Doc Brown's Chemistry  Advanced Level Inorganic Chemistry Periodic Table Revision Notes – Transition Metals

 

 Appendix 8 Complex ion stability, entropy changes and stability constants (Kstab)

This page discusses the relative stability of a wide range of transition metal ion complexes. The stability is measured in terms of the stability constant Kstab for the equilibrium expression for the formation of complex ion on interaction of the hydrated metal ion and another ligand with which it undergoes a ligand substitution reaction. Other than a variety of the 3d block transition metals and particular ligands, other factors dealt with include change in ion charge due to oxidation state, monodentate, bidentate and polydentate ligands (chelation) are considered for a particular transition metal ion.

(c) doc b GCSE/IGCSE Periodic Table Revision Notes * (c) doc b GCSE/IGCSE Transition Metals Revision Notes

INORGANIC Part 10 3d block TRANSITION METALS sub–index: 10.1–10.2 Introduction 3d–block Transition Metals * 10.3 Scandium * 10.4 Titanium * 10.5 Vanadium * 10.6 Chromium * 10.7 Manganese * 10.8 Iron * 10.9  Cobalt * 10.10 Nickel * 10.11 Copper * 10.12 Zinc * 10.13 Other Transition Metals e.g. Ag and Pt * Appendix 1. Hydrated salts, acidity of hexa–aqua ions * Appendix 2. Complexes & ligands * Appendix 3. Complexes and isomerism * Appendix 4. Electron configuration & colour theory * Appendix 5. Redox equations, feasibility, Eø * Appendix 6. Catalysis * Appendix 7. Redox equations * Appendix 8. Stability Constants and entropy changes * Appendix 9. Colorimetric analysis and complex ion formula * Appendix 10 3d block – extended data * Appendix 11 Some 3d–block compounds, complexes, oxidation states & electrode potentials * Appendix 12 Hydroxide complex precipitate 'pictures', formulae and equations

Advanced Level Inorganic Chemistry Periodic Table Index * Part 1 Periodic Table history * Part 2 Electron configurations, spectroscopy, hydrogen spectrum, ionisation energies * Part 3 Period 1 survey H to He * Part 4 Period 2 survey Li to Ne * Part 5 Period 3 survey Na to Ar * Part 6 Period 4 survey K to Kr and important trends down a group * Part 7 s–block Groups 1/2 Alkali Metals/Alkaline Earth Metals * Part 8  p–block Groups 3/13 to 0/18 * Part 9 Group 7/17 The Halogens * Part 10 3d block elements & Transition Metal Series * Part 11 Group & Series data & periodicity plots * All 11 Parts have their own sub–indexes near the top of the pages

Appendix 8. Complex ion stability, entropy changes and stability constants (Kstab)

  • Equilibrium expression for transition metal complex ion formation are readily written out, but as the case of weak acid ionisation Kc expressions, the concentration of the solvent water is considered a constant and omitted from the expression e.g.

    • Since [ ] used in depicting complex ion structure, I've used { } as well to denote concentration.

    • [Fe(H2O)6]3+(aq) + SCN(aq) ==> [Fe(H2O)5SCN]2+(aq) + H2O(l)

    • Kstab = {[Fe(H2O)5SCN]2+(aq)} / {[Fe(H2O)6]3+(aq)} {SCN(aq)} = 1.4 x 102 mol–1dm3

    • In this section {} = concentration, since [] used in complex formulae.

    • The bigger the value of Kstab the more stable the complex i.e. the equilibrium is more favoured to the right.

    • Kstab values vary considerably and are often quoted as lg Kstab = log10 (Kstab).

    • i.e. for the above equilibrium: log10 (1.4 x 102) = lg Kstab = 2.1

  • Complex formation by polydentate ligands is sometime called chelation or sequestration.

  • What governs the values of Kstab?

    • The variation of the stability constant with change in ligand is illustrated with the zinc ion.

      • The data set for zinc compares five different monodentate ligands.

      • The zinc complex ion data was one of the more comprehensive set of values I could find and it make no difference that zinc is NOT a true transition element to illustrate the point.

        • The stability constant equilibrium expression for the

        • Kstab = {[ZnL4]2+/2–(aq)} / {[Zn(H2O)4]2+(aq)} {[L(aq)]4} mol–4dm12

    • Ligand substitution reaction to give new complex ion Kstab lg Kstab
      [Zn(H2O)4]2+(aq) + 4CN(aq) ==> Zn(CN)4]2–(aq) + 4H2O(l) 5.0 x 1016 16.7
      [Zn(H2O)4]2+(aq) + 4NH3(aq) ==> Zn(NH3)4]2+(aq) + 4H2O(l) 3.8 x 109 9.58
      [Zn(H2O)4]2+(aq) + 4Cl(aq) ==> [ZnCl4]2–(aq) + 4H2O(l) 1.0 0.0
      [Zn(H2O)4]2+(aq) + 4Br(aq) ==> [ZnBr4]2–(aq) + 4H2O(l) 10–1 –1.0
      [Zn(H2O)4]2+(aq) + 4CN(aq) ==> [ZnI4]2–(aq) + 4H2O(l) 10–2 –2.0
    • The very value for the tetracyanozincate(II) in reflects the strong of central metal ion (Zn2+) – ligand (CN) bond.

    • The lower Kstab value for ammonia indicates on average a weaker dative covalent bond.

    • The ligand bonds are even weaker for the halide ions possibly due to their larger radius, since there is a steady decrease in Kstab as the halide radius increases, making the Zn–X dative covalent bond longer and weaker.

  • Why Kstab values are much higher for polydentate ligands? 'The chelate effect'

    • Here we consider ligand substitution with a bidentate of multidentate ligand to give more stable complexes and in doing so we compare the stability constants for a typical monodentate ligand, bidentate ligand and a hexadentate ligand.

    • Consider three Ni(II) complexes formed from the hexaaqua ion [Ni(H2O)6]2+:

    • (1) a monodentate ligand e.g. ammonia

      • [Ni(H2O)6]2+(aq) + 6NH3(aq) [Ni(NH3)6]2+(aq) + 6H2O(l)

      • Kstab = {[Ni(NH3)6]2+(aq)} / {[Ni(H2O)6]2+(aq)} {[NH3(aq)]6} = 4.8 x 107 mol–6 dm18

      • lg Kstab is 7.7, a typical value for a complex with a monodentate ligand (compared to the ligand water) and the Ni–NH3 bonds would appear to be stronger than the Ni–OH2 bonds.

    • (2) a bidentate ligand e.g. 1,2–diamoethane (en = H2N–CH2–CH2–NH2)

      • [Ni(H2O)6]2+(aq) + 3en(aq) [Ni(en)3]2+(aq) + 6H2O(l)

      • Kstab = {[Ni(en)3]2+(aq)} / {[Ni(H2O)6]2+(aq)} {]en(aq)]3} = 2.0 x 1018 mol–3 dm9

      • lg Kstab is 18.3, and considerably higher than for the monodentate ligand like ammonia.

    • (3) a polydentate ligand e.g. EDTA (a hexadentate ligand)

      • [Ni(H2O)6]2+(aq) + EDTA4–(aq) [Ni(EDTA)]2–(aq) + 6H2O(l) 

      • Kstab = {[Ni(EDTA)]2–(aq)} / {[Ni(H2O)6]2+(aq)} {EDTA4–(aq)} = 1 x 1019 mol–1 dm3

      • lg Kstab is 19, and even higher than for the bidentate ligand, EDTA is a hexadentate ligand.

    • (4) Theoretically adding EDTA to any of the other complexes mentioned above would cause the displacement of the original ligand e.g.
      • [Ni(NH3)6]2+(aq) + EDTA4–(aq) [Ni(EDTA)]2–(aq) + 6NH3(aq)
        • very much to the right since Kstab (Ni2+/EDTA) >> Kstab (Ni2+/NH3)
      • [Ni(en)3]2+(aq) + EDTA4–(aq) [Ni(EDTA)]2–(aq) + 3en(aq)
      • a bit more to the right than the left since Kstab (Ni2+/EDTA) just > Kstab (Ni2+/en).
    • The Kstab is much higher for polydentate ligands e.g. reactions (2/3), compared to monodentate ligands e.g. reaction (1), because of the considerable entropy increase in 'freeing' six small water molecules in reactions (2/3).

    • In (1) six small molecules displace six other small molecules (all monodentate ligands), which is likely to involve a much smaller entropy change, in (2) three larger bidentate ligands displace six smaller ligand molecules, but in (3) six small molecules are displaced by one larger hexadentate ligand molecule.

    • In general the more water molecules of water freed the greater the entropy increase because there are more ways of distributing the particles and more ways of distributing the energy of the system.

      • Therefore the Kstab is greater i.e. the equilibrium is much further to right in terms of the ligand displacement reaction.

    • In general, but with lots of exceptions!, Kstab values tend to be in the order polydentate > bidentate > monodentate ligands.

Advanced Inorganic Chemistry Page Index and Links


 

Some examples of 3d–block element complex formation and the values of Kstab

NOTE (i) By convention, the term [ H2O(l) ]n is omitted from equilibrium expressions, because water is the medium, and the bulk of the solution, therefore its concentration effectively remains constant. (ii) I'm afraid again, Kstab values to differ depending on the source!

Scandium

Titanium

Vanadium

Advanced Inorganic Chemistry Page Index and LinksChromium

  • Both the hexa–aqua ions of chromium(II) and chromium(III) readily complex with EDTA

    • [Cr(H2O)6]2+(aq) + EDTA4–(aq) ===> [Cr(EDTA)]2–(aq) + 6H2O(l)

      • Kstab = {[Cr(EDTA)3]2–(aq)} / {[Cr(H2O)6]2+(aq)} [EDTA4–(aq)]

      • Kstab = 1.0 x 1013 mol–1 dm3 [lg(Kstab) = 13.0]

    • [Cr(H2O)6]3+(aq) + EDTA4–(aq) ===> [Cr(EDTA)](aq) + 6H2O(l)

      • Kstab = {[Cr(EDTA)3](aq)} / {[Cr(H2O)6]3+(aq)} [EDTA4–(aq)]

      • Kstab = 1.0 x 1024 mol–1 dm3 [lg(Kstab) = 24.0]

    • Note that the more highly charged Cr3+(aq) ion complexes more strongly than the Cr2+(aq) ion.

Advanced Inorganic Chemistry Page Index and LinksManganese

  • The hexa–aqua manganese(II) ion readily forms complexes with polydentate ligands.

  • (i) [Mn(H2O)6]2+(aq) + 3en(aq) ===> [Mn(en)3]2+(aq) + 6H2O(l)   (en = H2NCH2CH2NH2)

    • Kstab = {[Mn(en)3]2+(aq)} / {[Mn(H2O)6]2+(aq)} [en(aq)]3

    • Kstab = 5.0 x 105 mol–3 dm9 [lg(Kstab) = 5.7]

  • (ii) [Mn(H2O)6]2+(aq) + EDTA4–(aq) ===> [Mn(EDTA)]2–(aq) + 6H2O(l)

    • Kstab = {[Mn(EDTA)3]2–(aq)} / {[Mn(H2O)6]2+(aq)} [EDTA4–(aq)]

    • Kstab = 1.0 x 1014 mol–1 dm3 [lg(Kstab) = 14.0]

  • The higher Kstab value for EDTA reflects the greater entropy change. A simplistic, but not illegitimate argument, shows that in (i) a net gain of 3 particles, but in (ii) 5 more particles are formed.

Advanced Inorganic Chemistry Page Index and LinksIron

  • Complex with thiocyanate ion

  • Fe(H2O)6]3+(aq) + SCN(aq) ==> [Fe(H2O)5SCN]2+(aq) + H2O(l)

    • Kstab = {[Fe(H2O)5SCN]2+(aq)} / {[Fe(H2O)6]3+(aq)} {SCN(aq)}

    • Kstab = 1.4 x 102 mol–1dm3 [lg(Kstab) = 2.1]

  • Initial monosubstituted complex with the fluoride ion ligand

  • [Fe(H2O)6]3+(aq) + F(aq) ==> [Fe(H2O)5F]2+(aq) + H2O(l)

    • Kstab = {[Fe(H2O)5F]2+(aq)} / {[Fe(H2O)6]3+(aq)} {F(aq)}

    • Kstab = 2.4 x 105 mol–1dm3  [lg(Kstab) = 5.38]

  • Fe3+ ions give an anionic complex in concentrated chloride ion solutions

    • [Fe(H2O)6]2+(aq) + 4Cl(aq) ==> [FeCl4](aq) + 2H2O(l)

    • formation of the tetrachlroroferrate(III) anion.

    • Kstab = {[FeCl4](aq)} / {[Fe(H2O)6]2+(aq)} [Cl(aq)]4

    • Kstab = 8 x 10–1 mol–4 dm12 [lg(Kstab) = –0.097]

  • Both the hexa–aqua ions of iron(II) and iron(III) readily complex with EDTA

    • [Fe(H2O)6]2+(aq) + EDTA4–(aq) ===> [Fe(EDTA)]2–(aq) + 6H2O(l)

      • Kstab = {[Fe(EDTA)3]2–(aq)} / {[Fe(H2O)6]2+(aq)} [EDTA4–(aq)]

      • Kstab = 2.0 x 1013 mol–1 dm3 [lg(Kstab) = 14.3]

    • [Fe(H2O)6]3+(aq) + EDTA4–(aq) ===> [Fe(EDTA)](aq) + 6H2O(l)

      • Kstab = {[Fe(EDTA)3](aq)} / {[Fe(H2O)6]3+(aq)} [EDTA4–(aq)]

      • Kstab = 1.3 x 1025 mol–1 dm3 [lg(Kstab) = 25.1]

    • Note that the more highly charged Fe3+(aq) ion complexes more strongly than the Fe2+(aq) ion.

  • Iron(III) ions complex with the bidentate ligand, the 1,2–diaminoethane molecule (en)

    • [Fe(H2O)6]3+(aq) + 3en(aq) ==> [Fe(en)3]3+(aq) + 6H2O(l)

    • Kstab = {[Fe(en)3]3+(aq)} / {[Fe(H2O)6]3+(aq)} [en(aq)]3

    • Kstab = 3.98 x 109 mol–3 dm9 [lg(Kstab) = 9.6]

Advanced Inorganic Chemistry Page Index and LinksCobalt

  • Both the hexa–aqua ions of cobalt(II) and cobalt(III) readily complex with EDTA

    • [Co(H2O)6]2+(aq) + EDTA4–(aq) ===> [Co(EDTA)]2–(aq) + 6H2O(l)

      • Kstab = {[Co(EDTA)3]2–(aq)} / {[Co(H2O)6]2+(aq)} [EDTA4–(aq)]

      • Kstab = 2.0 x 1016 mol–1 dm3 [lg(Kstab) = 16.3]

    • [Co(H2O)6]3+(aq) + EDTA4–(aq) ===> [Co(EDTA)](aq) + 6H2O(l)

      • Kstab = {[Co(EDTA)3](aq)} / {[Co(H2O)6]3+(aq)} [EDTA4–(aq)]

      • Kstab = 1.0 x 1036 mol–1 dm3 [lg(Kstab) = 36.0]

    • Note that the more highly charged Co3+(aq) ion complexes more strongly than the Co2+(aq) ion (see also below with the ammonia complexes).

  • Comparison of the stability of the hexammine complexes

    • [Co(H2O)6]2+(aq) + 6NH3(aq) ==> [Co(NH3)6]2+(aq) + 6H2O(l)

      • Kstab = {[Co(NH3)6]2+(aq)} / {[Co(H2O)6]2+(aq)} [NH3(aq)]6

      • Kstab = 7.7 x 104 mol–6 dm18  [lg(Kstab) = 4.9]

    • [Co(H2O)6]3+(aq) + 6NH3(aq) ==> [Co(NH3)6]3+(aq) + 6H2O(l)

      • Kstab = {[Co(NH3)6]2+(aq)} / {[Co(H2O)6]2+(aq)} [NH3(aq)]6

      • Kstab = 4.5 x 1033 mol–6 dm18  [lg(Kstab) = 33.7]

  • The cobalt(II) ion complexes with 1,2–diaminoethane, a bidentate ligand

    • [Co(H2O)6]2+(aq) + 3en(aq) ==> [Co(en)3]2+(aq) + 6H2O(l)

    • Kstab = {[Co(en)3]2+(aq)} / {[Co(H2O)6]2+(aq)} [en(aq)]3

Advanced Inorganic Chemistry Page Index and LinksNickel

  • [Ni(H2O)6]2+(aq) + 6NH3(aq) rev [Ni(NH3)6]2+(aq) + 6H2O(l)

    • [Ni(H2O)6]2+(aq) + 6NH3(aq) ==> [Ni(NH3)6]2+(aq) + 6H2O(l)

      • Kstab = {[Ni(NH3)6]2+(aq)} / {[Ni(H2O)6]2+(aq)} [NH3(aq)]6

      • Kstab = 4.8 x 107 mol–6 dm18  [lg(Kstab) = 7.7]

  • The hexa–aquanickel(II) ion also forms complexes with other amine ligands

    • e.g. the bidentate ligand 1,2–diaminoethane (H2N–CH2–CH2–NH2, often abbreviated to en)

    • [Ni(H2O)6]2+(aq) + 3en(aq) ===> [Ni(en)3]2+(aq) + 6H2O(l)

      • Kstab = {[Ni(en)3]2+(aq)} / {[Ni(H2O)6]2+(aq)} [en(aq)]3

      • Kstab = 2.0 x 1018 mol–3 dm9 [lg(Kstab) = 18.3]

  • The complex with EDTA is also readily formed.

    • [Ni(H2O)6]2+(aq) + EDTA4–(aq) ===> [Ni(EDTA)]2–(aq) + 6H2O(l)

      • Kstab = {[Ni(EDTA)3]2–(aq)} / {[Ni(H2O)6]2+(aq)} [EDTA4–(aq)]

      • Kstab = 1.0 x 1019 mol–1 dm3 [lg(Kstab) = 19.0]

    • Note that Kstab for the same ion tend to increase the greater the chelating power of an individual ligand in terms of the ligand bond formed – mainly due to the increase in entropy as more particles are formed by the polydentate ligands

    • e.g. for the same nickel(II) ion Kstab(EDTA) > Kstab(en) > Kstab(NH3)

  • Ni2+ forms the tetrachloronickelate(II) ion, [NiCl4]2–, a tetrahedral anionic complex with the chloride ion (Cl).

    • [Ni(H2O)6]2+(aq) + 4Cl(aq) ==> [NiCl4]2–(aq) + 6H2O(l)

    • Kstab = {[NiCl4]2–(aq)} / {[Ni(H2O)6]2+(aq)} [Cl(aq)]4

    • Kstab = ? mol4 dm–12  [lg(Kstab) = ?]

  • Ni2+ forms the tetracyanonickelate(II) ion, [Ni(CN)4]2–, a square planar anionic complex with the cyanide ion (CN).

    • [Ni(H2O)6]2+(aq) + 4CN(aq) ==> [NiCN4]2–(aq) + 6H2O(l)

    • Kstab = {[NiCN4]2–(aq)} / {[Ni(H2O)6]2+(aq)} [CN(aq)]4

    • Kstab = 2 x 1031 mol4 dm–12  [lg(Kstab) = 31.3]

  • Its likely that the more bulky chloride ion (radius Cl > C) 'forces' the formation of the tetrahedral shape rather than a square planar shaped complex.

Advanced Inorganic Chemistry Page Index and LinksCopper

  • [Cu(H2O)6]2+(aq) + 4NH3(aq) rev [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)

    • Kstab = [ [Cu(NH3)4(H2O)2]2+(aq) ] / [ [Cu(H2O)6]2+(aq) ] [ NH3 (aq) ]4 = 1.0 x 1012 mol–4 dm12 (lg Kstab = 12)

Zinc

 


Scandium * Titanium * Vanadium * Chromium * Manganese * Iron * Cobalt * Nickel * Copper * Zinc * Silver & Platinum

Introduction 3d–block Transition Metals * Appendix 1. Hydrated salts, acidity of hexa–aqua ions * Appendix 2. Complexes & ligands * Appendix 3. Complexes and isomerism * Appendix 4. Electron configuration & colour theory * Appendix 5. Redox equations, feasibility, Eø * Appendix 6. Catalysis * Appendix 7. Redox equations * Appendix 8. Stability Constants and entropy changes * Appendix 9. Colorimetric analysis and complex ion formula * Appendix 10 3d block – extended data * Appendix 11 Some 3d–block compounds, complexes, oxidation states & electrode potentials * Appendix 12 Hydroxide complex precipitate 'pictures', formulae and equations

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