Worked Solutions
Chapter 6
Question 5
Write the formula of each of the following compounds: (a) sodium chloride (b) potassium
oxide (c) magnesium bromide (d) calcium oxide (e) aluminium oxide.
Answer:
a) The electronic structure of sodium is: Na 2, 8, 1. The electronic structure
of chlorine is: Cl 2, 8, 7 To obey the octet rule and end up with eight
electrons in the outer level, sodium loses one electron to form the sodium
ion, Na+, and chlorine gains one electron to form the chloride ion, Cl-. So
that the total charge is zero, one sodium ion and one chloride ion are
required. Therefore the formula of sodium chloride is NaCl.
b) The electronic structure of potassium is: K 2, 8, 8, 1. The electronic
structure of oxygen is: O 2, 6. To obey the octet rule and end up with
eight electrons in the outer level, potassium loses one electron to form the
potassium ion K+, and oxygen gains two electrons to form the oxide ion,
O2-. So that the total charge is zero, two potassium ions and one oxide ion
are required. Therefore the formula of potassium oxide is K2O.
c) The electronic structure of magnesium is: Mg 2, 8, 2. The electronic
structure of bromine is: Br 2, 8, 18, 7. To obey the octet rule and end up
with eight electrons in the outer level, magnesium loses two electrons to
form the magnesium ion Mg2+, and bromine gains one electron to form
the bromide ion, Br-. So that the total charge is zero, one magnesium ion
and two bromide ions are required. Therefore the formula of magnesium
bromide is MgBr2.
d) The electronic structure of calcium is: Ca 2, 8, 8, 2. The electronic
structure of oxygen is: O 2, 6. To obey the octet rule and end up with
eight electrons in the outer level, calcium loses two electrons to form the
calcium ion Ca++, and oxygen gains two electrons to form the oxide ion,
O2-. So that the total charge is zero, one calcium ion and one oxide ion
are required. Therefore the formula of calcium oxide is CaO.
e) The electronic structure of aluminium is: Al 2, 8, 3. The electronic
structure of oxygen is: O 2, 6. To obey the octet rule and end up with
eight electrons in the outer level, aluminium loses three electrons to form
the calcium ion Al3+, and oxygen gains two electrons to form the oxide
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ion, O2-. So that the total charge is zero, two aluminium ions and three
oxide ions are needed. Therefore the formula of aluminium oxide is
Al2O3.
Question 6
Write the formula of each of the following compounds: (a) aluminium hydroxide (b) calcium
carbonate (c) magnesium nitrate (d) sodium hydrogencarbonate (e) potassium sulfite
(f) aluminium sulfate.
Answer:
a) The electronic structure of aluminium is: Al 2, 8, 3. The aluminium ion is
Al3+ and the hydroxide ion is OH-. One aluminium ion and three
hydroxide ions are required so that the total charge is zero. Therefore the
formula of aluminium hydroxide is Al(OH)3.
b) The electronic structure of calcium is: Ca 2, 8, 8, 2. The calcium ion is
Ca2+ and the carbonate ion is CO32-. One calcium ion and one carbonate
ion are required so that the total charge is zero. Therefore the formula of
calcium carbonate is CaCO3.
c) The electronic structure of magnesium is: Mg 2, 8, 2. The magnesium ion
is Mg2+ and the nitrate ion is NO3-. One magnesium ion and two nitrate
ions are required so that the total charge is zero. Therefore the formula of
magnesium nitrate is Mg(NO3)2.
d) The electronic structure of sodium is: Na 2, 8, 1. The sodium ion is Na+
and the hydrogencarbonate ion is HCO3-. One sodium ion and one
hydrogencarbonate ion are required so that the total charge is zero.
Therefore the formula of sodium hydrogencarbonate is NaHCO3.
e) The electronic structure of potassium is: K 2, 8, 8, 1. The potassium ion
is K+ and the sulfite ion is SO32-. Two potassium ion and one sulfite ion
are required so that the total charge is zero. Therefore the formula of
potassium sulfite is K2SO3.
f) The electronic structure of aluminium is: Al 2, 8, 3. The aluminium ion is
Al3+ and the sulfate ion is SO42-. Two aluminium ions and three sulfate
ions are required so that the total charge is zero. Therefore the formula of
aluminium sulphate is Al2(SO4)3.
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Question 7
What is the valency of each of the following elements: sodium, hydrogen, lithium, fluorine,
magnesium, sulfur, iodine, aluminium, bromine, oxygen, potassium, nitrogen, arsenic, boron,
silicon.
Answer:
Element
Valency Element
Valency Element
Valency
Sodium
1
Sulfur
2
Potassium
1
Hydrogen
1
Iodine
1
Nitrogen
3
Lithium
1
Aluminium
3
Arsenic
3
Fluorine
1
Bromine
1
Boron
3
Magnesium
2
Oxygen
2
Silicon
4
Question 8
Using valencies, write the formula of each of the following compounds: (a) beryllium
chloride (b) hydrogen fluoride (c) calcium chloride (d) hydrogen sulfide.
Answer:
a) Beryllium has a valency of 2, while chlorine has a valency of 1. A
beryllium atom will therefore form two bonds, and chlorine will form one
bond. Two chlorine atoms are therefore required to bond with one atom
of beryllium. Thus the formula of beryllium chloride is BeCl2.
b) Hydrogen and fluorine each has a valency of 1, so each will form one
bond. One hydrogen atom is therefore required to bond with one atom of
fluorine. Thus the formula of hydrogen fluoride is HF.
c) Calcium has a valency of 2, while chlorine has a valency of 1. A calcium
atom will therefore form two bonds, and chlorine will form one bond.
Two chlorine atoms are therefore required to bond with one atom of
calcium. Thus the formula of calcium chloride is CaCl2.
d) Sulfur has a valency of 2, while hydrogen has a valency of 1. A sulfur
atom will therefore form two bonds, and hydrogen will form one bond.
Two hydrogen atoms are therefore required to bond with one atom of
sulfur. Thus the formula of hydrogen sulfide is H2S.
Question 42
Use electronegativity values to predict the type of bonding, whether non-polar covalent, polar
covalent or ionic in (a) CH4 (b) SiO2 (c) KI (d) HF (e) HBr (f) LiCl (g) PH3 (h) H2O (i) MgO
(j) SO2.
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a) The electronegativity difference between carbon and hydrogen is 2.5 –
2.1 = 0.4, which is < 1.7 but > 0. The bonding in CH4 is predicted to be
polar covalent.
b) The electronegativity difference between oxygen and silicon is 3.5 – 1.8
= 1.7. The bonding in SiO2 is predicted to be strongly polar covalent.
c) The electronegativity difference between potassium and iodine is 2.7 –
0.8 = 1.9, which is > 1.7. The bonding in KI is predicted to be ionic.
d) The electronegativity difference between hydrogen and fluorine is 4.0 –
2.1 = 1.9, which is > 1.7. The bonding in HF is predicted to be ionic.
e) The electronegativity difference between hydrogen and bromine is 2.8 –
2.1 = 0.7, which is < 1.7 but > 0. The bonding in HBr is predicted to be
polar covalent.
f) The electronegativity difference between lithium and chlorine is 3.0 – 1.0
= 2.0, which is > 1.7. The bonding in LiCl is predicted to be ionic.
g) The electronegativity difference between phosphorous and hydrogen is
2.1 – 2.1 = 0. The bonding in PH3 is predicted to be non-polar covalent.
h) The electronegativity difference between hydrogen and oxygen is 3.5 –
2.1 = 1.4, which is < 1.7 but > 0. The bonding in H2O is predicted to be
polar covalent.
i)
The electronegativity difference between magnesium and oxygen 3.5 –
1.2 = 2.3, which is > 1.7. The bonding in MgO is predicted to be ionic.
j)
The electronegativity difference between sulfur and oxygen is 3.5 – 2.5 =
1.0, which is < 1.7 but > 0. The bonding in SO2 is predicted to be polar
covalent.
Question 43
Write the formula of each of the following ionic compounds: (a) potassium oxide (b) lithium
fluoride (c) magnesium oxide (d) aluminium fluoride (e) calcium bromide (f) potassium
carbonate (g) aluminium hydroxide (h) sodium sulfate (i) magnesium nitrate (j) potassium
hydrogencarbonate.
Answer:
(a) The electronic structure of potassium is: K 2, 8, 8, 1. The electronic structure of
oxygen is: O 2, 6. To obey the octet rule and end up with eight electrons in the outer
level, potassium loses one electron to form the potassium ion K+, and oxygen gains
two electrons to form the oxide ion, O2-. So that the total charge is zero, two
potassium ions and one oxide ion are required. Therefore the formula of potassium
oxide is K2O.
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(b) The electronic structure of lithium is: Li 2, 1. The electronic structure of fluorine is: F
2, 7. Lithium loses one electron to form the lithium ion, Li+, and fluorine gains one
electron to form the fluoride ion, F-. So that the total charge is zero, one lithium ion
and one fluoride ion are required. Therefore the formula of lithium fluoride is LiF.
(c) The electronic structure of magnesium is: Mg 2, 8, 2. The electronic structure of
oxygen is: O 2, 6. Magnesium loses two electrons to form the calcium ion Mg2+, and
oxygen gains two electrons to form the oxide ion, O2-. So that the total charge is zero,
one magnesium ion and one oxide ion are required. Therefore the formula of
magnesium oxide is MgO.
(d) The electronic structure of aluminium is: Al 2, 8, 3. The electronic structure of
fluorine is: F 2, 7. Aluminium loses three electrons to form the aluminium ion Al3+,
and fluorine gains one electron to form the fluoride ion, F-. So that the total charge is
zero, one aluminium ion and three fluoride ions are required. Therefore the formula
of aluminium fluoride is AlF3
(e) The electronic structure of calcium is: Ca 2, 8, 8, 2. The electronic structure of
bromine is: Br 2, 8, 18, 7. Calcium loses two electrons to form the calcium ion Ca2+,
and bromine gains one electron to form the bromide ion, Br-. So that the total charge
is zero, one calcium ion and two bromide ions are required. Therefore the formula of
calcium bromide is CaBr2
(f) The electronic structure of potassium is: K 2, 8, 8, 1. The potassium ion is K+ and the
carbonate ion is CO32-. Two potassium ions and one carbonate ion are required so that
the total charge is zero. Therefore the formula of potassium carbonate is K2CO3
(g) The electronic structure of aluminium is: Al 2, 8, 3. The aluminium ion is Al3+ and
the hydroxide ion is OH-. One aluminium ion and three hydroxide ions are required
so that the total charge is zero. Therefore the formula of aluminium hydroxide is
Al(OH)3.
(h) The electronic structure of sodium is: Na 2, 8, 1. The sodium ion is Na+ and the
sulfate ion is SO42-. Two sodium ions and one sulfate ion are required so that the total
charge is zero. Therefore the formula of sodium sulfate is Na2SO4
(i) The electronic structure of magnesium is: Mg 2, 8, 2. The magnesium ion is Mg2+ and
the nitrate ion is NO3-. One magnesium ion and two nitrate ions are required so that
the total charge is zero. Therefore the formula of magnesium nitrate is Mg(NO3)2.
(j) The electronic structure of potassium is: K 2, 8, 8, 1. The potassium ion is K+ and the
hydrogencarbonate ion is HCO3-. One potassium ion and one hydrogencarbonate ion
are required so that the total charge is zero. Therefore the formula of potassium
hydrogencarbonate is KHCO3.
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Question 44
What is the valency of each of the following elements: oxygen, calcium, potassium, bromine,
iodine, aluminium, magnesium, sulfur, chlorine, sodium, hydrogen?
Answer:
Element
Valency Element
Valency Element
Valency
Oxygen
2
Iodine
1
Chlorine
1
Calcium
2
Aluminium
3
Sodium
1
Potassium
1
Magnesium
2
Hydrogen
1
Bromine
1
Sulfur
2
Question 45
Using valencies write the formulas of (a) hydrogen sulfide (b) calcium fluoride (c) hydrogen
fluoride (d) beryllium chloride (e) silicon sulfide.
Answer:
(a) Sulfur has a valency of 2, while hydrogen has a valency of 1. A sulfur atom will
therefore form two bonds, and hydrogen will form one bond. Two hydrogen atoms
are therefore required to bond with one atom of sulfur. Thus the formula of hydrogen
sulphide is H2S.
(b) Calcium has a valency of 2, while fluorine has a valency of 1. A calcium atom will
therefore form two bonds, and fluorine will form one bond. Two fluorine atoms are
therefore required to bond with one atom of calcium. Thus the formula of calcium
fluoride is CaF2.
(c) Hydrogen and fluorine each has a valency of 1, so each will form one bond. One
hydrogen atom is therefore required to bond with one atom of fluorine. Thus the
formula of hydrogen fluoride is HF.
(d) Beryllium has a valency of 2, while chlorine has a valency of 1. A beryllium atom
will therefore form two bonds, and chlorine will form one bond. Two chlorine atoms
are therefore required to bond with one atom of beryllium. Thus the formula of
beryllium chloride is BeCl2.
(e) Silicon has a valency of 4, while sulfur has a valency of 2. In this case, the 4:2 ratio
of valencies is reduced to the simpler 2:1 ratio. A silicon atom will therefore form
two bonds, and sulfur will form one bond. Two sulfur atoms are therefore required to
bond with one atom of silicon. Thus the formula of silicon sulfide is SiS2.
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