SCH4U: Exam Review - Answers to non-calculation/non-text problems
Unit 2: Chemical Equilibrium
a)
1a)
Solubility Equilibrium
least soluble
soluble
AgI
most
Ag2CrO4
1b)
AgI(s) �Ag+(aq) + I-(aq)
b)
3a)
Acid-Base Equilibrium
Neutral
Ag2CO3
Acidic
KNO3
AgCl
AgNO3
Basic
N2H5Cl
NH4I
LiC6H5COO
LiCN
3b)
Acidic
1+
� N H (aq) + H O1+(aq)
N2H1+
2 4
5 (aq) + H2O(l)
1+ 3
NH4 (aq) + H2O(l) � NH3(aq) + H3O (aq)
Basic
C6H5COO1-(aq) + H O(l) � C 6H 5COOH(aq)
CN1-(aq) + H2O(l)
HCN(aq) + OH1-(aq)
+ OH1-(aq)
Unit 3: Atomic Structure and Classification of Solids
1)
Bohr
introduced the idea of electrons with quantized energy resulting in
energy levels of fixed energy
DeBroglie
wave/particle duality - electrons can be considered as both having
particle and wave like properties
Heisenberg
uncertainty principle - the exact location of an electron cannot be
known with any degree of certainty, lead to the concept of orbitals
Schrodinger
developed wave equation that can be used to predict regions of
probability of finding electrons
Pauli
Pauli exclusion principle - only two spin paired electrons can occupy
the same orbital
2a)
2b)
2c)
An orbital is a region or volume of space around an atomic nucleus in which
there is a 95% chance of finding an electron.
Maximum # electrons = 2n2
n=2
2 sublevels: 2s and 2p
2s = one spherical shaped “s” orbital
2p = three dumbbell shaped “p” orbitals
along x,y and z axes
n=4
4 sublevels: 4s, 4p, 4d, 4f
4s = one spherical shaped “s” orbital
4p = three dumbbell shaped “p” orbitals
along x,y and z axes
4d = five “d” orbitals
4f = seven “f” orbitals
-23a)
4)
3b)
27
Co 1s2 2s2 2p6 3s2 3p6 4s2 3d7
n
l
ml
ms
1s
1
0
0
±½
2s
2
0
0
±½
2p
2
1
-1
+½
2
1
0
+½
2
1
+1
+½
Fe2+ 1s2 2s2 2p6 3s2 3p6 3d6
5)
Compound
General formula
Shape
Polarity
PCl3
AX3E
trig. Pyramidal
polar
SiF4
AX4
tetrahedral
non-polar
CO32-
AX3
planar triangle
N/A
H2S
AX2E2
bent
slightly polar
XeCl4
AX4E2
square planar
non-polar
PO43-
AX4
tetrahedral
N/A
O3
AX2E
bent
non-polar
CO
AX
linear
polar
NO31-
AX3
planar triangle
N/A
SF4
AX4E
see-saw
polar
6)
Fluoromethane is a polar molecular solid while methane is a non-polar molecular
solid. The DDF between CH3F molecules will be much stronger than the LDF
between CH4 molecules. As a result, CH3F will have a higher BP than methane.
7)
Iodine is a non-polar molecule and will therefore be more soluble in a non-polar
solvent like tetrachloromethane rather than the very polar water.
8)
ethanal: 3
o
a,b,d) C1 = sp hybridization, tetrahedral arrangement, 109.5
C2 = sp2 hybridization, planar triangular arrangement, 120o
c)
[C1(2sp3 ) -ó- H(1s)] x3
C1(2sp3 ) -ó- C2(2sp 2)
C2(2sp2) -ó- H(1s)
C2(2sp2) -ó- O(2p)
C2(2p) -ð- O(2p)
-39)
Substance
type of solid
identity
A
molecular (polar)
methanol
B
network
quartz
C
molecular (non-polar)
sulphur
D
metallic
aluminum
E
ionic
potassium chloride
F
molecular (non-polar)
methane
G
ionic
calcium oxide
Unit 4: Organic Chemistry
1)
2)
a) 3-ethyl-4methyl-2-hexene
c) 3,4,5-trimethylheptane
e) isopropylcyclopentane
g) 2,4-cyclopentadiene-1-ol
i) 1,3-dichlorobenzene (m-dichlorbenzene)
k) 1-ethoxypropane
m) 2-hexanone
o) ethylbenzoate
q) N-ethyl-N,2,4-trimethyl-3-pentanamine
b) 3,5-dimethylcyclohexene
d) 3-methyl-1,3-pentadiene
f) 3-methylbutanal
h) cyclohexanone
j) 2,3-butanediol
l) methylpropanoate
n) butanoic acid
p) N,3-dimethyl-2-butanamine
r) N-isopropylbutanamide
-43)
a) 3-methyl-1-butyne + 2Cl2 = 1,1,2,2-tetrachloro-3-methylbutane
b) methane + excess O2
= CO2 + 2H2O
c) 2-methyl-3-pentanol + KMnO4 = 2-methyl-3-pentanone
d) methylpropene + HBr = 2-bromo-methylpropane
e) 2-methylbutanal + KMnO 4 = 2-methylbutanoic acid
f) cyclopentene + H2O (H2SO 4 catalyst) = cyclopentanol
g) dimethyl-1,3-propandiol + KMnO 4 = dimethylpropanedial = dimethylpropanedioic acid
h) ethanoic acid + phenol = phenylethanoate + water
a)
c)
d)
e)
f)
g)
h)
4)
propanone
2-propanol
2-chloropropane
propanoic acid
propane
dipole-dipole
H-bonding
dipole-dipole
H-bonding
LDF
highly soluble
highly soluble
slightly soluble
highly soluble
insoluble
lowest BP
highest BP
propane < 2-chloropropane < propanone < 2-propanol < propanoic acid
-5Unit 5 - Electrochemistry
1)
red ½ rxn: Cl2 + 2e1- = 2Cl1-
Er = 1.36 V
ox ½ rxn: Zn = Zn2+ + 2e1cell:
Cl2 + Zn = Zn2+ + 2Cl1-
Eox = 0.76 V
red ½ rxn: Pb2+ + 2e1- = Pb
Er = - 0.13 V
ox ½ rxn: 2Na = 2Na1+ + 2e1-
Eox = 2.71 V
cell:
Ecell = 2.12 V
Pb2+ + 2Na = 2Na1+ + Pb
Ecell = 2.58 V
red ½ rxn: Au3+ + 3e1- = Au
Er = 1.50 V
ox ½ rxn: 3Ag = 3Ag1+ + 3e1-
Eox = - 0.80 V
cell:
Au3+ + 3Ag = Au + 3Ag1+
Ecell = 0.70 V
2)
red ½ rxn: 2Ag1+ + 2e1- = 2Ag E = 0.80 V
r
2+
1E
= 2.37 V
=
Mg
+
2e
ox ½ rxn: Mg
=
1+
2+0x
cell:
2Ag + Mg
2Ag + Mg Ecell = 3.17 V
3)
a)
b)
c)
1+
12+
�
5H2C2O4 + 6H + 2MnO4
10CO2 + 2Mn + 8H2O
� 2CrO 2- + 3Br1- + 5H O + 2OH1312Cr(OH)6 + 3BrO
4
2
� 6K O + N
2KNO3 + 10K
2
2