Chemistry 401

Intermediate Inorganic Chemistry

University of Rhode Island

Fall 2007

Exam 3

1. Find the point group for the following: a) SF6 b) SeF4 c) pentacyanonitrosylferrate(III).

a) SF6: ,   Oh.

b) SeF4: ,   C2v.

c) pentacyanonitrosylferrate(III): ,   C4v.

2. Find the LFSE (in terms of Dq and P) and the spin-only magnetic moment for the following: a) hexacyanoferrate(II) ion; b) diaquadichlorocobalt(II); c) hexaamminechromium(III) ion.

a) hexacyanoferrate(II) ion

Fe(CN)64–: Fe2+ is d6 and CN is a strong ligand, so the complex will be low spin. The electron configuration is t2g6, which has LFSE = 24Dq – 2P. There are 0 unpaired spins so μ = 0 μB

b) diaquadichlorocobalt(II)

Co(H2O)2Cl2: Co2+ is d7, H2O and Cl are both weak ligands, and the complex must be in a tetrahderal geometry (no cis or trans information is given) so the complex will be high spin. The electron configuration is e4t23, which has LFSE = 12Dq. There are 3 unpaired spins so μ = [3(3+2)]½ = 3.87 μB

c) hexaamminechromium(III) ion

Cr(NH3)63+: Cr3+ is d3 so the ligand strength is irrelevant. The electron configuration is t2g3, which has LFSE = 12Dq. There are 3 unpaired spins so μ = [3(3+2)]½ = 3.87 μB

3. Which of the following complexes will be stable by the EAN rule: a) tetracarbonylnickel(0); b) bis(η1-cyclopentadienyl)bis(η5-cyclopentadienyl)titanium(IV); c) trichloro-η2-etheneplatinate(II) ion. Explain your reasoning.

a) tetracarbonylnickel(0) : Ni0 is d10 and each CO donates 2 e, so the total count is 10 + 4(2) = 18, stable.

b) bis(η1-cyclopentadienyl)bis(η5-cyclopentadienyl)titanium(IV) : Ti4+ is d0, each η1-cyclopentadienyl donates 2 e, and each η5-cyclopentadienyl donates 6 e so the total count is 0 + 2(2) + 2(6) = 16, probably stable.

c) trichloro-η2-etheneplatinate(II) ion : Pt2+ is d8, each Cl donates 2 e, and η2-ethene donates 2 e so the total count is 8 + 3(2) + 2 = 16, probably stable.

4. Which of the following complexes will be Jahn-Teller active: a) dichloroargentate(I) ion; b) hexaaquamanganese(III) ion c) hexanitritovanadate(III) ion. For those complexes that are Jahn-Teller active, predict the nature of the distortion.

a) dichloroargentate(I) ion: Cl–Ag–Cl, this is a linear ion so is not Jahn-teller active.

b) hexaaquamanganese(III) ion: octahedral Mn3+ with the weak ligand water is t2g3eg1, which is Jahn-Teller active. No prediction on the nature of the distortion can be made.

c) hexanitritovanadate(III) ion: octahedral V3+ with the strong ligand nitrite ion is t2g2, which is Jahn-Teller active. The complex is predicted to exhibit an axial elongation.

5. Explain why four coordinate Cu2+ complexes are never found with pure tetrahedral geometry.

Cu2+ with a pure Td geometry would have an electron configuration of e4t25, which is Jahn-Teller active. Consequently, the complex must distort away from the tetrahedral geometry (probably by twisting to a D2d structure).