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How does a potential-energy curve describe a covalent bond, and what controls bond length and bond energy?

Topic 2.2 Intramolecular Force and Potential Energy: interpret a potential-energy versus internuclear-distance curve to define bond length and bond energy, and explain how bond order, atomic size and charge affect bond strength.

A focused answer to AP Chemistry Topic 2.2, covering the potential-energy versus internuclear-distance curve, equilibrium bond length, bond energy, and how bond order, atomic radius and ionic charge control bond strength, with full worked reasoning.

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  1. What this topic is asking
  2. The potential-energy curve
  3. Bond length and bond energy
  4. What controls bond strength
  5. Try this

What this topic is asking

The College Board (Topic 2.2) wants you to read a potential-energy versus internuclear-distance curve for a bond, to define bond length and bond energy from it, and to explain how three factors, bond order, atomic size and ionic charge, control how strong a bond is. The unifying tool is Coulomb's law.

The potential-energy curve

The shape encodes a balance of two Coulombic effects. As atoms approach, the attraction between each nucleus and the other atom's electrons lowers the energy. But if they get too close, the like-charged nuclei (and inner electrons) repel strongly, raising the energy sharply. The minimum is the distance where these balance, the most stable arrangement, and that is where the bond sits.

Bond length and bond energy

Bond length is the equilibrium internuclear distance, and bond energy is the energy needed to break one mole of those bonds. The two are linked: a stronger bond is generally a shorter bond, because a deeper potential well pulls the nuclei closer. So if you know one quantity's trend you can usually predict the other.

What controls bond strength

Three factors recur:

  • Bond order (covalent): triple > double > single in strength, and triple < double < single in length. More shared pairs means more electron density between the nuclei and a stronger attraction.
  • Atomic size: smaller atoms bond at shorter distances, so their nuclei attract the shared electrons more strongly (a shorter bond is generally stronger).
  • Ionic charge and size (ionic): by Coulomb's law, Fq1q2r2F \propto \dfrac{q_1 q_2}{r^2}, larger ionic charges and smaller ions (shorter rr) give a stronger attraction and a higher lattice energy.

The thread through all three is Coulomb's law: bond strength grows with the magnitude of the interacting charges and shrinks with distance. This is why NN\text{N}\equiv\text{N} (bond order 3) is one of the strongest bonds known, why MgO\text{MgO} (2+2+ and 22- ions) has a far higher lattice energy than NaCl\text{NaCl} (1+1+ and 11-), and why bond and lattice energies underpin the energetics you meet later in thermochemistry. Reading the potential-energy curve is the qualitative version of this same idea, and the College Board often pairs a curve with a "compare and explain" prompt.

Try this

Q1. State what the depth of the minimum on a potential-energy curve represents. [1 point]

  • Cue. The bond energy (the energy required to break the bond).

Q2. Predict which has the stronger interaction, NaF\text{NaF} or MgO\text{MgO}, and justify with Coulomb's law. [2 points]

  • Cue. MgO\text{MgO}; its ions carry charges of 2+2+ and 22- versus 1+1+ and 11- for NaF\text{NaF}, and the larger charges give a stronger Coulombic attraction.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)4 marksSection II (short FRQ). (a) On a potential-energy versus internuclear-distance graph for a diatomic molecule, identify what the minimum point represents. (b) Compare the bond length and bond energy of a C-C single bond and a C=C double bond. (c) Explain the comparison in terms of bond order. (d) Predict which bond, NN\text{N}\equiv\text{N} or O=O\text{O}=\text{O}, is stronger and why.
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A 4-point FRQ on potential-energy curves and bond strength.

(a) Minimum (1 point): the lowest point gives the equilibrium bond length (the internuclear distance of minimum potential energy), and its depth below zero is the bond energy.
(b) Compare (1 point): the C=C double bond is shorter and stronger (higher bond energy) than the C-C single bond.
(c) Explain (1 point): a double bond has a higher bond order (2 versus 1), meaning more shared electron pairs and greater electrostatic attraction pulling the nuclei closer, so it is shorter and stronger.
(d) Predict (1 point): NN\text{N}\equiv\text{N} is stronger because it has a higher bond order (triple, 3) than O=O\text{O}=\text{O} (double, 2), so more shared pairs and a stronger attraction.

Markers reward identifying the minimum as equilibrium bond length and bond energy, the length-strength comparison, and a bond-order justification.

AP 2021 (style)1 marksSection I (multiple choice). On a potential-energy versus internuclear-distance curve, what happens to potential energy as two atoms are pushed closer than the equilibrium bond length? (A) it keeps decreasing (B) it stays constant (C) it rises sharply (D) it becomes zero. Justify your choice.
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A 1-point conceptual MCQ. The answer is (C).

At distances shorter than the equilibrium bond length, the positively charged nuclei (and the inner electrons) repel strongly, so the potential energy rises steeply. The minimum of the curve is the equilibrium distance; moving closer increases repulsion, while moving farther apart weakens the attraction and also raises the energy back toward zero.

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