Worked solutions to textbook questions 17
Chapter 7 Covalent molecules, networks and layers
E1.
a Give the electronic configuration for an atom of beryllium.
b How many electrons are in the outer shell of an atom of beryllium in the molecule BeH2?
AE1.
a 1s22s2
b 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)
A1.
/ hydrogen fluoride (HF)
water (H2O)
/ tetrachloromethane (CC14)
phosphine (PH3)
/ butane (C4H10)
carbon dioxide (CO2)
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 1
b 2
c 3
d 4
e 1
f 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.
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 CCl4
b NBr3
c SiO2
d HF
e 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 0
b 1
c 0
d 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
c
d
e
Q8.
Describe the shape of each of the molecules given in Question 7.
A8.
a V-shaped
b linear
c tetrahedral
d pyramidal
e 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 O
b C
c N
d N
e F
f F
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 P–F
b 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 NH3
b CHCl3
c CH3Cl
d F2O
e HBr
f H2S
g HF
h H2O
i H2
A13.
i a to h
ii a, g, h
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
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 A lot of energy is required to disrupt the strong covalent bonding in three dimensions in diamond.
ii 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.
b i Diamond is hard because it has strong covalent bonds throughout the lattice, with all atoms being held in fixed positions.
ii 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.
c i Diamond is a non-conductor of electricity because all of its electrons are localised in covalent bonds and are not free to move.
ii Graphite is able to conduct electricity because it has delocalised electrons between its layers of carbon atoms.
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 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.
b 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
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 3 bonding pairs, 2 non-bonding pairs
b 4 bonding pairs, 9 non-bonding pairs
c 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.
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 / fQ22.
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.
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 SO3 i non-polar; ii dispersion forces
b SiCl4 i non-polar; ii dispersion forces
c CF4 i non-polar; ii dispersion forces
d NF3 i polar; ii dipole–dipole attraction
e CH3NH2 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 andCO2 are both non-polar, and so intermolecular forces operating between these molecules will be dispersion forces.
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 / Melting temperature (°C)Fluorine (F2) / –220
Chlorine (Cl2) / –101
Bromine (Br2) / –7
Iodine (I2) / 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.)
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 CCl4
b 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 dispersion forces
b dispersion forces, dipole–dipole attractions
c dispersion forces, hydrogen bonding
d dispersion forces
e dispersion forces, hydrogen bonding
f 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.
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 / Melting temperature (°C)Nitrogen / –210
Phosphorus / 44
Silicon / 1410
Sulfur / 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?