Covalent bonding — bond length, bond energy, bond order: Visual walkthrough

We use hydrogen, $H_{2}$ : the simplest possible bond — two protons, two electrons. Everything generalises.

Step 1 — Set up the one number we care about: separation $r$

WHAT. Put two hydrogen atoms in empty space. The only thing that matters is how far apart their nuclei sit. Call that distance $r$ (for radius of separation), measured in picometres (pm): $1 pm = 1 0^{- 12}$ m, a trillionth of a metre.

WHY $r$ and nothing else? Because a bond is a tug-of-war along the line joining two nuclei. Sideways motion, rotation — none of it changes whether they bind. The whole story is one number ( $r$ ) versus one number (the total energy $E$ ). That is why we can draw the entire physics on a flat 2-D graph.

PICTURE. Two nuclei on a ruler. $r$ is just the ruler reading between them.

Figure — Covalent bonding — bond length, bond energy, bond order

Step 2 — Force one: the nuclei shove each other away

WHAT. Each nucleus is a proton: a positive charge. Two positives repel. This repulsion energy grows as they get closer.

WHY this shape? Two charges $q_{1}, q_{2}$ a distance $r$ apart store electrical potential energy

$E_{repel} = + k \frac{q _{1} q _{2}}{r}$

$q_{1} q_{2}$ — both positive, so their product is positive → energy is positive (costs energy to hold them near).
$k$ — a fixed constant of nature (Coulomb's constant); it just sets the units.
$\frac{1}{r}$ — the crucial part: as $r \to 0$ , $\frac{1}{r} \to \infty$ . The repulsion blows up when the nuclei nearly touch.

The tool here is the Coulomb $1/ r$ law, chosen because it is the rule for how much energy two charges store — nothing exotic, just static electricity.

PICTURE. A curve that is huge on the left (small $r$ ) and sinks toward zero on the right (large $r$ ), always above the axis.

Step 3 — Force two: the shared electron glue pulls inward

WHAT. When the two atomic orbitals overlap in phase (their electron waves add up, not cancel), extra electron density piles up between the nuclei. That negative puddle sits between two positive nuclei and yanks both toward the middle — an attractive, energy-lowering effect.

WHY does density pile up in the middle? An electron is described by a wave $ψ$ . When two atoms' waves meet crest-to-crest (in phase), they add. The chance of finding the electron somewhere is $∣ ψ ∣^{2}$ , and in the middle region the added waves make $∣ ψ ∣^{2}$ bigger. More negative charge exactly where it can pull on both protons. (This constructive overlap is the heart of Sigma and Pi Bonds and Molecular Orbital Theory.)

WHY is attraction energy negative? Lowering energy = more stable. Pulling opposite charges together releases energy, so it sits below the zero line. It is strongest (deepest) at moderate $r$ and fades to zero at large $r$ .

PICTURE. A curve that stays below the axis, dipping down, then climbing back toward zero on both far sides.

Step 4 — Add the two curves: the well appears

WHAT. The real energy is repulsion (Step 2, positive) plus attraction (Step 3, negative), added point-by-point at every $r$ :

$E (r) = up on the left E_{repel} (r) + down in the middle E_{attract} (r)$

WHY does adding them make a dip?

At large $r$ : repulsion has already faded (its $1/ r$ is small), attraction still reaches out → the sum is negative → the curve falls as atoms approach. This is the atoms "wanting" to bond.
At small $r$ : nuclear repulsion's $1/ r$ explodes and overwhelms everything → the sum shoots up.
In between the two effects trade places, so the sum must pass through a lowest point — a valley, or potential well.

PICTURE. Grey repulsion curve + cyan attraction curve → their amber sum, showing a clear valley.

Step 5 — Give the valley a formula: the Morse curve

WHAT. We fit the valley with a compact, well-tested model — the Morse potential:

$E (r) = D_{e} (1 - e^{- a (r - r_{e})})^{2} - D_{e}$

Read each piece:

$r$ — the separation (our x-axis).
$r_{e}$ — a constant, the separation where the valley bottoms out (we'll prove this next). The "e" stands for equilibrium.
$D_{e}$ — a constant, the depth of the valley (positive number of energy).
$a$ — a constant that sets how steep/narrow the walls are.
$e^{- a (r - r_{e})}$ — the exponential is chosen because it naturally does two jobs at once: it explodes when $r < r_{e}$ (steep repulsive wall) and dies to zero when $r ≫ r_{e}$ (flat, no interaction far away). No polynomial does both so cleanly — that is why Morse used an exponential.
$(1 - \dots)^{2}$ — the square keeps the whole first term $\geq 0$ , so subtracting $D_{e}$ makes the lowest possible value exactly $- D_{e}$ .
$- D_{e}$ at the end — shifts the whole curve down so the valley bottom sits at $- D_{e}$ and the far-away value sits at $0$ (matching our chosen zero from Step 1).

WHY a model at all? Because with a formula we can use one tool — the derivative — to locate the bottom exactly, instead of eyeballing the graph.

PICTURE. The Morse curve with $r_{e}$ , $D_{e}$ , and the zero line all labelled.

Step 6 — Find the bottom: where the slope is zero

WHAT. The bottom of any valley is the flat spot — where the slope (the derivative $\frac{d E}{d r}$ , "how fast energy changes as we nudge $r$ ") equals zero.

WHY the derivative? The derivative is the slope of the curve at each point. Downhill → negative slope; uphill → positive slope; the turning point between them → slope exactly $0$ . So "solve $\frac{d E}{d r} = 0$ " is literally "find where the curve stops falling and starts rising." That is the tool built for this exact question.

Differentiate the Morse formula (chain rule):

$\frac{d E}{d r} = 2 D_{e} (1 - e^{- a (r - r_{e})}) \cdot a e^{- a (r - r_{e})}$

Term by term:

$2 D_{e} a$ — positive constants, never zero, so they can't kill the slope.
$e^{- a (r - r_{e})}$ — an exponential is never zero either.
$(1 - e^{- a (r - r_{e})})$ — this is the only piece that can hit zero.

Set it to zero:

$1 - e^{- a (r - r_{e})} = 0 \Rightarrow e^{- a (r - r_{e})} = 1 \Rightarrow - a (r - r_{e}) = 0 \Rightarrow r = r_{e}$

So the slope is zero exactly at $r = r_{e}$ . The constant we called the equilibrium separation genuinely is the valley bottom — the bond length. ✔

PICTURE. The Morse curve with a horizontal tangent line (slope $= 0$ ) touching the bottom at $r = r_{e}$ .

Step 7 — Read off the depth: the bond energy

WHAT. Plug the two special separations into $E (r)$ .

At $r = r_{e}$ : the exponent is $e^{0} = 1$ , so $(1 - 1)^{2} = 0$ , giving

$E (r_{e}) = D_{e} \cdot 0 - D_{e} = - D_{e} .$

At $r \to \infty$ : $e^{- a (r - r_{e})} \to 0$ , so $(1 - 0)^{2} = 1$ , giving

$E (\infty) = D_{e} \cdot 1 - D_{e} = 0.$

The energy needed to climb out of the well (to pull the atoms fully apart) is the top minus the bottom:

$B E = E (\infty) - E (r_{e}) = 0 - (- D_{e}) = D_{e} .$

WHY is this the bond energy? Breaking a bond = dragging the atoms from the valley floor up to the "infinitely apart" plateau. The vertical climb is $D_{e}$ . Because you spend energy climbing, $B E$ is positive — which is exactly why bond energies are always tabulated as positive numbers.

PICTURE. The well with a vertical amber arrow of height $D_{e}$ from the floor up to the zero line, labelled "= bond energy $B E$ ".

Step 8 — Edge & degenerate cases (never leave a gap)

Case A — Higher bond order (more electron glue). Add a second or third shared pair (double, triple bond). More negative charge between the nuclei → stronger inward pull → the valley moves left (shorter $r_{e}$ ) and gets deeper (bigger $D_{e}$ ). This is the master trend of the parent note, now visible as two wells side by side. (More pairs come from filling bonding orbitals — see Molecular Orbital Theory.)

Case B — No net glue (no bond). If overlap is out of phase, the waves cancel in the middle, density is removed from between the nuclei, and repulsion is never overcome. There is no valley at all — the curve only ever falls as $r$ grows. Two atoms just fly apart. This is the degenerate case: bond order $0$ , no bond length, no bond energy.

Case C — Fractional bond order (delocalisation). In benzene the glue is smeared over a ring, giving each C–C an effective $1.5$ pairs. Its well sits between the single and double wells → length $139$ pm, between $154$ and $134$ . (This is Resonance and Delocalisation in one picture.)

PICTURE. Three (or "no-well") curves overlaid: deeper/left as bond order rises, and a flat repulsive-only curve for "no bond."

The one-picture summary

Everything above, compressed: the repulsion piece and the attraction piece combine into a single well; its horizontal position is the bond length $r_{e}$ ; its depth is the bond energy $B E = D_{e}$ ; and raising the bond order slides the well left-and-down.

Recall Feynman retelling — the whole walkthrough in plain words

Two tiny magnets that hate each other (the nuclei — both positive, both pushing apart). If nothing else happened, the closer you brought them the harder they'd shove: that's the wall on the left of our graph, the $1/ r$ repulsion.

Now smear a blob of sticky electron-glue right in the gap between them. That negative blob grabs both positive magnets and hauls them inward — that's the dip, the attraction, drawn below the line.

Add the shove and the pull together and you get a valley: too close and the shove wins (energy shoots up), too far and neither reaches (energy climbs back to zero), so somewhere in between the energy bottoms out. The spot at the bottom is how far apart they choose to sit — the bond length. The depth of the valley is how much effort it takes to scrape off the glue and rip them apart — the bond energy.

We wrote the valley as the Morse formula only so we could use the slope trick: the flat bottom is where the slope is zero, and solving that gives $r = r_{e}$ exactly. Plugging the top and bottom of the valley in shows the climb is $D_{e}$ — the bond energy.

Finally: pile in more glue (double, triple bond) and the valley gets deeper and moves left — stronger and shorter. Use anti-glue (out-of-phase waves) and there's no valley at all — no bond. That's the entire chapter, standing on one graph.

Quick self-check

Which valley feature gives the bond energy?

The depth of the well,

D_{e}

(the vertical climb from the floor up to the zero line).

Which feature gives the bond length?

The horizontal position of the well's minimum,

r_{e}

Why is

\frac{d E}{d r} = 0

the right condition for the bond length?

The bottom of the valley is the flat point where the slope stops being downhill and turns uphill.

What happens to the well as bond order rises?

It gets deeper and shifts to shorter

r

— stronger, shorter bond.

What does the "no valley" curve represent?

Out-of-phase overlap: no glue between the nuclei, so no bond (bond order 0).