Chem 481. Chapter 1. Atomic Stucture And Periodic Table Worksheet With Answers Page 4
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7
8
9
1015.
Suggest a reason why the covalent radius of germanium (1.22 Å) is almost the same as that of silicon (1.17 Å), even
though germanium has 18 more electrons than silicon.
10
Between Si and Ge lies the 3d
cofiguration due to the first transition series. This leads to much greater Z* because the d
electrons are so poor at shielding. Hence, Ge is smaller than expected for an element in a higher shell than Si. This effect
has been called the Scandide contraction.
16.
Suggest a reason why the covalent radius of hafnium (1.44 Å) is less than that of zirconium (1.45 Å), even though the
element above it in the periodic table.
14
Between Nb and Ta lies the 4f
configuration due to the Lanthanides. This leads to much greater Z* because the f
electrons are so poor at shielding. Hence, Ta is smaller than expected for an element in a higher shell than Nb. This effect
has been called the Lanthanide contraction.
17.
Using Slater’s rules, calculate the effective nuclear charge on an electron in each of the orbitals in an atom of
potassium.
2
2
6
2
6
1
Potassium has the configuration: 1s
2s
2p
3s
3p
4s
. The problem actually simplifies to treating each principal quantum
number (shell), since Slater’s rules do not distinguish among s and p electrons:
K: n = 1
Z* = 19 – [0.30] = 18.7 (remember that within the 1s level, each electron shields the other by only 0.30)
Z* = 19 – [(2 × 0.85) + (7 × 0.35)] = 14.85
K: n = 2
Z* = 19 – [2 + (8 × 0.85) + (7 × 0.35)] = 7.75
K: n = 3
Z* = 19 – [10 + (8 × 0.85)] = 2.2
K: n = 4
18.
Using Slater’s rules, calculate the relative effective nuclear charge on one of the 3d electrons compared to that on one of
the 4s electrons for an atom of manganese.
2
2
6
2
6
5
2
Manganese has the configuration: (1s
)(2s
2p
)(3s
3p
)(3d
)(4s
). We thus contrast the two electrons as follows:
Z* = 25 – [18 + (4 × 0.35)] = 5.6
Mn: 3s
Z* = 25 – [10 + (13 × 0.85) + 0.35] = 3.6
Mn: 4s
19.
The second ionization energies of some Period 4 elements are given in the table below. Identify the orbital from which
ionization occurs and account for the trend in values.
In fact, these values are not easy to account for a priori. We need to make some assumptions. Slater’s rules give a very
nice rationalization so long as we use the electron configuration suggested by removal of the electrons from the normal
Aufbau sequence, remembering to remove 4s electrons before 3d electrons, i.e. the electron configurations for the 1+ ions
are:
+
+
+
+
+
+
Ca
Sc
Ti
V
Cr
Mn
11.87
12.80
13.58
14.15
16.5
15.64
2
2
6
2
2
2
6
2
2
2
6
2
2
2
6
2
2
2
6
2
2
2
6
2
(1s
)(2s
2p
)(3s
3p
(1s
)(2s
2p
)(3s
3p
(1s
)(2s
2p
)(3s
3p
(1s
)(2s
2p
)(3s
3p
(1s
)(2s
2p
)(3s
3p
(1s
)(2s
2p
)(3s
3p
6
1
6
1
1
6
2
1
6
3
1
6
5
6
5
1
)(4s
)
)(3d
)(4s
)
)(3d
)(4s
)
)(3d
)(4s
)
)(3d
)
)(3d
)(4s
)
1
1
1
1
5
1
Ionize 4s
Ionize 4s
Ionize 4s
Ionize 4s
Ionize 3d
Ionize 4s
20–[10+8×0.85]
21–[10+9×0.85]
22–[10+10×0.85]
23–[10+11×0.85]
24–[18+4×0.35]
25–[10+13×0.85]
3.2
3.35
3.5
3.65
4.6
3.95
This gives a very nice trend – not exactly a linear fit, but nonetheless a good correlation between the measured data and
the (very crude) estimation provided by Slater’s rules. Thus the anomalous value for chromium is due to the fact that the
electron must come out of a 3d rather than a 4s electron.
20.
Compare the first ionization energies of strontium, barium and radium. Relate the irregularity to the lanthanide
contraction. Data: IP of Sr = 5.695; IP of Ba = 5.211; IP of Ra = 5.278.
2
2
2
IP of Sr = 5.695 for [Kr]5s
; IP of Ba = 5.211 for [Xe]6s
; IP of Ra = 5.278 for [Rn] 7s
. We are comparing ionizations from
exactly the same valence electron configurations in each case. However, the 4f level is filled between Ba and Ra, which are
very poor at shielding higher-lying electrons from the additional 14 protons added at the same time as the f electrons. Thus
Ra experiences a higher effective nuclear charge than Ba. Thus, despite the fact that its valence electron lies in a higher
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