CHEM10312 Basic Inorganic Chemistry - Exemplar materials (Exam paper 2010-11)
Contents Question 1 a.................................................................................................................................................. 2
Question 1b .................................................................................................................................................. 3
Question 1c................................................................................................................................................... 4
Question 1d .................................................................................................................................................. 7
Question 2a................................................................................................................................................... 8
Question 2b .................................................................................................................................................. 9
Question 2c................................................................................................................................................. 11
Question 3a................................................................................................................................................. 13
Question 3b ................................................................................................................................................ 14
Question 3c................................................................................................................................................. 15
Question 4a................................................................................................................................................. 17
Question 4b ................................................................................................................................................ 19
Question 1 a
Commentary
1a: first example is wrong. The student is forgetting that edge, corner and face sites are only
fractionally occupied.
Second example is correct. Lost a half mark for not showing working (8 x 1/8th + 6 x ½ ) It’s
ALWAYS advisable to show your working.
Question 1b
Question 1c
Commentary
The first answer got full marks. The second answer confused hard/soft, a related, but slightly
different concept. This question is about models, and how deviations from the model (hard
spheres, electrostatic bonding) reduce its predictive ability. NaI is poor agreement because I is
compressible, and the bonding isn’t 100% ionic. F is pretty much incompressible; the bonding
isn’t 100% ionic either, but it’s much closer to the model.
Commentary
Lithium is too small to comfortably accommodate 6 waters in its co-ordination sphere; the
water-water core repulsions prevent this.
Commentary: First answer is all fine, but lost a mark for not mentioning that the ring is
important in maximizing the entropy gain. If it wasn’t cyclic, it would have to lose a lot of
entropy upon binding, which would reduce the magnitude of the positive entropy of
complexation. Because it’s cyclic, it has very little freedom to lose, so the net increase in
entropy on freeing the waters is much greater.
Commentary
Both of these are OK for 2 marks. First one is better.
Question 1d
Commentary
Answer 1di is an enormous error: A LEWIS ACID is an electron ACCEPTOR. LEWIS BASE
is electron pair donor. The Water is HARD BASE, so will form more stable complexes with
hard acids. The higher the charge density, the harder, and the more stable, so high values of .
Mg2+ is hardest (smallest, highest charge i.e. high charge density), Ag+ is softest (largest,
lowest charge i.e. low charge density). The H3As is a very soft base (large, polarizable), so
stability follows opposite order.
Question 2a
Question 2b
Commentary
2nd
answer is better here. Fewer words, but more accurate.
Commentary
A perfect answer in both cases. Right number, and right working.
Commentary
First answer: Read The Question: “Use the information” means we expect to see how you got
to answer. So 2nd
attempt gets a few more marks for using the weight info. Note: A good idea to
write down stuff: the weights are hidden info provided in the periodic table. Bring the info to
your mental work space by writing it down. The second answer also does well by providing a
balanced equation, which gives the answer to the identity of the products, since you should
know that SiO bonds are strong. Additionally considering the valency of the atoms leads to the
inevitable conclusion that the Me2SiO compound is a polymer (or cyclic compound). But you
were told in the question anyway. Make clear in the answer by writing the repeat unit out.
Question 2c
Commentary
Both of these answers are OK, but they miss the point that side-on -overlap will be relatively
weaker for the more distant P atoms than for the N atoms. 3 marks, so look for more than one
reason. Lone pair repulsion is only one point.
Commentary
Both correct, and both show working, and ‘use the data’ so full marks.
Commentary
Right drawing, right relation stated, full marks. Easy to spot this as an
analogue of P4O10, in notes and tutorial, since S is in the same group
as O.
A reasonable guess for the product would also be P4S6, analogue of
P4O6; that would get some marks, but you’d realise your mistake
when you saw it didn’t fit the provided data.
Question 3a
Commentary
Both are quite well-answered; where there are dropped marks, it is down to using oct Crystal
Field diagram for a tetrahedral case, and some errors in computing oxidation state, caused
probably by not writing enough stages down. Show thinking clearly: part (iv): two bariums;
barium is in group 2, so they will each carry a 2+ charge, 4+ overall, so overall charge on the
Mn complex must be 4- to balance this. CN is –ve charge, 6 of them is 6-, so for the ion to have
a 4- charge, it must be Mn2+. It’s easy when you do it in stages, and write it down.
Next: Mn is in group 7, so d-conf is group number-oxidation state, 7-2 = d5
Also note: Usually, it’s high spin for Mn(II), d5, but with cyanide ligands, everything is low-
spin.
Question 3b
Commentary
A perfect answer. Right choices, right reasons. Except perhaps too clever in the first part, the
distortion is from a 2E ground state (regular octahedral d
9) to a
2A ground state (distorted oct
d9), not the other way about. And it is possible to confuse this nomenclature with the arrow
with an electronic excitation. None of this was asked in this question.
Question 3c
Commentary
Yes, perfect
Question 4a
Commentary
This is very generously marked as 4 out of 4. There is no clear explanation of sigma having a
maximum along the internuclear axis, pi having a node at that axis. The statement “So the ligand
donates into * antibonding orbitals” is wrong. Also, the ligand does indeed have p-electrons to
donate, in its filled -orbitals, so the following sentence is wrong as well, but the overlap between
them and the metal when it binds end-on is very slight. In contrast, its empty * orbitals are the ones
which can accept d-electron density from the metal, as correctly shown. The diagrams save the marks
for the candidate, with phase shown, they are all correct.
Commentary
NO+ can be hypothetically generated from CN
by adding one proton to the nucleus of each atom (C
becomes N, N becomes O, the charges work), so it will be a poorer sigma donor (the nitrogen will keep
a better hold of its electrons with its extra proton in the nucleus, so less keen to donate them to metal.
Could also use electronegativities, but the above argument is the reason behind the electronegativities.
Commentary
Simple charge argument works here too, though the more fundamental reasoning (same number of
electrons, greater number of protons in the nucleus) would have been welcome! Key point: do not
panic if faced with something you don’t know, just think rationally to get an answer. Positive attracts
negative; that’s it.
Question 4b
Commentary
Also, N is the less electronegative atom; that is the reason that the HOMO of the diatomic,
hence the position of the “lone pair” is based on N.
Commentary
First example is all fine, second clearly gets the idea, since they must know what the ligand
charges are in order to get the right metal ‘charge’ (i.e. oxidation state), but lost marks because
they did not answer the question as asked, which asked for ligand charges. It was asked this
way to try to help the sitter, since ligand charges have to be determined in order to get oxidation
state, then d-electron configuration (group number – oxidation state), then term symbol.
Note the Fe(III), second example, is high spin here, 6A, because the ligands are not high in the
spectrochemical series.
First answer is also better because working is shown for magnetic moments.
Top Related