AP Chem 2.2: Intramolecular Force and Potential Energy

Why do two atoms stop at a particular distance?

In lesson 2.4.2 you dragged two hydrogen atoms together: the energy dropped as they approached, reached its lowest point at 74 pm, and shot up when they got closer still. That potential energy curve势能曲线 is the central picture of AP topic 2.2. Expect to read one, sketch one, and decide which of several curves on the same axes belongs to which molecule.

The three regions of the curve

The horizontal axis is the distance between the two nuclei, and the vertical axis is the potential energy of the system. Zero is defined as the two atoms being far apart, with no interaction.

Far apart (right side): the atoms don’t affect each other, and the energy is close to 0.
Moving closer: each nucleus starts attracting the other atom’s electrons; attraction outweighs repulsion, and the energy drops below 0.
The minimum: attraction and repulsion balance, and the system is most stable. The distance at the minimum is the bond length键长 (the equilibrium internuclear distance). How far the minimum sits below 0 is the bond energy键能: the energy needed to pull the atoms apart again, which equals the energy released when the bond forms.
Too close (left side): repulsion between the two nuclei (and between the two electron clouds) grows rapidly, and the energy shoots up.

For $\ce{H2}$ , the minimum is at 74 pm and −436 kJ/mol, so the H–H bond length is 74 pm and the bond energy 436 kJ/mol.

The shape of the curve also tells you the net force between the atoms. Farther apart than the bond length, separating the atoms further would raise the energy, so the net force is attractive and pulls them together. Closer than the bond length, moving them closer would raise the energy, so the net force is repulsive and pushes them apart. At the bond length, attraction and repulsion balance and the net force is zero. Judge by which way the curve rises, not by whether the energy is positive or negative: a little to the left of the bond length the energy is still below 0, yet the net force is already repulsive.

Bond order: more shared electrons, shorter and stronger bonds

Bond	C–C	C=C	C≡C	N–N	N=N	N≡N
Bond order	1	2	3	1	2	3
Length (pm)	154	134	120	147	124	110
Energy (kJ/mol)	348	614	839	163	418	941

A higher bond order means more shared electrons, each attracted by both nuclei, pulling the nuclei closer together and making them harder to separate: a shorter bond, with the minimum farther left, and a stronger bond, with a deeper minimum. $\ce{N2}$ (triple bond, 110 pm, 941 kJ/mol), $\ce{O2}$ (double bond, 121 pm, 495 kJ/mol), and $\ce{F2}$ (single bond, 143 pm, 155 kJ/mol) follow the same pattern.

Atom size: larger atoms, longer and usually weaker bonds

Bond	H–F	H–Cl	H–Br	H–I
Length (pm)	92	127	141	161
Energy (kJ/mol)	567	431	366	299

From F to I the halogen atom gets larger, the bonding electrons are farther from the halogen’s nucleus, and they’re attracted less strongly: a longer bond, with the minimum farther right, and a weaker bond, with a shallower minimum.

H is the smallest atom, so H–H is the shortest bond (74 pm). Replace one H with the much larger Cl and the bond grows to 127 pm (H–Cl); with two Cl atoms it is 199 pm (Cl–Cl).

The halogens themselves follow the same trend with one exception. Cl–Cl (242), Br–Br (193), and I–I (151 kJ/mol) get weaker in turn, but F–F is only 155 kJ/mol, weaker than Cl–Cl. Fluorine atoms are so small that the lone pairs on the two atoms crowd each other, and their repulsion weakens the F–F bond.

Between ions: Coulomb’s law

The interaction between a cation and an anion can also be understood with Coulomb’s law (lesson 2.4.1):

E \propto \frac{q_1 q_2}{r}

Larger ion charges mean a stronger interaction. The charge product for $\ce{Mg^2+}$ and $\ce{O^2-}$ is 4 times that for $\ce{Na+}$ and $\ce{Cl-}$ .
Smaller ions sit closer together, so they interact more strongly. That’s why the ionic bonding in NaF is stronger than in NaI.

U2.3 uses this to compare the lattice energies and melting points of ionic solids.

Compare a few sets of curves first, then switch to “Match the curves” and pair each curve with its bond using the position and depth of its minimum:

Try it: compare several potential-energy curves

The position (bond length) and depth (bond energy) of each minimum are data; the shape of each curve is schematic. Zero energy means the atoms are far apart, with no interaction.

Bond lengths and energies: average values from Brown et al., Chemistry: The Central Science (bonds.ts).

Example 1: Matching curves to molecules

Three potential-energy curves, A, B, and C, have minima at A (110 pm, −941 kJ/mol), B (121 pm, −495 kJ/mol), and C (143 pm, −155 kJ/mol). Which belongs to $\ce{N2}$ , $\ce{O2}$ , and $\ce{F2}$ ? Justify your answer.

Step-by-step solution

Identify: the two readings on each curve. The x-coordinate of the minimum is the bond length, and the depth of the minimum is the bond energy (the energy needed to break 1 mol of bonds):
- A: 110 pm, 941 kJ/mol
- B: 121 pm, 495 kJ/mol
- C: 143 pm, 155 kJ/mol
Identify: the bond order of each molecule. From the Lewis diagrams: $\ce{N2}$ has a triple bond (N≡N), $\ce{O2}$ a double bond (O=O), and $\ce{F2}$ a single bond (F–F).
Compare: how bond order relates to length and energy. The higher the bond order, the shorter the bond (minimum farther left) and the stronger it is (deeper minimum):
$\begin{aligned} \text{length} &: \text{A} < \text{B} < \text{C} \\ \text{energy} &: \text{A} > \text{B} > \text{C} \end{aligned}$
Match them.
- A (shortest, deepest) → $\ce{N2}$
- B (middle) → $\ce{O2}$
- C (longest, shallowest) → $\ce{F2}$
Connect to the reason. The more electrons shared between two nuclei, the more strongly both nuclei are attracted, so they are pulled closer (shorter bond) and are harder to separate (larger bond energy).

Example 2: How do the curves for H–Cl and H–Br differ?

If the potential-energy curves for H–Cl and H–Br are drawn on the same axes, how do they differ? Explain in terms of atomic structure.

Step-by-step solution

Identify: what differs between the two bonds. Both contain H; the difference is the halogen. Br has one more electron shell than Cl, so it has a larger atomic radius.
Compare the position of the minimum. The bonding electrons are farther from the Br nucleus, so the minimum lies farther right (a longer bond, in pm):
$\begin{aligned} \text{H–Br} &: 141 \\ \text{H–Cl} &: 127 \end{aligned}$
Compare the depth of the minimum. Electrons farther from the nucleus are attracted less strongly, so the minimum is shallower (a smaller bond energy, in kJ/mol):
$\begin{aligned} \text{H–Br} &: 366 \\ \text{H–Cl} &: 431 \end{aligned}$
Connect to atomic structure. Larger Br atom → bonding electrons farther from the nucleus → weaker attraction → longer, weaker bond.
What the two curves share.
- Right end: the atoms are far apart and don’t interact, so both curves approach 0.
- Left end: the nuclei are pushed too close and repel, so both curves rise steeply.

Practice: which bond has the deeper potential-energy curve (larger bond energy)?

C–C or C=C
N≡N or N=N
H–F or H–I
Cl–Cl or Br–Br
F–F or Cl–Cl
O=O or F–F
H–Cl or H–Br
I–I or Br–Br
N–N or N≡N
H–H or Cl–Cl

Common mistakes

Mistake 1: “The higher the minimum, the stronger the bond.” Lower energy means more stable. The deeper (more negative) the minimum, the more energy it takes to break the bond, so the stronger it is.

Mistake 2: “The closer two atoms get, the more stable they are.” Too close, and the repulsion between the nuclei grows rapidly and the energy shoots up. The most stable distance is where attraction and repulsion balance: the bond length.

Mistake 3: “Forming a bond absorbs energy.” It’s the reverse: as two atoms move from far apart down to the minimum, the energy falls, so forming a bond releases energy. Breaking a bond absorbs energy.

Quick check

The minimum of a diatomic molecule's potential-energy curve is at 127 pm and −431 kJ/mol. Which statement is correct?

The N≡N bond is shorter than the N–N bond. Which is the best explanation?

Estimating the separation of each pair of ions from the sum of their radii, which pair has the strongest Coulombic attraction, and why?

The minimum of the potential-energy curve for $\ce{H2}$ is at 74 pm. When the nuclei of the two H atoms are 50 pm apart, which statement is correct?

Each diagram shows two potential-energy curves: the solid curve is $\ce{H2}$ and the dashed curve is $\ce{Cl2}$ . Which diagram is most reasonable?

AP-style free response

The potential-energy curve for $\ce{H2}$ has its minimum at 74 pm and −436 kJ/mol; when the two hydrogen atoms are 300 pm apart, the potential energy is close to 0.

(a) Use the curve to give the H–H bond length and bond energy.

(b) Explain why the potential energy rises steeply when the internuclear distance is less than 74 pm.

(c) The Cl–Cl bond length is 199 pm and its bond energy 242 kJ/mol. Describe how the curve for $\ce{Cl2}$ would look on the same axes, and explain the difference from $\ce{H2}$ in terms of atomic structure.

(d) Explain why the potential energy is close to 0 when the hydrogen atoms are 300 pm apart.

Model answer and scoring

(a) Bond length 74 pm (the position of the minimum); bond energy 436 kJ/mol (how far the minimum lies below 0).

(b) When the atoms get too close, repulsion between the two positively charged nuclei (and between the electron clouds) grows rapidly and outweighs the attraction of the nuclei for the electrons, so the potential energy rises steeply.

(c) The $\ce{Cl2}$ minimum lies farther right, at 199 pm, and is shallower (−242 kJ/mol). Chlorine atoms are much larger than hydrogen atoms (two more electron shells), so the bonding electrons are farther from both nuclei and attracted less strongly, giving a longer, weaker bond.

(d) At 300 pm the atoms barely interact, with no significant attraction or repulsion, and the zero of potential energy is defined as the atoms being far apart with no interaction.

Scoring: (a) both values, with where they appear on the graph; (b) “repulsion between the nuclei”; (c) both “farther right” and “shallower”, explained by larger atoms with electrons farther from the nuclei; (d) “almost no interaction” and the definition of zero.

Summary

Potential-energy curve: on the right the atoms are far apart, at energy 0; as they approach, attraction wins and the energy falls; the position of the minimum is the bond length and its depth the bond energy; too close, repulsion between the nuclei makes the energy shoot up.
Farther apart than the bond length, the net force is attractive; closer, it is repulsive; at the bond length, the two balance.
Higher bond order means shorter, stronger bonds (minimum farther left and deeper).
Larger atoms mean longer, usually weaker bonds (minimum farther right and shallower); F–F being weaker than Cl–Cl is the exception.
Between ions, $E \propto q_1 q_2 / r$ : larger charges and smaller ions mean stronger interactions.

AP tip: Potential-energy-curve questions typically ask you to mark the bond length and bond energy on a graph, decide which curve belongs to which molecule (from the position and depth of the minimum), or explain why the left side rises steeply (repulsion between the nuclei). Explain bond length and bond energy with bond order (how many electrons are shared) and atom size (how far the electrons are from the nucleus).

Intramolecular Force and Potential Energy分子内作用力与势能曲线

Learning goals

Before you start

Why do two atoms stop at a particular distance?

The three regions of the curve

Bond order: more shared electrons, shorter and stronger bonds

Atom size: larger atoms, longer and usually weaker bonds

Between ions: Coulomb’s law

Example 1: Matching curves to molecules

Example 2: How do the curves for H–Cl and H–Br differ?

Common mistakes

Quick check

AP-style free response

Summary

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