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Tutorial #
A.
Name
Which element is described by each of the following? (Place each part letter (a, b, c, etc.) in the correct element box in the
blank periodic table below).
a) smallest atomic radius in group 15. (Nitrogen)
b) largest atomic radius in period 3. (Sodium)
c) highest first ionization energy in group 14. (Carbon)
d) group 13 element that forms the most basic oxide. (Thallium)
e) period 4 transition element that forms a 3+ diamagnetic ion. (Scandium)
f) period 5 element that forms 3+ ion with a “pseudo-rare-gas” configuration. (Indium)
g) whose ground state electron configuration is [Ne]3s23p2. (Silicon)
h) ground state electron configuration is [Kr]5s24d6. (Ruthenium)
i) least metallic in Group 15. (Nitrogen)
j) a period 4 transition metal element that forms a 2+ ion with a half-filled d sublevel. (Manganese)
k) name any two of the above elements for 1 bonus mark!!
Tutorial #
B.
Name
Which element is described by each of the following? (Place each part letter (a, b, c, etc.) in the correct element box in the
blank periodic table below).
a) smallest atomic radius in group 17. (Fluorine)
b) smallest atomic radius in period 4. (Krypton)
c) lowest first ionization energy in group 18. (Radon)
d) group 14 element that forms the most basic oxide. (Lead)
e) period 4 transition element that forms a 2+ diamagnetic ion. (Zinc)
f) period 4 element that forms 3+ ion with a “pseudo-rare-gas” configuration. (Gallium)
g) whose ground state electron configuration is [Ar]4s13d10. (Copper)
h) ground state electron configuration is [Kr]5s24d6. (Ruthenium)
i) least metallic in Group 13. (Boron)
j) a period 4 transition metal element that forms a 2+ ion with a half-filled d sublevel. (Manganese)
k) name any two of the above elements for 1 bonus mark!!
Tutorial #
C.
Name
Which element is described by each of the following? (Place each part letter (a, b, c, etc.) in the correct element box in the
blank periodic table below).
a) smallest atomic radius in group 17. (Fluorine)
b) largest atomic radius in period 3. (Sodium)
c) highest first ionization energy in group 18. (Helium)
d) group 14 element that forms the most basic oxide. (Lead)
e) period 4 transition element that forms a 3+ diamagnetic ion. (Scandium)
f) period 5 element that forms 3+ ion with a “pseudo-rare-gas” configuration. (Indium)
g) whose condensed electron configuration is [Ne]3s23p2. (Silicon)
h) ground state electron configuration is [Kr]5s24d6. (Ruthenium)
i) least metallic in Group 14. (Carbon)
j) a period 4 transition metal element that forms a 2+ ion with a half-filled d sublevel. (Manganese)
k) name any two of the above elements for 1 bonus mark!!
A2.
a)
Write down the principle and orbital angular momentum quantum numbers (n and
l, respectively) for each of these orbitals and state how many electrons each sub-level
can accommodate.
3s - n=3, l=0 : 2 electrons
3p - n=3, l=1 : 6 electrons
3d - n=3, l=2 : 10 electrons
b)
Below is are plotted the radial probability distribution functions for the 3s, 3p and
3d orbitals. Note that the angular nodes are not shown in these types of plots.
For a one electron atom, such as hydrogen, the 3s, 3p and 3d orbitals are degenerate.
However, for a many-electron atom the energy levels split. Briefly explain, referring
to this figure, how penetration affects the energy splitting of the 3s, 3p and 3d
orbitals for a many-electron atom.
From the diagrams above it can be seen that the 3s electrons penetrate very close to the
nucleus (the smallest and second smallest maxima), closer than the 3p electrons, which
penetrate closer than the 3d electrons. In a many electron atom, since the 3s electron spends
time closer to the nucleus than the 3p electron it is lower in energy or more stable due to the
greater electron-nucleus attraction. Similarly, the 3p electrons are more stable than the 3d
electrons. The orbitals within the same shell are, therefore, no longer degenerate in a manyelectron atom.
B2.
a)
Write the complete electron configuration for argon (Ar), which has 18 electrons.
1s22s22p63s23p6
b)
Both Ni (28 electrons) and Cu (29 electrons) have [Ar] as the “core” electron
configuration. Write condensed electron configurations and orbital diagrams
(showing all electrons other than the first 18) for both Ni and Cu.
Ni [Ar]4s23d8 or [Ar]3d84s2
[Ar]
3d
4s
Cu [Ar]4s13d10 or [Ar]3d104s1
[Ar]
3d
B3.
4s
a) Briefly explain i) the main problem associated with the Rutherford model of the atom and
ii) how Bohr’s model overcame this problem.
i) The Rutherford model of the atom had electrons orbiting the nucleus. However,
classical physics predicted that a charged particle travelling in a circular orbit would
emit light, losing energy, and spiral into the nucleus. The atom and thus matter would
be unstable.
ii) Bohr used the idea of quantization to overcome the problem with Rutherford’s
model. He postulated that the electrons in an atom are confined only to specific energy
levels and cannot exist in states between these allowed energy levels.
b) Briefly describe the problems associated with the Bohr model of the atom.
The Bohr model of the atom states that the electron orbits the nucleus in a clearly
defined orbital. However, the uncertainty principle states that if we know the energy
with any certainty it is impossible to know the position with any certainty.
Another problem with the Bohr model is that it only works for the hydrogen atom or
hydrogen-like ions (with one electron).
C2.
a)
Write the complete electron configuration for argon (Ar), which has 18 electrons.
1s22s22p63s23p6
b)
Both vanadium (V, 23 electrons) and chromium (Cr, 24 electrons) have [Ar] as the
“core” electron configuration. Write condensed electron configurations and orbital
diagrams (showing all electrons other than the first 18) for both V and Cr.
V
[Ar]4s23d3 or [Ar]3d34s2
[Ar]
3d
Cr
4s
[Ar]4s13d5 or [Ar]3d54s1
[Ar]
3d
C3.
4s
Below is are plotted the radial probability distribution functions for the 3s, 3p and 3d
orbitals. Note that the angular nodes are not shown in these types of plots.
For a one electron atom, such as hydrogen, the 3s, 3p and 3d orbitals are degenerate.
However, for a many-electron atom the energy levels split. Briefly explain, referring to this
figure, how penetration affects the energy splitting of the 3s, 3p and 3d orbitals for a manyelectron atom.
From the diagrams above it can be seen that the 3s electrons penetrate very close to the
nucleus (the smallest and second smallest maxima), closer than the 3p electrons, which
penetrate closer than the 3d electrons. In a many electron atom, since the 3s electron spends
time closer to the nucleus than the 3p electron it is lower in energy or more stable due to the
greater electron-nucleus attraction. Similarly, the 3p electrons are more stable than the 3d
electrons. The orbitals within the same shell are, therefore, no longer degenerate in a manyelectron atom.