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2017 FALL Semester Midterm Examination For General Chemistry II (CH103)

Date: October 18 (Wed), Time Limit: 19:00 ~ 21:00 Write down your information neatly in the space provided below; print your Student ID in the upper right corner of every page.

Professor Name Class Student I.D. Number Name

Problem points Problem points TOTAL pts

1 /9 6 /7

/100

2 /15 7 /14

3 /5 8 /13

4 /10 9 /10

5 /7 10 /10 ** This paper consists of 15 sheets with 10 problems (pages 13: fundamental constants, page 14: periodic table, page 15: claim form). Please check all page numbers before taking the exam. Write down your work and answers in the sheet.

Please write down the unit of your answer when applicable. You will get 30% deduction for a missing unit.

NOTICE: SCHEDULES on RETURN and CLAIM of the MARKED EXAM PAPER.

(채점답안지 분배 및 이의신청 일정)

1. Period, Location, and Procedure 1) Return and Claim Period: October 23 (Mon, 7:00 ~ 8:00 p.m.) 2) Location: Room for quiz session 3) Procedure:

Rule 1: Students cannot bring their own writing tools into the room. (Use a pen only provided by TA) Rule 2: With or without claim, you must submit the paper back to TA. (Do not go out of the room with it)

If you have any claims on it, you can submit the claim paper with your opinion. After writing your opinions on the claim form, attach it to your mid-term paper with a stapler. Give them to TA.

2. Final Confirmation 1) Period: October 26 (Thu) – October 27 (Fri) 2) Procedure: During this period, you can check the final score of the examination on the website.

** For further information, please visit General Chemistry website at www.gencheminkaist.pe.kr.

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1. (total 9 points) A buffer solution can be made by partially neutralizing a weak acid solution with

strong base. Suppose 20.0 mL of 0.200 M NaOH is added to 25.0 mL of 0.500 M CH2ClCOOH (Ka =

1.40×10-3)

(a) Write the neutralization reaction and find its equilibrium constant.

(Answer)

(b) Calculate the pH of the resulting solution.

(Answer)

(c) What further added volume of NaOH would be required to make a buffer of greatest effectiveness?

(Answer)

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2. (total 15 points) Phosphonocarboxylic acid (H3PC) effectively inhibits the replication of the herpes

virus. Structurally, it is a combination of phosphoric acid and acetic acid. It can donate three protons.

The equilibrium constant values are Ka1 = 1.0 X 10–2, Ka2 = 7.8 X 10–6, and Ka3 = 2.0 X 10–9.

phosphonocarboxylic acid

(a) Calculate the concentrations of all four forms of the acid in 0.10 M H3PC in aqueous solution and

pH.

(Answer)

(b) Calculate pH range where H2PC– is the most dominant species among four forms of the acid in

solution.

(Answer)

(c) Calculate the approximate pH value when H2PC– fraction is the highest.

(Answer)

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(d) Enough H3PC is added to blood (pH 7.40) to make its total concentration 1.0 X 10–5 M. The pH of

the blood does not change. Determine the concentrations of all four forms of the acid in this mixture.

(Answer)

3. (5 points) Calculate the mass of AgCl that can dissolve in 100 mL of 0.150 M NaCl solution.

(Ksp(AgCl) = 1.6×10-10)

(Answer)

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4. (total 10 points) For each of the following ionic compounds, state whether the solubility will increase,

decrease, or remain unchanged as a solution at pH 7 is made acidic and describe the reason.

(a) Ca3(PO4)2

(Answer)

(b) PbI2

(Answer)

(c) SrCO3

(Answer)

(d) MnS

(Answer)

(e) MgF2

(Answer)

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5. (7 points) The reaction of OH– with HCN in aqueous solution at 25 oC has a forward rate constant

kf of 3.7×10-9 L mol–1 s–1. Using this information and the measured acid ionization constant of HCN,

Ka = 6.17×10–10, calculate the reverse rate constant kr in the first-order rate law rate = kr [CN–] for the

transfer of hydrogen ions to CN– from surrounding water molecules.

(Answer)

6. (total 7 points) The rate constant of the elementary reaction

BH4-(aq) + NH4

+(aq) → BH3NH3(aq) + H2(g)

is k = 1.94×10-4 L mol–1 s–1 at 30.0 oC, and the reaction has an activation energy of 161 kJ mol-1.

(a) Compute the rate constant of the reaction at a temperature of 40.0 oC.

(Answer)

(b) After equal concentrations of BH4-(aq) and NH4

+(aq) are mixed at 30.0 oC, 1.00×104 s is required

for half of them to be consumed. How long will it take to consume half of the reactants if an identical

experiment is performed at 40.0 oC?

(Answer)

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7. (total 14 points) Acetylcholinesterase (AchE) is an enzyme that hydrolyses acetylcholine, one of

major neurotransmitters in our body.

NO

OAchEH2O

NOH

+HO

O

Acetylcholine Choline Acetic acid By denoting the free enzyme as E, substrate as S, product as P, and the enzyme-substrate complex

as ES, one can simplify the catalytic mechanism as following:

E + S ES E + Pk1

k-1

k2

(a) Express the rate law for ES (i.e. 𝑑𝑑[𝐸𝐸𝐸𝐸]𝑑𝑑𝑑𝑑

= ?) in terms of k1, k-1, k2, [ES], [E], and [S].

(Answer)

(b) Apply the steady-state approximation on the a)’s result. From the definition of Michaelis constant

𝐾𝐾𝑚𝑚 = 𝑘𝑘−1+𝑘𝑘2𝑘𝑘1

and the condition [E]total = [E] + [ES], express the rate law for P (i.e. 𝑑𝑑[𝑃𝑃]𝑑𝑑𝑑𝑑

= ?) in terms of

k2, Km, [E]total, and [S].

(Answer)

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(c) The turnover number kcat is defined as the number of substrate molecules converted into product

per enzyme molecule per second. It can be calculated from the relation 𝑘𝑘𝑐𝑐𝑐𝑐𝑑𝑑 = 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚[𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑚𝑚𝑡𝑡

. Using the

definitions and 𝑉𝑉𝑚𝑚𝑐𝑐𝑚𝑚 = lim[𝐸𝐸]→∞

𝑑𝑑[𝑃𝑃]𝑑𝑑𝑑𝑑

, find the relation between kcat and k2 from b)’s result.

(Answer)

(d) The alternative form of Michaelis-Menten equation can be written down as following. 𝑑𝑑[𝑃𝑃]𝑑𝑑𝑑𝑑

= 𝑘𝑘𝐸𝐸[𝐸𝐸][𝑆𝑆]

The diffusion-controlled limit for the rate constant kE is 108 to 109 M-1s-1 , and enzymes whose rates

are diffusion-limited are said to have reached “catalytic perfection”.

AchE has kcat = 1.4 × 104𝑠𝑠−1, Km = 8.9 × 10−5𝑀𝑀. Is AchE at the “catalytic perfection”? Explain your

reasoning.

(Answer)

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8. (total 13 points) NupJoogi is trying to identify a compound with the molar mass of 88.1. This

corresponds to the molecular formula of C4H8O2.

(a) NupJoogi was able to obtain a clean IR spectrum of his compound.

The following table recites characteristic vibrational frequencies and infrared absorption intensities

of selected chemical bond stretchings.

Frequency (cm-1) Bond Stretching Relative Intensity

3650 – 3200 O – H Medium

3300 – 2700 C – H Medium

1820 – 1630 C = O Strong

1680 – 1600 C = C Weak

1250 – 1000 C – O Strong

Using the data from the table, circle three possible structures of NupJoogi’s compound.

(Answer)

HOOH

OH

OH

O

OH

O

OH

O

O

O

H O

O

O

O

OHO

H

OOHOH

OH

O

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(b) Next, NupJoogi is trying to build and use 1H NMR spectrometer to identify his compound. He has

a permanent magnet with a field strength of 7.0 T. Using the Zeeman effect equation 𝜇𝜇𝐻𝐻 = 𝑔𝑔𝐻𝐻𝑚𝑚𝐼𝐼𝜇𝜇𝑁𝑁

and E = −𝜇𝜇𝐻𝐻𝐵𝐵0, determine the frequency of the radio wave in megahertz (MHz) that he must apply

to induce nuclear magnetic resonance. (Nuclear g-factor of proton gH = 5.586, nuclear magneton

𝜇𝜇𝑁𝑁 = 𝑒𝑒ℏ2𝑚𝑚𝑝𝑝

)

(Answer)

(c) Finally, NupJoogi has successfully built his own NMR instrument and obtained the 1H NMR

spectrum of his compound.

Relative areas of each peaks were 2.0, 3.0, 3.0 from downfield to upfield.

Among the three candidates from (a), identify the structure of NupJoogi’s compound.

(Answer)

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9. (total 10 points) Predict the 1H-NMR spectra of the following compounds including ordering of

chemical shift from upfield (shielding) to downfield (deshielding) and multiplicity of peaks by spin-spin

splitting at each H indicated by Arabic numeral.

COCH2CH3

OH3C

(a)

1 2 3

(Answer)

(b)CH3CH2OCH2CH3

1 2 3 4

(Answer)

(c)CHCH3

ClCH3

1 2 3

(Answer)

1

2 3

(d) CH2CH2NO2

(Answer)

(e) HOCH2CH2CH3

1 2 3 4 (Answer)

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10. (total 10 points) After the light absorption occurs from the ground state of a dimeric molecule

(depicted as a vertical arrow), explain all possible emission and nonradiative relaxation processes

using the following energy curves.

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Claim Form for General Chemistry Examination Page ( / )

Class: , Professor Name: , I.D.# : , Name: If you have any claims on the marked paper, please write down them on this form and submit this with your paper in the assigned place. (And this form should be attached on the top of the marked paper with a stapler.) Please, copy this sheet if you need more before use.

By Student By TA

Question # Claims Accepted? Yes(√) or No(√)

Yes: □ No: □ Pts (+/-) Reasons

<The Answers>

Problem points Problem points TOTAL pts

1 3+3+3/9 6 4+3/7

/100

2 4+3+2+6/15 7 3+3+3+5/14

3 3+2/5 8 6+3+4/13

4 2x5/10 9 2x5/10

5 /7 10 2x5/10

전체 기준: 전개과정은 맞으나 답이나 unit 이 틀리면 -1

답은 맞으나 전개과정이 약간 틀렸을 때 -1

식을 전혀 쓰지 않고 (혹은 흔적이 전혀 없고) 답만 맞았을 때 -1 (3 pts), -2 (4 pts 이상)

1. (total 9 points) (a) (3 pts)

(b) (3 pts)

(c) (3 pts)

2. (total 15 points)

(a) (4 pts) each answer 1 pt; 𝐾𝐾𝑎𝑎1 = 𝑥𝑥2

0.1−𝑥𝑥

[H3PC] = 0.1 – x = 0.073 M, [H2PC–] = [H3O+]= x = 0.027 M, pH = 1.57, 𝐾𝐾𝑎𝑎2 ≈ 𝑦𝑦 [HPC2–] = y =7.8 X 10–6 M,

𝐾𝐾𝑎𝑎3 ≈0.027 𝑧𝑧

7.8 × 10−6

[PC3–] = z = 5.8 X 10–13 M (b) (2 pts) 2 < pH < 5.1 ( pKa1 < pH < pKa2)

(c) (3 pts) 3.55 ( pH = 0.5(pKa1 + pKa2))

(d) (6 pts) correct equation 2 pts; each answer 1 pt

3. (5 points) molar solubility 3 pts; gram solubility +2 pts

4. (total 10 points) each statement 1 pt; each reason 1 pt

(a) Increase, A higher concentration of H3O+ drives the dissolution by removing product PO43- ion as

HPO2-4 ion.

(b) Unchanged. The anion is an exceedingly weak base. It has litter interaction with H3O+ even at a

high concentration of H3O+.

(c) Increase. CO32- reacts with H3O+ to give HCO3

-.

(d) Increase. HS- from the dissolution of the MnS reacts with H3O+ to give H2S.

(e) Increase. F- reacts with H3O+ to give HF.

5. (7 points) kr/kf 3 pts; kr value 3 pts; unit (s-1) 1 pt

3.7 x 109 Lmol-1s-1 for gives kr= 6.0x104 s-1

6. (total 7 points) (a) (4 pts)

(b) (3 pts)

7. (total 14 points) (a) (3 pts)

Rate of formation of ES from E and S = k1[E][S]

Rate of dissociation of ES to E and S = k-1[ES]

Rate of conversion of ES to E and P = k2[ES]

𝑑𝑑[𝐸𝐸𝐸𝐸]𝑑𝑑𝑑𝑑

= 𝑘𝑘1[𝐸𝐸][𝐸𝐸] − 𝑘𝑘−1[𝐸𝐸𝐸𝐸] − 𝑘𝑘2[𝐸𝐸𝐸𝐸]

= 𝑘𝑘1[𝐸𝐸][𝐸𝐸] − (𝑘𝑘−1 + 𝑘𝑘2)[𝐸𝐸𝐸𝐸]

(b) (3 pts)

𝑑𝑑[𝐸𝐸𝐸𝐸]𝑑𝑑𝑑𝑑

= 𝑘𝑘1[𝐸𝐸][𝐸𝐸] − (𝑘𝑘−1 + 𝑘𝑘2)[𝐸𝐸𝐸𝐸] ≈ 0

⇒ [𝐸𝐸][𝐸𝐸][𝐸𝐸𝐸𝐸]

=𝑘𝑘−1 + 𝑘𝑘2

𝑘𝑘1= 𝐾𝐾𝑚𝑚

⇒ 𝐾𝐾𝑚𝑚[ES] = [𝐸𝐸][𝐸𝐸] = ([𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑡𝑡 − [𝐸𝐸𝐸𝐸])[𝐸𝐸] = [𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑡𝑡[𝐸𝐸] − [𝐸𝐸𝐸𝐸][𝐸𝐸]

⇒ (𝐾𝐾𝑚𝑚 + [𝐸𝐸])[𝐸𝐸𝐸𝐸] = [𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑡𝑡[𝐸𝐸]

⇒ [ES] = [𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑡𝑡[𝐸𝐸]𝐾𝐾𝑚𝑚 + [𝐸𝐸]

∴𝑑𝑑[𝑃𝑃]𝑑𝑑𝑑𝑑

= 𝑘𝑘2[𝐸𝐸𝐸𝐸] = 𝑘𝑘2[𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑡𝑡[𝐸𝐸]𝐾𝐾𝑚𝑚 + [𝐸𝐸]

(c) (3 pts)

𝑉𝑉𝑚𝑚𝑎𝑎𝑥𝑥 = lim[𝑆𝑆]→∞

𝑑𝑑[𝑃𝑃]𝑑𝑑𝑑𝑑

= lim[𝑆𝑆]→∞

𝑘𝑘2[𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑡𝑡[𝐸𝐸]𝐾𝐾𝑚𝑚 + [𝐸𝐸] = 𝑘𝑘2[𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑡𝑡

∴ 𝑘𝑘𝑐𝑐𝑎𝑎𝑡𝑡 = 𝑉𝑉𝑚𝑚𝑎𝑎𝑥𝑥

[𝐸𝐸]𝑡𝑡𝑡𝑡𝑡𝑡𝑎𝑎𝑡𝑡= 𝑘𝑘2

(d) (5 pts)

𝑑𝑑[𝑃𝑃]𝑑𝑑𝑑𝑑

= 𝑘𝑘𝐸𝐸 �[𝐸𝐸][𝐸𝐸][𝐸𝐸𝐸𝐸] �

[ES] = 𝑘𝑘𝐸𝐸𝐾𝐾𝑚𝑚[𝐸𝐸𝐸𝐸] = 𝑘𝑘2[𝐸𝐸𝐸𝐸]

⇒ 𝑘𝑘𝐸𝐸𝐾𝐾𝑚𝑚 = 𝑘𝑘2 = 𝑘𝑘𝑐𝑐𝑎𝑎𝑡𝑡

⇒ 𝑘𝑘𝐸𝐸 =𝑘𝑘𝑐𝑐𝑎𝑎𝑡𝑡𝐾𝐾𝑚𝑚

=1.4 × 104𝑠𝑠−1

8.9 × 10−5𝑀𝑀= 1.6 × 108𝑀𝑀−1𝑠𝑠−1

Therefore, it can be said that AchE is at “catalytic perfection”.

8. (total 14 points) (a) (6 pts) each possible structure 2 pts

HOOH

OH

OH

O

OH

O

OH

O

O

O

H O

O

O

O

OHO

H

OOHOH

OH

O

The IR spectrum has no O-H stretching peak, so all alcohols and carboxylic acids can be left out. Also, the candidate structure must possess C=O double bond since the presence of strong C=O stretching peak.

(b) (4 pts)

𝜇𝜇𝑁𝑁 =1.6022 × 10−19𝐶𝐶 × 6.6261 × 10−34𝐽𝐽 ∙ 𝑠𝑠

4𝜋𝜋 × 1.6726 × 10−27𝑘𝑘𝑘𝑘= 5.0509 × 10−27𝐽𝐽/𝑇𝑇

For proton nuclei, g = 5.586 and mI = ±1/2.

𝐸𝐸+ = −𝑘𝑘𝐻𝐻 �+12� 𝜇𝜇𝑁𝑁𝐵𝐵0

𝐸𝐸− = −𝑘𝑘𝐻𝐻 �−12� 𝜇𝜇𝑁𝑁𝐵𝐵0

⇒ Δ𝐸𝐸 = 𝐸𝐸− − 𝐸𝐸+ = 𝑘𝑘𝐻𝐻𝜇𝜇𝑁𝑁𝐵𝐵0 = 5.586 × 5.0509 × 10−27J/T × 7.0T = 1.98 × 10−25𝐽𝐽

From ΔE = hν,

ν =Δ𝐸𝐸ℎ

=1.98 × 10−25𝐽𝐽

6.626 × 10−34𝐽𝐽 ∙ 𝑠𝑠= 2.98 × 108𝑠𝑠−1 ≒ 300𝑀𝑀𝑀𝑀𝑧𝑧

(c) (4 pts)

The NMR spectrum is showing three sets of peaks, so 3-methoxypropanal can be crossed-out. The quadruplet peak at δ 4.10 (2H) is from the methylene (-CH2-) group adjacent to the methyl (-CH3) group, indicating the presence of ethyl (CH3CH2-) group. Hence, the molecule is ethyl acetate.

9. (total 10 points) each 2 pts

(a) 3 (t) – 1 (singlet) – 2 (q)

(b) 1 (t) – 2 (q) or 4 (t) – 3 (q)

(c) 1 (d) – 2 (septet) or 3 (d) – 2 (septet)

(d) 2 (t) – 3 (t) – 1 (singlet) or (multiplet)

(e) 4 (t) – 3 (sextet) – 2 (t) – 1 (singlet)

10. (total 10 points) each term 2 pts

Vibrational relaxation: excited vibrational levels return to lower vibrational levels by dissipating the

energy as heat

Fluoresence: Emission from transitions between states of the same spin

Phosphoresence: Emission from transitions between states of different spin

Internal conversion: transition between states of the same spin

Intersystem crossing: transition between states of different spin

1

2017 FALL Semester Final Examination For General Chemistry II (CH103)

Date: December 13 (Wed), Time Limit: 19:00 ~ 21:00 Write down your information neatly in the space provided below; print your Student ID in the upper right corner of every page.

Professor Name Class Student I.D. Number Name

Problem points Problem points TOTAL pts

1 /11 7 /5

/100

2 /8 8 /10

3 /6 9 /4

4 /8 10 /12

5 /17 11 /9

6 /10 ** This paper consists of 12 sheets with 11 problems (page 10: fundamental constants, page 11: periodic table, page 12: claim form). Please check all page numbers before taking the exam. Write down your work and answers in the sheet.

Please write down the unit of your answer when applicable. You will get 30% deduction for a missing unit.

NOTICE: SCHEDULES on RETURN and CLAIM of the MARKED EXAM PAPER. 1. Period, Location and Procedure

1) Return and Claim Period: December 15 (Friday, 12:00-14:00 p.m.) 2) Location: Creative Learning Bldg.(E11)

Class Room Class Class

A 202 C 210

B 208 D 211 3) Claim Procedure:

Rule 1: Students cannot bring their own writing tools into the room. (Use a pen only provided by TA) Rule 2: With or without claim, you must submit the paper back to TA. (Do not go out of the room with it)

(During the period, you can check the marked exam paper from your TA and should hand in the paper with a FORM for claims if you have any claims on it. The claim is permitted only on the period. Keep that in mind! A solution file with answers for the

examination will be uploaded on 12/16 on the web.) 2. Final Confirmation

1) Period: December 16(Sat) – 17(Sun) 2) Procedure: During this period, you can check the final score of the examination on the website again.

To get more information, visit the website at www.gencheminkaist.pe.kr.

2

1. (Total 11 pts) For the following cell reaction at 25 oC and 1 atm.

6Hg2+(aq) + 2Au(s) ⇄ 3Hg22+(aq) + 2Au3+(aq)

where half-reaction potentials are as follows:

Au3+ + 3e– Au, Eo = 1.42 V and

2Hg2+ + 2e– Hg22+, Eo = 0.905 V

(a) Write line notation of cell.

(Answer)

(b) Calculate the cell potential when all concentrations are 0.1 M (ignore activity coefficient). Is it

spontaneous reaction?

(Answer)

(c) Calculate the equilibrium constant (Keq).

(Answer)

(d) If 1.00 L of a 1.00 M Au(NO3)3 solution is added to 1.00 L of a 1.00 M Hg2(NO3)2 solution, calculate

the concentrations of Hg2+, Hg22+, and Au3+ at equilibrium.

(Answer)

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2. (Total 8 pts) In a galvanic cell, the first half-cell consists of gaseous chlorine bubbled over a platinum

electrode at a pressure of 1.00 atm into a 1.00 M solution of ZnCl2 and pH 0. The second half-cell has

a strip of solid gallium immerse in a 1.00 M Ga(NO3)3 solution. The initial cell potential for combination

of two half cells is measured to be 1.918 V at 25 oC, and as the cell operates, the concentration of

chloride ion is observed to increase.

The followings are half-reaction potentials:

Zn2+ + 2e– Zn, Eo = –0.7618 V

Cl2(g) + 2e– 2Cl–, Eo = 1.3583 V

2H+ + 2e– H2, Eo = 0.00 V

O2 + 4H+ + 4e– 2H2O, Eo = 1.229 V

(a) Write balance equations for the half-reactions at the anode and the cathode.

(Answer)

(b) Calculate the standard reduction potential of a Ga3+|Ga half-cell.

(Answer)

(c) The cell becomes an electrolytic cell when the concentration of 1.00 M of ZnCl2 at the first half-cell

is reduced to 0.1 M, pH is adjusted to 7, and gaseous chlorine bubble is stopped. Write balance

equations for the half-reactions at the first half-cell as the cathode and calculate cell potential.

(Answer)

4

3. (Total 6 pts) The following reduction potentials are measured at pH 0 in aqueous solution:

2 IO3– + 12 H+ + 10 e– I2(l) + 6 H2O Eo = 1.195 V

I2(l) + 2 e– 2 I– Eo = 0.5355 V

(a) Write chemical equation of iodine disproportionation.

(Answer)

(b) Calculate the standard potential difference. Does iodine disproportionate spontaneously in acidic

solution?

(Answer)

(c) Which is the stronger reducing agent at pH 0: I2(l) or I–?

(Answer)

4. (Total 8 pts) Following is a derivative of morphine.

HO

O

ONO

N

(a) Name all the functional groups in morphine.

(Answer)

(b) How many chiral centers does the compound have? Draw * on each chiral center of the above

structure.

(Answer)

5

5. (Total 17 pts) A compound C6H8O4 has two carboxylic acids as functional groups and has no C=C

bond.

(a) Write all possible structures for this compounds, including stereoisomers.

(Answer)

(b) Among the structures in the answer (a), identify pairs of compounds that cannot be distinguished

by physical methods other than optical rotation.

(Answer)

(c) The pairs in the answer (b) could be distinguished and separated by reacting with two equivalent

of the compound C4H11N to form amides. Draw possible structures for this compound and amide

product. .

(Answer)

6

6. (Total 10 pts) The following ionic radii (in angstroms) are estimated for the +2 ions of selected

elements of the first transition-metal series, based on the structures of their oxides: Ca2+(0.99),

Ti2+(0.71), V2+(0.64), Mn2+(0.80), Fe2+(0.75), Co2+(0.72), Ni2+(0.69), Cu2+(0.71), Zn2+(0.74). The oxides

take the rock salt structure, in which each metal ion is surrounded by oxygen atoms forming an

octahedral geometry.

(a) Draw a graph of ionic radius versus atomic number in this series

(Answer)

(b) Why does the minimum occur in V2+ ion?

(Answer)

(c) Why does another minimum occur in Ni2+ ion?

(Answer)

(d) Are these solids better described as high- or low-spin transition-metal complexes?

(Answer)

7

7. (5 pts) A complex, [Ti(OH2)6]3+, exhibits a strong peak with a maximum wavelength at 492 nm.

Estimate the crystal field stabilization energy of the complex in the unit of cm-1.

(Answer)

8. (Total 10 pts)

(a) An aqueous solution of Mn(NO3)2 is very pale pink, but an aqueous solution of K4[Mn(CN)6] is deep

blue. Explain why the two differ so much in the intensities of their colors.

(Answer)

(b) Predict which of the following compounds would be colorless in aqueous solution, and explain the

reason.

K2[Co(NCS)4], Zn(NO3)2, [Cu(NH3)4]Cl2, CdSO4, AgClO3, Cr(NO3)2

(Answer)

8

9. (Total 4 pts) Polymers with more than 10,000 molecular weight can be made by addition

polymerization or condensation polymerization.

(a) Which polymerization reaction produces high molecular weight polymers with 10% conversion?

(Answer)

(b) Which polymerization reaction mixture contains no monomers at 10% conversion?

(Answer)

10. (Total 12 pts) Identify what kind of polymers are obtained by condensation polymerization of the

following monomers (Assume that A functional group can react only with B (vice versa), and also

almost equal amount of A and B functional groups are used).

(a) A-A + B-B + B4

(Answer)

(b) A-B + B4

(Answer)

(c) AB2 + AB

(Answer)

(d) AB2 + A3

(Answer)

9

11. (Total 9 pts) Vinyl chloride monomer undergoes addition polymerization with peroxide radical

initiators.

H2C CH

Cl

R O O R

Vinyl chloride Peroxide Initiator

(a) Write the initiation reactions of this polymerization.

(Answer)

(b) Write the propagation reactions of this polymerization.

(Answer)

(c) Write the termination reactions of this polymerization.

(Answer)

10

11

12

Claim Form for General Chemistry Examination Page ( / )

Class: , Professor Name: , I.D.# : , Name: If you have any claims on the marked paper, please write down them on this form and submit this with your paper in the assigned place. (And this form should be attached on the top of the marked paper with a stapler.) Please, copy this sheet if you need more before use.

By Student By TA

Question # Claims Accepted? Yes(√) or No(√)

Yes: □ No: □ Pts (+/-) Reasons

<The Answers>

Problem points Problem points TOTAL pts

1 2+3+2+4/11 7 5/5

/100

2 2+2+4/8 8 6+4/10

3 2+2+2/6 9 2+2/4

4 5+3/8 10 3+3+3+3/12

5 8+5+4/17 11 3+2+4/9

6 2+3+3+2/10

전체 기준: 전개과정은 맞으나 답이나 unit 이 틀리면 -1

1. (Total 11 pts) (a) (2 pts) Au | Au3+(aq) || Hg2+(aq), Hg2

2+(aq) | Pt (or C or Au)

(b) (3 pts) Eo = Eocathode – Eo

anode = 0.905 – 1.42 = –0.515 V

E = (−0.515) − 0.05926

log �(0.1)3(0.1)2

(0.1)6 � = −0.525 𝑉𝑉, non-spntaneous reaction

(c) (2 pts)

(d) (4 pts)

2. (Total 8 pts) (a) (2 pts) Anode: Ga Ga3+ + 3e–

Cathode: Cl2 + 2e– 2Cl–

(b) (2 pts) Eocathode – Eo

anode = 1.3583 V – x V = 1.918 V

x = –0.5597 V

(c) (4 pts) We should consider the following reactions:

(i) E(Cl2|Cl–) = 1.3583 V – (0.0592 V/2)log((0.1)2/1) = 1.4175 V

Eocathode – Eo

anode = 1.4175 V – (–0.5597 V) > 0 : galvanic cell (X)

(ii) E(H+/H2) = 0.00 V – (0.0592 V /2)log(1/(10–7)2) = –0.4144 V

Eocathode – Eo

anode = –0.4144 V – (–0.5597 V) > 0 : galvanic cell (X)

(iii) E(Zn2+|Zn) = –0.7618 V – (0.0592 V/2)log(1/0.1) = –0.7914 V

Eocathode – Eo

anode = –0.7914 V – (–0.5597 V) = –0.2317 V : electrolytic cell

Therefore, Cathode: Zn2+ + 2e– Zn

Cell potential: –0.2317 V

3. (Total 6 pts) (a) (2 pts) 6I2 + 6H2O 10I– + 2IO3

– + 12H+

(b) (2 pts) E = 0.5355 V – 1.195 V = –0.6595 V: nonspontaneous reaction

(c) (2 pts) I– : more easily oxidized than I2 and must be a stronger reducing agent.

4. (Total 8 pts) (a) (5 pts) for each, 1 pt

amine, ether, ester, nitrile, phenol (or alcohol, hydroxyl)

(b) (3 pts) HOHO

OO

OONNOONN **

** ** ****

5

5. (Total 17 pts)

(a) (8 pts) 구조 1개당 0.5 pt; 16개 이상은 모두 8 pts

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

HOOC

COOH

HOOC

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

HOOC COOH HOOC COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

or or

HOOC COOHCOOHHOOC

or

COOHHOOC

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOHor or or

(b) (5 pts) 1 pt per each pair; 5 pts when more than 5 pairs

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

COOH

HOOC COOH HOOC COOHCOOH

COOH

COOH

COOH

COOHHOOC COOHHOOC

COOH

COOH

COOH

COOH

COOH

COOH

or

COOH

COOH

COOH

COOHor

COOH

COOH

(c) (4 pts) 각 2점씩 (chiral center표시가 없어도 2점)

NH2 NH2

and corresponding amides

6. (Total 10 pts) (a) (2 pts)

(b) (3 pts) size contraction +1 pt; size enlargement +2 pts

As the nuclear charge Z increases, the size of these +2 ions would be expected to contract due to

the effective nuclear charge (Zeff) increases.

The minimum occurs in V2+ ions, which has three d electrons. When the ligand field is weak, then

the metal ions have high-spin configurations, and the 4th electron to join the d subshell must go into a

higher energy eg orbital. This makes the Mn2+ ions bigger because the eg orbitals are further away

from the nucleus. Exactly such an increase in size is observed.

(c) (3 pts) size contraction +1 pt; size enlargement +2 pts

As the nuclear charge Z increases, the size of these +2 ions would be expected to contract due to

the effective nuclear charge (Zeff) increases.

The minimum occurs in Ni2+ ions, which has eight d electrons. When the ligand field is weak, then

the metal ions have high-spin configurations, and the 9th electron to join the d subshell must go into a

higher energy eg orbital. This makes the Cu2+ ions bigger because the eg orbitals are further away

from the nucleus. Exactly such an increase in size is observed.

(d) (2 pts)

The metal ions are in weak-field environments and are well described as high spin octahedral

complexes.

7. (5 pts) unit conversion +1 pts; d1 +1 pts; -2/5∆o +3 pts

∆𝑂𝑂 = 1492 𝑛𝑛𝑛𝑛

= 20325 𝑐𝑐𝑐𝑐−1

Crystal field stabilization of d1 configuration = -2/5 ∆o = -8130 cm-1

8. (10 pts) (a) (6 pts) high spin d5 +3 pts; low spin d5 +3 pts

(b) (4 pts) for each, 1 pt; d10 1 pt

9. (4 pts) (a) (2 pts) addition polymerization

(b) (2 pts) condensation polymerization

10. (12 pts) (a) (3 pts)

A-A + B-B + B4 : Crosslinked polymers

(가교된 고분자)

(b) (3 pts)

A-B + B4 : Branched polymers (more accurately, polymers with 2 branches)

(두 개의 곁가지를 (branch)를 갖는 고분자)

(c) (3 pts) “branched” 2 pts; “highly” 1 pt

AB2 + AB : Highly branched polymers (or Hyperbranched polymers)

(많은 곁가지를(branch) 갖는 고분자, 하이퍼브랜치 고분자)

(d) (3 pts)

AB2 + A3 : Crosslinked polymers

(가교된 고분자)

11. (9 pts) (a) (3 pts) initiation

(1 pt)

(2 pts)

(b) (2 pts) propagation

(c) termination

(c) (4 pts) termination

a. Coupling (2 pts)

b. Disproportionation (2 pts)