CHM2004 Inorganic Chemistry Exam Notes

Exam Information

  • Exam Code: CHM2004
  • Sections: A, B, and C
  • Materials Allowed: Answer books, Chemistry Data Sheet & Periodic Table, Graph Paper, Approved Calculators, Unassembled Molecular Model Kits
  • Date: Friday, 3rd May 2024
  • Time: 9:30 AM - 11:30 AM
  • Instructions:
    • Write on both sides of the answer paper.
    • All questions carry equal marks.
    • Use separate answer books for each section.
    • Answer FOUR questions: one from each section and one other.
    • Time: TWO HOURS to complete the paper.

Section A

Question 1

(a) Derivation of Crystal Field d-orbital splitting diagram for an octahedral transition metal complex. [6 MARKS]

  • In an octahedral complex, the six ligands approach the metal ion along the x, y, and z axes.
  • The d-orbitals are split into two sets: t{2g} (dxy, dxz, dyz) and eg (dx^2-y^2, dz^2).
  • The e_g orbitals point directly at the ligands, experiencing greater repulsion and thus higher energy.
  • The t_{2g} orbitals point between the ligands, experiencing less repulsion and thus lower energy.
  • The energy difference between the eg and t{2g} sets is denoted as \Delta_o (crystal field splitting energy).

(b) \Deltao for [Mn(OH2)_6]^{3+} was measured to be 21000 cm^{-1}.

(i) Explanation of how \Delta_o is normally measured for transition metal complexes. [2 MARKS]

  • \Delta_o is typically measured using UV-Vis spectroscopy.
  • The complex absorbs light, causing electronic transitions between the t{2g} and eg orbitals.
  • The energy of the absorbed light corresponds to \Delta_o.
  • The wavelength of maximum absorbance is related to \Deltao by the equation: \Deltao = \frac{hc}{\lambda}, where h is Planck's constant, c is the speed of light, and \_lambda is the wavelength.

(ii) P (pairing energy) for Mn^{3+} ion was found to be 28000 cm^{-1}. Predict an approximate magnetic moment for the complex [Mn(OH2)6]^{3+}, explaining clearly each step in your reasoning and showing all your workings. [4 MARKS]

  • Electronic configuration of Mn^{3+} is d^4.
  • Since \Delta_o < P, the complex is high spin.
  • Electronic configuration: t{2g}^3 eg^1.
  • Number of unpaired electrons, n = 4.
  • Spin-only magnetic moment, \mu_{eff} = \sqrt{n(n+2)} BM.
  • \mu_{eff} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.9 BM.

(iii) Predict approximate magnetic moments for the complexes [MnCl6]^{3-} and [Mn(CN)6]^{3-}, given that the f factors for Cl^- and CN^- are 0.78 and 1.70 respectively. [4 MARKS]

  • For [MnCl6]^{3-}, \Deltao = 21000 \times 0.78 = 16380 cm^{-1}. Since \Deltao < P, it is a high-spin complex with 4 unpaired electrons, so \mu{eff} \approx 4.9 BM.
  • For [Mn(CN)6]^{3-}, \Deltao = 21000 \times 1.70 = 35700 cm^{-1}. Since \Deltao > P, it is a low-spin complex. Electronic configuration: t{2g}^4 eg^0. Number of unpaired electrons, n = 2. Spin-only magnetic moment, \mu{eff} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 BM.

(iv) With reference to molecular orbital diagrams, explain why the f factor for Cl^- is much less than that for CN^-. [4 MARKS]

  • CN^- is a stronger field ligand than Cl^- because it can form both \sigma and \pi bonds with the metal.
  • Cl^- primarily forms \sigma bonds and has filled p orbitals that can interact with the metal d orbitals, leading to increased repulsion and a smaller \Delta_o.
  • The \pi-acceptor nature of CN^- stabilizes the metal d orbitals, leading to a larger \Delta_o and a higher f factor.
  • Molecular orbital diagrams show that the \pi interaction in CN^- leads to a greater splitting of the d orbitals compared to Cl^-.

Question 2

(a) Predict to which side each of the following equilibria would lie. In each case explain your answer and the basis of any underlying principles which you use in your reasoning.

(i) [Pt(NCS)2(PPh3)2] \rightleftharpoons [Pt(SCN)2(PPh3)2]. [4 MARKS]

  • The equilibrium lies to the side of [Pt(SCN)2(PPh3)_2].
  • This is due to the