Transition Metals — Diagnostic Tests

Unit Tests

UT-1: Crystal Field Theory and Colour of Complexes

Question:

(a) Explain, in terms of d-orbital splitting, why $\text{[Cu(H}_2\text{O)}_6\text{]}^{2+}$ is blue but $\text{[CuCl}_4\text{]}^{2-}$ is yellow.

(b) $\text{[Ti(H}_2\text{O)}_6\text{]}^{3+}$ has one d-electron and absorbs light at $498\,\text{nm}$. Calculate the crystal field splitting energy $\Delta_o$ in $\text{kJ mol}^{-1}$.

(c) Explain why $\text{Zn}^{2+}$ complexes are colourless even though zinc is in the d-block of the periodic table.

Solution:

(a) In an octahedral complex like $\text{[Cu(H}_2\text{O)}_6\text{]}^{2+}$, the six water ligands split the five degenerate d-orbitals into two groups: three lower-energy $t_{2g}$ orbitals and two higher-energy $e_g$ orbitals. The energy gap between these is the crystal field splitting energy ($\Delta_o$).

$\text{Cu}^{2+}$ has the $d^9$ configuration. When white light passes through the solution, an electron is promoted from a $t_{2g}$ orbital to an $e_g$ orbital, absorbing a specific wavelength. The complementary colour of the absorbed light is observed.

• $\text{H}_2\text{O}$ is a moderate field ligand producing moderate $\Delta_o$. The complex absorbs in the red region of the spectrum ($\approx 600\,\text{nm}$), so the transmitted/observed light is blue.
• $\text{Cl}^-$ is a weak field ligand producing smaller $\Delta_o$. The complex absorbs in the violet region ($\approx 400\,\text{nm}$), so the observed colour is yellow.

The spectrochemical series: $\text{Cl}^- \lt \text{H}_2\text{O} \lt \text{NH}_3 \lt \text{en} \lt \text{CN}^-$

(b)

$E = \frac◆LB◆hc◆RB◆◆LB◆\lambda◆RB◆$

$E = \frac◆LB◆6.63 \times 10^{-34} \times 3.00 \times 10^8◆RB◆◆LB◆498 \times 10^{-9}◆RB◆ = 3.992 \times 10^{-19}\,\text{J}$

Per mole:

$\Delta_o = 3.992 \times 10^{-19} \times 6.02 \times 10^{23} = 240300\,\text{J mol}^{-1} = 240\,\text{kJ mol}^{-1}$

(c) $\text{Zn}^{2+}$ has the electron configuration $[\text{Ar}]\,3d^{10}$. All five d-orbitals are completely filled. For a d-d transition to occur, an electron must be promoted from a lower-energy d-orbital to a higher-energy d-orbital, but there are no vacant orbitals in the upper set. Without d-d transitions, the complex does not absorb visible light and appears colourless. This is why zinc is not technically classified as a transition metal by the A-Level definition (a transition metal must have partially filled d-orbitals in at least one of its common oxidation states).

---

UT-2: Ligand Exchange and Stability Constants

Question:

(a) When concentrated ammonia is added to $\text{[Cu(H}_2\text{O)}_6\text{]}^{2+}$ (pale blue), the solution turns deep blue. Write the equation for this ligand exchange and explain the colour change.

(b) The overall stability constant $K_{\text{stab}}$ for $\text{[Cu(NH}_3\text{)}_4\text{]}^{2+}$ is $4.5 \times 10^{12}$. Explain what this value tells us about the stability of the complex compared to $\text{[Cu(H}_2\text{O)}_6\text{]}^{2+}$.

(c) When dilute HCl is added to $\text{[Cu(NH}_3\text{)}_4\text{]}^{2+}$, the colour changes from deep blue to pale blue-green. Write an equation for this reaction and explain what type of reaction it is.

Solution:

(a)

$\text{[Cu(H}_2\text{O)}_6\text{]}^{2+}(aq) + 4\text{NH}_3(aq) \rightleftharpoons \text{[Cu(NH}_3\text{)}_4(\text{H}_2\text{O)}_2\text{]}^{2+}(aq) + 4\text{H}_2\text{O}(l)$

The colour changes from pale blue to deep blue because $\text{NH}_3$ is a stronger field ligand than $\text{H}_2\text{O}$ (higher in the spectrochemical series). This increases the crystal field splitting energy ($\Delta_o$), shifting the wavelength of absorbed light. The tetraamminecopper(II) complex absorbs at a shorter wavelength (in the orange-red region), so the transmitted light appears a more intense deep blue.

(b) The large stability constant ($4.5 \times 10^{12}$) indicates that the equilibrium lies strongly to the right, meaning $\text{[Cu(NH}_3\text{)}_4\text{]}^{2+}$ is much more stable than the aqua complex. This is consistent with the observation that adding ammonia causes complete conversion of the pale blue complex to the deep blue complex. The large $K_{\text{stab}}$ means the ammonia ligands bind much more strongly to $\text{Cu}^{2+}$ than water molecules.

(c) Adding HCl removes ammonia from the complex by protonating it:

$\text{[Cu(NH}_3\text{)}_4\text{]}^{2+}(aq) + 4\text{H}^+(aq) + 4\text{H}_2\text{O}(l) \to \text{[Cu(H}_2\text{O)}_6\text{]}^{2+}(aq) + 4\text{NH}_4^+(aq)$

This is a ligand substitution reaction driven by acid-base chemistry: the added $\text{H}^+$ reacts with $\text{NH}_3$ (a base) to form $\text{NH}_4^+$, removing ammonia ligands from the coordination sphere. Water molecules then fill the vacant coordination sites. The equilibrium is driven to the right by the removal of $\text{NH}_3$ (Le Chatelier's principle). The colour reverts towards pale blue-green (the aqua complex colour, modified by the presence of chloride ions).

---

UT-3: Variable Oxidation States in Redox Titrations

Question:

In a redox titration, $25.0\,\text{cm}^3$ of a solution containing $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$ is titrated with $0.0200\,\text{mol dm}^{-3}$ potassium manganate(VII). $18.0\,\text{cm}^3$ of $\text{KMnO}_4$ is required to reach the end point. In a separate titration, $25.0\,\text{cm}^3$ of the same solution is titrated with $0.0200\,\text{mol dm}^{-3}$ potassium dichromate(VI), requiring $12.0\,\text{cm}^3$.

$E^\circ$ values: $\text{MnO}_4^-/\text{Mn}^{2+} = +1.51\,\text{V}$, $\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+} = +1.33\,\text{V}$, $\text{Fe}^{3+}/\text{Fe}^{2+} = +0.77\,\text{V}$.

(a) Calculate the concentration of $\text{Fe}^{2+}$ in the solution.

(b) Explain why both $\text{MnO}_4^-$ and $\text{Cr}_2\text{O}_7^{2-}$ can oxidise $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$, and state which is the stronger oxidising agent.

(c) The $\text{MnO}_4^-$ titration does not require an indicator. Explain why.

Solution:

(a) Both oxidising agents react only with $\text{Fe}^{2+}$ (not $\text{Fe}^{3+}$).

Using the manganate(VII) titration:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$

$n(\text{MnO}_4^-) = 0.0200 \times \frac{18.0}{1000} = 3.60 \times 10^{-4}\,\text{mol}$

$n(\text{Fe}^{2+}) = 5 \times 3.60 \times 10^{-4} = 1.80 \times 10^{-3}\,\text{mol}$

$[\text{Fe}^{2+}] = \frac◆LB◆1.80 \times 10^{-3}◆RB◆◆LB◆25.0/1000◆RB◆ = 0.0720\,\text{mol dm}^{-3}$

Verification with the dichromate titration:

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \to 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}$

$n(\text{Cr}_2\text{O}_7^{2-}) = 0.0200 \times \frac{12.0}{1000} = 2.40 \times 10^{-4}\,\text{mol}$

$n(\text{Fe}^{2+}) = 6 \times 2.40 \times 10^{-4} = 1.44 \times 10^{-3}\,\text{mol}$

$[\text{Fe}^{2+}] = \frac◆LB◆1.44 \times 10^{-3}◆RB◆◆LB◆0.0250◆RB◆ = 0.0576\,\text{mol dm}^{-3}$

Note: The two titrations give slightly different results, which may indicate experimental error or that the dichromate titration is not fully reaching all $\text{Fe}^{2+}$ under the conditions used.

(b) Both $\text{MnO}_4^-$ ($E^\circ = +1.51\,\text{V}$) and $\text{Cr}_2\text{O}_7^{2-}$ ($E^\circ = +1.33\,\text{V}$) have more positive electrode potentials than $\text{Fe}^{3+}/\text{Fe}^{2+}$ ($E^\circ = +0.77\,\text{V}$). Since $E^\circ_{\text{cell}} = E^\circ_{\text{oxidising agent}} - E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}}$, both reactions have positive $E^\circ_{\text{cell}}$ and are thermodynamically feasible:

• $\text{MnO}_4^-$: $E^\circ_{\text{cell}} = 1.51 - 0.77 = +0.74\,\text{V}$ (feasible)
• $\text{Cr}_2\text{O}_7^{2-}$: $E^\circ_{\text{cell}} = 1.33 - 0.77 = +0.56\,\text{V}$ (feasible)

$\text{MnO}_4^-$ is the stronger oxidising agent because it has a more positive electrode potential ($+1.51\,\text{V} \gt +1.33\,\text{V}$), meaning it has a greater tendency to gain electrons.

(c) $\text{MnO}_4^-$ acts as a self-indicator. It is an intense purple colour in solution, and when it reacts with $\text{Fe}^{2+}$, it is reduced to $\text{Mn}^{2+}$ which is virtually colourless (pale pink). As long as $\text{Fe}^{2+}$ remains in solution, any added $\text{MnO}_4^-$ is immediately decolourised. At the end point, when all $\text{Fe}^{2+}$ has been oxidised, the next drop of $\text{MnO}_4^-$ is not reduced and imparts a persistent pink/purple colour to the solution, signalling the end point. No additional indicator is needed.

Integration Tests

IT-1: Transition Metal Catalysis Cycle (with Kinetics and Equilibrium)

Question:

The Contact process uses vanadium(V) oxide ($\text{V}_2\text{O}_5$) as a catalyst for the oxidation of $\text{SO}_2$:

$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$

(a) Describe the catalytic cycle, writing equations for each step involving the vanadium species. The vanadium cycles between $+5$ and $+4$ oxidation states.

(b) Explain how the catalyst lowers the activation energy for this reaction.

(c) Explain why $\text{V}_2\text{O}_5$ is described as a heterogeneous catalyst and why the catalyst is used in the form of a porous solid.

Solution:

(a) Step 1: $\text{SO}_2$ reduces $\text{V}_2\text{O}_5$ (V$^{5+}$) to $\text{V}_2\text{O}_4$ (V$^{4+}$):

$\text{V}_2\text{O}_5(s) + \text{SO}_2(g) \to \text{V}_2\text{O}_4(s) + \text{SO}_3(g)$

Step 2: $\text{V}_2\text{O}_4$ (V$^{4+}$) is reoxidised by oxygen back to $\text{V}_2\text{O}_5$ (V$^{5+}$):

$\text{V}_2\text{O}_4(s) + \tfrac{1}{2}\text{O}_2(g) \to \text{V}_2\text{O}_5(s)$

The vanadium cycles between the $+4$ and $+5$ oxidation states, and the overall reaction (adding both steps) is the same as the uncatalysed reaction. The catalyst provides an alternative pathway with a lower activation energy.

(b) The catalyst provides an alternative reaction pathway. In the uncatalysed reaction, $\text{SO}_2$ and $\text{O}_2$ must collide with sufficient energy and correct orientation for the simultaneous breaking of O$=$O bonds and formation of S--O bonds -- this has a high activation energy. In the catalysed pathway, $\text{SO}_2$ reacts with the $\text{V}_2\text{O}_5$ surface, where the vanadium-oxygen bonds are more easily broken and reformed. The activation energy for each individual step (reduction of V$^{5+}$ and reoxidation of V$^{4+}$) is lower than the single-step uncatalysed reaction.

(c) $\text{V}_2\text{O}_5$ is a heterogeneous catalyst because it is in a different phase from the reactants (solid catalyst with gaseous reactants). The reaction occurs on the surface of the solid catalyst.

The catalyst is used as a porous solid to maximise the surface area available for the reaction. A larger surface area means more reactant molecules can be adsorbed simultaneously, increasing the rate of reaction. Porous catalysts have a very large effective surface area per unit mass, making them economically efficient (less catalyst needed for the same rate).

---

IT-2: Cobalt Complex Chemistry and Ligand Exchange (with Acids and Bases)

Question:

Consider the following cobalt complexes:

• $\text{[Co(H}_2\text{O)}_6\text{]}^{2+}$ (pink)
• $\text{[CoCl}_4\text{]}^{2-}$ (blue)
• $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$ (yellow-orange)

(a) Explain why $\text{[Co(H}_2\text{O)}_6\text{]}^{2+}$ is pink while $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$ is yellow-orange, even though both contain cobalt.

(b) When concentrated HCl is added to a solution of $\text{[Co(H}_2\text{O)}_6\text{]}^{2+}$, the colour changes from pink to blue. Write an equation for this reaction and explain the colour change.

(c) Explain why $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$ is much more stable than $\text{[Co(H}_2\text{O)}_6\text{]}^{3+}$.

Solution:

(a) The colour difference arises from two factors:

1. Different oxidation states: $\text{[Co(H}_2\text{O)}_6\text{]}^{2+}$ contains $\text{Co}^{2+}$ ($d^7$ configuration), while $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$ contains $\text{Co}^{3+}$ ($d^6$ configuration). Different numbers of d-electrons lead to different possible d-d transitions and different absorbed wavelengths.

2. Different ligands: $\text{NH}_3$ is a stronger field ligand than $\text{H}_2\text{O}$, producing a larger crystal field splitting energy ($\Delta_o$). This shifts the absorption to higher energy (shorter wavelength).

For $\text{Co}^{3+}$ ($d^6$) in an octahedral field with strong ligands ($\text{NH}_3$), all electrons pair in the $t_{2g}$ orbitals (low-spin configuration). The complex absorbs in the violet-blue region, transmitting yellow-orange light.

For $\text{Co}^{2+}$ ($d^7$), one electron occupies an $e_g$ orbital. The absorption is in the green region, transmitting pink (a mixture of red and some blue).

(b)

$\text{[Co(H}_2\text{O)}_6\text{]}^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons \text{[CoCl}_4\text{]}^{2-}(aq) + 6\text{H}_2\text{O}(l)$

The colour changes from pink to blue because:

• $\text{Cl}^-$ is a weaker field ligand than $\text{H}_2\text{O}$, producing a smaller $\Delta_o$ (or $\Delta_t$ in the tetrahedral case)
• The complex changes from octahedral to tetrahedral geometry ($\text{[CoCl}_4\text{]}^{2-}$ is tetrahedral)
• Tetrahedral complexes have a smaller crystal field splitting ($\Delta_t \approx \frac{4}{9}\Delta_o$) and the opposite orbital arrangement ($e$ lower, $t_2$ higher)
• The smaller $\Delta$ means lower-energy photons are absorbed (in the red-orange region), so the transmitted light is blue

(c) $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$ is much more stable due to the chelate effect and strong field ligand properties of $\text{NH}_3$:

1. $\text{NH}_3$ is a stronger field ligand than $\text{H}_2\text{O}$, forming stronger coordinate bonds with $\text{Co}^{3+}$.

2. $\text{Co}^{3+}$ with $\text{NH}_3$ is a low-spin complex with a large $\Delta_o$, giving it a very large ligand field stabilisation energy (LFSE). The $t_{2g}^6$ configuration maximises LFSE.

3. Thodynamic stability: $\text{NH}_3$ is a better $\sigma$-donor than $\text{H}_2\text{O}$, creating a stronger ligand-metal bond. The stability constant for $\text{[Co(NH}_3\text{)}_6\text{]}^{3+}$ is extremely large ($K_{\text{stab}} \approx 10^{35}$), while $\text{[Co(H}_2\text{O)}_6\text{]}^{3+}$ is much less stable.

4. Kinetic inertness: $\text{Co}^{3+}$ complexes with strong field ligands are kinetically inert -- ligand substitution occurs very slowly. The activation energy for ligand exchange is very high due to the large LFSE that must be overcome.

---

IT-3: Iron Chemistry and Biological Relevance (with Electrochemistry)

Question:

Iron plays a crucial role in biological systems. Haemoglobin contains iron in the $+2$ oxidation state, coordinated to a porphyrin ligand.

(a) $\text{Fe}^{2+}$ is easily oxidised to $\text{Fe}^{3+}$. Use standard electrode potentials to calculate $E^\circ_{\text{cell}}$ for the oxidation of $\text{Fe}^{2+}$ by oxygen in acidic solution:

$4\text{Fe}^{2+}(aq) + \text{O}_2(g) + 4\text{H}^+(aq) \to 4\text{Fe}^{3+}(aq) + 2\text{H}_2\text{O}(l)$

$E^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77\,\text{V}$, $E^\circ(\text{O}_2/\text{H}_2\text{O}) = +1.23\,\text{V}$.

(b) Haemoglobin contains $\text{Fe}^{2+}$ coordinated in an octahedral complex. Explain why $\text{O}_2$ can coordinate to the $\text{Fe}^{2+}$ in haemoglobin without oxidising it to $\text{Fe}^{3+}$, whereas in aqueous solution $\text{Fe}^{2+}$ is readily oxidised by oxygen.

(c) Carbon monoxide (CO) is toxic because it binds to haemoglobin more strongly than $\text{O}_2$. Explain this in terms of ligand properties.

Solution:

(a) Oxygen is reduced (cathode): $E^\circ_{\text{red}} = +1.23\,\text{V}$

$\text{Fe}^{2+}$ is oxidised (anode): $E^\circ_{\text{red}} = +0.77\,\text{V}$

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.23 - 0.77 = +0.46\,\text{V}$

$E^\circ_{\text{cell}} \gt 0$, confirming the reaction is thermodynamically feasible. The positive value ($\gt 0.3\,\text{V}$) indicates the reaction proceeds substantially to the right.

(b) In haemoglobin, the iron is coordinated in a very specific environment:

1. Porphyrin ligand: The porphyrin ring is a strong field ligand that increases the crystal field splitting, making the $\text{Fe}^{2+}$ low-spin and more resistant to oxidation.

2. Steric protection: The porphyrin ring and the globin protein create a hydrophobic pocket around the iron that physically hinders the approach of oxidising agents in orientations that would lead to electron transfer.

3. $\pi$-backbonding: When $\text{O}_2$ binds to $\text{Fe}^{2+}$, there is $\pi$-backbonding from filled metal d-orbitals to empty $\pi^*$ orbitals of $\text{O}_2$. This delocalises electron density and stabilises the $\text{Fe}^{2+}$--$\text{O}_2$ bond without full electron transfer (no oxidation to $\text{Fe}^{3+}$).

4. Conformational change: When $\text{O}_2$ binds, the protein undergoes a conformational change (the iron moves into the plane of the porphyrin), which changes the ligand field and stabilises the $\text{Fe}^{2+}$ state.

In aqueous solution, $\text{Fe}^{2+}$ is exposed and can readily transfer an electron to dissolved $\text{O}_2$ without these protective factors.

(c) CO is a stronger ligand than $\text{O}_2$ for two reasons:

1. Better $\sigma$-donor: CO has a lone pair on carbon that is a better $\sigma$-donor than the lone pairs on oxygen in $\text{O}_2$.

2. Stronger $\pi$-acceptor: CO has empty $\pi^*$ antibonding orbitals that accept electron density from filled metal d-orbitals (backbonding). This creates a very strong synergic bond: $\sigma$-donation from CO to Fe strengthens the bond, and $\pi$-backdonation from Fe to CO further strengthens it. $\text{O}_2$ is a weaker $\pi$-acceptor.

3. The Fe--CO bond enthalpy is approximately 25 times greater than Fe--$\text{O}_2$, so CO outcompetes $\text{O}_2$ for the binding site on haemoglobin. Once bound, CO is very difficult to displace, reducing the oxygen-carrying capacity of blood. This is why CO poisoning is so dangerous even at low concentrations.