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1
Chapter 7 Covalent molecules, networks and layers
E1.
a
b
Give the electronic configuration for an atom of beryllium.
How many electrons are in the outer shell of an atom of beryllium in the molecule
BeH2?
AE1.
a
b
1s22s2
4
E2.
The noble gases helium and neon do not form any compounds. The noble gas xenon,
however, does form a covalent molecular compound with fluorine. Suggest a reason
for this difference.
AE2.
The two valence electrons of helium fill helium’s outer shell (shell number 1 can
contain a maximum of 2 electrons). Similarly, the outer shell of neon, shell number 2,
is filled with neon’s 8 valence electrons. So, helium and neon have filled outer shells
and do not form compounds by sharing electrons with other non metals. Xenon also
has 8 electrons in its outer shell. However, the outer shell of xenon is shell number 5,
which can hold a maximum of 50 electrons. So, under appropriate circumstances,
xenon can form covalent compounds in which xenon shares electrons with another
non-metal, such as fluorine.
Q1.
Draw electron dot and valence structures for each of the following molecules:
fluorine (F2), hydrogen fluoride (HF), water (H2O), tetrachloromethane (CCl4),
phosphine (PH3), butane (C4H10), carbon dioxide (CO2)
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A1.
fluorine (F2)
hydrogen fluoride (HF)
tetrachloromethane (CC14)
water (H2O)
butane (C4H10)
phosphine (PH3)
carbon dioxide (CO2)
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Q2.
What is the maximum number of covalent bonds an atom of each of the following
elements can form?
a F
b O
c N
d C
e H
f Ne
A2.
a
b
c
d
e
f
1
2
3
4
1
0
Q3.
A knowledge of electronic configurations would lead you to predict that H2O is the
empirical formula for water. Why?
A3.
Oxygen has six outer-shell electrons and needs two more electrons to complete its
outer shell. It gains access to two more electrons by forming two covalent bonds with
two different hydrogen atoms. Each of the hydrogen atoms gains an electron and also
completes its outer shell.
Q4.
When oxygen forms covalent molecular compounds with other non-metals, the
valence structures that represent the molecules of these compounds all show each
oxygen atom with two lone non-bonding electron pairs. Why are there always two
lone pairs?
A4.
To complete its outer shell, the oxygen uses two of its outer-shell electrons to form
two single bonds or a double bond with suitable non-metal atoms. The remaining four
electrons in the outer shell are not required for bonding, as the outer shell is now
complete, and they arrange themselves as two lone pairs around the oxygen atom.
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Q5.
Suggest the most likely formula of the compound formed between the following pairs
of elements:
a C, Cl
b N, Br
c Si, O
d H, F
e P, F
A5.
a
b
c
d
e
CCl4
NBr3
SiO2
HF
PF3
Q6.
How many lone pairs would you expect atoms of the following elements to have
when they form covalent bonds with other non-metal atoms?
a H
b F
c C
d N
A6.
a
b
c
d
0
1
0
1
Q7.
Draw the valence structure for each of the following molecules:
a H2S
b HI
c CCl4
d PH3
e CS2
A7.
a
b
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c
d
e
Q8.
Describe the shape of each of the molecules given in Question 7.
A8.
a
b
c
d
e
V-shaped
linear
tetrahedral
pyramidal
linear
Q9.
Covalent bonds can form between the following pairs of elements in a variety of
compounds. Use the electronegativity values given in Table 7.4 (page 123) to identify
the atom in each pair that would have the larger share of bonding electrons.
a S and O
b C and H
c C and N
d N and H
e F and O
f P and F
A9.
a
b
c
d
e
f
O
C
N
N
F
F
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Q10.
The greater the differences in electronegativity between two atoms, the more polar is
the bond formed between them.
a Which of the examples in Question 9 would be the most polar bond?
b Which of the examples in Question 9 would be the least polar bond?
A10.
a
b
P–F
C–H
Q11.
Figure 7.29 (page 126) represents models for a number of molecules. Examine the
models and identify the polar molecules.
A11.
CH3OH; CH3F
Q12.
Which of the molecules in Question 11 would be capable of forming hydrogen bonds?
A12.
CH3OH. Hydrogen bonding is often called FON bonding because fluorine, oxygen
and nitrogen are almost always the other elements involved.
Q13.
Consider each of the following substances. In which ones are the molecules held to
one another by:
i dipole–dipole attraction?
ii hydrogen bonds?
a
b
c
d
e
f
g
h
i
NH3
CHCl3
CH3Cl
F2O
HBr
H2S
HF
H2O
H2
A13.
i
ii
a to h
a, g, h
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Q14.
In ice, each water molecule is surrounded, at equal distances, by four other water
molecules. In each case, there is an attraction between the positive hydrogen atom on
one water molecule and a lone pair associated with the oxygen atom of another water
molecule. Draw a diagram to show the arrangement of four water molecules around
another water molecule.
A14.
Because of hydrogen bonding, ice is less dense than liquid water, and so ice floats on
water. (For most liquids, the solid is denser than the liquid.) This is good news for
fish, but not good news for travellers on the Titanic!
A15.
‘Cloudy ammonia’ is often used as a cleaning solution in bathrooms. This solution
contains ammonia dissolved in water. Draw a diagram to represent hydrogen bonding
between a water molecule and an ammonia molecule.
A15.
E3.
The molecular formula of ethane and propane are C2H6 and C3H8, respectively. For
each, give the:
a empirical formula
b structural formula
c semistructural formula
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AE3.
a
CH3; C3H8
b
c
CH3CH3; CH3CH2CH3
Q16.
Explain the following properties of:
i diamond, and
ii graphite
in terms of their respective structures:
a high melting temperature
b hardness or softness
c ability or inability to conduct electricity
A16.
a
i
ii
b
i
ii
c
i
ii
A lot of energy is required to disrupt the strong covalent bonding in three
dimensions in diamond.
A large amount of energy is required to disrupt the strong covalent bonding
in two dimensions in graphite. Therefore, diamond and graphite both have
high melting temperatures.
Diamond is hard because it has strong covalent bonds throughout the lattice,
with all atoms being held in fixed positions.
Graphite is soft because there are weak dispersion forces between the layers
in graphite, so layers can be made to slide over each other easily.
Diamond is a non-conductor of electricity because all of its electrons are
localised in covalent bonds and are not free to move.
Graphite is able to conduct electricity because it has delocalised electrons
between its layers of carbon atoms.
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Q17.
Explain the following uses in terms of the structures of graphite and diamond:
a graphite is used as a lubricant
b diamond is often used as an edge on saws and a tip on drills
A17.
a
b
Graphite is used as a lubricant because it is greasy. The dispersion forces between
the layers in graphite enable the layers to slide over each other easily.
The strong covalent bonding throughout the lattice means that the carbon atoms
are fixed in place. This makes the diamond very hard and suitable as a material
for cutting other less hard materials.
Chapter review
Q18.
Draw electron dot formulas for each of the following molecules and identify the
number of bonding and non-bonding electrons in each molecule:
a HBr
b H2O2
c CF4
d C2H6
e PF3
f Cl2O
g CH4
h H2S
A18.
a
2 bonding electrons, 6 non-bonding electrons
b
6 bonding electrons, 8 non-bonding electrons
c
8 bonding electrons, 24 non-bonding electrons
d
14 bonding electrons, 0 non-bonding electrons
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e
6 bonding electrons, 20 non-bonding electrons
f
4 bonding electrons, 16 non-bonding electrons
g
8 bonding electrons
h
4 bonding electrons, 4 non-bonding electrons
Q19.
Identify the number of bonding and non-bonding electrons in the following
molecules:
a N2
b CHCl3
c O2
A19.
a
b
c
3 bonding pairs, 2 non-bonding pairs
4 bonding pairs, 9 non-bonding pairs
2 bonding pairs, 4 non-bonding pairs
Q20.
Draw valence structures for the following molecules. Underneath each structure
indicate whether the molecules are:
i symmetrical or non-symmetrical
ii polar or non-polar
a CO2
b PH3
c N2
A20.
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Q21.
Are the following molecules polar or non-polar? Draw valence structures or make
models to help you to decide.
a CS2
b Cl2O
c SiH4
d CH3Cl
e CH3CH3
f CCl4
A21.
d
e
f
Q22.
Use Table 7.4 (page 123) to determine which of the following molecules contains the
most polar bond:
a CO2
b H2O
c H2
d H2S
e NH3
A22.
The O–H bond in water is the most polar bond.
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Q23.
For each of the structures shown below, state whether:
i the molecule is polar or non-polar
ii the strongest intermolecular forces of attraction between molecules of each type
would be dispersion forces, hydrogen bonding or dipole–dipole attraction
A23.
a
b
c
d
e
SO3
SiCl4
CF4
NF3
CH3NH2
i non-polar; ii dispersion forces
i non-polar; ii dispersion forces
i non-polar; ii dispersion forces
i polar; ii dipole–dipole attraction
i polar; ii hydrogen bonding
Q24
Consider solid samples of the following compounds. In which cases will the only
forces between molecules in the samples be dispersion forces? (You should first
ascertain whether molecules of these compounds are polar or non-polar. You can do
this by drawing an accurate structural formula for each one.)
a tetrachloromethane CCl4(s)
b sulfur dioxide SO2(s)
c carbon dioxide CO2(s)
d hydrogen sulfide H2S(s)
A24.
Compounds CCl4 and CO2 are both non-polar, and so intermolecular forces operating
between these molecules will be dispersion forces.
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Q25.
The melting temperatures of four of the halogens are given in the table below. (Refer
to the periodic table on page 408 to establish where the halogens occur in the table.)
Describe and explain the trend in melting temperatures of these elements.
Halogen
Fluorine (F2)
Chlorine (Cl2)
Bromine (Br2)
Iodine (I2)
Melting temperature (°C)
–220
–101
–7
114
A25.
Melting temperatures increase down the table because the molecules increase in mass
and size and there are more electrons in the molecules; therefore, the strength of the
dispersion forces increases.
Q26.
Suppose that you had samples of the two compounds OF2 and CF4. Between
molecules of which sample would the intermolecular forces of attraction be greater?
Explain your answer.
A26.
CF4 has a slightly higher boiling temperature (–128°C) than OF2 (–145°C), indicating
that the forces between molecules in CF4 are stronger. OF2 is slightly polar; CF4 is
non-polar. OF2 molecules are held together by forces of dipole–dipole attraction and
dispersion forces. Although CF4 molecules are attracted by dispersion forces only, the
much larger size of CF4 molecules makes the dispersion forces stronger than the sum
of the dipole–dipole forces and the dispersion forces between OF2 molecules.
Q27.
The mass of a hydrogen fluoride molecule is similar to the mass of a neon atom. The
boiling temperatures of these substances are very different, however. That of
hydrogen fluoride is 19.5°C and that of neon is –246°C. Explain the difference in this
property of the two substances.
A27.
Neon exists as single atoms, with the only forces of attraction being dispersion forces.
As Ne atoms have very few electrons, the dispersion forces are extremely weak. Neon
therefore has a very low boiling temperature. Hydrogen fluoride molecules, however,
are very polar and so are held together by electrostatic attraction between permanent
dipoles. Because hydrogen is bonded to the very electronegative fluorine, the forces
between molecules are known as hydrogen bonds. These are relatively strong
intermolecular bonds and HF, therefore, has a much higher boiling temperature than
Ne. (The dispersion forces operating between HF molecules are extremely weak.)
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Q28.
At room temperature, CCl4 is a liquid whereas CH4 is a gas.
a Which substance has the stronger intermolecular attractions?
b Explain the difference in the strengths of the intermolecular attractions.
A28.
a
b
CCl4
CH4 and CCl4 are both non-polar and so are held together in a lattice only by
dispersion forces. CCl4 is the larger of these two molecules and has more
electrons, so the dispersion forces between CCl4 molecules will be greater than
those between CH4 molecules. As there are stronger dispersion forces between
molecules of CCl4 than for CH4, it takes more energy to vaporise CCl4.
Q29.
What are the forces of attraction between the following molecules?
a H2
b HCl
c NH3
d CH4
e H2O
f C2H6
A29.
a
b
c
d
e
f
dispersion forces
dispersion forces, dipole–dipole attractions
dispersion forces, hydrogen bonding
dispersion forces
dispersion forces, hydrogen bonding
dispersion forces
Q30.
Silicon carbide is a substance that is almost as hard as diamond and is used as a
commercial abrasive. It is made by heating silica and carbon to a very high
temperature. Silicon carbide consists of a tetrahedral covalent network lattice,
containing alternating silicon and carbon atoms. Draw a section of this lattice.
A30.
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Q31.
Look at the data for the melting temperatures of nitrogen, phosphorus, silicon and
sulfur in the table below. Which do you think exist as covalent molecular substances
and which are covalent network lattices?
Element
Nitrogen
Phosphorus
Silicon
Sulfur
Melting temperature (°C)
–210
44
1410
119
A31.
covalent molecular: nitrogen, phosphorus and sulfur; covalent network lattice: silicon
Q32.
The atoms in molecules of nitrogen, oxygen and fluorine are held together by covalent
bonds. How are the bonds in these molecules:
a similar?
b different?
A32.
a
b
The bonds are similar in that they all involve the sharing of electron pairs
between two atoms; that is, they are covalent bonds.
They differ in the number of electron pairs shared: one pair (fluorine), two pairs
(oxygen) and three pairs (nitrogen).
Q33.
The elements carbon and silicon have much in common. There are four electrons in
the outer shell in each of the atoms. Both form a dioxide. But, although carbon
dioxide is a gas at room temperature, silicon dioxide has the very high melting
temperature of 1700°C. Explain this significant difference in terms of their structure.
A33.
The atoms in CO2 and SiO2 are held together by covalent bonds. In the case of CO2,
this bonding is between atoms to form discrete molecules. The relatively low boiling
temperature of CO2 is a reflection of the weak dispersion forces between these CO2
molecules. In SiO2, however, the bonding is between silicon and oxygen atoms in a
covalent network lattice. There are strong covalent bonds holding every atom to
others in the lattice and, hence, a large amount of energy is needed to disrupt this
lattice. SiO2 has much higher melting and boiling temperatures than CO2.
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Q34.
Experimental evidence shows that the double bond between the two oxygen atoms in
O2 is much stronger than a single bond between two oxygen atoms in a compound
such as hydrogen peroxide (H2O2).
a Draw electron dot diagrams and structural formulas for O2 and H2O2.
b Explain why the oxygen double bond is stronger than the oxygen single bond.
c Why does oxygen not form a triple bond or three single covalent bonds?
A34.
a
b
c
In a single bond there are two electrons that are shared between the atoms,
whereas in a double bond there are four electrons. The net attraction is stronger
and more energy is needed to break the double bond than the single bond.
Oxygen has six outer-shell electrons and completes its outer shell by forming two
single covalent bonds or a double bond with another non-metal. There is no need
for the oxygen to form a triple bond or three single covalent bonds as this exceeds
its requirement of two electrons to complete its outer shell.
Q35.
Consider the following list of molecules:
N2, Cl2, O2, NH3, HCl, CH4, H2O, CO2, CCl4, CHCl3.
a Draw a valence structure for each of the following from the list, showing bonding
and non-bonding electron pairs:
i A molecule that contains one triple bond.
ii A molecule that contains one double bond.
iii A molecule that contains two double bonds.
b From the list, which substances contain:
i polar molecules?
ii symmetrical molecules?
iii molecules with hydrogen bonding between them?
A35.
a
i
N2
ii
O2
–NN–
iii CO2
b
i NH3, HCl, H2O, CHCl3
ii N2, Cl2, O2, CH4, CO2, CCl4
iii NH3, H2O
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Q36.
Water is a polar molecule. Explain how this fact shows that water is not a linear
molecule.
A36.
If water was a linear molecule, the polarity of the two O–H bonds would cancel each
other out and make the molecule non-polar. As water is polar, it cannot be a linear
molecule.
Q37.
Draw your own bonding concept map using the terms listed as a guide:
covalent bond, lattice, molecule, intermolecular bond, non-metal atom, electron,
melting temperature, electron, charge, shell, stable configuration, nucleus
A37.
An example of a concept map using these terms is shown.
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