σ vs π bonds — overlap, strength: Visual walkthrough

We will build up in numbered steps. Read them in order; each figure carries the argument.

Step 1 — What is an orbital, as a picture?

WHAT. Before we can talk about orbitals overlapping, we need to see one orbital. An orbital is just a region of space where an electron is likely to be found. We draw it as a shaded shape whose darkness means "more likely to find the electron here."

WHY. Every claim on this page — "head-on", "sideways", "overlap" — is a statement about shapes touching. So we must fix the two shapes we care about first.

PICTURE. Two kinds matter here:

An $s$ orbital: a single round ball centred on the nucleus.
A $p$ orbital: a dumbbell — two lobes on opposite sides of the nucleus, with a flat empty sheet through the middle (a node, where the electron is never found). We label the two lobes $+$ and $-$ ; these are just the sign of the wavefunction (like the crest and trough of a wave), not electric charge.

Figure — σ vs π bonds — overlap, strength

Step 2 — What does "a bond" mean in this picture?

WHAT. Bring two atoms close. Their orbitals reach into the same patch of space. Where both $ψ_{A}$ (atom A's wave) and $ψ_{B}$ (atom B's wave) are non-zero at once, we say the orbitals overlap.

WHY. A covalent bond is this shared region. The electrons in the overlap sit between the two nuclei and glue them together — their negative charge pulls both positive nuclei inward and cancels the nuclei's mutual push.

PICTURE. The green shaded almond in the figure is the overlap region — the only place where "A's wave" and "B's wave" are both present. Bigger, denser almond = more glue = stronger bond.

Step 3 — Turning "amount of overlap" into a number: $S$

WHAT. "Amount of overlap" sounds vague. We make it exact by multiplying the two waves at every point of space and adding up the result. That sum is the overlap integral $S$ .

WHY this tool and not another? We need a single number that is large when the lobes sit on top of each other and small when they barely touch. Multiplying $ψ_{A}$ by $ψ_{B}$ does exactly that: at a point where both waves are big, the product $ψ_{A} ψ_{B}$ is big; where either is zero, the product is zero and contributes nothing. Adding ("integrating") those products over all space gives one honest number for "how much do these two shapes share?" No other simple combination has this property.

PICTURE. The figure shows the same two orbitals sliced along the axis: the product curve $ψ_{A} ψ_{B}$ (plum) bulges wherever the two individual waves (orange, teal) both rise. The area under the plum bulge is $S$ .

Step 4 — Head-on: the σ overlap, and why $S$ is large

WHAT. Point the two $p$ dumbbells straight at each other along the line joining the nuclei (call that line the bond axis). The near lobes — say both $+$ — face and merge.

WHY it wins. The two lobes come at each other tip to tip, so their densest parts land in the same place, right on the axis between the nuclei. The product $ψ_{A} ψ_{B}$ is large exactly in the region that binds best. This is the biggest almond you can make → largest $S$ . Call it $S_{σ}$ .

PICTURE. Both $+$ lobes (orange) collide on the axis; the green overlap almond sits on the line between the nuclei. The result is cylindrically symmetric — spin the whole thing around the axis and nothing changes.

The same head-on story works for $s$ – $s$ (two balls touching) and $s$ – $p$ (ball meets a lobe tip). All are σ. Hybrid orbitals ( $s p, s p^{2}, s p^{3}$ — see Hybridisation (sp, sp2, sp3)) are just directional $s$ / $p$ mixtures pointing at the neighbour, so they too form σ.

Step 5 — Sideways: the π overlap, and why $S$ is smaller

WHAT. Now keep the two $p$ dumbbells parallel, standing side by side, perpendicular to the bond axis. The top lobes overlap a little; the bottom lobes overlap a little.

WHY it loses. The lobes are displaced off the axis — their densest parts do not sit on top of each other, they sit beside each other. The product $ψ_{A} ψ_{B}$ is spread thin, in two weak patches (one above, one below the axis), away from the internuclear line where binding is most effective. Smaller product, smaller almond → smaller $S$ . Call it $S_{π}$ .

PICTURE. Two parallel dumbbells; the green overlap is now two thin slivers, one above and one below the axis, and the axis itself runs through an empty sheet — the nodal plane.

Step 6 — The head-to-head comparison: $S_{σ} > S_{π}$

WHAT. Put Step 4 and Step 5 side by side at the same nucleus–nucleus distance and compare the almonds.

WHY. Same atoms, same distance, only the geometry of approach differs. So the difference in $S$ is caused purely by head-on vs sideways — a clean, fair comparison.

PICTURE. Left: fat central almond ( $S_{σ}$ ). Right: two thin off-axis slivers ( $S_{π}$ ). Eye-balling the shaded areas, the left is clearly bigger.

$S_{σ} > S_{π} ⟹ σ bond stronger than π bond$

$S_{σ}$ — head-on overlap number (Step 4). Large.
$S_{π}$ — sideways overlap number (Step 5). Small.
$⟹$ — "therefore," using the master principle of Step 2 (more overlap ⇒ bigger dissociation energy $D$ ⇒ stronger).

This inequality is the parent note's central result, now seen, not asserted. (The full energy story also involves the interaction element $H_{ab}$ — see Molecular Orbital Theory — but geometrically, bigger overlap ⇒ stronger, and that is what the pictures show.)

Step 7 — Why single, double, triple: σ first, then π's pile on

WHAT. When atoms bond, they pull as close as possible. The first orbitals to meet are the ones pointing along the axis → they form the σ. Only the leftover parallel $p$ orbitals remain, and they can only meet sideways → π.

WHY this order. Head-on is greedy: it grabs the axis and fixes the geometry. Whatever $p$ orbitals are left are now stuck perpendicular to that fixed axis, so their only option is lateral overlap.

PICTURE. N₂ (N≡N). Each N has three unpaired $p$ electrons ( $p_{x}, p_{y}, p_{z}$ ).

$p_{z}$ – $p_{z}$ point along the axis → 1 σ (Step 4 geometry).
$p_{x}$ – $p_{x}$ and $p_{y}$ – $p_{y}$ are perpendicular → 2 π (Step 5 geometry).
Total: 1σ + 2π = triple bond.

So single = 1σ, double = 1σ + 1π, triple = 1σ + 2π. Each extra bond is a weaker π stacked onto the one strong σ.

Step 8 — Edge case: rotate a double bond and watch π die

WHAT. Take a C=C double bond (1σ + 1π, as in ethene — each C is $s p^{2}$ , with one leftover $p$ ). Twist one end by $9 0^{\circ}$ .

WHY this case matters. It is the degenerate scenario where overlap goes to zero. It explains why double bonds are locked and why cis-trans isomerism exists at all.

PICTURE. At $0^{\circ}$ the two leftover $p$ orbitals are parallel → full sideways overlap → π intact. Rotate one $p$ toward $9 0^{\circ}$ and the top $+$ lobe of one now points at the empty side of the other: the overlap almond shrinks, and at exactly $9 0^{\circ}$ the two orbitals are perpendicular → $S_{π} = 0$ → the π bond is broken.

The σ, being cylindrically symmetric (Step 4), does not care about rotation — it survives.
The π, needing parallel orbitals (Step 5), is destroyed. This is why C=C cannot rotate freely.

Degenerate check: when one $p$ is turned $9 0^{\circ}$ , its $+$ lobe overlaps equal halves of the other orbital's $+$ and $-$ lobes. One half gives a $+ \times +$ (positive) contribution and the mirror half gives an equal $+ \times -$ (negative) contribution, so they cancel exactly and $S_{π} = 0$ . Zero overlap = zero π bond. The picture and the sign-bookkeeping agree.

The one-picture summary

Everything on this page compresses into one frame: the almond of overlap is fat and on-axis for σ, thin and off-axis for π, and that single visual difference is $S_{σ} > S_{π}$ — which by the master principle is "σ stronger than π," which is why bonds stack as 1σ + (extra π's), which is why double bonds are locked.

Recall Feynman: the whole walkthrough in plain words

An orbital is a fuzzy cloud where an electron lives. A bond is two clouds sharing the same patch of space — the more they share, the stronger the glue, because the shared electrons sit between the two nuclei and hold them together. We measured "how much they share" by multiplying the two clouds together everywhere and adding it up — call that number $S$ . If you shove the two dumbbell-shaped $p$ clouds tip-to-tip along the line between the atoms, their fat ends land right on top of each other, smack between the nuclei — huge shared patch, big $S$ : that's a σ bond, the strong one, and you can spin it freely because it's round about the axis. If instead you lay the dumbbells side by side, they only kiss along their edges, off to the side of the line — a thin shared patch, small $S$ : that's a π bond, the weak one. Atoms always grab the strong tip-to-tip stick first; only then can leftover clouds add weak side-by-side sticks. So a double bond is one strong plus one weak, a triple is one strong plus two weak. And if you try to twist a double bond, you rotate one side-by-side cloud until it's perpendicular to its partner — the sharing drops to nothing and the weak π snaps. That's the whole story: $S_{σ} > S_{π}$ , seen with your eyes.

Recall

On the picture, what number measures overlap and how is it built? ::: $S = \int ψ_{A} ψ_{B} d τ$ — multiply the two waves at every point and add up. Why is the σ almond bigger than the π slivers at the same distance? ::: Head-on lobes land on top of each other on the axis (dense product); sideways lobes are displaced off-axis (thin product). What happens to $S_{π}$ when you rotate a double bond by 90°? ::: It drops to zero — the parallel $p$ orbitals become perpendicular, the overlap cancels, the π breaks. Compose triple bond in σ/π and say which orbitals. ::: 1σ (axial $p_{z}$ – $p_{z}$ ) + 2π (lateral $p_{x}$ – $p_{x}$ , $p_{y}$ – $p_{y}$ ).