1

2015 FALL Semester Midterm Examination For General Chemistry II (CH103)

Date: October 21 (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 /12 6 /14

/100

2 /8 7 /14 3 /12 8 /8 4 /6 9 /9 5 /7 10 /10

** This paper consists of 14 sheets with 10 problems (pages 10,11: standard reduction potential, page 12: fundamental constants, page 13: periodic table, page 14: 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 26 (Mon, 6:30 ~ 7:30 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 29 (Thu) – October 30 (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.

2

Problem points Problem points TOTAL pts

1 3x4/12 6 8+2+4/14

/100

2 4+4/8 7 6+8/14 3 6+6/12 8 2+2+2+2/8 4 4+2/6 9 3x3/9 5 2+2+3/7 10 4+3+3/10

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 12 pts) (a) (3 pts) 0 mL of NaOH

(b) (3 pts) 100 mL of 0.05M NaOH

In the total volume 150ml, we now have 0.005mol less of CH3COOH and 0.005mol more of

CH3COO-

The titration has reached the half-equivalence point.

3

(c) (3 pts) 200ml of 0.05M NaOH

0.01mol of OH- has reacted with 0.01mol of CH3COOH. Neutralization is complete; and now in a

total volume of 250ml, there is 0.01mole of CH3COO- that undergoes hydrolysis to produce OH-

(d) (3 pts) 300ml of 0.05M NaOH

2. (Total 8 pts) For each, procedure 2 pts, answer 2 pts (a) (4 pts) Quinine = Q,

[QH+] = [OH-] = x, Kb = 𝑥𝑥2

0.0016−𝑥𝑥= 10−5.1,

(a1) Assuming x > x, x = √0.33 × 1.26 × 10−9 =2.04 × 10−5

pH = -log (2.04 × 10−5) = 4.7

Q (aq)

+ H2O (l)

QH+

(aq)

+ OH (aq)

QH+(aq) + H2O

(l)

Q

(aq)

+ H3O+

4

3. (Total 12 pts) (a) (6 pts) [Zn2+] = s, [OH-] = 2s

Ksp = 4s3 = 4.5 x 10-17

s = 2.2 x 10-6 M = [Zn2+]

Solubility in pure water = 2.2 x 10-4 g/L

Using Henderson-Hasselbalch equation,

pH = -log(6.2 x 10-8) + log (0.6M / 0.4 M) = 7.4

pOH = 14.0 – 7.4 = 6.6, [OH-] = 2.51 x 10-7

[Zn2+] = Ksp /[OH-]2 = 4.5 x 10-17 / (2.51 x 10-7)2 = 7.14 x 10-4 M

Solubility in the buffer solution = 0.0709 g/L

(b) (6 pts)

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4. (Total 6 pts) (a) (4 pts) Equation for cathode: MnO4- + 8H3O+ + 5e- -> Mn2+ + 12H2O

Equation for anode: Pb -> Pb2+ + 2e-

Overall reaction: 2MnO4- + 16H3O+ + 5Pb -> 2Mn2+ + 24H2O + 5Pb2+

(b) (2 pts) Standard cell potential difference = (Standard reduction potential of cathode) – (Standard

reduction potential of anode) = 1.49 V – ( -0.1263V) = 1.62 V

5. (Total 7 pts) equation 2 pts; Q 2 pts; Nernst 3 pts The half-cell reactions are

Note that n = 2 for the overall cell reaction, and the reaction quotient is shown as

𝑄𝑄 = [𝐻𝐻3𝑂𝑂]2

[𝐴𝐴𝐴𝐴+]2𝑃𝑃𝐻𝐻2

Because PH2 = 1 atm. The Nernst equation is

𝐸𝐸 = 𝐸𝐸𝑜𝑜 − 0.0592

2log𝑄𝑄 = 0.800 + 0.0592 𝑝𝑝𝐻𝐻 + 0.0592 log[𝐴𝐴𝐴𝐴+] = 0.915

pH = 2.94

6. (Total 14 pts) (a) (8 pts) oxidation 2 pts; reduction 2 pts; total 2 pts; E 2 pts

(b) (2 pts)

(c) (4 pts) statement 1 pt; equation 1 pt; K 2 pts

H2 + 2H2O

2H3O

+ + 2e-

(anode)2Ag+

+ 2e-

Ag(s)

(cathod)

6

7. (Total 14 pts) (a) (6 pts) each 3 pts

W = V × I × t = 1.5 V × 0.7 A × 3600 s = 3.8×103 J

# of electrons = I t / F

Mass of zinc = # of electrons / n × MZn = (0.7 × 3600 / 96485) × 1/2 × 65.4 = 0.85 g

(b) (8 pts)

Zn Zn2+ + 2

e- E0 = 0.7628

MnO4- + 8H3O

+ + 5 e- Mn2+

+ 12 H2O E0

= 1.491

2MnO4- + 5Zn

+ 16H3O+ 5Zn2+

+ 2Mn2+ + 24

H2O E0

= 1.491+(0.7628) = 2.2538

V

E = E0

- 0.0592n

log [Mn2+]2

[Zn2

+]5

[MnO4-]2

[H3O

+]16

0.059210

log (0.1)2

(0

.01)5

(0.1)2

(1

.0x10-2)16

= 2.2538 -

= 2.12 V

8. (Total 8 pts) each of the numbered equation, 2 pts

steady - state approximation

(1)

(2)

7

rate = = k [NO Cl] [Cl] (3)

∴ rate = (4)

8

9. (Total 9 pts) (a) (3 pts) k = A X exp(-Ea/RT) lnk1-lnk2 = ln(k1/k2) = Ea/R X (1/T2 – 1/T1)

Hence,

Ea = ln(k1/k2) X R X 1/(1/T2 – 1/T1) = ln(0.76/0.87) X 8.314 J/molK X 1/(1/1030K – 1/1000K)

= 38.6 kJmol-1

(b) (3 pts) from k = A X exp(-Ea/RT)

A = k / exp(-Ea/RT) = 0.76 s-1 / exp ((-38.6 X 1000 J/mol) / (8.314 J/molK X 1000K)) =

= 78.9 s-1

(c) (3 pts) from k = A X exp(-Ea/RT)

k at 1100 K = 78.9 s-1 X exp ((-38.6 X 1000 J/mol) / (8.314 J/molK X 1100K))

= 1.16 s-1

10. (Total 10 pts) (a) (4 pts) each, 2 pts

rate = k[H2][NO]2, Third order reaction.

0.38 M-2s-1

(c) (3 pts)

Mechanisms B and C

(d) (3 pts)

k = k1k2/k-1

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1. (Total 12 pts) To 50 mL of 0.20 M acetic acid (Ka = 1.76×10-5) at 25 °C, a chemist adds a 0.050 M

solution of NaOH in the following amounts (a) 0 mL, (b) 100 mL, (c) 200 mL, and (d) 300

mL. Compute the pH for each points.

(a) (Answer)

(b) (Answer)

(c) (Answer)

(d) (Answer)

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2. (Total 8 pts) The antimalarial properties of quinine (C20H24N2O2) saved thousands of lives during

construction of the Panama Canal. This substance is a classic example of the medicinal wealth of

tropical forests. Quinine has two N atoms. Both N atoms are basic, but the N (bond) of the 3o amine

group is far more basic (pKb = 5.1) than the N within the aromatic ring system (pKb = 9.7)

N

H3CO

NHO

Quinine

(a) A saturated solution of quinine in water is only 1.6ⅹ10-3 M. What is the pH of this solution?

(Answer)

(b) Because of its low solubility, quinine is given as the salt quinine hydrochloride (C20H24N2O2•HCl),

which is 120 times more soluble than quinine. What is the pH of 0.33 M quinine hydrochloride?

(Answer)

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3. (Total 12 pts)

(a) Compare the solubility of Zn(OH)2 (g/L) in pure water with that in a buffer solution containing

0.60 M Na2HPO4 and 0.40 M NaH2PO4 (Mw(Zn(OH)2) = 99.40, Ksp(Zn(OH)2) = 4.5×10-17,

Ka1(H3PO4) = 7.5×10−3, Ka2(H3PO4) = 6.2×10−8, Ka3(H3PO4) = 4.8×10−13).

(Answer)

(b) In a solution saturated with H2S, [H2S] is fixed at 0.1 M. Calculate the molar solubility of FeS(s)

in such a solution if it is buffered at pH 3.0 (Ka(H2S) = 9.1×10–8, Equilibrium constants for FeS

dissolution at 25 oC = 5×10-19).

(Answer)

12

4. (Total 6 pts) It is known that permanganate ion (MnO4-) can be reduced to the manganese ion(II)

(Mn2+) in acidic, aqueous solutions. Let’s suppose that this half-cell is connected to Pb2+ | Pb half-

cell in a galvanic cell, with [MnO4-] = [Mn2+] = [Pb2+] = [H3O+] = 1 M.

(a) Write the balanced equation for the overall cell reaction.

(Answer)

(b) Calculate the standard cell potential difference.

(Answer)

5. (7 pts) If the Ecell of the following cell is 0.915 V, what is the pH in the anode?

Pt(s) | H2(1.00 atm) | H3O+(aq) || Ag+(0.10 M) | Ag (s)

(Answer)

13

6. (Total 14 pts) Manganese is the twelfth-most-abundant element on the earth’s surface. Its most

important ore source is pyrolusite (MnO2). The preparation and uses of manganese and its

compounds (which range up to +7 in oxidation state) are intimately bound up with electrochemistry.

(a) The Winkler method is an analytical procedure for determining the amount of oxygen dissolved

in water. In the first step, Mn(OH)2(s) is oxidized by gaseous oxygen to Mn(OH)3(s) in basic

aqueous solution. Write the oxidation and reduction half-equations for this step, and write the

balanced overall equation. Then calculate the standard voltage that would be measured if this

reaction were carried out in an electrochemical cell.

(Answer)

(b) Calculate the equilibrium constant at 25 °C for the reaction in part (a).

(Answer)

(c) In the second step of the Winkler method, the Mn(OH)3 is acidified to give Mn3+, and iodide ion is

added. Will Mn3+ spontaneously oxidize I-? Write a balanced equation for its reaction with I-,

and calculate its equilibrium constant. Titration of the I2 produced completes the use of the Winkler

method.

(Answer)

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7. (Total 14 pts)

(a) Manganese(IV) is an even stronger oxidizing agent than manganese(III). It oxidizes zinc to Zn2+

in the dry cell. Such a battery has a cell voltage of 1.5 V. Calculate the electrical work done by this

battery in 1.00 hour if it produces a steady current of 0.70 A. Calculate the mass of zinc reacting in

this process.

(Answer)

(b) A galvanic cell is made from two half-cells. In the first, a platinum electrode is immersed in a

solution at pH 2.00 that is 0.100 M in both MnO4− and Mn2+. In the second, a zinc electrode is

immersed in a 0.0100 M solution of Zn(NO3)2. Calculate the cell voltage that will be measured.

(Answer)

15

8. (Total 8 pts) The mechanism for the decomposition of NO2Cl is

By making a steady-state approximation for [Cl], express the rate of appearance of Cl2 in terms of

the concentrations of NO2Cl and NO2.

(Answer)

9. (Total 9 pts) The rate constant of the first-order reaction 2N2O (g) → 2N2 (g) + O2 (g) is 0.76 s-1 at

1000 K and 0.87 s-1 at 1030 K.

(a) Calculate the activation energy (Ea) of the reaction.

(Answer)

(b) Calculate the pre-exponential factor (A, from the Arrhrenius equation) of the reaction.

(Answer)

(c) What would be the predicted rate constant at 1100 K?

(Answer)

16

10. (Total 10 pts) The following data were collected for the reaction between hydrogen and nitric

oxide at 700 oC.

22HH22((gg)) ++

22NNOO((gg))

22HH22OO((gg))

++ NN22((gg))

Experiment [H2] (M) [NO] (M) Initial rate (M/s)

1 0.010 0.025 2.4ⅹ10-6

2 0.0050 0.025 1.2ⅹ10-6

3 0.010 0.0125 6.0ⅹ10-7

(a) Determine the order of the reaction, and calculate the rate constant including unit.

(Answer)

(b) Which of the following mechanisms agree with the observed rate expression?

H2

+ NO

H2O + N

(slow)

N

+ NO

N2 + O

(fast)

O

+ H2

H2O

(fast)

2NO

N2O2

(fast)

N2O2

+ H2

N2O + H2O

(slow)

N2O

+ H2

N2 + H2O

(fast)

H2

+ 2NO

N2O + H2O

(slow)

N2O

+ H2

N2 + H2O

(fast)

Mechanism A Mechanism

B

Mechanism C

k1k-1k2

k3

(Answer)

(c) For the mechanism B above, provide the rate constant with respect to k1, k-1, k2, or k3. (Answer)

17

18

19

20

21

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

1

Problem points Problem points TOTAL pts

1 8+4/12 6 6+2/8

/100

2 2+3/5 7 3+4+2/9 3 2+2+2+2+3/11 8 4+1+3+4/12 4 9+3/12 9 4+4/8 5 4+2+4+2/12 10 4+3+2+2/11

전체 기준: 전개과정은 맞으나 답이나 unit 이 틀리면 -1

답은 맞으나 전개과정이 약간 틀렸을 때 -1

1. (Total 12 pts) (a) (8 pts) Drawing molecular orbital: 4 pts; energy level diagram and symbols: 4 pts

2

(b) (4 pts) 2 pts for each

σ

π

π∗

σ∗

σ

π

π∗

σ∗

groundelectronic

state

first excited

electronicstate

2. (Total 5 pts) (a) (2 pts) water: H2O

(b) (3 pts) permanent electric dipole moment

3. (Total 11 pts) (a) (2 pts) 1

(b) (2 pts) 4

(c) (2 pts) 7

(d) (2 pts) 6

(e) (3 pts) 7

3

4. (Total 12 pts) (a) (9 pts) for each step, 3 pts

peroxide is broken to give radicals

(b) (3 pts)

4

5. (Total 12 pts) (a) (4 pts) for each isomer, 2 pts

(b) (2 pts) for each isomer, 1 pt

All axial hydrogens should be along with z-axis.

All C-C and C-H bonds should be parallel with other existing bonds.

(c) (4 pts) for each isomer, 2 pts

(d) (2 pts) for each stable isomer, 1 pt

H

H

HH

HH

H

H

H

HH

H

H

HH

H

HHand or and

two axial

methyls&

one equatorial methyl

one axial methyls

& two

equatorial methyl

two axial

methyls&

one equatorial methyl

one axial methyls

& two

equatorial methyl

more stable more stable

6. (Total 8 pts) (a) (6 pts) for each step (total 3 steps), 2 pts

An acid-catalyzed mechanism via oxonium-ion intermediate.

O

H3C

+ H+ O

H3C

H

+ CH3OHH3C

HO

OCH3H

- H+

H3C

HO

OCH3

(b) (2 pts)

H3C

H3CO OH

H

HH

H

H

H

HH

H

H

HH

H

HH

H

H

H

HH

H

H

HHor

5

7. (Total 9 pts) (a) (3 pts) for each functional group (except sulfide), 1 pt

HN

ON

S

O

O

OH

Sulf ide

Penicillin G

Amide

AmideCarboxylic

acid

(b) (4 pts) for each *, 1 pt

HN

ON

S

O

O

OH

Ampicillin

NH2

*

* * *

(c) (2 pts)

Penicillin : 2*2*2 = 8

Ampicillin : 2*2*2*2 = 16

6

8. (Total 12 pts) (a) (4 pts) for each structure, 2 pts

PtClCl

H3N NH3

PtNH3Cl

H3N Cl

(b) (1 pts)

Left one.

(c) (3 pts)

Diamagnetic, with the bottom four levels in the square-planar configuration (all but dx2-y2) occupied

(d) (4 pts) for each complex, 2 pts

Red transmission corresponds to absorption of green light by K2[PtCl4]. A colorless solution has

absorptions at either higher or lower frequency than visible. Because Cl- is a weaker field ligand

than NH3, the absorption frequency should be higher for the Pt(NH3)42+ complex, putting it in the

ultraviolet region of the spectrum.

9. (Total 8 pts) # of unpaired electrons: [Fe(CN)6]3- one, [Fe(OH2)6]3+ five.

For each, 2 pts

CFSEs: [Fe(CN)6]3- -2 ∆o, [Fe(OH2)6]3+ 0.

For each, 2 pts

7

10. (Total 11 pts) (a) (4 pts) only trans- and cis-, 2 pts

(b) (3 pts) wavelength, 1 pt; color, 1 pt

Absorption wavelength: 209x103 J mol-1 = NAhc/λ.

L = 6.02x1023 mol-1 x 6.63x10-34 J s x 3.00x108 m s-1 / 209x103 J mol-1 = 573 nm

The complementary color of purple is yellow.

(c) (2 pts)

zero

(d) (2 pts)

Ammonia is a ligand generating a field weaker than the cyano ligand, then the absorption

wavelength will shift to the longer wavelength.

8

2015 FALL Semester Final Examination For General Chemistry II (CH103)

Date: December 16 (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 /12 6 /8

/100

2 /5 7 /9 3 /11 8 /12 4 /12 9 /8 5 /12 10 /11

** This paper consists of 12 sheets with 10 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.

(채점답안지 분배 및 이의신청 일정)

1. Period, Location, and Procedure 1) Return and Claim Period: December 18 (Fri, 12:00 ~ 14:00 p.m.) 2) Location: Creative Learning Bldg.(E11)

Class Room Class Room Class Room Class Room

A 202 B 205 C 210 D 211

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 marked final examination paper with a stapler. Give them to TA.

2. Final Confirmation 1) Period: December 19 (Sat) – December 20 (Sun) 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.

9

1. (Total 12 pts)

(a) Draw the molecular orbital and energy level diagram of the ethylene.

(Draw two lowest unoccupied MOs and two highest occupied MOs. Put them in the order of energy

and draw the shape of the molecular orbitals as in the textbook.)

(Answer)

(b) Indicate the electron configuration of the ground electronic state and the first excited electronic

state of ethylene.

(Answer)

10

2. (Total 5 pts)

(a) When you heat up a glass of milk using a microwave oven, what molecule is actually absorbing

the microwave energy?

(Answer)

(b) What kind of physical property the molecule above should have in order to absorb the

microwave energy?

(Answer)

3. (Total 11 pts) How many lines appear in the proton NMR spectrum? Include signals arising from

both the chemical shift and J couplings.

(a) C2H4 (b)

(Answer) (Answer)

(c) CH3CH2COCH2CH3 (d) CH3CHCl2

(e)

(Answer) (Answer)

(Answer)

11

4. (Total 12 pts)

(a) Consider the polymerization of vinyl chloride (CH2=CHCl) to polyvinyl chloride. Describe 3 steps

of addition polymerization that involves the free-radical chain reaction of vinyl chloride.

(Answer)

(b) Tropomyosin is a two-stranded ɑ-helical coiled coil 70 kd muscle protein. Estimate an

approximate length of the tropomyosin molecule (molecular weight of 1 amino acid is 110 dalton and

the rise per residue of an ɑ-helix is 1.5 Å)

(Answer)

12

5. (Total 12 pts) Cyclohexane is a cycloalkane with the molecular formula C6H12. On an industrial

scale, cyclohexane is produced by hydrogenation (addition of hydrogen gas) of benzene. In order to

release the ring strain in the hexagonal form, cyclohexane is known to have two conformations of

chair and boat forms.

(a) Draw the chair conformation of cyclohexane with all 12 hydrogens and (b) circle all axial

hydrogens in the chair conformation.

(Answer)

(c) Draw two possible chair conformations of 1,2,4-trimethyl cyclohexane (relative positions of the

methyl groups are ONLY considered, optical isomers are NOT considered, and hydrogens can be

omitted).

1,2,4-trimethyl cyclohexane

(Answer)

(d) For 1,2,4-trimethyl cyclohexane, given the fact that an axial methyl group makes the compound

unstable due to steric interactions with other axial hydrogens or methyl groups, which conformer is

more stable than the other?

(Answer)

13

6. (Total 8 pts) Epoxides can react with water or alcohols in the presence of acids.

O

H3C+ CH3OH

H+

H3C

HO OCH3

(a) Draw the reaction mechanism of the above reaction using the arrow notation. In your drawing,

specify the role of acid and provide possible intermediates.

(Answer)

(b) In addition to the given product, one more compound was isolated and identified as a

geometrical isomer. Draw the other isolated product.

(Answer)

14

7. (Total 9 pts) In nature, surprisingly complex organic molecules are often found in microorganisms

and plants. And not only that, they sometimes found to be useful in our real life. One of the

examples is Penicillin, the famous antibiotics extracted from Penicillium Fungi. Answer the following

questions.

(a) Circle all of the functional groups in the molecule below and name them. One of the functional

groups, sulfide, is already shown.

(Answer)

HN

ON

S

O

O

OH

Sulfide

Penicillin G

(b) Replacing one of the hydrogen atoms of –CH2- group to –NH2 (amine) group results another

famous antibiotics, Ampicillin. Mark all of the chiral carbon atoms in Ampicillin with asterisk mark (*).

(Answer)

HN

ON

S

O

O

OH

Ampicillin

NH2

(c) How many stereoisomers exist in Penicillin G and Ampicillin?

(Answer)

15

8. (Total 12 pts) Cisplatin, a simple square planar complex of platinum(Pt), has a molecular formula

of PtCl2(NH3)2.

(a) Draw the geometric isomers of cisplatin.

(Answer)

(b) In cisplatin, two ammonia ligands are in cis position to each other. What is the structure of

cisplatin?

(Answer)

(c) Is cisplatin diamagnetic or paramagnetic?

(Answer)

(d) The salt K2[PtCl4] is red, but [Pt(NH3)4]Cl2•H2O is colorless. In what regions of the spectrum do

the dominant absorptions for these compounds lie?

(Answer)

9. (Total 8 pts) Experiments can measure not only whether a compound is paramagnetic, but also

the number of unpaired electrons. It is found that the octahedral complex ion [Fe(CN)6]3- has fewer

unpaired electrons than the octahedral complex ion [Fe(OH2)6]3+. How many unpaired electrons are

present in each species? In each case, express the CFSE in terms of ∆o.

(Answer)

16

10. (Total 11 pts)

(a) Iron(III) forms octahedral complexes. Sketch the structures of all the distinct isomers of

[Fe(en)2Cl2]+. (en = ethylenediamine, the en ligand can be represented as N------N.)

(Answer)

(b) The complex [Ni(NH3)6]2+ has a ligand field splitting of 209 kJ mol-1 and forms a purple solution.

What is the wavelength and color of the absorbed light?

(Answer)

(c) The complex [Co(CN)6]3- is pale yellow. How many unpaired electrons are present in the

complex?

(Answer)

(d) If ammonia molecules replace the cyanide ions as ligands in the complex [Co(CN)6]3-, will the

wavelength of the radiation absorbed be longer or shorter?

(Answer)

17

18

19

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