CHEM1405 Answers to Problem Sheet 3
1.
The ideal gas law relates the pressure, P, and volume, V, to the temperature, T, and the
number of moles, n, of gas:
𝑷𝑽 = 𝒏𝑹𝑻
where the proportionality constant, R, is known as the gas constant.
With the balloon on the ground, the air inside a hot air balloon is heated and its
temperature increases. The pressure is constant (equal to the atmospheric pressure on
the day) as is the amount of gas trapped in the balloon. Hence, if T increases, the
volume must also increase.
If V increases, the density (m/V) must decrease. The hot air has a lower density and
hence it is more buoyant the balloon rises.
As the balloon rises, the pressure outside decreases and the air outside the balloon
becomes less dense. Eventually, the density inside and outside are the same and the
balloon stops rising unless the burner is turned back on.
2.
The number of moles of oxygen can be increased by:
(i)
(ii)
increasing the total pressure, or
increasing the proportion of the air that is oxygen (the partial pressure of
oxygen, 𝒑𝐎𝟐 .
As it is convenient to carry out surgical procedures at atmospheric pressure, the latter
approach is preferred. Oxygen-rich air is used.
3.
Elements with high electron affinity have a tendency to gain electrons and thus often
form ionic compounds in which they are anions. Examples include fluorine (forming
F-) and chlorine (forming Cl-).
Adding more than one electron to an element is always unfavourable but, if the energy
required is not too big, the gain in ionic bonding may be sufficient for this to occur.
Thus, oxygen forms O2- and nitrogen forms N3- in ionic compounds.
Elements with low ionization energies have a tendency to lose electrons and thus often
form ionic compounds in which they are cations. Examples include lithium (forming
Li-) and sodium (forming Na+). It takes more energy to remove more electrons but, if
the energy required is not too big, the gain in ionic bonding may be sufficient for this to
occur. Thus, magnesium forms Mg2+ and aluminium forms Al3+ in ionic compounds.
Elements with high electron affinity also tend to have high ionization energies. Thus,
when two such elements react with each other, they tend not to form ionic compounds
as these require a cation and an anion. Rather, covalent molecules are formed in which
electrons are shared rather than transferred.
Elements with low electron affinities also tend to have low ionization energies. Thus
when two such elements react with each other, they also tend not to form ionic
compounds. Instead, they form metals.
4.
(a)
(b)
(c)
(d)
(e)
sodium fluoride, NaF - ionic
calcium carbonate, CaCO3 - ionic
ethane, C2H6 – covalent (molecular)
diamond – covalent (network)
bismuth - metallic
5.
(a)
(b)
(c)
(d)
(e)
soluble in water – ionic bonding: NaF (CaCO3 has very low solubility in pure water)
conducts electricity in the solid state – metallic bonding: bismuth
conducts electricity when melted – metallic and ionic bonding: bismuth, NaF and CaCO3
soluble in carbon tetrachloride – covalent (molecular) bonding: ethane
high melting point – ionic, metallic and covalent (network) bonding: NaF,
CaCO3, diamond and bismuth
6.
(a)
carbonate ion, CO32–
O
O
C
O
O
C
O
O
C
O
O
(b)
benzene, C6H6
(c)
ethanol, CH3CH2OH. Two lone pairs on the oxygen atom
H
(d)
H
H
C
C
H
H
O
H
phosphoric acid, H3PO4, i.e. PO(OH)3
H
H
O
O
H
O
P
O
H
O
P
O
O
H
H
O
O
(e)
propylamine, CH3CH2CH2NH2. There is one lone pair on nitrogen – amines are
weaker bases than ammonia, NH3. Propylamine is a primary amine (one carbon
group on the nitrogen).
H
H
H
H
H
C
C
C
N
H
H
H
H
7.
Ion or molecule
Structure
No. of lone
pairs on the
central atom
No. of valence
electrons on central
atom
0
4 from C
4 × 1 from H
H
Methane
CH4
H
C
H
H
No. of 
bonds
non-polar
0
polar
0
polar
0
8
H
amide ion
NH2
Polar or
non-polar
N
H
2
5 from N
2 × 1 from H
1 from charge
8
ammonia
NH3
H
N
H
1
5 from N
3 × 1 from H
8
H
S
O
sulfur dioxide
SO2
O
S
O
O
S
O
O
2
1
6 from S
2 × 2 from O
10
polar
1
1
4 from C
2 from O
2 × 1 from H
O
formaldehyde
H2CO
0
C
H
hydrogen
cyanide
HCN
H
H
C
polar
1
polar
2
8
4 from C
1 from H
3 from N
0
N
8
8.
~109.5°
sp3
H
O
H
C
H
O
C
C
C
~120°
sp2
C
H
C
H
H
H
~109.5° ~109.5°
sp3
sp3
C
C
N
O
H
H
H
CH3