 20 · TOPIC 2.1

Which end of a bond attracts electrons more strongly?

You will be able to: Use electronegativity to assign partial charges without confusing them with ion charges.

Bonding, geometry and chemical reasoningFree study resourceReview editionTeacher review pending

Which end of a bond attracts electrons more strongly?

Two people can share a blanket unevenly. Similarly, atoms can share electrons unequally: in H–F, electron density is drawn toward F. The analogy is about unequal sharing, not electrons being a solid object.

A useful starting point: Why do salt, water and copper behave differently? →

Words and symbols before equations

Electronegativity χ
Relative tendency of a bonded atom to attract shared electrons; dimensionless.
Partial charge δ
A fractional imbalance of electron density, not a whole-ion charge.
Bond dipole
Separation of positive and negative charge within a bond.
χ(A) = 2.2; χ(B) = 3.2ABA δ+B δ−Orange marker: qualitative electron-density shift, not an electron.
Read this model snapshot. Δχ = 1. Electron density shifts toward B.
What this picture assumes

Atom A has χ = 2.2. The electron-density marker indicates only the direction of unequal sharing; it is not a calculated charge or electron position.

Read the picture in three steps

  1. Read the species and labels first. A Lewis line represents two electrons; a spatial stick indicates connectivity. Use the stated quantities and units for numerical comparisons.
  2. Δχ = 1. Electron density shifts toward B.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the chemistry

Compare both atoms on the same electronegativity scale. The more electronegative atom is the δ− end and its partner the δ+ end.

Equal atoms such as those in F₂ share symmetrically, so their bond has no permanent dipole. Increasing the electronegativity difference generally increases unequal sharing for comparable bonds.

Ionic and covalent character form a continuum. A numerical cutoff is a teaching convention, not a universal law. Electronegativity alone does not classify every substance, and a polar bond does not guarantee a polar molecule.

A worked example, step by step

Using χ(H) = 2.2 and χ(F) = 4.0, find the difference and label H–F.

  1. Define Δχ as the absolute difference on one dimensionless scale.
  2. Δχ =
  3. 4.0 − 2.2
  4. = 1.8.
  5. F attracts the shared electrons more strongly: Hδ+–Fδ−.
  6. This indicates a polar bond; it does not mean a free H⁺ and F⁻ pair exists in each HF molecule.
Common mix-up

A partial negative charge and an integer ionic charge are different descriptions.

CHECK THE IDEA

Does a bond between identical atoms have a permanent bond dipole?

Compare with an explanation

No. Both nuclei attract the shared electron density equally by symmetry.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Hold atom A at χ = 2.2. Change χ of atom B through smaller, equal and larger values. Predict where δ− should appear; the displayed imbalance is qualitative.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

χ(A) = 2.2; χ(B) = 3.2ABA δ+B δ−Orange marker: qualitative electron-density shift, not an electron.

Δχ = 1. Electron density shifts toward B.

Atom A has χ = 2.2. The electron-density marker indicates only the direction of unequal sharing; it is not a calculated charge or electron position.

Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using electron accounting, electrostatic interactions or spatial geometry. Identify what the representation cannot tell you.

Apply the idea to a fresh problem Practice →

Show what you understand.

Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.

1. In H–Cl, chlorine is more electronegative. Which label is appropriate?

Show answer and reasoning

Hδ+–Clδ−. Unequal sharing shifts density toward chlorine, making it δ−; both atoms cannot carry the same negative imbalance in neutral HCl.

2. Given χ(A)=2.5 and χ(B)=2.5, what does the simple model predict?

Show answer and reasoning

Symmetric sharing. Equal electronegativities imply no preferred direction of shared density in this model.

Original written challenge

4 points · self-check · not an official AP question

Compare A–B with χ(A)=2.0, χ(B)=3.0 and C–C with χ(C)=2.5. Compute both differences, label partial charges and state a limit of your conclusion.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: A–B has Δχ = 1.0.
  2. 1 point: A is δ+ and B is δ−.
  3. 1 point: C–C has Δχ = 0 and no permanent bond dipole.
  4. 1 point: Bond polarity alone does not determine a whole molecule’s polarity.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1What does δ mean?

A partial charge from unequal electron sharing.

RECALL 2Which atom is δ−?

The more electronegative bonded atom.

RECALL 3Why avoid a universal ionic cutoff?

Bonding lies on a continuum and material properties also matter.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Which end of a bond attracts electrons more strongly?

  • Δχ =
  • χB − χA
  • ; no units.
  • The higher-χ atom carries δ− in the bond.

Remember: A partial negative charge and an integer ionic charge are different descriptions.

Conditions: Atom A has χ = 2.2. The electron-density marker indicates only the direction of unequal sharing; it is not a calculated charge or electron position.

Refresh Kid · AP Chemistry Unit 2 · Objectives 2.1.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 2.1, objectives 2.1.A. CED effective Fall 2024, current official file checked September 16, 2026, together with the published clarifications. This is Unit 2: Compound Structure and Properties, Topics 2.1–2.7. The focused lesson breakdown is Refresh Kid’s editorial sequence. Models and original practice are teaching materials, not official AP questions. Numerical potential curves, ion comparisons and orbital-alignment indices state their approximations. Five- and six-domain shapes are included; d-orbital hybridization and molecular-orbital diagrams are not required here. GitHub’s 3D website examples, including the Three.js Mars camera-control example, informed the use of rotatable scenes. Our scientific geometry and viewer code are original; no repository artwork or tutorial code was copied. The self-hosted Three.js library retains its MIT license. Camera rotation does not alter chemistry. See also the official clarifications.

Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.

Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.

Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.

OPTIONAL LIVE SUPPORT

Want to work through this with a tutor?

Bring your question about Which end of a bond attracts electrons more strongly? Your explanation and answers remain free to access.

Request a chemistry tutor →Ask about this lesson on WhatsAppThe team can confirm teacher availability and next steps.