KEY
Last Name ____________________________ First Name _______________
Lab Sec. # ____; TA: _____________________; Lab day/time: ___________
thelifecurve.com
Dr. Toupadakis
Circle one
Fall 2010
CHEMISTRY 2C
Section A
FINAL EXAM
Instructions:
CLOSED BOOK EXAM! No books, notes, or additional scrap paper
are permitted. All information required is contained on the exam.
Place all work in the space provided. If you require additional space,
use the back of the exam. A scientific calculator may be used (if it
is a programmable calculator, its memory must be cleared before the
exam). This exam has 19 pages total.
(1)
Read each question carefully.
01. a
b
c
d
e
02. a
b
c
d
e
03. a
b
c
d
e
04. a
b
c
d
e
05. a
b
c
d
e
06. a
b
c
d
e
07. a
b
c
d
e
08. a
b
c
d
e
09. a
b
c
d
e
10. a
b
c
d
e
11. a
b
c
d
e
(2)
For Part I, II and III there is no partial credit. There is
12. a
b
c
d
e
13. a
b
c
d
e
(3)
partial credit for part IV.
The last three pages contain a periodic table and some useful
information. You may remove them for easy access.
14. a
b
c
d
e
(4)
If you finish early, RECHECK YOUR ANSWERS!
15. a
b
c
d
e
16. a
b
c
d
e
17. a
b
c
d
e
18. a
b
c
d
e
19. a
b
c
d
e
20. a
b
c
d
e
21. a
b
c
d
e
22. a
b
c
d
e
23. a
b
c
d
e
24. a
b
c
d
e
25. a
b
c
d
e
26. a
b
c
d
e
27. a
b
c
d
e
28. a
b
c
d
e
29. a
b
c
d
e
U.C. Davis is an Honor Institution
Possible Points
#
#
#
#
#
#
#
#
#
#
#
01–25
26-29
30
31
32
33
34
35
36
37
38
Total
(2 points each)
(6 points each)
(10 points total)
(06 points total)
(10 points total)
(10 points total)
(08 points total)
(10 points total)
(10 points total)
(12 points total)
(14 points total)
Score (164)
Earned Points
/ 50
/ 24
/ 10
/ 06
/ 10
/ 10
/ 08
/ 10
/ 10
/ 12
/ 14
/164 or
%
Toupadakis
F2010
Final Exam
page
Part I: Concepts (2 points each)
No partial credit is available
1.
The reaction: Cl2 + CH2 =CH2  ClCH2 CH2 Cl
is an example of:
(a)
a polymerization reaction
(b)
a substitution reaction
(c)
an elimination reaction
(d)
an isomerization reaction
(e)
an addition reaction
2.
A nucleus is held together by
(a)
an electric force
(b)
a gravitational force
(c)
a nuclear force
(d)
a magnetic force
(e)
none of the above
3.
P2O5 reacts with water. The product is:
(a) P4
(b) H5P3O10
(c) H3PO2
(d) H3PO3
(e) H3PO4
4.
If ∆ stands for the crystal field splitting and P for the electron pairing energy, how
many unpaired electrons does a d5 octahedral complex have if ∆ > P?
(a) 5
(b) 1
(c) 2
(d) 3
(e) 4
5.
The hybridization of the carbon atom marked with the asterisk (*) and the total # of
and
(a)
(b)
(c)
(d)
(e)
6.
bonds for the structure below will be:
sp2
5 +3
sp
11 + 3
sp2
11 + 3
2
sp
2 +3
sp
10 + 3
O
C
*
What is the coordination number and oxidation number of the metal ion in Pd(en)Cl2?
(a) CN = 3, ON = +2
(b) CN = 3, ON = +4
(c) CN = 3, ON = +2
(d) CN = 4, ON = 0
(e) CN = 4, ON = +2
Your Name
2
Toupadakis
F2010
Final Exam
page
3
7.
What is the correct IUPAC name for [Co(NH3)5Br]SO4 ?
(a)
Bromopentaamminecobalt(III) sulfate
(b)
Pentaamminebromocobalt(II) sulfate
(c)
Pentaamminebromocobalt(III) sulfide
(d)
Pentaamminebromocobalt(III) sulfate
(e)
Pentaamminebromocobaltate(III) sulfate
8.
As the atomic number of an element INCREASES, the ratio of neutrons to protons in
stable nuclei
(a)
does not relate to stability.
(b)
increases.
(c)
stays the same.
(d)
decreases.
(e)
none of the above.
9.
Choose the CORRECT statement:
(a) Mass defect of a nucleus is the gain in mass that occurs when isolated protons
and neutrons combine to form the nucleus.
(b) Binding energy of a nucleus is the mass defect converted into energy that is
gained during the nuclear reaction.
(c) Binding energy of a nucleus is the mass of the nucleus converted to energy.
(d) Mass defect of a nucleus is the loss in mass that occurs when isolated protons
and neutrons combine to form the nucleus.
(e) None of the above
10.
Identify the species X in the following nuclear equation:
2 11H
2X
(a) proton
(b) neutron
(c) “ ” particle
(d) electron
(e) none of the above
11.
What is the IUPAC name for the compound?
(a)
(b)
(c)
(d)
(e)
Your Name
4,5-dimethylhexane
2-isopropylpentane
2-methyl-3-propylbutane
2,3,4,5,6-pentamethylheptane
Octane
4
2 He
Toupadakis
F2010
Final Exam
page
12.
Choose the INCORRECT statement:
(a)
Both, lighter and heavier elements are less stable than midweight elements
near iron-56.
(b)
In nuclear fission heavy elements gain stability and release energy by
fragmenting to yield midweight elements.
(c)
In nuclear fusion light elements gain stability and release energy by fusing
together.
(d)
In a supercritical fission process, on average, less than one neutron causes
another fission event.
(e)
None of the above
13.
Hypophosphorous acid H3PO2 is a
(a)
monoprotic acid
(b)
diprotic acid
(c)
triprotic acid
(d)
strong acid
(e)
None of the above
14.
Assume that the mechanism of a reaction has 2 steps.
The CORRECT statement is:
(a)
There are 2 transition states.
(b)
There are 2 intermediates.
(c)
There are 3 activation energies.
(d)
The overall order of the reaction is 2.
(e)
The reaction is bimolecular.
15.
In the following electrochemical cell what is the oxidizing agent and how many
electrons are transferred?
Zn(s) | Zn2+(aq) || Fe3+(aq), Fe2+(aq) || Pt(s)
(a)
(b)
(c)
(d)
(e)
16.
Zn2+,
Zn,
Fe3+,
Fe2+,
Pt,
2e
2e
2e
2e
0e
Consider the chemical equation: H2O + H2O  H3O+ + OHThe number of moles of electrons transferred in the reaction represented by this
chemical equation is:
Your Name
4
Toupadakis
(a) 0
F2010
(b) 1
(c) 2
(d) 3
Final Exam
page
5
(e) 4
17. Write the formula of tetraaquadichlorochromium(III) chloride.
(a)
Cl[Cr(H2O)4Cl2]
(b)
[Cr(H2O)4Cl]Cl
(c)
[Cr(H2O)4Cl]Cl2
(d)
[Cr(H2O)4Cl2]Cl
(e)
[(H2O)4Cl2Cr]Cl
18.
Choose the CORRECT statement:
(a)
the rate constant increases
energy increases.
(b)
the rate constant increases
energy decreases.
(c)
the rate constant increases
energy decreases.
(d)
the rate constant increases
energy increases.
(e)
None of the above.
as the temperature increases and the activation
as the temperature decreases and the activation
as the temperature increases and the activation
as the temperature decreases and the activation
19.
Determine the number of different isomers for the complex, M(en)F2I2. When trying
to determine the number of isomers, keep the (en) group in the same position in all of
your drawings. M stands for the metal ion and en for ethylenediamine.
(a) 4 (b) 3 (c) 2 (d) 1 (e) 5
20.
From the following elements, pick the transition metal.
(a) Se
(b) Re
(c) Rn
(d) Ra
(e) Al
21.
The coordination compounds [Co(NO2)(NH3)5]SO4 and
[Co(ONO)(NH3)5]SO4 are:
(a)
Ionization isomers
(b)
Linkage isomers
(c)
Geometric isomers
(d)
Coordination isomers
(e)
The same chemical compound
22.
Which of the following is FALSE?
(a)
A fuel cell is a closed system.
(b)
A sacrificial anode can be used to prevent corrosion.
(c)
A primary cell battery cannot be recharged.
(d)
As a galvanic cell operates its potential for electrical work decreases.
(e)
None of the above.
Your Name
Toupadakis
F2010
Final Exam
page
23.
Choose the CORRECT statement:
(a)
A zero-order reaction is a reaction with a rate constant equal to zero.
(b)
The higher the value of an exponent in a rate law, the less sensitive is the
reaction rate to a concentration change of that component.
(c)
The rate constant does not depend on the presence of a catalyst.
(d)
Intermediates sometimes can be isolated.
(e)
None of the above
24.
213
Bi decays with emission of an -particle. The resulting nuclide is unstable and emits
a -particle. What is the final nuclide formed?
(a) 209Pb
25.
6
(b) 8He
(c)3Be
(d) 4Li
(e) 12C
In the coordination compound [M(CN)2(en)2]Br2, the charge of the metal and its
coordination number, respectively are:
(a) +1, 3
(b) +2, 3
(c) +2, 6
(d) +2, 4
(e) +4, 6
Your Name
Toupadakis
F2010
Final Exam
page
7
Part II: Short Calculations (6 points each)
No partial credit is available
26.
A current of 2.00 A passing for 5.00 h through a molten metal chloride MClx deposits
22.2 g of the metal. If the molar mass of the metal is 118.69 g/mol, then the value of
x is:
(a) 1
(b)
(5.00 h)
2
3600 s
2.00 C
h
s
(22.2 g M)
1 mol M
118 .69 g M
0.373 mol e
0.187 mol M
27.
(c) 3
=
(d) 4
1 mol e
96500 C
(e) 5
= 0.373 mol e-
= 0.187 mol M
2 mol e
1 mol M
=
2e
1 metal atom
 x=2
The activation energy of a reaction is 53 kJ/mol. Its rate doubles when the
temperature increases by 10oC? What was the temperature before it was
increased?
(a)
(b)
(c)
(d)
(e)
ln
k1
k2
ln
1
2
Your Name
280 K
298 K
33 K
290 K
640 K
E 1
R T2
1
T1
53000 J/mol
1
8.314 J / mol K (x 10) K
1
x K
 x = 298
Toupadakis
28.
F2010
Final Exam
page
8
Calculate the maximum kinetic energy of the beta particle emitted in the radioactive
decay of 6He. Assume that the beta particle has maximum energy when no other
emission is involved. Use the conversion, 1 amu = 931.5 MeV.
Isotopic mass of 6He = 6.01889 amu, Isotopic mass of 6Li = 6.01512 amu.
(a)
1.67 MeV
(b)
5.85 MeV
(c)
7.49 MeV
(d)
3.51 MeV
(e)
9.51 MeV
Loss of mass = (mass of 6He) - (mass of 6Li)
= (6.01889 amu) - (6.01512 amu)
= 0.00377 amu
Energy equivalent = (931.5 MeV/amu)( 0.00377 amu)
= 3.51 MeV
29.
The general formula of the series of compounds whose molecule contains two double
bonds, one ring and an ether group is:
(a)
(b)
(c)
(d)
(e)
CnH2nO
CnH2n+1O
CnH2n+2O
CnH2n-4O
CnH2n-6O
Saturated hydrocarbon, CnH2n+2
Two double bonds, CnH2n+2-4 or CnH2n-2
One ring, CnH2n-2-2 or CnH2n-4
One ether group, CnH2n-4O
Therefore: CnH2n-4O
Your Name
Toupadakis
F2010
Final Exam
page
Part III. Short Answer
Fill in the blanks with the appropriate answers
30.
[10 points possible--each is worth 2 pts]
Identify the following molecules as chiral or achiral. Also, designate the molecule as
R, S, fac, mer, cis, trans, or none of the above.
Molecule
Chiral (optically active) or
Achiral (optically inactive)?
Designation
Chiral
Cis
Achiral
NONE
Achiral
Fac
Chiral
S
Achiral
NONE
H2
C
H2
N
H2C
CH2
H2N
NH 2
Co
H2C
N
H2
Br
Cl
B
A
B
M
A
B
A
OH
H
HOOC
C
CH3
CH3
H3C
H3C
Your Name
C
CH3
9
Toupadakis
31.
F2010
Final Exam
[6 points possible--each is worth 2 pts] Draw the following:
trans-1,2-dibromocyclobutane
Br
H
H
Br
-
CO
mer-tribromotricarbonylferrate(II) ion
Br
CO
Fe
Br
Br
CO
3-chloro-2-hexene
Cl
32.
[10 points possible--each is worth 2 pts]
Identify the correct functional group with the molecule listed.
Molecule
Functional Group
Ether
O
NH2
Amine
NH 2
CHO
H
C
Aldehyde
H
CH 3
O
Ketone
NH
O
Your Name
Amide
page 10
Toupadakis
F2010
Final Exam
page
11
33. (10 pts--each is worth 1 pt) Fill in the following table.
Coordination
number
Metal d
electrons
6
[Cr(H2O)4Cl2]+
4
2+
[Cu(NH3)4]
d
3
d
9
Metal
oxidation
number
Can it be
chiral?
Possible
geometries
+3
No
octahedral
no
Sq. Planer
or
Tetrahedral
+2
34. (8 pts total--each is worth 2 pts) Consider the following molecule:
A
B
C
HC C
O
a. (2 Pts) What is the hybridization of atom A?
sp
b. (2 Pts) What is the hybridization of atom B?
sp3
c. (2 Pts) What is the hybridization of atom C?
sp2
d. (4 Pts) What type of atomic orbitals overlap to form the C=O double bond?
A. -bond. C ( ? ) and O ( ? )
B. -bond C ( ? ) and O ( ? )
A. -bond. C ( sp2 ) and O ( p )
B. -bond C ( p ) and O ( p )
Your Name
Toupadakis
35.
F2010
Final Exam
page 12
[10 points—each formula is worth 1 pt] Fill in the blanks with the appropriate
chemical formula(s).
Sodium oxide + H2O

Sodium peroxide + H2O

Potasium superoxide + H2O 
P4O10
+
H 2O

Li
+
H2O

Na
+
Cl2

Your Name
NaOH
NaOH + H2O2
KOH + H2O2 + O2
H3PO4
LiOH + H2
NaCl
Toupadakis
F2010
Final Exam
page 13
Part IV. Long Answer Calculations
Show your work for partial credit. Put answers in the spaces provided.
36. [10 Points]
Consider a reaction that is represented by the following general chemical equation:
A(g) + B(g) → 2 C(g) + 3 B(g)
©AT2010F
From the following data:
a)
b)
c)
d)
Determine the overall reaction order (explain or show your work).
Determine the value and units of the rate constant.
Determine the rate of the reaction for [A] = 3.0x10-2 M and [B] = 2.0x10-3 Μ.
Could this reaction be a one-step reaction? Explain.
Experiment
Initial [A], M
Initial [B], M
Initial Rate (M/s)
1
0.10
0.10
0.0050
2
0.40
0.10
0.080
3
0.10
0.20
0.0050
a)
Overall reaction order is:
Experiment 1 and 2  R
[A]2
Experiment 1 and 3  R
0
[B]
R
[A]2
R = k [A]2
Therefore: Overall reaction order = 2
b)
Rate constant k is:
From experiment 1: R = k [A]2  0.0050 = k(0.10)2 
c)
Rate of reaction is:
R = k [A]2  R = (0.50)( 3.0x10-2)2
d)
k = 0.50 M-1s-1

Rate = 4.5x10-4 M/s
No. If this reaction was a one-step reaction then its rate law expression would
be: R = [A][B]. Because it is different from the experimentally determined
rate law, the reaction is a multistep reaction.
Your Name
Toupadakis
F2010
Final Exam
37. (12 pts)
Consider the reaction described by the following chemical equation:
page 14
©AT2010F
2NO + Br2  2NOBr
During the reaction an intermediate is detected, and the rate law determined by
experiment is:
Rate = k [NO]2[Br2]
1.
2.
Could this reaction be a one-step reaction? Explain why.
For this reaction the following mechanism has been suggested:
Step 1:
NO + Br2
k1
NOBr2
Fast equilibrium
k-1
Step 2:
k2
NOBr2 + NO
2NOBr
Slow step
a.
Demonstrate by showing your work that this mechanism is acceptable
because it satisfies the two requirements.
b.
Besides the fact that this mechanism satisfies the two requirements,
what other additional support is there according to the data given in
this problem? Explain briefly.
…………………………………………………………………………………………………………………………………………………….
1.
Even though the rate law determined by experiment agrees with a one-step
reaction mechanism, the detection of an intermediate drops this possibility
because by definition a one-step reaction mechanism does not involve any
intermediates.
2a.
First requirement
The sum of the steps must give the balanced equation of the reaction.
Step 1:
NO + Br2
k1
NOBr2
k-1
Step 2:
NOBr2 + NO
NO + Br2 + NOBr2 + NO
2NOBr
NOBr2 + 2NOBr
Overall chemical equation: 2NO + Br2  2NOBr
Your Name
Toupadakis
F2010
Final Exam
page 15
Second requirement
The mechanism has to agree with the experimentally determined rate
law.
Overall rate = rate of slow step = rate of 2nd step = k2 [NO][NOBr2]
First step is fast equilibrium  Rate (forward) = Rate (reverse)
k1 [NO][Br2] = k-1[NOBr2]  [NOBr2] = (k1 /k-1)[NO][Br2]
Overall rate = k2[NO][NOBr2] = k2[NO](k1 /k-1)[NO][Br2]
= (k1 /k-1)k2[NO]2 [Br2] = k[NO]2[Br2]
Where k = (k1 /k-1)k2
2b.
The detection of an intermediate.
38. (14 points) Consider the reaction represented by the following chemical equation:
2Cr(s) + 3H2O(l) + 3OCl–(aq)  2Cr3+(aq) + 3Cl–(aq) + 6OH–(aq)
A.
Make a drawing of a cell that uses this reaction and indicate on the diagram
the following:
(a)
(b)
(c)
B.
Your Name
(a)
(b)
(c)
(d)
The anode and the cathode.
The direction in which the anions and cations migrate through the salt
bridge.
The direction in which the electrons migrate through the external
circuit.
Write the anode half reaction.
Write the cathode half reaction.
Calculate Eo for the cell.
Write the cell notation.
Toupadakis
F2010
Final Exam
page 16
A.
B.
(a)
Anode half reaction
Cr (s)  Cr3+(aq) + 3 e- Eo = +0.74 V
(b)
Cathode half reaction
OCl-(aq) + H2O(l) + 2e-  Cl-(aq) + 2OH-(aq)
(c)
Eo = +0.94V
Eo for the cell
2Cr (s)

2Cr3+(aq) + 6 e-
Eo = +0.74 V
3OCl-(aq) + 3H2O(l) + 6e-

3Cl-(aq) + 6OH-(aq)
Eo = +0.94V
2Cr(s) + 3H2O(l) + 3OCl–(aq)  2Cr3+(aq) + 3Cl–(aq) + 6OH–(aq)
Eocell = (+0.74 V) + (+0.94V) = + 1.68 V
(d)
Cell notation
Cr(s)|Cr3+(aq)|| OH–(aq), OCl–(aq), Cl–(aq), |Pt(s)
Your Name
Toupadakis
F2010
Final Exam
page 17
Potentially Useful Information: (You may remove this page for ease of access)
Constants:
R = 8.3145 J / mol K
NA = 6.022 x 1023 1 atm = 760 torr
c = 3.00 x 108 m/s
R = 0.0821 L atm / mol K
h = 6.626 x 10-34 J s 1 nm = 10-9 m 1 L = 1000 cm3
F = 96,485 C/ mol e1 amu = 1.6605x10-27 kg
1 g = 6.02 x 1023 amu
1 MeV = 1.6 x 10-13 J
Equations:
Slope = Δy/Δx
G
H
PV
nRT
rate k[ A]0
rate
rate
T S
ln
G
k1
k2
[ A]t
ln[ A]t
1
[ A]t
k[ A] 2
o
Ecell
1 A
1C
1s
E 1
R T2
k
k[ A]
Ecell
ToF – 32 = 1.8 ToC
TK = ToC + 273.15
G
1
T1
1
[ A]0
2
t1
2
t1
2
0.693
k
1
k[ A]0
Ecell = Eox + Ered
0.0592
log Keq at 25oC.
n
o
nFE cell
G
1 mol e
where N = # of remaining atoms
620
r
96,485 C
400
430
v
o
b
y
G
ln Nt = ln No – t
800
g
580
560
Your Name
Ae
ΔE = Δm c2
ch arg e (C) current ( A) x time (s)
N
k
[ A]0
2k
t1
kt ln[ A]0
kt
RT ln K eq
c
h
kt [ A]0
0.0592
o
log Q and Ecell
n
Rate of decay =
G
RT ln Q
Ea
RT
490
nFE cell
Toupadakis
F2010
Final Exam
page 18
Potentially Useful Information: (You may remove this page for ease of access)
Cr2O7
Electrode Half Reaction
(aq) + 14H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l)
2-
O2(g) + 4H+(aq) + 4e-  2 H2O(l)
Pt2+(aq) + 2 e-  Pt (s)
-
E˚/volts
+1.33
+1.23
+1.20
-
Br2(l) + 2e  2Br (aq)
+1.09
OCl (aq) + H2O(l) + 2e  Cl-(aq) + 2OH (aq)
+0.94
Ag+(aq) + e–  Ag(s)
+0.80
I2 (s) + 2e-  2 I- (aq)
+0.54
-
-
2+
-
–
Cu (aq) + 2e  Cu(s)
Fe3+(aq) + 1 e-  Fe2+ (aq)
+0.34
Sn4+(aq) + 2e-  Sn2+(aq)
2H+(aq) + 2 e-  H2 (g)
+0.15
Sn2+(aq) + 2 e-  Sn (s)
–0.14
3+
-
+0.77
0.00
2+
V (aq) + e  V (aq)
3+
-
-0.26
2+
Cr (aq) + e  Cr (aq)
Fe2+(aq) + 2 e-  Fe (s)
-0.424
Cr3+(aq) + 3 e-  Cr (s)
Zn2+(aq) + 2 e-  Zn (s)
–0.74
Cr
2+
–0.44
–0.76
-
(aq) + 2 e  Cr (s)
-.90
Mn2+(aq) + 2e-  Mn(s)
-1.18
V2+ (aq) + 2e-  V (s)
-1.19
3+
-
Al (aq) + 3e  Al (s)
-1.66
Mg2+ (aq) + 2e-  Mg (s)
-2.36
Li+ (aq) + 1e-  Li (s)
-3.040
Spectrochemical Series
CN- > CO > NO2- > en > NH3 > SCN- > H2O > ONO- > ox2- > OH- > F- > SCN- > Cl- > Br- > I-
Visible Region of the Electromagnetic Spectrum
Violet
400 nm
Your Name
Blue
475
Green
510
Yellow
570
Orange
590
Red
650 nm
Toupadakis
F2010
Final Exam
page 19
Potentially Useful Information: (You may remove this page for ease of access)
Periodic Table
Key
1
Atomic Number
Symbol
Atomic Mass
Electronegativity
H
2
He
1.008
2.20
3
4
5
6
7
8
9
Li
Be
B
C
N
O
F
Ne
6.941
0.98
9.012
1.57
10.81
2.04
12.01
2.55
14.01
3.04
16.00
3.44
19.00
3.98
20.18
-
11
Na
Mg
Al
Si
14
15
16
S
Cl
Ar
22.99
0.93
24.31
1.31
26.98
1.61
28.09
1.90
30.97
2.19
32.06
2.58
35.45
3.16
39.95
-
19
12
21
Ti
22
23
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.10
0.82
40.08
1.00
44.96
1.36
47.90
1.54
50.94
1.63
52.00
1.66
54.94
1.55
55.85
1.83
58.93
1.88
58.70
1.91
63.55
1.90
65.38
1.65
69.72
1.81
72.59
2.01
74.92
2.18
78.96
2.55
79.90
2.96
83.80
-
36
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
52
53
I
Xe
85.47
0.82
87.62
0.95
88.91
1.22
91.22
1.33
92.91
1.6
95.94
2.16
(98)
1.9
101.1
2.2
102.9
2.28
106.4
2.20
107.9
1.93
112.4
1.69
114.8
1.78
118.7
1.96
121.8
2.05
127.6
2.1
126.9
2.66
131.3
-
Ba
Lu
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
82
83
Bi
Po
At
Rn
132.9
0.79
137.3
0.89
175.0
1.27
178.5
1.3
180.9
1.5
183.9
2.36
186.2
1.9
190.2
2.2
192.2
2.20
195.1
2.28
197.0
2.54
200.6
2.00
204.4
2.04
207.2
2.33
209.0
2.02
(209)
2.0
(210)
2.2
(222)
-
-
87
88
103
104
105
106
107
Ra
Lr
Unq
Unp
Unh
Uns
(223)
0.7
(226)
0.9
(260)
-
-
-
-
-
57
59
60
61
80
81
84
85
109
Une
-
62
63
64
65
66
67
68
69
70
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
138.9
1.10
140.1
1.12
140.9
1.13
144.2
1.14
(145)
1.13
150.4
1.17
152.0
1.2
157.3
1.20
158.9
1.2
162.5
1.22
164.9
1.23
167.3
1.24
168.9
1.25
173.0
1.1
89
Your Name
58
79
54
Cs
Fr
78
51
35
Tc
77
50
34
Mo
76
49
33
Nb
75
48
32
Zr
74
47
31
Y
73
46
30
39
72
45
29
38
71
44
28
Sr
56
43
27
Rb
55
42
26
18
Sc
41
25
17
Ca
40
24
P
K
37
20
13
4.003
10
Ac
Th
90
Pa
91
92
U
Np
93
Pu
94
Am
95
Cm
96
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
(227)
1.1
232.0
1.3
(231)
1.5
238.0
1.38
(237)
1.36
(244)
1.28
(243)
1.3
(247)
1.3
(247)
1.3
(251)
1.3
(252)
1.3
(257)
1.3
(258)
1.3
(259)
1.3
86