A-level chemistry question
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Another of your well-written questions, Kahynickel. I particularly like your table testing the student's familiarity with Group VII chemistry.
However, if I may be frank, I have some reservations and doubts on your part (c) question. The student may not know how exactly to interpret your question. Is it asking about Group II reactivity versus Group VII reactivity, or asking about why Group II metals have increased reactivity down the group, while Group VII halogens have decreased reactivity down the group.If the latter, then the expected answer should include :When Group II metals react, they lose electrons and are oxidized. Down the group, due to increasing atomic radius and increasing shielding effect, electrostatic attraction between the positively charged nucleus and valence shell electrons decreases, hence going down the group, Group II metals have smaller (ie. less endothermic and thus more favourable) ionization energies and thus have increased reactivities.When Group VII halogens react, they gain electrons and are reduced. Down the group, due to increasing atomic radius and increasing shielding effect, electrostatic attraction between the positively charged nucleus and valence shell electrons decreases, hence going down the group, Group VII halogens have smaller (ie. less exothermic and thus less favourable) electron affinities and thus have decreased reactivities. -
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.
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Originally posted by Kahynickel:
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.
Keep up the good work.
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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.
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Originally posted by Kahynickel:
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.
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)."
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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.
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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. -
Thank you very much Sir.
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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?
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Originally posted by Kahynickel:
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.