CHM2004 May 2024 Exam Notes

Section A

Question 1

(a) Crystal Field Theory for Octahedral Complexes

• Derivation of d-orbital splitting: In an octahedral transition metal complex, the six ligands approach the central metal ion along the x, y, and z axes. This interaction raises the energy of the metal d-orbitals. However, not all d-orbitals are affected equally.
• d-orbital splitting: The $d<em>{x^2-y^2}$ and $d</em>{z^2}$ orbitals, which point directly towards the ligands, experience a stronger repulsion and are raised in energy to a greater extent. These form the $e_g$ set.
• The $d<em>{xy}$, $d</em>{xz}$, and $d<em>{yz}$ orbitals, which point between the ligands, experience less repulsion and are raised in energy to a lesser extent. These form the $t</em>{2g}$ set.
• Crystal Field Splitting Energy ($\Delta<em>o$): The energy difference between the $e</em>g$ and $t<em>{2g}$ sets is denoted as $\Delta</em>o$ (delta octahedral). The $e<em>g$ set is destabilized by +0.6$\Delta</em>o$, and the $t<em>{2g}$ set is stabilized by -0.4$\Delta</em>o$.

(b) [Mn(OH2)6]3+ Complex

• (i) Measuring $\Delta<em>o$: $\Delta</em>o$ is typically measured using UV-Vis spectroscopy. The complex absorbs light corresponding to the energy required to promote an electron from the $t<em>{2g}$ to the $e</em>g$ set. The wavelength of maximum absorbance ($\lambda<em>{max}$) is related to $\Delta</em>o$ by the equation: $\Delta<em>o = \frac{hc}{\lambda</em>{max}}$, where h is Planck's constant and c is the speed of light.
• (ii) Predicting Magnetic Moment:
  • Given: $\Delta_o$ = 21000 cm-1, P (pairing energy) = 28000 cm-1 for Mn3+.
  • Since $\Delta_o$ < P, the complex is high spin.
  • Mn3+ has a d4 electronic configuration.
  • High-spin configuration: $(t<em>{2g})^3 (e</em>g)^1$
  • Number of unpaired electrons, n = 4.
  • The spin-only magnetic moment ($\mu<em>{eff}$) is calculated using the formula: $\mu</em>{eff} = \sqrt{n(n+2)}$ BM (Bohr Magnetons).
  • $\mu_{eff} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9$ BM
• (iii) Predicting Magnetic Moments for [MnCl6]3- and [Mn(CN)6]3-
  • For [MnCl6]3-: Cl- has an f factor of 0.78. Since Cl- is a weak field ligand, it will result in a high-spin complex. The magnetic moment will be similar to that of [Mn(OH2)6]3+ (approximately 4.9 BM).
  • For [Mn(CN)6]3-: CN- has an f factor of 1.70. Since CN- is a strong field ligand, it will result in a low-spin complex.
    • Low-spin configuration: $(t<em>{2g})^4 (e</em>g)^0$
    • Number of unpaired electrons, n = 2.
    • $\mu_{eff} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83$ BM
• (iv) Explanation of f Factor Difference
  • The f factor reflects the ligand's ability to cause d-orbital splitting. CN- is a strong-field ligand, while Cl- is a weak-field ligand. This difference arises from their electronic structures and bonding interactions with the metal.
  • Cl-: As a $\pi$-donor ligand, Cl- donates electron density to the metal through its filled p-orbitals. This donation increases the electron density around the metal, leading to increased repulsion between the metal d-electrons and the ligands, therefore reducing $\Delta_o$.
  • CN-: As a $\pi$-acceptor ligand, CN- accepts electron density from the metal d-orbitals through its empty $\pi<em>$ orbitals (backbonding). This reduces electron density around the metal, decreasing repulsion and increasing $\Delta_o$. The molecular orbital diagram shows that the interaction between the metal d-orbitals and the CN- $\pi</em>$ orbitals is stronger than the interaction with Cl-, leading to a larger f factor for CN- and a larger splitting.

Question 2

(a) Predicting Equilibrium Positions

• (i) [Pt(NCS)2(PPh3)2] ⇌ [Pt(SCN)2(PPh3)2]
  • This equilibrium involves linkage isomerism. The "harder" Pt2+ ion will prefer to bond to the harder N end of the thiocyanate (NCS-), placing the equilibrium towards the left. Hence, the equilibrium lies to the left due to the preference of Pt(II) for bonding to the harder nitrogen atom rather than the softer sulfur atom.
• (ii) [Pd(NH3)4]2+ + 2 H2NCH2CH2NH2 ⇌ [Pd(H2NCH2CH2NH2)2]2+ + 4 NH3
  • This is a chelate effect scenario. Ethylenediamine (H$_2$NCH$_2$CH$_2$NH$_2$) is a bidentate ligand, while ammonia is a monodentate ligand. The chelate effect favors the formation of complexes with chelated ligands due to increased entropy. Therefore, the equilibrium lies to the right.
  • The change in entropy can be explained by considering the number of species in each sides. There are 3 species in the reactants and 5 in the products, thus, the products are favored.
• (iii) [Pd(PtBu3)3] + PtBu3 ⇌ [Pd(PtBu3)4]
  • This equilibrium is influenced by steric factors. PtBu3 (tri-tert-butylphosphine) is a bulky ligand. Palladium(0) is a large metal center, so the addition of the fourth PtBu3 ligand is unlikely to be possible as it would lead to significant steric hindrance. Therefore, the equilibrium lies to the left.

(b) Metal-Metal Bonding in [Mo2Cl8]2-

• (i) Molecular Orbital Diagram

  • Atomic Orbitals: The Mo-Mo bond is formed by the interaction of the d-orbitals of two molybdenum atoms. The relevant d-orbitals are $d<em>z^2$, $d</em>{xz}$, $d<em>{yz}$, $d</em>{xy}$, and $d_{x^2-y^2}$.
  • Sigma ($\sigma$) Bond: The $d_z^2$ orbitals on each Mo atom overlap along the internuclear axis to form a $\sigma$ bonding and a $\sigma*$ antibonding molecular orbital.
  • Pi ($\pi$) Bonds: The $d<em>{xz}$ and $d</em>{yz}$ orbitals on each Mo atom overlap side-by-side to form two $\pi$ bonding and two $\pi*$ antibonding molecular orbitals.
  • Delta ($\delta$) Bond: The $d<em>{xy}$ and $d</em>{x^2-y^2}$ orbitals overlap forming a $\delta$ bonding and a $\delta*$ antibonding molecular orbital.

• (ii) Bond Order and Structure

  • [Mo2Cl8]2- has 8 d-electrons from the two Mo(II) ions (each Mo contributes 4 d-electrons).
  • Filling the molecular orbitals: $\sigma^2 \pi^4 \delta^2$
  • Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2
  • Bond Order = (8 - 0) / 2 = 4
  • The structure of [Mo2Cl8]2- features a quadruple bond between the two molybdenum atoms. The chloride ligands are arranged around each molybdenum in a square planar geometry (eclipsed conformation).

Section B

Question 3

(a) Catalytic Reaction Mechanisms

• (i) Importance of Understanding Mechanisms:
  • Optimization: Understanding the mechanism allows for the optimization of reaction conditions (temperature, pressure, catalyst concentration) to improve yield and selectivity.
  • Catalyst Design: Mechanistic insights guide the design of more efficient and selective catalysts. By understanding the key steps, catalysts can be tailored to enhance desired pathways and minimize side reactions.
  • Troubleshooting: Understanding the process makes it easier to troubleshoot catalyst deactivation or poisoning.
• (ii) Experiments and Techniques for Elucidating Mechanisms
  • Kinetic Studies: Determining the rate law and reaction order with respect to each reactant and catalyst provides information about the rate-determining step and intermediates involved.
  • Spectroscopic Techniques:
    • NMR Spectroscopy: Used to identify intermediates and monitor their concentrations during the reaction. Multinuclear NMR (e.g., 31P NMR for phosphine ligands) is particularly useful.
    • IR Spectroscopy: Helps to identify functional groups and bonding modes of ligands and intermediates.
    • UV-Vis Spectroscopy: Used to study electronic transitions and monitor changes in oxidation states of the metal center.
    • EPR Spectroscopy: Useful for detecting paramagnetic intermediates with unpaired electrons.
  • Isotopic Labeling: Using isotopes (e.g., deuterium, 13C) to track the fate of atoms during the reaction helps to reveal bond-breaking and bond-forming steps.
  • Stoichiometric Reactions: Studying the reactions of individual steps in the catalytic cycle with well-defined complexes provides insight into their feasibility and role in the overall mechanism.
  • Computational Methods: DFT calculations can be used to model reaction pathways, predict the stability of intermediates, and calculate activation energies.
  • X-ray Crystallography: Determining the structures of isolated intermediates provides valuable information about their bonding and geometry.

(b) Rhodium Complex and Hydroformylation

• (i) Isomerization Mechanism: The mechanism involves a series of migratory insertions and $\beta$-hydride eliminations that shift the double bond along the carbon chain.
  • Coordination of the alkene to the Rh center.
  • Migratory insertion of the alkene into a Rh-H bond, forming an alkyl complex.
  • $\beta$-hydride elimination to form a new alkene isomer and regenerate the Rh-H bond. This process repeats to move the double bond along the chain.
• (ii) Undesirable Isomerization:
  • Product Distribution: Isomerization leads to a mixture of different alkene isomers, resulting in a complex product mixture.
  • Reduced Selectivity: Hydroformylation is most efficient when the alkene double bond is at the terminal position. Isomerization shifts the double bond to internal positions, leading to the formation of branched aldehydes, which are often less desirable.
  • Separation Challenges: Separating the mixture of aldehydes can be difficult and costly, reducing the overall efficiency of the industrial process.

(c) Zirconium Complex and Polymerization of Ethylene

• Mechanism: The zirconium complex acts as a Ziegler-Natta catalyst for the polymerization of ethylene. The mechanism involves coordination of ethylene to the Zr center, followed by migratory insertion of ethylene into the growing polymer chain. This process repeats, leading to the synthesis of a long polymer chain.
  • Initiation: Coordination of ethylene to the vacant coordination site on the Zr center.
  • Propagation: Migratory insertion of the coordinated ethylene into the Zr-C bond of the growing polymer chain. This extends the chain by one ethylene unit and regenerates a vacant coordination site.
  • Chain Termination: Chain termination can occur through $\beta$-hydride elimination, leading to the formation of a terminal alkene and a Zr-H species. Alternatively, chain transfer to another monomer or a chain transfer agent can occur.

Section C

Question 4

(a) Stepwise Formation Constants for [Cu(OH2)6]2+ with NH3

• The stepwise formation constants (K1, K2, K3, K4) for the complexation of NH3 with [Cu(OH2)6]2+ decrease sequentially. This trend is primarily due to statistical factors and steric hindrance.
  • Statistical Factors: As more NH3 ligands bind to the Cu2+ ion, there are fewer water molecules available to be replaced, making each subsequent substitution less favorable.
  • Steric Hindrance: The increasing number of NH3 ligands around the Cu2+ ion creates steric congestion, making it more difficult for additional NH3 ligands to approach and bind.
• The value for K5 is significantly lower (negative log K5) because of the Jahn-Teller distortion in copper(II) complexes. [Cu(NH3)4(OH2)2]2+ exhibits a tetragonal distortion, with two water molecules at longer distances along the z-axis. Replacing these elongated bonds is much more difficult, hence the large drop in K5.

(b) Stepwise Formation Constants for [Cu(OH2)6]2+ with NH2CH2CH2NH2

• The values for ethylenediamine (en) are much higher than ammonia due to the chelate effect. Ethylenediamine is a bidentate ligand, and its binding is entropically more favorable than the binding of two separate ammonia molecules.
  • When ethylenediamine binds, it displaces two water molecules but only requires one molecule to bind to the metal center. This results in a net increase in the number of free particles in solution, leading to a more positive (more favorable) entropy change compared to ammonia.
  • The chelate effect is the enhanced stability of complexes containing chelate rings compared to complexes containing only monodentate ligands.

(c) Synthesis of cis- and trans-[PtCl2(NH3)(PH3)]

• Synthesis of cis-[PtCl2(NH3)(PH3)]:
  • React [PtCl4]2- with NH3 first to obtain [PtCl3(NH3)]-. trans effect of Cl- will direct the incoming NH3 to the cis position.
  • Then react [PtCl3(NH3)]- with PH3 to obtain cis-[PtCl2(NH3)(PH3)]. Because Cl- has a stronger trans effect, PH3 will take the trans position of one of the Cl- ions.
• Synthesis of trans-[PtCl2(NH3)(PH3)]: Phosphine exhibits a strong trans effect and directs the substitution trans to itself. The order of trans effect is PH3 > Cl- > NH3.
  • React [PtCl4]2- with PH3 first to obtain [PtCl3(PH3)]-. trans effect of PH3 will direct the incoming PH3 to the trans position.
  • Then react [PtCl3(PH3)]- with NH3 to obtain trans-[PtCl2(NH3)(PH3)]. Because PH3 has a stronger trans effect, NH3 will take the trans position of one of the PH3 ions.

(d) Rate of Cl- Attack on Complexes A and B

• The 1000-fold difference in the rate of Cl- attack on complex A compared to complex B suggests a significant difference in the electronic or steric environment around the metal center i.e trans effect.
  • Complex A likely has ligands that facilitate the incoming Cl- nucleophile, either electronically (through trans effect) or sterically (by providing less hindrance).
  • Complex B likely has bulky or electron-donating ligands that hinder the approach of Cl- or stabilize the starting complex, making substitution more difficult.

(e) $\Delta^{\ddagger}$V and $\Delta^{\ddagger}$H for Substitution Reactions

• [V(OH2)6]2+
  • $\Delta^{\ddagger}$H = 68.6 kJ mol-1 and $\Delta^{\ddagger}$V = -4.1 cm3 mol-1
  • The negative $\Delta^{\ddagger}$V suggests an associative mechanism. The incoming ligand "L" approaches the metal center, increasing the coordination number in the transition state and reducing the volume.
  • The positive $\Delta^{\ddagger}$H indicates that bond-making with the incoming ligand is more important than bond-breaking with the leaving water ligand in the transition state.
• [Co(OH2)6]2+
  • $\Delta^{\ddagger}$H = 43.5 kJ mol-1 and $\Delta^{\ddagger}$V = 6.1 cm3 mol-1
  • The positive $\Delta^{\ddagger}$V suggests a dissociative mechanism. The leaving water molecule dissociates from the metal center, decreasing the coordination number in the transition state and increasing the volume.
  • The lower $\Delta^{\ddagger}$H compared to the vanadium complex indicates that bond-breaking with the leaving water ligand is more important than bond-making with the incoming ligand in the transition state.
  • The difference in mechanisms is related to the electronic structures of the metal ions. V2+ (d3) tends to favor associative mechanisms, while Co2+ (d7) tends to favor dissociative mechanisms.

(f) Reaction of [Co(SCN)(OH2)5]2+ with Fe2+(aq)

• The detection of [Fe(SCN)(OH2)5]2+ indicates an inner-sphere mechanism.
  • In an inner-sphere mechanism, the reaction proceeds through a bridging ligand (SCN- in this case). The Fe2+ ion attacks the SCN- ligand coordinated to the Co2+ ion, forming a bridged intermediate [Co-SCN-Fe].
  • Electron transfer occurs through the bridging ligand, leading to the formation of Fe3+ and Co2+.
  • The bridging ligand then migrates to the Fe3+ ion, forming [Fe(SCN)(OH2)5]2+.
  • The observation of [Fe(SCN)(OH2)5]2+ confirms that the SCN- ligand acted as a bridge for electron transfer.

Question 5

(a) Metal-Carbon Bond Formation in s- and p-block Metals

• (i) Methods of Formation
  • Reaction with Organolithium or Grignard Reagents:
    • $MCl + RLi \rightarrow R-M + LiCl$ (M = s- or p-block metal, R = organic group)
    • $MCl + RMgX \rightarrow R-M + MgXCl$ (X = Cl, Br, I)
    • Example: $SiCl<em>4 + 4MeMgCl \rightarrow SiMe</em>4 + 4MgCl_2$
  • Transmetallation: Transfer of an organic group from one metal to another.
    • $M + R_2Hg \rightarrow R-M + RHg$ where M is an alkali metal.
  • Direct Reaction with Alkenes or Alkynes:
    • Some s-block metals (especially Li) can directly react with unsaturated hydrocarbons to form organometallic compounds.
• (ii) Group 1 Metals as Reducing Agents
  • Group 1 metals have low ionization energies, meaning it takes little energy to remove an electron, making them good reducing agents.
  • Reactions to form deeply colored compounds:
    • The reaction of group 1 metals with polyaromatic hydrocarbons (e.g., naphthalene) involves the transfer of an electron from the metal to the aromatic system, forming a radical anion. This radical anion is intensely colored due to the absorption of light in the visible region.
    • $K + Naphthalene \rightarrow K^+[Naphthalene]^{-.}$
  • Application: The organometallic complex formed from the reaction of potassium with naphthalene (potassium naphthalide) is a powerful reducing agent used in organic synthesis.

(b) Boron Chemistry

• (i) Balanced Equations
  • (a) BCl3 and ethanol:
    • $BCl<em>3 + 3 EtOH \rightarrow B(OEt)</em>3 + 3 HCl$
    • Boron trichloride reacts with ethanol to form triethyl borate and hydrogen chloride. The driving force is the formation of strong B-O bonds.
  • (b) BCl3 and pyridine in a hydrocarbon solution:
    • $BCl<em>3 + Pyridine \rightarrow Cl</em>3B-Pyridine$
    • BCl3 acts as a Lewis acid and coordinates with the nitrogen atom of pyridine, which acts as a Lewis base.
  • (c) BBr3 and F3BN(CH3)3:
    • $BBr<em>3 + F</em>3BN(CH<em>3)</em>3 \rightarrow BF<em>3 + Br</em>3BN(CH<em>3)</em>3$
    • This is a Lewis acid-base exchange reaction. Boron tribromide is a stronger Lewis acid (due to the larger size and lower electronegativity of bromine compared to fluorine) and displaces BF3 from the adduct with trimethylamine.
• (ii) Boron Reagents in Olefin Oxidation
  • Hydroboration:
    • Boron-containing reagents (e.g., BH3 or boranes) can be used to hydroborate olefins (alkenes), adding a B-H bond across the double bond to form an alkylborane.
  • Oxidation:
    • The resulting alkylborane can then be oxidized with hydrogen peroxide (H2O2) in the presence of a base (e.g., NaOH) to form an alcohol. The oxidation reaction replaces the B-C bond with a C-O bond, resulting in the formation of an alcohol.

Question 6

(a) First Ionization Energy of Group 13 Elements

• The general trend is a decrease in ionization energy down the group due to increasing atomic size and increased shielding of the nuclear charge by inner electrons. However, there are some exceptions:
  • B (801 kJ/mol) to Al (577 kJ/mol): Decrease as expected due to increasing atomic size and shielding.
  • Al (577 kJ/mol) to Ga (579 kJ/mol): Small increase. This is due to the poor shielding of the nuclear charge by the 3d electrons in Ga. The 3d electrons are not as effective at shielding the valence electrons, leading to a slightly higher effective nuclear charge and a slightly higher ionization energy.
  • Ga (579 kJ/mol) to In (558 kJ/mol): Decrease, as expected, but less pronounced.
  • In (558 kJ/mol) to Tl (589 kJ/mol): Notable Increase. This is attributed to the lanthanide contraction and the relativistic effects on the 6s electrons in Tl. The lanthanide contraction results in a smaller-than-expected size for Tl, and the relativistic effects stabilize the 6s electrons, making them more difficult to remove.

(b) Phosphorus Chemistry

• (i) Synthesis of Electrochemical Grade LiPF6 from Calcium Fluorapatite
  • Step 1: Production of Phosphoric Acid (H3PO4):
    • React calcium fluorapatite with sulfuric acid to produce phosphoric acid and calcium sulfate.
      $Ca<em>5F(PO</em>4)<em>3 + 5H</em>2SO<em>4 + 10H</em>2O \rightarrow 3H<em>3PO</em>4 + 5CaSO<em>4 \cdot 2H</em>2O + HF$
    • The HF is a byproduct that needs to be managed.
  • Step 2: Production of Phosphorus Pentachloride (PCl5):
    • React phosphoric acid with chlorine gas and carbon to produce phosphorus pentachloride, water, and carbon monoxide.
      $2H<em>3PO</em>4 + 10Cl<em>2 + 10C \rightarrow 2PCl</em>5 + 6H_2O + 10CO$
  • Step 3: Production of Phosphorus Pentafluoride (PF5):
    • React phosphorus pentachloride with hydrogen fluoride to produce phosphorus pentafluoride and hydrogen chloride.
      $PCl<em>5 + 5HF \rightarrow PF</em>5 + 5HCl$
  • Step 4: Production of Lithium Hexafluorophosphate (LiPF6):
    • React phosphorus pentafluoride with lithium fluoride to produce lithium hexafluorophosphate.
      $PF<em>5 + LiF \rightarrow LiPF</em>6$
    • Stringent purification is required to obtain electrochemical grade LiPF6.
• (ii) Hydrolysis of Phosphorus Pentachloride
  • With one equivalent of water:
    • $PCl<em>5 + H</em>2O \rightarrow POCl_3 + 2HCl$
    • The product is phosphorus oxychloride (POCl3), which has a tetrahedral structure.
  • With excess water:
    • $POCl<em>3 + 3H</em>2O \rightarrow H<em>3PO</em>4 + 3HCl$
    • Overall reaction:
    • $PCl<em>5 + 4H</em>2O \rightarrow H<em>3PO</em>4 + 5HCl$
    • The product is phosphoric acid (H3PO4), which has a tetrahedral structure.

(c) Silicon Chemistry

• (i) Stability of CF4 and SiF4 to Hydrolysis
  • CF4 is resistant to hydrolysis due to the following reasons:
    • Strong C-F Bonds: The C-F bond is very strong and kinetically stable.
    • Lack of d-orbitals: Carbon does not have available d-orbitals to expand its coordination number and accommodate the formation of a transition state for nucleophilic attack by water.
  • SiF4 is readily hydrolyzed because:
    • Weaker Si-F Bonds: The Si-F bond is weaker than the C-F bond.
    • Availability of d-orbitals: Silicon has available d-orbitals that can be used to expand its coordination number and form a stable intermediate with water. Water can attack the silicon center, and fluoride ions can be eliminated.
    • $SiF<em>4 + 2H</em>2O \rightarrow SiO_2 + 4HF$
• (ii) Preparation of Cross-Linked Siloxane Polymer (Silicone Tubing)
  • Step 1: Synthesis of a Siloxane Precursor:
    • Start with a dichlorosilane, such as dimethyldichlorosilane ((CH3)2SiCl2).
  • Step 2: Hydrolysis and Polymerization:
    • Hydrolyze the dichlorosilane with water. This results in the formation of silanol groups (Si-OH).
    • $(CH<em>3)</em>2SiCl<em>2 + 2H</em>2O \rightarrow (CH<em>3)</em>2Si(OH)_2 + 2HCl$
    • The silanol groups then undergo condensation polymerization to form a linear siloxane polymer.
    • $n(CH<em>3)</em>2Si(OH)<em>2 \rightarrow [((CH</em>3)<em>2SiO)</em>n] + nH_2O$
  • Step 3: Cross-Linking:
    • To form a cross-linked polymer, include a small amount of a trifunctional silane (e.g., methyltrichlorosilane, CH3SiCl3) or a tetrafunctional silane (e.g., silicon tetrachloride, SiCl4) in the reaction mixture.
    • These compounds introduce branching points in the polymer chain.
    • The cross-linking process involves the formation of Si-O-Si bonds between different polymer chains, creating a three-dimensional network structure and producing a silicone elastomer.