Universitas Diponegoro
Jurusan Teknik Kimia – Fakultas Teknik
Kampus Undip Tembalang
Semarang
GENERAL CHEMISTRY ASSINGMENT
Tugas Kimia Umum Kelompok B (Jadwal Kuliah Rabu Pagi jam 10.30)
Albertus Adrian Sutanto [21030110130079]
albertusadrian.co.cc
albertusadrian.wordpress.com
LEANGLE © 2010
1. Thomson was able to determine the mass/charge ratio of the electron but not its
mass. How did Millikan’s experiment allow determination of the electron’s mass?
Robert Millikan measured the charge of the electron. He did so by observing the
movement of tiny droplets of the "highest grade clock oil" in an apparatus that
contained electrically charged plates and an x-ray source.
Figure 1 Millikan's oil-drop experiment for measuring an electron's charge
Here is a description of the basis of Millikan’s experiment:
X-rays knocked electrons from gas molecules in the air, and as an oil droplet fell
through a hole in the positive (upper) plate, the electrons stuck to the drop, giving it
a negative charge. With the electric field off, Millikan measured the mass of the
droplet from its rate of fall. By turning on the field and varying its strength, he could
make the drop fall more slowly, rise, or pause suspended. From these data, Millikan
calculated the total charge of the droplet.
After studying many droplets, Millikan calculated that the various charges of the
droplets were always some whole-number multiple of a minimum charge. He
reasoned that different oil droplets picked up different numbers of electrons, so this
minimum charge must be that of the electron itself. The value, which he calculated
over 95 years ago, is within 1 % of the modern value of the electron's charge, 1.602x 10-19 C. Using the electron's mass/charge ratio from work by Thomson and
others and this value for the electron's charge, the electron’s mass can be calculated :
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𝑀𝑎𝑠𝑠 𝑜𝑓 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛 =
𝑚𝑎𝑠𝑠
kg
𝑥 𝑐ℎ𝑎𝑟𝑔𝑒 = −5.686𝑥10−12
𝑐ℎ𝑎𝑟𝑔𝑒
C
−1.602𝑥10−19
= 9.109𝑥10−31
(Silberberg, page 46-47)
2. How can ionic compounds be neutral if they consist of positive and negative
ions?
Ionic compounds can be neutral because they have equal number of positive
and negative charge but not necessarily equal numbers of positive and negative
ions.
For the example is BaCl2. BaCl2 is composed of Ba2+ cation and Cl- anion. BaCl2 is
neutral although Ba has +2 ions and Cl has -1 ion, because Ba bonds two Cl atoms so
BaCl2 possess no net charge.
(Silberberg, page 58 & Istadi, page 12)
3. Rank the following photons in terms of increasing energy: (a). blue (?=453 nm),
(b) red (?=660 nm), and (c). yellow (?=595 nm).
The energy of the photons can be calculated using the following formula :
ℎ𝑐
𝐸=
𝜆
(Silberberg, page 263)
(a) Blue (𝜆 = 453𝑛𝑚)
6.626𝑥10−34 𝐽𝑠 3𝑥108 𝑚𝑠 −1
𝐸=
= 4.388𝑥10−19 𝐽
453𝑥10−9 𝑚
(b) Red (𝜆 = 660𝑛𝑚)
6.626𝑥10−34 𝐽𝑠 3𝑥108 𝑚𝑠 −1
= 3.012𝑥10−19 𝐽
660𝑥10−9 𝑚
(c) Yellow (𝜆 = 595𝑛𝑚)
𝐸=
6.626𝑥10−34 𝐽𝑠 3𝑥108 𝑚𝑠 −1
𝐸=
= 3.341𝑥10−19 𝐽
595𝑥10−9 𝑚
𝐸𝑏𝑙𝑢𝑒 > 𝐸𝑦𝑒𝑙𝑙𝑜𝑤 > 𝐸𝑟𝑒𝑑
4. Are the following quantum number combinations allowed? If not, show two ways
to correct them: (a). n=1, l=0, ml=0; (b). n=2, l=2, ml=+1; (c). n=7, l=1, ml=+2; (d).
n=3, l=1, ml=-2
(a)
n=1
l=0
ml=0
Allowed → 1𝑠1 𝑜𝑟 1𝑠 2
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(b)
n=2
l=2
ml=+1
(c)
n=7
l=1
ml=+2
(d)
n=3
l=1
ml=-2
NOT allowed :
n=2 → n=3 → 3𝑑 4 𝑜𝑟 3𝑑 9
or
l=2 → l=1 → 2𝑝3 𝑜𝑟 2𝑝6
NOT allowed :
ml =+2 → ml =0 → 7𝑝2 𝑜𝑟 7𝑝5
or
ml =+2 → ml =+1 → 7𝑝3 𝑜𝑟 7𝑝6
NOT allowed :
l=1 → l=2 → 3𝑑1 𝑜𝑟 3𝑑 6
or
ml =-2 → ml =-1 → 3𝑝1 𝑜𝑟 3𝑝4
(Istadi, page 39-41)
5. Write a full set of quantum numbers for the following: (a). outermost electron in
an Li atom; (b). The electron gained when a Br atom becomes a Br- ion; (c). The
electron lost when a Cs atom ionizes; (d). the highest energy electron in the
ground-state B atom.
(a)
3𝐿𝑖
1𝑠 2 2𝑠1
𝑛 = 2; 𝑙 = 0; 𝑚𝑙 = 0; 𝑠 = + 1 2
(b) 𝐵𝑟 −
18
𝐴𝑟 4𝑠 2 3𝑑10 4𝑝6
(c)
36
𝐾𝑟 5𝑠 2 4𝑑10 5𝑝6 6𝑠1
55 𝐶𝑠
𝑛 = 4; 𝑙 = 1; 𝑚𝑙 = +1; 𝑠 = − 1 2
𝑛 = 6; 𝑙 = 0; 𝑚𝑙 = 0; 𝑠 = + 1 2
1
→ 𝑛𝐵= 4;1𝑠𝑙 2=2𝑠
1;2 𝑚2𝑝
1 +1; 𝑠 = −
𝑙 =
(d)
𝑛 = 1; 𝑙2= 0; 𝑚𝑙 = 0; 𝑠 = − 1 2
5
→ 𝑛 = 4; 𝑙 = 1; 𝑚𝑙 = +1; 𝑠 = − 1 2
(Istadi, page 39-41)
6. Write the condensed ground-state electron configuration of these transition
metal ions, and state which are paramagnetic: (a). Mo3+; (b). Au+; (c). Mn2+; (d).
Hf2+
(a)
42𝑀𝑜 36
𝐾𝑟 5𝑠 1 4𝑑5 → 𝑀𝑜3+ 36 𝐾𝑟 4𝑑3
Paramagnetic
(b)
79𝐴𝑢 54
𝑋𝑒 6𝑠 1 4𝑓 14 5𝑑10 → 𝐴𝑢+ 54 𝑋𝑒 4𝑓 14 5𝑑10
Diamagnetic
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(c)
25𝑀𝑛 18
𝐴𝑟 4𝑠 2 3𝑑 5 → 𝑀𝑛2+ 18 𝐴𝑟 3𝑑 5
Paramagnetic
(d)
𝐻𝑓 54 𝑋𝑒 6𝑠 2 4𝑓 14 5𝑑 2 → 𝐻𝑓 2+ 54 𝑋𝑒 4𝑓 14 5𝑑2
72
Paramagnetic
(Istadi, page 73)
7. There are some exceptions to the trends of first and successive ionization
energies. For each of the following pairs, explain which ionization energy would
be higher: (a). IE1 of Ga or IE1 of Ge; (b). IE2 of Ga or IE2 of Ge; (c). IE3 of Ga or
IE3 or Ge; (d). IE4 of Ga or IE4 of Ge.
(a) 𝐺𝑎 18 𝐴𝑟 4𝑠 2 3𝑑10 4𝑝1
𝐺𝑒 18 𝐴𝑟 4𝑠 2 3𝑑10 4𝑝2
In each case, an electron is removed from a 4p orbital. Because the Zeff of
Ge is greater, IE1 of Ge will be higher.
(b) 𝐺𝑎+ 18 𝐴𝑟 4𝑠 2 3𝑑10
𝐺𝑒 + 18 𝐴𝑟 4𝑠 2 3𝑑10 4𝑝1
IE2 would be higher for Ga+, because an electron must be removed from a
full 4s orbital. For Ge+ the electron is removed from the 4p orbital.
(c) 𝐺𝑎2+ 18 𝐴𝑟 4𝑠 1 3𝑑10
𝐺𝑒 2+ 18 𝐴𝑟 4𝑠 2 3𝑑10
In each case, an electron is removed from a 4s orbital. Because the Zeff of
Ge2+ is greater, IE3 of Ge2+ will be higher.
(d) 𝐺𝑎3+ 18 𝐴𝑟 3𝑑10
𝐺𝑒 3+ 18 𝐴𝑟 4𝑠1 3𝑑10
IE4 would be higher for Ga3+, because an electron must be removed from a
full 3d orbital. For Ge3+ the electron is removed from the 4s orbital.
(Silberberg, page 311)
8. For single bonds between similar types of atoms, how does the strength of the
bond relate to the sizes of the atoms? Explain scientifically.
The strength of a covalent bond depends on the magnitude of the mutual
attraction between bonded nuclei and shared electrons. So, the sizes of the atoms
influence the strength of the bond. If the sizes of atoms become greater, the strength
of the bond becomes weaker because the distance between atoms become longer so
the interaction between bonded nuclei and shared electron becomes weaker.
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For the example, see the following picture :
Figure 2 Bond Length and Covalent Radius
In the picture above, we can see that the radius (or the size) of the atoms
influence the distance between two bonded atoms. The longer size of the atoms
make the attraction force between electrons and nuclei weaker. From the picture
above we can conclude that the strength of the bond of F2>Cl2>Br2.
(Silberberg, page 341-342)

REFERENCES
Istadi, Aprilina Purbasari. 2008. Buku Ajar Kimia Umum. Semarang : Jurusan Teknik
Kimia Fakultas Teknik Universitas Diponegoro.
Silberberg, Martin S. 2006. Chemistry : The Molecular Nature of Matter and Change
Fourth Edition. New York : The MacGraw-Hill Companies.
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always seeking the reasons why the work is not finished. And people who
find success are always looking for reasons why the work can be completed.”
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