Inorganic Chemistry I June 2010 Answers
1I.1 Molecular structure a) Answer ALL parts of this question: i) Sketch and label the molecular orbital energy level diagram for CO. On your diagram, include drawings of the MOs and electron occupancies. Indicate if significant MO mixing is expected, highlighting which molecular orbitals are likely to be involved in mixing.
ANSWER: i) If no MO mixing is taken into account, the following diagram is obtained:
Allowing MO mixing, the following changes should be expected:
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Inorganic Chemistry I June 2010 Answers
The photoelectron spectrum is the key to realising that the 5σ is higher than the 1 π and therefore MO mixing is expected. Although two diagrams are herein shown, it would be OK if the students go directly to the one where mixing has taken place (and highlight the orbitals involved in mixing).
Correct orientation/set of axis ( 1 mark ); relative energy of C and O atomic orbitals ( 1 mark ); right combinations of AOs to yield the correct number of MOs, plus drawing of MOs ( 4 marks ); right electron occupancy ( 1 mark ); correct MO mixing including a simple diagram showing which orbitals mix and difference in energy ( 3 marks ). ii) Use your diagram to briefly explain why CO is a good π-acceptor in organometallic chemistry.
For a molecule to be a pi acceptor it needs to have empty orbitals of the right symmetry to accept electrons from the filled d atomic orbital on the metal ( 1 mark ). From the MO diagram, it can be seen that the lowest unoccupied MO is of pi symmetry; since this is relatively low in energy, it can be used to accept pi electron density from the metal ( 1 mark ). Since the 2p MO has mostly antibonding character, addition of electrons to this orbital will reduce the CO bond order weakening the bond ( 1 mark ). b) Using VSEPR theory, state the pseudostructures and structures of the following compounds. Include drawings of both structures in your answer and highlight any deviations from ideal structures.
XeF
2
SF
4
2

ANSWER :
The predicted structures from VSEPR are:
XeF
2
Inorganic Chemistry I June 2010 Answers
SF
4
Pseudostructure: Trigonal bipyramidal Trigonal bipyramidal
Structure:
(
Linear
2 marks ) (
See-saw shaped
2 marks )
In XeF
2
the lone pairs occupy the equatorial positions to minimize repulsions. The molecule is therefore linear with no significant distortions of the F-Xe-F angle (of 180 degrees). In SF
4 the lone pair occupies an equatorial position to minimise repulsions. The S-F bonds are bent away from the lone pair generating a considerable distortion of the F-S-F angles. ( 2 marks ) c) Answer ALL parts of this question: i) Which of the following molecules possess a centre of symmetry: HCN, C
2
H
2
and BF
3
.
Illustrate your answer by showing the centre of symmetry on a drawing of the molecule(s) that possess one.
ANSWER:
Only C
2
H
2
has a centre of symmetry . ( 1 mark )
( 1 mark ) ii) Al
2
Cl
6
has the structure shown below (each Al atom is tetrahedral). Mark all the rotational axis of symmetry and determine its point group.
ANSWER:
There are three C
2
rotational axis of symmetry (orthogonal to each other – i.e. along the x, y and z axis.
3

( 1 mark )
Inorganic Chemistry I June 2010 Answers
Point group: D
2h
( 1 mark ) iii) Show that BF
3
possesses an S
3
axis but NF
3
does not.
BF
3
is a planar molecule and therefore there is an S
3
axis as shown below:
( 1 mark )
In contrast, the trigonal pyramidal NF
3
molecule does not have a plane of symmetry perpendicular to the C
3
rotational axis and therefore there is no S
3
axis (see below):
( 1 mark ) d) Draw the structure of trans -N
2
F
2
. Use an appropriate hybridisation scheme to account for the geometry around the nitrogen atoms and use valence bond theory to describe the bonding in this molecule.
ANSWER :
Using VSEPR it is possible to predict that the structure of trans-N
2
F
2
is planar as shown below ( 2 marks ):
This structure is consistent with an sp
2
hybridisation of the N atoms:
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Inorganic Chemistry I June 2010 Answers
On each N atom, one of the sp
2
orbitals is used to form a σ bond with F, another one is occupied by a lone pair (in a trigonal planar position) and the third one is used to form a σ bond with the other N atom. ( 2 marks ) In addition, on each N atom there is one p orbital (perpendicular to the plane of the molecule) which leads to the formation of a π bond (giving a double bond between the two N atoms). ( 2 marks )
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Inorganic Chemistry I June 2010 Answers
(a) (i) Down the group (Cl
2
to Br
2
to I
2
) bond energy decreases since the principal quantum number is increasing and therefore the orbitals used in the bond are larger and therefore the overlap is less good and weaker [3] . F
2
is much weaker than expected since the bond is very short and the lone pairs on the neighbouring F atoms are close to each other resulting in lots of lp-lp repulsions destabilising the bond
[2] . (An MO discussion with electron-pairs in pi* orbitals repelling each other is also acceptable).
(ii) oxidation states: ClF
3
(+3) and ClF
5
(+5) [1] . Hybridisation: ClF
3
(sp
3 d) and ClF
5
(sp
3 d
2
) [3] . ClF
7 does not exist since it is too sterically crowded to fit 7 F around one Cl and also Cl is more electronegative than I it is difficult to oxidise it to the formal +7 oxidation state [2] .
(b) (i) Ga has 3 valence electrons and can form 3 bonds with other atoms giving 6 electrons around the group 13 atom, which is therefore electron deficient as it is 2 short from an octet. [2] In both Ga
2
Cl
6
and
Ga
2
H
6
the Ga is sp
3
hybridised and tetrahedral [1] . In Ga
2
H
6
there are 4 terminal 2e-2c sigma bonds, and two 2e-3c Ga-H-Ga bridging bonds [2] . However in Ga
2
Cl
6 to relieve the electron deficiency GaCl
3 dimerises with donation (sigma) from the lone pair on the adjacent GaCl
3
Cl atom into the empty Ga sp
3 orbital (draw picture). Hence all bonds are 2c-2e in this compound [2]
(ii) SO
3
is a trigonal planar molecule [2] and therefore is flat and has a very open structure allowing the
δ+ sulfur atom to be easily attacked by a nucleophile
[1] . SO
4
2-
however is tetrahedral in shape and the sulfur atom is surrounded by 4 electronegative oxygen atoms [2] . These 4 oxygen atoms prevent attack of the central sulfur atom by a nucleophile due to steric hindrance / electronic repulsions and therefore provide kinetic stability [2] . Hence sulfates are very inert.
(iii) Draw all Lewis structures (2 resonance forms for N
2
O and NO
2
and one for NO
2
+
) [5] . NO
2 contains an unpaired electron (radical) and therefore will dimerise at low temperatures in the gas phase in order to pair up these electrons [2]
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Inorganic Chemistry I June 2010 Answers
a) Discuss any TWO of the following:
(i) Dinitrogen (N
2
) is isoelectronic with CO. With this in mind, show the nature of the bonding between N
2
and a d-block metal centre. Draw both geometrical isomers of the square-planar complex [RhCl(PPh
3
)
2
(N
2
)], and label them clearly.
( 2 marks )
N
2
is also expected to be a pi-acceptor ligand. ( 1 mark ) (credit will be given for eta-2 bonding mode)
Cis- and trans-[RhCl(PPh
3
)
2
(N
2
)]geometrical isomers ( 1 mark for drawing , 1 mark for labelling )
(ii) Solid-state structures of two tungsten complexes of the chemical formulae:
[WBr
3
(C
5
Me
5
)(CO)
2
] and [W
2
Br
4
(CO)
8
] are shown below (figure 1). Provide the formal
IUPAC chemical names of these compounds. Explain the terms: “  ” and “  ” in their nomenclature.
Figure 1.
[from: H. Blackburn, H-B Kraatz, R. Poli and R. C. Torralba, Polyhedron , 1995 , 14 , 2225; and F. A. Cotton, L. R. Falvello, and J. H. Meadows, Inorg. Chem ., 1985 , 24 , 514.] tri b romodi c arbonyl(
 5
p entamethylcyclopentadienyl)tungsten(IV) and ( 1 mark )
(dibromo)(
 2
-dibromo)hexacarbonylditungsten(II) ( 2 marks )
Explain the nomenclature (2 x 1 mark)
(iii) Crystal field splitting parameters (
 o
) for [Mn(CN)
6
]
4-
and [Fe(CN)
6
]
3-
are 30,000 and 35,000 cm
-1
, respectively. The Mn complex is high-spin, whereas the Fe complex is low-spin. Use the Crystal Field theory to explain these differences.
Mn
2+
vs Fe
3+
, both d
5
and have the same number and type of ligands ( 1 mark ), but the latter has higher oxidation state, which accounts for the larger
 o
( 1 mark ). Explanation: higher charge causes the ligands to be attracted towards the metal more closely, and so interact more strongly with the d-orbitals ( 1 mark ). Complexes tends to be low-spin if
 o
> P (pairing energy), and vice versa ( 1 mark ). In this case, we can estimate that 30,000 cm
-1
< P < 35,000 cm
-1
( 1 mark ).
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Inorganic Chemistry I June 2010 Answers b) Answer ALL parts of this question, with reference to the following:
Ethylenediamine tetraacetic acid (H
4
-EDTA) and glycine (gly) form stable complexes with M
3+
ions of d-block elements (equations 1 and 2, respectively). Some formation constants are given in Table
1.
Table 1.
Selected formation constants.
Complex log

[Fe(EDTA)]
-
25.7
[Fe(gly)
3
]
[Co(EDTA)]
-
10.0
41.3
(i) Draw the structures of BOTH the EDTA and glycine complexes of iron, showing clearly how the ligand coordinates.
(2 marks)
1 mark for each correct structure
(ii) Identify any isomers that may exist in the complexes you have drawn in part i).
(2 marks)
[M(gly)
3
] exists as optical isomers,

and

-forms
(iii) Compare and contrast EDTA and glycine as ligands in these complexes. Explain the term
„denticity‟ in your answer.
(3 marks)
Both contain N and O-donor atoms (1 mark)
EDTA is hexadentate, whereas glycine are bidentate (2 x 1 mark).
(iv) Write an equation showing an equilibrium process, from which

2
for [M (gly)
2
(OH
2
)
2
]
+
is derived.
(2 marks)

2
= [M(gly)
2
(OH
2
)
2
]
+
[H
+
]/{ [M(gly)(OH
2
)
4
]
2+
[gly] (1 mark only if H
+
not included or incorrect charges)
(v) Considering log

values given in Table 1, compare relative stabilities of the three metal complexes. Explain these observations.
(6 marks)
Difference between Fe complexes primarily due to chelate effect (1 mark) + good explanation of chelate effect (2 mark).
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Inorganic Chemistry I June 2010 Answers
Co EDTA complex is markedly more stable than Fe EDTA complex – both are in the oxidation state, and only next to each other in the d-block (small difference in size), the difference may be attributed to the (thermodynamic) stability of d
5
vs d
6
ion. Since we can assume both are high-spin – there is no CFSE for Fe as there is in the Co. (3 marks)
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