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

Level 1 — Recognition

Goal: read a ligand and name its type. No arithmetic yet, just "count the teeth."

Problem 1.1

State the donor atom in each ligand and its denticity: $C l^{-}$ , $N H_{3}$ , $H_{2} O$ , $C N^{-}$ .

Recall Solution 1.1

The donor atom is the atom carrying the lone pair that actually reaches the metal. Denticity = how many such atoms bite the same metal at once.

Ligand	Donor atom	Denticity
$C l^{-}$	Cl	1 (monodentate)
$N H_{3}$	N	1 (monodentate)
$H_{2} O$	O	1 (monodentate)
$C N^{-}$	C (usually)	1 (monodentate)

Why all 1? Each of these has lone pairs on essentially one reachable atom. Even $C N^{-}$ — which has a lone pair on both C and N — can only stretch one of those to a single metal at a time, so its denticity to one metal is 1.

Problem 1.2

Classify ethylenediamine (en) and oxalate $C_{2} O_{4}^{2 -}$ . Give donor atoms, denticity, and whether they chelate.

Recall Solution 1.2

en = $H_{2} N - C H_{2} - C H_{2} - N H_{2}$ . Two nitrogen donor atoms, both reach the same metal → bidentate. Because both grab one metal it closes a ring → chelating.
oxalate = $^{-} O O C - C O O^{-}$ . Two oxygen donor atoms (one from each carboxylate) grip one metal → bidentate, chelating.

Look at the figure below: both wrap the metal into a 5-membered ring (metal counts as one corner).

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

Level 2 — Application

Goal: use denticity to compute the coordination number.

Problem 2.1

Find the coordination number of $[P t (e n) C l_{2}]$ .

Recall Solution 2.1

en: bidentate → denticity 2, and there is 1 en → $2 \times 1 = 2$ .
$C l^{-}$ : monodentate → denticity 1, and there are 2 → $1 \times 2 = 2$ . $CN = 2 + 2 = 4$

Problem 2.2

Find the coordination number of $[F e (C_{2} O_{4})_{3}]^{3 -}$ .

Recall Solution 2.2

oxalate: bidentate (denticity 2), and there are 3 of them. $CN = 2 \times 3 = 6$

Problem 2.3

Find the coordination number of $[C a (EDTA)]^{2 -}$ .

Recall Solution 2.3

EDTA $^{4 -}$ is hexadentate: 2 nitrogen + 4 carboxylate-oxygen donors = 6 donor atoms, and there is 1 EDTA. $CN = 6 \times 1 = 6$ See EDTA Titrations for where this "one ligand fills all six sites" fact gets used.

Level 3 — Analysis

Goal: distinguish look-alikes — ambidentate vs bidentate, chelating vs bridging.

Problem 3.1

$N O_{2}^{-}$ has both N and O with lone pairs. Is it bidentate? Classify it precisely and explain what two isomers it can create in $[C o (N H_{3})_{5} (N O_{2})]^{2 +}$ .

Recall Solution 3.1

$N O_{2}^{-}$ is monodentate but ambidentate: it has two different donor atoms (N and O) but binds through only one at a time to a single metal.

Bind via N → nitro, ligand written $- N O_{2}$ . The complex $[C o (N H_{3})_{5} (N O_{2})]^{2 +}$ is yellow.
Bind via O → nitrito, ligand written $- O N O$ . The complex $[C o (N H_{3})_{5} (O N O)]^{2 +}$ is red.

These two are linkage isomers (see Linkage Isomerism) — same formula, different donor atom. Its denticity is 1, not 2.

Problem 3.2

The same $S C N^{-}$ ligand appears in two situations: (a) $[P d (S C N)_{2} (P P h_{3})_{2}]$ , binding one Pd per SCN; (b) an $M - S C N - M^{'}$ unit connecting two metals. State the denticity and role of $S C N^{-}$ in each.

Recall Solution 3.2

(a) Terminal: $S C N^{-}$ binds one metal via S (thiocyanato) or N (isothiocyanato). Denticity to that metal = 1, ambidentate, not chelating (a chelate needs a ring on one metal, and one bond makes no ring).
(b) Bridging: $S C N^{-}$ uses S to one metal and N to a different metal. It now uses 2 donor atoms, but they touch two different metals, so it is a bridging ligand — not a chelate (no ring closes on a single centre).

Key phrase: denticity is defined "to the same metal." Case (b) uses two atoms but not on one metal, so it isn't "bidentate to a metal" in the chelating sense.

Level 4 — Synthesis

Goal: combine denticity with thermodynamics — quantify the chelate effect.

Problem 4.1

Two reactions of an aqueous $N i^{2 +}$ , both forming 6 Ni–N bonds: $(A) [N i (H_{2} O)_{6}]^{2 +} + 6 N H_{3} \to [N i (N H_{3})_{6}]^{2 +} + 6 H_{2} O$ $(B) [N i (H_{2} O)_{6}]^{2 +} + 3 e n \to [N i (e n)_{3}]^{2 +} + 6 H_{2} O$ Count the change in number of independent particles for each. Which has the more positive $Δ S^{\circ}$ , and hence the larger $K$ ? (This is the chelate effect.)

Recall Solution 4.1

Count particles on each side (only free/independent species matter):

	Reactants	Products	$Δ (particles)$
(A)	$1 + 6 = 7$	$1 + 6 = 7$	$0$
(B)	$1 + 3 = 4$	$1 + 6 = 7$	$+ 3$

Reaction (A) releases as many particles as it consumes → $Δ S^{\circ} \approx 0$ .
Reaction (B) releases 3 more particles than it consumes → $Δ S^{\circ} > 0$ (more disorder).

Since the bonds are the same kind (Ni–N), $Δ H^{\circ}$ is nearly equal for both. Using $ln K = - \frac{Δ H ^{\circ}}{R T} + \frac{Δ S ^{\circ}}{R},$ the larger $Δ S^{\circ}$ of (B) makes $ln K$ larger → (B) has the larger $K$ → the chelate $[N i (e n)_{3}]^{2 +}$ is more stable. That entropy-driven bonus is the chelate effect. ✔

Problem 4.2

At $T = 298 K$ , suppose reaction (B) has $Δ H^{\circ} = - 12.0 kJ mol^{- 1}$ and $Δ S^{\circ} = + 90.0 J mol^{- 1} K^{- 1}$ . Compute $Δ G^{\circ}$ and $ln K$ .

Recall Solution 4.2

Step 1 — $Δ G^{\circ}$ (convert entropy term to kJ: $T Δ S^{\circ} = 298 \times 90.0 = 26820 J = 26.82 kJ$ ): $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = - 12.0 - 26.82 = - 38.82 kJ mol^{- 1}$ Why negative and large: the favourable enthalpy and the positive entropy both push $Δ G^{\circ}$ down. That double push is the fingerprint of a chelate.

Step 2 — $ln K$ from $Δ G^{\circ} = - R T ln K$ , so $ln K = - Δ G^{\circ} / (R T)$ . Use $R = 8.314 J mol^{- 1} K^{- 1}$ , $R T = 8.314 \times 298 = 2477.6 J mol^{- 1}$ : $ln K = \frac{38820}{2477.6} = 15.67$ A large positive $ln K$ ⇒ very stable complex.

Level 5 — Mastery

Goal: reason where naive rules break — degenerate cases, ambidentate isomer counts, ring geometry.

Problem 5.1

EDTA $^{4 -}$ is hexadentate, yet in some crowded complexes it binds a metal using only 5 of its 6 arms (one carboxylate dangles free). What is the coordination number contributed by EDTA in that case, and does the definition of denticity still hold?

Recall Solution 5.1

Denticity counts donor atoms actually bonding, not the maximum the ligand could offer. If only 5 arms bond, EDTA contributes CN = 5 here (pentadentate in this particular complex), and a solvent molecule such as $H_{2} O$ typically fills the 6th octahedral site.

Why the definition still holds: "denticity = donor atoms simultaneously bonded to the same metal" is context-dependent. EDTA's potential denticity is 6, but its actual denticity in a given complex can be lower. This mirrors the general rule: count what bonds, not what could.

Problem 5.2

The dithiocyanate ligand $S C N^{-}$ is ambidentate. In the complex $[C r (N H_{3})_{5} (S C N)]^{2 +}$ how many linkage isomers exist, and name them? Then generalise: why does $C l^{-}$ produce no linkage isomers?

Recall Solution 5.2

$S C N^{-}$ has two different donor atoms (S and N), so it gives $2$ linkage isomers:

S-bound → thiocyanato: $[C r (N H_{3})_{5} (S C N)]^{2 +}$
N-bound → isothiocyanato: $[C r (N H_{3})_{5} (N C S)]^{2 +}$

Generalise: linkage isomerism needs an ambidentate ligand — two chemically different donor atoms. $C l^{-}$ has only one kind of donor atom (Cl), so there is nothing to switch → 0 linkage isomers. Same reasoning: $N H_{3}$ (only N) and $H_{2} O$ (only O) can't be ambidentate.

Problem 5.3

Rank the expected chelate-ring stability for rings of size 3, 4, 5, 6, 8 membered (metal counts as one vertex). Explain the two competing effects that make the ranking non-monotonic.

Recall Solution 5.3

Expected stability ordering (most → least stable): $5 \approx 6 > 7 - 8 > 4 > 3$ Two competing effects:

Bond-angle strain (an enthalpy penalty): 3- and 4-membered rings force the donor atoms into sharp, strained angles — very unstable.
Entropic floppiness / reach (an entropy + probability penalty): 7+ membered rings have long, floppy chains, so both donor atoms rarely reach the same metal at once.

The sweet spot where both penalties are minimal is 5- and 6-membered rings — which is exactly why en (5-membered) and acac (6-membered) are workhorse chelators. See the ring-strain sketch:

Recall wrap-up

Recall One-line answers to every numeric result on this page

2.1 CN $[P t (e n) C l_{2}]$ ::: 4
2.2 CN $[F e (C_{2} O_{4})_{3}]^{3 -}$ ::: 6
2.3 CN $[C a (E D T A)]^{2 -}$ ::: 6
4.1 Extra particles released (chelate route B) ::: +3
4.2 $Δ G^{\circ}$ ::: $- 38.82 kJ mol^{- 1}$ ; $ln K$ ::: 15.67
5.1 EDTA acting through 5 arms → CN ::: 5
5.2 Linkage isomers of $S C N^{-}$ complex ::: 2; $C l^{-}$ ::: 0
5.3 Most stable ring sizes ::: 5 and 6 membered

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