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
Problem
points
1
2
3
4
5
Student I.D. Number
Problem
/12
/8
/12
/6
/7
Name
TOTAL pts
points
6
7
8
9
10
/14
/14
/8
/9
/100
/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.
1
<The Answers>
Problem
1
2
3
4
5
points
Problem
3x4/12
4+4/8
6+6/12
4+2/6
2+2+3/7
6
7
8
9
10
points
TOTAL pts
8+2+4/14
6+8/14
2+2+2+2/8
3x3/9
/100
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 CH 3 COOH and 0.005mol more of
CH 3 COO-
The titration has reached the half-equivalence point.
2
(c) (3 pts) 200ml of 0.05M NaOH
0.01mol of OH- has reacted with 0.01mol of CH 3 COOH. Neutralization is complete; and now in a
total volume of 250ml, there is 0.01mole of CH 3 COO- 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, Q (aq) + H2O (l)
[QH+] = [OH-] = x,
𝑥𝑥 2
Kb =
0.0016−𝑥𝑥
+
QH (aq) + OH (aq)
= 10−5.1 ,
(a1) Assuming x << 0.0016, x = √0.0016 × 7.94 × 10−6 = 1.13 × 10−4 . Continuing by reinserting this
approximate value,
𝑥𝑥 2
0.0016−0.000113
= 10−5.1 . x = 1.08 × 10−4 . Iterating this calculation once again, we
obtain x = 1.09 × 10−4 . pOH = 4.0. pH = 14 – pOH = 10.0.
(without re-inserting approximate values: -1 pt)
(a2) Solving the quadratic equation to give x = 1.09 × 10−4 . pOH = 4.0. pH = 14 – pOH = 10.0.
(b) (4 pts)
[Q] = [H 3 O+] = x,
+
QH (aq) + H2O (l)
Ka = =
𝑥𝑥 2
0.33−𝑥𝑥
pH = -log (2.04 × 10−5 ) = 4.7
+
Q (aq) + H3O
= 10−8.9 . Given 0.33 >> x, x = √0.33 × 1.26 × 10−9 =2.04 × 10−5
3
3. (Total 12 pts)
(a) (6 pts) [Zn2+] = s, [OH-] = 2s
K sp = 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)
4
4. (Total 6 pts)
(a) (4 pts) Equation for cathode: MnO 4 - + 8H 3 O+ + 5e- -> Mn2+ + 12H 2 O
Equation for anode: Pb -> Pb2+ + 2eOverall reaction: 2MnO 4 - + 16H 3 O+ + 5Pb -> 2Mn2+ + 24H 2 O + 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
-
+
2H3O + 2e
Ag(s)
H2 + 2H2O
+
2Ag + 2e
(anode)
(cathod)
Note that n = 2 for the overall cell reaction, and the reaction quotient is shown as
𝑄𝑄 =
Because P H2 = 1 atm. The Nernst equation is
pH = 2.94
𝐸𝐸 = 𝐸𝐸 𝑜𝑜 −
[𝐻𝐻3𝑂𝑂]2
[𝐴𝐴𝐴𝐴+]2 𝑃𝑃𝐻𝐻2
0.0592
log 𝑄𝑄 = 0.800 + 0.0592 𝑝𝑝𝑝𝑝 + 0.0592 log[𝐴𝐴𝐴𝐴+ ] = 0.915
2
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
5
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 × M Zn = (0.7 × 3600 / 96485) × 1/2 × 65.4 = 0.85 g
(b) (8 pts)
-
.
E0 = 0 7628
+
Zn2 + 2 e
Zn
-
-
+
MnO4 + 8H3O + 5 e
-
+
2MnO4 + 5Zn + 16H3O
+
Mn2 + 12 H2O
.
E0 = 1 491
.
.
+
+
5Zn2 + 2Mn2 + 24 H2O E0 = 1 491+(0 7628)
= 2.2538 V
+
+
.
Mn2 ]2 [Zn2 ]5
[
0
0592
=
o
l g
E E0
+ 16
2
n
[MnO4 ] [H3O ]
. 2 . 5
.
1
= 2.2538 - 0 0592 log (0 ) (0 01-)
. 2 .
2 16
10
(0 1) (1 0x10 )
= 2.12 V
8. (Total 8 pts) each of the numbered equation, 2 pts
steady - state approximation
(1)
(2)
6
rate =
∴ rate =
= k [NO Cl] [Cl]
(3)
(4)
7
9. (Total 9 pts)
(a) (3 pts) k = A X exp(-E a /RT)
lnk 1 -lnk 2 = ln(k 1 /k 2 ) = E a /R X (1/T 2 – 1/T 1 )
Hence,
E a = ln(k 1 /k 2 ) X R X 1/(1/T 2 – 1/T 1 ) = 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(-E a /RT)
A = k / exp(-E a /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(-E a /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[H 2 ][NO]2, Third order reaction.
0.38 M-2s-1
(c) (3 pts)
Mechanisms B and C
(d) (3 pts)
k = k 1 k 2 /k -1
8
1. (Total 12 pts) To 50 mL of 0.20 M acetic acid (K a = 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)
9
2. (Total 8 pts) The antimalarial properties of quinine (C 20 H 24 N 2 O 2 ) 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 (pK b = 5.1) than the N within the aromatic ring system (pK b = 9.7)
N
HO
H3CO
N
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 (C 20 H 24 N 2 O 2 •HCl),
which is 120 times more soluble than quinine. What is the pH of 0.33 M quinine hydrochloride?
(Answer)
10
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 Na 2 HPO 4 and 0.40 M NaH 2 PO 4 (Mw(Zn(OH) 2 ) = 99.40, K sp (Zn(OH) 2 ) = 4.5×10-17,
K a1 (H 3 PO 4 ) = 7.5×10−3, K a2 (H 3 PO 4 ) = 6.2×10−8, K a3 (H 3 PO 4 ) = 4.8×10−13).
(Answer)
(b) In a solution saturated with H 2 S, [H 2 S] 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 (K a (H 2 S) = 9.1×10–8, Equilibrium constants for FeS
dissolution at 25 oC = 5×10-19).
(Answer)
11
4. (Total 6 pts) It is known that permanganate ion (MnO 4 -) 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 halfcell in a galvanic cell, with [MnO 4 -] = [Mn2+] = [Pb2+] = [H 3 O+] = 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 E cell of the following cell is 0.915 V, what is the pH in the anode?
Pt(s) | H 2 (1.00 atm) | H 3 O+(aq) || Ag+(0.10 M) | Ag (s)
(Answer)
12
6. (Total 14 pts) Manganese is the twelfth-most-abundant element on the earth’s surface. Its most
important ore source is pyrolusite (MnO 2 ). 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 I 2 produced completes the use of the Winkler
method.
(Answer)
13
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 MnO 4 − and Mn2+. In the second, a zinc electrode is
immersed in a 0.0100 M solution of Zn(NO 3 ) 2 . Calculate the cell voltage that will be measured.
(Answer)
14
8. (Total 8 pts) The mechanism for the decomposition of NO 2 Cl is
By making a steady-state approximation for [Cl], express the rate of appearance of Cl 2 in terms of
the concentrations of NO 2 Cl and NO 2 .
(Answer)
9. (Total 9 pts) The rate constant of the first-order reaction 2N 2 O (g) → 2N 2 (g) + O 2 (g) is 0.76 s-1 at
1000 K and 0.87 s-1 at 1030 K.
(a) Calculate the activation energy (E a ) 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)
15
10. (Total 10 pts) The following data were collected for the reaction between hydrogen and nitric
oxide at 700 oC.
2H2 ( g) ++ 2NO(g)
2H2 O( g) ++ N2 (g)
Experiment [H 2 ] (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?
Mechanism B
Mechanism A
H2 + NO
N + NO
O + H2
H2O + N
(slow)
N2 + O
(f ast)
(f ast)
H2O
k1
k -1
2NO
N2O2 + H2
N2O + H2
k2
k3
N2O2
(f ast)
N2O + H2O (slow)
N2 + H2O (f ast)
Mechanism C
H2 + 2NO
N2O + H2O (slow)
N2O + H2
N2 + H2O
(f ast)
(Answer)
(c) For the mechanism B above, provide the rate constant with respect to k 1 , k -1 , k 2 , or k 3 .
(Answer)
16
17
18
19
20
Claim Form for General Chemistry Examination
Class:
, Professor Name:
, I.D.# :
Page (
/
)
, 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
Question #
Claims
Accepted?
Yes: □
Pts (+/-)
By TA
Yes(√) or No(√)
No: □
Reasons
21
<The Answers>
Problem
1
2
3
4
5
points
Problem
8+4/12
2+3/5
2+2+2+2+3/11
9+3/12
4+2+4+2/12
6
7
8
9
10
points
TOTAL pts
6+2/8
3+4+2/9
4+1+3+4/12
4+4/8
/100
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
1
(b) (4 pts) 2 pts for each
σ∗
σ∗
π∗
π∗
π
π
σ
σ
ground
electronic
state
first excited
electronic
state
2. (Total 5 pts)
(a) (2 pts) water: H 2 O
(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
2
4. (Total 12 pts)
(a) (9 pts) for each step, 3 pts
peroxide
is broken to give radicals
(b) (3 pts)
3
5. (Total 12 pts)
(a) (4 pts) for each isomer, 2 pts
(b) (2 pts) for each isomer, 1 pt
H
H
H
H
H
H
or
H
H
H
H
H
H H
H H
H
H
H H
H H
H
H
H
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 H
H H
H
H
H
H
H H
and
H
H
H
H
two axial methyls
& one equatorial methyl
H H
or
and
H
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.
H
O
+ H
H3C
H3C
HO
-
O
+
+ CH OH
3
H3C
OCH3
H
+
H
HO
H3C
OCH3
(b) (2 pts)
H3CO
OH
H3C
4
7. (Total 9 pts)
(a) (3 pts) for each functional group (except sulfide), 1 pt
Amide
Sulf ide
S
H
N
N
O
OH
O
Carboxylic acid
O
Amide
Penicillin G
(b) (4 pts) for each *, 1 pt
NH2
H
N
*
*
S
*
N
O
*
OH
O
O
Ampicillin
(c) (2 pts)
Penicillin : 2*2*2 = 8
Ampicillin : 2*2*2*2 = 16
5
8. (Total 12 pts)
(a) (4 pts) for each structure, 2 pts
Cl
H3N
Cl
Pt
NH3
NH3
Cl
H3N
Pt
Cl
(b) (1 pts)
Left one.
(c) (3 pts)
Diamagnetic, with the bottom four levels in the square-planar configuration (all but d x2-y2 ) occupied
(d) (4 pts) for each complex, 2 pts
Red transmission corresponds to absorption of green light by K 2 [PtCl 4 ]. A colorless solution has
absorptions at either higher or lower frequency than visible. Because Cl- is a weaker field ligand
than NH 3 , the absorption frequency should be higher for the Pt(NH 3 ) 4 2+ complex, putting it in the
ultraviolet region of the spectrum.
9. (Total 8 pts)
# of unpaired electrons: [Fe(CN) 6 ]3- one, [Fe(OH 2 ) 6 ]3+ five.
For each, 2 pts
CFSEs: [Fe(CN) 6 ]3- -2 ∆ o , [Fe(OH 2 ) 6 ]3+ 0.
For each, 2 pts
6
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 = N A hc/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.
7
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
Problem
points
1
2
3
4
5
Student I.D. Number
Problem
/12
/5
/11
/12
/12
Name
TOTAL pts
points
6
7
8
9
10
/8
/9
/12
/100
/8
/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)
3)
Location: Creative Learning Bldg.(E11)
Class
Room
Class
Room
Class
Room
Class
Room
A
202
B
205
C
210
D
211
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.
8
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)
9
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) C 2 H 4
(b)
(Answer)
(Answer)
(c) CH 3 CH 2 COCH 2 CH 3
(d) CH 3 CHCl 2
(e)
(Answer)
(Answer)
(Answer)
10
4. (Total 12 pts)
(a) Consider the polymerization of vinyl chloride (CH 2 =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)
11
5. (Total 12 pts) Cyclohexane is a cycloalkane with the molecular formula C 6 H 12 . 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)
12
6. (Total 8 pts) Epoxides can react with water or alcohols in the presence of acids.
O
+
+ CH OH
3
H
H3C
HO
OCH3
H3C
(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)
13
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)
Sulfide
S
H
N
N
O
OH
O
O
Penicillin G
(b) Replacing one of the hydrogen atoms of –CH 2 - group to –NH 2 (amine) group results another
famous antibiotics, Ampicillin. Mark all of the chiral carbon atoms in Ampicillin with asterisk mark (*).
(Answer)
NH2
S
H
N
N
O
OH
O
O
Ampicillin
(c) How many stereoisomers exist in Penicillin G and Ampicillin?
(Answer)
14
8. (Total 12 pts) Cisplatin, a simple square planar complex of platinum(Pt), has a molecular formula
of PtCl 2 (NH 3 ) 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 K 2 [PtCl 4 ] is red, but [Pt(NH 3 ) 4 ]Cl 2 •H 2 O 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(OH 2 ) 6 ]3+. How many unpaired electrons are
present in each species? In each case, express the CFSE in terms of ∆ o .
(Answer)
15
10. (Total 11 pts)
(a) Iron(III) forms octahedral complexes. Sketch the structures of all the distinct isomers of
[Fe(en) 2 Cl 2 ]+. (en = ethylenediamine, the en ligand can be represented as N------N.)
(Answer)
(b) The complex [Ni(NH 3 ) 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)
16
17
18
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