Chemistry 401

Intermediate Inorganic Chemistry

University of Rhode Island

Fall 2012

Exam 3

For each of the following compounds or ions: 1) draw the structure and give the point group; 2) indicate the appropriate bonding theory to predict if the compound or ion will be stable or not, briefly explaining your prediction (hint: quantitative predictions are the best); 3) if the compound or ion is paramagnetic, predict the spin-only magnetic moment in units of Bohr-Magnetons; 4) determine if the compound or ion will be Jahn-Teller active and, if so, predict the nature of the distortion.

1. hexacyanoferrate(III) ion

1) structure = Point group = Oh

2) Ligand field theory predicts a stable compound for this d5, t2g5 ion in a strong field with LFSE = 20Dq – 2P

3) There is one unpaired spin so the predicted magnetic moment is μ = [1(1+2)]½ = 1.73 μB.

4) The electronic state is degenerate so is Jahn-Teller active. An axial elongation is predicted.

2. bichlorobiaquacobalt(II) (hint: this is a deep blue color)

1) structure = Point group = C2v. Both the lack of cis or trans designation and the hint about the color (a blue color means a small Δt) indicate a tetrahedral distribution of the ligands about the metal ion.

2) Ligand field theory predicts a stable compound for this d7, e4t23 ion in a weak field with LFSE = 12Dqt

3) There are three unpaired spins so the predicted magnetic moment is μ = [3(3+2)]½ = 3.87 μB.

4) The electronic state is nondegenerate so is not Jahn-Teller active.

3. hexacarbonylchromium(0)

1) structure = Point group = Oh.

2) Valence bond theory predicts a stable compound: for this d6 metal the EAN count is 6 e from Cr and 2 × 6 = 12 e from the ligands for a total of 18 e.

3) There are no unpaired spins.

4) The electronic state is nondegenerate so is not Jahn-Teller active.

4. trans-chlorobis(hydrido)tris(triphenylphosphine)rhodium(III) (hint: the PPh3 ligand freely rotates)

1) structure = Point group = C2v.

2) Ligand field theory predicts a stable compound for this d6, t2g6 compound in a strong field with LFSE = 24Dq – 2P

3) There are no unpaired spins.

4) The electronic state is nondegenerate so is not Jahn-Teller active.

5. cis-tetraamminebinitritonickel(II)

1) structure = Point group = C2v.

2) Ligand field theory predicts a stable compound for this d8, t2g6eg2 compound in a strong field with LFSE = 12Dq

3) There are two unpaired spins so the predicted magnetic moment is μ = [2(2+2)]½ = 2.83 μB.

4) The electronic state is nondegenerate so is not Jahn-Teller active.

6. bis(η1-cyclopentadienyl)bis(η5-cyclopentadienyl)titanium(IV)

1) structure = Point group = C2v.

2) Valence bond theory predicts that this compound is probably stable: for this d0 metal the EAN count is 0 e from Ti, 2 × 2 = 4 e from the η1-cp ligands, and 2 × 6 = 12 e from the η5-cp ligands for a total of 16 e.

3) There are no unpaired spins.

4) The electronic state is nondegenerate so is not Jahn-Teller active.

7. pentacyanonitrosylmanganate(III) ion

1) structure = Point group = C4v.

2) Ligand field theory predicts a stable compound for this d4, t2g4 compound in a strong field with LFSE = 16Dq – P

3) There are two unpaired spins so the predicted magnetic moment is μ = [2(2+2)]½ = 2.83 μB.

4) The electronic state is degenerate so is Jahn-Teller active. An axial compression is predicted.

8. cis-bibromobichlorocuprate(II) ion

1) structure = Point group = C2v.

2) Ligand field theory predicts a stable compound for this d9, t2g6eg3 ion in a weak field with LFSE = 6Dq

3) There is one unpaired spin so the predicted magnetic moment is μ = [1(1+2)]½ = 1.73 μB.

4) The electronic state is degenerate so is Jahn-Teller active. An axial elongation is predicted, hence the ion is square planar.

9. fac-triamminetriaquachromium(III) ion

1) structure = Point group = C3v.

2) Ligand field theory predicts a stable compound for this d3, t2g3 ion in a weak field with LFSE = 12Dq

3) There are three unpaired spins so the predicted magnetic moment is μ = [3(3+2)]½ = 3.87 μB.

4) The electronic state is not degenerate so is not Jahn-Teller active.

10. bi(η6-benzene)molybdenum(0)

1) structure = Point group = D6h if the benzene rings are eclipsed and D6d if the benzene rings are staggered.

2) Valence bond theory predicts that this compound is stable: for this d6 metal the EAN count is 6 e from Mo and 2 × 6 = 12 e from the η6-benzene ligands for a total of 18 e.

3) There are no unpaired spins.

4) The electronic state is nondegenerate so is not Jahn-Teller active.