QUESTIONS AND ANSWERS ON ATOMS AND ATOMIC STRUCTURE
1.1 exercise 1 - atomic symbols
Deduce the number of protons, neutrons
and electrons in the following species:
1.
11H
2.
178O
3.
42He2+
4.
13254Xe
5.
2713Al3+
6.
23592U
7.
11H+
8.
4521Sc3+
9.
3717Cl-
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
1.1 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.
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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?
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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
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9.
Deduce, giving reasons, the relative molecular mass of compound A, which has
the following mass spectrum:
1.1
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. N3-
5. Ar
6. Sc3+
7. Mn2+
8. Fe3+
9. V3+
Write the electronic configuration of the following using the shorthand
arrow and box method:
10. Cl-
11. Fe
12. Br
Write the electronic configuration of the following using the shorthand
orbital method:
13. Cr
14. Ga3+
15. Pb2+
1.1 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 to 1.1
Exercises
1.1 Exercise 1
1. 1p, 0n, 1e 2. 8p,
9n, 8e 3. 2p, 2n, 0e
4. 54p, 78n, 54e 5. 13p,
14n, 10e 6. 92p, 143n, 92e
7. 1p, 0n, 0e 8. 21p,
24n, 18e 9. 17p, 20n, 18e
10. 6p, 8n, 6e
11. 39K+ 12. 16O2- 13. 3H 14. 208Pb2+
15. 127I-
1.1 Exercise 2
1. 28.29 2. 107.96 3. 10.85 4. 69.80
5. 91.4 (approx)
6. two isotopes approximately equally
abundant
7. 22 neutrons.
Other isotopes are lighter, and not very abundant
8. 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
1.1 Exercise 3
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4. 1s22s22p6 5. 1s22s22p63s23p6 6. 1s22s22p63s23p6
7. 1s22s22p63s23p63d5 8. 1s22s22p63s23p63d5 9. 1s22s22p63s23p63d2
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3s
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11.
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12.
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13. [Ar]4s13d5 14. [Ar]3d10 15. [Xe]
6s24f145d10
1.1 Exercise 4
1. Number of protons increases, shielding
stays the same, so attraction of outer electrons to nucleus increases
2. 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. 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
4. More shells, so more shielding, so
attraction of outer electrons to the nucleus decreases
5. No shielding in 1st period
so electrons closely held than in other periods, and more protons than hydrogen
so greater attraction to nucleus
6. Less electrons, so less electron
repulsion
7. 1st electron removed from
3s, second electron removed from 2p so much less shielding
8. Number of protons increases, shielding
stays the same, so attraction of outer electrons to nucleus increases and they
move closer
9. More shells, so more shielding, so
attraction of outer electrons to the nucleus decreases and they are pushed
further away
10. Less electrons, so less repulsion, so
electrons can get closer to the nucleus
11. More electrons, so more repulsion, so
electrons are pushed further away
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