Chemical Bonding and Molecular Structure · TOPIC LESSON
Bond Energies and Potential Curves
Read potential wells, force and vibration correctly, then connect them to bond parameters and reaction-energy estimates.
JEE Advanced · NEET UG50 min study guide9 indexed sections
BY THE END, YOU CAN
Interpret energy-zero, minimum, well-depth and force signs.
Apply harmonic and Morse models within stated limits.
Distinguish D_e, D₀, stiffness and thermal bond enthalpy.
Explain bond length, angle, order and dipole with appropriate comparisons.
Estimate gas-phase reaction enthalpy from supplied bond energies.
BEFORE YOU START
Differentiation and simple exponentials.
Energy conservation, vectors and unit conversion.
Lewis structures, geometry and chemical equations.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Choose the energy zero before reading a bond curve
A diatomic potential-energy curve plots energy against internuclear separation r for a specified electronic state. A common convention sets U(∞)=0 for separated atoms. A bound state then lies in a negative-energy well. The negative potential energy means stabilization relative to that reference, not a negative amount of energy required to break the bond.
At large separation the interaction approaches the chosen zero. At intermediate separation attraction lowers the energy. At very small separation strong repulsive effects involving overlapping electron distributions and nuclear repulsion raise it steeply. A stable equilibrium separation r_e occurs at a local minimum.
The well depth D_e is the positive difference between the minimum and the separated-atom limit: D_e=U(∞)−U(r_e). If the minimum is −4.0 eV and the limit is zero, the well depth is +4.0 eV. Changing the plotted energy zero shifts the entire curve but does not change forces or energy differences.
A curve is not a literal orbit followed by atoms and is not a graph of energy versus time. The nuclei can vibrate on the potential, and the allowed quantum states depend on both the curve and their masses. Different electronic states can have different curves, so one drawing is not every possible state of a molecule.
ORIGINAL PRACTICE · 1
A potential minimum is −300 kJ mol⁻¹ with separated atoms at zero. Is the classical well depth +300 or −300 kJ mol⁻¹?
Reveal answer & reasoning
+300 kJ mol⁻¹.
D_e=0−(−300); energy must be supplied to reach the separated-atom reference.
02 / UNDERSTAND & APPLY
2. Force is minus the slope, not the energy itself
For a one-dimensional separation coordinate, the generalized radial force is F_r=−dU/dr, with positive direction chosen toward increasing separation. On the compressed side of a usual bonding minimum, U decreases as r increases, so its slope is negative and the force is positive: the atoms are pushed apart.
On the stretched side, U increases toward its asymptote as r increases, so the slope is positive and the force is negative: attraction tends to reduce separation. At the minimum dU/dr=0, so the net force is zero even though the attractive and repulsive contributions individually need not vanish.
A stationary point is stable when small displacements lead to restoring forces; mathematically d²U/dr²>0 at a smooth minimum. A maximum has negative curvature and is unstable along that coordinate. Zero first derivative alone is therefore not enough to establish stability.
Energy and force have different units. If U is energy per pair in joules and r is metres, the derivative gives newtons. If U is in kJ mol⁻¹ and r in picometres, the numerical slope must be converted from molar energy and length before it can be interpreted as force on one pair.
ORIGINAL PRACTICE · 1
On a stretched-bond branch where dU/dr>0, what is the sign of radial force under the increasing-r convention?
Reveal answer & reasoning
Negative.
F_r=−dU/dr, so the force tends to reduce the separation.
03 / UNDERSTAND & APPLY
3. Curvature controls small-amplitude vibration
Near a minimum, Taylor expansion gives U(r)≈U(r_e)+½k(r−r_e)², where k=(d²U/dr²) at r_e is the force constant. The linear term vanishes because the slope at equilibrium is zero. A larger k means a steeper, narrower local well and greater resistance to a small displacement.
Differentiation gives F_r≈−k(r−r_e), the restoring-force law. With k in N m⁻¹ and displacement in m, the force is in N. This approximation is local; extending a parabola to arbitrary separation would predict infinite dissociation energy, which is not physically suitable for a real breakable bond.
For a diatomic harmonic oscillator, angular frequency ω=√(k/μ), with reduced mass μ=m₁m₂/(m₁+m₂). Ordinary frequency is ν=ω/(2π). Increasing reduced mass at fixed force constant lowers vibrational frequency; isotope substitution can therefore shift spectra without a large change in electronic bonding.
Well depth and curvature are related only through a specified potential model. Two different curves can have the same depth and different stiffness, or the same stiffness and different depths. A bond dissociation energy alone does not uniquely determine its vibrational frequency.
ORIGINAL WORKED EXAMPLE
Restoring force with SI units
A model bond has k=200 N m⁻¹ and is stretched by 5.0 pm from equilibrium. Find the harmonic restoring force.
Convert displacement: 5.0 pm=5.0×10⁻¹² m.
F=−kΔr=−200×5.0×10⁻¹²=−1.0×10⁻⁹ N.
The negative sign indicates a force toward smaller separation.
Answer −1.0×10⁻⁹ N in the increasing-r direction convention.
ORIGINAL PRACTICE · 1
If reduced mass doubles while k is unchanged, how does harmonic frequency change?
Reveal answer & reasoning
It becomes 1/√2 of the original frequency.
ν is proportional to μ⁻¹/² when the force constant is fixed.
04 / UNDERSTAND & APPLY
4. An anharmonic model can approach a dissociation limit
A useful model is U(r)=D_e[(1−e^(−a(r−r_e)))²−1], with D_e>0 and a>0. This version sets separated atoms at zero and the minimum at −D_e. The parameter a has inverse-length units, so the exponent is dimensionless. Other books add D_e to the whole expression and put the minimum at zero; that is a reference change, not a different force.
The repulsive side rises steeply while the stretched side approaches zero, unlike a symmetric harmonic parabola. Differentiation gives F_r=2D_e a[e^(−2a(r−r_e))−e^(−a(r−r_e))]. Near the minimum k=2D_ea² when D_e is energy per pair and consistent length units are used.
Anharmonicity means the potential is not exactly quadratic. In a quantum treatment, high vibrational levels become more closely spaced as dissociation is approached, rather than remaining equally spaced forever. The Morse model is useful but not exact, and its far-distance tail does not reproduce every real interaction law.
The plotted curve in this lesson uses explicitly hypothetical parameters. It illustrates minima, force directions, zero choice and finite dissociation; it is not a measured potential for a named molecule. Real potential-energy surfaces for polyatomic molecules involve multiple bond lengths and angles, not just one coordinate.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Original calculated hypothetical curve, not molecular measurement. Energy is per pair in electronvolts; positive radial force means increasing separation. No quantum levels are inferred from these parameters.
ORIGINAL WORKED EXAMPLE
Read a model energy algebraically
For the stated Morse form, find U when e^(−a(r−r_e))=1/2. If D_e=4.0 eV, what is the energy relative to separated atoms?
Substitute one half: U=D_e[(1−1/2)²−1].
The bracket is 1/4−1=−3/4.
U=−0.75×4.0=−3.0 eV. The system is 1.0 eV above the well minimum but 3.0 eV below the asymptote.
Answer U=−3D_e/4=−3.0 eV for the supplied depth.
ORIGINAL PRACTICE · 1
Why is a harmonic parabola unsuitable for predicting finite bond dissociation at very large r?
Reveal answer & reasoning
It rises without bound as separation increases.
A breakable molecular bond instead approaches a finite separated-fragment energy limit in the relevant dissociation channel.
05 / UNDERSTAND & APPLY
5. D_e and D_0 refer to different starting levels
Even the lowest quantum vibrational state lies above the classical minimum. In the harmonic approximation its zero-point energy is E_ZPE=½hν. Therefore the dissociation energy from the lowest vibrational level, D₀, is smaller than the well depth D_e: D₀=D_e−E_ZPE in the simple one-channel picture.
The difference matters when comparing a potential minimum with a spectroscopic dissociation measurement. A molecule in an excited vibrational level needs still less additional energy to reach the same dissociation limit, although selection rules and kinetics affect how it gets there.
A tabulated bond dissociation enthalpy at a stated temperature is not automatically identical to either D_e or D₀. Thermal contributions, zero-point energies, pressure conventions and the specified fragments matter. Keep the symbol and physical process attached to the number.
Useful conversions include 1 eV per particle=96.485 kJ mol⁻¹ and 1 pm=10⁻¹² m. Multiplying by Avogadro’s constant converts per-particle energy to molar energy; it does not change whether the process is bond breaking or bond formation.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Independent hypothetical level illustration. The ground-level energy is supplied, not calculated from the preceding Morse parameters because no reduced mass is specified.
ORIGINAL WORKED EXAMPLE
Separate well depth, ground-state dissociation and units
A hypothetical curve has D_e=4.0 eV and zero-point energy 0.10 eV above its minimum. Find D₀ and its approximate molar equivalent using 96.485 kJ mol⁻¹ per eV.
D₀=4.0−0.10=3.90 eV.
Multiply 3.90×96.485=376.2915 kJ mol⁻¹.
This conversion concerns the stated ground-state dissociation energy, not a separately measured room-temperature enthalpy.
Answer D₀=3.90 eV, approximately 376 kJ mol⁻¹.
ORIGINAL PRACTICE · 1
Can shifting U(r) upward by a constant alter F_r?
Reveal answer & reasoning
No.
The derivative of a constant is zero; force and energy differences are unchanged.
06 / UNDERSTAND & APPLY
6. Bond parameters describe length, angle, order, polarity and strength
Bond length is an equilibrium or experimentally averaged internuclear distance, depending on the measurement. Bond angle describes the angle between bonds sharing an atom. A double bond between the same pair of elements is usually shorter and stronger than a comparable single bond, and a triple bond usually shorter still, but different chemical environments can complicate simple rankings.
Bond order describes bonding multiplicity in a model. Lewis structures use single, double and triple bonds; delocalized systems can have fractional average bond orders. Equivalent bonds in carbonate do not alternate permanently between one long single and one short double bond. Resonance structures are contributors to one delocalized state, not molecules rapidly switching drawings.
Bond enthalpy or dissociation energy quantifies a specified breaking process, while a force constant describes local stiffness. Bond dipole describes charge separation and contributes vectorially to molecular dipole. A short bond, a strong bond and a highly polar bond are related concepts only in context; none is a universal synonym for the others.
Angles depend on electron-domain geometry, lone pairs, multiple bonds and molecular environment. Tetrahedral, trigonal-planar and linear reference angles are about 109.5°, 120° and 180°, but real molecules can deviate. A measured molecular dipole combines geometry and charge distribution, so polar bonds can cancel in a symmetric molecule.
ORIGINAL WORKED EXAMPLE
Vector addition of two equal bond dipoles
A planar model has two equal bond-dipole vectors of magnitude 1.0 D separated by 120°. Find their resultant magnitude.
For two equal vectors, μ_result=2μ cos(θ/2).
With θ=120°, cos60°=0.5.
μ_result=2×1.0×0.5=1.0 D; its direction bisects the two vectors.
Answer 1.0 D along the angle bisector in the stated vector model.
ORIGINAL PRACTICE · 1
Why can a linear molecule with two equal opposing polar bonds have zero molecular dipole?
Reveal answer & reasoning
The bond-dipole vectors cancel.
Molecular polarity depends on the vector sum, not just the presence of polar bonds.
07 / UNDERSTAND & APPLY
7. Bond-energy estimates use broken minus formed
Breaking a bond into the specified gas-phase fragments requires energy; forming the same bond from those fragments releases it. A common estimate is ΔH_rxn≈ΣD(bonds broken)−ΣD(bonds formed). A negative result indicates that forming product bonds releases more energy than breaking reactant bonds requires.
Mean bond enthalpies are averages over different molecular environments. They are useful for estimates, but a C–H bond in every molecule does not have precisely the same dissociation enthalpy. Successive bond removals from one molecule can also differ because the fragments change after each step.
The usual tabulated calculation assumes gas-phase molecules. If a reactant or product is liquid or solid in the stated reaction, phase-change terms may be needed. Ionic lattice energy is a collective solid-state quantity and should not be inserted as if it were one ordinary covalent bond in a molecule.
Atom balance comes first. Count only the bonds whose net changes are represented, or count all bonds consistently on both sides. A stoichiometric coefficient multiplies the number of molecules and their bond counts. The method is a Hess-law energy cycle, not a literal claim that the reaction mechanism first breaks every bond at once.
ORIGINAL WORKED EXAMPLE
Estimate a gas-phase reaction enthalpy
For H₂(g)+Br₂(g) → 2HBr(g), use supplied bond enthalpies H–H=436, Br–Br=193 and H–Br=366 kJ mol⁻¹. Estimate ΔH.
Break one H–H and one Br–Br: 436+193=629 kJ mol⁻¹ of reaction.
Form two H–Br bonds: 2×366=732 kJ mol⁻¹ released.
ΔH≈629−732=−103 kJ mol⁻¹. A liquid Br₂ reactant would require an additional phase correction.
Answer Approximately −103 kJ mol⁻¹ for the stated gas-phase reaction.
ORIGINAL PRACTICE · 1
Does a stronger product bond contribute a positive or negative term to the broken-minus-formed estimate?
Reveal answer & reasoning
A more negative contribution through the subtracted formation term.
Bond formation releases energy; the positive dissociation magnitude is subtracted.
08 / UNDERSTAND & APPLY
8. Avoid turning a trend into a universal identity
Higher bond order often shortens bonds for a comparable atom pair, but comparing different elements also changes atomic size, orbital overlap and electron repulsion. A short bond across one pair cannot automatically be ranked against a longer bond across a different pair using length alone.
A deeper well means greater classical separation energy from its minimum, while a steeper curvature means greater small-displacement stiffness. These features can vary independently in a general family of curves. In the Morse model they are linked by the additional parameter a, demonstrating why D_e alone is insufficient.
The minimum marks zero net force, not zero potential energy, zero kinetic energy or absence of quantum motion. At finite temperature, vibration samples a range of separations; anharmonicity can make average separation differ from r_e.
For multiple dissociation channels or changing electronic states, a one-curve diagram may not capture the full process. The school-level curve is powerful when its coordinate, state and energy reference are explicitly stated.
ORIGINAL PRACTICE · 1
Two Morse models have the same D_e but the second has twice the parameter a. How do their near-minimum force constants compare?
Reveal answer & reasoning
The second has four times the force constant.
k=2D_ea², so doubling a at fixed D_e multiplies k by four.
09 / UNDERSTAND & APPLY
9. Read the graph before calculating
Mark the separated-fragment reference, minimum location, well depth and any vibrational level. Determine force from slope and stability from curvature. Then decide whether the requested quantity is D_e, D₀, a finite-temperature enthalpy, stiffness or molecular geometry.
For numerical work, distinguish energy per pair from energy per mole and picometres from metres. For reaction estimates, balance atoms, count broken and formed bonds, preserve signs and account for phases.
All plotted parameters and numerical exercises here are supplied original models. They illustrate how to reason from curves and bond parameters rather than pretending to provide measured data for an unspecified molecule.
KEEP THE BIG IDEAS
Quick revision
A negative potential minimum gives a positive dissociation well depth.
Force is minus the potential slope; curvature determines local stability.
Harmonic stiffness is a local property, not complete dissociation energy.
Bond parameters must retain their definitions and chemical context.
Broken-minus-formed bond energies estimate a specified reaction and phase.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.