transition metal rr(2)

8/12/2019 Transition Metal RR(2)

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"eer56ambert 6aw alc4

Grams of Cu+2 and Experimental % Cu Calc:

% Yield Calc:

% Error Calc:

Discussion:!n this experiment we synthesized two transition metal compounds from the same set of reactants

by manipulatin& the limitin& rea&ent to dri7e the reaction toward one product or the other. 8urthermore,we had to calculate the percent yield and percent copper composition of both.

8rom our experiment we obtained percent yields of +.+02 and (*.-2 for #u!$%"py and u!"pyrespecti7ely. !f comparin& our percent yield to the standards in 9o&el:s textbook of practical or&anicchemistry they would be considered excellent for the (*.-2 and 7ery &ood for the +.+02. Somepossible sources could be loss of product throu&h the 7arious transfers from flask to beaker or throu&hfiltration. 3lso, another source of error could be that there was still some moisture in the final productwhen we measured it thus throwin& off the percent accuracy as well. 8inally, one last source of error isproduct impurity. There could be some traces of the kinetic product in the thermodynamic sample orthermodynamic product in the kinetic sample, and thus throw off our percent yield.

!n terms of 2 copper composition we obtained experimental 7alues of *0.02 and %*.-2 foru!"py and #u!$%"py respecti7ely. These experimental 7alues when compared to the theoretical 7alues


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ha7e a acceptable mar&inal percent error of %.)2 and .02 respecti7ely. There are se7eral sources oferror that could ha7e caused the discrepancies between our experimental and actual 7alues. 8irst, becausewe did not test for purity of the product, there could be contamination from traces of kinetic product in thethermodynamic sample or contamination from traces of thermodynamic product in the kinetic sample.This will throw our results off because each has a different percent copper composition. Second, becausethe readin& on the spectrophotometer had to be read by humans there is a certain de&ree of uncertaintywhether the absorbance read .1% or say .11 and that can affect the concentration that we calculate.8urthermore, any fin&erprints or residue on the cu7ettes could also affect the spectrophotometer readin&

and thus our percei7ed concentration that was used to calculate 2 copper composition.The other important concept in coordination chemistry is the color of the compound. The two

products that we obtains were yellow and red. The specific colors of each product maes sense chemically

speain! for se"eral reasons. #irst$ color results from the electrons absorbin! li!ht of certain freuenciesto excite their electrons to a hi!her ener!y le"el$ and reflectin! li!ht of freuencies it can&t use. The

different colors we see results is due to the ma!nitude of splittin! between the d orbitals. The ma!nitude

of splittin! is caused by the types of fields produced by the li!ands attached to the metals. 'ea fieldli!ands such as the halo!ens cause a small amount of splittin! while stron! field such as compounds

containin! nitro!en ha"e stron! amounts of splittin!. This is why Cu()*py is red and Cu(, 2*py is yellow.

(n Cu()*py each copper is surrounded by - iodine and only one nitro!en from *py$ while Cu(,2 *py issurrounded by 2 iodines and 2 nitro!ens from *py. Therefore$ Cu(,2 *py should ha"e a hi!er splittin! of

d orbitals and thus absorb li!ht of a hi!her freuency than Cu()*py. This is in fact what we see as Cu(*pyis red which means that the compound absorbed li!ht of the complementary wa"elen!th !reen. n the

other hand Cu(,2 *py is yellow$ which means that it absorbed purple li!ht. /urple li!ht has a hi!herfreuency than !reen li!ht which a!rees with our conclusion that Cu(,2*py has !reater field splittin! than

Cu(*py. #urthermore$ the fact that the splittin! in Cu(,2*py is !reater than the splittin! of Cu(*py

mirrors the fact that Cu(,2*py is the inetic product and Cu(bpy is the thermodynamic product. *ecauseCu(*py is the thermodynamic product$ it is more stable$ which is reflected in its weaer field splittin!$

lower free ener!y. 'hile on the hand$ since Cu(,2*py is the inetic product$ it is more unstable$ and

therefore has a stron!er field splittin!$ hi!her free ener!y.

Transition Metal Post Lab Questions

0. 1ow many bonds would each of the followin! chelates form with a metal ion

a. 3iethylenetriamine) Two

4125 C125 C125 41 5 C125 C125 412b. 6alen)four

c. 7cetylacetone) two


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d. Glutamic 7cid$ an amino acid side chain found in proteins) one

2.

a. 1ow many d electrons does Cu+ha"e Cu2+

Cu+has 08 d electrons and Cu+2has 9 d electrons.

b. 3raw the d orbital splittin! for the metal centers in the compounds you made$ with the

correct oxidation state for copper. emember$ your compounds were 4T octahedral. 6ee

pp!. 9;8)9;< in =umbdahl for help drawin! the correct d orbital dia!ram.

c. 7ccordin! to the dia!ram from part b.$ should your compound be colored 'hy or why

not

Yes our compounds should be colored because they do not ha"e a full d electron set.

Therefore$ electrons can absorb li!ht of a certain wa"elen!th to excite themsel"es and

reflect the rest which results in the colors we see.

d. b"iously it has a "ery stron! color. 'hy is this the case

The stron! color results from the electrons absorbin! li!ht of certain freuencies to excite

their electrons to a hi!her ener!y le"el$ and reflectin! li!ht of freuencies it can&t use. The

u!5"py

#u!$%"py


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different colors we see results is due to the ma!nitude of splittin! between the d orbitals.

The ma!nitude of splittin! is caused by the types of fields produced by the li!ands attached

to the metals. 'ea field li!ands such as the halo!ens cause a small amount of splittin!

while stron! field such as compounds containin! nitro!en ha"e stron! amounts of splittin!.

This is why Cu()*py is red and Cu(,2*py is yellow. (n Cu()*py each copper is

surrounded by - iodine and only one nitro!en from *py$ while Cu(,2 *py is surrounded

by 2 iodines and 2 nitro!ens from *py. Therefore$ Cu(,2 *py should ha"e a hi!er splittin!

of d orbitals and thus absorb li!ht of a hi!her freuency than Cu()*py. This is in fact whatwe see as Cu(*py is red which means that the compound absorbed li!ht of the

complementary wa"elen!th !reen. n the other hand Cu(,2 *py is yellow$ which means

that it absorbed purple li!ht. /urple li!ht has a hi!her freuency than !reen li!ht which

a!rees with our conclusion that Cu(,2*py has !reater field splittin! than Cu(*py.

-. 3raw an ener!y dia!ram comparin! the thermodynamic and inetic products. >abel the followin!:

reactant$ thermodynamic and inetic product$ acti"ation ener!y Ea,$ chan!e in enthalpy ?1,.

@*ecause this is a dia!ram of free ener!y$ but the uestion ass for labelin! of ?1$ we assume that

?G

Reaction oordinate

8r

ee.n

er&y

Reactants

ProductsProducts

atalyzed Reaction

;ncatalyzed Reaction


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the ?1 is proportional to ?G.

a. 'hat can you do to mae the inetic product instead of the thermodynamic product The

thermodynamic product instead of the inetic product

ther than chan!in! the ratio of the reactants and picin! a limitin! rea!ent that forces the

formation of the thermodynamic or inetic product$ if we wanted to mae the inetic product

instead of the thermodynamic product$ we can do se"eral other thin!s. #irst$ we can run the

reaction at lower temperatures so that there is not enou!h ener!y to o"ercome the hi!her acti"ation

ener!y reuired for the thermodynamic product. 7lso by allowin! the reaction to run for only brief

periods of time$ you will be able to pre"ent your inetic product from spontaneously turnin! into

the thermodynamic one. (f you wanted to mae your thermodynamic product instead of the inetic

product$ you should run the reaction at hi!her temperatures so that the acti"ation ener!y for the

thermodynamic product can be surpassed. #urthermore$ allowin! the reaction to run for extended

periods of time will cause any inetic product that does form to turn into the thermodynamic

product spontaneous because it is more stable.

A. Figure 1shows the different B dia!rams for your two complexes. >abel each dia!ram as

belon!in! to the thermodynamic or inetic product$ based on the color of the product.

. >etDs say you wanted to mae a networ similar to what you did in lab$ but you donDt want to use

iodine. The other options you ha"e a"ailable in your lab that could be used as li!ands are the

followin!:


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'hich one could you use as a "iable substitute for ()

1ydroxide would probably be the next best thin! as a substitute for (odine because it most closely

resembles the number of lone pairs of iodine and therefore can chelate more than once to lin metals in a

networ. The other two only ha"e one lone pair free and therefore cannot form networ metals becausethey can only bind to one metal atom.

transition metal rr(2)

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