Chemistry 30 Review and Answers
1.
Lab Safety.
a)
What is the WHMIS symbol for a flammable substance?
b)
What is the household symbol which means "the contents of the container are flammable"?
c)
What is the symbol for a radioactive substance?
d)
What are the three WHMIS symbols which warn of poisons, and what is the difference between
them?
Class D, Division 1 - Poisonous and infectious material. Acute toxin; immediate and serious
toxic effects.
Class D, Division 2 - Poisonous and infectious material. Chronic toxin; other toxic effects;
may cause cancer or other diseases because of long term exposure.
Class D, Division 3 - Poisonous and infectious material. Biohazard; may cause infections and
other effects
2.
Significant Digits
a)
Perform the following additions and subtractions:
i)
ii)
iii)
iv)
b)
233 g / 15.36 g/mol
1.59 mol / 0.12 L
13.1 mol x 44.01 g/mol
22 L x 2.145 mol/L
15.2 mol
13 mol/L
577 g
47 mol
Nomenclature
a)
Give the formulas for the following substances:
i)
ii)
iii)
iv)
v)
vi)
vii)
viii)
ix)
x)
b)
magnesium hydroxide
sodium hydrogen carbonate
aluminum oxide
calcium oxide hexahydrate
phosphorus pentaoxide
iron (III) phosphate
hexane
ethanol
2-methyl-2-pentene
benzene
Mg(OH)2
NaHCO3
Al2O3
CaO ▪ 6 H2O
PO5
FePO4
C6H14
C2H5OH
C6H12
C6H6
Give the name for the following compounds:
i) NaCl
ii) NH3
iii) CaSO4
iv) N2O4
v) C2H2
vi) Ca(OH)2 ·5H2O
vii) MnO2
viii) CO2
ix) NH4NO3
x) C8H18
4.
41.4 g
100.1 L
2300 L
0.91 mol
Perform the following multiplications and divisions
i)
ii)
iii)
iv)
3.
22.26 g + 19.1 g
100.1 L + 0.025 L
2500 L - 155 L
1.236 mol - 0.33 mol
sodium chloride
ammonia
calcium sulfate
dinitrogen tetraoxide
ethyne
calcium hydroxide pentahydrate
manganese (IV) oxide
carbon dioxide
ammonium nitrate
octane
The Mole
a)
Determine the molar mass of
i)
ii)
iii)
iv)
v)
vi)
NaOH
Cu(NO3)2
SnCl4
C2H5OH
O2
Mg(OH)2 ▪ 9 H2O
40.00 g/mol
187.57 g/mol
260.51 g/mol
46.08 g/mol
32.00 g/mol
220.50 g/mol
2
b)
Calculate the number of moles in
(i) 65.0 grams of Ag
(ii) 423 g of N
(iii) 325 grams of Sn
c)
0.603 mol
30.2 mol
2.74 mol
Calculate the mass of
(i) 2.35 moles of Cu atoms
(ii) 0.17 moles of Mg atoms
(iii) 2.5 x 103 mol of chlorine gas
d)
149 g
4.1 g
1.8 x 105 g
Calculate the concentration of the following
(i)
(ii)
0.025 mol of NaCl in 50.0 mL of water
25.00 mol of Ca(OH)2 in 15.0 L
0.50 mol/L
1.67 mol/L
(iii)
10.00 g of Mg3(PO4)2 in 500.0 mL
262.84 g/mol
(iv)
55.00 g of C6H11O6 in 355.0 mL
0.07609 mol/L
179.17 g/mol
0.8647 mol/L
e)
Determine the following:
(i)
(ii)
(iii)
(iv)
Moles of CaCO3 in 300.0 mL of 1.5 x 10-4 mol/L solution
4.5 x 10-5 mol
Mass of O2 in 50.00 L of 1.00 x 10-2 mol/L solution
16.0 g
Volume of 1.30 mol/L solution of CuCl which contains 0.025 mol CuCl.
0.0192 L
-3
Volume of 6.50 x 10 mol/L solution of NaNO3 which contains 15.00 g NaNO3
85.00 g/mol
27.1 L
3
5.
Chemical Equations
i)
ii)
Balance the following equations. Use whole numbers only.
a)
2 Mg +
b)
2 CuS +
1 O2 →
3 O2
c)
1 (NH4)2S +
d)
1 P2O5 +
2 MgO
2
→
CuO +
2 SO2
1 Pb(NO3)2 →
1 H 2O
→
2 NH4NO3 +
2 PO2 +
1 O2 +
1 PbS
1 H2
Write and balance the following equations. Include the phase of each species.
a)
The complete combustion of butane gas, C4H10
2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O
b)
(g)
The synthesis of sulphur dioxide gas from its elements.
1 S8 (s) + 8 O2 (g) → 8 SO2 (g)
c)
Aqueous silver nitrate reacts with a solution of calcium chloride to produce aqueous
calcium nitrate and a precipitate of silver chloride.
2 AgNO3 (aq) + 1 CaCl2 (aq) → 1 Ca(NO3)2 (aq) + 2 AgCl
d)
When heated, solid calcium carbonate decomposes to form carbon dioxide gas and
solid calcium oxide.
1 CaCO3 (s) → 1 CO2 (g) + 1 CaO
6.
(s)
(s)
Solutions
a)
Discuss the principle involved in determining whether two substances will be miscible or
immiscible.
LIKE DISSOLVES LIKE; polar solvents dissolve polar and ionic solutes; non-polar
solvents dissolve non-polar solutes. Polar and non-polar substances are immiscible.
b)
Write equations for the solvation and dissociation of the following substances in water and
determine the concentration of the ions in solution:
i)
ii)
0.025 mol/L KNO3
KNO3 (s) →
K1+(aq) +
NO31-(aq)
0.025 mol/L
0.025 mol/L
0.025 mol/L
1.14 mol/L AlCl3
AlCl3 (s) →
Al3+(aq)
1.14 mol/L
+
1.14 mol/L
3 Cl1-(aq)
3.42 mol/L
4
iii)
2.25 x 10-4 mol/L Na3PO4
Na3PO4 (s) →
3 Na1+(aq)
2.25 x 10-4 mol/L
iv)
6.75 x 10-4 mol/L
5.38 mol/L Fe2(SO4)3
Fe2(SO4)3 (s) → 2 Fe3+(aq)
5.38 mol/L
c)
d)
PO43+(aq)
+
+
10.8 mol/L
2.25 x 10-4 mol/L
3 SO42-(aq)
16.1 mol/L
Calculate the following:
i)
The volume of 1.15 mol/L HCl to make 45.0 mL of 0.750 mol/L HCl.
ii)
The volume of 21 mol/L CH3COOH to make 1.00 L of 2.00 mol/L
CH3COOH.
0.095 L
0.0293 L
Write the chemical equation, the overall ionic equation and net ionic equations for the reactions
of the following solutions (make sure to check whether a precipitate will form).
i)
NaOH + HCl
NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)
Na1+(aq) + OH1-(aq) + H1+(aq) + Cl1-(aq)
H1+(aq) + OH1-(aq)
→ H2O(l) + Na1+(aq) + Cl1-(aq)
→ H2O(l)
ii)
CuSO4 + FeCl3
iii)
K2CO3 + Sr(NO3)2
No Reaction
K2CO3 (aq) + Sr(NO3)2 (aq) → SrCO3 (s) + 2 KNO3 (aq)
2 K1+(aq) + CO32-(aq) + Sr2+(aq) + 2 NO31-(aq)
Sr2+(aq) + CO32-(aq)
iv)
→ SrCO3 (s) + 2 K1+(aq) + 2 NO31-(aq)
→ SrCO3 (s)
Na2SO3 + CaBr2
Na2SO3 (aq) + CaBr2 (aq)
→ 2 NaBr(aq) + CaSO3 (s)
2 Na1+(aq) + SO32-(aq) + Ca2+(aq) + 2 Br1-(aq)
Ca2+(aq) + SO32-(aq)
→ CaSO3 (s)
5
→ CaSO3 (s) + 2 Na1+(aq) + 2 Br1-(aq)
v)
MgSO4 + K3PO4
3 MgSO4 (aq) + 2 K3PO4 (aq) → 3 K2SO4 (aq) + Mg3(PO4)2 (s)
1+
23 Mg2+(aq) + 3 SO42-(aq) + 6 K1+(aq) + 2 PO43-(aq) → Mg3(PO4)2 (s) + 6 K (aq) + 3 SO4 (aq)
3 Mg2+(aq) + 2 PO43-(aq)
e)
→ Mg3(PO4)2 (s)
Devise a procedure that will separate the following pairs of ions.
i)
OH- and Br- add anything except Ag1+, Pb2+, Hg22+, Cu1+, the alkali ions or NH41+.. It will
precipitate the
OH1- add Ag1+, Pb2+, Hg22+, Cu1+ to precipitate the Br1-
ii)
Cu2+ and Ca2+
- add SO42- to precipitate the Ca2+, leaving the Cu2+ in solution
- all S2- to precipitate the Cu2+.
- (this could also be done in reverse.)
7.
Thermodynamics
a)
Calculate the amount of heat required:
i)
ii)
iii)
iv)
v)
b)
c)
Calcu
To
To
To
To
To
heat 5.00 kg of copper from 20. °C to 400.°C.
cool 350.0 g of water from 20°C to 0°C.
boil 45.0 g of water.
freeze 1.50 kg of water.
change 875 g of ice at -25ºC to steam at 400ºC.
(45 kJ + 292 kJ + 368 kJ + 1980 kJ + 543 kJ)
741 kJ
- 29 kJ
102 kJ
- 501 kJ
3230 kJ
ΔH for the following reactions:
→ SO2 (g) + 1/2 O2 (g)
+ 98 kJ
i)
SO3 (g)
ii)
NO(g) + 1/2 O2(g)
→ NO2(g)
- 57.3 kJ
iii)
C3H8(g) + 5 O2(g)
→ 3 CO2(g) + 4 H2O(g)
- 2046 kJ
ΔS is positive or negative for each reaction in b), and indicate the
conditions under which the reaction may proceed spontaneously (Δ
- the first equation has a positive entropy and enthalpy. It will be spontaneous at high
temperature.
- the second equation has a negative entropy and enthalpy. It will be spontaneous at low
temperature.
- The third equation has a positive entropy and a positive enthalpy. It will be spontaneous at
all temperatures.
6
d)
For the reaction
C(s) + O2 (g) → CO2 (g) + 394 kJ
Indicate the effects and the reasons for the effect of:
e)
i)
Crushing the coal into a powder.
increases surface area; increases the number of collisions; increases the rate of
reaction.
ii)
Increasing the temperature.
increases the number and energy of collisions; increases the rate of reaction.
iii)
Increasing the concentration of oxygen.
increases the number of collisions; increases the rate of reaction.
iv)
Addition of a catalyst.
lowers activation energy; increases the rate of reaction.
v)
Addition of an inhibitor.
increases the activation energy; decreases the rate of reaction
What is the rate-determining step?
in a multi-step reaction it has the highest activation energy of all the steps. It
determines how fast the total reaction can go.
f)
On the grids below, plot energy diagrams and answer the questions for each of the
following cases:
i)
Potential energy of reactants: 300 kJ
Potential energy of products: 150 kJ
Potential energy of activated complex: 400 kJ
Activation energy:
100 kJ
H:
-150 kJ
Exo- or endo- thermic?
exothermic
ii)
Potential energy of reactants: 100 kJ
Potential energy of products:
150 kJ
Potential energy of activated complex: 350 kJ
Activation energy: 250 kJ
H: + 50 kJ
Exo- or endo- thermic?
endothermic
iii)
Potential energy of reactants: 100 kJ
Potential energy of products: 200 kJ
Potential energy of activated complex: 450 kJ
Activation energy: 350 kJ
H:
+ 100 kJ
Exo- or endo- thermic?
endothermic
7
If an inhibitor is added to c), use a dotted line to indicate what would be the likely
effect on the diagram. Why does this happen ?
inhibitors raise activation energy by preventing the reactants from forming the activated complex.
g)
For the following reactions:
i)
ii)
iii)
iv)
v)
determine the order for each reactant
determine the overall order of the reaction
write the rate equation
calculate the rate constant
rewrite the rate equation with the rate constant
1.
A + B  C
Experiment
Initial [A] (mol/L)
Initial [B] (mol/L)
1
2
3
0.0100
0.0200
0.0100
0.0300
0.0300
0.0600





as [A] doubles, rate increases 4 times
as [B] doubles, rate does not change
Rate = k[A]2[B]0
k= 2.4
Rate = 2.4[A]2
2.
Experiment
1
2
3
4






Initial Rate
(mol/(L·s))
2.40 x 10-4
9.60 x 10-4
2.40 x 10-4
second order
zero order
A + B + C  E + F
Initial [A]
(mol/L)
0.050
0.100
0.050
0.050
Initial [B]
(mol/L)
0.050
0.050
0.100
0.050
as [A] doubles rate is unchanged
as [B] doubles rate increases 4 times
as [C] doubles rate doubles
Rate = k[A]0[B]2[C]1
k = 6.4
Rate = 6.4[B]2[C]
Initial [C]
(mol/L)
0.050
0.050
0.050
0.100
zero order
second order
first order
8
Initial Rate
(mol/(L·s))
8.00 x 10-4
8.00 x 10-4
3.20 x 10-3
1.60 x 10-3
8.
Equilibrium.
a)
What is Le Chatalier's Principle?
If a system in equilibrium is subjected to a change; processes occur that tend to
counteract the imposed change and the system reaches a new state of equilibrium
b)
For the equilibrium reaction
2 SO2 (g) + O2 (g)
 2 SO3 (g) + 196 kJ
What is the effect of:
i)
ii)
iii)
iv)
v)
vi)
c)
d)
Increasing the temperature.
Increasing the pressure.
Increasing the volume.
Increasing the [SO2]
Increasing the [SO3]
Addition of a catalyst.
favours
favours
favours
favours
favours
favours
reactants
products
reactants
products
reactants
neither
Determine the Keq of the following reactions:
[NO2]2
[NO]2[O2]
i)
2 NO (g) + O2 (g)  2 NO2 (g)
Keq =
ii)
4 HCl (g) + O2 (g)  2 H2O(g) + 2 Cl2 (g)
Keq = [H2O]2[Cl2]2
[HCl]4[O2]
iii)
2 NOCl (g)  2 NO (g) + Cl2 (g)
[NO]2[Cl]
[NOCl]2
iv)
Fe3O4 (s) + H2 (g)  3 FeO (s) + H2O (g)
v)
Ag2S (s)  2 Ag1+(aq) + S2-(aq)
Keq =
Keq =
[H2O]
[H2]
Ksp = [Ag1+]2[S2-]
Perform the following calculations:
i)
The keq for the following reaction is 1.84 x 10-2 :
2 HI(g)  H2 (g)
+ I2 (g)
If 0.100 mol of HI is placed in a 1.00 L container and allowed to reach equilibrium, find
the concentration of each species at equilibrium.
use the ICE box to solve the problem, where the concentration of H 2 and I2 are set to x
and the concentration of HI is (0.100 mol/L - 2x). Solve for x to get the following:
[H2] and [I2] = 0.0107 mol/L
[HI] = 0.0787 mol/L
9
ii)
For the reaction
2 NO(g) + O2(g)  2 NO2(g)
Calculate the Keq if the concentration of each species at equilibrium are:
[NO] = 0.100 mol/L
[O2] = 0.0140 mol/L [NO2] = 0.100 mol/L
Keq = 71.4
iii)
Calcium carbonate, CaCO3, has a solubility in water of 7.1 x 10-5
mol/L at 20ºC. Calculate the ksp value of calcium carbonate.
Ksp = 5.0 x 10-9
iv)
Calculate the concentrations of silver ions and sulfate ions in a saturated aqueous
solution of silver sulfate, Ag2SO4 in which the ksp is 1.2 x 10-5.
[Ag1+] = 2.88 x 10-2 mol/L
9.
[SO42-] = 1.44 x 10-2 mol/L
Acids and Bases
a)
For each of the following, which one is the stronger acid?
i)
ii)
iii)
b)
c)
HIO3 or HI
HS- or H2S
Using the half-reactions reactions from appendix C, write the acid/base equations for the pairs
from question a), indicating which is the acid and which is the base on the left-hand side of the
equation.
i)
HNO3 + NO21- → NO31- + HNO2
acid
base
ii)
HI +
IO31- → I1- + HIO3
acid
base
iii)
H2S
acid
+
S2- →
base
2 HS1-
Compare and contrast between strength and concentration.
-
d)
HNO2 or HNO3
are completely independent terms
strength refers to the degree of dissociation
concentration refers to the number of moles per litre
Calculate the pH of each of the following solutions:
i)
ii)
iii)
iv)
v)
[H3O1+] = 3.0 x 10-3 mol/L
[OH1-] = 6.0 x 10-4 mol/L
[Ba(OH)2 = 0.0020 mol/L
HNO3, 250.0 mL containing 1.26 g.
pure water
10
2.52
10.78
11.60
7.00
1.097
e)
Calculate the hydronium ion and hydroxyl ion concentrations of
i)
100. mL of solution containing 0.60 g of NaOH.
[H1+] = 0.15 mol/L
[OH1-] = 6.7 x 10-14 mol/L
ii)
a blood sample with pH 7.400.
[H1+] = 3.98 x 10-8 mol/L
[OH1-] = 2.51 x 10-7 mol/L
orange juice with a pH 3.650.
[H1+] = 2.24 x 10-4 mol/L
[OH1-] = 4.47 x 10-11 mol/L
iii)
f)
Titration
10.
i)
Find the [HCl] if 25.00 mL of the acid is neutralized by 10.00 mL
of 0.150 mol/L NaOH.
CA = 0.0600 mol/L
ii)
Find the [Ca(OH)2] if 125 mL of the base is completely neutralized by 62.5 mL of HCl
with a pH of 2.60.
CB = 6.3 x 10-4 mol/L
Atomic Theory
a)
Describe the nature and role of the proton, neutron, and electron in the atom.
Proton
Neutron
Electron
Location
in the nucleus.
Cannot leave
in the nucleus.
Cannot leave
in orbit about
nucleus. Can be
added to or taken
away from atom
Charge
positive (1+)
neutral (0)
negative (1-)
Mass
large
equal to proton
insignificant
Role in the Atom
- number of protons
determines atomic
number.
- gives the identity of
the atom
- with protons
determines the mass
number of the atom;
determines the
isotope
- mediates strong
nuclear force; holds
nucleus together
- with protons
determines atomic
charge
- responsible for
chemical and physical
properties of an
element.
11
b)
Write an electron configuration, orbital representation and Lewis diagram for
Electron Configurations
(i) O
(ii) Ne
1s22s22p4
1s22s22p6
(iii) Na+
(iv) H+
1s22s22p6
(v) Ca2+
(vi) Al
(vii) S2-
1s22s22p63s23p6
1s22s22p63s23p1 (or 1s22s22p63s13p2 with hybridization)
1s22s22p63s23p6
(viii) Si
Orbital diagrams
1s0 (no electrons)
1s22s22p63s23p2 (or 1s22s22p63s13p3 with hybridization)
(i) O
1s
↑↓
2s
↑↓
2p
↑↓ ↑ ↑
(ii) Ne
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓
(iii) Na+
3s
3p
↑↓ ↑↓ ↑↓
(iv) H+
no electrons
(v) Ca2+
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓
(vi) Al
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑
↑
(vii) S2-
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑↓
↑↓ ↑↓ ↑↓
(viii) Si
↑↓
↑↓
↑↓ ↑↓ ↑↓
↑
↑
12
↑
↑
↑
the following:
Electron-dot diagrams
(i) O
(ii) Ne
(iii) Na1+
(iv) H1+
(v) Ca2+
(vi) Al
(vii) S2-
(viii) Si
c)
List the diatomic elements, as well as the two other polyatomic elements.
H2 N2 O2 F2 Cl2 Br2 I2 At2 S8 P4
d)
e)
Determine the number of protons and electrons in the following:
(i) Ca2+
(ii) Ar
(iii) P3-
20 protons, 18 electrons
18 protons, 18 electrons
(iv) Mn7+
25 protons, 18 electrons
15 protons, 18 electrons
Write the symbols for the following atoms or ions; indicate the type of ion:
i) 13 protons, 10 electrons
ii) 80 protons, 78 electrons
iii) 54 protons, 54 electrons
iv) 34 protons, 36 electrons
Al3+ cation
Hg2+ cation
Xe
atom
Se2- anion
13
f)
Determine the number of protons and neutrons in the following:
i)
238
U
92 protons, 146 neutrons
92
ii)
C
14
6 protons, 8 neutrons
6
12.
iii)
21
iv)
47
10
22
Ne
10 protons, 11 neutrons
Ti
22 protons, 25 neutrons
Chemical Bonding
a)
Draw orbital diagrams, Lewis diagrams and structural diagrams of the following molecules:
i)
H2O
Shape: Angular
Electronegativity difference: H - O bond = | 2.1 - 3.5 | = 1.4
polar covalent bond
Bond dipole goes from the H (positive end) to the O (negative end)
Because the molecule is angular the bond dipoles do not cancel; this is a polar molecule. The
oxygen end is negative and the hydrogen end is positive.
14
ii)
HF
Shape: Only two atoms (linear)
Electronegativity difference: H - F bond = | 2.1 - 4.0 | = 1.9
ionic bond
The ionic bond means there is no bond dipole. The H has a positive charge and the fluorine
has a negative charge.
iii)
NH3
Shape: Trigonal Pyramidal
Electronegativity difference: H - N bond = | 2.1 - 3.0 | = 0.9
polar covalent bond
Bond dipole goes from the H (positive end) to the N (negative end)
Because the molecule is trigonal pyramidal the bond dipoles do not cancel; this is a polar
molecule. The nitrogen end is negative and the hydrogen end is positive.
15
iv)
CH4
Shape: Tetrahedral
Electronegativity difference: H - C bond = | 2.1 - 2.5 | = 0.4
polar covalent bond
Bond dipole goes from the H (positive end) to the C (negative end)
Because the molecule is tetrahedral the bond dipoles all point toward the centre and cancel;
this is a non-polar molecule.
v)
CO2
Shape: Linear
Electronegativity difference: O - C bond = | 3.5 - 2.5 | = 1.0
polar covalent bond
Bond dipole goes from the C (positive end) to the O (negative end)
Because the molecule is linear the bond dipoles all point out from the centre and cancel; this is
a non-polar molecule.
16
vi)
C2H6
Shape: Tetrahedral about each carbon
Electronegativity difference:
H - C bond = | 2.1 - 2.5 | = 0.4
C - C bond = | 2.5 - 2.5 | = 0.0
polar covalent bond
covalent bond
Bond dipole goes from the H (positive end) to the C (negative end)
Because the molecule is tetrahedral about each carbon the bond dipoles all point toward the
centre and cancel; this is a non-polar molecule.
17
vii)
CH3Cl
Shape: Tetrahedral
Electronegativity difference:
H - C bond = | 2.1 - 2.5 | = 0.4
Cl - C bond = | 3.0 - 2.5 | = 0.5
polar covalent bond
polar covalent bond
Bond dipole goes from the H (positive end) to the C (negative end) and from the C (positive
end) to the Cl (negative end)
The molecule is tetrahedral, but the H - C bond dipoles all point toward the centre and the C Cl bond dipole points out from the centre; the bond dipoles do not cancel so this is a polar
molecule.
18
viii)
CaH2
Shape: Linear
Electronegativity difference: H - Ca bond = | 2.1 - 1.0 | = 1.1
polar covalent bond
Bond dipole goes from the Ca (positive end) to the H (negative end)
Because the molecule is linear the bond dipoles all point out from the centre and cancel; this is
a non-polar molecule.
e)
What kind of intermolecular force is involved in Au ? Why does this force occur ? What
physical properties result from this force ?
Au is a metal, so metallic bonding occurs between atoms of gold. Metallic bonding is
based on the existence of loosely held outer electrons which become delocalized; that is, they
are free to move randomly from atom to atom in the metal. These electrons result in metals
being malleable and ductile. Metals can also conduct electricity and heat. The more
delocalized the electrons the greater these properties and the lower the boiling point and
melting point (bottom right hand side of transition metals). As electrons become less
delocalized the metals become harder, more brittle and conduct less well, but their melting and
boiling points increase (top left hand side of transition metals).
f)
What kind of intermolecular force is involved in diamond ? Why does this force occur ? What
physical properties result from this force ?
Diamonds are made of pure carbon joined together in 3 dimensions with network covalent
bonds. The covalent bonds are the same that occur in molecular substances, but network
covalent molecules have upwards of 1020 atoms or more, making them crystalline
macromolecules. Because the covalent bond is so strong these molecules are very hard and
very brittle and have very high melting and boiling points. Diamond cannot conduct electricity,
but some forms of these molecules can (graphite, for instance).
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g)
Which of the following would have the highest boiling point ? Explain your answer.
i)
SeCl2 and SCl2
Both central atoms are group 16 with two chemical bonds and two lone pair electrons,
so are angular in shape. The electronegativity difference of the Se-Cl bond is 0.6 and
that of the S-Cl bond is 0.5 (making both bonds polar covalent), so SCl 2 is slightly more
polar, but not by much. Since they have similar polarity, but SeCl2 contains more
electrons (London dispersion forces are basic to all molecules), the SeCl2 likely has the
highest boiling point.
ii)
CH4 and SiH4
Both central atoms are in group 14, each bonded to 4 atoms with no lone pair
electrons, so are tetrahedral in shape. The electronegativity difference in each case
makes the bonds polar covalent, but that is not a factor because the tetrahedral shape
means the dipoles cancel out. This makes each molecule non-polar and London
dispersion is the primary intermolecular force. Since the strength of this force is based
on the number of electrons, SiH4 will have the greatest boiling point.
iii)
BeBr2 and BeCl2
Beryllium is in group two. It has two bonding electrons (because of electron
promotion) and no lone pair electrons. In each case the Be is bonded to two atoms
making a linear shape. The electronegativity difference in each case makes the bonds
polar covalent, but that is not a factor because the linear shape means the dipoles
cancel out. This makes each molecule non-polar and London dispersion is the primary
intermolecular force. Since the strength of this force is based on the number of
electrons, BeBr2 will have the greatest boiling point.
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