Topic 17

Topic 17 on a page

Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.

Definitions to learn word for word

See Transition metals and their electrons, Ligands and complex ions and Two kinds of catalyst.

Electrons and oxidation numbers

Rule Detail
Filling 4s before 3d; exceptions Cr [Ar] 3d⁵ 4s¹ and Cu [Ar] 3d¹⁰ 4s¹
Ions lose 4s first, then 3d: Fe²⁺ [Ar] 3d⁶, Fe³⁺ [Ar] 3d⁵, Cu²⁺ [Ar] 3d⁹
Not transition metals Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰)
Variable oxidation number 4s and 3d close in energy, so ionisation energies rise gradually
Highest oxidation number (Sc to Mn) number of 4s + 3d electrons

Shapes

Coordination number Shape Angle Examples
6 octahedral 90° [Cu(H₂O)₆]²⁺, [Ni(NH₃)₆]²⁺, [Cr(OH)₆]³⁻
4 tetrahedral (large Cl⁻) 109.5° [CuCl₄]²⁻, [CoCl₄]²⁻
4 square planar 90° cis-platin, [Pt(NH₃)₂Cl₂]

Ligands that count: H₂O, NH₃, OH⁻, Cl⁻ (one bond each); NH₂CH₂CH₂NH₂ (two); EDTA⁴⁻ (six).

Colour

The d-orbitals split by ligands, with a d electron promoted by absorbing light.
Energy levels of the d-orbitals in an isolated ion and in a complex, with a d electron promoted by absorbed light.

Vanadium and chromium

Oxidation number Vanadium Chromium
+6 Cr₂O₇²⁻ orange; CrO₄²⁻ yellow
+5 VO₂⁺ yellow
+4 VO²⁺ blue
+3 V³⁺ green Cr³⁺ green; [Cr(OH)₆]³⁻ dark green
+2 V²⁺ violet Cr²⁺ blue

Observations with NaOH and NH₃

Ion A few drops NaOH Excess NaOH A few drops NH₃ Excess NH₃
Cr³⁺ green ppt dark green solution green ppt purple solution
Mn²⁺ off-white ppt, browns in air no change off-white ppt no change
Fe²⁺ green ppt, browns in air no change green ppt no change
Fe³⁺ orange-brown ppt no change orange-brown ppt no change
Co²⁺ blue ppt, pink on standing no change blue ppt straw solution, darkens in air
Ni²⁺ green ppt no change green ppt blue solution
Cu²⁺ blue ppt no change blue ppt deep blue solution
Zn²⁺ white ppt colourless solution white ppt colourless solution

Three kinds of reaction:

Details are on pages 11, 12 and 13.

Ligand exchange, stability, haemoglobin

Catalysts

Catalyst Type How it works
V₂O₅(s), contact process heterogeneous V +5 → +4 (SO₂ → SO₃), then +4 → +5 (O₂)
Pt/Rh(s), catalytic converter heterogeneous CO and NO adsorbed, bonds weakened, react; CO₂ and N₂ desorbed
Fe²⁺(aq), I⁻ with S₂O₈²⁻ homogeneous Fe²⁺ → Fe³⁺ → Fe²⁺; intermediate Fe³⁺
Mn²⁺(aq), MnO₄⁻ with C₂O₄²⁻ homogeneous, autocatalysis product catalyses; slow, fast, slowing

Standard electrode potentials used in this topic

Couple E° / V
V²⁺/V −1.18
Zn²⁺/Zn −0.76
Cr³⁺/Cr²⁺ −0.41
V³⁺/V²⁺ −0.26
CrO₄²⁻/Cr(OH)₃ (alkali) −0.13
Sn⁴⁺/Sn²⁺ +0.15
VO²⁺/V³⁺ +0.34
I₂/I⁻ +0.54
Fe³⁺/Fe²⁺ +0.77
H₂O₂/OH⁻ (alkali) +0.88
VO₂⁺/VO²⁺ +1.00
O₂/H₂O +1.23
Cr₂O₇²⁻/Cr³⁺ +1.33
S₂O₈²⁻/SO₄²⁻ +2.01

The ten traps

  1. Writing Fe²⁺ as [Ar] 3d⁴ 4s²: remove the 4s electrons first.
  2. Defining a transition metal by the atom: it is the ion with an incompletely filled d sub-shell.
  3. Counting ligands, not dative bonds, for the coordination number.
  4. Drawing the bond from H, not O, in an aqua complex; leaving out brackets or the charge.
  5. Saying the electron emits the colour. The colour seen is the light not absorbed.
  6. Saying the first green in the vanadium series is V³⁺. It is a mixture.
  7. Treating yellow to orange (chromate to dichromate) as redox. Cr is +6 in both.
  8. Writing NH₃ as a ligand in the first step. A few drops of ammonia are a base.
  9. Calling the dissolving in excess ammonia amphoteric. It is ligand exchange.
  10. Explaining chelate stability by "more bonds" instead of more particles and a positive ΔSsystem.

Mixed test

Try these with no notes, then check the answers.

  1. Write the configuration of Co²⁺ (Z = 27) and explain why cobalt is a transition metal.
  2. Give the oxidation number of manganese in MnO₄⁻ and the coordination number of nickel in [Ni(NH₂CH₂CH₂NH₂)₃]²⁺.
  3. Explain why [Ni(H₂O)₆]²⁺ is green but [Zn(H₂O)₆]²⁺ is colourless.
  4. Will acidified dichromate(VI) oxidise Fe²⁺? Use E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V and E°(Fe³⁺/Fe²⁺) = +0.77 V, and write the equation.
  5. Describe what you would see when sodium hydroxide is added to Zn²⁺(aq) dropwise and then in excess. Write both equations.
  6. Describe what happens when excess ammonia is added to Ni²⁺(aq), and write the equation starting from [Ni(H₂O)₆]²⁺.
  7. Pink [Co(H₂O)₆]²⁺ turns blue in concentrated hydrochloric acid. Give the equation and the changes in coordination number and shape.
  8. Explain why [Cu(en)₃]²⁺ is more stable than [Cu(H₂O)₆]²⁺, given [Cu(HX2O)X6]X2++3 en→[Cu(en)X3]X2++6 HX2O\ce{[Cu(H2O)6]^2+ + 3en -> [Cu(en)3]^2+ + 6H2O}.
  9. Why is carbon monoxide toxic?
  10. A student starts the preparation of chromium(II) ethanoate with 4.00 g of K₂Cr₂O₇ (MrM_\text{r} = 294.2) and obtains 2.60 g of [Cr₂(CH₃COO)₄(H₂O)₂] (MrM_\text{r} = 376.0). Calculate the percentage yield.

Answers cover these until you have tried the questions

  1. Co²⁺ has 25 electrons: [Ar] 3d⁷ 4s² loses the 4s electrons, giving [Ar] 3d⁷. Cobalt forms stable ions (Co²⁺, 3d⁷) with an incompletely filled d sub-shell, so it is a transition metal.
  2. MnO₄⁻: x − 8 = −1, so +7. Three bidentate ligands, each forming 2 dative bonds: coordination number 6.
  3. Ni²⁺ is 3d⁸: ligands split the d-orbitals, an electron absorbs visible light and is promoted, and the remaining light is seen (green). Zn²⁺ is 3d¹⁰: the d sub-shell is full, no electron can be promoted, so no visible light is absorbed.
  4. E°cell = +1.33 − (+0.77) = +0.56 V, positive, so yes. CrX2OX7X2−+14 HX++6 FeX2+→2 CrX3++6 FeX3++7 HX2O\ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 6Fe^3+ + 7H2O}
  5. A white precipitate forms, which dissolves in excess to a colourless solution. [Zn(HX2O)X6]X2++2 OHX−→[Zn(OH)X2(HX2O)X4]+2 HX2O\ce{[Zn(H2O)6]^2+ + 2OH- -> [Zn(OH)2(H2O)4] + 2H2O}; [Zn(OH)X2(HX2O)X4]+2 OHX−→[Zn(OH)X4]X2−+4 HX2O\ce{[Zn(OH)2(H2O)4] + 2OH- -> [Zn(OH)4]^2- + 4H2O}
  6. A green precipitate forms with a few drops, and dissolves in excess ammonia to a blue solution. [Ni(HX2O)X6]X2++6 NHX3→[Ni(NHX3)X6]X2++6 HX2O\ce{[Ni(H2O)6]^2+ + 6NH3 -> [Ni(NH3)6]^2+ + 6H2O}
  7. [Co(HX2O)X6]X2++4 ClX−⇌[CoClX4]X2−+6 HX2O\ce{[Co(H2O)6]^2+ + 4Cl- <=> [CoCl4]^2- + 6H2O}. Coordination number 6 to 4; octahedral to tetrahedral.
  8. Particles: 1 + 3 = 4 on the left, 1 + 6 = 7 on the right, so ΔSsystem is positive. Six dative bonds are broken and six made, so ΔH ≈ 0. ΔStotal is positive, the exchange is feasible and K is large.
  9. CO replaces O₂ on the Fe²⁺ of haemoglobin by ligand exchange. It binds more strongly, so it is not easily released and the blood cannot carry oxygen.
  10. n(K₂Cr₂O₇) = 4.00 ÷ 294.2 = 0.013596 mol, which gives 0.013596 mol of product. Theoretical mass = 0.013596 × 376.0 = 5.112 g. Yield = 2.60 ÷ 5.112 × 100 = 50.9%.