Homework Problems Section 1 Exercise 1 Decide if valence bond theory is able to explain shapes of the following species. Explain your decision using appropriate electron box diagrams. PdCl 4 2 - (square planar) NiCl 4 2 - (tetrahedral) Cr(CO) 6 (octahedral) ZnCl 4 2 - (tetrahedral) Ag(NH 3 ) 2 + (linear) Au(PPh 3 ) 3 + (trigonal planar) Cu(NH 3 ) 4 2+ (tetrahedral) Answer a) No. One would need to move an electron from one half-filled d-orbital into the other half-filled d-orbital. This would require to reverse the spin of the electron which is quantum-mechanically not allowed. b) Yes, because s and p orbitals are empty and can be sp 3 hybridized: c) No, because one would need to move and spin-pair one s and two d electrons under spin-reversal. d) Yes, because empty s and p orbitals are available for hybridization e) Yes, because empty s and p orbitals are available for sp hybridization which is able to explain linear shape. f) Yes, because empty s and p orbitals are available for sp 2 hybridization which is able to explain trigonal planar shape. g) Yes, because empty s and p orbitals are available for sp 3 hybridization which is able to explain tetrahedral shape. Exercise 2 We assume a hypothetical cubic crystal field. Predict how the energies of the metal d-orbitals split in this crystal field? Explain your decision briefly. What is the symmetry of these orbitals? Answer Exercise 3 The octahedral crystal field Δ O for a d 4 metal complex is larger than the spin pairing energy. How many unpaired electrons would you expect in this complex? Answer I would expect two unpaired electrons. Exercise 4 Verify by applying the symmetry adapted linear combination of atomic orbitals (SALC) method that the ligand group orbitals suitable for sigma bonding have the symmetry types A g , E g , and T 1u for an octahedral complex. Answer Exercise 5 Can the 4p orbitals of a period 4 transition metal also make pi-bonding with the ligands in an octahedral complex? Explain your decision briefly? Answer Yes, the 4p orbitals have T 1u symmetry which is suitable to make pi-bonding with the T 1u pi-ligand group orbitals. Exercise 6 Which of the following ligands would you expect to be π-acceptors? Explain your decision briefly. a) NO + b) H - c) Cl - d) CH 3 - Answer Only NO + , because it is the only ligand that has π * -orbitals. Exercise 7 A general trigonal planar complex of the composition M(CN)3 is given. How many π- and π * orbitals of the ligands will be involved in the π-bonding? Answer 3*4=12 orbitals Exercise 8 The energies of the metal d orbitals of the above complex are -24 eV. The energies of the π-orbitals of the ligand are -27 eV, the energies of the π*-orbitals are -10 eV. Would you expect the ligand to act as a π-donor or as a π-acceptor? Why? Answer It would act as a pi-donor, because the energy of the pi-orbitals is much closer to the energy of the metal d orbitals. Exercise 9 Two octahedral complexes MX 6 and MY 6 are given. The energy of the ligand HOMOs suitable for sigma-bonding is about the same, but the orbital overlap between the metal and the ligand Y is significantly greater than the orbital overlap between the ligand X and the metal. For which complex would you expect a larger Δ O ? Answer Δ O will be greater for the orbitals with the ligand Y. This is because the stronger interactions raise the energy of the e g * orbitals, and thus the energy difference between these orbitals and the t 2g orbitals becomes larger. Exercise 10 Construct a qualitative molecular orbital diagram of a square pyramidal complex using the symmetry adapted linear combination of atomic orbitals approach. Consider sigma-bonding only. Answer Exercise 11 Strong π-donating ligands lead to a decrease of the octahedral ligand field Δo. Illustrate this by sketching the relevant part of MO diagram. Answer Dr. Kai Landskron ( Lehigh University ). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate .