F1
(a) Noble gas atoms have an outer shell structure ns2np6 , where n is the principal quantum number of the outer
shell.
(b) 4s2 4p6 . Since krypton is in the fourth period of the table, the principal quantum number of each electron in its
outer shell is n = 4.
(c) The outer shell of the bromine atom will have the structure 4s2 4p5 , because the atom contains one fewer
electron than the krypton atom. Bromine is in group 7, the one next to the group of noble gas elements (group 0).
It is, like krypton, in the fourth period because in both cases the principal quantum number of the outer shell is
n = 4.
FLAP P8.4
The periodic table and chemical bonding
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S570 V1.1
F2
(a) In most transition elements, a d subshell is incomplete when the s subshell of the outer shell contains at least
one electron (and usually two). To put it less formally, you could say that in the transition elements, inner d
subshells begin to fill while an outer s subshell already contains at least one electron.
(b) In most of the lanthanides, the 4f subshells begin to fill when there are two electrons in the 6s subshell.
(c) In most of the actinides, the 5f subshells begin to fill when there are two electrons in the 7s subshell.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
F3
Oxygen has the electronic configuration 1s2 2s2 2p4 . It can attain the noble gas electronic configuration (neon) by
acquiring two electrons, that is, by forming an O2− ion. Oxygen in magnesium oxide therefore has a charge of −2
(where +1 is defined as the electric charge of the proton). Because chemical compounds are electrically neutral,
the charges of the oxygen ion and the magnesium ion must balance each other. The formula of the oxide is
therefore MgO, corresponding to Mg 2+O2−.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
F4
The empirical formula is SO3. The valency of oxygen is two, so as the valency of sulphur is six, the ratio of
sulphur atoms to oxygen atoms is 11:13.
FLAP P8.4
The periodic table and chemical bonding
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S570 V1.1
F5
(a) and (b) CCl4 and F 2 O are covalent compounds. The low melting and boiling temperatures suggest discrete
molecular compounds and, in the case of CCl 4 , this is corroborated by the low conductivity of the liquid.
For CCl 4 and F2 O we can write the following Lewis structures
.. .
..
O× F
.
.. .
.
..
×
×
..
. .
.. .
.
××
××
..
.. . × .. .
.. × C Cl
..
Cl
..
Cl
×
..
..
..
Cl
..F
XVI
(c) Rb2 O is an ionic compound. This is apparent from its high melting temperature and from its structure: each
oxygen is surrounded by eight rubidium ‘units’ while each rubidium ‘unit’ is surrounded by four oxygen ‘units’.
Such structures are characteristic of ionic compounds.
In the case of Rb2 O, the ions Rb+ and O 2− have noble gas electronic configurations (krypton and neon,
respectively). We can therefore write Rb2 O as (Rb +)2O2−.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route
through the module and to proceed directly to Ready to study? in Subsection 1.3.
Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the
Closing items.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it
here.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
R1
Every atom consists of a positively charged central nucleus surrounded by negatively charged electrons.
An argon atom with Z = 18 has a nuclear charge that is 18 times greater than that on a hydrogen nucleus
(i.e. 18e, where e is the charge on the proton). In order that it should be electrically neutral, an argon atom must
have 18 electrons ‘orbiting’ its nucleus. Apart from revealing the charge on the nucleus and the number of
electrons in the neutral atom, the atomic number provides other items of information regarding X-ray spectra,
etc. but these may be seen as consequences of the charge on the nucleus.
Consult the relevant terms in the Glossary for further information.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
R2
A lithium ion is positively charged, whereas a helium atom is neutral: lithium has three protons in its nucleus,
whereas helium has only two. The superscript on the symbol for the lithium ion (Li+) records the single excess
positive charge.
Consult atomic number and ion in the Glossary for further information.
FLAP P8.4
The periodic table and chemical bonding
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
s
2
T1
d
10
f
14
7p
6d
5f
7th period
7s
6p
5d
4f
energy increasing
From Figure 3 you can see that
after 6s, the level of next highest
energy is 4f, followed by 5d. The 4f
level can accommodate up to
fourteen electrons (in common with
all f states). The procedure we have
been following therefore suggests
that lanthanum should have the
structure [Xe]6s2 4f1 because this
atom contains one more electron
than a barium atom and this should
go into the next highest energy
state, the 4f state.
p
6
6s
5p
5th period
3d
4th period
4p
4s
3p
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
4d
5s
3s
Figure 34Subshell energy level
diagram.
6th period
2p
2s
1s
3rd period
2nd period
1st period
S570 V1.1
Similarly, the electronic configuration of praseodymium should be [Xe]6s24f3 because the additional three
electrons should all be accommodated in the 4f state. Using identical reasoning, the structures of the remaining
elements should be: terbium, [Xe]6s24f9; ytterbium, [Xe]6s24f14 and lutetium, [Xe]6s2 4f145d1.
Comment
With only one exception these predictions are correct; the exception is lanthanum, which in fact has the
configuration [Xe]6s2 5d1 .
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
T2
(a) The elements with atomic numbers 16 and 28 have electronic
configurations 1s22s2 2p6 3s2 3p4 and 1s2 2s2 2p6 3s2 3p6 4s2 3d8 , respectively.
These configurations are determined by using Figure 4, which gives the
order in which the subshells are filled. In order to find the configuration
of the elements, it is simply necessary to bear in mind that s shells can
accommodate up to two electrons, whereas p shells can accommodate
up to six and d shells accommodate up to ten (this maximum occupancy
is given at the top of Figure 3).
(b) Element 16 has an outer electronic configuration s 2 p x so it is a
typical element; by contrast, element 28 has an outer electronic
configuration s2 dx, which means that it is a transition element.
8s
3s
2s
7s
7p
6s
6p
6d
5s
5p
5d
5f
4s
4p
4d
4f
3p
3d
2p
1s
Figure 44Order of ascending energy for
the subshells.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
s
2
(c) The last subshells to be filled in
element 16 are 3s and 3p, so Figure
3 tells you that the element is in
period 3. In the same way, the last
subshells to be filled in element 28
are 4s and 3d, so it is in period 4.
d
10
f
14
7p
6d
5f
7th period
7s
6p
5d
4f
energy increasing
(d) Element 16 falls in group 6 of
the periodic table: it is sulphur.
Element 28 is nickel.
p
6
6th period
6s
5p
4d
5th period
3d
4th period
5s
4p
4s
3p
3s
Figure 34Subshell energy level
diagram.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
2p
2s
1s
3rd period
2nd period
1st period
S570 V1.1
T3
One atom of oxygen combines with two atoms of hydrogen, so the valency of oxygen is two.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
T4
(a) SiH4 . If the valency of silicon is four, one atom of silicon can combine with four atoms of hydrogen.
(b) SiO2 . One atom of silicon can combine with two atoms of oxygen.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
T5
The configuration of Na+ is 1s 2 2s2 2p6 . This corresponds to the electronic configuration of neon.
FLAP P8.4
The periodic table and chemical bonding
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T6
The electronic configuration of Cl− is 1s 2 2s2 2p6 3s2 3p6 . This corresponds to the configuration of the noble gas
argon. The chloride ion now has 18 electrons (each with a single negative charge), but there are still only 17
positively charged protons in the nucleus. Consequently, the chlorine now has one excess negative charge, as the
minus sign of Cl− illustrates.
FLAP P8.4
The periodic table and chemical bonding
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T7
The electronic configuration of magnesium is 1s22s2 2p6 3s2 . Loss of the two 3s2 electrons would produce a Mg++
ion (written as Mg2+) with the electronic configuration of neon, and two excess positive charges. So, in order to
be neutral, magnesium fluoride must be MgF 2 , that is, Mg2+(F−)2. This second formula records the fact that for
every magnesium ion in the compound, there are two fluoride ions.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
T8
The four molecules are as given below.
H
H
Cl
H
C
H
H
Cl
Cl
C
Cl
Cl
H
O
H
The valencies are one for hydrogen and chlorine, two for oxygen and four for carbon. In each case, the number
of lines issuing from an atom is equal to its valency.
Comment
You may not have predicted the correct shape for the water molecule. The reasons for this shape are too
complex to be discussed in this module.
FLAP P8.4
The periodic table and chemical bonding
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THE OPEN UNIVERSITY
S570 V1.1
T9
Carbon has four outer electrons (2s2 2p2 ). It can attain the electronic configuration of neon (2s2 2p6 ) by gaining
another four electrons from four shared electron pairs:
H
.
.×
×
.
H×C
×
.
H
H
In this structure, each hydrogen again attains the helium electronic configuration.
FLAP P8.4
The periodic table and chemical bonding
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1
T10
Calcium chloride (CaCl 2 ) is ionic, iodine bromide (IBr) is covalent and CaMg2 is metallic.
The properties listed are characteristic of (a) an ionic substance, (b) a covalent substance, and (c) a metallic
substance. CaCl2 is a combination of elements from the extreme left and extreme right of the periodic table, so
the electronegativity difference will be large and this will be the ionic compound. IBr will be covalent because it
is a combination of elements of high electronegativity from the extreme right of the table. CaMg 2 will be a
metallic alloy because it is a combination of metallic elements with low electronegativity from the left of the
table.
FLAP P8.4
The periodic table and chemical bonding
COPYRIGHT © 1998
THE OPEN UNIVERSITY
S570 V1.1