Ligands — classification (mono, bi, poly, ambidentate, chelating); denticity: Visual walkthrough

We only reference tools from these vault notes as we need them: Lewis Acids and Bases, Coordination Number, Stability Constants of Complexes, Crystal Field Theory, EDTA Titrations, Werner's Coordination Theory.

Step 1 — What is grabbing what?

WHAT. Look at figure s01. The blue disc is the metal ion. The little yellow "lobe" sticking out of each ligand is the lone pair — think of it as an outstretched hand.

WHY. Before we can count how many hands grab, we must agree that bonding here means "a lone pair reaching from ligand to metal." Every arrow you see later is one such reach.

PICTURE. The arrow points from the donor atom to the metal — electrons flow that way. In water, the arrow starts on oxygen, never on hydrogen, because oxygen is the atom holding the lone pairs.

Figure — Ligands — classification (mono, bi, poly, ambidentate, chelating); denticity

Step 2 — Counting hands: denticity

WHAT. In figure s02, on the left one ligand reaches with one hand (denticity $= 1$ , called monodentate). On the right, one ligand reaches with two hands (denticity $= 2$ , called bidentate).

WHY "same metal, same time"? Because a ligand might own several lone pairs but only use one. Denticity counts the hands actually gripping this metal now — not the hands it could theoretically use elsewhere.

PICTURE. Notice the bidentate ligand on the right is a single connected molecule (the grey backbone joins its two hands). That backbone is the whole secret of this page — keep your eye on it.

Step 3 — When two hands close a ring: the chelate

WHAT. Follow the closed loop in figure s03: start at the metal $M$ , walk along the first bond to nitrogen $N$ , along the backbone $C - C$ , to the second nitrogen $N$ , and back along the second bond to $M$ . That closed path is the ring.

WHY count ring members? Ring size controls strain. Count the vertices: $M, N, C, C, N$ = 5 atoms → a 5-membered ring. Five- and six-membered rings sit at natural bond angles, so they barely strain — that is why ethylenediamine (en) is a textbook-stable chelate.

PICTURE. The green loop is drawn smooth and relaxed. Imagine squeezing it to a 3-membered triangle: the bonds would bend painfully. Imagine stretching it to an 8-membered floppy loop: the second hand can barely reach back to the metal. Five/six is the "just right."

Step 4 — The fair experiment: same bonds, different packaging

WHAT. Figure s04 shows the two competitors for $N i^{2 +}$ :

Left: $[N i (N H_{3})_{6}]^{2 +}$ — six separate ammonia ( $N H_{3}$ ) molecules, so six Ni–N bonds from six loose pieces.
Right: $[N i (e n)_{3}]^{2 +}$ — three ethylenediamine molecules, each bidentate, so again six Ni–N bonds — but from only three pieces, each holding with both hands.

WHY this is fair. Six Ni–N bonds on both sides. So the bond strength (the energy released per bond, which we will call enthalpy in Step 6) is essentially identical. Anything left over that makes one side better cannot be bond strength — it must be something else. That "something else" is the whole point.

PICTURE. Count pieces gripping the metal: six on the left, three on the right. Hold that count.

Step 5 — Count the particles before and after (why entropy enters)

For the chelate: $[N i (H_{2} O)_{6}]^{2 +} + 3 en ⟶ [N i (en)_{3}]^{2 +} + 6 H_{2} O$

Term by term, right where it appears:

$[N i (H_{2} O)_{6}]^{2 +}$ — the metal already caged by six water molecules (1 particle).
$3 en$ — three free bidentate ligands (3 particles). Left total $= 1 + 3 = 4$ particles.
$[N i (en)_{3}]^{2 +}$ — the new chelated complex (1 particle).
$6 H_{2} O$ — the six waters kicked loose and now free (6 particles). Right total $= 1 + 6 = 7$ particles.

WHAT. Figure s05 tallies it: 4 particles in, 7 particles out. The number of free, independent particles grew.

WHY this matters. More free particles = more ways for the system to be arranged = more disorder. In physics/chemistry, "amount of disorder" has a name: entropy, symbol $S$ . The reaction increases entropy, so $Δ S > 0$ (the $Δ$ , "delta," just means "change = after minus before").

Now do the same tally for the six-ammonia case: $[N i (H_{2} O)_{6}]^{2 +} + 6 N H_{3} ⟶ [N i (N H_{3})_{6}]^{2 +} + 6 H_{2} O$ Left $= 1 + 6 = 7$ ; right $= 1 + 6 = 7$ . Particle count unchanged → almost no entropy gain. This is the asymmetry the chelate exploits.

Step 6 — Turning "more disorder" into "more stable": the equations

Now we introduce exactly three tools, each because we need it.

WHY subtract $T Δ S$ ? Because a reaction is favoured both by releasing energy (negative $Δ H$ ) and by increasing disorder (positive $Δ S$ ). The minus sign means a positive $Δ S$ lowers $Δ G$ — helping the reaction go.

WHAT (combine the two tools). Set the two expressions for $Δ G^{\circ}$ equal: $- R T ln K = Δ H^{\circ} - T Δ S^{\circ}$ Divide every term by $- R T$ : $ln K = - \frac{Δ H ^{\circ}}{R T} + \frac{Δ S ^{\circ}}{R}$

$- \frac{Δ H ^{\circ}}{R T}$ — the bond-strength contribution. Same for both competitors (Step 4).
$+ \frac{Δ S ^{\circ}}{R}$ — the disorder contribution. Bigger for the chelate (Step 5).

WHY this finishes the proof. Both complexes share the first term. The chelate has a larger second term (larger $Δ S^{\circ}$ ). Therefore the chelate has a larger $ln K$ , hence a larger $K$ , hence more stability. This entropy-driven bonus is the chelate effect.

PICTURE. Figure s06 is a bar chart: two nearly equal enthalpy bars, then a tall extra entropy bar only on the chelate side stacking up to a taller total $ln K$ .

Step 7 — Edge cases: when the "hug" fails

PICTURE. Figure s07 shows all three failures side by side: (A) one hand used of two, (B) two metals bridged, (C) a strained tiny triangle ring — none earn the green "stable" glow.

The one-picture summary

Figure s08 compresses the whole walkthrough: donor atoms → hands → count = denticity → both hands on one metal = chelate ring → same bonds but fewer starting pieces → more freed particles → bigger $Δ S$ → (via $ln K = - \frac{Δ H ^{\circ}}{R T} + \frac{Δ S ^{\circ}}{R}$ ) bigger $K$ → more stable.

Recall Feynman retelling — explain the whole walkthrough to a 12-year-old

The metal is a hand waiting to be held. Each ligand is a person with a spare hand (its lone pair) who can hold it. Now race two teams that make the same number of grips. Team A is six separate people, each holding with one hand. Team B is three people, each hugging with both arms. Both make six grips, so the grips are equally strong — that is the enthalpy being equal. Here is the twist. To hold the metal, both teams first have to shove aside six water molecules that were already holding it. Team A brings six people in and pushes six waters out: crowd size stays the same. Team B brings only three people in but still pushes six waters out — so the room ends up more crowded and messier than before. Nature loves messier — that extra mess is entropy, and it makes Team B's arrangement harder to undo. "Harder to undo" is exactly what we mean by more stable, and the formula $ln K = - \frac{Δ H ^{\circ}}{R T} + \frac{Δ S ^{\circ}}{R}$ turns "more mess" straight into "bigger stability number $K$ ." That is the chelate effect — and it only works if both arms hug the same metal in a comfy 5- or 6-membered ring. One arm only, two different metals, or a cramped ring, and the bonus vanishes.

Recall Quick self-test

Why is $Δ H^{\circ}$ roughly equal for $[N i (N H_{3})_{6}]^{2 +}$ and $[N i (e n)_{3}]^{2 +}$ ? ::: Both make six Ni–N bonds of essentially the same strength. Where does the chelate's extra stability come from? ::: A larger positive $Δ S^{\circ}$ — more free particles are released — which raises $ln K$ via the $+ Δ S^{\circ} / R$ term. Does $S C N^{-}$ get the chelate bonus? ::: No — it is ambidentate/monodentate, uses one donor at a time, closes no ring. Which ring sizes give the strongest chelate effect? ::: 5- and 6-membered, minimal strain and geometrically comfortable.