Inorganic Chemistry I August 2010 Answers
Answers and marking scheme
1I.1 Molecular Structure
a) With the aid of molecular orbital energy level diagrams rationalise the bond
strength values shown below. Include drawings of the relevant molecular orbitals in
your answer.
Molecule
B2
C2
N2
Bond strength
(kJ mol-1)
297
607
945
ANSWER:
Students should draw a general MO energy level diagram for diatomics where MO
mixing is significant (i.e. the one that applies to these three bond (this was discussed at
length during the lectures).
Marks given in the following way: correct orientation – i.e. if pz is used to form the
sigma bond it means that the molecule is “sitting” on this axis (1 mark); correct order
of the energy levels including the mixing (2 marks); drawings of MOs (4 marks).
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Inorganic Chemistry I August 2010 Answers
Once the electrons are placed on the right MOs, the following information regarding
bond order (and number of electrons in bonding/antibonding MOs) can be obtained:
B2
C2
N2
Valence electrons
6
8
10
Electrons in bonding MO
4
6
8
Electron in antibonding MO
2
2
2
Bond Order
1
2
3
Clearly, the strongest bond should be for N2 (consistent with the experimental values)
since is the only one in the series that has a BO of 3. This is followed by C2 (BO = 2)
and finally B2 (BO = 1). This is consistent with the experimental values provided. (3
marks)
b) Answer ALL parts of this question:
i) Using VSEPR, state the pseudostructure and structure of the following compounds.
Include a drawing of the structures in your answer.
SnCl2
[H3O]+
BBr3
ANSWER:
The predicted structures from VSEPR are:
SnCl2
BBr3
[H3O] +
Pseudostruct: Trigonal planar
Trigonal planar
Tetrahedral
Structure:
Bent
Trigonal planar
Trigonal pyramidal
(2 marks)
(2 marks)
(2 marks)
ii) Give a suitable hybridisation scheme for the central atom in each case.
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Inorganic Chemistry I August 2010 Answers
ANSWER:
SnCl2 sp2
BBr3
sp2
+
[H3O] sp3
(3 marks)
c) [PtCl4]2– and POCl3 belong to the D4h and C3v point groups respectively. Determine the
structures of these two molecules and explain your reasoning.
ANSWER:
[PtCl4] 2– is square planar
(1 mark)
POCl3 is tetrahedral
(1 mark)
Since [PtCl4] 2– belongs to the D4h point group it should have a σh mirror plane. From
the two possible geometries that this molecule could have (namely tetrahedral or
square planar) only the latter could have a σh. In addition to this, a molecule belonging
to a D4h point group should have a C4 and 4C2. This is consistent with a square planar
geometry (the C4 is perpendicular to the plane of the molecule; two C2 along the Cl-PtCl bonds; 2 C2 bisecting the bonds and parallel to D4h point group). POCl3 belongs to
the C3v point group, hence, it should have 3σv mirror planes but no σh mirror plane. In
addition, it should have a C3 axis only. This is consistent with a tetrahedral molecule in
which one of the substituents is different to the other three (the C3 axis is located along
the O-P bond; three planes parallel to C3 each containing Cl-P-O and bisecting the ClP-Cl bonds). (4 marks)
d) Answer ALL parts of this question:
i) Using VSEPR predict whether CO2 is linear or bent.
ANSWER:
CO2 is linear (there are no lone pairs around the carbon and all other electron
are involved in bonding – i.e. an AX2 type moelcule):
O
C
O
(2 Marks)
ii) What hybridization is appropriate for the C atom?
ANSWER:
The following hybridization will allow us to rationalize the structure:
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Inorganic Chemistry I August 2010 Answers
The two remaining 2p orbitals remain as such.
(2 marks)
iii) Using VB, show a bonding scheme using the hybridization you have suggested.
ANSWER:
The two sp hybrid orbitals can form a  bond with each oxygen atom (which is sp2). The
two remaining p orbitals on the carbon, can form  bonds with a p orbital from each
oxygen.
(2 marks)
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Inorganic Chemistry I August 2010 Answers
1I2. PERIODICITY AND INORGANIC REACTIVITY ANSWERS
(a) (i) Boron (1s2 2s2 2p1) has
three valence electrons and
can therefore bond to three
other atoms. The predicted
shape for this molecule
(VSEPR) is trigonal planar
with a sp2 hybridised boron
atom. This gives three B-Cl
2c-2e sigma bonds. Boron
now has 6e around it and is
therefore electron deficient
(2e short of its octet). This is
compensated for by back donation of a pair of electrons from each Cl (2pz) into the empty
pz on the boron atom. [6 Marks]
(ii)
BF3.NMe3 + BCl3  BCl3.NMe3 + BF3 [1 Mark]
BCl3.NMe3 + BF3  no reaction [1 Mark]
The boron trihalides are Lewis acids and the NMe3 is a Lewis base [1]. BCl3 is more
acidic than BF3 [1].This is because the pi-orbital overlap and hence relief of electron
deficiency is best for B-F (2p-2p) rather than B-Cl (2p-3p) due to the size/energy
mismatch and longer bond for B-Cl [1].
(b)
(i) H2O expected to be bent and based on tetrahedral with two lone pairs [1] hence X
would be sp3 hybridised [1] H2O slightly off ideal sp3 angle due to lp-lp repulsions >
bp-bp repulsions [1] H2S close to 90 deg since S not hybridised much at all [1]– little or
no contribution from s orbital –[1] therefore H 1s overlaps with S 3p orbitals [1] which
are orientated at 90 deg to each other [1].
(ii) SF6 < SeF6 < TeF6. [2] Each are octahedral compounds [1] SF6 is kinetically inert.
The octahedral arrangement of the 6F around S makes it very difficult for reagents to
attack the S atom [2]. TeF6 and SeF6 are octahedral molecules like SF6 but the group 16
elements are descended they get larger and therefore more accessible to attack [2].
(iii) M + H2O -> M(OH)2 + H2 [1]
Expect reactivity with water to increase down the group (as with Group 1) [1]. Mg is
passivated and kinetically inert at ambient temperature due to ready formation of
protective oxide coating on the metal surface [2]
2Mg + O2 -> 2MgO [1]
Be is even less reactive with water than Mg [1] again because of its protective oxide
coating which it readily forms in air due to the very strong Be-O bonds [1]
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Inorganic Chemistry I August 2010 Answers
1I3.
a)
Coordination Chemistry
Discuss any TWO of the following:
(i)
Explain the origin of colour from charge-transfer processes between
metals and ligands and how they differ from d-d transitions. Give one
example each of MLCT and LMCT complexes.
Charge-transfer occurs when there is an electronic transition which results in a
displacement of charge from one atom to another. This can occur in transition metal
complexes between metals and ligands. (1 mark). Unlike d-d transitions, they are not
symmetry-forbidden and are often 10,000 times stronger (1 mark)
MLCT: electron transfer from filled metal orbital to empty ligand’s empty orbital
(usually *). Most commonly encountered in compounds containing pi-acceptor
ligands, such as CO, CN- and bipy (any reasonable example will be accepted).
LMCT: is the opposite, when the ligand donated electrons into empty metal orbitals,
found in intensely coloured d0 metal complexes, e.g. K[MnO4], [CrO4] 2-, etc. (3
marks)
(ii)
Give the formal name for the complex K2Mn(SO4)2.6H2O, according to
the IUPAC nomenclature. Given that the complex has a magnetic moment
of 5.9 B.M., determined the electronic configuration of the complex.
Illustrate how this complex undergoes ionisation isomerism.
Potassium disulfatomanganate(II) [2 marks], d5, high-spin.[1 mark]
Ionisation isomerism: K2[Mn(SO4)2].6H2O ↔ [Mn(H2O)2(SO4)] .4H2O + K2SO4 [2
marks] could also displace both sulphates.
(iii) Sigma donor ligands have only one pair of electrons to donate to the metal centre
(1 mark), while pi-acceptor ligands have (additionally) available empty pi* orbitals to
accept electrons from filled d-orbitals from the metal (1 mark). Any reasonable
examples (1 mark).
(2 x 1 mark)
b)
Answer ALL parts of this question, with reference to the following:
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Inorganic Chemistry I August 2010 Answers
Values of ligand field splitting parameter (o, in cm-1) for octahedral iron
complexes are given in the following Table.
ion
Fe2+
Fe3+
(i)
6 H2O
10400
14300
Ligand set
6 CN–
32800
35000
Comment on the difference in o observed between [Fe(H2O)6]2+ and
[Fe(H2O)6]3+, and that between [Fe(H2O)6]2+ and [Fe(CN)6]4–.
The greater the charge on the metal, the greater is o. If the charge is large,
then this should cause the ligands to be attracted towards the metal more
closely, and so interact more strongly with the d-orbitals. [1 mark] Increase in
oxidation state from+2 to +3 normally cases an increase in o of roughly 50%,
as is observed for [Fe(H2O)6] 2+ and [Fe(H2O)6] 3+ [1 mark].
o is also dependent on the position of the ligand in the Spectrochemical Series.
[1 mark] In this case, CN- is a strong field ligand (pi-acceptor), while H2O is a
weak field ligand, thus o of [Fe(H2O)6] 2+ is smaller than [Fe(CN)6] 4–.[1 mark]
(ii)
Draw the crystal field splitting diagrams for [Fe(H2O)6]2+ and
[Fe(H2O)6]3+ in their ground state electronic configuration (both high
spin). Calculate the crystal field stabilisation energies (in kJ mol-1) for
both of these complexes, use P to denote Pairing energy.
d6 (t2g4eg2) - 1 mark.
CFSE (Fe2+) = 4 x –0.4o + 2 x 0.6o + P = –0.4o + P [1 mark] = –49.8 + P
kJ mol-1 [1 mark]
(10,000 cm-1 = 119.7 kJ mol-1)
[plus 1 mark for working]
(iii) Predict the magnetic moment you will expect for complex [Fe(CN)6]3–.
This is expected to be low-spin due to large o (1 mark), so expect d5 (t2g5eg0) =
1unpaired electron (1 mark), therefore SO = √n(n+2) = 1.73 B.M. (1 mark).
(iv)
Illustrate stepwise and overall stability constants for the following
reaction:
[Fe(H2O)6]3+ ↔ [Fe(CN)6]3–
[Fe(H2O)6]3+ + CN– ↔ [Fe(CN)(H2O)5]2+
[Fe(CN)(H2O)5]2+ + CN– ↔ [Fe(CN)2(H2O)4]+
[Fe(CN)2(H2O)4] + + CN– ↔ [Fe(CN)3(H2O)3]
[Fe(CN)3(H2O)3] + CN– ↔ [Fe(CN)4(H2O)2]–
[Fe(CN)4(H2O)2]– + CN– ↔ [Fe(CN)5(H2O)]2–
[Fe(CN)5(H2O)]2– + CN– ↔ [Fe(CN)6]3–
illustrate: 1 mark]
1
2
3
4
5
6
[any to
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Inorganic Chemistry I August 2010 Answers
Stepwise, Kn = [Fe(CN)n(H2O)6-n]/{ [Fe(CN)n-1(H2O)7-n] [ CN–]}
Overall, n = [Fe(CN)n(H2O)6-n]/ {[Fe (H2O)6] [CN–]n} [1 mark]
therefore6 = K1 x K2 x….K6 [1 mark]
[1 mark]
half marks if general formula is given, instead of the example given, i.e. [M] and [L].
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