Inorganic Chemistry 422
Exam 3
April 28, 2005

All work must be shown for any credit to be given.

(3 pts.)    1. Sketch the splitting pattern for the 3d orbitals in an octahedral field.

(5 pts.)    2. Which 3d orbitals make up the t_2g set?  Which 3d orbitals make up the e_2g set?

t_2g set: d_xy, d_xz, d_yz

e_2g set: d_x2_-y2, d_z2

(3 pts.)    3. Sketch the splitting pattern for the 3d orbitals in an tetrahedral field.

(4 pts.)    4. Explain why high and low spin complexes are possible with octahedral fields but not with tetrahedral fields.

Low spin cases in octahedral symmetry is the result of the fact that D_o is larger than the energy involved in pairing the electrons in the t_2g orbitals. This results in the electrons concentrating in the lower energy leveles. With the tetrahedral symmetry D_t is considerably smaller than D_o and even smaller or equal to the pairing energy. This results in the electrons not pairing, but filling one electron in each orbital until all five 3d orbitals contains 1 electron.

(8 pts.)    5. Calculate the Ligand Field Stabilization Energy for the high spin and low spin d⁵ case.

(a.)     LFSE high spin d⁵ = (3(-4Dq) + 2(6Dq)) = 0 Dq

(b.)    LFSE low spin d⁵ = (5(-4Dq) + 0(6Dq)) = -20 Dq

(12 pts.)  6. Give the correct IUPAC name for each of the following:

(a.)      [Cr(H₂O)₆]Cl₃ hexaaqua chromium(III) chloride

(b.)      Na₂[CuCl₄] sodium tetrachlorocopperate(II) or sodium tetrachlorocuprate(II)

(c.)      [Co(NH₃)₄Cl₂] dichlorodiaminecobalt(II)

(12 pts.)  7. Give the formula for the coordination species shown (use [ ] to indicate coordination sphere):

(a.)    tetraaquocopper(II) sulfate [Cu(H₂O)₄]SO₄

(b.)    potassium hexafluoroferrate(III) K₃[FeF₆]

(c.)    diiodotetraaminecobalt(II) [Co(NH₃)₄I₂]

(4 pts.)    8. Calculate the effective atomic number (EAN) of the Ru⁺² ion in [Ru(NH₃)₄(HSO₃)₂] (assume that all the ligands are monodentate).

EAN = 44 - 2 + 6(2) = 54, the atomic number of the noble gas Xe

(6 pts.)    9. Calculate the number of unpaired electrons in each of the following:

(a.)     high spin d⁵  __5___ unpaired electrons

(b.)    low spin d⁵  __1___ unpaired electrons

(5 pts.)    10. Give the rate law expression for the following reaction:

                        ML₆  +  L’  à  ML₆L’ slow

                        ML₆L’  à  ML₅L’  +  L           fast

rate = k[ML₆][L']

(4 pts.)    11. In terms of reaction kinetics, compare inert and labile.

labile => exchange t_1/2 less than 1 min. at RT

inert => exchange t_1/2 more than 1 min. at RT

(4 pts.)    12. Describe one experimental technique used to determine a structure of an inorganic complex.

Your choice from NMR, IR, X-ray Diffraction, Mossbauer, etc.

(5 pts.)    13. Draw the structure of a real first row transition metal carbonyl

Again, your choice of carbonyls: such as:

Name ________________________________

Inorganic Chemistry 422
Take Home Exam 3
April 28, 2005

All work must be shown for any credit to be given.

(10 pts.)  . Give the microstates for a p³ case and determine the spectral term symbols.

L

2

1

1

-1

2

1

1

-1

0

0

0

0

0

0

0

0

-1

-2

-1

-2

+1

ab

ab

a

ab

ab

b

a

b

a

a

b

b

b

a

a

b

ml

0

a

ab

ab

b

ab

ab

a

b

a

b

a

b

a

b

a

b

-1

a

a

b

b

a

b

b

a

a

a

b

b

ab

ab

ab

ab

S

+1/2

+1/2

+1/2

+1/2

-1/2

-1/2

-1/2

-1/2

+3/2

-3/2

+1/2

+1/2

+1/2

-1/2

-1/2

-1/2

+1/2

+1/2

-1/2

-1/2

S

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

a=> s = +1/2 and b=> s = -1/2

2D	+2	+1	0	-1	-2
+1/2	x	x	x	x	x
-1/2	x	x	x	x	x

2P	+1	0	-1
+1/2	x	x	x
-1/2	x	x	x

4S	0
+3/2	x
+1/2	x
-1/2	x
-3/2	x

the terms are: ²D; ²P; and ⁴S

term symbols => ^{2S + 1}L; ground state => highest S and then highest L

thus the ground state is the ⁴S

(15 pts.) . For [Cr(H₂O)₆]⁺³ absorption bands are observed at 17,400, 24,600, and 37,900 cm^-1. Assign the bands and calculate or estimate Dq, B’, and the bending.

d³ case

Using the right hand diagram, left side:

n₁ = 17,400 cm^-1 A₂ ---> T₂ => 10Dq

n₂ = 24,600 cm^-1 A₂ ---> T₁(F) => 18Dq - c

n₃ = 37,900 cm^-1 A₂ ---> T₁(P) => 12Dq + 15B + c

n₁ = 17,400 cm^-1 = 10 Dq

Dq = 1740 cm^-1

n₂ = 24,600 cm^-1 = 18 Dq - c

c = (18 Dq - 24,600) cm^-1 = (18*1740 -24,600) cm^-1

c = 6720 cm^-1

n₃ = 37,900 cm^-1 = 6Dq + 15B + c

37,900 cm^-1 = 12Dq + 15B + c

37,900 cm^-1 = 12(1720 cm^-1) + 15B + 6720 cm^-1

15B = 37,900 cm^-1 - 6(1720 cm^-1) - 6720 cm^-1

B = 687 cm^-1

thus Dq = 1740 cm^-1; B = 687 cm^-1; c = 6720 cm^-1

               I neither received nor gave assistance on this take home exam, nor did I see others giving or receiving assistance.

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