Ligands — classification (mono, bi, poly, ambidentate, chelating); denticity: Worked examples

The scenario matrix

Before any example, here is the full map of cases a denticity/classification problem can throw at you. Each later example is tagged with the cell it covers.

Cell	Case class	What makes it tricky	Covered by
A	Plain monodentate	baseline — 1 donor, 1 grip	Ex 1
B	Simple chelate (bidentate)	2 donors, 1 ring, 1 metal	Ex 2
C	Mixed ligands → coordination number	multiply denticity × count, then sum	Ex 3
D	"Looks bidentate" but is ambidentate	2 donor atoms, only 1 used at a time	Ex 4
E	"Looks chelating" but is bridging	2 donors, but on two different metals	Ex 5
F	Zero / degenerate input	a would-be donor atom that has no lone pair free	Ex 6
G	Limiting case — huge polydentate cap	denticity can't exceed available coordination sites	Ex 7
H	Real-world word problem	translate a story into denticity + stability	Ex 8
I	Exam twist — stability comparison	chelate effect made quantitative	Ex 9

Read the matrix as a checklist: by Example 9 every row is ticked.

Intuition The vault topics this page leans on (so you needn't leave)

A few linked topics carry ideas we use below. Here is the one-line version of each, so this page stands on its own:

Coordination number — the total number of donor-atom→metal bonds around one metal. We build it in Ex 1, 3, 4, 5, 7.
Lewis base — a species that donates a lone pair; every ligand is one. The metal is the Lewis acid (lone-pair acceptor). Used in Ex 6.
Stability constant $β$ — a big number measuring how far the "metal + ligands → complex" reaction goes to completion; bigger $β$ = more stable complex. Central to Ex 9 (defined in full there).
Werner's theory — the idea that each metal has a fixed number of coordination sites (its secondary valence) that ligands must fill; this is the "ceiling" in Ex 7.
EDTA Titrations — using hexadentate EDTA to trap metal ions quantitatively; background for Ex 7 and 8.
Linkage Isomerism — two complexes with the same formula but an ambidentate ligand bound through different atoms; used in Ex 4.
Crystal Field Theory — explains colour differences (e.g. yellow vs red linkage isomers); mentioned in passing.

Example 1 — Cell A: the plain monodentate baseline

Worked example Statement

Classify the ligand $N H_{3}$ in $[C u (N H_{3})_{4}]^{2 +}$ , and give its contribution to the coordination number.

Forecast: Guess first — how many "hands" does one ammonia molecule use to hold the metal?

Step 1 — Find the donor atom. Why this step? Denticity counts donor atoms, so we must first ask which atom carries the lone pair. In $N H_{3}$ , nitrogen has one lone pair; the three hydrogens have none. So the donor atom is nitrogen.

Step 2 — Count how many donor atoms one molecule uses on the same metal. Why this step? This count is the denticity. One nitrogen → 1 donor atom → monodentate.

Step 3 — Contribution to coordination number. Why this step? Coordination number = total donor atoms bonded to the metal. Each $N H_{3}$ gives 1; there are 4 of them → $4 \times 1 = 4$ .

Verify: The complex is written $[C u (N H_{3})_{4}]^{2 +}$ — four ligand slots, each a monodentate $N H_{3}$ , so coordination number $= 4$ . Consistent. ✓

Example 2 — Cell B: a simple chelate (see figure)

Worked example Statement

Classify ethylenediamine (en, $H_{2} N - C H_{2} - C H_{2} - N H_{2}$ ) in $[C o (e n)_{3}]^{3 +}$ . What ring does it form and is it chelating?

Forecast: Two nitrogens on one molecule — does it use one or both to grab the same cobalt?

Step 1 — Locate donor atoms. Why: both nitrogens carry a lone pair → two candidate donors on one molecule.

Step 2 — Check "same metal, simultaneously". Why: denticity only counts donors that bind one metal at once. The $- C H_{2} - C H_{2} -$ backbone is short and flexible enough that both N atoms reach the same cobalt → bidentate.

Step 3 — Is there a ring? Why: a chelate must close a ring including the metal. Trace: Co → N → C → C → N → back to Co. That is 5 atoms in the loop → a 5-membered chelate ring, so en is chelating. Look at the red loop in the figure below.

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

Figure (Ex 2): the black zig-zag is the ethylenediamine molecule — the two outer black circles labelled N are the donor nitrogens, joined by two black C atoms. The metal M (our shorthand for the central metal, here cobalt) sits at the bottom. Follow the red closed loop: it runs M → N → C → C → N → back to M, threading through five atoms — that is the 5-membered chelate ring. The two red arrows are the coordinate (lone-pair) bonds from each nitrogen into the same metal; notice both arrows point at the SAME M, which is exactly what "bidentate to one metal" means.

Step 4 — Coordination number. Why: 3 en ligands, each bidentate → $3 \times 2 = 6$ .

Verify: 5-membered ring (M, N, C, C, N) is exactly the "most stable" ring size from the parent note; coordination number 6 matches an octahedral $C o^{3 +}$ . ✓ See Stability Constants of Complexes for why this extra ring stability shows up as a big stability constant $β$ .

Example 3 — Cell C: mixed ligands, build the coordination number

Worked example Statement

Find the coordination number of $[C r (C_{2} O_{4})_{2} (H_{2} O)_{2}]^{-}$ .

Forecast: There are two kinds of ligand with different denticities — will they add up to 4, 6, or 8?

Step 1 — Classify each ligand. Why: different ligands contribute different numbers of grips, so classify before counting.

Oxalate $C_{2} O_{4}^{2 -}$ : two O donors reach one Cr → bidentate.
Water $H_{2} O$ : one O donor → monodentate.

Step 2 — Multiply denticity × how many of that ligand. Why: coordination number is the total donor–metal bonds, so each ligand type contributes (denticity × count). This next line is the key arithmetic: $oxalate: 2 \times 2 = 4, water: 1 \times 2 = 2$

Step 3 — Sum. Why: total grips = sum over all ligand types. The key result: $4 + 2 = 6$

Verify: Coordination number $= 6$ , an octahedral Cr(III) — chemically standard. ✓

Example 4 — Cell D: it looks bidentate but it's ambidentate

Worked example Statement

A student claims $[C o (N H_{3})_{5} (N O_{2})]^{2 +}$ has $N O_{2}^{-}$ as a bidentate ligand (denticity 2). Is this right? Give the true denticity and name the phenomenon.

Forecast: $N O_{2}^{-}$ has one N and two O atoms — surely at least two can donate?

Step 1 — List candidate donors. Why: to see the trap, we must acknowledge it really does have multiple lone-pair atoms — N and the two O's.

Step 2 — Apply "simultaneously to the same metal". Why: denticity is not "how many atoms could donate" but "how many do, at once, to one metal." In $[C o (N H_{3})_{5} (N O_{2})]^{2 +}$ the ligand attaches through only one atom (either N, giving nitro; or O, giving nitrito). → denticity = 1, monodentate.

Step 3 — Name it. Why: a monodentate ligand with two different possible donor atoms used one-at-a-time is ambidentate. Choosing N vs O gives linkage isomers (yellow N-bound vs red O-bound; the colour difference is explained by Crystal Field Theory).

Verify: Coordination number is 5 ( $N H_{3}$ ) + 1 ( $N O_{2}^{-}$ ) $= 6$ , octahedral — which only works if $N O_{2}^{-}$ occupies one site, confirming denticity 1. ✓ See Linkage Isomerism.

Example 5 — Cell E: it looks chelating but it's bridging

Worked example Statement

In the dimer $[(N H_{3})_{5} C o - N H_{2} - C o (N H_{3})_{5}]^{5 +}$ the amide $N H_{2}^{-}$ connects the two cobalts. Is $N H_{2}^{-}$ here chelating? What is its role, and its denticity per metal?

Forecast: It uses its donor to touch two metals — does that make it bidentate/chelating?

Step 1 — Count how many metals it touches. Why: the word "chelate" demands a ring on one metal. Here the nitrogen donates to two different cobalts.

Step 2 — Apply "same metal". Why: since the two grips are on different metals, no ring closes around a single centre → not chelating. Its role is bridging.

Step 3 — Denticity per metal. Why: denticity is defined per metal centre. To each cobalt it offers one donor → monodentate to each metal, even though it bridges two.

Verify: Each Co is $[5 N H_{3} + 1 bridge] = 6$ -coordinate. The bridge counts as one site on each metal, exactly what "monodentate per metal" predicts. ✓

Example 6 — Cell F: the zero / degenerate case

Worked example Statement

Ammonium ion $N H_{4}^{+}$ is offered as a ligand. What is its denticity — and can it act as a ligand at all?

Forecast: It has a nitrogen like $N H_{3}$ — same donor, right?

Step 1 — Check for a free lone pair. Why: a ligand must donate a lone pair (it is a Lewis base — a lone-pair donor). So the first question is always "is there a spare, unshared pair of electrons?"

Step 2 — Track the lone pair explicitly. Why: this is where the trap hides. Neutral nitrogen brings 5 valence electrons. In ammonia $N H_{3}$ it forms 3 N–H bonds (using 3 of its electrons, one per bond), leaving 2 electrons as one lone pair — that pair is what makes $N H_{3}$ a ligand. Now go to $N H_{4}^{+}$ : nitrogen forms a 4th N–H bond, and the electrons it uses for that 4th bond are exactly the two electrons that used to be the lone pair. Because it donated that pair away to grab an extra proton ( $H^{+}$ ), the whole species is left one electron short and carries a $+ 1$ formal charge — that "+" is the visible receipt that the lone pair is gone.

Step 3 — Count usable donor atoms. Why: denticity counts donors with an available lone pair. Nitrogen in $N H_{4}^{+}$ has zero free lone pairs → denticity = 0 → it cannot coordinate; it is only a spectator counter-ion.

Verify: Formal-charge bookkeeping: N valence $= 5$ ; in $N H_{4}^{+}$ it has $0$ non-bonding electrons and shares $8$ bonding electrons (half = 4), so formal charge $= 5 - 0 - 4 = + 1$ — matching the "+" and confirming no lone pair remains. Denticity $0$ . ✓

Example 7 — Cell G: the limiting case (denticity meets the ceiling)

Worked example Statement

$E D T A^{4 -}$ has 6 donor atoms (2 N + 4 O). If we try to fit it onto a metal that is only 4-coordinate (square planar $P t^{2 +}$ , say), can it still act as a hexadentate ligand? What is the effective denticity, i.e. the limiting behaviour?

Forecast: Six teeth on a four-slot mouth — do all six bite?

Step 1 — Recall the physical ceiling. Why: by Werner's theory, each metal offers only a fixed number of coordination sites (its secondary valence). A donor atom can bond only if there is an empty site waiting for it — you cannot bond a 7th tooth to a metal that has only 6 slots. Once every site is filled, the coordination sphere is saturated and any extra donor atoms simply have nowhere to go and dangle unused.

Step 2 — State the limiting rule this forces. Why: combine "ligand wants to use all its donors" with "metal has a hard cap of sites." Introduce a handy shorthand: $min (a, b)$ is ordinary words in symbol form — it just means "take whichever of the two numbers $a$ and $b$ is smaller" (for example $min (6, 4) = 4$ ). Using it, the effective denticity in a real complex is: $effective denticity = min (ligand’s donor count, available metal sites)$ The $min$ (the "pick-the-smaller" rule) appears precisely because whichever runs out first — teeth or slots — stops the binding.

Step 3 — Apply. Why: metal offers 4 sites, ligand has 6 → $min (6, 4) = 4$ . EDTA can bind at most 4 donors here and lets the other two dangle → it acts tetradentate on a 4-site metal.

Step 4 — Contrast the intended case. Why: on an octahedral 6-site metal (e.g. $C a^{2 +}$ , $F e^{3 +}$ ), $min (6, 6) = 6$ → full hexadentate cage, coordination number 6.

Verify: $min (6, 6) = 6$ for octahedral (matches $[F e (E D T A)]^{-}$ , CN 6) and $min (6, 4) = 4$ for square-planar — both obey the ceiling. ✓ See EDTA Titrations.

Example 8 — Cell H: real-world word problem

Worked example Statement

Hard water contains $C a^{2 +}$ ( $1.0 \times 1 0^{- 3}$ mol) in 1 L. You dose it with $E D T A^{4 -}$ to trap the calcium as $[C a (E D T A)]^{2 -}$ (a 1:1 complex). (a) How many moles of EDTA are needed? (b) What volume of a $0.010 mol L^{- 1}$ EDTA solution delivers that amount? (c) Why does one EDTA molecule do the job where you'd otherwise need six monodentate ligands?

Forecast: Guess the EDTA : Ca mole ratio, and roughly how many millilitres of the dosing solution you'll need, before reading on.

Step 1 — Write the balanced reaction. Why: the stoichiometry (the mole ratio) is the bridge from "moles of calcium" to "moles of EDTA," so we must fix it first. $C a^{2 +} + E D T A^{4 -} \to [C a (E D T A)]^{2 -}$ One EDTA reacts with one $C a^{2 +}$ → the ratio is 1 : 1.

Step 2 — (a) Moles of EDTA. Why: with a 1:1 ratio, moles of EDTA needed simply equal moles of calcium present — no multiplier. The key line: $n_{E D T A} = n_{C a^{2 +}} = 1.0 \times 1 0^{- 3} mol$

Step 3 — (b) Volume of dosing solution. Why: concentration links moles to volume through $n = c V$ , so to get the volume we rearrange to $V = n / c$ . Plugging in the numbers: $V = \frac{n _{E D T A}}{c} = \frac{1.0 \times 1 0 ^{- 3} mol}{0.010 mol L ^{- 1}} = 0.10 L = 100 mL$

Step 4 — (c) Why one molecule does a six-atom job. Why: this connects the arithmetic back to denticity, the theme of the page. EDTA is hexadentate — its six donor atoms fill all six octahedral sites of one $C a^{2 +}$ at once. Six separate monodentate ligands would need six molecules; folding those six teeth into a single molecule is exactly what makes the trap both 1:1 and extremely stable (the chelate effect).

Verify: (a) $n_{E D T A} = 1.0 \times 1 0^{- 3}$ mol equals $n_{C a}$ , as a 1:1 complex demands. (b) $V = 1 0^{- 3} /0.010 = 0.10$ L $= 100$ mL; units cancel as $mol \div (mol L^{- 1}) = L$ . ✓ See EDTA Titrations.

Example 9 — Cell I: exam twist (stability made quantitative)

Worked example Statement

For $N i^{2 +}$ the stability constant (written $β$ ) of $[N i (N H_{3})_{6}]^{2 +}$ has $lo g β \approx 8.6$ ; that of $[N i (e n)_{3}]^{2 +}$ has $lo g β \approx 18.3$ . Both complexes make 6 Ni–N bonds. (a) By roughly how many times is the en complex more stable? (b) Which thermodynamic term explains it?

First, the symbols (nothing is assumed):

$β$ = the stability constant, the equilibrium constant for building the whole complex from the free metal and all its ligands. A bigger $β$ means the reaction runs further toward the complex → a more stable complex.
$lo g β$ = the base-10 logarithm of $β$ . Because stability constants are enormous (billions and up), chemists tabulate their logs. Key fact about logs: if $lo g β_{A} - lo g β_{B} = d$ , then $β_{A} / β_{B} = 1 0^{d}$ — a difference in logs becomes a ratio of the actual constants. That is why we subtract the two log values and then raise 10 to the difference.

Forecast: Same bonds — will $β$ differ by 10×? 1000×? A billion×?

Step 1 — (a) Difference in $lo g β$ . Why: by the log fact above, subtracting the two logs is the first move toward the ratio. This is the key line: $Δ lo g β = 18.3 - 8.6 = 9.7$

Step 2 — (a) Convert the log-difference into a ratio. Why: a difference of logs is the log of a ratio, so we undo the log by raising 10 to it. Key result: $\frac{β _{e n}}{β _{N H_{3}}} = 1 0^{9.7} \approx 5 \times 1 0^{9}$ The chelate is about five billion times more stable.

Step 3 — (b) Which thermodynamic term explains it? Why: to see the cause, recall the identity from the parent note. First define every symbol:

$Δ G^{\circ}$ = standard Gibbs free-energy change; more negative $Δ G^{\circ}$ ↔ larger $β$ (more stable).
$Δ H^{\circ}$ = standard enthalpy change (roughly, the energy released making bonds).
$Δ S^{\circ}$ = standard entropy change (a measure of increased disorder / number of free particles).
$T$ = absolute temperature in kelvin; $R$ = the gas constant ( $8.314 J K^{- 1} mol^{- 1}$ ).

Combining $Δ G^{\circ} = - R T ln β$ with $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ gives the parent-note identity: $ln β = - \frac{Δ H ^{\circ}}{R T} + \frac{Δ S ^{\circ}}{R}$ Both routes make the same 6 Ni–N bonds, so $Δ H^{\circ}$ is nearly equal → the first term is nearly the same for both. The extra stability of the en complex must come from the entropy term: 3 en molecules release 6 waters (fewer particles go in than come out → $Δ S^{\circ} > 0$ ), pushing $ln β$ up. This is the chelate effect.

Verify: $1 0^{9.7} \approx 5.0 \times 1 0^{9}$ , an entropy-driven ratio (not stronger bonds). ✓ See Stability Constants of Complexes for the full number tables.

Recall

Recall Did every cell get covered? (peek after answering)

Match each example to its matrix cell, then check. Ex1 ::: A (plain monodentate) Ex2 ::: B (simple chelate) Ex3 ::: C (mixed → coordination number) Ex4 ::: D (ambidentate, not bidentate) Ex5 ::: E (bridging, not chelating) Ex6 ::: F (zero/degenerate — no free lone pair) Ex7 ::: G (limiting — denticity capped by metal sites) Ex8 ::: H (real-world word problem) Ex9 ::: I (exam twist — quantitative chelate effect)