Exemplar Inorganic

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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

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Inorganic Chem

Transcript of Exemplar Inorganic

Page 1: Exemplar Inorganic

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

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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.

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Question 1b

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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.

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Commentary

Lithium is too small to comfortably accommodate 6 waters in its co-ordination sphere; the

water-water core repulsions prevent this.

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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.

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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.

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Question 2a

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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.

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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.

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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.

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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.

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Question 3a

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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.

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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.

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Question 3c

Commentary

Yes, perfect

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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.

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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.

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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.

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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.