Chem 481. Chapter 1. Atomic Stucture And Periodic Table Worksheet With Answers Page 3
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1
2
3
4
5
6
7
8
9
103+
2+
3+
2+
9.
Give the ground state electron configuration of (a) C, (b) F, (c) Ca, (d) Ga
, (e) Bi, (f) Pb
, (g) Sc, (h) V
, (i) Mn
, (j)
2+
3+
3+
2+
4+
Cr
, (k) Cu, (l) Gd
, (m) W, (n) Eu
, (o) Eu
, (p) Mo
.
2
2
2
5
2
3+
10
14
10
2
3
2+
14
10
2
1
2
3+
C: [He]2s
2p
; F: [He]2s
2p
; Ca: [Ar]4s
; Ga
: [Ar]3d
; Bi: [Xe]4f
5d
6s
6p
; Pb
: [Xe]4f
5d
6s
; Sc: [Ar]3d
4s
; V
:
2
2+
4
10
1
3+
7
14
4
2
3+
6
2+
7
4+
2
[Ar]3d
; Cr
: [Ar]3d
; Cu: [Ar] 3d
4s
; Gd
: [Xe]4f
; W: [Xe]4f
5d
6s
; Eu
: [Xe]4f
; Eu
: [Xe]4f
; Mo
: [Kr]4d
.
10.
Which atom should have the larger covalent radius, (a) potassium or calcium; (b) fluorine or chlorine? Give your
reasoning.
(a) Potassium should be larger than calcium. They are both in the same period of the PT, and radii decrease from left to
right. To prove this, we can calculate Z*. K: Z* = 19 – [10 + (8 × 0.85)] = 2.2
Z* = 20 – [10 + (8 × 0.85) +
Ca:
(0.35)] = 2.85. Thus we expect the forces operating on the valence shell in calcium to be greater than those in potassium,
leading to a smaller radius in calcium.
(b) Chlorine should be larger than fluorine. We would expect Z* to be very similar, since the two atoms are in the same
group of the periodic table. However, the valence electrons in chlorine are 3p electrons, as opposed to the 2p in fluorine.
Thus the electrons occupy a shell with a larger Bohr radius, and being further from the nucleus they will be less attracted
to the nucleus and the atom will be larger.
11.
Rationalize the first four ionization energies of Al (5.98, 18.82, 28.44, 119.96 eV) on the basis of ground-state electronic
configurations.
2
1
+
2
2+
1
3+
2
5
Al (5.98) [Ne]3s
3p
; Al
(18.82) [Ne]3s
; Al
(28.44) [Ne]3s
; Al
(153.77) [He]2s
2p
. It is progressively harder to remove
an electron from the more positively charged ions. However, there is a huge jump in removing the fourth electron, since
this electron comes from a core orbital that experiences a much greater Z*.
12.
Consider the data below for the radii of some atoms and their most common ions.
+
Na 1.91 Å
Na
1.02 Å
2-
S 1.27 Å
S
1.84 Å
+
Use the concept of nuclear shielding to compare these numbers. For example, compare the size of Na with Na
, S with
2-
+
2-
S
, Na with S and Na
with S
. Make explicit use of the electron configurations of these atoms and ions in your
arguments.
1
+
Sodium, [Ne]3s
is much larger than Na
, [Ne], since the latter has lost the voluminous 3s valence shell, where the loosely-
1
4
bound electron is free to move over a large volume. Also, Na is larger than S, [Ne]3s
3p
, since the Z* experienced for S is
greater than for Na due to the only partial shielding the electrons in the same shell provide for each other, but the number
2–
of protons has increased steadily. S is smaller than S
because the additional 2 electrons added to the neutral atom cause
2
6
2–
the whole 3s
3p
electron cloud to experience a lower Z*, and the whole valence cloud expands outward. Similarly, S
is
+
bigger than Na
, because the former has a deficiency of electrons over protons, while the latter has an excess of electrons
over protons.
13. Compare the reactivity with water both among the alkali elements (Li, Na, K, etc.), and with that of the coinage
elements (Cu, Ag, Au). Rationalize this chemical behaviour using the concept of nuclear shielding. Look at a plot of t he
first ionization energy of these elements to aid you in your reasoning.
If you look at the IP's, Cu, Ag and Au are much higher than Li, Na and K. This is due to the much greater Z* experienced
by the coinage metals, because they come after the filling of the 3d subshell. Thus while Li, Na and K react explosively
with water to release hydrogen gas, none of the coinage metals are oxidized by water. Indeed, copper is used to sheathe
roofs and the hulls of ships - it had better not react!
14. Compare the f i rst ionization energies of calcium to that of zinc. Explain the difference in terms using concepts of
shielding. Use Slaters’s rules explicitly.
st
2
st
10
2
1
IP of Ca = 6.11 eV for [Ar]4s
; 1
IP of Zn is 9.39 for [Ar]3d
4s
. Between Ca and Zn, 10 additional protons are added,
and 10 electrons are added to the 3d level. If we remember that d orbital electrons are poor at shielding other electrons, we
expect that Zn will experience a higher Z* than Ca. This is consistent with the larger ionization energy of Zn (i.e. it is
harder to remove an electron from Zn than from Ca.) To “prove” this we calculate the Z* by Slater’s rules:
Z* = 20 – [10 + (8 × 0.85) + (0.35)] = 2.85
Ca:
Z* = 30 – [10 + (18 × 0.85) + (0.35)] = 4.35
Zn:
Indeed Z* is considerably larger for the Zn valence electron.
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