A2 Chemistry 9701: Ni(en)3. Bidentate Ligands, Shape, Isomerism and Kstab

1,2-diaminoethane, $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ (commonly abbreviated "en"), is a bidentate ligand. In aqueous solution, nickel(II) ions normally exist as the octahedral complex $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ . Adding excess en to this solution replaces all six water ligands, forming $[\text{Ni}(\text{en})_3]^{2+}$ :

$[\text{Ni}(\text{H}_2\text{O})_6]^{2+}(aq) + 3\text{en}(aq) \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+}(aq) + 6\text{H}_2\text{O}(l)$

(a) State what is meant by the term bidentate ligand, and use this to explain why three molecules of en (rather than six) are needed to replace the six water ligands in this ligand-exchange reaction. [2]

(b) State the coordination number of nickel, and the shape, of the $[\text{Ni}(\text{en})_3]^{2+}$ ion. [1]

(c) $[\text{Ni}(\text{en})_3]^{2+}$ exists as two non-superimposable mirror-image forms (it shows optical isomerism), whereas $[\text{Ni}(\text{NH}_3)_6]^{2+}$ , which also contains six monodentate nitrogen-donor ligands, does not. Suggest a reason for this difference. [2]

(d) Write an expression for the stability constant, $K_{stab}$ , of the ligand-exchange reaction shown above, and state, with a reason, whether you would expect $K_{stab}$ for this reaction to be large or small. [3]

Part (a): Bidentate ligands and why three en molecules are needed

A bidentate ligand is a ligand that can form two coordinate (dative covalent) bonds to a central metal ion, by donating two separate lone pairs of electrons from two different donor atoms within the same molecule. In en, $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ , each of the two nitrogen atoms carries a lone pair, so a single en molecule can occupy two of the coordination positions around the metal ion.

Since $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ has six monodentate water ligands (six coordinate bonds in total), replacing all of them while keeping the same coordination number of 6 requires ligands that together supply six coordinate bonds. Because each en supplies 2 bonds, exactly $6\div2=3$ en molecules are needed, not six, since six en molecules would supply twelve bonds, far more than the nickel ion’s coordination sphere can accommodate.

Part (b): Coordination number and shape

Nickel forms six coordinate bonds in total (two from each of the three en ligands), so:

Coordination number $=6$
Shape $=$ octahedral

Part (c): Why optical isomerism arises here but not for $[\text{Ni}(\text{NH}_3)_6]^{2+}$

A complex shows optical isomerism (is chiral) only if it has no plane of symmetry, so that its mirror image cannot be superimposed on the original.

In $[\text{Ni}(\text{en})_3]^{2+}$ , the three bidentate en ligands each span two adjacent coordination positions, wrapping around the octahedral nickel centre in a twisted, propeller-like arrangement. This arrangement has no plane of symmetry, so the complex and its mirror image are two distinct, non-superimposable structures, the two optical isomers (enantiomers).

By contrast, $[\text{Ni}(\text{NH}_3)_6]^{2+}$ contains six identical, monodentate ligands, each independently occupying one coordination position with no “linking” between positions. This gives the ion a high degree of symmetry (it possesses planes of symmetry), so its mirror image can be superimposed onto the original structure. It is achiral, and no optical isomerism occurs.

Part (d): Stability constant expression and its expected magnitude

For the ligand-exchange equilibrium $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}(aq) + 3\text{en}(aq) \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+}(aq) + 6\text{H}_2\text{O}(l)$ water is the solvent (present in large, effectively constant excess) and is not included in the $K_{stab}$ expression: $K_{stab}=\frac{[[\text{Ni}(\text{en})_3]^{2+}]}{[[\text{Ni}(\text{H}_2\text{O})_6]^{2+}][\text{en}]^3}$

Expected magnitude: large. This is an example of the chelate effect. Counting the free aqueous particles (excluding the solvent water) on each side of the equation: the left-hand side has $1+3=4$ particles ( $1$ complex ion $+3$ en molecules), while the right-hand side releases $6$ water molecules in addition to the $1$ new complex ion. Because more particles are released than are consumed in forming the chelate complex, the reaction increases the total disorder of the system, so $\Delta S$ is positive. A positive $\Delta S$ makes $\Delta G^{\ominus}=\Delta H^{\ominus}-T\Delta S^{\ominus}$ more negative, and since $\Delta G^{\ominus}=-RT\ln K_{stab}$ , a more negative $\Delta G^{\ominus}$ corresponds to a larger $K_{stab}$ . Chelating (multidentate) ligands therefore generally form substantially more stable complexes than the equivalent number of monodentate ligands.

Final answers

(a) A bidentate ligand forms two coordinate bonds via two donor lone pairs; three en molecules ( $3\times2=6$ bonds) replace six monodentate $\text{H}_2\text{O}$ ligands.
(b) Coordination number $=6$ ; shape $=$ octahedral.
(c) The three chelating en ligands give an asymmetric arrangement with no plane of symmetry (chiral), unlike the highly symmetric $[\text{Ni}(\text{NH}_3)_6]^{2+}$ (achiral).
(d) $K_{stab}=\dfrac{[[\text{Ni}(\text{en})_3]^{2+}]}{[[\text{Ni}(\text{H}_2\text{O})_6]^{2+}][\text{en}]^3}$ , expected to be large, due to the chelate effect (favourable increase in entropy).