Download CHEM1002 Worksheet 13 – Answers to Critical Thinking Questions

CHEM1002 Worksheet 13 – Answers to Critical Thinking Questions
The worksheets are available in the tutorials and form an integral part of the learning outcomes and experience
for this unit.
Model 1: Coordination Compounds and Complex Ions
1.
See table below.
Model 2: The oxidation number of the transition metal in a coordination compound
1.
See table below.
Coordination compound
Species formed upon
dissolving in water
Oxidation
number
[Co(NH3)6]Cl3
[Co(NH3)6]3+(aq)
+ 3Cl-(aq)
+3
Y
Na2[CdCl4]
2Na+(aq)
+ [CdCl4]2-(aq)
+2
N
[PdCl2(OH2)2]
[PdCl2(OH2)2]
+2
Y
Na[MnO4]
Na+(aq)
+ [MnO4(aq)]-(aq)
+7
N
[Ru(NH3)5(OH2)](NO3)2
[Ru(NH3)5(OH2)]2+(aq)
+ 2NO3-(aq)
+2
Y
[Cr(en)3]Cl3
[Cr(en)3]3+(aq)
+ 3Cl-(aq)
+3
Y
[Cr(NH3)5Cl]Cl2
[Cr(NH3)5Cl]2+(aq)
+ 2Cl-(aq)
+3
Y
Cs2[Zn(OH)4]
2Cs+(aq)
+ [Zn(OH)4]2-(aq)
+2
N
[Ni(en)2(OH2)2]Br2
[Ni(en)2(OH2)2]2+(aq)
+ 2Br-(aq)
+2
Y
en = ethylenediamine = NH2CH2CH2NH2
d electron configuration
Paramagnetic?
(Y/N)
Revision
Revision
1.
D.
HNO3 is the only strong acid. For this acid, [H3O+] = 0.001 M and so pH = -log[H3O+] = -log(0.001) = 3.0.
The other acids are weak acids and so [H3O+] would need to be calculated and would be lower than the
concentration of the acid.
2.
A.
The indicator should change colour as near as possible to the equivalence point. Indicators change colour at a
pH close to their pKa value. From the titration curve, the equivalence point is 8.5 so the indicator of choice
would be phenolphthalein with pKa = 9.6.
3.
E.
The solution is initially basic and the pH lowers during the titration. The titration involves addition of acid to
a base.
At the equivalence point (the mid point of the steep part of the curve), the pH is ~5.5. This indicates that a
weak acid is present at this point. This weak acid is the conjugate acid of the weak base that was originally
present.
4.
C.
From the titration curve, 50 mL of acid is needed to reach the equivalence point (when all of the base initially
present has reacted). When 25 mL of acid has been added, ½ of the weak base will have reacted to produce
an equal amount of its conjugate base. When [base] = [acid], the Henderson-Hasellbalch equation gives:
pH = pKa + log[base]/[acid] = pKa + log(1) = pKa
5.
B.
HNO3 is a strong acid. The remaining acids are weak acid. A higher value of Ka indicates more dissociation
of the acid in solution and so a stronger acid. pKa = -logKa so a lower value of pKa indicates a stronger acid.
6.
A.
Conjugate acid-base pairs differ by one H+. The conjugate base of HSO3- is therefore SO32-.
7.
A.
It becomes increasingly difficult to remove a H+ from a polyprotic acid. Ka1 corresponds to removing a H+
from H3PO4. Ka2 corresponds to removing a H+ from H2PO4-. Ka3 corresponds to removing a H+ from HPO4-.
Hence, Ka1 > Ka2 > Ka3.
8.
C.
Electronegativity increases across each row of the Periodic Table due to the increase in the nuclear charge.
Hence, N < O < F.
Electronegativity decreases down each row of the Periodic Table due to the increase in the size of the atom.
Hence, As < P < N.
9.
C.
Due to the increase in the shell number, the valence electrons must orbit further from the nucleus and so
atomic radii increase down each group. Due to the increase in the nuclear charge, electrons are held
increasing tightly and atomic radii decrease across a period.
10.
D.
Face-centred cubic is a close packed arrangement. It has the most dense packing possible with 74% packing
efficiency. Body centred cubic is not close packed but is more densely packed, with 68% packing efficiency,
than the simple cubic arrangement which has only 52% packing efficiency.
11.
B.
A body centred cubic unit cell has 1 unshared atom at the centre of the cube and 8 atoms on the corners
which are shared by 8 cells and so contribute ⅛.
12.
E.
Liquid to gas is vaporisation.
13.
C.
Molecules have more freedom to move and disperse energy when in the liquid phase compared to the solid.
Entropy is a measure of the dispersal of energy.
14.
E.
Point A is in the solid region and point B is in the liquid
region of the phase diagram. Moving from A to B thus results
in a change from solid to liquid.
Point C is on the line between liquid and gas. The line
represents all of the temperatures and pressures at which these
two phase co-exist.
Pressure
At point D, the substance co-exists as a solid, liquid and gas.
This is the triple point. The critical point is the highest
temperature at which the liquid phase can exist.
solid
liquid
B
A
C
D
gas
Temperature
15.
C.
The critical point is the highest temperature at which the liquid can exist. Thus, CH4 cannot be liquefied
above -82 °C. As the critical points of CH3Cl and SO2 are above 25 °C, they can be liquefied at this
temperature by applying pressure.
16.
E.
Entropy is a measure of the dispersal of energy and usually corresponds to the amount of disorder. The
entropy of a chemical system will usually increase when:
•
a molecule is broken down into two or more smaller fragments, as the energy is dispersed in more
fragments,
•
a solid changes to a liquid, as molecules have more freedom to move in the liquid phase, and
•
a liquid changes into a gas, as molecules have complete freedom to move in the gas phase.
Revision
17.
B.
Coordination isomerism involves swapping a coordinated ligand for a counter ion. trans-[Cr(OH2)4BrCl]Cl is
a coordination isomer of trans-[Cr(OH2)4Cl2]Br have the same formula but only differ in this way.
cis-[Cr(OH2)4Cl2]Br is a geometrical isomer of trans-[Cr(OH2)4Cl2]Br as it differs in the arrangement of the
bonds in space.
trans-[Cr(OH2)4Br2]Cl and trans-[CrBr2(OH2)4]Cl are not isomers of trans-[Cr(OH2)4Cl2]Br as they have
different chemical formula.
trans-[CrCl2(OH2)4]Br is identical to trans-[Cr(OH2)4Cl2]Br. Only the order that the ligands have been
written in the complex ion has been changed.
18.
B.
Octahedral complexes have a coordination numbers of 6. Square planar and tetrahedral complexes both have
coordination numbers of 4.
19.
E.
Fe is in Group 8: it has 8 valence electrons. In the transition metal cations, any remaining electrons occupy
the 3d orbitals and the 4s are empty. As Fe3+ has lost 3 electrons, it has 8 – 3 = 5 valence electrons remaining
and so its configuration is [Ar] 4s0 3d5.
20.
D.
For PbI2(s), Ksp = [Pb2+(aq)][I-(aq)]2 or [Pb2+(aq)] = Ksp / [I-(aq)]2. Ksp is an equilibrium constant: its value
does not change.
If the concentration of I–(aq) ions is doubled, then [Pb2+(aq)] must decrease by a factor of 4.