Universitas Diponegoro

Jurusan Teknik Kimia – Fakultas Teknik

Kampus Undip Tembalang

Semarang

Albertus Adrian Sutanto [21030110130079]

albertusadrian.co.cc

albertusadrian.wordpress.com

GENERAL CHEMISTRY ASSINGMENT

Tugas Kimia Umum Kelompok B (Jadwal Kuliah Rabu Pagi jam 10.30)

LEANGLE © 2010

General Chemistry Assignment | albertusadrian.co.cc| LEANGLE ©2010 1

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 :

General Chemistry Assignment | albertusadrian.co.cc| LEANGLE ©2010 2

𝑀𝑎𝑠𝑠 𝑜𝑓 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = 𝑚𝑎𝑠𝑠

𝑐ℎ𝑎𝑟𝑔𝑒 𝑥 𝑐ℎ𝑎𝑟𝑔𝑒 = −5.686𝑥10−12

kg

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 :

𝐸 = ℎ𝑐

𝜆

(Silberberg, page 263) (a) Blue (𝜆 = 453𝑛𝑚)

𝐸 = 6.626𝑥10−34 𝐽𝑠 3𝑥108𝑚𝑠−1

453𝑥10−9𝑚 = 4.388𝑥10−19 𝐽

(b) Red (𝜆 = 660𝑛𝑚)

𝐸 = 6.626𝑥10−34 𝐽𝑠 3𝑥108𝑚𝑠−1

660𝑥10−9𝑚 = 3.012𝑥10−19 𝐽

(c) Yellow (𝜆 = 595𝑛𝑚)

𝐸 = 6.626𝑥10−34 𝐽𝑠 3𝑥108𝑚𝑠−1

595𝑥10−9𝑚 = 3.341𝑥10−19 𝐽

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

n=1 l=0 ml=0

Allowed → 1𝑠1 𝑜𝑟 1𝑠2 (a)

General Chemistry Assignment | albertusadrian.co.cc| LEANGLE ©2010 3

n=2 l=2 ml=+1 n=7 l=1 ml=+2 n=3 l=1 ml=-2

(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.

(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+

NOT allowed : (b) n=2 → n=3 → 3𝑑4 𝑜𝑟 3𝑑9

or l=2 → l=1 → 2𝑝3 𝑜𝑟 2𝑝6

NOT allowed : (c) ml =+2 → ml =0 → 7𝑝2 𝑜𝑟 7𝑝5

or ml =+2 → ml =+1 → 7𝑝3 𝑜𝑟 7𝑝6

NOT allowed : (d) l=1 → l=2 → 3𝑑1 𝑜𝑟 3𝑑6

or ml =-2 → ml =-1 → 3𝑝1 𝑜𝑟 3𝑝4

(a) 𝐿𝑖3 1𝑠2 2𝑠1 𝑛 = 2; 𝑙 = 0; 𝑚𝑙 = 0; 𝑠 = + 1

2

(c) 𝐶𝑠55 𝐾𝑟 36

5𝑠2 4𝑑10 5𝑝6 6𝑠1 𝑛 = 6; 𝑙 = 0; 𝑚𝑙 = 0; 𝑠 = + 12

→ 𝑛 = 4; 𝑙 = 1; 𝑚𝑙 = +1; 𝑠 = − 12

(b) 𝐵𝑟− 𝐴𝑟 18 4𝑠2 3𝑑10 4𝑝6 𝑛 = 4; 𝑙 = 1; 𝑚𝑙 = +1; 𝑠 = − 1

2

(d) 𝐵5 1𝑠2 2𝑠2 2𝑝1 𝑛 = 1; 𝑙 = 0; 𝑚𝑙 = 0; 𝑠 = − 1

2

→ 𝑛 = 4; 𝑙 = 1; 𝑚𝑙 = +1; 𝑠 = − 12

(a) 𝑀𝑜42 𝐾𝑟 36

5𝑠1 4𝑑5 → 𝑀𝑜3+ 𝐾𝑟 36 4𝑑3

Paramagnetic (b) 𝐴𝑢79

𝑋𝑒 54 6𝑠1 4𝑓14 5𝑑10 → 𝐴𝑢+ 𝑋𝑒 54

4𝑓14 5𝑑10

Diamagnetic

General Chemistry Assignment | albertusadrian.co.cc| LEANGLE ©2010 4

(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.

(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.

(c) 𝑀𝑛25 𝐴𝑟 18

4𝑠2 3𝑑5 → 𝑀𝑛2+ 𝐴𝑟 18 3𝑑5

Paramagnetic

(d) 𝐻𝑓

72

𝑋𝑒 54 6𝑠2 4𝑓14 5𝑑2 → 𝐻𝑓2+ 𝑋𝑒 54

4𝑓14 5𝑑2

Paramagnetic

(a) 𝐺𝑎 𝐴𝑟 18 4𝑠2 3𝑑10 4𝑝1 𝐺𝑒 𝐴𝑟 18

4𝑠2 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+ 𝐴𝑟 18

4𝑠2 3𝑑10 𝐺𝑒+ 𝐴𝑟 18 4𝑠2 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+ 𝐴𝑟 18 4𝑠1 3𝑑10 𝐺𝑒2+ 𝐴𝑟 18

4𝑠2 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+ 𝐴𝑟 18

3𝑑10 𝐺𝑒3+ 𝐴𝑟 18 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.

General Chemistry Assignment | albertusadrian.co.cc| LEANGLE ©2010 5

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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