A-level chemistry question

last SG · 2012-05-06T04:13:39+00:00

Thank you for your invaluable suggestions.I would like this question to be structure-one rather than essay-type since CIE exam papers prefers structured ques...

UltimaOnline SG

UltimaOnline with your support and guidance I have prepared two mocks. I would like to send a copy of each of these to you so that it may serve as a practice test for students there.

Would you like to give your e-mail address for this purpose to my mailbox if you intend to.

Regards.

Hi Kahynickel,

Sure, my email address is [email protected]. Thanks for your generous sharing of your well written mock exams which you've put in much effort towards.

hoay SG

Please check the email. I have emailed the papers on [email protected]

hoay SG

You would be welcome to have it Hoay I will email it to you soon. You seem to be asking too many questions.

Kahynickel SG

Please check part (b) and (c) of this question.

This question is about group VII elements.

(a) Describe, with the aid of suitable equations, the formation of HF, HCl and HI when hydrogen reacts with flourine, chlorine and iodine. Include the necessary conditions required in each reaction.

HF……………………………………………..…………………………………………………………………….....

……………………………………………………………………………………………………………………….....

Reaction: …………………………………………………………………………………………………………………………….

HCl……………………………………………………………………………………………………………………………………

………………………………………………………………………………………………………………………………………...

Reaction: …………………………………………………………………………………………………………………………….

HI……………………………………………………………………………………………………………………………………...

………………………………………………………………………………………………………………………………………...

Reaction: …………………………………………………………………………………………………………………………….

(b) The difference in the above reactions can be explained by the considering the following physical parameters.

Use the words increases, decreases or remains constant to complete the information regarding the halogens you have used in (a).

Answers are written in bold letters.

From flourine to iodine

Atomic radius increases

Shielding effect increases

Effective nuclear charge decreases

Strength of IMFs increases

Reactivity decreases

c) On the other hand, the reactivity of group I and II elements increases down the group. Choose two parameters from those listed in (b) and explain how they account for the greater reactivity of group I and II elements tha group VII.

Effective nuclear charge and shielding effect

Down the group effective nuclear charge decreases so shielding effect increases more significantly as compared to group VII elements because of more electrons in the valence shell in halogens (ns², np⁵)as compared to group II elements (ns²) hence the rise in atomic radii in much larger in group I elements as compared to group VII elements which results in the greater reactivity of group I elements than group VII elements.

d) Concentrated sulfuric acid has many uses in the industry. Some of them are given below.

reagent dehydrating agent acid catalyst oxidizing agent

Use the above words to complete the following table which enlists the detail of the action of concentrated sulfuric acid on the following compounds. In some cases more than one word may be used.

Observation on adding concentrated H₂SO₄

Product(s) formed responsible for the observation

Role(s) of concentrated H₂SO₄

CH₃CH₂CN

+ heat at 90^oC

C₂H₅COOH

MgCl₂

HCl

Colorless gas

Violet vapours

Yellow solid

NaBr

Acid and oxidizing agent

Kahynickel SG

Sorry for the random chart.

folloiwng is the complete one with answers in black. Others will be given in the question.

Observation on Product formed role of sulfuric acid

adding H2SO4

CH3CH2CN heat at 90oC not applicable CH3CH2COOH reagent

MgCl2 misty fumes HCl acid only

KI colorless gas HI as acid and

violet vapours I2(g) oxidizing agent

yellow solid S(s)

NaBr colorless gas HBr acid and

reddish-brown vapours Br2 oxidizing agent

UltimaOnline SG

Another of your well-written questions, Kahynickel. I particularly like your table testing the student's familiarity with Group VII chemistry.

Kahynickel SG

I intend to ask ”group VII elements are less reactive than group II elements”. Having completed part (b) listed parameters students are supposed to choose two parameters from part (b) and then explain in terms of their chosen parameters the increased reactivity of group II as compared to group VII. It is for this reason that I included in my answer the increased nuclear charge effect in favour of group VII elements as compared decreasing nuclear charge in group II- I presented the electronic configuration argument and did not mention electron affinity and ionization potential viewpoint.

But it seems that you have correctly pointed out the short-coming in choosing the parameters in part(b) it must have included “electron affinity” parameter atleast.

Consequently, the two inevitable alterations are as follows.

(i) In Part(b) “Electron affinity” less exothermic will be given in the question.

(ii) In the language of part (c) it will be now:

By choosing one or more parameters from part (b) and an additional parameter suitable for group II how you would compare the reactivities of group II elements with group VII elements.

UltimaOnline SG

But it seems that you have correctly pointed out the short-coming in choosing the parameters in part(b) it must have included “electron affinity” parameter atleast.

Consequently, the two inevitable alterations are as follows.

(i) In Part(b) “Electron affinity” less exothermic will be given in the question.

(ii) In the language of part (c) it will be now:

By choosing one or more parameters from part (b) and an additional parameter suitable for group II how you would compare the reactivities of group II elements with group VII elements.

Keep up the good work.

Kahynickel SG

I recently got Singapore A-level papers from Nov 2004 to Nov 2010 in exchange with the 2010 Prelims papers from a Singapore student. I still need Nov 2011 papers and June papers.

The questions are excellent. One of the things that I have noticed is that papers-setters sometime extend a multiple choice question into a full-length question for theory paper. The first question of my first mock that I e-mailed you was an example of this mind-set.

The question was:

When two liquids are mixed, heat may be evolved if intermolecular bonds formed are stronger

than those broken, even if there is no chemical reaction'

Which pair of liquids, when mixed, will give out heat?

A CH₂Cl₂ and (CH₃)₂CO

B CHCI₃ and C₆H₁₄

C CCl₄ and (CH₃)₂CO

D CCl₄ and CH₃CH₂OH

The answer was A.

In D CCl4 has VDW forces while ethanol has H-bonding so they don’t mix therefore no heat is evolved. H-bonds are stronger than VDW forces in CCl4 consequently heat must be added to make them mix. You can say that in the context of this question intermolecular forces are not broken. [Is this right]

In C propanone has permanent –dipoles as compared to VDW forces in CCl4 again the same explanation as in D.

B is also same as that of C.

In A the intermolecular forces are same in both cases. Propanone has, however, stronger permanent-dipoles forces than CH₂Cl₂ because of the high electronegative O atom against the low-electronegative C-atom. The intermolecular bonds /forces formed here are strong permanent dipoles (I don’t think that VDW forces have a role to play here albeit CH2Cl2 has more electron than propanone.) so heat is given out.

In addition for the rest of the pairs to be soluble heat will be added but only A is the one which gives out heat.

I am waiting for your comment as I have to finalize the question in a long format.

UltimaOnline SG

I recently got Singapore A-level papers from Nov 2004 to Nov 2010 in exchange with the 2010 Prelims papers from a Singapore student. I still need Nov 2011 papers and June papers.

The question was:

When two liquids are mixed, heat may be evolved if intermolecular bonds formed are stronger

than those broken, even if there is no chemical reaction'

Which pair of liquids, when mixed, will give out heat?

A CH₂Cl₂ and (CH₃)₂CO

B CHCI₃ and C₆H₁₄

C CCl₄ and (CH₃)₂CO

D CCl₄ and CH₃CH₂OH

The answer was A.

In C propanone has permanent –dipoles as compared to VDW forces in CCl4 again the same explanation as in D.

B is also same as that of C.

In addition for the rest of the pairs to be soluble heat will be added but only A is the one which gives out heat.

I am waiting for your comment as I have to finalize the question in a long format.

Btw and Fyi, the Singapore-Cambridge exams are only offered in November. We do not have June papers.

I've little to add to your excellent explanations on this particular MCQ question, except on two minor points :

1) For options B, C and D, although the two species listed are generally not miscible (due to reasons you've given), but due to favourable entropy, they will be a slight measure of mixing. Moreover, the question specified, "when the two species are mixed", which you may take it to mean "forcibly mixed" together, eg. by mechanical means. In which case, your structured or open-eneded question, may require the student to take this into consideration.

For options B, C and D, when mixing does occur, no heat is evolved, because the newly formed inter-species (solute-solvent) interactions are permanent dipole - induced dipole van der Waals (Debye) forces, which are weaker than the pre-existing permanent dipole - permanent dipole van der Waals (Keesom) forces (option B & C) or the hydrogen bonding (option D), that must be overcome as part of the mixing process.

2) For option A, as you've pointed out, since the new interactions are of the same type as the old interactions, ie. permanent dipole - permanent dipole van der Waals (Keesom) forces, we cannot be sure, without carrying out the relevant experiments, that A will definitely give out heat. But because this is an MCQ question, we can eliminate B, C, D as the answer, and therefore A is the only remaining viable answer. But if you were to phrase your structured or open-ended question to guide the process, eg. "In the table above, which experiment would be the most likely to have an exothermic reading?" and "For the experiment you identified, suggest with explanation at the molecular level, why this experiment might have an exothermic reading."

Then perhaps the molecular geometries and sterics might play a part. Dichloromethane is tetrahedral, while propanone is trigonal planar. Perhaps the student can score credit if he/she writes along the lines of, "due to sterics consideration of the molecular geometries involved, the newly formed permanent dipole - permanent dipole van der Waals (Keesom) forces between dichloromethane and propanone, might be stronger than the pre-existing permanent dipole - permanent dipole van der Waals (Keesom) forces between dichloromethane and dichloromethane, (but not necessarily stronger than the pre-existing permanent dipole - permanent dipole van der Waals (Keesom) forces between propanone and propanone)."

Kahynickel SG

My query pertains to the following question which apperaed in A-level [CIE] paper 41 Q.4.

Ethanolamine and phenylamine are two organic bases that are industrially important.
Ethanolamine is a useful solvent with basic properties, whilst phenylamine is an important
starting material in the manufacture of dyes and pharmaceuticals.
The following table lists some of their properties, together with those of propylamine.

formula Mr B.p/oC solubility in water

propylamine C3H7NH2 59 48 fairly soluble

ethanolamine HOCH2CH2NH2 61 170 very soluble

(a) Suggest why the boiling point of ethanolamine is much higher than that of propylamine. Draw a diagram to illustrate your answer.

The Cambridge marking scheme enlists the following answer.

hydrogen bonding [mark 1]

diag: NH2CH2CH2OH---OHCH2CH2NH2 or NH2CH2CH2OH---NH2CH2CH2OH

(i.e. H-bond from OH group to either OH or NH2) [mark 2]

My answer

This is a comparison question. Both propylamine and ethanoloamine have hydrogen bonding but prophyl amine has a dominant non-polar propyl chain which minimizes the effectivess of hydrogen bonding. In ethnoalmine there are two hydrogen bonds per molecule via nitrogen and Oxygen atoms resulting in extensive hydrogen bondind as comapred to propylamine.

NH2CH2CH2OH.........NH2CH2CH2OH.........NH2CH2CH2OH

Both HYDROGEN BONDS needs to be shown and not anyone of them.

UltimaOnline SG

Your answer is perfectly fine, and in complete accordance to the Cambridge mark scheme. Personally, I would say that "ethanolamine has more extensive intermolecular hydrogen bonding compared to propylamine, because ethanolamine (per molecule) has two functional groups (NH2 and OH) capable of both donating and accepting hydrogen bonds, while propaylamine (per molecule) only has one functional group capable of donating and accepting hydrogen bonds."

I would advise students to draw 3 ethanolamine molecules as you suggested, illustrating at least 2 of the 3 different *additional* types of intermolecular hydrogen bonding that ethanolamine has, over propylamine.

Kahynickel SG

Thank you very much Sir.

Kahynickel SG

One more thing to ask regarding this ethanoalmine-propylamine question.

If a candidate writes:

Ethanolamine has hydrogen bonding and hydrogen bonding is the strongest intermolecular force therefore has the higher boiling point than propylamine.

NH2CH2CH2OH---NH2CH2CH2OH

Will this be accepted as this makes no comparison with propylamine?

UltimaOnline SG

One more thing to ask regarding this ethanoalmine-propylamine question.

If a candidate writes:

Ethanolamine has hydrogen bonding and hydrogen bonding is the strongest intermolecular force therefore has the higher boiling point than propylamine.

NH2CH2CH2OH---NH2CH2CH2OH

Will this be accepted as this makes no comparison with propylamine?

This answer will not be accepted because it wrongly implies that propylamine has no intermolecular hydrogen bonds.

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