Second Test Solutions

7/29/2019 Second Test Solutions

http://slidepdf.com/reader/full/second-test-solutions 1/12

Name:

1

CHEM 3111/6170 Second testTopics, metal chemistry, coordination compounds,

electronic spectroscopy of transition metal compounds

October 29th, 2003

50 minutes to complete

Maximum score 76 points

1) Classify all of the following ligands according to their bonding modality as some combination of

σ donor, π acceptor and π donor. (6 points)

Phenanthroline σ donor and π acceptor

NH3 σ donor

H2O σ donor and π donor

Br - σ donor and π donor

CO σ donor and π acceptor

CH3- σ donor

2) Rank the following groups of ligands according to their ability to produce large values of ∆. Put

the ligand giving the largest ∆ first. No partial credit. (4 points)

a) phen, H2O, Cl

-

phen > H2O > Cl

-

b) CO, NH3, I- CO > NH3 > I

-

3) Very briefly explain why most octahedral coordination complexes of Co(III) are diamagnetic,

but CoF63- is paramagnetic. (4 points)

Most octahedral cobalt complexes are low spin and hence diamagnetic. CoF63- is

paramagnetic as it is high spin. The metal is high spin because the fluoride ligands produce

a very small ∆ value (π donors) and B is quite high because the complex is not very

covalent.



Name:

2

4) Very few Ti(II) compounds have been synthesized and studied using optical spectroscopy.

However, dissolution of TiCl2 in molten aluminum chloride is thought to give rise to species

similar to octahedral TiCl64-. The optical spectrum of this melt shows the presence of two

absorption bands at ν1 = 7600 cm-1 and ν2 = 14,500 cm-1.

a) Using one of the attached Tanabe-Sugano diagrams assign these spectroscopic transitions. ( 4points)

ν1 3T2g <-

3T1g

ν2 3T1g <-

3T1g

b) Estimate values for ∆ and B' for TiCl64- (6 points)

ν1/ν2 ~ 1.91. This corresponds to ∆/B ~ 16.7

For ν2 E/B ~ 28.0 if ∆/B ~ 16.7.

This gives B = 518 cm-1

and ∆ = 8648 cm-1

c) B for gas phase free Ti2+ is 718 cm-1. Explain why the free ion value is different from the one

you have just determined. (2 points)

B for the complex is lower than for the free ion as in the complex the electrons are partially

delocalized onto the ligands and this reduces the amount of electron electron repulsion.

5) For each of the following colored compound indicate the primary mechanism that is responsible

for the color of the compound. Your options are A) spin forbidden d-d transition, B) d-dtransition, C) Ligand to metal charge transfer, D) Metal to ligand charge transfer and E) charge

transfer between metals in different oxidation states.

Pale pink aqueous Mn2+ (2 points) A

Intensely purple MnO4- solutions. (2 points) C

Prussian blue ( Fe4[Fe(CN)6]3.xH2O ) (2 points) E

Yellow orange [Co(en)3]3+

(2 points) B

Yellow-orange CrO4-(2 points) C

Blue CoCl42-

(2 points) B



Name:

3

6) NiCl42- is paramagnetic, Ni(CN)42- is diamagnetic and BPh4

- (Ph= pheyl) is also diamagnetic.

Using general chemical principles plus the observed magnetic properties, deduce the shapes of

these three species. (6 points)

NiCl42- Tetrahedral

Ni(CN)42- Square planar

BPh4- Tetrahedral

7) Draw all the possible isomers for a complex of the type Co(en)X2Y2. Indicate which ones are

enantiomers. (8 points)

Cl

Co

Cl

N NH3

NH3 N

+

+

Co N Cl

Cl N

NH3

NH3

+

Co N Cl

NH3 N

Cl

NH3

+

Co N NH3

Cl N

Cl

NH3

The last two are related as enantiomers (they are chiral)

8) For each of the following groups identify one ion that is most likely to form a Jahn-Teller distorted six coordinate complex with the ligand NH3. (6 points)

a) Cu2+, Co3+, Cr 3+, Ti4+

b) Ca2+

, Zn2+, Mn2+, Mn3+

c) Fe

3+

, Co3+

, Cr2+

, Cr 3+

9) Sodium dissolves in dry liquid ammonia to give a blue solution. Write a balanced equationshowing the reaction that gives rise to the blue colored product. (2 points)

Na(s) -(Am liq)--> Na+

(am) + e-(am)



Name:

4

10) Using your general knowledge of the factors that lead to metal-metal bonding in transition metal

compounds deduce which one of the compounds in each of the following groups is metal-metal

bonded. (4 points)

a) CrCl2, MoCl2, TiCl2

b) NbO, FeO, MgO

11) For each of the following complexes calculate the ligand field stabilization energy. In order to do

this you may have to determine if the species is high or low spin. (12 points)

a) [Mn(OH2)6]2+

t2g3eg

2LFSE = 0

b) [Co(NH3)6]3+

t2g6eg

0LFSE = 6 * 2/5∆ο = 12/5 ∆o

c) [Fe(CN)6]3+

t2g5eg

0LFSE = 5 * 2/5∆ο = 2 ∆o

d) Tetrahedral [FeCl4]2-

e3t2

3LFSE = (3*3/5∆t - 3*2/5∆t) = 3/5 ∆t



Name:

5

d2

Tanabe Sugano diagram



Name:

6

d3




Name:

7

d4




Name:

8

d5




Name:

9

d6




Name:

10

d7




Name:

11

d8




Name:

12

Second Test Solutions

Documents

Transcript of Second Test Solutions