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Exercise 1 - atomic symbols
Deduce the number of protons, neutrons and electrons in the following species:
1.
1
2.
17
8O
3.
4
4.
132
54Xe
5.
27
3+
13Al
6.
235
92U
7.
1
8.
45
3+
21Sc
9.
37
17Cl
1H
2+
2He
+
1H
10. 146C
Use the periodic table to write symbols for the following species:
11. 19 protons, 20 neutrons, 18 electrons
12. 8 protons, 8 neutrons, 10 electrons
13. 1 proton, 2 neutrons, 1 electron
14. 82 protons, 126 neutrons, 80 electrons
15. 53 protons, 74 neutrons, 54 electrons
EXERCISE 2 – ram, rmm and mass spectra
Deduce the relative atomic masses of the following elements.
1. Silicon (28Si 92.21%, 29Si 4.70%, 30Si 3.09%)
2. Silver (107Ag 51.88%, 109Ag 48.12%)
3. Boron (10B 19.7%, 11B 80.7%)
4. Gallium (69Ga 60.2%, 71Ga 39.8%)
5. Zirconium:
Assume in all cases the relative isotopic mass is equal to the mass number.
-------6.
Bromine has two isotopes, with mass numbers 79 and 81. Its relative atomic mass is often given as 80. What does that tell you
about the relative abundance of the two isotopes?
7.
Most argon atoms have a mass number 40. How many neutrons does this isotope have? The relative isotopic mass of this
isotope is 39.961, but the relative atomic mass of argon is 39.948. What can you deduce about the other isotopes of argon?
---------8.
For each of the following stages in a mass spectrometer, state which part of the spectrometer is responsible for it and how it
works:
a) ionization
b) acceleration
c) deflection
d) detection
-------------------9. Deduce, giving reasons, the relative molecular mass of compound A, which has the following mass spectrum:
100
80
relative
abundance
60
29
57
40
15
20
43
72
Exercise 3 - electronic configuration
Write the electronic configuration of the following using the arrow and box
method:
1. C
2. Cu
3. Mg+
Write the electronic configuration of the following using the orbital method:
4. N35. Ar
6. Sc3+
7. Mn2+
8. Fe3+
9. V3+
Write the electronic configuration of the following using the shorthand arrow and
box method:
10. Cl11. Fe
12. Br
Write the electronic configuration of the following using the shorthand orbital
method:
13. Cr
14. Ga3+
15. Pb2+
HIGHER LEVEL Exercise 4 – ionisation energies
1. Why does the first ionisation energy of atoms generally increase across a period?
2. Why is the first ionisation energy of boron less than that of beryllium?
3. Why is the first ionisation energy of oxygen less than that of nitrogen?
4. Why do first ionisation energies decrease down a group?
5. Why does helium have the highest first ionisation energy of all the elements?
6. Why is the second ionisation energy of an atom always greater than the first?
7. Why is the second ionisation energy of sodium much greater than the first?
8. Why does atomic size decrease across a period?
9. Why does atomic size increase down a group?
10. Why are cations always smaller than the corresponding atoms?
11. Why are anions always larger than the corresponding atoms?
Answers
Exercise 1
1.
4.
7.
10.
11.
15.
1p, 0n, 1e
2. 8p, 9n, 8e
3. 2p, 2n, 0e
54p, 78n, 54e
5. 13p, 14n, 10e
6. 92p, 143n, 92e
1p, 0n, 0e
8. 21p, 24n, 18e
9. 17p, 20n, 18e
6p, 8n, 6e
39K+
12. 16O213. 3H
14. 208Pb2+
127I-
Exercise 2
1.
5.
6.
7.
8.
28.29
2. 107.96
3. 10.85
4. 69.80
91.4 (approx)
two isotopes approximately equally abundant
22 neutrons. Other isotopes are lighter, and not very abundant
a) electron gun – fires electrons at atom, knocking out other electrons
b) electric field – attracts ions towards it until all are traveling at same speed
c) magnetic field – moving charges are deflected according to m/z ratio
d) detector – ions land on it and create current proportional to abundance
9. 72 – peak with largest m/z ratio must be molecular ion peak
Exercise 3
1.
2.
3.
1s
↑↓
2s
↑↓
↑
2p
↑
1s
↑↓
2s
↑↓
↑↓
2p
↑↓
↑↓
3s
↑↓
1s
↑↓
2s
↑↓
↑↓
2p
↑↓
↑↓
3s
↑
↑↓
3p
↑↓
4s
↑
↑↓
↑↓
↑↓
3d
↑↓
↑↓
4. 1s22s22p6
5. 1s22s22p63s23p6
6. 1s22s22p63s23p6
7. 1s22s22p63s23p63d5 8. 1s22s22p63s23p63d5 9. 1s22s22p63s23p63d2
10.
[Ne]
3s
↑↓
↑↓
3p
↑↓
↑↓
[Ar]
4s
↑↓
↑↓
↑
3d
↑
↑
↑
[Ar]
4s
↑↓
↑↓
↑↓
3d
↑↓
↑↓
↑↓
11.
12.
↑↓
4p
↑↓
↑
13. [Ar]4s13d5 14. [Ar]3d10 15. [Xe] 6s24f145d10
Exercise 4
1.
2.
Number of protons increases, shielding stays the same, so attraction of outer electrons to
nucleus increases
Outermost electron in B is 2p, outermost electron in Be is 2s, 2p electron in B better
shielded than 2s electron in Be, so it is less attracted to nucleus
↑↓
3.
4.
5.
6.
7.
8.
9.
10.
11.
2p electron is paired in O but unpaired in N, so in O there is more repulsion in the orbital
which makes the electron easier to remove
More shells, so more shielding, so attraction of outer electrons to the nucleus decreases
No shielding in 1st period so electrons closely held than in other periods, and more
protons than hydrogen so greater attraction to nucleus
Less electrons, so less electron repulsion
1st electron removed from 3s, second electron removed from 2p so much less shielding
Number of protons increases, shielding stays the same, so attraction of outer electrons to
nucleus increases and they move closer
More shells, so more shielding, so attraction of outer electrons to the nucleus decreases
and they are pushed further away
Less electrons, so less repulsion, so electrons can get closer to the nucleus
More electrons, so more repulsion, so electrons are pushed further away